solution manual for fundamentals of electric circuits

1117
Chapter 1, Solution 1 (a) q = 6.482x10 17 x [-1.602x10 -19 C] = -0.10384 C (b) q = 1. 24x10 18 x [-1.602x10 -19 C] = -0.19865 C (c) q = 2.46x10 19 x [-1.602x10 -19 C] = -3.941 C (d) q = 1.628x10 20 x [-1.602x10 -19 C] = -26.08 C Chapter 1, Solution 2 (a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e -t + 10e -2t ) nA (d) i=dq/dt = 1200 120 S S cos t pA (e) i =dq/dt = e t t 4 80 50 1000 50 ( cos sin ) A P t Chapter 1, Solution 3 (a) C 1) (3t ³ q(0) i(t)dt q(t) (b) mC 5t) (t 2 ³ q(v) dt s) (2t q(t) (c) q(t) 20 cos 10t /6 q(0) (2sin(10 / 6) 1) C t S S P ³ (d) C 40t) sin 0.12 t (0.16cos40 e 30t - ³ t) cos 40 - t 40 sin 30 ( 1600 900 e 10 q(0) t 40 sin 10e q(t) -30t 30t - Chapter 1, Solution 4 mC 4.698 ³ ³ S S S 06 . 0 cos 1 6 5 t ʌ 6 cos 6 5 dt t ʌ 6 5sin idt q 10 0

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Page 1: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 1 (a) q = 6.482x1017

x [-1.602x10-19 C] = -0.10384 C

(b) q = 1. 24x1018

x [-1.602x10-19 C] = -0.19865 C

(c) q = 2.46x1019

x [-1.602x10-19 C] = -3.941 C

(d) q = 1.628x1020

x [-1.602x10-19 C] = -26.08 C

Chapter 1, Solution 2

(a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e-t + 10e-2t) nA (d) i=dq/dt = 1200 120cos t pA (e) i =dq/dt = e tt4 80 50 1000 50( cos sin ) At

Chapter 1, Solution 3

(a) C 1)(3tq(0)i(t)dt q(t)

(b) mC 5t)(t 2q(v)dt s)(2tq(t)

(c) q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ct

(d) C 40t) sin 0.12t(0.16cos40e 30t-

t)cos 40-t40sin30(1600900

e10q(0)t40sin10eq(t)-30t

30t-

Chapter 1, Solution 4

mC 4.69806.0cos165

t6cos6

5dt t 6 5sinidtq10

0

Page 2: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 5

µCmC )e1(21

e21-mC dteidtq

4

2

0

2t-2t-

490

Chapter 1, Solution 6

(a) At t = 1ms, mA 402

80dtdq

i

(b) At t = 6ms, mA 0dtdqi

(c) At t = 10ms, mA 20-4

80dtdqi

Chapter 1, Solution 7

8t6 25A,6t2 25A,-2t0 A,25

dtdq

i

which is sketched below:

Page 3: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 8

C 15 µ1102

110idtq

Chapter 1, Solution 9

(a) C 101

0dt 10idtq

(b) C 22.5251015

152

1510110idtq3

0

(c) C 30101010idt5

0q

Chapter 1, Solution 10

q ixt x x x8 10 15 10 1203 6 C Chapter 1, Solution 11

q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J

Chapter 1, Solution 12

For 0 < t < 6s, assuming q(0) = 0,

q t idt q tdt tt t

( ) ( ) .0 3 0 150

2

0

At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s,

Page 4: Solution Manual for Fundamentals of Electric Circuits

q t idt q dt tt t

( ) ( )6 18 54 18 56 6

4

66

At t=10, q(10) = 180 �– 54 = 126 For 10<t<15s,

q t idt q dt tt t

( ) ( ) ( )10 12 126 12 24610 10

At t=15, q(15) = -12x15 + 246 = 66 For 15<t<20s,

q t dt qt

( ) ( )0 1515

Thus,

q t

ttt

( )

.

,

1518 5412 24666

2 C, 0 < t < 6s C, 6 < t < 10s C, 10 < t < 15s

C 15 < t < 20s

The plot of the charge is shown below.

0 5 10 15 200

20

40

60

80

100

120

140

t

q(t)

Page 5: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 13

kJ 2.486-

216sin41

12008sin82

1200

1)-2x cos 2xcos (since,1)dt -t8cos2(1200

dtt 4cos1200vidtw

2

0

2

0

2

2

0

22

0

tt

Chapter 1, Solution 14

(a) C 2.1312e2110

2et10dte-110idtq0.5-

1

0

0.5t-1

0

0.5t-

(b) p(t) = v(t)i(t) p(1) = 5cos2 10(1- e-0.5) = (-2.081)(3.935) = -8.188 W Chapter 1, Solution 15

(a)

C 1.2971e5.1

e23

dt3eidtq

2-

2

0

2t2

0

2t-

(b) We90

)(

t4vip

e305e6dtdi5

v 2t-2t

(c) J 22.53

0

4t-3

0

4t- e490dt e-90pdtw

Page 6: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 16

mJ 916.7

)(

(

(((

,

4

3

22

4

3

2

3

2

22

1

1

0

3

4

3

3

2

2

1

1

0

3t

4t-t162502829

1225032

2503

250

dtt)42502t

-t42502

250t

3250

25t)mJ-t)(1001040t)dt2510010t)dt2510(25t)dt10v(t)i(t)dtw

4t3 V10t -403t1 V 101t0 V10t

v(t)4t2 mA25t -1002t0 mA25t

i(t)

Chapter 1, Solution 17

p = 0 -205 + 60 + 45 + 30 + p3 = 0 p3 = 205 �– 135 = 70 W Thus element 3 receives 70 W. Chapter 1, Solution 18

p1 = 30(-10) = -300 W p2 = 10(10) = 100 W p3 = 20(14) = 280 W p4 = 8(-4) = -32 W p5 = 12(-4) = -48 W

Chapter 1, Solution 19

p I x x xs s0 4 2 6 13 2 5 10 0 3 AI

Page 7: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 20

Since p = 0 -30 6 + 6 12 + 3V0 + 28 + 28 2 - 3 10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 V0 = 18 V

Chapter 1, Solution 21

nA8.

(.

C/s100.8 C/s10611084

electron) / C1061photon

electron81

secphoton

104

8-1911

1911

tq

i

Chapter 1, Solution 22

It should be noted that these are only typical answers.

(a) Light bulb 60 W, 100 W (b) Radio set 4 W (c) TV set 110 W (d) Refrigerator 700 W (e) PC 120 W (f) PC printer 18 W (g) Microwave oven 1000 W (h) Blender 350 W

Chapter 1, Solution 23

(a) W12.51201500

vpi

(b) kWh 1.125. kWh60451.5J60451051 3ptw

(c) Cost = 1.125 10 = 11.25 cents

Page 8: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 24

p = vi = 110 x 8 = 880 W Chapter 1, Solution 25

cents 21.6 cents/kWh 930hr 64 kW 1.2 Cost

Chapter 1, Solution 26

(a) mA 80.10h

hA80i

(b) p = vi = 6 0.08 = 0.48 W (c) w = pt = 0.48 10 Wh = 0.0048 kWh

Chapter 1, Solution 27

kC 43.2 36004 3 T33dt idt q

36005 4 4h T Let (a)T

0

kJ 475.2

..

.)((

36001625036004033600

250103

dt3600

t50103vidtpdtW b)

36004

0

2

0

T

0

tt

T

cents 1.188

(

cent 9kWh3600475.2 Cost

Ws)(J kWs, 475.2W c)

Page 9: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 28

A 0.2512030 (a)

VPi

$31.54(

262.8 $0.12CostkWh 262.8Wh 2436530ptW b)

Chapter 1, Solution 29

cents 39.6

.

3.3 cents 12CostkWh 3.30.92.4

hr6030kW 1.8hr

6045)1540(20kW21ptw

Chapter 1, Solution 30

Energy = (52.75 �– 5.23)/0.11 = 432 kWh Chapter 1, Solution 31

Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 = $526.60

Chapter 1, Solution 32

cents 39.6

.

3.3 cents 12CostkWh 3.30.92.4

hr6030kW 1.8hr

6045)1540(20kW21ptw

Page 10: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 33

C 631032000idtqdtdqi

Chapter 1, Solution 34

(b) Energy = = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 pt

= 10,000 kWh (c) Average power = 10,000/24 = 416.67 W

Chapter 1, Solution 35

kWh 10.4

)((

28007200

24002120012200210006400W a) dttp

W/h 433.3(h 24kW 10.4 b)

Chapter 1, Solution 36

days 6,667,(

A 4

day / 24hh000160

0.001A160Ah tb)

40hA160i (a)

Chapter 1, Solution 37

J 901.2...

12180WC 180106021105q 1920

qv

Page 11: Solution Manual for Fundamentals of Electric Circuits

Chapter 1, Solution 38

P = 10 hp = 7460 W W = pt = 7460 30 60 J = 13.43 106 J

Chapter 1, Solution 39

A 16.667120

102vpivip

3

Page 12: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 1

v = iR i = v/R = (16/5) mA = 3.2 mA

Chapter 2, Solution 2

p = v2/R R = v2/p = 14400/60 = 240 ohms

Chapter 2, Solution 3

R = v/i = 120/(2.5x10-3) = 48k ohms Chapter 2, Solution 4

(a) i = 3/100 = 30 mA (b) i = 3/150 = 20 mA

Chapter 2, Solution 5

n = 9; l = 7; b = n + l �– 1 = 15 Chapter 2, Solution 6

n = 12; l = 8; b = n + l �–1 = 19 Chapter 2, Solution 7

7 elements or 7 branches and 4 nodes, as indicated. 30 V 1 20 2 3 +++-

2A 30 60 40 10 4

+ -

Page 13: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 8

dc

b

a

9 A

i3i2

12 A

12 A

i1

8 A At node a, 8 = 12 + i1 i1 = - 4A At node c, 9 = 8 + i2 i2 = 1A At node d, 9 = 12 + i3 i3 = -3A Chapter 2, Solution 9

Applying KCL, i1 + 1 = 10 + 2 i1 = 11A 1 + i2 = 2 + 3 i2 = 4A i2 = i3 + 3 i3 = 1A Chapter 2, Solution 10 At node 1, 4 + 3 = i1 i1 = 7A At node 3, 3 + i2 = -2 i2 = -5A

3

2-2A

3A

1

4A

i2i1

Page 14: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 11

Applying KVL to each loop gives -8 + v1 + 12 = 0 v1 = 4v -12 - v2 + 6 = 0 v2 = -6v 10 - 6 - v3 = 0 v3 = 4v -v4 + 8 - 10 = 0 v4 = -2v Chapter 2, Solution 12 For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v For loop 2, -10 +15 -v2 = 0 v2 = 5v For loop 3, -v1 +v2 +v3 = 0 v3 = 30v

+ 15v -

loop 3

loop 2

loop 1 + 20v

-

+ 10v - �– 25v + + v2 -

+ v1

-

+ v3

-

Chapter 2, Solution 13

2A

I2 7A I4 1 2 3 4 4A

I1 3A I3

Page 15: Solution Manual for Fundamentals of Electric Circuits

At node 2, 3 7 0 102 2I I A

12

2 A

2 5 A

At node 1, I I I I A1 2 1 22 2

At node 4, 2 4 2 44 4I I

At node 3, 7 7 4 3 3I I I

Hence, I A I A I A I1 2 3 412 10 5 2, , , A

V

11

8

Chapter 2, Solution 14 + + - 3V V1 I4 V2 - I3 - + 2V - + - + V3 - + + 4V I2 - I1 V4 + -

5V

For mesh 1,

V V4 42 5 0 7 For mesh 2,

4 0 4 73 4 3V V V V For mesh 3,

3 0 31 3 1 3V V V V V For mesh 4,

V V V V V1 2 2 12 0 2 6 Thus, V V V V V V V1 2 3 48 6 11 V7, , ,

Page 16: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 15

+ +

+ 12V 1 v2 - - 8V + - v1 - 3 + 2 - v3 10V

- + For loop 1,

8 12 0 42 2v v V

V

V

For loop 2,

v v3 38 10 0 18 For loop 3,

v v v1 3 112 0 6 Thus, v V v V v1 2 36 4, , V18 Chapter 2, Solution 16

+ v1 -

+ - + -

6V -+ loop 1

loop 2 12V

10V

+ v1

- + v2 -

Applying KVL around loop 1,

�–6 + v1 + v1 �– 10 �– 12 = 0 v1 = 14V

Applying KVL around loop 2,

12 + 10 �– v2 = 0 v2 = 22V

Page 17: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 17 + v1 -

+

- +

-+ 10V

12V

24V

loop 2

+ v3

- v2

-loop 1

-+

It is evident that v3 = 10V Applying KVL to loop 2, v2 + v3 + 12 = 0 v2 = -22V Applying KVL to loop 1, -24 + v1 - v2 = 0 v1 = 2V Thus, v1 = 2V, v2 = -22V, v3 = 10V Chapter 2, Solution 18

Applying KVL, -30 -10 +8 + I(3+5) = 0

8I = 32 I = 4A -Vab + 5I + 8 = 0 Vab = 28V

Page 18: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 19

Applying KVL around the loop, we obtain

-12 + 10 - (-8) + 3i = 0 i = -2A Power dissipated by the resistor: p 3 = i2R = 4(3) = 12W Power supplied by the sources: p12V = 12 (- -2) = 24W p10V = 10 (-2) = -20W

p8V = (- -2) = -16W Chapter 2, Solution 20

Applying KVL around the loop,

-36 + 4i0 + 5i0 = 0 i0 = 4A Chapter 2, Solution 21

10

+-45V -

+

+ v0 -

Apply KVL to obtain -45 + 10i - 3V0 + 5i = 0 But v0 = 10i, 3v0-45 + 15i - 30i = 0 i = -3A P3 = i2R = 9 x 5 = 45W 5

Page 19: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 22 4

+ v0 -

10A

2v0 6 At the node, KCL requires that

00 v210

4v

= 0 v0 = �–4.444V

The current through the controlled source is i = 2V0 = -8.888A and the voltage across it is

v = (6 + 4) i0 = 10 111.114

v0

Hence, p2 vi = (-8.888)(-11.111) = 98.75 W Chapter 2, Solution 23 8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3 The circuit is reduced to that shown below. ix 1 + vx - 6A 2 3

Page 20: Solution Manual for Fundamentals of Electric Circuits

Applying current division,

i A A v ix x2

2 1 36 2 1 2( ) , Vx

The current through the 1.2- resistor is 0.5ix = 1A. The voltage across the 12- resistor is 1 x 4.8 = 4.8 V. Hence the power is

pvR

W2 24 8

12192

..

Chapter 2, Solution 24

(a) I0 = 21 RR

Vs

0V I0 43 RR = 43

43

21

0

RRRR

RRV

4321

430

RRRRRR

VsV

(b) If R1 = R2 = R3

= R4 = R,

1042

RR2V

V

S

0 = 40

Chapter 2, Solution 25

V0 = 5 x 10-3 x 10 x 103 = 50V

Using current division,

I20 )5001.0(205

x5 0.1 A

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW

Page 21: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 26

V0 = 5 x 10-3 x 10 x 103 = 50V

Using current division,

I20 )5001.0(205

x5 0.1 A

V20 = 20 x 0.1 kV = 2 kV p20 = I20 V20 = 0.2 kW Chapter 2, Solution 27

Using current division,

i1 = )20(64

4 8 A

i2 = )20(64

6 12 A

Chapter 2, Solution 28

We first combine the two resistors in parallel 1015 6 We now apply voltage division,

v1 = )40(614

14 20 V

v2 = v3 = )40(614

6 12 V

Hence, v1 = 28 V, v2 = 12 V, vs = 12 V

Page 22: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 29

The series combination of 6 and 3 resistors is shorted. Hence i2 = 0 = v2

v1 = 12, i1 = 4

12 3 A

Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2 Chapter 2, Solution 30

i +v-

8

6

i1

9A 4

By current division, )9(126

12i 6 A

4 x 3 = 11 i4v,A369i 12 V p6 = 12R = 36 x 6 = 216 W Chapter 2, Solution 31

The 5 resistor is in series with the combination of 5)64(10 . Hence by the voltage division principle,

)V20(55

5v 10 V

by ohm's law,

64

1064

vi 1 A

pp = i2R = (1)2(4) = 4 W

Page 23: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 32

We first combine resistors in parallel.

3020 50

30x20 12

4010 50

40x10 8

Using current division principle,

A12)20(2012ii,A8)20(

1288ii 4321

)8(5020i1 3.2 A

)8(5030i2 4.8 A

)12(5010i3 2.4A

)12(5040i4 9.6 A

Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below

i

+v-

9A 1S

i

+ v -

4S

4S

1S

9A 2S

SS 36 259

3x6 and 25 + 25 = 4 S

Using current division,

)9(

211

1i 6 A, v = 3(1) = 3 V

Page 24: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 34 By parallel and series combinations, the circuit is reduced to the one below:

-+

+ v1

-

8i1

)132(10 625

1510x

)64(15 625

1515x 28V 6 6)66(12

Thus i1 = 68

28 2 A and v1 = 6i1 = 12 V

We now work backward to get i2 and v2.

+ 6V

-

1A

1A 6

-+ +

12V -

12

8 i1 = 2A 28V 6 0.6A

+ 3.6V

-

4

+ 6V

-

1A

1A

15

6

-+ +

12V -

12

8 i1 = 2A 28V 6

Thus, v2 = ,123)63(1513 i2 = 24.0

13v2

p2 = i2R = (0.24)2 (2) = 0.1152 W i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W Chapter 2, Solution 35

i

20 + V0

- i2

a b

5

30 70

I0i1

+ V1

-

-+

50V

Page 25: Solution Manual for Fundamentals of Electric Circuits

Combining the versions in parallel,

3070 21100

30x70 , 1520 25

5x20 4

i = 421

50 2 A

vi = 21i = 42 V, v0 = 4i = 8 V

i1 = 70v1 0.6 A, i2 =

20v2 0.4 A

At node a, KCL must be satisfied i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A Hence v0 = 8 V and I0 = 0.2A Chapter 2, Solution 36

The 8- resistor is shorted. No current flows through the 1- resistor. Hence v0 is the voltage across the 6 resistor.

I0 = 44

16324 1 A

v0 = I0 0I263 2 V

Page 26: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 37

Let I = current through the 16 resistor. If 4 V is the voltage drop across the R6 combination, then 20 - 4 = 16 V in the voltage drop across the 16 resistor.

Hence, I = 1616 1 A.

But I = 1R616

20 4 = R6R6

R6 R = 12

Chapter 2, Solution 38

Let I0 = current through the 6 resistor. Since 6 and 3 resistors are in parallel. 6I0 = 2 x 3 R0 = 1 A The total current through the 4 resistor = 1 + 2 = 3 A. Hence vS = (2 + 4 + 32 ) (3 A) = 24 V

I = v

10

S 2.4 A

Chapter 2, Solution 39

(a) Req = 0R 0

(b) Req = RRRR2R

2R R

(c) Req = R2R2)RR()RR( R

(d) Req = )R21R(R3)RRR(R3

= R

23R3

R23Rx3

R

(e) Req = R3R3R2RR3

R2R

= R3R

32R3

R32Rx3

R32 R

116

Page 27: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 40

Req = 23)362(43 5

I = 5

10qRe

10 2 A

Chapter 2, Solution 41

Let R0 = combination of three 12 resistors in parallel

121

121

121

R1

o

Ro = 4

)R14(6030)RR10(6030R 0eq

R74

)R14(603050 74 + R = 42 + 3R

or R = 16 Chapter 2, Solution 42

(a) Rab = 25

20x5)128(5)30208(5 4

(b) Rab = 5.22255442)46(1058)35(42 6.5

Chapter 2, Solution 43

(a) Rab = 8450400

2520x54010205 12

(b) 302060 10660

301

201

601 1

Rab = )1010(80100

2080 16

Page 28: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 44

(a) Convert T to Y and obtain

Rx x x

120 20 20 10 10 20

1080010

80

R R2 380020

40

The circuit becomes that shown below. R1 a R3 R2 5 b R1//0 = 0, R3//5 = 40//5 = 4.444 R Rab 2 0 4 444 40 4 444 4/ /( . ) / / .

(b) 30//(20+50) = 30//70 = 21 Convert the T to Y and obtain

Rx x x

120 10 10 40 40 20

40140040

35

R2140020

70 , R3140010

140

The circuit is reduced to that shown below. 11 R1 R2 R3 30 21

21

15

Combining the resistors in parallel

Page 29: Solution Manual for Fundamentals of Electric Circuits

R1//15 =35//15=10.5, 30//R2=30//70 = 21 leads to the circuit below. 11 10.5 21 140

21 21 Coverting the T to Y leads to the circuit below. 11 10.5 R4 R5 R6 21 R

x x xR4 6

21 140 140 21 21 2121

632121

301

R5

6321140

4515.

10.5//301 = 10.15, 301//21 = 19.63 R5//(10.15 +19.63) = 45.15//29.78 = 17.94 Rab 11 17 94 28 94. . Chapter 2, Solution 45

(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8

Rab 5 50 4 8 59 8. . (b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give 30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus Rab 5 12 8 15 32 5. .

Page 30: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 46

(a) Rab = 2060407030 80

206040100

70x30

154021 76

(b) The 10- , 50- , 70- , and 80- resistors are shorted.

3020 1250

30x20

40 60 24100

60x40

Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Chapter 2, Solution 47

205 425

20x5

6 3 29

3x6

8 a b

10

2 4 Rab = 10 + 4 + 2 + 8 = 24

Page 31: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 48

(a) Ra = 3010

100100100R

RRRRRR

3

133221

Ra = Rb = Rc = 30

(b) 3.10330

310030

50x2050x3020x30R a

,15520

3100R b 6250

3100R c

Ra = 103.3 , Rb = 155 , Rc = 62

Chapter 2, Solution 49

(a) R1 = 436

1212RRR

RR

cba

ca

R1 = R2 = R3 = 4

(b) 18103060

30x60R1

6100

10x60R 2

3100

10x30R 3

R1 = 18 , R2 = 6 , R3 = 3 Chapter 2, Solution 50 Using = 3RR Y = 3R, we obtain the equivalent circuit shown below:

3R 30mA

R

R 3R

3R 3R

30mA 3R/2

Page 32: Solution Manual for Fundamentals of Electric Circuits

RR3 R43

R4RxR3

)R4/(3)R4/()RxR3(R3

RR

23R3

R23Rx3

R23R3R

43R

43R3

P = I2R 800 x 10-3 = (30 x 10-3)2 R R = 889 Chapter 2, Solution 51

(a) 153030 and 12)50/(20x302030

Rab = )39/(24x15)1212(15 9.31

b

15

a

20

30

b

a

20

30 30 30

12

12

(b) Converting the T-subnetwork into its equivalent network gives Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Rb'c' = 350/(10) = 35 , Ra'c' = 350/(20) = 17.5

Also 21)100/(70x307030 and 35/(15) = 35x15/(50) = 10.5

Rab = 25 + 5.315.1725)5.1021(5.17 Rab = 36.25

b�’

c�’ c�’ b

a

15 35 17.5

25 70 a�’

b

a

30

25

5

10

15

20

30

Page 33: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 52 (a) We first convert from T to .

R3

R2

b

a

100

100 100

100

100

100

100

b

a

200

200

100

100 100

100

100

100

100

100 R1

800100

80000100

100x200200x200200x1001R

R2 = R3 = 80000/(200) = 400

But 80500

400x100400100

We connect the to Y.

Ra = Rc = 3

400960

000,648008080

800x80

Rc

Rb

b

aRa

100

100

100

100

100

b

a

800

80

80 100

100

100

100

100

Rb = 320

96080x80

We convert T to .

R3�’

R2�’

b

a

500/3

500/3

R1�’

b

a

500/3

500/3

320/3

100

100

Page 34: Solution Manual for Fundamentals of Electric Circuits

75.293)3/(320

)3/(000,94

3320

3320x100

3320x100100x100

'1R

33.313100

)3/(000,94RR 13

'2

796.108)3/(1440

)3/(500x)3/(940)3/(500)30/(940

Rab = 36.511

6.217x75.293)796.108x2(75.293 125

(b) Converting the Ts to s, we have the equivalent circuit below.

100 a

b

300

300 100

100

300

100 300

300

100

300

100

100

100

b

a 100

100

100

100

100

100 ,75)400/(100x300100300 100)450/(150x300)7575(300

Rab = 100 + )400/(300x100200100300100 Rab = 2.75

100

300

300

100

300

100

100

Page 35: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 53 (a) Converting one to T yields the equivalent circuit below:

20

b�’

c�’

a�’

b 80

a 20

5

4

60

30

Ra'n = ,4501040

10x40 ,5100

50x10R n'b 20100

50x40R n'c

Rab = 20 + 80 + 20 + 6534120)560()430( Rab = 142.32

(a) We combine the resistor in series and in parallel.

2090

60x30)3030(30

We convert the balanced s to Ts as shown below:

b

10

10 10

10

10

30 30

30

30

30

30 b

a

20

20 10

a

Rab = 10 + 40202010)102010()1010( Rab = 33.33

Chapter 2, Solution 54

(a) Rab 50 100 150 100 150 50 100 400 130/ /( ) / / (b) Rab 60 100 150 100 150 60 100 400 140/ /( ) / /

Page 36: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 55 We convert the T to .

50

I0

-+

24 V

b

a

35

70

140

Req

-+

24 V 60

b

a

20

40

10

20

I0

60

70

Req

Rab = 3540

140040

20x1010x4040x20R

RRRRRR

3

133221

Rac = 1400/(10) = 140 , Rbc = 1400/(40) = 35 357070 and 160140 140x60/(200) = 42

Req = 0625.24)4235(35 I0 = 24/(Rab) = 0.9774A Chapter 2, Solution 56 We need to find Req and apply voltage division. We first tranform the Y network to .

c

35

16

30

37.5 a b

30 45

Req

+100 V

-20

35

16

30

10

12

15

Req

+ 100 V

-

20

Rab = 5.3712450

1215x1212x1010x15

Rac = 450/(10) = 45 , Rbc = 450/(15) = 30 Combining the resistors in parallel,

Page 37: Solution Manual for Fundamentals of Electric Circuits

30||20 = (600/50) = 12 ,

37.5||30 = (37.5x30/67.5) = 16.667

35||45 = (35x45/80) = 19.688

Req = 19.688||(12 + 16.667) = 11.672

By voltage division,

v = 10016672.11

672.11 = 42.18 V

Chapter 2, Solution 57

4

ec

f

a

bd

36

7

27

2

18

28 10

1 14

Rab = 1812216

126x88x1212x6

Rac = 216/(8) = 27 , Rbc = 36

Rde = 7856

84x88x22x4

Ref = 56/(4) = 14 , Rdf = 56/(2) = 28 Combining resistors in parallel,

,368.7382802810 868.5

437x36736

7.230

3x27327

Page 38: Solution Manual for Fundamentals of Electric Circuits

4

0.5964 3.977 1.829

4

7.568 14

2.7

14

18

5.868 7.568

829.1567.26

7.2x18867.57.218

7.2x18R an

977.3567.26868.5x18R bn

5904.0567.26

7.2x868.5R cn

)145964.0()368.7977.3(829.14R eq

5964.14346.11829.5 12.21 i = 20/(Req) = 1.64 A

Chapter 2, Solution 58 The resistor of the bulb is 120/(0.75) = 160 2.25 A 1.5 A 40

0.75 A

80 160 + 90 V - +

120-+

VS Once the 160 and 80 resistors are in parallel, they have the same voltage 120V. Hence the current through the 40 resistor is 40(0.75 + 1.5) = 2.25 x 40 = 90 Thus vs = 90 + 120 = 210 V

Page 39: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 59 Total power p = 30 + 40 + 50 + 120 W = vi

or i = p/(v) = 120/(100) = 1.2 A

Chapter 2, Solution 60

p = iv i = p/(v) i30W = 30/(100) = 0.3 A i40W = 40/(100) = 0.4 A i50W = 50/(100) = 0.5 A Chapter 2, Solution 61 There are three possibilities

(a) Use R1 and R2: R = 35.429080RR 21 p = i2R i = 1.2A + 5% = 1.2 0.06 = 1.26, 1.14A p = 67.23W or 55.04W, cost = $1.50

(b) Use R1 and R3:

R = 44.4410080RR 31 p = I2R = 70.52W or 57.76W, cost = $1.35

(c) Use R2 and R3: R = 37.4710090RR 32 p = I2R = 75.2W or 61.56W, cost = $1.65

Note that cases (b) and (c) give p that exceed 70W that can be supplied. Hence case (a) is the right choice, i.e. R1 and R2

Chapter 2, Solution 62

pA = 110x8 = 880 W, pB = 110x2 = 220 W Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60

Page 40: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 63

Use eq. (2.61),

Rn = 04.010x25

100x10x2RII 3

3

mm

mI

In = I - Im = 4.998 A p = (I 9992.0)04.0()998.4R 22

n 1 W Chapter 2, Solution 64

When Rx = 0, i R = A10x 1110

110

When Rx is maximum, ix = 1A 1101

110R xR

i.e., Rx = 110 - R = 99 Thus, R = 11 , Rx = 99 Chapter 2, Solution 65

k1mA1050R

IV

R mfs

fsn 4 k

Chapter 2, Solution 66

20 k /V = sensitivity = fsI1

i.e., Ifs = A50V/k201

The intended resistance Rm = k200)V/k20(10IV

fs

fs

(a) k200A50V50R

iV

R mfs

fsn 800 k

(b) p = )k800()A50(RI 2n

2fs 2 mW

Page 41: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 67

(a) By current division, i0 = 5/(5 + 5) (2 mA) = 1 mA V0 = (4 k ) i0 = 4 x 103 x 10-3 = 4 V

(b) .k4.2k6k4 By current division,

mA19.1)mA2(54.21

5i '0

)mA19.1)(k4.2(v'0 2.857 V

(c) % error = 0

'00

vvv

x 100% = 100x4143.1

28.57%

(d) .k6.3k30k4 By current division,

mA042.1)mA2(56.31

5i '0

V75.3)mA042.1)(k6.3(v'0

% error = 4

100x25.0%100xv

vv

0

'0 6.25%

Chapter 2, Solution 68

(a) 602440

i = 2416

40.1 A

(b) 24116

4i ' 0.09756 A

(c) % error = %100x1.009756.01.0

2.44%

Page 42: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 69

With the voltmeter in place,

Sm2S1

m20 V

RRRRRR

V

where Rm = 100 k without the voltmeter,

SS21

20 V

RRRR

V

(a) When R2 = 1 k , k101100RR 2m

V0 = )40(

30101100101100

1.278 V (with)

V0 = )40(301

1 1.29 V (without)

(b) When R2 = 10 k , k091.91101000RR m2

V0 = )40(30091.9

091.9 9.30 V (with)

V0 = )40(3010

1010 V (without)

(c) When R2 = 100 k , k50RR m2

)40(3050

50V0 25 V (with)

V0 = )40(30100

100 30.77 V (without)

Chapter 2, Solution 70

(a) Using voltage division, v Va

1212 8

25 15( )

v Vb10

10 1525 10( )

v v vab a b V15 10 5

Page 43: Solution Manual for Fundamentals of Electric Circuits

(b) + 8k 15k 25 V - a b

12k 10k o

v v V v v va b ab a b V0 10 0 10 1, , 0 Chapter 2, Solution 71

Vs RL

R1

+−

iL

Given that vs = 30 V, R1 = 20 , IL = 1 A, find RL.

v i R R Rvi

Rs L L Ls

L( )1 1

301

20 10

Page 44: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 72

The system can be modeled as shown. 12A +

9V R R R - The n parallel resistors R give a combined resistance of R/n. Thus,

9 12129

12 159

20xRn

nxR x

Chapter 2, Solution 73 By the current division principle, the current through the ammeter will be

one-half its previous value when R = 20 + Rx 65 = 20 + Rx Rx = 45 Chapter 2, Solution 74

With the switch in high position,

6 = (0.01 + R3 + 0.02) x 5 R3 = 1.17 At the medium position,

6 = (0.01 + R2 + R3 + 0.02) x 3 R2 + R3 = 1.97 or R2 = 1.97 - 1.17 = 0.8 At the low position, 6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 + R2 + R3 = 5.97 R1 = 5.97 - 1.97 = 4

Page 45: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 75

MR 100

VS -+

12

(a) When Rx = 0, then

Im = Ifs = mRR

t R2 = mfs

2

RIE = k9.19100

10x1.02

3

(b) For half-scale deflection, Im = mA05.02Ifs

Im = xm RRR

E Rx = k2010x05.0

2)RR(IE

3mm

20 k

Chapter 2, Solution 76

For series connection, R = 2 x 0.4 = 0.8

8.0)120(

RV 22

p 18 k (low)

For parallel connection, R = 1/2 x 0.4 = 0.2

p = 2.0)120(

RV

22

72 kW (high)

Chapter 2, Solution 77

(a) 5 = 202020201010

i.e., four 20 resistors in parallel.

(b) 311.8 = 300 + 10 + 1.8 = 300 + 8.12020 i.e., one 300 resistor in series with 1.8 resistor and a parallel combination of two 20 resistors.

(c) 40k = 12k + 28k = k50k56k2424 i.e., Two 24k resistors in parallel connected in series with two 50k resistors in parallel.

(d) 42.32k = 42l + 320

= 24k + 28k = 320 = 24k = 20300k56k56

i.e., A series combination of 20 resistor, 300 resistor, 24k resistor and a parallel combination of two 56k resistors.

Page 46: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 78

The equivalent circuit is shown below:

R

+ V0

--+

VS (1- )R

V0 = S0S VR)1(VR)1(R

R)1(

R)1(VV

S

0

Chapter 2, Solution 79

Since p = v2/R, the resistance of the sharpener is R = v2/(p) = 62/(240 x 10-3) = 150 I = p/(v) = 240 mW/(6V) = 40 mA Since R and Rx are in series, I flows through both. IRx = Vx = 9 - 6 = 3 V Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75

Chapter 2, Solution 80 The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:

V -+

Case 1

R1 -+ V R2

Hence ,R

Vp2

2

1

1

2

RR

pp

)12(4

10pRR

12

12p 30 W

Case 2

Page 47: Solution Manual for Fundamentals of Electric Circuits

Chapter 2, Solution 81

Let R1 and R2 be in k . 5RR 21eqR (1)

12

2

S

0

RR5R5

VV

(2)

From (1) and (2), 40R5

05. 10 2 = 2

22 R5

R5R5 or R2 = 3.33 k

From (1), 40 = R1 + 2 R1 = 38 k Thus R1 = 38 k , R2 = 3.33 k

Chapter 2, Solution 82 (a) 10

1 2

10

40 80 R12

R12 = 80 + 6

5080)4010(10 88.33

(b)

3 20

10

10

80

40

R13 1

R13 = 80 + 501010020)4010(10 108.33

Page 48: Solution Manual for Fundamentals of Electric Circuits

20 10

10

1

4

80

(c) R14 40

R14 = 2008020)104010(080 100 Chapter 2, Solution 83 The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible

compared with 24 V. Ignoring this voltage amp, we can calculate the current through the devices.

I1 = mA5V9mW

V1

1 45p

I2 = mA2024mW

V2

2 480p

60 mA i2 = 20 mA

R1

iR2

iR1

i1 = 5 mA

R2 -+

24 V By applying KCL, we obtain and mA402060I

1R mA35540I2R

Hence, R1RI 1 = 24 - 9 = 15 V

mA40V15R1 375

V9RI 2R 2 mA35V9

2R 257.14

Page 49: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 1.

8

v2

2

40 v1

6 A 10 A

At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V

P8 = 8

143.98v 22

1 10.45 W

P4 = 4vv 2

21 94.37 W

P2 = 2286.10

2v 21

2 52.9 W

Chapter 3, Solution 2

At node 1,

2

vv6

5v

10v 2111 60 = - 8v1 + 5v2 (1)

At node 2,

2

vv63

4v 212 36 = - 2v1 + 3v2 (2)

Solving (1) and (2), v1 = 0 V, v2 = 12 V

Page 50: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 3 Applying KCL to the upper node,

10 = 60v

230v

20v

10v 0oo0 v0 = 40 V

i1 = 10

0v 4 A , i2 =

20v0 2 A, i3 =

30v0 1.33 A, i4 =

60v0 67 mA

Chapter 3, Solution 4 At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node.

k4

vk6v20

k2v30 000 v0 = 20 V

i1

10 10

2A

i4 i3

v1 v2

i2

4 A 5 5 5 A

Page 51: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 6

i1 + i2 + i3 = 0 02

10v6v

412v 002

or v0 = 8.727 V

Chapter 3, Solution 7 At node a,

babaaa VV

VVVV3610

10153010

(1)

At node b,

babbba VV

VVVV72240

59

2012

10 (2)

Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V Chapter 3, Solution 8

i2

5

2

i13 v1 i3

1

+�–

4V0

3V�–+

+

V0

�–

i1 + i2 + i3 = 0 05

v4v1

3v5v 0111

But 10 v52v so that v1 + 5v1 - 15 + v1 - 0v

58

1

or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V

Page 52: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 9

+ v1

�–

i2

6 i13 v1 i3

8 +�–

+ v0 �–

2v0 12V

�–+

At the non-reference node,

6

v2v8v

3v 011112

(1)

But -12 + v0 + v1 = 0 v0 = 12 - v1 (2) Substituting (2) into (1),

6

24v38v

3v 11112

v0 = 3.652 V

Chapter 3, Solution 10 At node 1,

8v

41

vv 112 32 = -v1 + 8v2 - 8v0 (1)

1

4A 2i0

i0

v1 v0

8 2

v2 4

Page 53: Solution Manual for Fundamentals of Electric Circuits

At node 0,

00 I2

2v

4 and 8v

I 10 16 = 2v0 + v1 (2)

At node 2,

2I0 = 4

v1

v 212v and

8v1

0I v2 = v1 (3)

From (1), (2) and (3), v0 = 24 V, but from (2) we get

io = 624242

22v4 o

= - 4 A

Chapter 3, Solution 11

i3

6

vi1 i24 3

�–+

10 V 5 A Note that i2 = -5A. At the non-reference node

6v5

4v10 v = 18

i1 = 4

v10 -2 A, i2 = -5 A

Chapter 3, Solution 12

i3

40

v1 v210 20

�–+ 5 A

50 24 V

Page 54: Solution Manual for Fundamentals of Electric Circuits

At node 1, 40

0v20

vv10

v24 1211 96 = 7v1 - 2v2 (1)

At node 2, 50v

20vv 2215 500 = -5v1 + 7v2 (2)

Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V

40v

i 11 1.072 A, v2 =

50v2 2.041 A

Chapter 3, Solution 13

At node number 2, [(v2 + 2) �– 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) �– 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

Chapter 3, Solution 14

1 2

4

v0 v1

5 A

8

20 V �–+

40 V �–+

At node 1, 1

v405

2vv 001 v1 + v0 = 70 (1)

At node 0, 8

20v4v

52

vv 0001 4v1 - 7v0 = -20 (2)

Solving (1) and (2), v0 = 20 V

Page 55: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 15

1 2

4

v0 v1

5 A

8

20 V �–+

40 V �–+

Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1) At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2) At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3) Substituting (1) and (3) into (2),

2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 = 1156

v1 = v2 + 10 = 1154

i0 = 6vi = 29.45 A

P65 = 61154Gv

R

221

21v

144.6 W

P55 = 51156Gv

222 129.6 W

P35 = 3)2(Gvv 22

3L 12 W

Page 56: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 16

2 S

+ v0 �–

13 V �–+

i0

1 S 4 S

8 Sv1 v2 v3

2 A At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 �– v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But

v1 = v2 + 2v0 and v0 = v2. Hence

v1 = 3v2 (2) v3 = 13V (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Solution 17

60 V

i0

3i0

2

10

4

60 V �–+

8

Page 57: Solution Manual for Fundamentals of Electric Circuits

At node 1, 2

vv8v

4v60 2111 120 = 7v1 - 4v2 (1)

At node 2, 3i0 + 02

vv10

v60 212

But i0 = .4

v60 1

Hence

02

vv10

v604

v603 2121 1020 = 5v1 - 12v2 (2)

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = 4

v60 1 1.73 A

Chapter 3, Solution 18

+ v3 �–

+ v1 �–

10 V

�– + v1

v2

2 2

4 8

v3

5 A

(a) (b)

At node 2, in Fig. (a), 5 = 2

vv2

vv 3212 10 = - v1 + 2v2 - v3 (1)

At the supernode, 8v

4v

2vv

2vv 313212 40 = 2v1 + v3 (2)

From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

Page 58: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 19

At node 1,

32112131 4716

48235 VVVVVVVV

(1)

At node 2,

32132221 270

428VVV

VVVVV (2)

At node 3,

32132313 724360

42812

3 VVVVVVVV (3)

From (1) to (3),

BAVVVV

360

16

724271417

3

2

1

Using MATLAB,

V 267.12 V, 933.4 V, 10267.12933.410

3211 VVVBAV

Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence

040414 321

321 VVVVVV

(1)

. V1 . V2 2 V3 4 1 4

Page 59: Solution Manual for Fundamentals of Electric Circuits

Between nodes 1 and 3,

12012 1331 VVVV (2) Similarly, between nodes 1 and 2, (3) iVV 221

But i . Combining this with (2) and (3) gives 4/3V.V (4) 2/6 12 V

Solving (1), (2), and (4) leads to

V15 V,5.4 V,3 321 VVV Chapter 3, Solution 21

4 k

2 k

+ v0 �–

3v0v1 v2v3

1 k

+

3 mA

3v0

+ v2 �–

+ v3 �–

+

(b) (a) Let v3 be the voltage between the 2k resistor and the voltage-controlled voltage source. At node 1,

2000

vv4000

vv10x3 31213 12 = 3v1 - v2 - 2v3 (1)

At node 2,

1v

2vv

4vv 23121 3v1 - 5v2 - 2v3 = 0 (2)

Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V

Page 60: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 22

At node 1, 8

vv3

4v

2v 011012

24 = 7v1 - v2 (1)

At node 2, 3 + 1

v5v8

vv 2221

But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 (2) Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V

Chapter 3, Solution 23

At the supernode, 5 + 2 = 5v

10v 21 70 = v1 + 2v2 (1)

Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2) Solving (1) and (2), v1 = 18 V, v2 = 26 V

+ v1

�–5 10

v1 v2

8 V

�– +

5 A 2 A

+

v2

�– (b)(a)

Page 61: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 24

6mA 1 k 2 k 3 k V1 V2 + io - 30V 15V - 4 k 5 k + At node 1,

212111 2796

246

130 VVVVVV

(1)

At node 2,

211222 311530

253)15(

6 VVVVVV (2)

Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25 Using nodal analysis,

1

v0

2

4

2

10V �–+

�–+

40V �–+

i0

20V

40v

2v10

2v40

1v20 0000 v0 = 20V

i0 = 1

v20 0 = 0 A

Page 62: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 26 At node 1,

32121311 24745

5103

2015

VVVVVVVV

(1)

At node 2,

554

532221 VVVIVV o (2)

But 10

31 VVIo . Hence, (2) becomes

321 31570 VVV (3) At node 3,

32132331 52100

5510

103 VVV

VVVVV (4)

Putting (1), (3), and (4) in matrix form produces

BAVVVV

10045

5213157247

3

2

1

Using MATLAB leads to

96.1982.4835.9

1BAV

Thus, V 95.1 V, 982.4 V, 835.9 321 VVV Chapter 3, Solution 27 At node 1, 2 = 2v1 + v1 �– v2 + (v1 �– v3)4 + 3i0, i0 = 4v2. Hence, 2 = 7v1 + 11v2 �– 4v3 (1) At node 2, v1 �– v2 = 4v2 + v2 �– v3 0 = �– v1 + 6v2 �– v3 (2) At node 3,

2v3 = 4 + v2 �– v3 + 12v2 + 4(v1 �– v3)

Page 63: Solution Manual for Fundamentals of Electric Circuits

or �– 4 = 4v1 + 13v2 �– 7v3 (3) In matrix form,

402

vvv

71341614117

3

2

1

,1767134

1614117

1107134

1604112

1

,66744

101427

2 2864134

0612117

3

v1 = ,V625.01761101 v2 = V375.0

176662

v3 = .V625.11762863

v1 = 625 mV, v2 = 375 mV, v3 = 1.625 V. Chapter 3, Solution 28 At node c,

dcbcbccd VVVVVVVV

211505410

(1)

At node b,

cbabbcba VVVVVVVV

2445848

45 (2)

At node a,

dbabaada VVVVVVVV

427300845

16430

(3)

At node d,

dcacddda VVVVVVVV

72515010204

30 (4)

In matrix form, (1) to (4) become

Page 64: Solution Manual for Fundamentals of Electric Circuits

BAV

VVVV

d

c

b

a

15030450

72054027

024121150

We use MATLAB to invert A and obtain

17.29736.1

847.714.10

1BAV

Thus, V 17.29 V, 736.1 V, 847.7 V, 14.10 dcba VVVV Chapter 3, Solution 29 At node 1,

42121141 45025 VVVVVVVV (1) At node 2,

32132221 4700)(42 VVVVVVVV (2) At node 3,

4324332 546)(46 VVVVVVV (3) At node 4,

43144143 5232 VVVVVVVV (4) In matrix form, (1) to (4) become

BAV

VVVV

2605

51011540

04711014

4

3

2

1

Using MATLAB,

7076.0309.2209.17708.0

1BAV

i.e. V 7076.0 V, 309.2 V, 209.1 V, 7708.0 4321 VVVV

Page 65: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 30

v1

10

20

80

40

1

v0

I0

2I0

2

v2

4v0

120 V

+�–�–

+

�– +

100 V At node 1,

20

vv410

v10040

vv 1o121 (1)

But, vo = 120 + v2 v2 = vo �– 120. Hence (1) becomes 7v1 �– 9vo = 280 (2) At node 2,

Io + 2Io = 80

0v o

80v

40v120v

3 oo1

or 6v1 �– 7vo = -720 (3)

from (2) and (3), 720

280vv

7697

o

1

554497697

844077209280

1 , 67207206

28072

Page 66: Solution Manual for Fundamentals of Electric Circuits

v1 = ,16885

84401 vo = V13445

67202

Io = -5.6 A

Chapter 3, Solution 31 1

i0

2v0 v3

1

2

4 10 V

�–+

v1 v2

+ v0 �–

4

1 A At the supernode,

1 + 2v0 = 1

vv1

v4v 3121 (1)

But vo = v1 �– v3. Hence (1) becomes, 4 = -3v1 + 4v2 +4v3 (2) At node 3,

2vo + 2

v10vv

4v 3

312

or 20 = 4v1 + v2 �– 2v3 (3)

At the supernode, v2 = v1 + 4io. But io = 4

v 3 . Hence,

v2 = v1 + v3 (4) Solving (2) to (4) leads to, v1 = 4 V, v2 = 4 V, v3 = 0 V.

Page 67: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 32

v3

v1 v2

10 V

loop 2

�– +

20 V

+ �–

12 V

loop 1

�–+

+ v3 �–

+ v1 �–

10 k

4 mA

5 k

(b)(a) We have a supernode as shown in figure (a). It is evident that v2 = 12 V, Applying KVL to loops 1and 2 in figure (b), we obtain,

-v1 �– 10 + 12 = 0 or v1 = 2 and -12 + 20 + v3 = 0 or v3 = -8 V Thus, v1 = 2 V, v2 = 12 V, v3 = -8V. Chapter 3, Solution 33

(a) This is a non-planar circuit because there is no way of redrawing the circuit with no crossing branches.

(b) This is a planar circuit. It can be redrawn as shown below.

1

2

3

4

5 12 V

�–+

Page 68: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 34 (a) This is a planar circuit because it can be redrawn as shown below,

7

10 V �–+

1 2

3

4

6

5 (b) This is a non-planar circuit. Chapter 3, Solution 35

5 k i1

i2

+ v0 �–2 k

30 V �–+

�–+20 V

4 k Assume that i1 and i2 are in mA. We apply mesh analysis. For mesh 1,

-30 + 20 + 7i1 �– 5i2 = 0 or 7i1 �– 5i2 = 10 (1) For mesh 2, -20 + 9i2 �– 5i1 = 0 or -5i1 + 9i2 = 20 (2) Solving (1) and (2), we obtain, i2 = 5. v0 = 4i2 = 20 volts.

Page 69: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 36

i1 i2

2

4 10 V

+ �–

12 V �–+

I2

6 I1

i3 Applying mesh analysis gives,

12 = 10I1 �– 6I2

-10 = -6I1 + 8I2

or 2

1

II

4335

56

,114335

,94536

1 753

652

,119I 1

1 11

7I 22

i1 = -I1 = -9/11 = -0.8181 A, i2 = I1 �– I2 = 10/11 = 1.4545 A.

vo = 6i2 = 6x1.4545 = 8.727 V.

Chapter 3, Solution 37

5

4v0 2

1

3

3 V �–+

+�–

+ v0 �–

i2

i1

Page 70: Solution Manual for Fundamentals of Electric Circuits

Applying mesh analysis to loops 1 and 2, we get,

6i1 �– 1i2 + 3 = 0 which leads to i2 = 6i1 + 3 (1) -1i1 + 6i2 �– 3 + 4v0 = 0 (2) But, v0 = -2i1 (3)

Using (1), (2), and (3) we get i1 = -5/9. Therefore, we get v0 = -2i1 = -2(-5/9) = 1.111 volts Chapter 3, Solution 38

6

2v0 8

3

12 V �–+ +

�–

+ v0 �–

i2

i1

We apply mesh analysis.

12 = 3 i1 + 8(i1 �– i2) which leads to 12 = 11 i1 �– 8 i2 (1) -2 v0 = 6 i2 + 8(i2 �– i1) and v0 = 3 i1 or i1 = 7 i2 (2)

From (1) and (2), i1 = 84/69 and v0 = 3 i1 = 3x89/69 v0 = 3.652 volts Chapter 3, Solution 39 For mesh 1, 0610210 21 III x But . Hence, 21 III x

212121 245610121210 IIIIII (1) For mesh 2,

2112 43606812 IIII (2) Solving (1) and (2) leads to -0.9A A, 8.0 21 II

Page 71: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 40

2 k

i2

6 k

4 k

2 k 6 k

i3

i1

�–+

4 k

30V Assume all currents are in mA and apply mesh analysis for mesh 1.

30 = 12i1 �– 6i2 �– 4i3 15 = 6i1 �– 3i2 �– 2i3 (1) for mesh 2,

0 = - 6i1 + 14i2 �– 2i3 0 = -3i1 + 7i2 �– i3 (2) for mesh 2,

0 = -4i1 �– 2i2 + 10i3 0 = -2i1 �– i2 + 5i3 (3) Solving (1), (2), and (3), we obtain, io = i1 = 4.286 mA. Chapter 3, Solution 41 10

i3

ii2

2

1

8 V �–+

6 V

+ �–

i3

i2

i1

5 4

0

Page 72: Solution Manual for Fundamentals of Electric Circuits

For loop 1, 6 = 12i1 �– 2i2 3 = 6i1 �– i2 (1) For loop 2, -8 = 7i2 �– 2i1 �– i3 (2) For loop 3,

-8 + 6 + 6i3 �– i2 = 0 2 = 6i3 �– i2 (3) We put (1), (2), and (3) in matrix form,

283

iii

610172016

3

2

1

,234610172016

240620182036

2

38210872316

3

At node 0, i + i2 = i3 or i = i3 �– i2 = 234

2403823 = 1.188 A

Page 73: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 42 For mesh 1, (1) 2121 3050120305012 IIIIFor mesh 2, (2) 321312 40100308040301008 IIIIIIFor mesh 3, (3) 3223 50406040506 IIIIPutting eqs. (1) to (3) in matrix form, we get

BAIIII

68

12

50400401003003050

3

2

1

Using Matlab,

44.040.048.0

1BAI

i.e. I1 = 0.48 A, I2 = 0.4 A, I3 = 0.44 A Chapter 3, Solution 43

30

30

20

20

20

80 V �–+

80 V �–+

i3

i2

i1

a

+

Vab �–

30 b For loop 1, 80 = 70i1 �– 20i2 �– 30i3 8 = 7i1 �– 2i2 �– 3i3 (1)

Page 74: Solution Manual for Fundamentals of Electric Circuits

For loop 2, 80 = 70i2 �– 20i1 �– 30i3 8 = -2i1 + 7i2 �– 3i3 (2) For loop 3, 0 = -30i1 �– 30i2 + 90i3 0 = i1 + i2 �– 3i3 (3) Solving (1) to (3), we obtain i3 = 16/9

Io = i3 = 16/9 = 1.778 A Vab = 30i3 = 53.33 V. Chapter 3, Solution 44 6 V

1

4 i3 i2

i1

3 A

+

6 V �–+

2

5 i2 i1

Loop 1 and 2 form a supermesh. For the supermesh, 6i1 + 4i2 - 5i3 + 12 = 0 (1) For loop 3, -i1 �– 4i2 + 7i3 + 6 = 0 (2) Also, i2 = 3 + i1 (3) Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 �– i3 = -1.7333 A

Page 75: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 45 4 8

1 i1

2 6

3 �–+

i4i3

i2

30V For loop 1, 30 = 5i1 �– 3i2 �– 2i3 (1) For loop 2, 10i2 - 3i1 �– 6i4 = 0 (2) For the supermesh, 6i3 + 14i4 �– 2i1 �– 6i2 = 0 (3) But i4 �– i3 = 4 which leads to i4 = i3 + 4 (4) Solving (1) to (4) by elimination gives i = i1 = 8.561 A. Chapter 3, Solution 46

For loop 1, 12811081112 2121 iiii (1)

For loop 2, 02148 21 ovii

But v , 13io

21121 706148 iiiii (2) Substituting (2) into (1),

1739.012877 222 iii A and 217.17 21 ii A

Page 76: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 47 First, transform the current sources as shown below.

- 6V + 2 I3 V1 8 V2 4 V3 4 8 I1 2 I2 + + 20V 12V - - For mesh 1,

321321 47100821420 IIIIII (1) For mesh 2,

321312 2760421412 IIIIII (2) For mesh 3,

321123 7243084146 IIIIII (3) Putting (1) to (3) in matrix form, we obtain

BAIIII

36

10

724271417

3

2

1

Using MATLAB,

8667.1,0333.0 ,5.28667.10333.02

3211 IIIBAI

But

1 120

20 4 10 V4V

I V 1I

2 1 22( ) 4.933 VV I I Also,

32 3 2

1212 8 12.267V

8V

I V I

Page 77: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 48 We apply mesh analysis and let the mesh currents be in mA.

3k I4 4k 2k 5k Io 1k I3 - I1 I2 6V + + 12 V + 10k - 8V - For mesh 1,

421421 454045812 IIIIII (1) For mesh 2,

43214312 2101380210138 IIIIIIII (2) For mesh 3,

432423 5151060510156 IIIIII (3) For mesh 4,

014524 4321 IIII (4) Putting (1) to (4) in matrix form gives

BAI

IIII

0684

145245151002101314015

4

3

2

1

Using MATLAB,

6791.7087.8217.7

1BAI

The current through the 10k resistor is Io= I2 �– I3 = 0.2957 mA

Page 78: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 49

3

(a)

2i0

i20

16 V �–+

1 2

i3

i2i1

i1

2

16V 2

i2

+ v0 �–

+ v0 or�–

i1

1

�–+

2 (b)

For the supermesh in figure (a), 3i1 + 2i2 �– 3i3 + 16 = 0 (1) At node 0, i2 �– i1 = 2i0 and i0 = -i1 which leads to i2 = -i1 (2) For loop 3, -i1 �–2i2 + 6i3 = 0 which leads to 6i3 = -i1 (3) Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.

Page 79: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 50

10

4 2 i3

i2

i1

3i0

60 V �–+

8 i3 i2

For loop 1, 16i1 �– 10i2 �– 2i3 = 0 which leads to 8i1 �– 5i2 �– i3 = 0 (1) For the supermesh, -60 + 10i2 �– 10i1 + 10i3 �– 2i1 = 0 or -6i1 + 5i2 + 5i3 = 30 (2) Also, 3i0 = i3 �– i2 and i0 = i1 which leads to 3i1 = i3 �– i2 (3) Solving (1), (2), and (3), we obtain i1 = 1.731 and i0 = i1 = 1.731 A Chapter 3, Solution 51

5 A

+

v0

1

4

2

i3

i2

i1

20V �–+40 V

�–+

8

Page 80: Solution Manual for Fundamentals of Electric Circuits

For loop 1, i1 = 5A (1) For loop 2, -40 + 7i2 �– 2i1 �– 4i3 = 0 which leads to 50 = 7i2 �– 4i3 (2) For loop 3, -20 + 12i3 �– 4i2 = 0 which leads to 5 = - i2 + 3 i3 (3) Solving with (2) and (3), i2 = 10 A, i3 = 5 A And, v0 = 4(i2 �– i3) = 4(10 �– 5) = 20 V. Chapter 3, Solution 52

i3

i2

2

4

8

i3

i2

i1

3A

+ v0 �–

2V0 +�–

�–+

VS For mesh 1, 2(i1 �– i2) + 4(i1 �– i3) �– 12 = 0 which leads to 3i1 �– i2 �– 2i3 = 6 (1) For the supermesh, 2(i2 �– i1) + 8i2 + 2v0 + 4(i3 �– i1) = 0 But v0 = 2(i1 �– i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,

i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.

Page 81: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 53

i3

i2

2

4

8

i3

i2

i1

3A

+ v0 �–

2V0 +�–

�–+

VS For mesh 1, 2(i1 �– i2) + 4(i1 �– i3) �– 12 = 0 which leads to 3i1 �– i2 �– 2i3 = 6 (1) For the supermesh, 2(i2 �– i1) + 8i2 + 2v0 + 4(i3 �– i1) = 0 But v0 = 2(i1 �– i2) which leads to -i1 + 3i2 + 2i3 = 0 (2) For the independent current source, i3 = 3 + i2 (3) Solving (1), (2), and (3), we obtain,

i1 = 3.5 A, i2 = -0.5 A, i3 = 2.5 A.

Page 82: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 54

Let the mesh currents be in mA. For mesh 1, 2121 22021012 IIII (1)

For mesh 2, (2) 321312 3100310 IIIIII

For mesh 3, 3223 2120212 IIII (3)

Putting (1) to (3) in matrix form leads to

BAIIII

12102

210131

012

3

2

1

Using MATLAB,

mA 25.10,mA 5.8 ,mA 25.525.105.825.5

3211 IIIBAI

Chapter 3, Solution 55

dI1 i2

a 0

b c

1A

4A

i1

i3

I4

1A

I2

I3

8 V

12 4

2

6

10 V

+

+ �–

4 A

I4

I3

I2

Page 83: Solution Manual for Fundamentals of Electric Circuits

It is evident that I1 = 4 (1) For mesh 4, 12(I4 �– I1) + 4(I4 �– I3) �– 8 = 0 (2) For the supermesh 6(I2 �– I1) + 10 + 2I3 + 4(I3 �– I4) = 0 or -3I1 + 3I2 + 3I3 �– 2I4 = -5 (3) At node c, I2 = I3 + 1 (4) Solving (1), (2), (3), and (4) yields, I1 = 4A, I2 = 3A, I3 = 2A, and I4 = 4A At node b, i1 = I2 �– I1 = -1A At node a, i2 = 4 �– I4 = 0A At node 0, i3 = I4 �– I3 = 2A Chapter 3, Solution 56 + v1 �–

2

2

2

2

2

i3

i2

i1 12 V

�–+

+

v2 �–

For loop 1, 12 = 4i1 �– 2i2 �– 2i3 which leads to 6 = 2i1 �– i2 �– i3 (1) For loop 2, 0 = 6i2 �–2i1 �– 2 i3 which leads to 0 = -i1 + 3i2 �– i3 (2) For loop 3, 0 = 6i3 �– 2i1 �– 2i2 which leads to 0 = -i1 �– i2 + 3i3 (3)

Page 84: Solution Manual for Fundamentals of Electric Circuits

In matrix form (1), (2), and (3) become,

006

iii

311131112

3

2

1

= ,8311131112

2 = 24301131162

3 = 24011031612

, therefore i2 = i3 = 24/8 = 3A,

v1 = 2i2 = 6 volts, v = 2i3 = 6 volts Chapter 3, Solution 57

Assume R is in kilo-ohms. VVVVmAxkV 2872100100,72184 212

Current through R is

RR

RiViR

i RoR )18(3

3283

31,

This leads to R = 84/26 = 3.23 k Chapter 3, Solution 58 30

10 30

10

i3

i2

i1

120 V

�–+

30

Page 85: Solution Manual for Fundamentals of Electric Circuits

For loop 1, 120 + 40i1 �– 10i2 = 0, which leads to -12 = 4i1 �– i2 (1) For loop 2, 50i2 �– 10i1 �– 10i3 = 0, which leads to -i1 + 5i2 �– i3 = 0 (2) For loop 3, -120 �– 10i2 + 40i3 = 0, which leads to 12 = -i2 + 4i3 (3) Solving (1), (2), and (3), we get, i1 = -3A, i2 = 0, and i3 = 3A Chapter 3, Solution 59

i1

2I0

I0

10

20

40

+ v0 �–

100V �–+

120 V

�– +

+�–

4v0 i3

i2

80

i2 i3 For loop 1, -100 + 30i1 �– 20i2 + 4v0 = 0, where v0 = 80i3 or 5 = 1.5i1 �– i2 + 16i3 (1) For the supermesh, 60i2 �– 20i1 �– 120 + 80i3 �– 4 v0 = 0, where v0 = 80i3 or 6 = -i1 + 3i2 �– 12i3 (2) Also, 2I0 = i3 �– i2 and I0 = i2, hence, 3i2 = i3 (3)

From (1), (2), and (3), 130

12313223

06

10

iii

3

2

1

= ,5130

12313223

2 = ,28100

126132103

3 = 84030631

1023

I0 = i2 = 2/ = -28/5 = -5.6 A v0 = 8i3 = (-84/5)80 = -1344 volts

Page 86: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 60 0.5i0

i0

v1

1

4 8 10 V

10 V �–+

v2 2 At node 1, (v1/1) + (0.5v1/1) = (10 �– v1)/4, which leads to v1 = 10/7 At node 2, (0.5v1/1) + ((10 �– v2)/8) = v2/2 which leads to v2 = 22/7 P1 = (v1)2/1 = 2.041 watts, P2 = (v2)2/2 = 4.939 watts P4 = (10 �– v1)2/4 = 18.38 watts, P8 = (10 �– v2)2/8 = 5.88 watts Chapter 3, Solution 61

+ v0 �–

20 v2v1

+�–30 5v0

40

10 i0 is At node 1, is = (v1/30) + ((v1 �– v2)/20) which leads to 60is = 5v1 �– 3v2 (1) But v2 = -5v0 and v0 = v1 which leads to v2 = -5v1 Hence, 60is = 5v1 + 15v1 = 20v1 which leads to v1 = 3is, v2 = -15is i0 = v2/50 = -15is/50 which leads to i0/is = -15/50 = -0.3

Page 87: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 62

i1

4 k A

�–+

B8 k

100V �–+

i3i2

2 k 40 V We have a supermesh. Let all R be in k , i in mA, and v in volts. For the supermesh, -100 +4i1 + 8i2 + 2i3 + 40 = 0 or 30 = 2i1 + 4i2 + i3 (1) At node A, i1 + 4 = i2 (2) At node B, i2 = 2i1 + i3 (3) Solving (1), (2), and (3), we get i1 = 2 mA, i2 = 6 mA, and i3 = 2 mA. Chapter 3, Solution 63

A

5

10

i2i1

+�– 4ix

50 V �–+

For the supermesh, -50 + 10i1 + 5i2 + 4ix = 0, but ix = i1. Hence, 50 = 14i1 + 5i2 (1) At node A, i1 + 3 + (vx/4) = i2, but vx = 2(i1 �– i2), hence, i1 + 2 = i2 (2) Solving (1) and (2) gives i1 = 2.105 A and i2 = 4.105 A vx = 2(i1 �– i2) = -4 volts and ix = i2 �– 2 = 4.105 amp

Page 88: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 64

i0

i1

2 A

10

50 A 10 i2

+

0.2V0

4i0 +�–

100V �–+

i3

i2

i1

40 i1 i3B For mesh 2, 20i2 �– 10i1 + 4i0 = 0 (1) But at node A, io = i1 �– i2 so that (1) becomes i1 = (7/12)i2 (2) For the supermesh, -100 + 50i1 + 10(i1 �– i2) �– 4i0 + 40i3 = 0 or 50 = 28i1 �– 3i2 + 20i3 (3) At node B, i3 + 0.2v0 = 2 + i1 (4) But, v0 = 10i2 so that (4) becomes i3 = 2 �– (17/12)i2 (5) Solving (1) to (5), i2 = -0.674,

v0 = 10i2 = -6.74 volts, i0 = i1 - i2 = -(5/12)i2 = 0.281 amps Chapter 3, Solution 65

For mesh 1, 421 61212 III (1) For mesh 2, 54321 81660 IIIII (2) For mesh 3, 532 1589 III (3) For mesh 4, 5421 256 IIII (4) For mesh 5, 5432 8210 IIII (5)

Page 89: Solution Manual for Fundamentals of Electric Circuits

Casting (1) to (5) in matrix form gives

BAI

IIIII

10690

12

8211025011101580118166

010612

5

4

3

2

1

Using MATLAB leads to

411.2864.2733.1824.1673.1

1BAI

Thus, A 411.2 A, 864.1 A, 733.1 A, 824.1 A, 673.1 54321 IIIII

Chapter 3, Solution 66 Consider the circuit below.

2 k 2 k

+ + 20V I1 1 k I2 10V - - 1 k 1 k Io

1 k 2 k 2 k I3 I4 - 12V + We use mesh analysis. Let the mesh currents be in mA. For mesh 1, (1) 321420 IIIFor mesh 2, (2) 421 410 IIIFor mesh 3, (3) 431 412 IIIFor mesh 4, (4) 432 412 III

Page 90: Solution Manual for Fundamentals of Electric Circuits

In matrix form, (1) to (4) become

BAI

IIII

121210

20

411014011041

0114

4

3

2

1

Using MATLAB,

5.275.375.15.5

1BAI

Thus, mA 75.33II o Chapter 3, Solution 67

G11 = (1/1) + (1/4) = 1.25, G22 = (1/1) + (1/2) = 1.5 G12 = -1 = G21, i1 = 6 �– 3 = 3, i2 = 5-6 = -1

Hence, we have, 1

3vv

5.11125.1

2

1

25.1115.11

5.11125.1 1

, where = [(1.25)(1.5)-(-1)(-1)] = 0.875

24

)4286.1(1)1429.1(3)1429.1(1)7143.1(3

13

4286.11429.11429.17143.1

vv

2

1

Clearly v1 = 4 volts and v2 = 2 volts

Page 91: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 68

By inspection, G11 = 1 + 3 + 5 = 8S, G22 = 1 + 2 = 3S, G33 = 2 + 5 = 7S

G12 = -1, G13 = -5, G21 = -1, G23 = -2, G31 = -5, G32 = -2

i1 = 4, i2 = 2, i3 = -1

We can either use matrix inverse as we did in Problem 3.51 or use Cramer�’s Rule. Let us use Cramer�’s rule for this problem.

First, we develop the matrix relationships.

124

vvv

725231518

3

2

1

85721232514

,34725231518

1

87125

231418

,109715221548

32

v1 = 1/ = 85/34 = 3.5 volts, v2 = 2/ = 109/34 = 3.206 volts

v3 = 3/ = 87/34 = 2.56 volts

Page 92: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 69

Assume that all conductances are in mS, all currents are in mA, and all voltages are in volts. G11 = (1/2) + (1/4) + (1/1) = 1.75, G22 = (1/4) + (1/4) + (1/2) = 1, G33 = (1/1) + (1/4) = 1.25, G12 = -1/4 = -0.25, G13 = -1/1 = -1, G21 = -0.25, G23 = -1/4 = -0.25, G31 = -1, G32 = -0.25 i1 = 20, i2 = 5, and i3 = 10 �– 5 = 5 The node-voltage equations are:

55

20

vvv

25.125.0125.0125.0125.075.1

3

2

1

Chapter 3, Solution 70

G11 = G1 + G2 + G4, G12 = -G2, G13 = 0, G22 = G2 + G3, G21 = -G2, G23 = -G3, G33 = G1 + G3 + G5, G31 = 0, G32 = -G3 i1 = -I1, i2 = I2, and i3 = I1

Then, the node-voltage equations are:

1

2

1

3

2

1

5313

3212

2421

III

vvv

GGGG0GGGG0GGGG

Page 93: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 71

R11 = 4 + 2 = 6, R22 = 2 + 8 + 2 = 12, R33 = 2 + 5 = 7, R12 = -2, R13 = 0, R21 = -2, R23 = -2, R31 = 0, R32 = -2 v1 = 12, v2 = -8, and v3 = -20

Now we can write the matrix relationships for the mesh-current equations.

208

12

iii

7202122

026

3

2

1

Now we can solve for i2 using Cramer�’s Rule.

4087200282

0126,452

7202122

026

2

i2 = 2/ = -0.9026, p = (i2)2R = 6.52 watts Chapter 3, Solution 72

R11 = 5 + 2 = 7, R22 = 2 + 4 = 6, R33 = 1 + 4 = 5, R44 = 1 + 4 = 5, R12 = -2, R13 = 0 = R14, R21 = -2, R23 = -4, R24 = 0, R31 = 0, R32 = -4, R34 = -1, R41 = 0 = R42, R43 = -1, we note that Rij = Rji for all i not equal to j. v1 = 8, v2 = 4, v3 = -10, and v4 = -4

Hence the mesh-current equations are:

41048

iiii

51001540

04620027

4

3

2

1

Page 94: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 73

R11 = 2 + 3 +4 = 9, R22 = 3 + 5 = 8, R33 = 1 + 4 = 5, R44 = 1 + 1 = 2, R12 = -3, R13 = -4, R14 = 0, R23 = 0, R24 = 0, R34 = -1 v1 = 6, v2 = 4, v3 = 2, and v4 = -3

Hence,

3246

iiii

21001604

00830439

4

3

2

1

Chapter 3, Solution 74

R11 = R1 + R4 + R6, R22 = R2 + R4 + R5, R33 = R6 + R7 + R8, R44 = R3 + R5 + R8, R12 = -R4, R13 = -R6, R14 = 0, R23 = 0, R24 = -R5, R34 = -R8, again, we note that Rij = Rji for all i not equal to j.

The input voltage vector is =

4

3

2

1

VVV

V

4

3

2

1

4

3

2

1

85385

88766

55424

64641

VVV

V

iiii

RRRRR0RRRR0RR0RRRR0RRRRR

Page 95: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 75 * Schematics Netlist * R_R4 $N_0002 $N_0001 30 R_R2 $N_0001 $N_0003 10 R_R1 $N_0005 $N_0004 30 R_R3 $N_0003 $N_0004 10 R_R5 $N_0006 $N_0004 30 V_V4 $N_0003 0 120V v_V3 $N_0005 $N_0001 0 v_V2 0 $N_0006 0 v_V1 0 $N_0002 0

i3

i2i1

Clearly, i1 = -3 amps, i2 = 0 amps, and i3 = 3 amps, which agrees with the answers in Problem 3.44.

Page 96: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 76 * Schematics Netlist * I_I2 0 $N_0001 DC 4A R_R1 $N_0002 $N_0001 0.25 R_R3 $N_0003 $N_0001 1 R_R2 $N_0002 $N_0003 1 F_F1 $N_0002 $N_0001 VF_F1 3 VF_F1 $N_0003 $N_0004 0V R_R4 0 $N_0002 0.5 R_R6 0 $N_0001 0.5 I_I1 0 $N_0002 DC 2A R_R5 0 $N_0004 0.25

Clearly, v1 = 625 mVolts, v2 = 375 mVolts, and v3 = 1.625 volts, which agrees with the solution obtained in Problem 3.27.

Page 97: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 77 * Schematics Netlist * R_R2 0 $N_0001 4 I_I1 $N_0001 0 DC 3A I_I3 $N_0002 $N_0001 DC 6A R_R3 0 $N_0002 2 R_R1 $N_0001 $N_0002 1 I_I2 0 $N_0002 DC 5A

Clearly, v1 = 4 volts and v2 = 2 volts, which agrees with the answer obtained in Problem 3.51.

Page 98: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displaced on the pseudocomponents as shown. Thus,

,V15 V,5.4 V,3 321 VVV

.

Page 99: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 79 The schematic is shown below. When the circuit is saved and simulated, we obtain the node voltages as displaced. Thus,

V 88.26V V, 6944.0V V, 28.10V V, 278.5V dcba

Chapter 3, Solution 80 * Schematics Netlist * H_H1 $N_0002 $N_0003 VH_H1 6 VH_H1 0 $N_0001 0V I_I1 $N_0004 $N_0005 DC 8A V_V1 $N_0002 0 20V R_R4 0 $N_0003 4 R_R1 $N_0005 $N_0003 10 R_R2 $N_0003 $N_0002 12 R_R5 0 $N_0004 1 R_R3 $N_0004 $N_0001 2

Page 100: Solution Manual for Fundamentals of Electric Circuits

Clearly, v1 = 84 volts, v2 = 4 volts, v3 = 20 volts, and v4 = -5.333 volts Chapter 3, Solution 81

Clearly, v1 = 26.67 volts, v2 = 6.667 volts, v3 = 173.33 volts, and v4 = -46.67 volts which agrees with the results of Example 3.4.

Page 101: Solution Manual for Fundamentals of Electric Circuits

This is the netlist for this circuit. * Schematics Netlist * R_R1 0 $N_0001 2 R_R2 $N_0003 $N_0002 6 R_R3 0 $N_0002 4 R_R4 0 $N_0004 1 R_R5 $N_0001 $N_0004 3 I_I1 0 $N_0003 DC 10A V_V1 $N_0001 $N_0003 20V E_E1 $N_0002 $N_0004 $N_0001 $N_0004 3 Chapter 3, Solution 82

+ v0 �–

4

3 k

2 k

4 k 8 k

6 k

0

1 2 33v0

2i0

100V �–+

4A +

This network corresponds to the Netlist.

Page 102: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 83 The circuit is shown below.

+ v0 �–

4

3 k

2 k

4 k 8 k

6 k

1 3

20 V �–+ 30 2 A

0

50

0

1 2 3

3v0

2i0 2

100V �–+

4A +

20 70 When the circuit is saved and simulated, we obtain v2 = -12.5 volts Chapter 3, Solution 84

From the output loop, v0 = 50i0x20x103 = 106i0 (1) From the input loop, 3x10-3 + 4000i0 �– v0/100 = 0 (2) From (1) and (2) we get, i0 = 0.5 A and v0 = 0.5 volt. Chapter 3, Solution 85

The amplifier acts as a source. Rs + Vs RL - For maximum power transfer, 9sL RR

Page 103: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 86 Let v1 be the potential across the 2 k-ohm resistor with plus being on top. Then,

[(0.03 �– v1)/1k] + 400i = v1/2k (1) Assume that i is in mA. But, i = (0.03 �– v1)/1 (2) Combining (1) and (2) yields, v1 = 29.963 mVolts and i = 37.4 nA, therefore, v0 = -5000x400x37.4x10-9 = -74.8 mvolts

Chapter 3, Solution 87

v1 = 500(vs)/(500 + 2000) = vs/5 v0 = -400(60v1)/(400 + 2000) = -40v1 = -40(vs/5) = -8vs, Therefore, v0/vs = -8

Chapter 3, Solution 88 Let v1 be the potential at the top end of the 100-ohm resistor. (vs �– v1)/200 = v1/100 + (v1 �– 10-3v0)/2000 (1) For the right loop, v0 = -40i0(10,000) = -40(v1 �– 10-3)10,000/2000, or, v0 = -200v1 + 0.2v0 = -4x10-3v0 (2)

Substituting (2) into (1) gives, (vs + 0.004v1)/2 = -0.004v0 + (-0.004v1 �– 0.001v0)/20

This leads to 0.125v0 = 10vs or (v0/vs) = 10/0.125 = -80

Page 104: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 89 vi = VBE + 40k IB (1) 5 = VCE + 2k IC (2) If IC = IB = 75IB and VCE = 2 volts, then (2) becomes 5 = 2 +2k(75IB) which leads to IB = 20 A.

Substituting this into (1) produces, vi = 0.7 + 0.8 = 1.5 volts. 2 k IB 40 k

+ VBE

�–

-+

-+ vi 5 v

Chapter 3, Solution 90

1 k

100 k

IE

IB

+ VBE

�–

+ VCE �–

i2

i1 -+

18V -+

500 + V0 �–

vs For loop 1, -vs + 10k(IB) + VBE + IE (500) = 0 = -vs + 0.7 + 10,000IB + 500(1 + )IB which leads to vs + 0.7 = 10,000IB + 500(151)IB = 85,500IB But, v0 = 500IE = 500x151IB = 4 which leads to IB = 5.298x10-5 Therefore, vs = 0.7 + 85,500IB = 5.23 volts

Page 105: Solution Manual for Fundamentals of Electric Circuits

Chapter 3, Solution 91

We first determine the Thevenin equivalent for the input circuit.

RTh = 6||2 = 6x2/8 = 1.5 k and VTh = 2(3)/(2+6) = 0.75 volts 5 k

IC

1.5 k

IE

IB

+ VBE

�–

+ VCE �–

i2

i1 -+

9 V -+

400 + V0 �–

0.75 V For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500IB + 400(1 + )IB IB = 0.05/81,900 = 0.61 A v0 = 400IE = 400(1 + )IB = 49 mV For loop 2, -400IE �– VCE �– 5kIC + 9 = 0, but, IC = IB and IE = (1 + )IB VCE = 9 �– 5k IB �– 400(1 + )IB = 9 �– 0.659 = 8.641 volts

Chapter 3, Solution 92

IC

VC

10 k

IE

IB

+ VBE

�–

+ VCE �–

12V -+

4 k + V0 �–

I1 5 k

Page 106: Solution Manual for Fundamentals of Electric Circuits

I1 = IB + IC = (1 + )IB and IE = IB + IC = I1 Applying KVL around the outer loop, 4kIE + VBE + 10kIB + 5kI1 = 12 12 �– 0.7 = 5k(1 + )IB + 10kIB + 4k(1 + )IB = 919kIB IB = 11.3/919k = 12.296 A Also, 12 = 5kI1 + VC which leads to VC = 12 �– 5k(101)IB = 5.791 volts Chapter 3, Solution 93

i1 i2i

4

4

2

2

1

8

3v0

+ v2 �–

+ v1 �–

+

�–+

3v0

+2 i v2v1 i3

+ v0 �–

24V

(a) (b) From (b), -v1 + 2i �– 3v0 + v2 = 0 which leads to i = (v1 + 3v0 �– v2)/2 At node 1 in (a), ((24 �– v1)/4) = (v1/2) + ((v1 +3v0 �– v2)/2) + ((v1 �– v2)/1), where v0 = v2 or 24 = 9v1 which leads to v1 = 2.667 volts At node 2, ((v1 �– v2)/1) + ((v1 + 3v0 �– v2)/2) = (v2/8) + v2/4, v0 = v2 v2 = 4v1 = 10.66 volts Now we can solve for the currents, i1 = v1/2 = 1.333 A, i2 = 1.333 A, and i3 = 2.6667 A.

Page 107: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 1.

i io5

8

1

+

1 V 3

4)35(8 , 51

411i

101i

21io 0.1A

Chapter 4, Solution 2.

,3)24(6 A21i21i

,41i

21i 1o oo i2v 0.5V

5 4

i2

8

i1 io

6

1 A 2 If is = 1 A, then vo = 0.5 V Chapter 4, Solution 3.

R

+

vo

3R io

3R

3R

R

+

3R

+ 1 V

1.5R Vs

(b)(a)

Page 108: Solution Manual for Fundamentals of Electric Circuits

(a) We transform the Y sub-circuit to the equivalent .

,R43

R4R3R3R

2

R23R

43R

43

2v

v so independent of R

io = vo/(R) When vs = 1V, vo = 0.5V, io = 0.5A (b) When vs = 10V, vo = 5V, io = 5A (c) When vs = 10V and R = 10 ,

vo = 5V, io = 10/(10) = 500mA Chapter 4, Solution 4. If Io = 1, the voltage across the 6 resistor is 6V so that the current through the 3 resistor is 2A.

+

v1

3A

Is 2 4

i1 3A 1A

Is

2A

6 4 3

2 2

(a) (b)

263 , vo = 3(4) = 12V, .A34

vo1i

Hence Is = 3 + 3 = 6A If Is = 6A Io = 1 Is = 9A Io = 6/(9) = 0.6667A

Page 109: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 5.

If vo = 1V, V2131V1

3

10v322V 1s

If vs = 3

10 vo = 1

Then vs = 15 vo = 15x103 4.5V

vo3 2

+

6 6 6

v1

Vs

Chapter 4, Solution 6

Let sT

ToT V

RRR

VRRRR

RRR132

3232 then ,//

133221

32

132

32

32

32

1 RRRRRRRR

RRRRRRRRR

RRR

VV

kT

T

s

o

Page 110: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 7 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.

3Vx

5 5 + + 4V 15 VTh - 6 - + Vx -

From the figure,

V3)4(515

15,0 Thx VV

To find RTh, consider the circuit below:

3Vx

5 5 V1 V2 + 4V 15 1A - 6 + Vx - At node 1,

122111 73258616,

5153

54

VVxVVVV

VV

xx (1)

At node 2,

Page 111: Solution Manual for Fundamentals of Electric Circuits

9505

31 2121 VVVV

Vx (2)

Solving (1) and (2) leads to V2 = 101.75 V

mW 11.2275.1014

94

,75.1011

2

max2

xRV

pV

RTh

ThTh

Chapter 4, Solution 8. Let i = i1 + i2, where i1 and iL are due to current and voltage sources respectively.

6

i1 +

20V

i2

5 A 4 6

4 (a) (b)

i1 = ,A3)5(46

6 A246

202i

Thus i = i1 + i2 = 3 + 2 = 5A Chapter 4, Solution 9. Let i

2x1xx ii where i is due to 15V source and i is due to 4A source,

1x 2x

12

-4A

40 10

ix2

40

i

10

ix1

12

+

15V

(a) (b)

Page 112: Solution Manual for Fundamentals of Electric Circuits

For ix1, consider Fig. (a).

10||40 = 400/50 = 8 ohms, i = 15/(12 + 8) = 0.75

ix1 = [40/(40 + 10)]i = (4/5)0.75 = 0.6 For ix2, consider Fig. (b).

12||40 = 480/52 = 120/13

ix2 = [(120/13)/((120/13) + 10)](-4) = -1.92

ix = 0.6 �– 1.92 = -1.32 A

p = vix = ix2R = (-1.32)210 = 17.43 watts

Chapter 4, Solution 10. Let vab = vab1 + vab2 where vab1 and vab2 are due to the 4-V and the 2-A sources respectively.

+

vab2

10 +

3vab2

2 A

10

+

+

vab1

+

3vab1

4V

(a) (b) For vab1, consider Fig. (a). Applying KVL gives,

- vab1 �– 3 vab1 + 10x0 + 4 = 0, which leads to vab1 = 1 V For vab2, consider Fig. (b). Applying KVL gives,

- vab2 �– 3vab2 + 10x2 = 0, which leads to vab2 = 5

vab = 1 + 5 = 6 V

Page 113: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 11. Let i = i1 + i2, where i1 is due to the 12-V source and i2 is due to the 4-A source.

12V

4A

2 2

ix2

6

4A

32

i2

3

io

(a)

2

i1

6

+

(b) For i1, consider Fig. (a).

2||3 = 2x3/5 = 6/5, io = 12/(6 + 6/5) = 10/6

i1 = [3/(2 + 3)]io = (3/5)x(10/6) = 1 A For i2, consider Fig. (b), 6||3 = 2 ohm, i2 = 4/2 = 2 A

i = 1 + 2 = 3 A Chapter 4, Solution 12. Let vo = vo1 + vo2 + vo3, where vo1, vo2, and vo3 are due to the 2-A, 12-V, and 19-V sources respectively. For vo1, consider the circuit below.

5

5

+ vo1

io

2A2A

3

4

6 12

5

+ vo1

Page 114: Solution Manual for Fundamentals of Electric Circuits

6||3 = 2 ohms, 4||12 = 3 ohms. Hence,

io = 2/2 = 1, vo1 = 5io = 5 V

For vo2, consider the circuit below. 6 5 4 6 5

+ vo2

3 3

+

v1

+

12V

+ vo2

12

+

3

12V

3||8 = 24/11, v1 = [(24/11)/(6 + 24/11)]12 = 16/5

vo2 = (5/8)v1 = (5/8)(16/5) = 2 V For vo3, consider the circuit shown below. 4 5 5 4

2+ vo3

12

+

v2

+

19V

+

19V 6

+ vo3

12 3

7||12 = (84/19) ohms, v2 = [(84/19)/(4 + 84/19)]19 = 9.975

v = (-5/7)v2 = -7.125

vo = 5 + 2 �– 7.125 = -125 mV

,

Chapter 4, Solution 13 Let iiii 321o where i1, i2, and i3 are the contributions to io due to 30-V, 15-V, and 6-mA sources respectively. For i1, consider the circuit below.

Page 115: Solution Manual for Fundamentals of Electric Circuits

1 k 2 k 3 k + i1 30V - 4 k 5 k

3//5 = 15/8 = 1.875 kohm, 2 + 3//5 = 3.875 kohm, 1//3.875 = 3.875/4.875 = 0.7949 kohm. After combining the resistors except the 4-kohm resistor and transforming the voltage source, we obtain the circuit below. i1 30 mA 4 k 0.7949 k Using current division,

mA 4.973mA)30(7949.47949.0

1i

For i2, consider the circuit below. 1 k 2 k 3 k i2 - 15V 4 k 5 k +

After successive source transformation and resistance combinations, we obtain the circuit below: 2.42mA i2 4 k 0.7949 k

Using current division,

mA 4012.0mA)42.2(7949.47949.0

2i

Page 116: Solution Manual for Fundamentals of Electric Circuits

For i3, consider the circuit below.

6mA 1 k 2 k 3 k i3 4 k 5 k

After successive source transformation and resistance combinations, we obtain the circuit below: 3.097mA i3 4 k 0.7949 k

mA 5134.0mA)097.3(7949.47949.0

3i

Thus, mA 058.4321 iiiio Chapter 4, Solution 14. Let vo = vo1 + vo2 + vo3, where vo1, vo2 , and vo3, are due to the 20-V, 1-A, and 2-A sources respectively. For vo1, consider the circuit below. 6

20V

+

+

vo1

4 2

3

6||(4 + 2) = 3 ohms, vo1 = (½)20 = 10 V

Page 117: Solution Manual for Fundamentals of Electric Circuits

For vo2, consider the circuit below. 6 6

1A

2 4

+

vo2

4V

+ 2 4

3

+

vo2

3

3||6 = 2 ohms, vo2 = [2/(4 + 2 + 2)]4 = 1 V For vo3, consider the circuit below. 6

3

vo3 +

3

2A

2 4

3

+

vo3

2A

6||(4 + 2) = 3, vo3 = (-1)3 = -3 vo = 10 + 1 �– 3 = 8 V

Chapter 4, Solution 15. Let i = i1 + i2 + i3, where i1 , i2 , and i3 are due to the 20-V, 2-A, and 16-V sources. For i1, consider the circuit below.

io

4

3

i1

1

+

20V 2

Page 118: Solution Manual for Fundamentals of Electric Circuits

4||(3 + 1) = 2 ohms, Then io = [20/(2 + 2)] = 5 A, i1 = io/2 = 2.5 A For i3, consider the circuit below.

2||(1 + 3) = 4/3, vo�’ = [(4/3)/((4/3) + 4)](-16) = -4

i3 = vo�’/4 = -1 For i2, consider the circuit below.

3

i2

1 (4/3)

3

i2

1 2A

4

+

vo�’

4

3

i3

1

2

+ 16V

2A 2

2||4 = 4/3, 3 + 4/3 = 13/3 Using the current division principle.

i2 = [1/(1 + 13/2)]2 = 3/8 = 0.375

i = 2.5 + 0.375 - 1 = 1.875 A

p = i2R = (1.875)23 = 10.55 watts

Page 119: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 16. Let io = io1 + io2 + io3, where io1, io2, and io3 are due to the 12-V, 4-A, and 2-A sources. For io1, consider the circuit below.

5

io1

10

4

+

3 2

12V

10||(3 + 2 + 5) = 5 ohms, io1 = 12/(5 + 4) = (12/9) A For io2, consider the circuit below.

io2

i1

4A

10

2

4

3

5

2 + 5 + 4||10 = 7 + 40/14 = 69/7 i1 = [3/(3 + 69/7)]4 = 84/90, io2 =[-10/(4 + 10)]i1 = -6/9

For io3, consider the circuit below.

i2

2 A

io3

10 5 4

3 2

3 + 2 + 4||10 = 5 + 20/7 = 55/7

i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i2 = -5/9

io = (12/9) �– (6/9) �– (5/9) = 1/9 = 111.11 mA

Page 120: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 17. Let vx = vx1 + vx2 + vx3, where vx1,vx2, and vx3 are due to the 90-V, 6-A, and 40-V sources. For vx1, consider the circuit below.

30 10 20

12

+ vx1

10

20

3 A

io

30

+ vx1 60

+

90V

20||30 = 12 ohms, 60||30 = 20 ohms By using current division,

io = [20/(22 + 20)]3 = 60/42, vx1 = 10io = 600/42 = 14.286 V For vx2, consider the circuit below.

io�’

20 30

io�’

6A 60 30

10

+ vx2 + vx2

6A 20 12

10

io�’ = [12/(12 + 30)]6 = 72/42, vx2 = -10io�’ = -17.143 V

For vx3, consider the circuit below.

20

+ vx3 7.5

4A

io�”

+

40V30 30

+ vx3 60

10 10 10

io�” = [12/(12 + 30)]2 = 24/42, vx3 = -10io�” = -5.714

vx = 14.286 �– 17.143 �– 5.714 = -8.571 V

Page 121: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 18. Let ix = i1 + i2, where i1 and i2 are due to the 10-V and 2-A sources respectively. To obtain i1, consider the circuit below. 2

+

10i1

i12

+

10V

1 i1

4 5i1

1

+

4 10V

-10 + 10i1 + 7i1 = 0, therefore i1 = (10/17) A For i2, consider the circuit below.

io+

10i2 2

+

2V

1 io i2 1

4

2

10i2 2 A

+ 4

-2 + 10i2 + 7io = 0, but i2 + 2 = io. Hence,

-2 + 10i2 +7i2 + 14 = 0, or i2 = (-12/17) A

vx = 1xix = 1(i1 + i2) = (10/17) �– (12/17) = -2/17 = -117.6 mA Chapter 4, Solution 19. Let vx = v1 + v2, where v1 and v2 are due to the 4-A and 6-A sources respectively. ix v1 ix v2

+

v2

+

v1

8 2

4ix

6 A

+

82

4ix

4 A

+

(a) (b)

Page 122: Solution Manual for Fundamentals of Electric Circuits

To find v1, consider the circuit in Fig. (a).

v1/8 = 4 + (-4ix �– v1)/2 But, -ix = (-4ix �– v1)/2 and we have -2ix = v1. Thus,

v1/8 = 4 + (2v1 �– v1)/8, which leads to v1 = -32/3 To find v2, consider the circuit shown in Fig. (b).

v2/2 = 6 + (4ix �– v2)/8 But ix = v2/2 and 2ix = v2. Therefore,

v2/2 = 6 + (2v2 �– v2)/8 which leads to v2 = -16 Hence, vx = �–(32/3) �– 16 = -26.67 V Chapter 4, Solution 20. Transform the voltage sources and obtain the circuit in Fig. (a). Combining the 6-ohm and 3-ohm resistors produces a 2-ohm resistor (6||3 = 2). Combining the 2-A and 4-A sources gives a 6-A source. This leads to the circuit shown in Fig. (b).

i

6 4A32

i

6A 2 2

2A (a) (b) From Fig. (b), i = 6/2 = 3 A Chapter 4, Solution 21. To get io, transform the current sources as shown in Fig. (a). 6

+

vo

2 A 3 i

6 2 A

+

6V

io 3

+

12V

(a) (b)

Page 123: Solution Manual for Fundamentals of Electric Circuits

From Fig. (a), -12 + 9io + 6 = 0, therefore io = 666.7 mA To get vo, transform the voltage sources as shown in Fig. (b).

i = [6/(3 + 6)](2 + 2) = 8/3

vo = 3i = 8 V Chapter 4, Solution 22. We transform the two sources to get the circuit shown in Fig. (a). 5 5

10V

4 10 2A

i 101A

2A

(a)

104

+

(b) We now transform only the voltage source to obtain the circuit in Fig. (b). 10||10 = 5 ohms, i = [5/(5 + 4)](2 �– 1) = 5/9 = 555.5 mA

Page 124: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 23 If we transform the voltage source, we obtain the circuit below. 8 10 6 3 5A 3A 3//6 = 2-ohm. Convert the current sources to voltages sources as shown below. 10 8 2 + + 10V 30V -

- Applying KVL to the loop gives

A 10)2810(1030 II W82RIVIp

Page 125: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 24 Convert the current source to voltage source.

16 1

4 + 5 + 48 V 10 Vo

- + -

12 V - Combine the 16-ohm and 4-ohm resistors and convert both voltages sources to current Sources. We obtain the circuit below. 1 2.4A 20 5 2.4A 10 Combine the resistors and current sources. 20//5 = (20x5)/25 = 4 , 2.4 + 2.4 = 4.8 A Convert the current source to voltage source. We obtain the circuit below. 4 1 + + 19.2V Vo 10 - - Using voltage division,

8.12)2.19(1410

10oV V

Page 126: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 25. Transforming only the current source gives the circuit below.

12V

30 V

5

9 18 V

+ vo

4

2

i

+

+

+

+

30 V Applying KVL to the loop gives,

(4 + 9 + 5 + 2)i �– 12 �– 18 �– 30 �– 30 = 0

20i = 90 which leads to i = 4.5

vo = 2i = 9 V Chapter 4, Solution 26.

Transform the voltage sources to current sources. The result is shown in Fig. (a),

30||60 = 20 ohms, 30||20 = 12 ohms 10

+ + vx

12 10

96V i

20

+

60V

203A 2A

+ vx

60 306A

(a)

30

(b)

Page 127: Solution Manual for Fundamentals of Electric Circuits

Combining the resistors and transforming the current sources to voltage sources, we obtain the circuit in Fig. (b). Applying KVL to Fig. (b),

42i �– 60 + 96 = 0, which leads to i = -36/42

vx = 10i = -8.571 V Chapter 4, Solution 27. Transforming the voltage sources to current sources gives the circuit in Fig. (a).

10||40 = 8 ohms Transforming the current sources to voltage sources yields the circuit in Fig. (b). Applying KVL to the loop,

-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4

vx 12i = -48 V

12

+ vx 10 20405A 8A 2A

(a) 8 12 20

+

+

+ vx 40V 200V i

(b)

Page 128: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 28. Transforming only the current sources leads to Fig. (a). Continuing with source transformations finally produces the circuit in Fig. (d).

3 io 10 V

+

12 V

+

+

4 5 2

10

12 V (a) 4

5 io

+

12V

4

+

11V io

io

+

12V 10

4

2.2A10

io

+

22 V

(b)

+

12V 10

10

(c) (d) Applying KVL to the loop in fig. (d),

-12 + 9io + 11 = 0, produces io = 1/9 = 111.11 mA

Page 129: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 29. Transform the dependent voltage source to a current source as shown in Fig. (a). 2||4 = (4/3) k ohms

4 k

It is clear that i = 3 mA which leads to vo = 1000i = 3 V If the use of source transformations was not required for this problem, the actual answer could have been determined by inspection right away since the only current that could have flowed through the 1 k ohm resistor is 3 mA. Chapter 4, Solution 30 Transform the dependent current source as shown below.

ix 24 60 10 + + 12V 30 7ix - -

+

vo

1 k

+

vo

3 mA

2vo (4/3) k

(b)

i

+

3 mA

1.5vo

1 k

2 k

(a)

Combine the 60-ohm with the 10-ohm and transform the dependent source as shown below.

Page 130: Solution Manual for Fundamentals of Electric Circuits

ix 24 + 12V 30 70 0.1ix - Combining 30-ohm and 70-ohm gives 30//70 = 70x30/100 = 21-ohm. Transform the dependent current source as shown below.

ix 24 21 + + 12V 2.1ix - - Applying KVL to the loop gives

mA 8.2541.47

1201.21245 xxx iii

Chapter 4, Solution 31. Transform the dependent source so that we have the circuit in Fig. (a). 6||8 = (24/7) ohms. Transform the dependent source again to get the circuit in Fig. (b). 3

6

+ vx +

8

vx/3 12V (a)

(24/7) 3

i +

+ vx +

(8/7)vx 12V

(b)

Page 131: Solution Manual for Fundamentals of Electric Circuits

From Fig. (b),

vx = 3i, or i = vx/3. Applying KVL,

-12 + (3 + 24/7)i + (24/21)vx = 0

12 = [(21 + 24)/7]vx/3 + (8/7)vx, leads to vx = 84/23 = 3.625 V Chapter 4, Solution 32. As shown in Fig. (a), we transform the dependent current source to a voltage source,

15 10 +

50

+

5ix

40 60V

(a)

15

ix

50 50

(b)

+

ix 25

+ 60V 2.5ix

15

60V 0.1ix

(c)

Page 132: Solution Manual for Fundamentals of Electric Circuits

In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c), -60 + 40ix �– 2.5ix = 0, or ix = 1.6 A Chapter 4, Solution 33.

(a) RTh = 10||40 = 400/50 = 8 ohms

VTh = (40/(40 + 10))20 = 16 V (b) RTh = 30||60 = 1800/90 = 20 ohms

2 + (30 �– v1)/60 = v1/30, and v1 = VTh

120 + 30 �– v1 = 2v1, or v1 = 50 V

VTh = 50 V

Chapter 4, Solution 34. To find RTh, consider the circuit in Fig. (a).

3 A

RTh = 20 + 10||40 = 20 + 400/50 = 28 ohms (b)

v2

+

VTh

20

40

10 v1

+

40VRTh

20

40

10

(a)

To find VTh, consider the circuit in Fig. (b). At node 1, (40 �– v1)/10 = 3 + [(v1 �– v2)/20] + v1/40, 40 = 7v1 �– 2v2 (1) At node 2, 3 + (v1- v2)/20 = 0, or v1 = v2 �– 60 (2) Solving (1) and (2), v1 = 32 V, v2 = 92 V, and VTh = v2 = 92 V

Page 133: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 35. To find RTh, consider the circuit in Fig. (a).

RTh = Rab = 6||3 + 12||4 = 2 + 3 =5 ohms To find VTh, consider the circuit shown in Fig. (b).

RTh At node 1, 2 + (12 �– v1)/6 = v1/3, or v1 = 8 At node 2, (19 �– v2)/4 = 2 + v2/12, or v2 = 33/4 But, -v1 + VTh + v2 = 0, or VTh = v1 �– v2 = 8 �– 33/4 = -0.25

b

RTh = 5

10

+ vo a

+

4

+

+

v2

+

v1 12

v2+ VTh

3

2 A

v1

+

12V

6

(b)

a b

4 12 6 3

(a)

19V

VTh = (-1/4)V

vo = VTh/2 = -0.25/2 = -125 mV

Page 134: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 36. Remove the 30-V voltage source and the 20-ohm resistor.

a

b

10 +

VTh

+ 50V

a

b

40

RTh

10

40

(a) (b) From Fig. (a), RTh = 10||40 = 8 ohms From Fig. (b), VTh = (40/(10 + 40))50 = 40V

+

a

b

12

(c)

i

+

40V

8

30V The equivalent circuit of the original circuit is shown in Fig. (c). Applying KVL, 30 �– 40 + (8 + 12)i = 0, which leads to i = 500mA

Page 135: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 37 RN is found from the circuit below.

20 a 40 12 b

10)4020//(12NR IN is found from the circuit below.

2A

20 a + 40 120V 12 - IN b Applying source transformation to the current source yields the circuit below.

20 40 + 80 V - + 120V IN - Applying KVL to the loop yields

A 6667.060/4006080120 NN II

Page 136: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 38 We find Thevenin equivalent at the terminals of the 10-ohm resistor. For RTh, consider the circuit below. 1

4 5 RTh 16

541)164//(51ThR For VTh, consider the circuit below. 1

V1 4 V2 5 + 3A 16 VTh

+ -

12 V - At node 1,

21211 4548

4163 VV

VVV (1)

At node 2,

21221 95480

512

4VV

VVV (2)

Solving (1) and (2) leads to 2.192VVTh

Page 137: Solution Manual for Fundamentals of Electric Circuits

Thus, the given circuit can be replaced as shown below. 5 + + 19.2V Vo 10 - - Using voltage division,

8.12)2.19(510

10oV V

Chapter 4, Solution 39. To find RTh, consider the circuit in Fig. (a).

- 1 �– 3 + 10io = 0, or io = 0.4

RTh = 1/io = 2.5 ohms To find VTh, consider the circuit shown in Fig. (b).

[(4 �– v)/10] + 2 = 0, or v = 24 But, v = VTh + 3vab = 4VTh = 24, which leads to VTh = 6 V Chapter 4, Solution 40. To find RTh, consider the circuit in Fig. (a).

40V

a 10

(a)

+

1V50V

+

v1

+

3vab 10 io

b

a

(b)

+

440 V

+

3vab 20

2A

+

v

40

RTh

a b 10 20

+

v2

+ VTh

8 A

+

(b)

10

+

vab = VTh

b

(a)

Page 138: Solution Manual for Fundamentals of Electric Circuits

RTh = 10||40 + 20 = 28 ohms To get VTh, consider the circuit in Fig. (b). The two loops are independent. From loop 1,

v1 = (40/50)50 = 40 V For loop 2, -v2 + 20x8 + 40 = 0, or v2 = 200 But, VTh + v2 �– v1 = 0, VTh = v1 = v2 = 40 �– 200 = -160 volts This results in the following equivalent circuit. 28

vx = [12/(12 + 28)](-160) = -48 V Chapter 4, Solution 41 To find RTh, consider the circuit below 14 a 6 5 b

+

+

vx 12 -160V

NTh RR 4)614//(5 Applying source transformation to the 1-A current source, we obtain the circuit below.

Page 139: Solution Manual for Fundamentals of Electric Circuits

6 - 14V + 14 VTh a + 6V 3A 5 - b At node a,

V 85

3146

614Th

ThTh VVV

A 24/)8(Th

ThN R

VI

Thus, A 2 V,8,4 NThNTh IVRR

Chapter 4, Solution 42. To find RTh, consider the circuit in Fig. (a). 20 30

10

20 10

10 10

a

10 10

30 30 b

ba (a) (b) 20||20 = 10 ohms. Transform the wye sub-network to a delta as shown in Fig. (b). 10||30 = 7.5 ohms. RTh = Rab = 30||(7.5 + 7.5) = 10 ohms. To find VTh, we transform the 20-V and the 5-V sources. We obtain the circuit shown in Fig. (c).

Page 140: Solution Manual for Fundamentals of Electric Circuits

+

50V

10

10 10

10 +

10 V

a 10 + b

+

30V

i2i1

(c) For loop 1, -30 + 50 + 30i1 �– 10i2 = 0, or -2 = 3i1 �– i2 (1) For loop 2, -50 �– 10 + 30i2 �– 10i1 = 0, or 6 = -i1 + 3i2 (2) Solving (1) and (2), i1 = 0, i2 = 2 A Applying KVL to the output loop, -vab �– 10i1 + 30 �– 10i2 = 0, vab = 10 V

VTh = vab = 10 volts Chapter 4, Solution 43. To find RTh, consider the circuit in Fig. (a).

RTh

a b

5

+

vb

+

va

+ VTh

10

+

50V

10

5 10

a b

10

(a)

2 A

(b)

RTh = 10||10 + 5 = 10 ohms

Page 141: Solution Manual for Fundamentals of Electric Circuits

To find VTh, consider the circuit in Fig. (b).

vb = 2x5 = 10 V, va = 20/2 = 10 V But, -va + VTh + vb = 0, or VTh = va �– vb = 0 volts Chapter 4, Solution 44. (a) For RTh, consider the circuit in Fig. (a).

RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms For VTh, consider the circuit in Fig. (b). Applying KVL gives,

10 �– 24 + i(3 + 4 + 5 + 2), or i = 1

VTh = 4i = 4 V 3 1

+ a

+

VTh3 1 a 4 24V

+

b RTh4 10V 2 b 2

i 5 5

(b) (a) (b) For RTh, consider the circuit in Fig. (c).

3 1 3 1

+

24V

2

2A

c

b

5

+

VTh

vo4

2

5

4

RTh

c

b

(c) (d)

Page 142: Solution Manual for Fundamentals of Electric Circuits

RTh = 5||(2 + 3 + 4) = 3.214 ohms To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,

[(24 �– vo)/9] + 2 = vo/5, or vo = 15

VTh = vo = 15 V Chapter 4, Solution 45. For RN, consider the circuit in Fig. (a).

6 6

RN6 4 4A 6 4

IN

(a) (b)

RN = (6 + 6)||4 = 3 ohms For IN, consider the circuit in Fig. (b). The 4-ohm resistor is shorted so that 4-A current is equally divided between the two 6-ohm resistors. Hence, IN = 4/2 = 2 A Chapter 4, Solution 46. (a) RN = RTh = 8 ohms. To find IN, consider the circuit in Fig. (a). 10 60

4

+

30V 2A IN30

+

Isc

20V

(a) (b) IN = Isc = 20/10 = 2 A

(b) To get IN, consider the circuit in Fig. (b). IN = Isc = 2 + 30/60 = 2.5 A

Page 143: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 47 Since VTh = Vab = Vx, we apply KCL at the node a and obtain

V 19.1126/15026012

30ThTh

ThTh VVVV

To find RTh, consider the circuit below.

12 Vx a 2Vx 60

1A

At node a, KCL gives

4762.0126/601260

21 xxx

x VVV

V

5.24762.0/19.1,4762.01 Th

ThN

xTh R

VI

VR

Thus, A 5.2,4762.0,19.1 NNThTh IRRVV

Chapter 4, Solution 48. To get RTh, consider the circuit in Fig. (a).

+

VTh

10Io

+

Io

2A4

2

Io

2

4

+ +

V

10Io

1A

(a) (b) From Fig. (a), Io = 1, 6 �– 10 �– V = 0, or V = -4

RN = RTh = V/1 = -4 ohms

Page 144: Solution Manual for Fundamentals of Electric Circuits

To get VTh, consider the circuit in Fig. (b),

Io = 2, VTh = -10Io + 4Io = -12 V

IN = VTh/RTh = 3A Chapter 4, Solution 49. RN = RTh = 28 ohms To find IN, consider the circuit below,

3A

vo

io

20

+

40

10

Isc = IN 40V At the node, (40 �– vo)/10 = 3 + (vo/40) + (vo/20), or vo = 40/7 io = vo/20 = 2/7, but IN = Isc = io + 3 = 3.286 A Chapter 4, Solution 50. From Fig. (a), RN = 6 + 4 = 10 ohms

6 6

4 12V

+

4

2A

Isc = IN

(b) (a) From Fig. (b), 2 + (12 �– v)/6 = v/4, or v = 9.6 V

-IN = (12 �– v)/6 = 0.4, which leads to IN = -0.4 A Combining the Norton equivalent with the right-hand side of the original circuit produces the circuit in Fig. (c).

Page 145: Solution Manual for Fundamentals of Electric Circuits

i

4A5

0.4A 10

(c) i = [10/(10 + 5)] (4 �– 0.4) = 2.4 A Chapter 4, Solution 51. (a) From the circuit in Fig. (a),

RN = 4||(2 + 6||3) = 4||4 = 2 ohms

2

+

120V

+

6A

VTh

4

3

6

RTh

4

3

6

2 (a) (b) For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the circuit becomes that shown in Fig. (c).

i

2

+

12V

+

+ VTh

4 2

40V (c) Applying KVL to the circuit in Fig. (c),

-40 + 8i + 12 = 0 which gives i = 7/2

VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A

Page 146: Solution Manual for Fundamentals of Electric Circuits

(b) To get RN, consider the circuit in Fig. (d).

RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms

i

2

+

12V

+VTh

RN

4

3 2

6

(d) (e) To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in Fig. (e).

i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A Chapter 4, Solution 52. For RTh, consider the circuit in Fig. (a). a

Io

b

2 k3 k 20Io

RTh (a)

+ +

Io

a

b

2 k VTh 20Io

3 k

6V (b) For Fig. (a), Io = 0, hence the current source is inactive and

RTh = 2 k ohms

Page 147: Solution Manual for Fundamentals of Electric Circuits

For VTh, consider the circuit in Fig. (b).

Io = 6/3k = 2 mA

VTh = (-20Io)(2k) = -20x2x10-3x2x103 = -80 V Chapter 4, Solution 53. To get RTh, consider the circuit in Fig. (a). 0.25vo0.25vo

1/2

a 2

1A

+

vo

2

(b) b

+

vab

1/2

a

b

2

1A

+

vo

3

(a)

6 From Fig. (b),

vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0

vab = 3V

RN = vab/1 = 3 ohms To get IN, consider the circuit in Fig. (c). 0.25vo

6 a 2

+

vo

3

+

18V Isc = IN b (c)

[(18 �– vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V

But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A

Page 148: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 54

To find VTh =Vx, consider the left loop.

xoxo ViVi 2100030210003 (1) For the right loop, (2) oox iixV 20004050Combining (1) and (2), mA13000400010003 oooo iiii

222000 Thox ViV To find RTh, insert a 1-V source at terminals a-b and remove the 3-V independent source, as shown below.

1 k ix .

io + + + 2Vx 40io Vx 50 1V

- - -

mA210002

,1 xox

ViV

-60mAA501mA80

5040 x

ox

Vii

67.16060.0/11

xTh iR

Chapter 4, Solution 55. To get RN, apply a 1 mA source at the terminals a and b as shown in Fig. (a).

80I+

vab/1000 +

vab8 k

a I

50 k

1mA

b (a)

Page 149: Solution Manual for Fundamentals of Electric Circuits

We assume all resistances are in k ohms, all currents in mA, and all voltages in volts. At node a,

(vab/50) + 80I = 1 (1) Also,

-8I = (vab/1000), or I = -vab/8000 (2) From (1) and (2), (vab/50) �– (80vab/8000) = 1, or vab = 100

RN = vab/1 = 100 k ohms To get IN, consider the circuit in Fig. (b).

vab/1000

I

80I

a

50 k

+

vab

+

+

8 k

IN 2V

b (b) Since the 50-k ohm resistor is shorted,

IN = -80I, vab = 0 Hence, 8i = 2 which leads to I = (1/4) mA

IN = -20 mA Chapter 4, Solution 56. We first need RN and IN.

16V

2A 4

2 IN

1

+

+

20V

4 2 RN

a

b

1

(a) (b)

Page 150: Solution Manual for Fundamentals of Electric Circuits

To find RN, consider the circuit in Fig. (a).

RN = 1 + 2||4 = (7/3) ohms To get IN, short-circuit ab and find Isc from the circuit in Fig. (b). The current source can be transformed to a voltage source as shown in Fig. (c).

vo

i

RN

a

IN

1

+ 16V 2V

4

2 IN

+

+

20V

3

b (c) (d)

(20 �– vo)/2 = [(vo + 2)/1] + [(vo + 16)/4], or vo = 16/7

IN = (vo + 2)/1 = 30/7 From the Norton equivalent circuit in Fig. (d),

i = RN/(RN + 3), IN = [(7/3)/((7/3) + 3)](30/7) = 30/16 = 1.875 A Chapter 4, Solution 57. To find RTh, remove the 50V source and insert a 1-V source at a �– b, as shown in Fig. (a).

B A i

10 6

+

vx

0.5vx

a

b(a)

2

+

1V 3 We apply nodal analysis. At node A,

i + 0.5vx = (1/10) + (1 �– vx)/2, or i + vx = 0.6 (1) At node B,

(1 �– vo)/2 = (vx/3) + (vx/6), and vx = 0.5 (2)

Page 151: Solution Manual for Fundamentals of Electric Circuits

From (1) and (2), i = 0.1 and

RTh = 1/i = 10 ohms To get VTh, consider the circuit in Fig. (b).

10 6

+

vx

0.5vx

a

b(b)

2

+

v1 3 v2 +

VTh

50V At node 1, (50 �– v1)/3 = (v1/6) + (v1 �– v2)/2, or 100 = 6v1 �– 3v2 (3) At node 2, 0.5vx + (v1 �– v2)/2 = v2/10, vx = v1, and v1 = 0.6v2 (4) From (3) and (4),

v2 = VTh = 166.67 V

IN = VTh/RTh = 16.667 A

RN = RTh = 10 ohms Chapter 4, Solution 58. This problem does not have a solution as it was originally stated. The reason for this is that the load resistor is in series with a current source which means that the only equivalent circuit that will work will be a Norton circuit where the value of RN = infinity. IN can be found by solving for Isc.

voib

+

ib

R2

R1 Isc VS Writing the node equation at node vo,

ib + ib = vo/R2 = (1 + )ib

Page 152: Solution Manual for Fundamentals of Electric Circuits

But ib = (Vs �– vo)/R1

vo = Vs �– ibR1

Vs �– ibR1 = (1 + )R2ib, or ib = Vs/(R1 + (1 + )R2)

Isc = IN = - ib = - Vs/(R1 + (1 + )R2) Chapter 4, Solution 59. RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms To find VTh, consider the circuit below.

10

50

+ VTh

8A

i1 i2 20

40

i1 = i2 = 8/2 = 4, 10i1 + VTh �– 20i2 = 0, or VTh = 20i2 �–10i1 = 10i1 = 10x4 VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A Chapter 4, Solution 60. The circuit can be reduced by source transformations. 2A

5

10 18 V

+

12 V

+

10 V

+

Page 153: Solution Manual for Fundamentals of Electric Circuits

2A

b a

3A

2A

10

5

3A

10 V

+ 3.333a b 3.333 a b

Norton Equivalent Circuit Thevenin Equivalent Circuit Chapter 4, Solution 61. To find RTh, consider the circuit in Fig. (a). Let R = 2||18 = 1.8 ohms, RTh = 2R||R = (2/3)R = 1.2 ohms. To get VTh, we apply mesh analysis to the circuit in Fig. (d).

2 a

2

6

6

2

6

b

(a)

Page 154: Solution Manual for Fundamentals of Electric Circuits

1.8

1.8

a

b

1.8 RTh

18

18 18 2

a

b

(b)

2

2

(c)

+

+ 12V

i3

i2i12

6

6

a

b

2

6

12V

+

+ 12V

2

VTh

(d) -12 �– 12 + 14i1 �– 6i2 �– 6i3 = 0, and 7 i1 �– 3 i2 �– 3i3 = 12 (1) 12 + 12 + 14 i2 �– 6 i1 �– 6 i3 = 0, and -3 i1 + 7 i2 �– 3 i3 = -12 (2) 14 i3 �– 6 i1 �– 6 i2 = 0, and -3 i1 �– 3 i2 + 7 i3 = 0 (3) This leads to the following matrix form for (1), (2) and (3),

012

12

iii

733373337

3

2

1

Page 155: Solution Manual for Fundamentals of Electric Circuits

100733373337

, 12070331233127

2

i2 = / 2 = -120/100 = -1.2 A

VTh = 12 + 2i2 = 9.6 V, and IN = VTh/RTh = 8 A Chapter 4, Solution 62. Since there are no independent sources, VTh = 0 V To obtain RTh, consider the circuit below.

2

2vo

0.1io ix

v1

io

40

10

+

+

VS

+vo

1

20

At node 2,

ix + 0.1io = (1 �– v1)/10, or 10ix + io = 1 �– v1 (1) At node 1, (v1/20) + 0.1io = [(2vo �– v1)/40] + [(1 �– v1)/10] (2) But io = (v1/20) and vo = 1 �– v1, then (2) becomes,

1.1v1/20 = [(2 �– 3v1)/40] + [(1 �– v1)/10]

2.2v1 = 2 �– 3v1 + 4 �– 4v1 = 6 �– 7v1 or v1 = 6/9.2 (3) From (1) and (3),

10ix + v1/20 = 1 �– v1

10ix = 1 �– v1 �– v1/20 = 1 �– (21/20)v1 = 1 �– (21/20)(6/9.2)

ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.

Page 156: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 63. Because there are no independent sources, IN = Isc = 0 A RN can be found using the circuit below. 3

io10

+

+

vo

v1

0.5vo

20

1V Applying KCL at node 1, 0.5vo + (1 �– v1)/3 = v1/30, but vo = (20/30)v1 Hence, 0.5(2/3)(30)v1 + 10 �– 10v1 =v1, or v1 = 10 and io = (1 �– v1)/3 = -3 RN = 1/io = -1/3 = -333.3 m ohms Chapter 4, Solution 64.

With no independent sources, VTh = 0 V. To obtain RTh, consider the circuit shown below.

1

ix

io4

+

+ �–

10ix

vo

2

1V

ix = [(1 �– vo)/1] + [(10ix �– vo)/4], or 2vo = 1 + 3ix (1)

But ix = vo/2. Hence,

2vo = 1 + 1.5vo, or vo = 2, io = (1 �– vo)/1 = -1

Thus, RTh = 1/io = -1 ohm

Page 157: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 65 At the terminals of the unknown resistance, we replace the circuit by its Thevenin equivalent.

V 24)32(412

12,53212//42 ThTh VR

Thus, the circuit can be replaced by that shown below.

5 Io

+ + 24 V Vo

- - Applying KVL to the loop,

oooo I524V0VI524 Chapter 4, Solution 66. We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit in Fig. (a).

3

5

b a

RTh

2

2 10V +

i 20V

+

30V

3

+VTh

a b

+

5

(a) (b)

RTh = 2||(3 + 5) = 2||8 = 1.6 ohms By performing source transformation on the given circuit, we obatin the circuit in (b).

Page 158: Solution Manual for Fundamentals of Electric Circuits

We now use this to find VTh.

10i + 30 + 20 + 10 = 0, or i = -5

VTh + 10 + 2i = 0, or VTh = 2 V

p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts

Chapter 4, Solution 67. We need to find the Thevenin equivalent at terminals a and b. From Fig. (a),

RTh = 4||6 + 8||12 = 2.4 + 4.8 = 7.2 ohms From Fig. (b),

10i1 �– 30 = 0, or i1 = 3

+

4

12 8

6

i2

i1

+

+

30V

RTh

4

12 8

6

VTh

+

(b)(a)

20i2 + 30 = 0, or i2 = 1.5, VTh = 6i1 + 8i2 = 6x3 �– 8x1.5 = 6 V For maximum power transfer,

p = VTh2/(4RTh) = (6)2/[4(7.2)] = 1.25 watts

Page 159: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 68. This is a challenging problem in that the load is already specified. This now becomes a "minimize losses" style problem. When a load is specified and internal losses can be adjusted, then the objective becomes, reduce RThev as much as possible, which will result in maximum power transfer to the load.

-+

-+

Removing the 10 ohm resistor and solving for the Thevenin Circuit results in:

RTh = (Rx20/(R+20)) and a Voc = VTh = 12x(20/(R +20)) + (-8) As R goes to zero, RTh goes to zero and VTh goes to 4 volts, which produces the maximum power delivered to the 10-ohm resistor.

P = vi = v2/R = 4x4/10 = 1.6 watts Notice that if R = 20 ohms which gives an RTh = 10 ohms, then VTh becomes -2 volts and the power delivered to the load becomes 0.1 watts, much less that the 1.6 watts. It is also interesting to note that the internal losses for the first case are 122/20 = 7.2 watts and for the second case are = to 12 watts. This is a significant difference. Chapter 4, Solution 69. We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit below. 22 k v1 Assume that all resistances are in k ohms and all currents are in mA.

1mA 3vo

30 k40 k

+

vo

10 k

Page 160: Solution Manual for Fundamentals of Electric Circuits

10||40 = 8, and 8 + 22 = 30

1 + 3vo = (v1/30) + (v1/30) = (v1/15)

15 + 45vo = v1

But vo = (8/30)v1, hence,

15 + 45x(8v1/30) v1, which leads to v1 = 1.3636

RTh = v1/1 = -1.3636 k ohms To find VTh, consider the circuit below. 10 k 22 kvo v1

(100 �– vo)/10 = (vo/40) + (vo �– v1)/22 (1)

3vo

30 k40 k

+

vo

+

+

VTh

100V

[(vo �– v1)/22] + 3vo = (v1/30) (2)

Solving (1) and (2),

v1 = VTh = -243.6 volts

p = VTh2/(4RTh) = (243.6)2/[4(-1363.6)] = -10.882 watts

Page 161: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 70 We find the Thevenin equivalent across the 10-ohm resistor. To find VTh, consider the circuit below.

3Vx

5 5 + + 4V 15 VTh - 6 - + Vx -

From the figure,

V3)4(515

15,0 Thx VV

To find RTh, consider the circuit below:

3Vx

5 5 V1 V2 + 4V 15 1A - 6 + Vx - At node 1,

122111 73258616,

5153

54

VVxVVVV

VV

xx (1)

Page 162: Solution Manual for Fundamentals of Electric Circuits

At node 2,

9505

31 2121 VVVV

Vx (2)

Solving (1) and (2) leads to V2 = 101.75 V

mW 11.2275.1014

94

,75.1011

2

max2

xRV

pV

RTh

ThTh

Chapter 4, Solution 71. We need RTh and VTh at terminals a and b. To find RTh, we insert a 1-mA source at the terminals a and b as shown below. Assume that all resistances are in k ohms, all currents are in mA, and all voltages are in volts. At node a,

a

b

+

1mA 120vo

40 k1 k

+

vo

10 k

3 k

1 = (va/40) + [(va + 120vo)/10], or 40 = 5va + 480vo (1) The loop on the left side has no voltage source. Hence, vo = 0. From (1), va = 8 V.

RTh = va/1 mA = 8 kohms To get VTh, consider the original circuit. For the left loop,

vo = (1/4)8 = 2 V For the right loop, vR = VTh = (40/50)(-120vo) = -192 The resistance at the required resistor is

R = RTh = 8 kohms

p = VTh2/(4RTh) = (-192)2/(4x8x103) = 1.152 watts

Page 163: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 72. (a) RTh and VTh are calculated using the circuits shown in Fig. (a) and (b) respectively. From Fig. (a), RTh = 2 + 4 + 6 = 12 ohms From Fig. (b), -VTh + 12 + 8 + 20 = 0, or VTh = 40 V

12V 4 4 6 2 6 + +

VTh

+

RTh 2 8V 20V

+ (b)(a)

(b) i = VTh/(RTh + R) = 40/(12 + 8) = 2A (c) For maximum power transfer, RL = RTh = 12 ohms (d) p = VTh

2/(4RTh) = (40)2/(4x12) = 33.33 watts. Chapter 4, Solution 73 Find the Thevenin�’s equivalent circuit across the terminals of R.

10 25 RTh 20 5

833.1030/3255//2520//10ThR

Page 164: Solution Manual for Fundamentals of Electric Circuits

10 25 + + VTh - 60 V + + - Va Vb 20 5 - -

10)60(305,40)60(

3020

ba VV

V 3010400 baThbTha VVVVVV

W77.20833.104

304

22

max xRV

pTh

Th

Chapter 4, Solution 74. When RL is removed and Vs is short-circuited,

RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)]

RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)] When RL is removed and we apply the voltage division principle,

Voc = VTh = vR2 �– vR4

= ([R2/(R1 + R2)] �– [R4/(R3 + R4)])Vs = {[(R2R3) �– (R1R4)]/[(R1 + R2)(R3 + R4)]}Vs

pmax = VTh2/(4RTh)

= {[(R2R3) �– (R1R4)]2/[(R1 + R2)(R3 + R4)]2}Vs

2[( R1 + R2)( R3 + R4)]/[4(a)]

where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)

pmax =

[(R2R3) �– (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]

Page 165: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 75.

We need to first find RTh and VTh. R

+

+

2V

R

R

+

3V

+

VTh

vo

R R R

RTh

1V (a) (b) Consider the circuit in Fig. (a).

(1/RTh) = (1/R) + (1/R) + (1/R) = 3/R

RTh = R/3 From the circuit in Fig. (b),

((1 �– vo)/R) + ((2 �– vo)/R) + ((3 �– vo)/R) = 0

vo = 2 = VTh For maximum power transfer,

RL = RTh = R/3

Pmax = [(VTh)2/(4RTh)] = 3 mW

RTh = [(VTh)2/(4Pmax)] = 4/(4xPmax) = 1/Pmax = R/3

R = 3/(3x10-3) = 1 k ohms Chapter 4, Solution 76. Follow the steps in Example 4.14. The schematic and the output plots are shown below. From the plot, we obtain,

Page 166: Solution Manual for Fundamentals of Electric Circuits

V = 92 V [i = 0, voltage axis intercept]

R = Slope = (120 �– 92)/1 = 28 ohms

Page 167: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 77. (a) The schematic is shown below. We perform a dc sweep on a current source, I1, connected between terminals a and b. We label the top and bottom of source I1 as 2 and 1 respectively. We plot V(2) �– V(1) as shown.

VTh = 4 V [zero intercept]

RTh = (7.8 �– 4)/1 = 3.8 ohms

Page 168: Solution Manual for Fundamentals of Electric Circuits

(b) Everything remains the same as in part (a) except that the current source, I1, is connected between terminals b and c as shown below. We perform a dc sweep on I1 and obtain the plot shown below. From the plot, we obtain,

V = 15 V [zero intercept]

R = (18.2 �– 15)/1 = 3.2 ohms

Page 169: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 78.

he schematic is shown below. We perform a dc sweep on the current source, I1,

VTh = -80 V

Tconnected between terminals a and b. The plot is shown. From the plot we obtain,

[zero intercept]

RTh = (1920 �– (-80))/1 = 2 k ohms

Page 170: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 79. After drawing and saving the schematic as shown below, we perform a dc sweep on I1 connected across a and b. The plot is shown. From the plot, we get,

V = 167 V [zero intercept]

R = (177 �– 167)/1 = 10 ohms

Page 171: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 80. The schematic in shown below. We label nodes a and b as 1 and 2 respectively. We perform dc sweep on I1. In the Trace/Add menu, type v(1) �– v(2) which will result in the plot below. From the plot,

VTh = 40 V [zero intercept]

RTh = (40 �– 17.5)/1 = 22.5 ohms [slope]

Page 172: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 81. The schematic is shown below. We perform a dc sweep on the current source, I2, connected between terminals a and b. The plot of the voltage across I2 is shown below. From the plot,

VTh = 10 V [zero intercept]

RTh = (10 �– 6.4)/1 = 3.4 ohms.

Page 173: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 82.

VTh = Voc = 12 V, Isc = 20 A

RTh = Voc/Isc = 12/20 = 0.6 ohm. 0.6

i = 12/2.6 , p = i2R = (12/2.6)2(2) = 42.6 watts

i +

2 12V

Chapter 4, Solution 83.

VTh = Voc = 12 V, Isc = IN = 1.5 A

RTh = VTh/IN = 8 ohms, VTh = 12 V, RTh = 8 ohms

Page 174: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 84 Let the equivalent circuit of the battery terminated by a load be as shown below. RTh IL + + VTh - VL RL

-

For open circuit,

V 8.10, LocThL VVVR When RL = 4 ohm, VL=10.5,

7.24/8.10L

LL R

VI

But

4444.07.2

8.1012

L

LThThThLLTh I

VVRRIVV

Chapter 4, Solution 85 (a) Consider the equivalent circuit terminated with R as shown below. RTh a + + VTh Vab R - - b

ThTh

ThTh

ab VR

VRRR

V10

106

or ThTh VR 10660 (1)

where RTh is in k-ohm.

Page 175: Solution Manual for Fundamentals of Electric Circuits

Similarly,

ThThThTh

VRVR

301236030

3012 (2)

Solving (1) and (2) leads to

kRV ThTh 30 V, 24

(b) V 6.9)24(3020

20abV

Chapter 4, Solution 86. We replace the box with the Thevenin equivalent. RTh

+

v

i

R

+

VTh

VTh = v + iRTh When i = 1.5, v = 3, which implies that VTh = 3 + 1.5RTh (1) When i = 1, v = 8, which implies that VTh = 8 + 1xRTh (2) From (1) and (2), RTh = 10 ohms and VTh = 18 V. (a) When R = 4, i = VTh/(R + RTh) = 18/(4 + 10) = 1.2857 A (b) For maximum power, R = RTH

Pmax = (VTh)2/4RTh = 182/(4x10) = 8.1 watts Chapter 4, Solution 87. (a) im = 9.975 mA im = 9.876 mA

+

vm

Rs Rm Rs Rs Rm

Is Is

(a) (b)

Page 176: Solution Manual for Fundamentals of Electric Circuits

From Fig. (a),

vm = Rmim = 9.975 mA x 20 = 0.1995 V

Is = 9.975 mA + (0.1995/Rs) (1) From Fig. (b),

vm = Rmim = 20x9.876 = 0.19752 V

Is = 9.876 mA + (0.19752/2k) + (0.19752/Rs)

= 9.975 mA + (0.19752/Rs) (2) Solving (1) and (2) gives,

Rs = 8 k ohms, Is = 10 mA (b)

im�’ = 9.876 mA

8k||4k = 2.667 k ohms

im�’ = [2667/(2667 + 20)](10 mA) = 9.926 mA Chapter 4, Solution 88

To find RTh, consider the circuit below. RTh 5k A B

30k 20k

Rs

(b)

RsIs Rm

10k kRTh 445//201030

Page 177: Solution Manual for Fundamentals of Electric Circuits

To find VTh , consider the circuit below.

5k A B io +

30k 20k 4mA 60 V - 10k

V 72,48)60(2520,120430 BAThBA VVVVxV

Chapter 4, Solution 89 It is easy to solve this problem using Pspice. (a) The schematic is shown below. We insert IPROBE to measure the desired ammeter reading. We insert a very small resistance in series IPROBE to avoid problem. After the circuit is saved and simulated, the current is displaced on IPROBE as A99.99 .

(b) By interchanging the ammeter and the 12-V voltage source, the schematic is shown below. We obtain exactly the same result as in part (a).

Page 178: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 90.

Rx = (R3/R1)R2 = (4/2)R2 = 42.6, R2 = 21.3

which is (21.3ohms/100ohms)% = 21.3% Chapter 4, Solution 91.

Rx = (R3/R1)R2 (a) Since 0 < R2 < 50 ohms, to make 0 < Rx < 10 ohms requires that when R2

= 50 ohms, Rx = 10 ohms.

10 = (R3/R1)50 or R3 = R1/5

so we select R1 = 100 ohms and R3 = 20 ohms (b) For 0 < Rx < 100 ohms

100 = (R3/R1)50, or R3 = 2R1

So we can select R1 = 100 ohms and R3 = 200 ohms

Page 179: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 92. For a balanced bridge, vab = 0. We can use mesh analysis to find vab. Consider the circuit in Fig. (a), where i1 and i2 are assumed to be in mA.

2 k

5 k

i2

i1

+

+ vab

a b

3 k 6 k

220V 10 k 0 (a)

220 = 2i1 + 8(i1 �– i2) or 220 = 10i1 �– 8i2 (1) 0 = 24i2 �– 8i1 or i2 = (1/3)i1 (2) From (1) and (2),

i1 = 30 mA and i2 = 10 mA Applying KVL to loop 0ab0 gives

5(i2 �– i1) + vab + 10i2 = 0 V Since vab = 0, the bridge is balanced. When the 10 k ohm resistor is replaced by the 18 k ohm resistor, the gridge becomes unbalanced. (1) remains the same but (2) becomes

0 = 32i2 �– 8i1, or i2 = (1/4)i1 (3) Solving (1) and (3),

i1 = 27.5 mA, i2 = 6.875 mA

vab = 5(i1 �– i2) �– 18i2 = -20.625 V

VTh = vab = -20.625 V To obtain RTh, we convert the delta connection in Fig. (b) to a wye connection shown in Fig. (c).

Page 180: Solution Manual for Fundamentals of Electric Circuits

2 k

a RTh b

18 k

6 kR2

R1a RTh b

5 k 18 k R3

6 k3 k

(b) (c)

R1 = 3x5/(2 + 3 + 5) = 1.5 k ohms, R2 = 2x3/10 = 600 ohms,

R3 = 2x5/10 = 1 k ohm.

RTh = R1 + (R2 + 6)||(R3 + 18) = 1.5 + 6.6||9 = 6.398 k ohms

RL = RTh = 6.398 k ohms

Pmax = (VTh)2/(4RTh) = (20.625)2/(4x6.398) = 16.622 mWatts Chapter 4, Solution 93.

Rs Ro ix

Roix+

ix

+

VS

-Vs + (Rs + Ro)ix + Roix = 0

ix = Vs/(Rs + (1 + )Ro)

Page 181: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 94.

(a) Vo/Vg = Rp/(Rg + Rs + Rp) (1)

Req = Rp||(Rg + Rs) = Rg

Rg = Rp(Rg + Rs)/(Rp + Rg + Rs)

RgRp + Rg2 + RgRs = RpRg + RpRs

RpRs = Rg(Rg + Rs) (2)

From (1), Rp/ = Rg + Rs + Rp Rg + Rs = Rp((1/ ) �– 1) = Rp(1 - )/ (1a)

Combining (2) and (1a) gives, Rs = [(1 - )/ ]Req (3) = (1 �– 0.125)(100)/0.125 = 700 ohms

From (3) and (1a),

Rp(1 - )/ = Rg + [(1 - )/ ]Rg = Rg/

Rp = Rg/(1 - ) = 100/(1 �– 0.125) = 114.29 ohms

(b)

RTh

+

I RLVTh

VTh = Vs = 0.125Vg = 1.5 V

RTh = Rg = 100 ohms

I = VTh/(RTh + RL) = 1.5/150 = 10 mA

Page 182: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 95.

Let 1/sensitivity = 1/(20 k ohms/volt) = 50 A For the 0 �– 10 V scale,

Rm = Vfs/Ifs = 10/50 A = 200 k ohms For the 0 �– 50 V scale,

Rm = 50(20 k ohms/V) = 1 M ohm

+

VTh

RTh

Rm

VTh = I(RTh + Rm)

(a) A 4V reading corresponds to

I = (4/10)Ifs = 0.4x50 A = 20 A

VTh = 20 A RTh + 20 A 250 k ohms

= 4 + 20 A RTh (1)

(b) A 5V reading corresponds to

I = (5/50)Ifs = 0.1 x 50 A = 5 A

VTh = 5 A x RTh + 5 A x 1 M ohm

VTh = 5 + 5 A RTh (2) From (1) and (2)

0 = -1 + 15 A RTh which leads to RTh = 66.67 k ohms

From (1),

VTh = 4 + 20x10-6x(1/(15x10-6)) = 5.333 V

Page 183: Solution Manual for Fundamentals of Electric Circuits

Chapter 4, Solution 96. (a) The resistance network can be redrawn as shown in Fig. (a),

+

VTh

RTh

+

Vo

+

VTh

R

10 8 10

40

10

i2i1

+

60

8

9V R

(a) (b)

RTh = 10 + 10 + 60||(8 + 8 + 10||40) = 20 + 60||24 = 37.14 ohms Using mesh analysis, -9 + 50i1 - 40i2 = 0 (1) 116i2 �– 40i1 = 0 or i1 = 2.9i2 (2) From (1) and (2), i2 = 9/105

VTh = 60i2 = 5.143 V From Fig. (b),

Vo = [R/(R + RTh)]VTh = 1.8

R/(R + 37.14) = 1.8/5.143 which leads to R = 20 ohms (b) R = RTh = 37.14 ohms Imax = VTh/(2RTh) = 5.143/(2x37.14) = 69.23 mA Chapter 4, Solution 97.

4 k

4 k

+

+B

VTh

E

12V

Page 184: Solution Manual for Fundamentals of Electric Circuits

RTh = R1||R2 = 6||4 = 2.4 k ohms

VTh = [R2/(R1 + R2)]vs = [4/(6 + 4)](12) = 4.8 V Chapter 4, Solution 98. The 20-ohm, 60-ohm, and 14-ohm resistors form a delta connection which needs to be connected to the wye connection as shown in Fig. (b),

b

a

RTh

R2

R3

30

R1

b

a RTh

20

60

30

14

(a) (b)

R1 = 20x60/(20 + 60 + 14) = 1200/94 = 12.97 ohms

R2 = 20x14/94 = 2.98 ohms

R3 = 60x14/94 = 8.94 ohms

RTh = R3 + R1||(R2 + 30) = 8.94 + 12.77||32.98 = 18.15 ohms

To find VTh, consider the circuit in Fig. (c).

16 V

I1

IT

IT

+

b

a +

VTh

20

60

30

14

(c)

Page 185: Solution Manual for Fundamentals of Electric Circuits

IT = 16/(30 + 15.74) = 350 mA

I1 = [20/(20 + 60 + 14)]IT = 94.5 mA

VTh = 14I1 + 30IT = 11.824 V

I40 = VTh/(RTh + 40) = 11.824/(18.15 + 40) = 203.3 mA

P40 = I402R = 1.654 watts

Page 186: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 1. (a) Rin = 1.5 M (b) Rout = 60 (c) A = 8x104

Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 0.1V Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Solution 4.

v0 = Avd = A(v2 - v1)

v2 - v1 = V2010x24

Av

50

If v1 and v2 are in mV, then

v2 - v1 = -20 mV = 0.02 1 - v1 = -0.02

v1 = 1.02 mV Chapter 5, Solution 5.

+ v0

-

-

vd

+

R0

Rin

I

vi -+

Avd +-

Page 187: Solution Manual for Fundamentals of Electric Circuits

-vi + Avd + (Ri - R0) I = 0 (1) But vd = RiI, -vi + (Ri + R0 + RiA) I = 0

vd = i0

ii

R)A1(RRv

(2)

-Avd - R0I + v0 = 0

v0 = Avd + R0I = (R0 + RiA)I = i0

ii0

R)A1(Rv)ARR(

45

54

i0

i0

i

0 10)101(100

10x10100R)A1(R

ARRvv

45

9

10101

10001,100000,100 0.9999990

Chapter 5, Solution 6.

- vd

+ + vo

-

R0

Rin

I

vi

+ -

Avd +-

Page 188: Solution Manual for Fundamentals of Electric Circuits

(R0 + Ri)R + vi + Avd = 0 But vd = RiI, vi + (R0 + Ri + RiA)I = 0

I = i0

i

R)A1(Rv

(1)

-Avd - R0I + vo = 0 vo = Avd + R0I = (R0 + RiA)I Substituting for I in (1),

v0 = i0

i0

R)A1(RARR

vi

= 65

356

10x2x10x21501010x2x10x250

mV10x2x001,20010x2x000,200

6

6

v0 = -0.999995 mV Chapter 5, Solution 7. 100 k

1 210 k

-+

+ Vd -

+ Vout

-

Rout = 100

Rin AVd +-

VS

Page 189: Solution Manual for Fundamentals of Electric Circuits

At node 1, (VS �– V1)/10 k = [V1/100 k] + [(V1 �– V0)/100 k] 10 VS �– 10 V1 = V1 + V1 �– V0

which leads to V1 = (10VS + V0)/12 At node 2, (V1 �– V0)/100 k = (V0 �– AVd)/100

But Vd = V1 and A = 100,000, V1 �– V0 = 1000 (V0 �– 100,000V1)

0= 1001V0 �– 100,000,001[(10VS + V0)/12]

0 = -83,333,334.17 VS - 8,332,333.42 V0

which gives us (V0/ VS) = -10 (for all practical purposes) If VS = 1 mV, then V0 = -10 mV Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV

Chapter 5, Solution 8. (a) If va and vb are the voltages at the inverting and noninverting terminals of the op

amp.

va = vb = 0

1mA = k2v0 0 v0 = -2V

(b)

10 k

2V

+ -+ va -

10 k

ia

+ vo -

+ vo -

va

vb

ia

2 k

2V -+

1V -+

- +

(b) (a)

Page 190: Solution Manual for Fundamentals of Electric Circuits

Since va = vb = 1V and ia = 0, no current flows through the 10 k resistor. From Fig. (b), -va + 2 + v0 = 0 va = va - 2 = 1 - 2 = -1V Chapter 5, Solution 9. (a) Let va and vb be respectively the voltages at the inverting and noninverting terminals of the op amp va = vb = 4V At the inverting terminal,

1mA = k2

0v4 v0 = 2V

Since va = vb = 3V, -vb + 1 + vo = 0 vo = vb - 1 = 2V

+ vb -

+ vo -

+ -

(b) 1V

Chapter 5, Solution 10.

Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence

vs = vo 2v

101010 o

s

o

vv

= 2

Page 191: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 11.

8 k

vb = V2)3(510

10

io

b

a

+

5 k

2 k

4 k

+

vo

10 k

+

3 V

At node a,

8

vv2v3 oaa 12 = 5va �– vo

But va = vb = 2V,

12 = 10 �– vo vo = -2V

�–io = mA142

822

4v0

8vv ooa

i o = -1mA

Chapter 5, Solution 12. 4 k

b

a

+

2 k

1 k

+

vo

4 k

+

1.2V

Page 192: Solution Manual for Fundamentals of Electric Circuits

At node b, vb = ooo v32v

32v

244

At node a, 4

vv1

v2.1 oaa , but va = vb = ov32

4.8 - 4 x ooo vv32v

32 vo = V0570.2

78.4x3

va = vb = 76.9v

32

o

is = 7

2.11

v2. a1

p = vsis = 1.2 7

2.1 -205.7 mW

Chapter 5, Solution 13. By voltage division,

i1 i2

90 k

10 k

b

a

+

100 k

4 k

50 k

+ io

+

vo

1 V

va = V9.0)1(10090

vb = 3

vv

150o

o50

But va = vb 9.03

v0 vo = 2.7V

io = i1 + i2 = k150

vk10

v oo 0.27mA + 0.018mA = 288 A

Page 193: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 14. Transform the current source as shown below. At node 1,

10

vv20

vv5

v10 o1211

But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo 40 = 7v1 - 2vo (1)

20 k

vo

10 k

+

v1 +

v2

5 k

10 k

+

vo

10V

At node 2, 0v,10

vv20

vv2

o221 or v1 = -2vo (2)

From (1) and (2), 40 = -14vo - 2vo vo = -2.5V Chapter 5, Solution 15

(a) Let v1 be the voltage at the node where the three resistors meet. Applying KCL at this node gives

3321

3

1

2

1 11Rv

RRv

Rvv

Rv

i oos (1)

At the inverting terminal,

111

10Riv

Rv

i ss (2)

Combining (1) and (2) leads to

2

3131

33

1

2

11RRR

RRiv

Rv

RR

RR

is

oos

(b) For this case,

k 92- k 25

40204020 xiv

s

o

Page 194: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 16 10k ix 5k va iy - vb + vo + 2k 0.5V - 8k

Let currents be in mA and resistances be in k . At node a,

oaoaa vvvvv

31105

5.0 (1)

But

aooba vvvvv8

1028

8 (2)

Substituting (2) into (1) gives

148

81031 aaa vvv

Thus,

A 28.14mA 70/15

5.0 ax

vi

A 85.71mA 148

46.0)

810(6.0)(6.0

102xvvvv

vvvvi aaao

aoboy

Chapter 5, Solution 17.

(a) G = 5

12RR

vv

1

2

i

o -2.4

(b) 5

80vv

i

o = -16

(c) 5

2000vv

i

o -400

Page 195: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 18. Converting the voltage source to current source and back to a voltage source, we have the circuit shown below:

3202010 k

1 M

3v2

32050

1000v io

17200

v1

ov -11.764

Chapter 5, Solution 19. We convert the current source and back to a voltage source.

3442

5 k

vo

0V

(4/3) k 10 k

+

4 k

+

(2/3)V

(20/3) k

+

vo

+

50 k

+

2vi/3

Page 196: Solution Manual for Fundamentals of Electric Circuits

32

k34x4

k10vo -1.25V

k10

0vk5

vi oo

o -0.375mA

Chapter 5, Solution 20. 8 k

+

vs

+

4 k

+

vo

+

2 k4 k

a b

9 V At node a,

4

vv8

vv4v9 baoaa 18 = 5va �– vo - 2vb (1)

At node b,

2

vv4

vv obba va = 3vb - 2vo (2)

But vb = vs = 0; (2) becomes va = �–2vo and (1) becomes

-18 = -10vo �– vo vo = -18/(11) = -1.6364V

Page 197: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 21.

Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes

va = 3 x 3 - 2v0 = 9 - 2vo

Substituting this into (1), 18 = 5 (9-2vo) �– vo �– 6 leads to

vo = 21/(11) = 1.909V Chapter 5, Solution 22.

Av = -Rf/Ri = -15.

If Ri = 10k , then Rf = 150 k . Chapter 5, Solution 23

At the inverting terminal, v=0 so that KCL gives

121

000R

R

vv

Rv

RRv f

s

o

f

os

Chapter 5, Solution 24

v1 Rf

R1 R2 - vs + - + + R4 R3 vo v2 - We notice that v1 = v2. Applying KCL at node 1 gives

f

os

ff

os

Rv

Rv

vRRRR

vvRvv

Rv

21

21

1

2

1

1

1 1110)(

(1)

Page 198: Solution Manual for Fundamentals of Electric Circuits

Applying KCL at node 2 gives

ss v

RRR

vRvv

Rv

43

31

4

1

3

1 0 (2)

Substituting (2) into (1) yields

sf

fo vRRR

RRR

RR

RR

Rv243

3

2

43

1

3 1

i.e.

243

3

2

43

1

3 1RRR

RRR

RR

RR

Rkf

f

Chapter 5, Solution 25.

vo = 2 V

+

+

va

+

vo

-va + 3 + vo = 0 which leads to va = vo + 3 = 5 V. Chapter 5, Solution 26

+ vb - io + +

0.4V 5k - 2k vo 8k -

V 5.08.0/4.08.028

84.0 ooob vvvv

Hence,

mA 1.05

5.05 kkvo

oi

Page 199: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 27. (a) Let va be the voltage at the noninverting terminal. va = 2/(8+2) vi = 0.2vi

ia0 v2.10v20

1 1000v

G = v0/(vi) = 10.2

(b) vi = v0/(G) = 15/(10.2) cos 120 t = 1.471 cos 120 t V Chapter 5, Solution 28.

+

+

At node 1, k50vv

k10v0 o11

But v1 = 0.4V, -5v1 = v1 �– vo, leads to vo = 6v1 = 2.4V Alternatively, viewed as a noninverting amplifier, vo = (1 + (50/10)) (0.4V) = 2.4V io = vo/(20k) = 2.4/(20k) = 120 A

Page 200: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 29 R1 va + vb - + + vi R2 R2 vo - R1 -

obia vRR

Rvv

RRR

v21

1

21

2 ,

But oiba vRR

Rv

RRR

v21

1

21

2v

Or

1

2

RR

vv

i

o

Chapter 5, Solution 30. The output of the voltage becomes vo = vi = 12 k122030 By voltage division,

V2.0)2.1(6012

12vx

k202.0

k20v

i xx 10 A

k2004.0

Rvp

2x 2 W

Page 201: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 31. After converting the current source to a voltage source, the circuit is as shown below:

12 k

2

vo6 k

+

+ 6 k

3 k 1 v1

vo 12 V At node 1,

12

vv6

vv3

v o1o1112 48 = 7v1 - 3vo (1)

At node 2,

xoo1 i6

0v6

vv v1 = 2vo (2)

From (1) and (2),

1148vo

k6v

i ox 0.7272mA

Chapter 5, Solution 32. Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting amplifier.

xv10501 (4 mV) = 24 mV

k203060

Page 202: Solution Manual for Fundamentals of Electric Circuits

By voltage division,

vo = mV122

vv

2020o

o20

ix = k40

mV24k2020

vx 600nA

p = 3

62o

10x6010x144

Rv

204nW

Chapter 5, Solution 33. After transforming the current source, the current is as shown below:

1 k This is a noninverting amplifier.

3 k

vi

va+

+

2 k

4 k

vo

4 V

iio v23v

211v

Since the current entering the op amp is 0, the source resistor has a OV potential drop. Hence vi = 4V.

V6)4(23vo

Power dissipated by the 3k resistor is

k3

36Rv2

o 12mW

k1

64R

vvi oa

x -2mA

Page 203: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 34

0R

vvR

vv

2

in1

1

in1 (1)

but

o43

3a v

RRRv (2)

Combining (1) and (2),

0vRRv

RRvv a

2

12

2

1a1

22

11

2

1a v

RRv

RR1v

22

11

2

1

43

o3 vRRv

RR1

RRvR

22

11

2

13

43o v

RRv

RR1R

RRv

vO = )vRv()RR(R

RR221

213

43

Chapter 5, Solution 35.

10RR1

vv

Ai

f

i

ov Rf = 9Ri

If Ri = 10k , Rf = 90k

Page 204: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 36 V abTh V

But abs VRR

R

21

1v . Thus,

ssabTh vRR

vRRR

VV )1(1

2

1

21

To get RTh, apply a current source Io at terminals a-b as shown below.

v1 + v2 - a + R2

vo io R1 - b Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes through R1 and consequently R2 . Thus, vo=0 and

0o

oTh i

vR

Chapter 5, Solution 37.

33

f2

2

f1

1

fo v

RR

vRR

vRR

v

)3(3030)2(

2030)1(

1030

vo = -3V

Page 205: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 38.

44

f3

3

f2

2

f1

1

fo v

RR

vRR

vRR

vRR

v

)100(5050)50(

1050)20(

2050)10(

2550

= -120mV Chapter 5, Solution 39

This is a summing amplifier.

2233

22

11

5.29)1(5050

2050)2(

1050

vvvR

Rv

R

Rv

R

Rv fffo

Thus, V 35.295.16 22 vvvo

Chapter 5, Solution 40

R1 R2 va + R3 vb - + + v1 + - v2 Rf vo - + v3 R - - Applying KCL at node a,

)111(03213

3

2

2

1

1

3

3

2

2

1

1

RRRv

Rv

Rv

Rv

Rvv

Rvv

Rvv

aaaa (1)

Page 206: Solution Manual for Fundamentals of Electric Circuits

But

of

ba vRRR

vv (2)

Substituting (2) into (1)gives

)111(3213

3

2

2

1

1

RRRRRRv

Rv

Rv

Rv

f

o

or

)111/()(3213

3

2

2

1

1

RRRRv

Rv

Rv

R

RRv fo

Chapter 5, Solution 41. Rf/Ri = 1/(4) Ri = 4Rf = 40k The averaging amplifier is as shown below: Chapter 5, Solution 42

v1 R2 = 40 k

v2 R3 = 40 k

v3 R4 = 40 k

v4

10 k

+

R1 = 40 k

vo

k 10R31R 1f

Page 207: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 43. In order for

44

f3

3

f2

2

f1

1

fo v

RR

vRR

vRR

vRR

v

to become

4321o vvvv41v

41

RR

i

f 4

124

RR i

f 3k

Chapter 5, Solution 44. R4

At node b, 0R

vvR

vv

2

2b

1

1b

21

2

2

1

1

b

R1

R1

Rv

Rv

v (1)

b

a

R1 v1

R2 v2

R3

vo +

At node a, 4

oa

3

a

Rvv

Rv0

34

oa R/R1

vv (2)

But va = vb. We set (1) and (2) equal.

21

1112

34

o

RRvRvR

R/R1v

or

vo = 1112213

43 vRvRRRR

RR

Page 208: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 45. This can be achieved as follows:

21o v2/R

Rv3/R

Rv

22

f1

1

f vRR

vRR

i.e. Rf = R, R1 = R/3, and R2 = R/2 Thus we need an inverter to invert v1, and a summer, as shown below (R<100k ).

R/3

R/2 v2

R

+

R v1

-v1

R

+

vo Chapter 5, Solution 46.

33

f2

2

x1

1

f32

1o v

RR

)v(RR

vRR

v21)v(

31

3v

v

i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10k , a solution is given below.

v1

30 k

20 kv3

10 k

+

10 k v2

-v2

10 k

+

30 k

vo

Page 209: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 47.

If a is the inverting terminal at the op amp and b is the noninverting terminal, then,

V6vv,V6)8(13

3v bab and at node a,4

vv2

v10 oaa

which leads to vo = �–2 V and io = k4

)vv(k5

v oao = �–0.4 �– 2 mA = �–2.4 mA

Chapter 5, Solution 48. Since the op amp draws no current from the bridge, the bridge may be treated separately as follows: v1

v2

i2

i1

+

For loop 1, (10 + 30) i1 = 5 i1 = 5/(40) = 0.125 A For loop 2, (40 + 60) i2 = -5 i2 = -0.05 A But, 10i + v1 - 5 = 0 v1 = 5 - 10i = 3.75mV 60i + v2 + 5 = 0 v2 = -5 - 60i = -2mV As a difference amplifier,

mV)2(75.32080vv

RR

v 121

2o

= 23mV

Page 210: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 49. R1 = R3 = 10k , R2/(R1) = 2 i.e. R2 = 2R1 = 20k = R4

Verify: 11

22

43

21

1

2o v

RR

vR/R1R/R1

RR

v

1212 vv2v2v5.01)5.01(2

Thus, R1 = R3 = 10k , R2 = R4 = 20k

Chapter 5, Solution 50. (a) We use a difference amplifier, as shown below:

,vv2vvRR

v 12121

2o i.e. R2/R1 = 2

R1 v2

R2

v1

R2

+

R1

vo

If R1 = 10 k then R2 = 20k (b) We may apply the idea in Prob. 5.35.

210 v2v2v

21 v2/R

Rv2/R

R

22

f1

1

f vRR

vRR

i.e. Rf = R, R1 = R/2 = R2

Page 211: Solution Manual for Fundamentals of Electric Circuits

We need an inverter to invert v1 and a summer, as shown below. We may let R = 10k .

R/2

R/2 v2

R

+

R v1

-v1

R

+

vo

Chapter 5, Solution 51.

We achieve this by cascading an inverting amplifier and two-input inverting summer as shown below: Verify: vo = -va - v1 But va = -v2. Hence

R

R v1

R

+

R v2

va

R

+

vo

vo = v2 - v1.

Page 212: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 52 A summing amplifier shown below will achieve the objective. An inverter is inserted to invert v2. Let R = 10 k . R/2 R v1 R/5 v3 - + vo v4 R R R v2 - R/4 + Chapter 5, Solution 53. (a)

vb

va

R1 v2

R2

v1

R2

+

R1 vo At node a,

(1)

At node b, 221

2b v

RRR

v (2)

But va = vb. Setting (1) and (2) equal gives

21

o1122

21

2

RRvRvR

vRR

R

Page 213: Solution Manual for Fundamentals of Electric Circuits

io2

112 vv

RR

vv

i

o

vv

1

2

RR

(b)

At node A, 2/Rvv

Rvv

2/Rvv

1

aA

g

AB

1

A1

v1

R2

vb

va

vB

vA

R1/2 v2

R1/2 Rg

R2

+

vo

+

R1/2 R1/2

vi

+

or aAABg

1A1 vvvv

R2R

vv (1)

At node B, g

bB

1

AB

1

B2

Rvv

2/Rvv

2/Rvv

or bBABg

1B2 vv)vv(

R2R

vv (2)

Subtracting (1) from (2),

abABABg

1AB12 vvvvvv

R2R2

vvvv

Since, va = vb,

2v

vvR2R

12

vv iAB

g

112

Page 214: Solution Manual for Fundamentals of Electric Circuits

or

g

1

iAB

R2R

1

12v

vv (3)

But for the difference amplifier,

AB1

2o vv

2/RR

v

or o2

1AB v

R2R

vv (4)

Equating (3) and (4),

g

1

io

2

1

R2R

1

12v

vR2R

g

11

2

i

o

R2R

1

1RR

vv

(c) At node a, 2/R

vvR

vv

2

Aa

1

a1

A2

1a

2

1a1 v

RR2

vRR2

vv (1)

At node b, B2

1b

2

1b2 v

RR2

vRR2

vv (2)

Since va = vb, we subtract (1) from (2),

2v

)vv(R

R2v i

AB2

112v

or i1

2AB v

R2R

vv (3)

At node A,

2/Rvv

Rvv

2/Rvv oA

g

AB

2

Aa

oAABg

2Aa vvvv

R2R

vv (4)

Page 215: Solution Manual for Fundamentals of Electric Circuits

At node B, 2/R0v

Rvv

2/Rvv B

g

ABBb

BABg

2Bb vvv

R2R

vv (5)

Subtracting (5) from (4),

oBAABg

2AB vvvvv

RR

vv

og

2AB v

R2R

1vv2 (6)

Combining (3) and (6),

og

2i

1

2 vR2

R1v

RR

g

2

1

2

i

o

R2R

1RR

vv

Chapter 5, Solution 54. (a) A0 = A1A2A3 = (-30)(-12.5)(0.8) = 300 (b) A = A1A2A3A4 = A0A4 = 300A4

But dB60ALog20 10 3ALog10

A = 103 = 1000 A4 = A/(300) = 3.333 Chapter 5, Solution 55. Let A1 = k, A2 = k, and A3 = k/(4) A = A1A2A3 = k3/(4) 42ALog20 10

A = 101.2ALog102 1 = 125.89

k3 = 4A = 503.57 k = 956.57. 75033 Thus A1 = A2 = 7.956, A3 = 1.989

Page 216: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 56. There is a cascading system of two inverting amplifiers.

sso v6v612

412

v

mAv3k2

vi s

so

(a) When vs = 12V, io = 36mA (b) When vs = 10 cos 377t V, io = 30 cos 377t mA

Chapter 5, Solution 57

The first stage is a difference amplifier. Since R1/R2 = R3/R4,

mA10)41(50

100)vv(

RR

v 121

2o

The second stage is a non-inverter.

)given(mV40mA1040R

1v40R

1v oo

Which leads to, R = 120 k

Chapter 5, Solution 58. By voltage division, the input to the voltage follower is:

V45.0)6.0(13

3v1

Thus

15.3v7v5

10v

210

v 111o

k4v0

i oo 0.7875mA

Page 217: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 59. Let a be the node between the two op amps. va = vo

The first stage is a summer

oosa vv2010

v510

v

1.5vs = -2vs or

2v

5.1vs

o -1.333

Chapter 5, Solution 60. Transform the current source as shown below: 4 k Assume all currents are in mA. The first stage is a summer

v1

3

5 k

3 k

+

10 k

+

+

io

5is

2 k

osos1 v5.2i10v4

10i5

510

v (1)

By voltage division,

Page 218: Solution Manual for Fundamentals of Electric Circuits

oo1 v21

v33

3v (2)

Alternatively, we notice that the second stage is a non-inverter.

11o v2v33

1v

From (1) and (2), 0 3voso v5.2i10v5. o = 10is

3i10

i2 soov

35

ii

s

o 1.667

Chapter 5, Solution 61.

Let v01 be the voltage at the left end of R5. The first stage is an inverter, while the second stage is a summer.

11

201 v

RR

v

015

40 v

RR

v 23

4 vRR

v1 = 151RR42 v

RR2

3

4 vRR

Chapter 5, Solution 62. Let v1 = output of the first op amp v2 = output of the second op amp The first stage is a summer

i1

21 v

RR

v �– of

2 vRR

(1)

The second stage is a follower. By voltage division

Page 219: Solution Manual for Fundamentals of Electric Circuits

143

42o v

RRR

vv o4

431 v

RRR

v (2)

From (1) and (2),

i1

2o

4

3 vRR

vRR

1 of

2 vRR

i1

2o

f

2

4

3 vRR

vRR

RR

1

4

2

4

31

2

i

o

RR

RR

1

1RR

vv

4321

42

RRRRRR

Chapter 5, Solution 63. The two op amps are summer. Let v1 be the output of the first op amp. For the first stage,

o3

2i

1

21 v

RR

vRR

v (1)

For the second stage,

io

41

5

4o v

RR

vRR

v (2)

Combining (1) and (2),

i6

4o

3

2

5

4i

1

2

5

4o v

RR

vRR

RR

vRR

RR

v

i6

4

51

42

53

42o v

RR

RRRR

RRRR

1v

53

42

6

4

31

42

i

o

RRRR

1

RR

RRRR

vv

Page 220: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 64

G4 G G3 G1 1 G 2 - - + 0V + v 0V + + vs G2 vo

- -

At node 1, v1=0 so that KCL gives

GvvGvG os 41 (1) At node 2,

GvvGvG os 32 (2) From (1) and (2),

ososos vGGvGGvGvGvGvG )()( 43213241 or

43

21

GGGG

vv

s

o

Chapter 5, Solution 65

The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so that its output is

mV -18mV)6(1030

'ov

The third op amp is a noninverter so that

mV 6.21'4048

84040

' oooo vvvv

Page 221: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 66.

)2(10100

)4(2040

20100

)6(25110

vo

204024 -4V Chapter 5, Solution 67.

vo = )2.0(2080

)5.0(20408080

8.02.3 2.4V Chapter 5, Solution 68.

If Rq = , the first stage is an inverter.

mV30)10(5

15Va

when Va is the output of the first op amp. The second stage is a noninverting amplifier.

)30)(31(v26

1v ao -120mV

Chapter 5, Solution 69.

In this case, the first stage is a summer

ooa v5.130v1015

)10(5

15v

For the second stage,

oaao v5.1304v4v26

1v

120v7 o 7120

ov -17.143mV

Page 222: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 70.

The output of amplifier A is

9)2(1030

)10(1030

vA

The output of amplifier B is

14)4(1020

)3(1020

vB

40 k

V2)14(1060

60vb

vA

vB

a

b

60 k

20 k

+

10 k

vo

At node a, 40

vv20

vv oaaA

But va = vb = -2V, 2(-9+2) = -2-vo

Therefore, vo = 12V

Page 223: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 71 20k 5k 100k

- 40k +

+ v2 2V 80k - - 10k + + vo 20k - - 10k + v1 + - v3 + 3V 50k - 30k

8)30501(,8)2(

520,3 1321 vvvv

V 10)1020(80

10040

10032 vvvo

Page 224: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 72.

Since no current flows into the input terminals of ideal op amp, there is no voltage drop across the 20 k resistor. As a voltage summer, the output of the first op amp is v01 = 0.4 The second stage is an inverter

012 v100150v

)4.0(5.2 -1V Chapter 5, Solution 73.

The first stage is an inverter. The output is

V9)8.1(1050v01

The second stage is 012 vv -9V

Chapter 5, Solution 74. Let v1 = output of the first op amp v2 = input of the second op amp.

The two sub-circuits are inverting amplifiers

V6)6.0(10100

1v

V8)4.0(6.1

322v

k20

86k20vv 21

oi 100 A

Chapter 5, Solution 75. The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are involved as shown to measure vo and i respectively. Once the circuit is saved, we click Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:

Page 225: Solution Manual for Fundamentals of Electric Circuits

i = 200 A

(vo/vs) = -4/2 = -2 The results are slightly different than those obtained in Example 5.11.

Chapter 5, Solution 76. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as

io = -374.78 A

Page 226: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 77. The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as

io = -374.78 A

Chapter 5, Solution 78. The circuit is constructed as shown below. We insert a VIEWPOINT to display vo. Upon simulating the circuit, we obtain,

vo = 667.75 mV

Page 227: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 79. The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display vo. After saving and simulating the circuit, we obtain,

vo = -14.61 V

Chapter 5, Solution 80. The schematic is shown below. VIEWPOINT is inserted to display vo. After simulation, we obtain,

vo = 12 V

Page 228: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 81. The schematic is shown below. We insert one VIEWPOINT and one IPROBE to measure vo and io respectively. Upon saving and simulating the circuit, we obtain,

vo = 343.37 mV

io = 24.51 A

Page 229: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 82.

The maximum voltage level corresponds to

11111 = 25 �– 1 = 31 Hence, each bit is worth (7.75/31) = 250 mV

Chapter 5, Solution 83. The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 = 20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms, then,

-vo = (Rf/R1)v1 + --------- + (Rf/R6)v6

= v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6

(a) |vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies, [v1 v2 v3 v4 v5 v6] = [100110]

(b) |vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV (c) This corresponds to [1 1 1 1 1 1]. |vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V

Page 230: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 84. For (a), the process of the proof is time consuming and the results are only approximate, but close enough for the applications where this device is used. (a) The easiest way to solve this problem is to use superposition and to solve

for each term letting all of the corresponding voltages be equal to zero. Also, starting with each current contribution (ik) equal to one amp and working backwards is easiest.

2R R R R

+

ik 2R

v2 v4

+

v3

+

+

2R 2R

v1

R

For the first case, let v2 = v3 = v4 = 0, and i1 = 1A. Therefore, v1 = 2R volts or i1 = v1/(2R). Second case, let v1 = v3 = v4 = 0, and i2 = 1A. Therefore, v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not (1/4th), so where is the difference? (21/85) = 0.247 which is a really good approximation for 0.25. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A. Therefore, v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not (1/8th), so where is the difference? (1/8.5) = 0.11765 which is a really good approximation for 0.125. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.

Page 231: Solution Manual for Fundamentals of Electric Circuits

Therefore, v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not (1/16th), so where is the difference? (1/16.25) = 0.06154 which is a really good approximation for 0.0625. Since this is a practical electronic circuit, the result is good enough for all practical purposes. Please note that a goal of a lot of electronic design is to come up with practical circuits that are economical to design and build yet give the desired results.

(b) If Rf = 12 k ohms and R = 10 k ohms,

-vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)]

= 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4] For [v1 v2 v3 v4] = [1 0 11],

|vo| = 0.6[1 + 0.25 + 0.125] = 825 mV

For [v1 v2 v3 v4] = [0 1 0 1],

|vo| = 0.6[0.5 + 0.125] = 375 mV

Chapter 5, Solution 85.

Av = 1 + (2R/Rg) = 1 + 20,000/100 = 201 Chapter 5, Solution 86.

vo = A(v2 �– v1) = 200(v2 �– v1)

(a) vo = 200(0.386 �– 0.402) = -3.2 V (b) vo = 200(1.011 �– 1.002) = 1.8 V

Chapter 5, Solution 87.

The output, va, of the first op amp is,

va = (1 + (R2/R1))v1 (1) Also, vo = (-R4/R3)va + (1 + (R4/R3))v2 (2)

Page 232: Solution Manual for Fundamentals of Electric Circuits

Substituting (1) into (2), vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2 Or, vo = (1 + (R4/R3))v2 �– (R4/R3 + (R2R4/R1R3))v1 If R4 = R1 and R3 = R2, then, vo = (1 + (R4/R3))(v2 �– v1) which is a subtractor with a gain of (1 + (R4/R3)). Chapter 5, Solution 88. We need to find VTh at terminals a �– b, from this,

vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh

= 220VTh Now we use Fig. (b) to find VTh in terms of vi.

30 k

80 k

20 k

40 k

30 k

b

a

vi +

b

a

vi

20 k

40 k 80 k

(a) (b)

va = (3/5)vi, vb = (2/3)vi

VTh = vb �– va (1/15)vi

(vo/vi) = Av = -220/15 = -14.667

Page 233: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 89. If we use an inverter, R = 2 k ohms,

(vo/vi) = -R2/R1 = -6

R = 6R = 12 k ohms Hence the op amp circuit is as shown below. 12 k

2 k

+ +

+

vo

vi Chapter 5, Solution 90. Transforming the current source to a voltage source produces the circuit below, At node b, vb = (2/(2 + 4))vo = vo/3

20 k

io

b

a

2 k

5 k

++

vo

+

4 k

5is At node a, (5is �– va)/5 = (va �– vo)/20 But va = vb = vo/3. 20is �– (4/3)vo = (1/3)vo �– vo, or is = vo/30 io = [(2/(2 + 4))/2]vo = vo/6 io/is = (vo/6)/(vo/30) = 5

Page 234: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 91.

io

i2

i1

is

+vo

R1

R2

io = i1 + i2 (1) But i1 = is (2) R1 and R2 have the same voltage, vo, across them. R1i1 = R2i2, which leads to i2 = (R1/R2)i1 (3) Substituting (2) and (3) into (1) gives, io = is(1 + R1/R2) io/is = 1 + (R1/R2) = 1 + 8/1 = 9 Chapter 5, Solution 92

The top op amp circuit is a non-inverter, while the lower one is an inverter. The output at the top op amp is

v1 = (1 + 60/30)vi = 3vi while the output of the lower op amp is v2 = -(50/20)vi = -2.5vi Hence, vo = v1 �– v2 = 3vi + 2.5vi = 5.5vi

vo/vi = 5.5

Page 235: Solution Manual for Fundamentals of Electric Circuits

Chapter 5, Solution 93. R3

+

vi

+

vL

vb

R4

RL

io

iL

+

va R1

R2

+

vo

At node a, (vi �– va)/R1 = (va �– vo)/R3 vi �– va = (R1/R2)(va �– vo) vi + (R1/R3)vo = (1 + R1/R3)va (1) But va = vb = vL. Hence, (1) becomes vi = (1 + R1/R3)vL �– (R1/R3)vo (2)

io = vo/(R4 + R2||RL), iL = (RL/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL)) Or, vo = iL[(R2 + RL)( R4 + R2||RL)/R2 (3) But, vL = iLRL (4) Substituting (3) and (4) into (2),

vi = (1 + R1/R3) iLRL �– R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL

= [((R3 + R1)/R3)RL �– R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL = (1/A)iL

Page 236: Solution Manual for Fundamentals of Electric Circuits

Thus,

A =

L2

L24

32

L21L

3

1

RRRR

RRRRR

RRRR

1

1

Page 237: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 1.

t3t3 e6e25dtdv

Ci 10(1 - 3t)e-3t A

p = vi = 10(1-3t)e-3t 2t e-3t = 20t(1 - 3t)e-6t W

Chapter 6, Solution 2.

2211 )120)(40(

21

Cv21

w

w2 = 221 )80)(40(

21

2Cv

1

22

21 8012020www 160 kW Chapter 6, Solution 3.

i = C5

16028010x40

dtdv 3 480 mA

Chapter 6, Solution 4.

)0(vidtC1

vt

o

1tdt4sin621

= 1 - 0.75 cos 4t Chapter 6, Solution 5.

v = t

o)0(vidt

C1

For 0 < t < 1, i = 4t,

t

o6 t410x201

v dt + 0 = 100t2 kV

v(1) = 100 kV

Page 238: Solution Manual for Fundamentals of Electric Circuits

For 1 < t < 2, i = 8 - 4t,

t

16 )1(vdt)t48(10x201

v

= 100 (4t - t2 - 3) + 100 kV

Thus v (t) = 2t1,kV)2tt4(100

1t0,kVt1002

2

Chapter 6, Solution 6.

610x30dtdv

Ci x slope of the waveform.

For example, for 0 < t < 2,

310x210

dtdv

i = mA15010x210

x10x30dtdv

36C

Thus the current i is sketched below.

t (msec)

150

12 10 2

8

6

4

-150

i(t) (mA) Chapter 6, Solution 7.

t

o

33o 10dt10tx4

10x501

)t(vidtC1

v

= 1050t2 2

0.04k2 + 10 V

Page 239: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 8.

(a) tt BCeACedtdv

C 600100 600100i (1)

BABCACi 656001002)0( (2)

BAvv 50)0()0( (3) Solving (2) and (3) leads to A=61, B=-11

(b) J 5250010421

)0(21 32 xxxCvEnergy

(c ) From (1),

A 4.264.241041160010461100 60010060031003 tttt eeexxxexxxi Chapter 6, Solution 9.

v(t) = t

o

tt Vet120dte1621

1

v(2) = 12(2 + e-2) = 25.62 V

p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)

p(2) = 72(2-e-4) = 142.68 W Chapter 6, Solution 10

dtdv

xdtdv

Ci 3102

s4t316t,-64s 3t116,s10,16 tt

v

Page 240: Solution Manual for Fundamentals of Electric Circuits

s4t3,16x10-s 3t10,

s10,1016

6

6 tx

dtdv

s4t3kA, 32-s 3t10,

s10,kA 32)(

tti

Chapter 6, Solution 11.

v = t

o)0(vidt

C1

For 0 < t < 1,

t

o

36 t10dt10x40

10x41

v kV

v(1) = 10 kV For 1 < t < 2,

kV10)1(vvdtC1

vt

1

For 2 < t < 3,

t

2

36 )2(vdt)10x40(

10x41

v

= -10t + 30kV Thus

v(t) = 3t2,kV30t10

2t1,kV101t0,kVt10

Chapter 6, Solution 12.

4sin)(4(60x10x3dtdv

Ci 3 t)

= - 0.7e sin 4 t A P = vi = 60(-0.72) cos 4 t sin 4 t = -21.6 sin 8 t W

W = t dt t

o81

o8sin6.21pdt

= 88

6.cos

21 8/1o = -5.4J

Page 241: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 13.

Under dc conditions, the circuit becomes that shown below:

i250

20

+

60V

+

v1

i1

30

10

+

v2

i2 = 0, i1 = 60/(30+10+20) = 1A v1 = 30i2 = 30V, v2 = 60-20i1 = 40V

Thus, v1 = 30V, v2 = 40V Chapter 6, Solution 14. (a) Ceq = 4C = 120 mF

(b) 304

C4

C1

eq

Ceq = 7.5 mF

Chapter 6, Solution 15. In parallel, as in Fig. (a), v1 = v2 = 100

C2+

v2

C1 +

100V

+

v2

C2

+ v1

C1 +

v1

+

100V

(b)(a)

Page 242: Solution Manual for Fundamentals of Electric Circuits

w20 = 262 100x10x20x21Cv

21 0.1J

w30 = 26 100x10x30x21 0.15J

(b) When they are connected in series as in Fig. (b):

,60100x5030V

CCC

v21

21 v2 = 40

w20 = 26 60x10x30x21

36 mJ

w30 = 26 4010x302

xx1 24 mJ

Chapter 6, Solution 16

F 203080

8014 CCCx

Ceq

Chapter 6, Solution 17. (a) 4F in series with 12F = 4 x 12/(16) = 3F 3F in parallel with 6F and 3F = 3+6+3 = 12F 4F in series with 12F = 3F i.e. Ceq = 3F

(b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6F

131

61

21

C1

eq

Ceq = 1F

Page 243: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 18.

For the capacitors in parallel = 15 + 5 + 40 = 60 F 1

eqC

Hence 101

601

301

201

C1

eq

Ceq = 10 F Chapter 6, Solution 19. We combine 10-, 20-, and 30- F capacitors in parallel to get 60 F. The 60 - F capacitor in series with another 60- F capacitor gives 30 F. 30 + 50 = 80 F, 80 + 40 = 120 F The circuit is reduced to that shown below.

12 120

12 80

120- F capacitor in series with 80 F gives (80x120)/200 = 48 48 + 12 = 60 60- F capacitor in series with 12 F gives (60x12)/72 = 10 F Chapter 6, Solution 20.

3 in series with 6 = 6x3/(9) = 2 2 in parallel with 2 = 4 4 in series with 4 = (4x4)/8 = 2 The circuit is reduced to that shown below:

20

1 6

2

8

Page 244: Solution Manual for Fundamentals of Electric Circuits

6 in parallel with 2 = 8 8 in series with 8 = 4 4 in parallel with 1 = 5 5 in series with 20 = (5x20)/25 = 4 Thus Ceq = 4 mF

Chapter 6, Solution 21. 4 F in series with 12 F = (4x12)/16 = 3 F 3 F in parallel with 3 F = 6 F 6 F in series with 6 F = 3 F 3 F in parallel with 2 F = 5 F 5 F in series with 5 F = 2.5 F

Hence Ceq = 2.5 F Chapter 6, Solution 22. Combining the capacitors in parallel, we obtain the equivalent circuit shown below:

a b

40 F 60 F 30 F

20 F Combining the capacitors in series gives C , where 1

eq

101

301

201

601

C11eq

C = 10 F 1eq

Thus

Ceq = 10 + 40 = 50 F

Page 245: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 23.

(a) 3 F is in series with 6 F 3x6/(9) = 2 F v4 F = 1/2 x 120 = 60V v2 F = 60V

v6 F = (3 )6036

20V

v3 F = 60 - 20 = 40V

(b) Hence w = 1/2 Cv2 w4 F = 1/2 x 4 x 10-6 x 3600 = 7.2mJ w2 F = 1/2 x 2 x 10-6 x 3600 = 3.6mJ w6 F = 1/2 x 6 x 10-6 x 400 = 1.2mJ w3 F = 1/2 x 3 x 10-6 x 1600 = 2.4mJ

Chapter 6, Solution 24.

20 F is series with 80 F = 20x80/(100) = 16 F

14 F is parallel with 16 F = 30 F (a) v30 F = 90V

v60 F = 30V v14 F = 60V

v20 F = 60x8020

8048V

v80 F = 60 - 48 = 12V

(b) Since w = 2Cv21

w30 F = 1/2 x 30 x 10-6 x 8100 = 121.5mJ w60 F = 1/2 x 60 x 10-6 x 900 = 27mJ w14 F = 1/2 x 14 x 10-6 x 3600 = 25.2mJ w20 F = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ w80 F = 1/2 x 80 x 10-6 x 144 = 5.76mJ

Chapter 6, Solution 25.

(a) For the capacitors in series,

Q1 = Q2 C1v1 = C2v2 1

2

2

1

CC

vv

Page 246: Solution Manual for Fundamentals of Electric Circuits

vs = v1 + v2 = 21

2122

1

2 vC

CCvv

CC

s21

12 v

CCC

v

Similarly, s21

21 v

CCC

v

(b) For capacitors in parallel

v1 = v2 = 2

2

1

1

CQ

CQ

Qs = Q1 + Q2 = 22

2122

2

1 QC

CCQQ

CC

or

Q2 = 21

2

CCC

s21

11 Q

CCC

Q

i = dtdQ s

21

11 i

CCC

i , s21

22 i

CCC

i

Chapter 6, Solution 26.

(a) Ceq = C1 + C2 + C3 = 35 F (b) Q1 = C1v = 5 x 150 C = 0.75mC

Q2 = C2v = 10 x 150 C = 1.5mC Q3 = C3v = 20 x 150 = 3mC

(c) w = J150x35x21

222

eq vC1 = 393.8mJ

Page 247: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 27.

(a) 207

201

101

51

C1

C1

C1

C1

321eq

Ceq = F720 2.857 F

(b) Since the capacitors are in series,

Q1 = Q2 = Q3 = Q = Ceqv = V200x720 0.5714mV

(c) w = J200x720x

21

222

eq vC1 57.143mJ

Chapter 6, Solution 28. We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.

Ca

Cb

Cc

50 F 20 F

301

401

301

301

101

401

101

C1

a

= 102

401

101

403

Ca = 5 F

302

101

12001

3001

4001

C1

6

Cb = 15 F

Page 248: Solution Manual for Fundamentals of Electric Circuits

154

401

12001

3001

4001

C1

c

Cc = 3.75 F Cb in parallel with 50 F = 50 + 15 = 65 F Cc in series with 20 F = 23.75 F

65 F in series with 23.75 F = F39.1775.88

75.23x65

17.39 F in parallel with Ca = 17.39 + 5 = 22.39 F Hence Ceq = 22.39 F Chapter 6, Solution 29. (a) C in series with C = C/(2) C/2 in parallel with C = 3C/2

2C3 in series with C =

5C3

2C5

2C3Cx

5C3 in parallel with C = C +

5C3 1.6 C

(b) 2C

Ceq 2C

C1

C21

C21

C1

eq

Ceq = C

Page 249: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 30.

vo = t

o)0(iidt

C1

For 0 < t < 1, i = 60t mA,

kVt100tdt6010x3

10v 2t

o6

3

o

vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA,

vo = t

1 o6

3

)1(vdt)t60120(10x3

10 t

= [40t �– 10t2 kV10] 1 = 40t �– 10t2 - 20

2t1,kV20t10t40

1t0,kVt10)t(v

2

2

o

Chapter 6, Solution 31.

5t3,t10503t1,mA201t0,tmA20

)t(is

Ceq = 4 + 6 = 10 F

)0(vidtC1v

t

oeq

For 0 < t < 1,

t

o6

3

t2010x10

10v dt + 0 = t2 kV

For 1 < t < 3,

t

1

3

kV1)1t(2)1(vdt201010v

kV1t2 For 3 < t < 5,

t

3

3

)3(vdt)5t(101010v

Page 250: Solution Manual for Fundamentals of Electric Circuits

kV11t5tkV55t 2t3

2

5t3,kV11t5t

3t1,kV1t21t0,kVt

)t(v2

2

dtdv10x6

dtdvCi 6

11

5t3,mA30123t1,mA121t0,tmA12

dtdv10x4

dtdvCi 6

21

5t3,mA20t83t1,mA81t0,tmA8

Chapter 6, Solution 32.

(a) Ceq = (12x60)/72 = 10 F

13001250501250)0(301012

10 2

00

21

26

3

1t

tttt eevdte

xv

23025020250)0(301060

10 2

00

22

26

3

2t

tttt eevdte

xv

(b) At t=0.5s,

03.138230250,15.84013001250 12

11 evev

J 235.4)15.840(101221 26

12 xxxw F

J 1905.0)03.138(102021 26

20 xxxw F

J 381.0)03.138(104021 26

40 xxxFw

Page 251: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 33

Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals. 3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors. Therefore, we get: VTh = 7.5 V, CTh = 2.5 F

Chapter 6, Solution 34.

i = 6e-t/2

2/t3 e21)6(10x10

dtdiLv

= -30e-t/2 mV v(3) = -300e-3/2 mV = -0.9487 mV p = vi = -180e-t mW

p(3) = -180e-3 mW = -0.8 mW

Chapter 6, Solution 35.

dtdiLv

)2/(6.010x60

t/iVL

3

200 mH

Page 252: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 36.

V)t2sin)(2)(12(10x41

dtdiLv 3

= - 6 sin 2t mV p = vi = -72 sin 2t cos 2t mW But 2 sin A cos A = sin 2A

p = -36 sin 4t mW Chapter 6, Solution 37.

t100cos)100(4x10x12dtdiLv 3

= 4.8 cos 100t V p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t

w = t200sin6.9pdtt

o

200/11

o

Jt200cos200

6.9 200/11o

mJ)1(cos48 96 mJ Chapter 6, Solution 38.

dtte2e10x40dtdiLv t2t23

= 240 0t,mVe)t1( t2

Page 253: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 39

)0(iidtL1i

dtdiLv t

0

1dt)4t2t3(10x200

1i t0

23

1)t4tt(5t

023

i(t) = 5t3 + 5t2 + 20t + 1 A Chapter 6, Solution 40

dtdi

xdtdi

Lv 31020

ms 4t310t,40-ms 3t110t,-20

ms 10,10 tti

ms 4t3,10x10ms 3t1,10x10-

ms 10,1010

3

3

3 tx

dtdi

ms 4t3V, 200ms 3t1V, 200-

ms 10,V 200 tv

which is sketched below. v(t) V 200 0 1 2 3 4 t(ms) -200

Page 254: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 41.

t

o

t2t

03.0dt2120

21)0(ivdt

L1i

= Aetet tto

t 745103021 22 ..10

At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A

L21w i2 = 35.72J

Chapter 6, Solution 42.

t

o

t

o1dt)t(v

51)0(ivdt

L1i

For 0 < t < 1, t

01t21dt

510i A

For 1 < t < 2, i = 0 + i(1) = 1A

For 2 < t < 3, i = 1t2)2(idt1051 2

t

= 2t - 3 A For 3 < t < 4, i = 0 + i(3) = 3 A

For 4 < t < 5, i = t

4

t4 3t2)4(idt10

51

= 2t - 5 A

Thus,

2 1 , 0 11 , 1 2

( ) 2 3 , 2 33 , 3 42 5, 4

t A tA t

i t t A tA tt t 5

Page 255: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 43.

w = L )(Li21)t(Li

21idt

2t

010x60x10x80x21 33

= 144 J Chapter 6, Solution 44.

t

ot

t

oo 1dt)t2cos104(51tivdt

L1i

= 0.8t + sin 2t -1

Chapter 6, Solution 45.

i(t) = t

o)0(i)t(v

L1

For 0 < t < 1, v = 5t

t

o3 t510x101i dt + 0

= 0.25t2 kA

For 1 < t < 2, v = -10 + 5t

t

13 )1(idt)t510(10x101i

( t

1kA25.0dt)1t5.0

= 1 - t + 0.25t2 kA

2t1,kAt25.0t1

1t0,kAt25.0)t(i

2

2

Page 256: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 46. Under dc conditions, the circuit is as shown below:

2

+

vC 3 A

iL

4

By current division,

)3(24

4iL 2A, vc = 0V

L21w L

22L )2(

21

21i 1J

C21w c )v)(2(

21v2

c 0J

Chapter 6, Solution 47. Under dc conditions, the circuit is equivalent to that shown below: R

+ vC

5 A

iL

2

,2R

10)5(2R

2iL 2R

R10Riv Lc

Page 257: Solution Manual for Fundamentals of Electric Circuits

2

262

cc )2R(R100x10x80Cv

21w

232

1L )2R(100x10x2Li

21w

If wc = wL,

2

3

2

26

)2R(100x10x2

)2Rx(R100x10x80 80 x 10-3R2 = 2

R = 5

Chapter 6, Solution 48. Under dc conditions, the circuit is as shown below:

+

vC2 +

+

vC1

iL1

6

4

iL2 30V

6430ii

2L1L 3A

1L1C i6v 18V

2Cv 0V

Page 258: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 49.

(a) 36544165Leq 7H (b) 41266112Leq 3H (c) 446324Leq 2H

Chapter 6, Solution 50.

63124510Leq

= 10 + 5||(3 + 2) = 10 + 2.5 = 12.5 mH

Chapter 6, Solution 51.

101

301

201

601

L1 L = 10 mH

4535x10102510Leq

= 7.778 mH

Chapter 6, Solution 52. 3//2//6 = 1H, 4//12 = 3H After the parallel combinations, the circuit becomes that shown below. 3H a 1H 1 H b Lab = (3+1)//1 = (4x1)/5 = 0.8 H

Page 259: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 53.

)48(6)128(58106Leq 416)44(816 Leq = 20 mH

Chapter 6, Solution 54.

126010)39(4Leq 34)40(124 Leq = 7H Chapter 6, Solution 55.

(a) L//L = 0.5L, L + L = 2L

LLLLLx

LLLLLeq 4.15.025.025.0//2

(b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 0.5L

Chapter 6, Solution 56.

3L

L31LLL

Hence the given circuit is equivalent to that shown below: L

L/3 L

L/3

Page 260: Solution Manual for Fundamentals of Electric Circuits

L35L

L35Lx

L32LLLeq L

85

Chapter 6, Solution 57.

Let dtdiLeqv (1)

221 vdtdi4vvv (2)

i = i1 + i2 i2 = i �– i1 (3)

3

vdtdi

ordtdi

3v 2112 (4)

and

0dtdi

5dtdi2v 2

2

dtdi

5dtdi2v 2

2 (5)

Incorporating (3) and (4) into (5),

3

v5

dtdi7

dtdi

5dtdi5

dtdi2v 21

2

dtdi7

351v2

dtdi

835v2

Substituting this into (2) gives

dtdi

835

dtdi4v

dtdi

867

Comparing this with (1),

867Leq 8.375H

Page 261: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 58.

dtdi3

dtdiLv 3 x slope of i(t).

Thus v is sketched below:

6

t (s)

7 5 1 4 3 2

-6

v(t) (V) 6 Chapter 6, Solution 59.

(a) dtdiLLv 21s

21

s

LLv

dtdi

,dtdiL11v

dtdiL22v

,vLL

Lv s

21

11 s

21

2L v

LLL

v

(b) dtdi

Ldtdi

Lvv 22

112i

21s iii

21

21

21

21s

LLLL

vLv

Lv

dtdi

dtdi

dtdi

dtdtdi

LLLL

L1vdt

L1i s

21

21

111 s

21

2 iLL

L

Page 262: Solution Manual for Fundamentals of Electric Circuits

dtdtdi

LLLL

L1vdt

L1i s

21

21

222 s

21

1 iLL

L

Chapter 6, Solution 60

8

155//3eqL

tteqo ee

dtd

dtdi

Lv 22 1548

15

t

tt

ttt

ooo eeeidttvLI

i0

2

0

22

0

A 5.15.05.12)15(512)0()(

Chapter 6, Solution 61.

(a) is = i1 + i2 i )0(i)0(i)0( 21s

6 i)0(i4 2 2(0) = 2mA (b) Using current division:

t2s1 e64.0i

203020i 2.4e-2t mA

1s2 iii 3.6e-2t mA

(c) mH1250

20x302030

3t231 10xe6

dtd10x10

dtdiLv -120e-2t V

3t232 10xe6

dtd10x12

dtdiLv -144e-2t V

(d) 6t43mH10 10xe3610x30x

21w

Je8.021

t

t4

= 24.36nJ

2/1t6t43

mH30 10xe76.510x30x21w

= 11.693nJ

2/1t6t43

mH20 10xe96.1210x20x21w

= 17.54 nJ

Page 263: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 62.

(a) mH 4080

60202560//2025 xLeq

ttt

eqeq ieidte

xidttv

Li

dtdi

Lv0

333

3

)0()1(1.0)0(121040

10)0()(1

Using current division,

iiiii41,

43

8060

21

01333.0)0(01.0)0(75.0)0(43)0(1 iiii

mA 67.2125e- A )08667.01.0(41 3t-3

2tei

mA 33.367.2125)0(2i

(b) mA 6575e- A )08667.01.0(43 3t-3

1tei

mA 67.2125e- -3t2i

Chapter 6, Solution 63. We apply superposition principle and let

21 vvvo where v1 and v2 are due to i1 and i2 respectively.

63,230,2

2 111 t

tdtdi

dtdi

Lv

64,442,020,4

2 222

ttt

dtdi

dtdi

Lv

Page 264: Solution Manual for Fundamentals of Electric Circuits

v1 v2 2 4 0 3 6 t 0 2 4 6 t -2 -4 Adding v1 and v2 gives vo, which is shown below. vo(t) V 6 2 0 2 3 4 6 t (s) -2 -6 Chapter 6, Solution 64.

(a) When the switch is in position A, i=-6 =i(0) When the switch is in position B,

8/1/,34/12)( RLi

A 93)]()0([)()( 8/ tt eeiiiti (b) -12 + 4i(0) + v=0, i.e. v=12 �– 4i(0) = 36 V (c) At steady state, the inductor becomes a short circuit so that v= 0 V

Page 265: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 65.

(a) 22115 )4(x5x

21iL

21w 40 W

220 )2)(20(

21w 40 W

(b) w = w5 + w20 = 80 W

(c) 3t20011 10xe502005

dtdvLi

= -50e-200tA

3t20022 10xe50)200(20

dtdvLi

= -200e-200tA

3t20022 10xe50)200(20

dtdvLi

= -200e-200t A (d) i = i1 + i2 = -250e-200t A

Chapter 6, Solution 66. mH60243640601620Leq

dtdiLv

t

o)0(ivdt

L1i

121 dt + 0 mA t

o3 t4sin10x60

tot4cos50i 50(1 - cos 4t) mA

mH244060

mV)t4cos1)(50(dtd10x24

dtdiLv 3

= 4.8 sin 4t mV

Page 266: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 67.

viRC1vo dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3

t50sin10210v

3

o dt

vo = 100 cos 50t mV Chapter 6, Solution 68.

viRC1vo dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5

vo = t

ot20dt10

51

The op amp will saturate at vo = 12 -12 = -2t t = 6s

Chapter 6, Solution 69. RC = 4 x 106 x 1 x 10-6 = 4

dtv41dtv

RC1v iio

For 0 < t < 1, vi = 20, t

oo dt2041v -5t mV

For 1 < t < 2, vi = 10, t

1o 5)1t(5.2)1(vdt1041v

= -2.5t - 2.5mV

For 2 < t < 4, vi = - 20, t

2o 5.7)2t(5)2(vdt2041v

= 5t - 17.5 mV

For 4 < t < 5m, vi = -10, 5.2)4t(5.2)4(vdt1041v

t

4o

= 2.5t - 7.5 mV

For 5 < t < 6, vi = 20, t

5o 5)5t(5)5(vdt2041v

= - 5t + 30 mV

Page 267: Solution Manual for Fundamentals of Electric Circuits

Thus vo(t) is as shown below:

2 5

6

5

751 432

5

0 Chapter 6, Solution 70.

One possibility is as follows:

50RC1

Let R = 100 k , F2.010x100x50

13C

Chapter 6, Solution 71. By combining a summer with an integrator, we have the circuit below:

dtvCR

1dtvCR

1dtvCR

1v 22

22

11

o

+

For the given problem, C = 2 F, R1C = 1 R1 = 1/(C) = 1006/(2) = 500 k R2C = 1/(4) R2 = 1/(4C) = 500k /(4) = 125 k R3C = 1/(10) R3 = 1/(10C) = 50 k

Page 268: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 72.

The output of the first op amp is

i1 vRC1v dt =

t

o63 2t100idt

10x2x10x101

= - 50t

io vRC1v dt =

t

o63 dt)t50(10x5.0x10x20

1

= 2500t2 At t = 1.5ms, 62

o 10x)5.1(2500v 5.625 mV Chapter 6, Solution 73.

Consider the op amp as shown below: Let va = vb = v

At node a, R

vvR

v0 o 2v - vo = 0 (1)

v +

vo

b +

vi C

R

v R

a

R

+

R

At node b, dtdvC

Rvv

Rvv oi

Page 269: Solution Manual for Fundamentals of Electric Circuits

dtdvRCvv2v oi (2)

Combining (1) and (2),

dt

dv2

RCvvv oooi

or

io vRC2v dt

showing that the circuit is a noninverting integrator. Chapter 6, Solution 74.

RC = 0.01 x 20 x 10-3 sec

secmdtdv2.0

dtdv

RCv io

4t3,V23t1,V21t0,V2

vo

Thus vo(t) is as sketched below:

3

t (ms)

2 1

-2

vo(t) (V) 2

Page 270: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 75.

,dt

dvRCv i

0 5.210x10x10x250RC 63

)t12(dtd5.2vo -30 mV

Chapter 6, Solution 76.

,dt

dvRCv i

o RC = 50 x 103 x 10 x 10-6 = 0.5

5t5,55t0,10

dtdv

5.0v io

The input is sketched in Fig. (a), while the output is sketched in Fig. (b).

t (ms)

5

5

0 10

(b)

15

-10

t (ms)

vi(t) (V)

5

5

0 10

(a)

15

vo(t) (V) Chapter 6, Solution 77. i = iR + iC

oF

0i v0dtdC

Rv0

R0v

110x10CR 66F

Page 271: Solution Manual for Fundamentals of Electric Circuits

Hence dt

dvvv o

oi

Thus vi is obtained from vo as shown below: �–dvo(t)/dt �– vo(t) (V)

4

-4

t (ms)

1

4

0 2 3

vi(t) (V)

3

8

2 1

t (ms)

-8

4 -4

4

-4

t (ms)

1

4

0 2 3

Chapter 6, Solution 78.

ooo

2

vdtdv2

t2sin10dtvd

Thus, by combining integrators with a summer, we obtain the appropriate analog computer as shown below:

Page 272: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 79.

We can write the equation as

)(4)( tytfdtdy

which is implemented by the circuit below. 1 V t=0 C R R R R/4 R dy/dt - - -

+ -y + + R dy/dt

f(t)

R

t = 0

-dvo/dt

dvo/dt

d2vo/dt2

d2vo/dt2

-sin2t

2vo

C C

R

R/10

R

R

R

R/2

R

R

+

+

+

+

+

+

sin2t

+vo

R

Page 273: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 80.

From the given circuit,

dt

dvk200k1000v

k5000k1000)t(f

dtvd o

o2o

2

or

)t(fv2dt

dv5

dtvd

oo

2o

2

Chapter 6, Solution 81

We can write the equation as

)(252

2

tfvdtvd

which is implemented by the circuit below. C C R R - R R/5 d2v/dt2 + - -dv/dt + v - + d2v/dt2 R/2 f(t)

Page 274: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 82 The circuit consists of a summer, an inverter, and an integrator. Such circuit is shown below. 10R R R R - + - vo + R C=1/(2R) R - + + vs - Chapter 6, Solution 83.

Since two 10 F capacitors in series gives 5 F, rated at 600V, it requires 8 groups in parallel with each group consisting of two capacitors in series, as shown below:

+

600

Answer: 8 groups in parallel with each group made up of 2 capacitors in series.

Page 275: Solution Manual for Fundamentals of Electric Circuits

Chapter 6, Solution 84.

tqI I x t = q

q = 0.6 x 4 x 10-6

= 2.4 C

62.4 10 150

(36 20)q x

C nv

F

Chapter 6, Solution 85. It is evident that differentiating i will give a waveform similar to v. Hence,

dtdiLv

2t1,t48

1t0,t4i

2t1,L4

1t0,L4dtdiLv

But, 2t1,mV5

1t0,mV5v

Thus, 4L = 5 x 10-3 L = 1.25 mH in a 1.25 mH inductor Chapter 6, Solution 86. (a) For the series-connected capacitor

Cs = 8C

C1....

CC11

1

For the parallel-connected strings,

F3

1000x108C10

C10C sseq 1250 F

Page 276: Solution Manual for Fundamentals of Electric Circuits

(b) vT = 8 x 100V = 800V

262Teq )800(10x1250

21vC

21w

= 400J

Page 277: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 1.

Applying KVL to Fig. 7.1.

0RidtiC1 t

-

Taking the derivative of each term,

0dtdi

RCi

or RCdt

idi

Integrating,

RCt-

I)t(i

ln0

RCt-0eI)t(i

RCt-0eRI)t(Ri)t(v

or RCt-0eV)t(v

Chapter 7, Solution 2.

CR th where is the Thevenin equivalent at the capacitor terminals. thR

601280||120R th -3105.060 ms30

Chapter 7, Solution 3.

(a) ms 10102105,510//10 63 xxxCRkR ThTh (b) 6s3.020,208)255//(20 xCRR ThTh

Chapter 7, Solution 4.

eqeqCR

where 21

21eq CC

CCC ,

21

21eq RR

RRR

)CC)(RR(CCRR

2121

2121

Page 278: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 5.

4)-(t-e)4(v)t(v where 24)4(v , 2)1.0)(20(RC

24)-(t-e24)t(v 26-e24)10(v V195.1

Chapter 7, Solution 6.

Ve4)t(v25210x2x10x40RC,ev)t(v

V4)24(210

2)0(vv

t5.12

36/to

o

Chapter 7, Solution 7.

t-e)0(v)t(v , CR th

where is the Thevenin resistance across the capacitor. To determine , we insert a 1-V voltage source in place of the capacitor as shown below.

thR thR

8 i2 i

i1

10 0.5 V

+

v = 1

+

+

1 V

1.0101

i1 , 161

85.01

i2

8013

161

1.0iii 21

1380

i1

R th

138

1.01380

CR th

)t(v V20 813t-e

Page 279: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 8.

(a) 41

RC

dtdv

Ci-

Ce-4))(10(Ce0.2- -4t-4t mF5

C41

R 50

(b) 41

RC s25.0

(c) )100)(105(21

CV21

)0(w 3-20C mJ250

(d) 02t-20

20R e1CV

21

CV21

21

w

21

ee15.0 00 8t-8t-

or 2e 08t

)2(ln81

t 0 ms6.86

Chapter 7, Solution 9.

t-e)0(v)t(v , CR eq

82423||68||82R eq

2)8)(25.0(CR eq

)t(v Ve20 2t-

Page 280: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 10.

10

10 mF +

v

i

15 io

iT

4

A215

)3)(10(ii10i15 oo

i.e. if i , then A3)0( A2)0(io A5)0(i)0(i)0(i oT

V502030)0(i4)0(i10)0(v T across the capacitor terminals.

106415||104R th 1.0)1010)(10(CR -3

th -10tt- e50e)0(v)t(v

)e500-)(1010(dtdv

Ci 10t-3-C

Ci Ae5- -10t By applying the current division principle,

CC i-0.6)i-(1510

15)t(i Ae3 -10t

Chapter 7, Solution 11.

Applying KCL to the RL circuit,

0Rv

dtvL1

Differentiating both sides,

0vLR

dtdv

0dtdv

R1

Lv

LRt-eAv

Page 281: Solution Manual for Fundamentals of Electric Circuits

If the initial current is , then 0I

ARI)0(v 0

t-0 eRIv ,

RL

t

-dt)t(v

L1

i

t-

t-0 eL

RI-i

t-0 eRI-i

t-0 eI)t(i

Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).

3

i(0-)

+

12 V 2 H 4

(a) (b)

A43

12)0(i

Since the current through an inductor cannot change abruptly, A4)0(i)0(i)0(i

When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).

5.042

RL

Hence, t-e)0(i)t(i Ae4 -2t

Page 282: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 13.

thRL

where is the Thevenin resistance at the terminals of the inductor. thR

37162120||8030||70R th

37102 -3

s08.81

Chapter 7, Solution 14

Converting the wye-subnetwork to delta gives 16 R2 80mH R1 R3 30

17010

1700,3450

1700,8520/170020

105050202010321 RR

xxxR

30//170 = (30x170)/200 = 25.5 , 34//16=(34x16)/50 =10.88

sx

RLx

RTh

Th m 14.3476.251080,476.25

38.12138.3685)88.105.25//(85

3

Page 283: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 15

(a) sRL

RTh

Th 25.020/5,2040//1012

(b) ms 5.040/)1020(,408160//40 3xRL

RTh

Th

Chapter 7, Solution 16.

eq

eq

RL

(a) LLeq and 31

31312

31

312eq RR

RR)RR(RRR

RRRR

31312

31

RR)RR(R)RR(L

(b) where 21

21eq LL

LLL and

21

21213

21

213eq RR

RR)RR(RRR

RRRR

)RR)RR(R()LL()RR(LL

2121321

2121

Chapter 7, Solution 17.

t-e)0(i)t(i , 161

441

RL

eq

-16te2)t(i

16t-16t-

o e2)16-)(41(e6dtdi

Li3)t(v

)t(vo Ve2- -16t

Page 284: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 18. If , the circuit can be redrawn as shown below. 0)t(v

+

vo(t)

i(t)

Req0.4 H

56

3||2R eq , 31

65

52

RL

-3tt- ee)0(i)t(i

3t-o e-3)(

52-

dtdi

-L)t(v Ve2.1 -3t

Chapter 7, Solution 19.

i1 i2

i2 i1

i/2 10 40

+

1 V i

To find we replace the inductor by a 1-V voltage source as shown above. thR

0i401i10 21 But 2iii 2 and 1iii.e. i2i2i 21

301

i0i201i10

30i1

R th

s2.0306

RL

th

)t(i Ae2 -5t

Page 285: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 20.

(a). L50R501

RL

dtdi

Lv-

Le-50))(30(Le150- -50t-50t H1.0 L50R 5

(b). 501

RL

ms20

(c). 22 )30)(1.0(21

)0(iL21

w J45

(d). Let p be the fraction

02t-00 e1IL

21

pIL21

3296.0e1e1p -0.450(2)(10)- i.e. p %33

Chapter 7, Solution 21.

The circuit can be replaced by its Thevenin equivalent shown below.

Rth

+

Vth 2 H

V40)60(4080

80Vth

R3

80R80||40R th

R38040

RV

)(i)0(iIth

th

1380R

40)2(

21

IL21

w2

2

340

R1380R

40

R 33.13

Page 286: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 22.

t-e)0(i)t(i , eqR

L

5120||5R eq , 52

)t(i Ae10 -2.5t Using current division, the current through the 20 ohm resistor is

2.5t-o e-2

5i-

-i)(205

5i

oi20)t(v Ve04- -2.5t

Chapter 7, Solution 23.

Since the 2 resistor, 1/3 H inductor, and the (3+1) resistor are in parallel, they always have the same voltage.

-1.5)0(i5.113

222

i-

The Thevenin resistance at the inductor�’s terminals is thR

34

)13(||2R th , 41

3431

RL

th

0t,e-1.5e)0(i)t(i -4tt-

4t-oL e/3)-1.5(-4)(1

dtdi

Lvv

ov 0t,Ve2 -4t

Lx v13

1v 0t,Ve5.0 -4t

Chapter 7, Solution 24.

(a) )t(v u(t)5-

(b) )5t(u)3t(u10)3t(u)t(u-10)t(i

= )5t(u10)3t(u20)t(u10-

Page 287: Solution Manual for Fundamentals of Electric Circuits

(c) )3t(u)2t(u)2t(u)1t(u)1t()t(x )4t(u)3t(u)t4(

= )4t(u)4t()3t(u)3t()2t(u)2t()1t(u)1t( = )4t(r)3t(r)2t(r)1t(r

(d) )1t(u)t(u5)t-(u2)t(y = )1t(u5)t(u5)t-(u2

Chapter 7, Solution 25.

v(t) = [u(t) + r(t �– 1) �– r(t �– 2) �– 2u(t �– 2)] V Chapter 7, Solution 26.

(a) )t(u)1t(u)t(u)1t(u)t(v1

)t(v1 )1t(u)t(u2)1t(u

(b) )4t(u)2t(u)t4()t(v2 )4t(u)4t()2t(u)4t(-)t(v2

)t(v2 4)r(t2)r(t2)u(t2

(c) 6)u(t4)u(t44)u(t2)u(t2)t(v3 )t(v3 6)u(t44)u(t22)u(t2

(d) )2t(ut1)u(t-t)2t(u)1t(u-t)t(v4

)2t(u)22t()1t(u)11t-()t(v4 )t(v4 2)u(t22)r(t1)u(t1)r(t-

Chapter 7, Solution 27.

v(t) is sketched below.

1

2

4 3 2 0 1 t -1

v(t)

Page 288: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 28. i(t) is sketched below.

1

4 3 2 0 1 t

-1

i(t)

Chapter 7, Solution 29

x(t) (a) 3.679 0 1 t (b) y(t) 27.18

0 t

Page 289: Solution Manual for Fundamentals of Electric Circuits

(c) )1(6536.0)1(4cos)1(4cos)( tttttz , which is sketched below. z(t) 0 1 t -0.653 ( )t Chapter 7, Solution 30.

(a) 1t210

02 t4dt)1t(t4 4

(b) cos)t2cos(dt)5.0t()t2cos( 5.0t-1-

Chapter 7, Solution 31.

(a) 16-2t

4t--

4t- eedt)2t(e 22 -910112

(b) 115)t2cos(e5dt)t(t2cos)t(e)t(5 0tt-

-t- 7

Chapter 7, Solution 32.

(a) 11)(111

tdduttt

(b) 5.42

)1(0)1( 41

4

1

21

0

4

0

ttdttdtdttr

(c ) 16)6()2()6( 22

5

1

2ttdttt

Page 290: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 33.

)0(idt)t(vL1

)t(it

0

0dt)2t(201010

10)t(i

t

03-

-3

)t(i A)2t(u2

Chapter 7, Solution 34.

(a) )1t()1t(01)1t()1t()1t(u

)1t(u)1t()1t(u )1t(udtd

(b) )6t(u)2t(01)6t(u)2t()6t(r

)2t(u)6t(u)2t(u )6t(rdtd

(c) )3t(5366.0)3t(u t4cos4)3t(3x4sin)3t(u t4cos4

)3t(t4sin)3t(u t4cos4)3t(u t4sindtd

Chapter 7, Solution 35.

(a) 35t-eA)t(v , -2A)0(v )t(v Ve2- 35t-

(b) 32teA)t(v , 5A)0(v )t(v Ve5 32t

Page 291: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 36.

(a) 0t,eBA)t(v -t

1A , B10)0(v or B -1 )t(v 0t,Ve1 -t

(b) 0t,eBA)t(v 2t

-3A , B-3-6)0(v or -3B )t(v 0t,Ve13- 2t

Chapter 7, Solution 37. Let v = vh + vp, vp =10.

4/041 t

hh Aevvhv

tAev 25.010 8102)0( AAv tev 25.0810

(a) s4 (b) 10)(v V (c ) te 25.0810v

Chapter 7, Solution 38

Let i = ip +ih

)(03 3 tuAeiii thhh

Let 32

)(2)(3,0),( ktutkuitku ppi

Page 292: Solution Manual for Fundamentals of Electric Circuits

)(32 tui p

)()32

( 3 tuAei t

If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus

)()1(32 3 tuei t

Chapter 7, Solution 39.

(a) Before t = 0,

)20(14

1)t(v V4

After t = 0, t-e)(v)0(v)(v)t(v

8)2)(4(RC , 4)0(v , 20)(v 8t-e)208(20)t(v

)t(v Ve1220 8t-

(b) Before t = 0, 21 vvv , where is due to the 12-V source and is due to the 2-A source.

1v 2v

V12v1 To get v , transform the current source as shown in Fig. (a). 2

V-8v2 Thus,

812v V4 After t = 0, the circuit becomes that shown in Fig. (b).

2 F 2 F 4

12 V

+

+ v2

8 V

+

3 3

(a) (b)

Page 293: Solution Manual for Fundamentals of Electric Circuits

t-e)(v)0(v)(v)t(v 12)(v , 4)0(v , 6)3)(2(RC

6t-e)124(12)t(v )t(v Ve812 6t-

Chapter 7, Solution 40.

(a) Before t = 0, v V12 .

After t = 0, t-e)(v)0(v)(v)t(v 4)(v , 12)0(v , 6)3)(2(RC

6t-e)412(4)t(v )t(v Ve84 6t-

(b) Before t = 0, v V12 .

After t = 0, t-e)(v)0(v)(v)t(v After transforming the current source, the circuit is shown below.

t = 0

4

2

+

12 V 5 F

12)0(v , 12)(v , 10)5)(2(RC

v V12 Chapter 7, Solution 41.

0)0(v , 10)12(1630

)(v

536

)30)(6()1)(30||6(CR eq

t-e)(v)0(v)(v)t(v

5t-e)100(10)t(v

)t(v V)e1(10 -0.2t

Page 294: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 42.

(a) t-

oooo e)(v)0(v)(v)t(v

0)0(vo , 8)12(24

4)(vo

eqeqCR , 34

4||2R eq

4)3(34

4t-

o e88)t(v )t(vo V)e1(8 -0.25t

(b) For this case, 0)(vo so that

t-oo e)0(v)t(v

8)12(24

4)0(vo , 12)3)(4(RC

)t(vo Ve8 12t- Chapter 7, Solution 43. Before t = 0, the circuit has reached steady state so that the capacitor acts like an open circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage source.

40v

2i5.0 o , 80v

i o

Hence, 645

320v

40v

280v

21

ooo

80v

i o A8.0

After t = 0, the circuit is as shown in Fig. (b).

t-CC e)0(v)t(v , CR th

To find , we replace the capacitor with a 1-V voltage source as shown in Fig. (c). thR

0.5i vC

+

0.5i

i

1 V 80

(c)

Page 295: Solution Manual for Fundamentals of Electric Circuits

801

80v

i C , 80

5.0i5.0io

1605.0

80i1

Ro

th , 480CR th

V64)0(vC 480t-

C e64)t(v

480t-CC e64

4801

-3dt

dv-Ci-i5.0

)t(i Ae8.0 480t- Chapter 7, Solution 44.

23||6R eq , 4RC t-e)(v)0(v)(v)t(v

Using voltage division,

V10)30(63

3)0(v , V4)12(

633

)(v

Thus, 4t-4t- e64e)410(4)t(v

4t-e41-

)6)(2(dtdv

C)t(i Ae3- -0.25t

Chapter 7, Solution 45.

For t < 0, 0)0(v0)t(u5vs

For t > 0, , 5vs 45

)5(124

4)(v

1012||47R eq , 5)21)(10(CR eq

t-e)(v)0(v)(v)t(v

)t(v V)e1(25.1 5t-

5t-e

51-

45-

21

dtdv

C)t(i

)t(i Ae125.0 5t-

Page 296: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 46.

30566)(,0)0(,225.0)62( xivvsxCR sTh

)1(30)]()0([)()( 2// tt eevvvtv V Chapter 7, Solution 47.

For t < 0, , 0)t(u 0)1t(u , 0)0(v For 0 < t < 1, 1)1.0)(82(RC

0)0(v , 24)3)(8()(v t-e)(v)0(v)(v)t(v

t-e124)t(v For t > 1, 17.15e124)1(v 1-

30)(v024-)v(6- 1)--(te)3017.15(30)t(v

1)--(te83.1430)t(v Thus,

)t(v1t,Ve83.1430

1t0,Ve124-1)(t-

t-

Chapter 7, Solution 48.

For t < 0, , 1-t)(u V10)0(v For t > 0, , 0-t)(u 0)(v

301020R th , 3)1.0)(30(CR th t-e)(v)0(v)(v)t(v

)t(v Ve10 3t-

3t-e1031-

)1.0(dtdv

C)t(i

)t(i Ae31- 3t-

Page 297: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 49.

For 0 < t < 1, , 0)0(v 8)4)(2()(v

1064R eq , 5)5.0)(10(CR eq t-e)(v)0(v)(v)t(v

Ve18)t(v 5t- For t > 1, 45.1e18)1(v -0.2 , 0)(v

)1t(-e)(v)1(v)(v)t(v Ve45.1)t(v 5)1t(-

Thus,

)t(v1t,Ve45.1

1t0,Ve185)1t(-

5t-

Chapter 7, Solution 50.

For the capacitor voltage, t-e)(v)0(v)(v)t(v

0)0(v For t < 0, we transform the current source to a voltage source as shown in Fig. (a).

1 k1 k

+

v

+

30 V 2 k

(a)

V15)30(112

2)(v

k12||)11(R th

41

1041

10CR 3-3th

0t,e115)t(v -4t

Page 298: Solution Manual for Fundamentals of Electric Circuits

We now obtain i from v(t). Consider Fig. (b). x

ix

1/4 mF

v 1 k

1 k

iT

30 mA 2 k

(b)

Tx imA30i

But dtdv

CRv

i3

T

Ae-15)(-4)(1041

mAe15.7)t(i 4t-3-4t-T

mAe15.7)t(i -4tT

Thus,

mAe5.75.730)t(i -4tx

)t(i x 0t,mAe35.7 -4t Chapter 7, Solution 51.

Consider the circuit below.

fter the switch is closed, applying KVL gives

R-di

L

R

i

t = 0

+

+

v

VS

A

dtLRiVS

di

RV

iR-dtdi

L S or

dtLRVi S

tegrating both sides, In

Page 299: Solution Manual for Fundamentals of Electric Circuits

tLR-

Riln I0

V )t(iS

t-RVIRVi

lnS0

S

t-

S0

S eRVIRVi

or

t-S0

S eRV

IRV

)t(i

which is the same as Eq. (7.60).

hapter 7, Solution 52.

C

A21020

)0(i , A2)(i t-e)(i)0(i)(i)t(i

)t A2 (i

hapter 7, Solution 53.

(a) Before t = 0,

C

2325

i A5

After t = 0, t-e)0(i)t(i L

224

R, 5)0(i

)t(i Ae5 2t-

(b) Before t = 0, the inductor acts as a short circuit so that the 2 and 4 resistors are short-circuited.

)t(i A6 After t = 0, we have an RL circuit.

e)0(i)t(i , 23

R t- L

)t(i Ae6 3t2-

Page 300: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 54. (a) Before t = 0, i is obtained by current division or

)2(44

4)t(i A1

After t = 0, t-e)(i)0(i)(i)t(i

eqRL

, 712||44R eq

21

75.3

1)0(i , 76

)2(34

3)2(

12||4412||4

)(i

t2-e76

176

)t(i

)t(i Ae671 2t-

(b) Before t = 0, 32

10)t(i A2

After t = 0, 5.42||63R eq

94

5.42

RL

eq

2)0(i To find , consider the circuit below, at t = when the inductor becomes a short circuit,

)(i

v

2

24 V

6

+

+

i

2 H 10 V

3

9v3v

6v24

2v10

A33v

)(i 4t9-e)32(3)t(i

)t(i Ae3 4t9-

Page 301: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a).

+

v 4io io

+

io 0.5 H

+

0.5 H

+

v +

i 3 8

2 2 24 V 20 V

(a) (b)

24i0i424i3 ooo

oi4)t(v V96 A482v

i

For t > 0, consider the circuit in Fig. (b).

t-e)(i)0(i)(i)t(i

48)0(i , A228

20)(i

1082R th , 201

105.0

RL

th

-20t-20t e462e)248(2)t(i )t(i2)t(v Ve924 -20t

Chapter 7, Solution 56.

105||206R eq , 05.0RL

t-e)(i)0(i)(i)t(i i(0) is found by applying nodal analysis to the following circuit.

Page 302: Solution Manual for Fundamentals of Electric Circuits

0.5 H

+

12

5

20

6

+

v

vx i

20 V

2 A

12v6

v20v

12v

5v20

2 xxxxx

A26

v)0(i x

45||20

6.1)4(64

4)(i

-20t0.05t- e4.06.1e)6.12(6.1)t(i

Since ,

20t-e-20)()4.0(21

dtdi

L)t(v

)t(v Ve4- -20t Chapter 7, Solution 57.

At , the circuit has reached steady state so that the inductors act like short circuits.

0t

+

6 i

5

i1 i2

20 30 V

31030

20||5630

i , 4.2)3(2520

i1 , 6.0i2

A4.2)0(i1 , A6.0)0(i2

Page 303: Solution Manual for Fundamentals of Electric Circuits

For t > 0, the switch is closed so that the energies in L and flow through the closed switch and become dissipated in the 5 and 20 resistors.

1 2L

1t-11 e)0(i)t(i ,

21

55.2

RL

1

11

)t(i1 Ae4.2 -2t

2t-22 e)0(i)t(i ,

51

204

RL

2

22

)t(i2 Ae6.0 -5t

Chapter 7, Solution 58.

For t < 0, 0)t(vo

For t > 0, , 10)0(i 531

20)(i

431R th , 161

441

RL

th

t-e)(i)0(i)(i)t(i )t(i Ae15 -16t

16t-16t-

o e-16)(5)(41

e115dtdi

Li3)t(v

)t(vo Ve515 -16t Chapter 7, Solution 59.

Let I be the current through the inductor. For t < 0, , 0vs 0)0(i

For t > 0, 63||64Req , 25.065.1

RL

eq

1)3(42

2)(i

t-e)(i)0(i)(i)t(i -4te1)t(i

)-4)(-e)(5.1(dtdi

L)t(v 4t-o

)t(vo Ve6 -4t

Page 304: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 60.

Let I be the inductor current. For t < 0, 0)0(i0)t(u

For t > 0, 420||5Req , 248

RL

eq

4)(i t-e)(i)0(i)(i)t(i

2t-e14)t(i

2t-e21-

)4-)(8(dtdi

L)t(v

)t(v Ve16 -0.5t Chapter 7, Solution 61.

The current source is transformed as shown below.

4

20u(-t) + 40u(t)

+

0.5 H

81

421

RL

, 5)0(i , 10)(i t-e)(i)0(i)(i)t(i

)t(i Ae510 -8t

8t-e)8-)(5-(21

dtdi

L)t(v

)t(v Ve20 -8t Chapter 7, Solution 62.

16||3

2RL

eq

For 0 < t < 1, so that 0)1t(u

Page 305: Solution Manual for Fundamentals of Electric Circuits

0)0(i , 61

)(i

t-e161

)t(i

For t > 1, 1054.0e161

)1(i 1-

21

61

31

)(i 1)--(te)5.01054.0(5.0)t(i

1)--(te3946.05.0)t(i Thus,

)t(i1tAe3946.05.0

1t0Ae161

-1)(t-

t-

Chapter 7, Solution 63.

For t < 0, , 1)t-(u 25

10)0(i

For t > 0, , 0-t)(u 0)(i

420||5R th , 81

45.0

RL

th

t-e)(i)0(i)(i)t(i )t(i Ae2 -8t

8t-e)2)(8-(

21

dtdi

L)t(v

)t(v Ve8- -8t Chapter 7, Solution 64.

Let i be the inductor current. For t < 0, the inductor acts like a short circuit and the 3 resistor is short-circuited so that the equivalent circuit is shown in Fig. (a).

v 6

(b)

i

3

2

+

10

(a)

i

3

6

+

io

10

Page 306: Solution Manual for Fundamentals of Electric Circuits

A667.16

10)0(ii

For t > 0, 46||32R th , 144

RL

th

To find , consider the circuit in Fig. (b). )(i

610

v2v

3v

6v10

65

2v

)(ii t-e)(i)0(i)(i)t(i

Ae165

e65

610

65

)t(i t-t-

ov is the voltage across the 4 H inductor and the 2 resistor

t-t-t-o e

610

610

e-1)(65

)4(e6

106

10dtdi

Li2)t(v

)t(vo Ve1667.1 -t Chapter 7, Solution 65.

Since )1t(u)t(u10vs , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in the figure below.

For 0 < t < 1, , 0)0(i 25

10)(i

vs

-10

10

1

t

420||5R th , 21

42

RL

th

t-e)(i)0(i)(i)t(i

Page 307: Solution Manual for Fundamentals of Electric Circuits

Ae12)t(i -2t 729.1e12)1(i -2

For t > 1, since 0v0)(i s

)1t(-e)1(i)t(i Ae729.1)t(i )1t(-2

Thus,

)t(i1tAe729.1

1t0Ae12)1t(2-

2t-

Chapter 7, Solution 66.

Following Practice Problem 7.14, t-

T eV)t(v

-4)0(vVT , 501

)102)(1010(CR 6-3f

-50te-4)t(v 0t,e4)t(v-)t(v -50t

o

50t-3

o

oo e

10104

R)t(v

)t(i 0t,mAe4.0 -50t

Chapter 7, Solution 67.

The op amp is a voltage follower so that vvo as shown below.

R

vo

vo+

+

vo C

v1 R

R

Page 308: Solution Manual for Fundamentals of Electric Circuits

At node 1,

o1o111o v

32

vR

vvR

0vR

vv

At the noninverting terminal,

0R

vvdt

dvC 1oo

ooo1oo v

31v

32vvv

dtdv

RC

RC3v

dtdv oo

3RCt-To eV)t(v

V5)0(vV oT , 100

3)101)(1010)(3(RC3 6-3

)t(vo Ve5 3100t-

Chapter 7, Solution 68. This is a very interesting problem and has both an important ideal solution as well as an important practical solution. Let us look at the ideal solution first. Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero. As soon as the switch closes, the output tries to go to a voltage such that the input to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary to reach this condition. An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So vo will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be. Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it. This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts.

vC(t) = Vop amp max(1 �– e-t/(RoutC)) volts, for all values of vC less than 8 V,

= 8 V when t is large enough so that the 8 V is reached. Chapter 7, Solution 69.

Let v be the capacitor voltage. x

For t < 0, 0)0(vx

Page 309: Solution Manual for Fundamentals of Electric Circuits

For t > 0, the 20 k and 100 k resistors are in series since no current enters the op amp terminals. As t , the capacitor acts like an open circuit so that

1348

)4(1010020

10020)(vx

k12010020R th , 3000)1025)(10120(CR 3-3th

t-xxxx e)(v)0(v)(v)t(v

3000t-x e1

1348

)t(v

)t(v120100

)t(v xo Ve11340 3000t-

Chapter 7, Solution 70.

Let v = capacitor voltage. For t < 0, the switch is open and 0)0(v . For t > 0, the switch is closed and the circuit becomes as shown below.

2

1

C R

+ v +

vo

+

vS

s21 vvv (1)

dtdv

CR

v0 s (2)

where (3) vvvvvv soos

From (1),

0RCv

dtdv s

RCvt-

)0(vdtvRC

1-v s

s

Since v is constant,

Page 310: Solution Manual for Fundamentals of Electric Circuits

1.0)105)(1020(RC -63

mVt-200mV0.1

t20-v

From (3),

t20020vvv so

ov mV)t101(20 Chapter 7, Solution 71.

Let v = voltage across the capacitor. Let v = voltage across the 8 k resistor. o

For t < 2, so that 0v 0)2(v . For t > 2, we have the circuit shown below.

+

vo

10 k

+

20 k

+

v 100 mF

10 k

+ io

4 V 8 k

Since no current enters the op amp, the input circuit forms an RC circuit.

1000)10100)(1010(RC 3-3 )2t(-e)(v)2(v)(v)t(v

1000)2t(-e14)t(v As an inverter,

1e2vk20k10-

v 1000)2t(-o

8v

i oo A1e25.0 1000)2t(-

Page 311: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 72.

The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below.

C

+ v

+

io

3u(t) R

Hence,

t-e)(v)0(v)(v)t(v V2-)0(v , V3)(v , 1.0)1010)(1010(RC -63

-10t-10t e53e3)--2(3)t(v

10t-6-o e)10-)(5-)(1010(

dtdv

Ci

oi 0t,mAe5.0 -10t Chapter 7, Solution 73.

Consider the circuit below.

Rf

+

vo

+ v

+

R1

+

C v2v1 v3

v1

At node 2,

dtdv

CR

vv

1

21 (1)

At node 3,

Page 312: Solution Manual for Fundamentals of Electric Circuits

f

o3

Rvv

dtdv

C (2)

But and 0v3 232 vvvv . Hence, (1) becomes

dtdv

CR

vv

1

1

dtdv

CRvv 11

or CR

vCR

vdtdv

1

1

1

which is similar to Eq. (7.42). Hence,

0tevvv0tv

)t(v t-1T1

T

where and 1)0(vvT 4v1 2.0)1020)(1010(CR 6-3

1

0te340t1

)t(v 5t-

From (2),

)e15)(1020)(1020(dtdv

CR-v 5t-6-3fo

0t,e-6v -5to

ov V)t(ue6- -5t Chapter 7, Solution 74. Let v = capacitor voltage.

Rf

+

vo

+ v

+

R1

+

C v2v1 v3

v1

Page 313: Solution Manual for Fundamentals of Electric Circuits

For t < 0, 0)0(vFor t > 0, . Consider the circuit below. A10is

Rv

dtdv

Cis (1) t-e)(v)0(v)(v)t(v (2)

It is evident from the circuit that

1.0)1050)(102(RC 36

is

+

vo

C Rf

+R

is

At steady state, the capacitor acts like an open circuit so that i passes through R. Hence,

s

V5.0)1050)(1010(Ri)(v 36s

Then,

Ve15.0)t(v -10t (3)

But fsof

os Ri-v

Rv0

i (4)

Combining (1), (3), and (4), we obtain

dtdv

CRvRR-

v ff

o

dtdv

)102)(1010(v51-

v 6-3o

-10t-2-10to e10-)5.0)(102(e0.1-0.1v

1.0e2.0v -10to

ov V1e21.0 -10t

Page 314: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 75. Let v = voltage at the noninverting terminal. 1

Let 2v = voltage at the inverting terminal.

For t > 0, 4vvv s21

o1

s iR

v0, k20R1

Ri-v oo (1)

Also, dtdv

CRv

i2

o , k10R 2 , F2C

i.e. dtdv

CRv

Rv-

21

s (2)

This is a step response.

t-e)(v)0(v)(v)t(v , 1)0(v

where 501

)102)(1010(CR 6-32

At steady state, the capacitor acts like an open circuit so that i passes through

. Hence, as o

2R t

2o

1

s

R)(v

iRv-

i.e. -2)4(2010-

vRR-

)(v s1

2

-50te2)(1-2)t(v

-50te3-2)t(v But os vvvor t50-

so e324vvv

ov Ve36 t50-

mA-0.220k

4-Rv-

i1

so

or dtdv

CRv

i2

o mA0.2-

Page 315: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 76. The schematic is shown below. For the pulse, we use IPWL and enter the corresponding values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and simulating the circuit, we select Trace/Add and display �–V(C1:2). The plot of V(t) is shown below.

Chapter 7, Solution 77. The schematic is shown below. We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation. The plot of V is shown below. Note from the plot that V(0) = 12 V and V( ) = -24 V which are correct.

Page 316: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 78. (a) When the switch is in position (a), the schematic is shown below. We insert

IPROBE to display i. After simulation, we obtain,

i(0) = 7.714 A from the display of IPROBE.

Page 317: Solution Manual for Fundamentals of Electric Circuits

(b) When the switch is in position (b), the schematic is as shown below. For inductor I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i( ) = 12A, which is correct.

Page 318: Solution Manual for Fundamentals of Electric Circuits
Page 319: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 79. When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2,

3/80,3/84//)53(A, 5.035

4)(LR

Ri ThTho

A 5.45.0)]()0([)()( 80/3/ tt

oooo eeiiiti Chapter 7, Solution 80.

(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are short-

circuited so that 0)0()0()0( 21 ovii but the current through the 4-H inductor is iL(0) =30/10 = 3A. (b) When the switch is in position B,

5.04/2,26//3LR

R ThTh

A 330)]()0([)()( 25.0// ttt

LLLL eeeiiiti

(c) A 0)(93)(,A 2

51030)( 21 Liii

V 0)()( oL

o vdtdi

Ltv

Chapter 7, Solution 81. The schematic is shown below. We use VPWL for the pulse and specify the attributes as shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S. By inserting a current marker at one termial of LI, we automatically obtain the plot of i after simulation as shown below.

Page 320: Solution Manual for Fundamentals of Electric Circuits

hapter 7, Solution 82.

C

6-

-3

10100103

CRRC 30

Page 321: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 83.

sxxxRCvv 51010151034,0)0(,120)( 66

)1(1206.85)]()0([)()( 510// tt eevvvtv Solving for t gives st 16.637488.3ln510 speed = 4000m/637.16s = 6.278m/s

Chapter 7, Solution 84.

Let Io be the final value of the current. Then 50/18/16.0/),1()( / LReIti t

o

. ms 33.184.0

1ln

501

)1(6.0 50 teII too

Chapter 7, Solution 85.

(a) s24)106)(104(RC -66

Since t-e)(v)0(v)(v)t(v

1t-1 e)(v)0(v)(v)t(v (1)

2t-2 e)(v)0(v)(v)t(v (2)

Dividing (1) by (2),

)tt(

2

1 12e)(v)t(v)(v)t(v

)(v)t(v)(v)t(v

lnttt2

1120

)2(ln241203012075

ln24t0 s63.16

(b) Since , the light flashes repeatedly every tt0

RC s24

Page 322: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 86.

t-e)(v)0(v)(v)t(v 12)(v , 0)0(v

t-e112)t(v 0t-

0 e1128)t(v

31

ee1128

00 t-t-

)3(lnt0 For , k100R

s2.0)102)(10100(RC -63 s2197.0)3(ln2.0t0

For , M1R

s2)102)(101(RC -66 s197.2)3(ln2t0

Thus,

s197.2ts2197.0 0

Chapter 7, Solution 87.

Let i be the inductor current.

For t < 0, A2.1100120

)0(i

For t > 0, we have an RL circuit

1.0400100

50RL

, 0)(i

t-e)(i)0(i)(i)t(i -10te2.1)t(i

At t = 100 ms = 0.1 s,

-1e2.1)1.0(i A441.0 which is the same as the current through the resistor.

Page 323: Solution Manual for Fundamentals of Electric Circuits

Chapter 7, Solution 88.

(a) s60)10200)(10300(RC -123

As a differentiator, ms6.0s60010T

i.e. minT ms6.0

(b) s60RCAs an integrator,

s61.0T i.e. maxT s6

Chapter 7, Solution 89.

Since s1T1.0

s1RL

)101)(10200(10RL -63-6

mH200L

Chapter 7, Solution 90.

We determine the Thevenin equivalent circuit for the capacitor C . s

ips

sth v

RRR

v , psth R||RR

Rth

+

Vth Cs

The Thevenin equivalent is an RC circuit. Since

ps

sith RR

R101

v101

v

96

R91

R ps M32

Also,

Page 324: Solution Manual for Fundamentals of Electric Circuits

s15CR sth

where M6.0326)32(6

R||RR spth

6

6-

ths 100.6

1015R

C pF25

Chapter 7, Solution 91.

mA2405012

)0(io , 0)(i

t-e)(i)0(i)(i)t(i t-e240)t(i

R2

RL

0t-0 e24010)t(i

)24(lnt24e 0t0

R2

573.1)24(ln

5)24(ln

t 0

573.12

R 271.1

Chapter 7, Solution 92.

DR6-

R3-9-

ttt10510-

tt0102

10

104dtdv

Ci

s5ms2tms2mA8-ms2t0A20

)t(i

which is sketched below.

20 A

2 ms

5 s

-8 mA

t

i(t)

(not to scale)

Page 325: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a).

+

vL

6

10 H

+

v

(a)

6

+

6

VS

10 F

(b)

i(0-) = 12/6 = 2A, v(0-) = 12V

At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V (b) For t > 0, we have the equivalent circuit shown in Figure (b).

vL = Ldi/dt or di/dt = vL/L Applying KVL at t = 0+, we obtain,

vL(0+) �– v(0+) + 10i(0+) = 0

vL(0+) �– 12 + 20 = 0, or vL(0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, iC = Cdv/dt, or dv/dt = iC/C

iC(0+) = -i(0+) = -2

dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state.

i( ) = 0 A, v( ) = 0 V

Page 326: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 2. (a) At t = 0-, the equivalent circuit is shown in Figure (a). 25 k 20 k

iR +

+

v

iL

60 k

80V

(a) 25 k 20 k

iR +

iL 80V

(b)

60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA. By the current division principle,

iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA

vC(0-) = 0 At t = 0+,

vC(0+) = vC(0-) = 0

iL(0+) = iL(0-) = 1.5 mA

80 = iR(0+)(25 + 20) + vC(0-)

iR(0+) = 80/45k = 1.778 mA But, iR = iC + iL

1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA

Page 327: Solution Manual for Fundamentals of Electric Circuits

(b) vL(0+) = vC(0+) = 0

But, vL = LdiL/dt and diL(0+)/dt = vL(0+)/L = 0

diL(0+)/dt = 0

Again, 80 = 45iR + vC

0 = 45diR/dt + dvC/dt

But, dvC(0+)/dt = iC(0+)/C = 0.278 mohms/1 F = 278 V/s

Hence, diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45

diR(0+)/dt = -6.1778 A/s

Also, iR = iC + iL

diR(0+)/dt = diC(0+)/dt + diL(0+)/dt

-6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure

(b).

iR( ) = iL( ) = 80/45k = 1.778 mA

iC( ) = Cdv( )/dt = 0. Chapter 8, Solution 3. At t = 0-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR(0-) = 0. But, -vR(0-) + vC(0-) + 10 = 0, or vC(0-) = -10V. (a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,

the inductor current must still be equal to 0A, the capacitor has a voltage equal to �–10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, vR(0+) = 0 V.

(b) At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,

iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s. Now for the value of dvR(0+)/dt. Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s.

Page 328: Solution Manual for Fundamentals of Electric Circuits

40 40

+

10V

+

vC

10

2A

iL +

vR

+

vR

+ 10V

+

vC

10

(b) (a) (c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b).

iL( ) = 10(2)/(40 + 10) = 400 mA

vC( ) = 2[10||40] �–10 = 16 �– 10 = 6V

vR( ) = 2[10||40] = 16 V Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a).

i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. Hence, i(0+) = i(0-) = 5A v(0+) = v(0-) = 25V

3

5

i +

v

+

40V

(a)

Page 329: Solution Manual for Fundamentals of Electric Circuits

0.25 H3

iRiC

+

+ vL i

5 0.1F

4 A

40V (b) (b) iC = Cdv/dt or dv(0+)/dt = iC(0+)/C For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly,

iR = v/5 = 25/5 = 5A, i(0+) + 4 =iC(0+) + iR(0+)

5 + 4 = iC(0+) + 5 which leads to iC(0+) = 4

dv(0+)/dt = 4/0.1 = 40 V/s Chapter 8, Solution 5. (a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0).

iL(0-) = 0 and vC(0-) = 0.

For t = 0+, 4u(t) = 4. Consider the circuit below.

iL

iC + vL

1 H

+

v

4 0.25F +

vC

Ai

6

4A

Since the 4-ohm resistor is in parallel with the capacitor,

i(0+) = vC(0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6iL(0+) = 0V.

Page 330: Solution Manual for Fundamentals of Electric Circuits

(b) di(0+)/dt = d(vR(0+)/R)/dt = (1/R)dvR(0+)/dt = (1/R)dvC(0+)/dt

= (1/4)4/0.25 A/s = 4 A/s

v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0

Therefore dv(0+)/dt = 0 V/s

(c) As t approaches infinity, the circuit is in steady-state.

i( ) = 6(4)/10 = 2.4 A

v( ) = 6(4 �– 2.4) = 9.6 V Chapter 8, Solution 6. (a) Let i = the inductor current. For t < 0, u(t) = 0 so that

i(0) = 0 and v(0) = 0.

For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. vR(0+) = Ri(0+) = 0 V

Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V. (1)

(b) Since i(0+) = 0, iC(0+) = VS/RS

But, iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS) (2)

From (1), dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt (3) vR = iR or dvR/dt = Rdi/dt (4)

But, vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0 (5)

From (4) and (5), dvR(0+)/dt = 0 V/s

From (2) and (3), dvL(0+)/dt = dv(0+)/dt = Vs/(CRs) (c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit.

vR( ) = [R/(R + Rs)]Vs

vL( ) = 0 V

Page 331: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 7.

s2 + 4s + 4 = 0, thus s1,2 = 2

4x444 2

= -2, repeated roots.

v(t) = [(A + Bt)e-2t], v(0) = 1 = A

dv/dt = [Be-2t] + [-2(A + Bt)e-2t]

dv(0)/dt = -1 = B �– 2A = B �– 2 or B = 1.

Therefore, v(t) = [(1 + t)e-2t] V

Chapter 8, Solution 8.

s2 + 6s + 9 = 0, thus s1,2 = 2

3666 2

= -3, repeated roots.

i(t) = [(A + Bt)e-3t], i(0) = 0 = A

di/dt = [Be-3t] + [-3(Bt)e-3t]

di(0)/dt = 4 = B.

Therefore, i(t) = [4te-3t] A

Chapter 8, Solution 9.

s2 + 10s + 25 = 0, thus s1,2 = 2

101010 = -5, repeated roots.

i(t) = [(A + Bt)e-5t], i(0) = 10 = A

di/dt = [Be-5t] + [-5(A + Bt)e-5t]

di(0)/dt = 0 = B �– 5A = B �– 50 or B = 50.

Therefore, i(t) = [(10 + 50t)e-5t] A

Page 332: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 10.

s2 + 5s + 4 = 0, thus s1,2 = 2

16255 = -4, -1.

v(t) = (Ae-4t + Be-t), v(0) = 0 = A + B, or B = -A

dv/dt = (-4Ae-4t - Be-t)

dv(0)/dt = 10 = �– 4A �– B = �–3A or A = �–10/3 and B = 10/3.

Therefore, v(t) = (�–(10/3)e-4t + (10/3)e-t) V Chapter 8, Solution 11.

s2 + 2s + 1 = 0, thus s1,2 = 2

442 = -1, repeated roots.

v(t) = [(A + Bt)e-t], v(0) = 10 = A

dv/dt = [Be-t] + [-(A + Bt)e-t]

dv(0)/dt = 0 = B �– A = B �– 10 or B = 10.

Therefore, v(t) = [(10 + 10t)e-t] V

Chapter 8, Solution 12. (a) Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF (b) Critically damped when C = 6 mF (c) Underdamped when C < 6mF

Page 333: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 13. Let R||60 = Ro. For a series RLC circuit,

o = LC1 =

4x01.01 = 5

For critical damping, o = = Ro/(2L) = 5

or Ro = 10L = 40 = 60R/(60 + R)

which leads to R = 120 ohms

Chapter 8, Solution 14. This is a series, source-free circuit. 60||30 = 20 ohms

= R/(2L) = 20/(2x2) = 5 and o = LC1 =

04.01 = 5

o = leads to critical damping

i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}

v(0) = 6 = 2B �– 10A = 2B �– 20 or B = 13.

Therefore, i(t) = [(2 + 13t)e-5t] A

Chapter 8, Solution 15. This is a series, source-free circuit. 60||30 = 20 ohms

= R/(2L) = 20/(2x2) = 5 and o = LC1 =

04.01 = 5

o = leads to critical damping

i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}

v(0) = 6 = 2B �– 10A = 2B �– 20 or B = 13.

Therefore, i(t) = [(2 + 13t)e-5t] A

Page 334: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 16. At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit.

= R/(2L) = (40 + 60)/5 = 20 and o = LC1 =

5.2x1013

= 20

o = leads to critical damping

i(t) = [(A + Bt)e-20t], i(0) = 0 = A

di/dt = {[Be-20t] + [-20(Bt)e-20t]},

but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]

Hence, B = -9.6 or i(t) = [-9.6te-20t] A

Chapter 8, Solution 17.

.iswhich,20

412

10L2

R

10

251

411

LC1

240)600(4)VRI(L1

dt)0(di

6015x4V)0(v,0I)0(i

o

o

00

00

t268t32.3721

2121

t32.372

t68.21

2o

2

ee928.6)t(i

A928.6AtoleadsThis

240A32.37A68.2dt

)0(di,AA0)0(i

eAeA)t(i

32.37,68.23102030020s

getwe,60dt)t(iC1)t(v,Since t

0

v(t) = (60 + 64.53e-2.68t �– 4.6412e-37.32t) V

Page 335: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit.

5.02

1,2125.0

11RCxLC

o

936.125.04case dunderdampe 22d oo

Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V

)936.1sin936.1cos()sincos()( 215.0

21 tAtAetAtAetv tdd

t v(0) =0 = A1

)936.1cos936.1936.1sin936.1()936.1sin936.1cos)(5.0( 215.0

215.0 tAtAetAtAe

dtdv tt

066.2936.15.041

)40()()0(221 AAA

RCRIV

dtdv oo

Thus,

tetv t 936.1sin066.2)( 5.0 Chapter 8, Solution 19. For t < 0, the equivalent circuit is shown in Figure (a). 10 i

+

+

v

i L C

+

v

120V (a) (b)

i(0) = 120/10 = 12, v(0) = 0

Page 336: Solution Manual for Fundamentals of Electric Circuits

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = .

o = LC1 =

41 = 0.5 = d

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A

v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],

which leads to -v(0)/L = 0 = B

Hence, i(t) = 12cos0.5t A and v = 0.5

However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V

Chapter 8, Solution 20. For t < 0, the equivalent circuit is as shown below.

2

+ 12

+vC

i

v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit.

= R/(2L) = 2/(2x0.5) = 2

o = 1/ 2241x5.0/1LC

Since is less than o, we have an under-damped response.

24822od

i(t) = (Acos2t + Bsin2t)e-2t

i(0) = 6 = A

Page 337: Solution Manual for Fundamentals of Electric Circuits

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e- t

di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 �– 12] = 0

Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A Chapter 8, Solution 21. By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms.

12

+

+

v

t = 0 i

24

6 3 H

24V (1/27)F At t = 0-, i(0) = 0, v(0) = 24x24/36 = 16V For t > 0, we have a series RLC circuit. R = 30 ohms, L = 3 H, C = (1/27) F

= R/(2L) = 30/6 = 5

27/1x3/1LC/1o = 3, clearly > o (overdamped response)

s1,2 = 222o

2 355 = -9, -1

v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B (1)

i = Cdv/dt = C[-Ae-t - 9Be-9t]

i(0) = 0 = C[-A �– 9B] or A = -9B (2) From (1) and (2), B = -2 and A = 18.

Hence, v(t) = (18e-t �– 2e-9t) V

Page 338: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 22. = 20 = 1/(2RC) or RC = 1/40 (1)

22od 50 which leads to 2500 + 400 = o

2 = 1/(LC) Thus, LC 1/2900 (2) In a parallel circuit, vC = vL = vR But, iC = CdvC/dt or iC/C = dvC/dt

= -80e-20tcos50t �– 200e-20tsin50t + 200e-20tsin50t �– 500e-20tcos50t = -580e-20tcos50t

iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 F

R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms

L = 1/(2900x11.21) = 30.76 H

Chapter 8, Solution 23. Let Co = C + 0.01. For a parallel RLC circuit,

= 1/(2RCo), o = 1/ oLC

= 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF

o = 1/ 5.0x5.0 = 6.32 > (underdamped)

Co = C + 10 mF = 50 mF or 40 mF Chapter 8, Solution 24. For t < 0, u(-t) 1, namely, the switch is on.

v(0) = 0, i(0) = 25/5 = 5A For t > 0, the voltage source is off and we have a source-free parallel RLC circuit.

= 1/(2RC) = 1/(2x5x10-3) = 100

Page 339: Solution Manual for Fundamentals of Electric Circuits

o = 1/ 310x1.0/1LC = 100

o = (critically damped)

v(t) = [(A1 + A2t)e-100t]

v(0) = 0 = A1

dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000

But, dv/dt = [(A2 + (-100)A2t)e-100t]

Therefore, dv(0)/dt = -5000 = A2 �– 0 v(t) = -5000te-100t V

Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0.

(1/4)F

+

8

2

t=0, note this is a make before break

switch so the inductor current is

not interrupted.

1 H io(t)

+

vo(t)

30V

Figure 8.78 For Problem 8.25. At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit.

= 1/(2RC) = ¼

o = 1/ 241x1/1LC

Since is less than o, we have an under-damped response.

9843.1)16/1(422od

vo(t) = (A1cos dt + A2sin dt)e- t

Page 340: Solution Manual for Fundamentals of Electric Circuits

vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0.

dvo/dt = - (A1cos dt + A2sin dt)e- t + (- dA1sin dt + dA2cos dt)e- t

at t = 0, we get dvo(0)/dt = 0 = - A1 + dA2

Thus, A2 = ( / d)A1 = (1/4)(24)/1.9843 = 3.024

vo(t) = (24cos dt + 3.024sin dt)e-t/4 volts

Chapter 8, Solution 26.

s2 + 2s + 5 = 0, which leads to s1,2 = 2

2042 = -1 j4

i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2

i(0) = 2 = = 2 + A1, or A1 = 0

di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1

i(t) = 2 + sin4te-t A

Chapter 8, Solution 27.

s2 + 4s + 8 = 0 leads to s = 2j22

32164

v(t) = Vs + (A1cos2t + A2sin2t)e-2t

8Vs = 24 means that Vs = 3

v(0) = 0 = 3 + A1 leads to A1 = -3

dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t

0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3

v(t) = [3 �– 3(cos2t + sin2t)e-2t] volts

Page 341: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 28.

The characteristic equation is s2 + 6s + 8 with roots

2,42

323662,1s

Hence,

tts BeAeIti 42)(

5.1128 ss II

BAi 5.100)0( (1)

tt BeAe

dtdi 42 42

BABAdtdi 210422)0( (2)

Solving (1) and (2) leads to A=-2 and B=0.5.

tt eeti 42 5.025.1)( A Chapter 8, Solution 29.

(a) s2 + 4 = 0 which leads to s1,2 = j2 (an undamped circuit)

v(t) = Vs + Acos2t + Bsin2t

4Vs = 12 or Vs = 3

v(0) = 0 = 3 + A or A = -3

dv/dt = -2Asin2t + 2Bcos2t

dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 �– 3cos2t + sin2t) V

(b) s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4

i(t) = (Is + Ae-t + Be-4t)

4Is = 8 or Is = 2

i(0) = -1 = 2 + A + B, or A + B = -3 (1)

Page 342: Solution Manual for Fundamentals of Electric Circuits

di/dt = -Ae-t - 4Be-4t

di(0)/dt = 0 = -A �– 4B, or B = -A/4 (2)

From (1) and (2) we get A = -4 and B = 1

i(t) = (2 �– 4e-t + e-4t) A

(c) s2 + 2s + 1 = 0, s1,2 = -1, -1

v(t) = [Vs + (A + Bt)e-t], Vs = 3.

v(0) = 5 = 3 + A or A = 2

dv/dt = [-(A + Bt)e-t] + [Be-t]

dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3

v(t) = [3 + (2 + 3t)e-t] V Chapter 8, Solution 30.

22

222

1 800,500 oo ss

LR

ss2

6502130021

Hence,

mH 8.1536502

2002 xR

L

LCss oo

145.6232300 2221

F25.16)45.632(

12L

C

Page 343: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 31. For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL,

v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+) = 0 which leads to vL(0+) = 40x1 + 40 = 80

vL(0+) = 80 V, vC(0+) = 40 V 40 10 i1

0.5H

+

v 50V

+

+

vL

40 10

i +

v

50V

+

(a) (b) Chapter 8, Solution 32. For t = 0-, the equivalent circuit is shown below. 2 A

i

6

+ v

i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input.

= R/(2L) = 6/2 = 3, o = 1/ 04.0/1LC

s = 4j32593

Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]

Page 344: Solution Manual for Fundamentals of Electric Circuits

where Vf = final capacitor voltage = 50 V

v(t) = 50 + [(Acos4t + Bsin4t)e-3t]

v(0) = -12 = 50 + A which gives A = -62

i(0) = 0 = Cdv(0)/dt

dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5

v(t) = {50 + [(-62cos4t �– 46.5sin4t)e-3t]} V Chapter 8, Solution 33. We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a).

1 H

i

+

30V

+

v 4F

i

+

5

10

+

v

10 30V (a) (b)

i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V

For t > 0, we have a series RLC circuit.

= R/(2L) = 5/2 = 2.5

4/1LC/1o = 0.25, clearly > o (overdamped response)

s1,2 = 25.025.65.22o

2 = -4.95, -0.05

v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20.

v(0) = 10 = 20 + A1 + A2 (1)

Page 345: Solution Manual for Fundamentals of Electric Circuits

i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2

Hence, ½ = -4.95A1 �– 0.05A2 (2) From (1) and (2), A1 = 0, A2 = -10.

v(t) = {20 �– 10e-0.05t} V Chapter 8, Solution 34. Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit.

i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below.

+ Vx

(1/16)F

(¼) H

i This is a lossless, source-free, series RLC circuit.

= R/(2L) = 0, o = 1/ LC = 1/41

161 = 8, s = j8

Since is less than o, we have an underdamped response. Therefore,

i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1

di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we have i(t) = -10sin8t A

Page 346: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 35. At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input.

= R/(2L) = 2/2 = 1, o = 1/ LC = 1/ 5/1 = 5

s1,2 = 2j12

o2

v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12.

v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0.

But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]

0 = dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2

v(t) = {12 �– (4cos2t + 2sin2t)e-t V. Chapter 8, Solution 36. For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below.

20 V 2

0.2 F

i 10

+

+

5 H 10 +

v

15V

= R/(2L) = (2 + 5 + 1)/(2x5) = 0.8

o = 1/ LC = 1/ 2.0x5 = 1

Page 347: Solution Manual for Fundamentals of Electric Circuits

s1,2 = 6.0j8.02o

2

v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t]

Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15

i(0) = Cdv(0)/dt = 0

But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t]

0 = dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20

v(t) = {35 �– [(15cos0.6t + 20sin0.6t)e-0.8t]} V i = Cdv/dt = 0.2{[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t �– 20cos0.6t)e-0.8t]}

i(t) = [(5sin0.6t)e-0.8t] A

Chapter 8, Solution 37. For t = 0-, the equivalent circuit is shown below.

6

+

10V

+

i2

i1

+

v(0)

6

6

30V

18i2 �– 6i1 = 0 or i1 = 3i2 (1) -30 + 6(i1 �– i2) + 10 = 0 or i1 �– i2 = 10/3 (2)

From (1) and (2). i1 = 5, i2 = 5/3

i(0) = i1 = 5A

-10 �– 6i2 + v(0) = 0

Page 348: Solution Manual for Fundamentals of Electric Circuits

v(0) = 10 + 6x5/3 = 20

For t > 0, we have a series RLC circuit.

R = 6||12 = 4

o = 1/ LC = 1/ )8/1)(2/1( = 4

= R/(2L) = (4)/(2x(1/2)) = 4

= o, therefore the circuit is critically damped

v(t) = Vs +[(A + Bt)e-4t], and Vs = 10

v(0) = 20 = 10 + A, or A = 10

i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t]

i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80

i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t]

i(t) = [(5 �– 40t)e-4t] A

Chapter 8, Solution 38. At t = 0-, the equivalent circuit is as shown.

2 A

10

i

i1

5

+

v

10

Page 349: Solution Manual for Fundamentals of Electric Circuits

i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A

v(0) = 5i1(0) = 4V

For t > 0, we have a source-free series RLC circuit.

R = 5||(10 + 10) = 4 ohms

o = 1/ LC = 1/ )4/3)(3/1( = 2

= R/(2L) = (4)/(2x(3/4)) = 8/3

s1,2 = 2o

2 -4.431, -0.903

i(t) = [Ae-4.431t + Be-0.903t]

i(0) = A + B = 2 (1)

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333

Hence, -5.333 = -4.431A �– 0.903B (2)

From (1) and (2), A = 1 and B = 1.

i(t) = [e-4.431t + e-0.903t] A

Chapter 8, Solution 39. For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and 30u(t) = 0.

+

30V 20

+

+ v

20

30 0.5F 0.25H 30 60V (a) (b)

v(0) = (20/50)(60) = 24 and i(0) = 0

Page 350: Solution Manual for Fundamentals of Electric Circuits

For t > 0, the circuit is shown in Figure (b).

R = 20||30 = 12 ohms

o = 1/ LC = 1/ )4/1)(2/1( = 8

= R/(2L) = (12)/(0.5) = 24

Since > o, we have an overdamped response.

s1,2 = 2o

2 -47.833, -0.167 Thus, v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30 v(0) = 24 = 30 + A + B or -6 = A + B (1)

i(0) = Cdv(0)/dt = 0

But, dv(0)/dt = -47.833A �– 0.167B = 0 B = -286.43A (2)

From (1) and (2), A = 0.021 and B = -6.021

v(t) = 30 + [0.021e-47.833t �– 6.021e-0.167t] V Chapter 8, Solution 40. At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below.

+ v

6 +

+

12V24V

i 14 0.02 F 2 H

o = 1/ LC = 1/ 02.0x2 = 5

= R/(2L) = (6 + 14)/(2x2) = 5

Page 351: Solution Manual for Fundamentals of Electric Circuits

Since = o, we have a critically damped response.

v(t) = Vs + [(A + Bt)e-5t], Vs = 24 �– 12 = 12V

v(0) = 0 = 12 + A or A = -12

i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]}

i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90

Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]}

i(t) = {(3 �– 9t)e-5t} A

Chapter 8, Solution 41. At t = 0-, the switch is open. i(0) = 0, and

v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b).

1 H

+

20V

+

v

i 4

20 10 F

10 H

5A

(a)

5

0.04F (b)

o = 1/ LC = 1/ 25/1x1 = 5

= R/(2L) = (4)/(2x1) = 2

s1,2 = 2o

2 -2 j4.583 Thus, v(t) = Vs + [(Acos dt + Bsin dt)e-2t],

where d = 4.583 and Vs = 20

v(0) = 50/3 = 20 + A or A = -10/3

Page 352: Solution Manual for Fundamentals of Electric Circuits

i(t) = Cdv/dt = C(-2) [(Acos dt + Bsin dt)e-2t] + C d[(-Asin dt + Bcos dt)e-2t]

i(0) = 0 = -2A + dB

B = 2A/ d = -20/(3x4.583) = -1.455

i(t) = C{[(0cos dt + (-2B - dA)sin dt)]e-2t}

= (1/25){[(2.91 + 15.2767) sin dt)]e-2t}

i(t) = {0.7275sin(4.583t)e-2t} A

Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a).

i(0) = i(0) = 0, and v(0) = 4 �– 12 = -8V

+

v

6

+ 12V

1 H i 4V + +

+

v(0)

12V

1

5

0.04F

(a) (b) For t > 0, the circuit becomes that shown in Figure (b) after source transformation.

o = 1/ LC = 1/ 25/1x1 = 5

= R/(2L) = (6)/(2) = 3

s1,2 = 2o

2 -3 j4 Thus, v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12

v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

i(0) = -3A + 4B or B = 3

v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A

Page 353: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 43.

For t>0, we have a source-free series RLC circuit.

85.08222

xxLRLR

836649003022ood

mF 392.25.0836

1112 xL

CLC o

o

Chapter 8, Solution 44.

4

910

10100

11,500

121000

2 xLCxLR

o

o underdamped. Chapter 8, Solution 45.

o = 1/ LC = 1/ 5.0x1 = 2

= R/(2L) = (1)/(2x2x0.5) = 0.5

Since < o, we have an underdamped response.

s1,2 = 2o

2 -0.5 j1.323 Thus, i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4

i(0) = 1 = 4 + A or A = -3

v = vC = vL = Ldi(0)/dt = 0

di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]

di(0)/dt = 0 = 1.323B �– 0.5A or B = 0.5(-3)/1.323 = -1.134 Thus, i(t) = {4 �– [(3cos1.323t + 1.134sin1.323t)e-0.5t]} A

Page 354: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 46. For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below.

5 F8mH

+

v

i 2 k

6mA

= 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50

o = 1/ LC = 1/ 63 10x5x10x8 = 5,000

Since < o, we have an underdamped response.

s1,2 = 2o

2 -50 j5,000 Thus, i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA

i(0) = 0 = 6 + A or A = -6mA

v(0) = 0 = Ldi(0)/dt

di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]

di(0)/dt = 0 = 5,000B �– 50A or B = 0.01(-6) = -0.06mA Thus, i(t) = {6 �– [(6cos5,000t + 0.06sin5,000t)e-50t]} mA Chapter 8, Solution 47. At t = 0-, we obtain, iL(0) = 3x5/(10 + 5) = 1A

and vo(0) = 0.

For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input.

= 1/(2RC) = (1)/(2x5x0.01) = 10

o = 1/ LC = 1/ 01.0x1 = 10

Page 355: Solution Manual for Fundamentals of Electric Circuits

Since = o, we have a critically damped response.

s1,2 = -10

Thus, i(t) = Is + [(A + Bt)e-10t], Is = 3

i(0) = 1 = 3 + A or A = -2

vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]

vo(0) = 0 = B �– 10A or B = -20

Thus, vo(t) = (200te-10t) V

Chapter 8, Solution 48. For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2.

For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit.

= 1/(2RC) = (1)/(2x1x0.25) = 2

o = 1/ LC = 1/ 25.0x1 = 2

Since = o, we have a critically damped response.

s1,2 = -2

Thus, i(t) = [(A + Bt)e-2t], i(0) = -2 = A

v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t]

vo(0) = 2 = B + 4 or B = -2

Thus, i(t) = [(-2 - 2t)e-2t] A

and v(t) = [(2 + 4t)e-2t] V

Page 356: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 49. For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0.

For t > 0, we have a parallel RLC circuit with a step input.

= 1/(2RC) = (1)/(2x5x0.05) = 2

o = 1/ LC = 1/ 05.0x5 = 2

Since = o, we have a critically damped response.

s1,2 = -2

Thus, i(t) = Is + [(A + Bt)e-2t], Is = 3

i(0) = 6 = 3 + A or A = 3

v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t]

v(0)/L = 0 = di(0)/dt = B �– 2x3 or B = 6

Thus, i(t) = {3 + [(3 + 6t)e-2t]} A Chapter 8, Solution 50. For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit.

i

40 6A 3A

10 mF+

v 10

10 H

Is = 3 + 6 = 9A and R = 10||40 = 8 ohms

= 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25

o = 1/ LC = 1/ 01.0x4 = 5

Since > o, we have a overdamped response.

s1,2 = 2o

2 -10, -2.5

Page 357: Solution Manual for Fundamentals of Electric Circuits

Thus, i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9

i(0) = 3 = 9 + A + B or A + B = -6

di/dt = [-10Ae-10t] + [-2.5Be-2.5t],

v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A �– 2.5B or B = -4A

Thus, A = 2 and B = -8

Clearly, i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A Chapter 8, Solution 51. Let i = inductor current and v = capacitor voltage.

At t = 0, v(0) = 0 and i(0) = io.

For t > 0, we have a parallel, source-free LC circuit (R = ).

= 1/(2RC) = 0 and o = 1/ LC which leads to s1,2 = j o

v = Acos ot + Bsin ot, v(0) = 0 A

iC = Cdv/dt = -i

dv/dt = oBsin ot = -i/C

dv(0)/dt = oB = -io/C therefore B = io/( oC)

v(t) = -(io/( oC))sin ot V where o = LC Chapter 8, Solution 52.

RC21

300 (1)

LCood

1575.264300400400 2222 (2)

From (2),

F71.2851050)575.264(

132 xx

C

From (1),

833.5)3500(30021

21

xCR

Page 358: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 53.

C1 R2

+ v1

i2i1

+

C2

+

vo

R1

vS

i2 = C2dvo/dt (1) i1 = C1dv1/dt (2) 0 = R2i2 + R1(i2 �– i1) +vo (3) Substituting (1) and (2) into (3) we get, 0 = R2C2dvo/dt + R1(C2dvo/dt �– C1dv1/dt) (4) Applying KVL to the outer loop produces,

vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to v1 = vs �– vo �– R2C2dvo/dt (5) Substituting (5) into (4) leads to,

0 = R1C2dvo/dt + R1C2dvo/dt �– R1C1(dvs/dt �– dvo/dt �– R2C2d2vo/dt2)

Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt) Chapter 8, Solution 54. Let i be the inductor current.

dtdvv

i 5.04

(1)

dtdi

i2v (2)

Substituting (1) into (2) gives

Page 359: Solution Manual for Fundamentals of Electric Circuits

035.221

41

2 2

2

2

2

vdtdv

dtvd

dtvd

dtdv

dtdvv

v

199.125.1035.22 jsss

tBetAev tt 199.1sin199.1cos 25.125.1

v(0) = 2=A. Let w=1.199

)cossin()sincos(25.1 25.125.125.125.1 wtBewtAewwtBewtAedtdv tttt

085.2199.1

225.125.10)0( XBBwA

dtdv

V 199.1sin085.2199.1cos2 25.125.1 tetev tt

Chapter 8, Solution 55. At the top node, writing a KCL equation produces,

i/4 +i = C1dv/dt, C1 = 0.1

5i/4 = C1dv/dt = 0.1dv/dt i = 0.08dv/dt (1)

But, v = )idt)C/1(i2( 2 , C2 = 0.5 or -dv/dt = 2di/dt + 2i (2)

Substituting (1) into (2) gives,

-dv/dt = 0.16d2v/dt2 + 0.16dv/dt

0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0

Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25

v(t) = A + Be-7.25t (3) v(0) = 4 = A + B (4)

Page 360: Solution Manual for Fundamentals of Electric Circuits

From (1), i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25

But, dv/dt = -7.25Be-7.25t, which leads to,

dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 �– B = 4 + 3.448 = 7.448

Thus, v(t) = {7.45 �– 3.45e-7.25t} V Chapter 8, Solution 56. For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4

io

i 6

+

0.04F

i 20 0.25H Applying KVL to the larger loop,

-20 +6io +0.25dio/dt + 25 dt)ii( o = 0 Taking the derivative,

6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0 (1) For the smaller loop, 4 + 25 dt)ii( o = 0 Taking the derivative, 25(i + io) = 0 or i = -io (2) From (1) and (2) 6dio/dt + 0.25d2io/dt2 = 0

This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24

Page 361: Solution Manual for Fundamentals of Electric Circuits

io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A

As t approaches infinity, io( ) = 20/10 = 2 = A, therefore B = -2

Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A Chapter 8, Solution 57. (a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state.

v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit.

R 8 + 12 = 20 ohms, L = 1H, C = 4mF.

= R/(2L) = (20)/(2x1) = 10

o = 1/ LC = 1/ )36/1(x1 = 6

Since > o, we have a overdamped response.

s1,2 = 2o

2 -18, -2

Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. (b) i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B (1)

To get di(0)/dt, consider the circuit below at t = 0+.

+

vL

12

(1/36)F

+

v

i 8

1 H

-v(0) + 20i(0) + vL(0) = 0, which leads to,

Page 362: Solution Manual for Fundamentals of Electric Circuits

-16 + 20x2 + vL(0) = 0 or vL(0) = -24

Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s

Hence -24 = -2A �– 18B or 12 = A + 9B (2) From (1) and (2), B = 1.25 and A = 0.75

i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A

v(t) = 8i(t) = [6e-2t + 10e-18t] A Chapter 8, Solution 58.

(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4

V/s 85.0

)04()]0()0([)0(RCRiv

dtdv

(b) For , the circuit is a source-free RLC parallel circuit. 0t

2125.0

11,115.02

12

1xLCxxRC o

732.11422

od Thus,

)732.1sin732.1cos()( 21 tAtAetv t v(0) = 4 = A1

tAetAetAetAedtdv tttt 732.1cos732.1732.1sin732.1sin732.1732.1cos 2211

309.2732.18)0(221 AAA

dtdv

)732.1sin309.2732.1cos4()( ttetv t V

Page 363: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 59.

Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with

2,216/14

11,242

162 oo

xLCxLR

tt BteAeti 22)( i(0) = 2 = A

ttt BteBeAedtdi 222 22

4),032(412)]0()0([12)0(

BBAvRiL

BAdtdi

tt teeti 22 42)( t

ttt

ttt

tt

teedttedtevidtC

v0

2

00

22

0

2

0

)12(4

64166432)0(1

ttev 232 V

Chapter 8, Solution 60.

At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0) (1)

Applying nodal analysis,

4 = 0.5di1/dt + i1 + i2 (2) Also, i2 = [1di1/dt �– 1di2/dt]/3 or 3i2 = di1/dt �– di2/dt (3) Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt (4) From (2) and (3), di2/dt = di1/dt �– 3i2 = di1/dt �– 3(4 �– i1 �– 0.5di1/dt)

= di1/dt �– 12 + 3i1 + 1.5di1/dt Substituting this into (4),

d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)

Page 364: Solution Manual for Fundamentals of Electric Circuits

Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4

i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B] (5)

i2 = 4 �– i1 �– 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] �– [-Ae-t - 6Be-6t]

= [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B (6) From (5) and (6), A = -3.2 and B = -0.8

i1(t) = {4 + [-3.2e-t �– 0.8e-6t]} A

i2(t) = [1.6e-t �– 1.6e-6t] A Chapter 8, Solution 61. For t > 0, we obtain the natural response by considering the circuit below.

0.25F+

vC

a iL

4

1 H 6 At node a, vC/4 + 0.25dvC/dt + iL = 0 (1) But, vC = 1diL/dt + 6iL (2) Combining (1) and (2),

(1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0

d2iL/dt2 + 7diL/dt + 10iL = 0

s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5

Thus, iL(t) = iL( ) + [Ae-2t + Be-5t],

where iL( ) represents the final inductor current = 4(4)/(4 + 6) = 1.6

iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B (3)

diL/dt = [-2Ae-2t - 5Be-5t]

Page 365: Solution Manual for Fundamentals of Electric Circuits

and diL(0)/dt = 0 = -2A �– 5B or A = -2.5B (4)

From (3) and (4), A = -8/3 and B = 16/15

iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t]

v(t) = 6iL(t) = {9.6 + [-16e-2t + 6.4e-5t]} V

vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]}

vC = {9.6 + [-(32/3)e-2t + 1.0667e-5t]}

i(t) = vC/4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A Chapter 8, Solution 62. This is a parallel RLC circuit as evident when the voltage source is turned off.

= 1/(2RC) = (1)/(2x3x(1/18)) = 3

o = 1/ LC = 1/ 18/1x2 = 3

Since = o, we have a critically damped response.

s1,2 = -3

Let v(t) = capacitor voltage

Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0

But -10 + vR + v = 0 or vR = 10 �– v

Therefore vR = 10 �– [(A + Bt)e-3t] where A and B are determined from initial conditions.

Page 366: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 63. - R R v1 + vo vs v2 C C At node 1,

dtdv

CRvvs 11 (1)

At node 2,

dtdv

CRvv oo2 (2)

As a voltage follower, vvv 21 . Hence (2) becomes

dtdv

RCvv oo (3)

and (1) becomes

dtdv

RCvvs (4)

Substituting (3) into (4) gives

2

222

dtvd

CRdtdv

RCdtdv

RCvv oooos

or

sooo vvdtdv

RCdtvd

CR 22

222

Chapter 8, Solution 64.

C2

vs R1 2 1

v1

R2

+

C1

vo

Page 367: Solution Manual for Fundamentals of Electric Circuits

At node 1, (vs �– v1)/R1 = C1 d(v1 �– 0)/dt or vs = v1 + R1C1dv1/dt (1) At node 2, C1dv1/dt = (0 �– vo)/R2 + C2d(0 �– vo)/dt or �–R2C1dv1/dt = vo + C2dvo/dt (2) From (1) and (2), (vs �– v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt) or v1 = vs + (R1/R2)(vo + C2dvo/dt) (3) Substituting (3) into (1) produces,

vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1d{vs + (R1/R2)(vo + C2dvo/dt)}/dt

= vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt + (R1

2 C1C2/R2)[d2vo/dt2]

Simplifying we get,

d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt

Chapter 8, Solution 65. At the input of the first op amp,

(vo �– 0)/R = Cd(v1 �– 0) (1) At the input of the second op amp, (-v1 �– 0)/R = Cdv2/dt (2) Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following,

vo = -v2 or v2 = -vo (3) Combining (1), (2), and (3), eliminating v1 and v2 we get,

0v100dt

vdv

CR1

dtvd

o2o

2

o222o

2

Page 368: Solution Manual for Fundamentals of Electric Circuits

Which leads to s2 �– 100 = 0

Clearly this produces roots of �–10 and +10. And, we obtain,

vo(t) = (Ae+10t + Be-10t)V

At t = 0, vo(0+) = �– v2(0+) = 0 = A + B, thus B = �–A

This leads to vo(t) = (Ae+10t �– Ae-10t)V. Now we can use v1(0+) = 2V.

From (2), v1 = �–RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t)

v1(0+) = 2 = 0.1(20A) = 2A or A = 1

Thus, vo(t) = (e+10t �– e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the right-half-plane). Chapter 8, Solution 66. C2

R4

R2

+�–

C1

vS 1

2

R3

R1

vo Note that the voltage across C1 is v2 = [R3/(R3 + R4)]vo This is the only difference between this problem and Example 8.11, i.e. v = kv, where k = [R3/(R3 + R4)].

Page 369: Solution Manual for Fundamentals of Electric Circuits

At node 1,

(vs �– v1)/R1 = C2[d(v1 �– vo)/dt] + (v1 �– v2)/R2

vs/R1 = (v1/R1) + C2[d(v1)/dt] �– C2[d(vo)/dt] + (v1 �– kvo)/R2 (1) At node 2,

(v1 �– kvo)/R2 = C1[d(kvo)/dt] or v1 = kvo + kR2C1[d(vo)/dt] (2) Substituting (2) into (1), vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] �– (kvo/R2) + kC1[d(vo)/dt] �– (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) �– (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)] = vs/(kR1R2C1C2) If R1 = R2 10 kohms, C1 = C2 = 100 F, R3 = 20 kohms, and R4 = 60 kohms,

k = [R3/(R3 + R4)] = 1/3

R1R2C1C2 = 104 x104 x10-4 x10-4 = 1

(1/C2R1) + (1/ R2C2) + (1/R2C1) �– (1/ kR2C1) = 1 + 1 + 1 �– 3 = 3 �– 3 = 0 Hence, [d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = j

vo(t) = Vs + [Acost + B sint], Vs = 6

vo(0) = 0 = 6 + A or A = �–6

dvo/dt = �–Asint + Bcost, but dvo(0)/dt = 0 = B

Hence, vo(t) = 6(1 �– cost)u(t) volts.

Page 370: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 67. At node 1,

dt

)0v(dC

dt)vv(d

CR

vv 12

o11

1

1in (1)

At node 2, 2

o12 R

v0dt

)0v(dC , or

22

o1

RCv

dtdv

(2)

From (1) and (2),

2

o1

o11

o

22

111in R

vR

dtdv

CRdt

dvRCCR

vv

2

o1

o11

o

22

11in1 R

vR

dtdv

CRdt

dvRCCR

vv (3)

C1

From (2) and (3),

0Vvin

C2

R2

+v1

1 2 R1

vo

dtdv

RR

dtvd

CRdt

dvRCCR

dtdv

dtdv

RCv o

2

12

o2

11o

22

11in1

22

o

dtdv

CR1

RRCCv

dtdv

C1

C1

R1

dtvd in

111221

oo

2122

o2

But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1

210x10

2CR

2C1

C1

R1

4412212

dtdv

vdt

dv2

dtvd in

oo

2o

2

Page 371: Solution Manual for Fundamentals of Electric Circuits

Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = �–1, �–1

Therefore, vo(t) = [(A + Bt)e-t] + Vf

As t approaches infinity, the capacitor acts like an open circuit so that

Vf = vo( ) = 0

vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0

means that vo(0) = 0 which leads to A = 0.

vo(t) = [Bte-t]

dtdvo = [(B �– Bt)e-t] (4)

From (2), 0RC

)0(vdt

)0(dv

22

oo

From (1) at t = 0+,

dt)0(dv

CR

01 o1

1

which leads to 1RC1

dt)0(dv

11

o

Substituting this into (4) gives B = �–1

Thus, v(t) = �–te-tu(t) V

Page 372: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.

Page 373: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.

Page 374: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 70. The schematic is shown below.

After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.

Page 375: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.

Page 376: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 72. When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.

When the circuit is simulated, we obtain i(t) as shown below.

Page 377: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 73. (a) For t < 0, we have the schematic below. When this is saved and simulated, we

obtain the initial inductor current and capacitor voltage as

iL(0) = 3 A and vc(0) = 24 V.

(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as shown below.

Page 378: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 74.

10 5 + 20 V 2F 4H - Hence the dual circuit is shown below. 2H 4F 0.2 20A 0.1

Page 379: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b).

0.5 H 2 F

12A

24A

0.25

0.1

10 F

12A +

24V

10 H

10

0.25

24A

+

12V

0.5 F

10 H

(a)

4

0.1

(b)

Page 380: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b).

120 A

2 V

+

2 A

30

1/3

10

0.1

120 V

�– +

60 V

+ 60 A

4F

1 F 4 H 1 H

20

0.05 (a) 0.05

60 A 1 H

120 A

1/4 F

0.1

+

2V

1/30 (b)

Page 381: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b).

+

5 V

1/4 F

1 H

2

1

1/3 12 A

1

12 A

1/3

3 1/2

5 V

�– + 5 A

+

12V1/4 F

1 F 1/4 H

1 H

(a)

1

2

(b)

Page 382: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 78. The voltage across the igniter is vR = vC since the circuit is a parallel RLC type.

vC(0) = 12, and iL(0) = 0.

= 1/(2RC) = 1/(2x3x1/30) = 5

o 30/1x10x60/1LC/1 3 = 22.36

< o produces an underdamped response.

2o

22,1s = �–5 j21.794

vC(t) = e-5t(Acos21.794t + Bsin21.794t) (1)

vC(0) = 12 = A dvC/dt = �–5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(�–Asin21.794t + Bcos21.794t)e-5t] (2)

dvC(0)/dt = �–5A + 21.794B

But, dvC(0)/dt = �–[vC(0) + RiL(0)]/(RC) = �–(12 + 0)/(1/10) = �–120

Hence, �–120 = �–5A + 21.794B, leads to B (5x12 �– 120)/21.794 = �–2.753

At the peak value, dvC(to)/dt = 0, i.e.,

0 = A + Btan21.794to + (A21.794/5)tan21.794to �– 21.794B/5

(B + A21.794/5)tan21.794to = (21.794B/5) �– A

tan21.794to = [(21.794B/5) �– A]/(B + A21.794/5) = �–24/49.55 = �–0.484

Therefore, 21.7945to = |�–0.451|

to = |�–0.451|/21.794 = 20.68 ms

Page 383: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 79.

For critical damping of a parallel RLC circuit,

LCRCo1

21

Hence,

F434144425.0

4 2 xRL

C

Chapter 8, Solution 80.

t1 = 1/|s1| = 0.1x10-3 leads to s1 = �–1000/0.1 = �–10,000

t2 = 1/|s2| = 0.5x10-3 leads to s1 = �–2,000

2o

21s

2o

22s

s1 + s2 = �–2 = �–12,000, therefore = 6,000 = R/(2L)

L = R/12,000 = 60,000/12,000 = 5H

2o

22s = �–2,000

2o

2 = 2,000

2o

2000,6 = 2,000

2o

2 = 4,000

2 �– = 16x102o

6

2o = 2 �– 16x106 = 36x106 �– 16x106

o = 103 LC/120

C = 1/(20x106x5) = 10 nF

Page 384: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 81.

t = 1/ = 0.25 leads to = 4

But, 1/(2RC) or, C = 1/(2 R) = 1/(2x4x200) = 625 F

22od

232322

d2o 010x42(16)10x42( = 1/(LC)

This results in L = 1/(64 2x106x625x10-6) = 2.533 H

Chapter 8, Solution 82. For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below.

+

vo

C1

+

v

a

R2

R1 C2 At node a,

(vo �– v/R1 = (v/R2) + C2dv/dt

vo = v(1 + R1/R2) + R1C2 dv/dt

60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt

60 = 3v + 25dv/dt

v(t) = Vs + [Ae-3t/25]

where 3Vs = 60 yields Vs = 20

v(0) = 0 = 20 + A or A = �–20

v(t) = 20(1 �– e-3t/25)V

Page 385: Solution Manual for Fundamentals of Electric Circuits

Chapter 8, Solution 83. i = iD + Cdv/dt (1) �–vs + iR + Ldi/dt + v = 0 (2) Substituting (1) into (2),

vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0

LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs

d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC

Page 386: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 1.

(a) angular frequency = 103 rad/s

(b) frequency f = 2

= 159.2 Hz

(c) period T = f

= 1

6.283 ms

(d) Since sin(A) = cos(A �– 90 ),

vs = 12 sin(103t + 24 ) = 12 cos(103t + 24 �– 90 ) vs in cosine form is vs = 12 cos(103t �– 66 ) V

(e) vs(2.5 ms) = 12 )24)105.2)(10sin(( 3-3

= 12 sin(2.5 + 24 ) = 12 sin(143.24 + 24 ) = 2.65 V

Chapter 9, Solution 2.

(a) amplitude = 8 A (b) = 500 = 1570.8 rad/s

(c) f = 2

= 250 Hz

(d) Is = 8 -25 A

Is(2 ms) = )25)102)(500cos((8 3-

= 8 cos( 25 ) = 8 cos(155 ) = -7.25 A

Chapter 9, Solution 3.

(a) 4 sin( t �– 30 ) = 4 cos( t �– 30 �– 90 ) = 4 cos( t �– 120 ) (b) -2 sin(6t) = 2 cos(6t + 90 ) (c) -10 sin( t + 20 ) = 10 cos( t + 20 + 90 ) = 10 cos( t + 110 )

Page 387: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 4.

(a) v = 8 cos(7t + 15 ) = 8 sin(7t + 15 + 90 ) = 8 sin(7t + 105 ) (b) i = -10 sin(3t �– 85 ) = 10 cos(3t �– 85 + 90 ) = 10 cos(3t + 5 )

Chapter 9, Solution 5. v1 = 20 sin( t + 60 ) = 20 cos( t + 60 90 ) = 20 cos( t 30 ) v2 = 60 cos( t 10 )

This indicates that the phase angle between the two signals is 20 and that v1 lags v2.

Chapter 9, Solution 6.

(a) v(t) = 10 cos(4t �– 60 ) i(t) = 4 sin(4t + 50 ) = 4 cos(4t + 50 �– 90 ) = 4 cos(4t �– 40 ) Thus, i(t) leads v(t) by 20 .

(b) v1(t) = 4 cos(377t + 10 )

v2(t) = -20 cos(377t) = 20 cos(377t + 180 ) Thus, v2(t) leads v1(t) by 170 .

(c) x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t �– 90 )

X = 13 0 + 5 -90 = 13 �– j5 = 13.928 -21.04 x(t) = 13.928 cos(2t �– 21.04 ) y(t) = 15 cos(2t �– 11.8 ) phase difference = -11.8 + 21.04 = 9.24 Thus, y(t) leads x(t) by 9.24 .

Chapter 9, Solution 7. If f( ) = cos + j sin ,

)(fj)sinj(cosjcosj-sinddf

djfdf

Page 388: Solution Manual for Fundamentals of Electric Circuits

Integrating both sides ln f = j + ln A f = Aej = cos + j sin f(0) = A = 1 i.e. f( ) = ej = cos + j sin

Chapter 9, Solution 8.

(a) 4j3

4515+ j2 =

53.13-54515

+ j2

= 3 98.13 + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97

(b) (2 + j)(3 �– j4) = 6 �– j8 + j3 + 4 = 10 �– j5 = 11.18 -26.57

j4)-j)(3(220-8

+j125-

10 =

26.57-11.1820-8

+14425

)10)(12j5-(

= 0.7156 6.57 0.2958 j0.71

= 0.7109 + j0.08188 0.2958 j0.71

= 0.4151 j0.6281 (c) 10 + (8 50 )(13 -68.38 ) = 10+104 -17.38

= 109.25 �– j31.07 Chapter 9, Solution 9.

(a) 2 +8j54j3

= 2 +6425

)8j5)(4j3(

= 2 +89

3220j24j15

= 1.809 + j0.4944

(b) 4 -10 +632j1

= 4 -10 +6363.43-236.2

Page 389: Solution Manual for Fundamentals of Electric Circuits

= 4 -10 + 0.7453 -69.43 = 3.939 �– j0.6946 + 0.2619 �– j0.6978 = 4.201 �– j1.392

(c) 50480920-6108

= 064.3j571.2863.8j5628.1052.2j638.53892.1j879.7

= 799.5j0083.1

6629.0j517.13 =

86.99886.581.2-533.13

= 2.299 -102.67 = -0.5043 �– j2.243

Chapter 9, Solution 10.

(a) z 9282.64z and ,566.8z ,86 321 jjj

93.1966.10321 jzzz

(b) 499.7999.93

21 jzzz

Chapter 9, Solution 11.

(a) = (-3 + j4)(12 + j5) 21zz = -36 �– j15 + j48 �– 20 = -56 + j33

(b) 2

1

zz

= 5j124j3-

= 25144

)5j12)(4j3(- = -0.3314 + j0.1953

(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz

21 zz = (-3 + j4) �– (12 + j5) = -15 �– j

21

21

zzzz

= )j15(-)j1(9

= 22 115j)-15)(j1(9-

= 226

)14j16(9-

= -0.6372 �– j0.5575

Page 390: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 12.

(a) = (-3 + j4)(12 + j5) 21zz = -36 �– j15 + j48 �– 20 = -56 + j33

(b) 2

1

zz

= 5j124j3-

= 25144

)5j12)(4j3(- = -0.3314 + j0.1953

(c) = (-3 + j4) + (12 + j5) = 9 + j9 21 zz

21 zz = (-3 + j4) �– (12 + j5) = -15 �– j

21

21

zzzz

= )j15(-)j1(9

= 22 115j)-15)(j1(9-

= 226

)14j16(9-

= -0.6372 �– j0.5575 Chapter 9, Solution 13.

(a) 1520.02749.1)2534.08425.0()4054.04324.0 jjj(

(b) 0833.215024

3050o

o

(c) (2+j3)(8-j5) �–(-4) = 35 +j14

Chapter 9, Solution 14.

(a) 5116.05751.01115

143j

jj

(b) 55.11922.17.213406.246

9.694424186)5983.1096.16)(8467(

)8060)(8056.13882.231116.62(j

jjjjjj

(c) 89.2004.256)120260(42 2 jjj

Page 391: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 15.

(a) j1-5-3j26j10

= -10 �– j6 + j10 �– 6 + 10 �– j15

= -6 �– j11

(b) 453016

10-4-3020 = 60 15 + 64 -10

= 57.96 + j15.529 + 63.03 �– j11.114 = 120.99 �– j4.415

(c)

j1j0jj1

j1j1j1j

0jj1

= 1 )j1(j)j1(j0101 22

= 1 )j1j1(1 = 1 �– 2 = -1

Chapter 9, Solution 16.

(a) -10 cos(4t + 75 ) = 10 cos(4t + 75 180 ) = 10 cos(4t 105 )

The phasor form is 10 -105 (b) 5 sin(20t �– 10 ) = 5 cos(20t �– 10 �– 90 )

= 5 cos(20t �– 100 ) The phasor form is 5 -100

(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t �– 90 )

The phasor form is 4 0 + 3 -90 = 4 �– j3 = 5 -36.87 Chapter 9, Solution 17.

(a) Let A = 8 -30 + 6 0 = 12.928 �– j4 = 13.533 -17.19

a(t) = 13.533 cos(5t + 342.81 )

Page 392: Solution Manual for Fundamentals of Electric Circuits

(b) We know that -sin = cos( + 90 ). Let B = 20 45 + 30 (20 + 90 )

= 14.142 + j14.142 �– 10.261 + j28.19 = 3.881 + j42.33 = 42.51 84.76

b(t) = 42.51 cos(120 t + 84.76 ) (c) Let C = 4 -90 + 3 (-10 �– 90 )

= -j4 �– 0.5209 �– j2.954 = 6.974 265.72

c(t) = 6.974 cos(8t + 265.72 ) Chapter 9, Solution 18.

(a) = )t(v1 60 cos(t + 15 ) (b) = 6 + j8 = 10 53.13 2V

)t(v2 = 10 cos(40t + 53.13 ) (c) = )t(i1 2.8 cos(377t �– /3) (d) = -0.5 �– j1.2 = 1.3 247.4 2I

)t(i 2 = 1.3 cos(103t + 247.4 ) Chapter 9, Solution 19.

(a) 3 10 5 -30 = 2.954 + j0.5209 �– 4.33 + j2.5 = -1.376 + j3.021 = 3.32 114.49

Therefore, 3 cos(20t + 10 ) �– 5 cos(20t �– 30 ) = 3.32 cos(20t + 114.49 )

(b) 4 -90 + 3 -45 = -j40 + 21.21 �– j21.21

= 21.21 �– j61.21 = 64.78 -70.89

Therefore, 40 sin(50t) + 30 cos(50t �– 45 ) = 64.78 cos(50t �– 70.89 ) (c) Using sin = cos( 90 ),

20 -90 + 10 60 5 -110 = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 �– j6.641 = 9.44 -44.7

Therefore, 20 sin(400t) + 10 cos(400t + 60 ) �– 5 sin(400t �– 20 ) = 9.44 cos(400t �– 44.7 )

Page 393: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 20. (a) oooo jj 399.4966.82139.383.32464.340590604V Hence,

)399.4377cos(966.8 otv (b) 5,90208010 ooo jI , i.e. ooI 04.1651.49204010

)04.165cos(51.49 oti

Chapter 9, Solution 21.

(a) oooo jF 86.343236.8758.48296.690304155

)86.3430cos(324.8)( ottf

(b) G ooo j 49.62565.59358.4571.2504908

)49.62cos(565.5)( ottg

(c) 40,9050101 oo

jH

i.e. ooo jH 6.1162795.0125.025.0180125.09025.0

)6.11640cos(2795.0)( otth Chapter 9, Solution 22.

Let f(t) =t

dttvdtdvtv )(24)(10

oVjVVjVF 3020,5,2410

ojjVjVjVF 97.921.440)1032.17)(6.1910(4.02010

)97.925cos(1.440)( ottf

Page 394: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 23.

(a) v(t) = 40 cos( t �– 60 ) (b) V = -30 10 + 50 60

= -4.54 + j38.09 = 38.36 96.8 v(t) = 38.36 cos( t + 96.8 )

(c) I = j6 -10 = 6 (90 10 ) = 6 80

i(t) = 6 cos( t + 80 )

(d) I = j2

+ 10 -45 = -j2 + 7.071 �– j7.071

= 11.5 -52.06 i(t) = 11.5 cos( t �– 52.06 )

Chapter 9, Solution 24.

(a)

1,010jV

V

10)j1(V

45071.75j5j1

10V

Therefore, v(t) = 7.071 cos(t + 45 ) (b)

4),9010(20j4

5jV

VV

80-204j4

54jV

96.110-43.33j580-20

V

Therefore, v(t) = 3.43 cos(4t �– 110.96 )

Page 395: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 25.

(a) 2,45-432j II

45-4)4j3(I

98.13-8.013.53545-4

j4345-4

I

Therefore, i(t) = 0.8 cos(2t �– 98.13 ) (b)

5,2256jj

10 III

225)65j2j-( I

56.4-745.056.26708.6

2253j6

225I

Therefore, i(t) = 0.745 cos(5t �– 4.56 ) Chapter 9, Solution 26.

2,01j

2jI

II

12j1

22jI

87.36-4.05.1j2

1I

Therefore, i(t) = 0.4 cos(2t �– 36.87 ) Chapter 9, Solution 27.

377,10-110j

10050jV

VV

10-110377100j

50377jV

10-110)45.826.380(V 45.92-289.0V

Therefore, v(t) = 0.289 cos(377t �– 92.45 ).

Page 396: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 28.

8

)t377cos(110R

)t(v)t(i s 13.75 cos(377t) A.

Chapter 9, Solution 29.

5.0j-)102)(10(j

1Cj

16-6Z

65-2)90-5.0)(254(IZV

Therefore v(t) = 2 sin(106t �– 65 ) V.

Chapter 9, Solution 30.

Z 2j)104)(500(jLj 3-

155-30902

65-60ZV

I

Therefore, i(t) = 30 cos(500t �– 155 ) A. Chapter 9, Solution 31.

i(t) = 10 sin( t + 30 ) = 10 cos( t + 30 90 ) = 10 cos( t 60 ) Thus, I = 10 -60

v(t) = -65 cos( t + 120 ) = 65 cos( t + 120 180 ) = 65 cos( t 60 )

Thus, V = 65 -60

5.660-1060-65

IV

Z

Since V and I are in phase, the element is a resistor with R = 6.5 .

Page 397: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 32. V = 180 10 , I = 12 -30 , = 2

642.9j49.11041530-1201180

IV

Z

One element is a resistor with R = 11.49 . The other element is an inductor with L = 9.642 or L = 4.821 H.

Chapter 9, Solution 33. 2

L2R vv110

2R

2L v110v

22L 85110v 69.82 V

Chapter 9, Solution 34.

if 0vo LC1

C1

L

)102)(105(1

33 100 rad/s

Chapter 9, Solution 35.

05sV2j)1)(2(jLj

2j-)25.0)(2(j

1Cj

1

)05)(901(0522j

2j2j22j

so VV 5 90

Thus, 5 cos(2t + 90 ) = )t(vo -5 sin(2t) V

Page 398: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 36.

Let Z be the input impedance at the source.

2010100200mH 100 3 jxxjLj

5002001010

11F10 6 jxxjCj

1000//-j500 = 200 �–j400 1000//(j20 + 200 �–j400) = 242.62 �–j239.84

ojZ 104.6225584.23962.2242

mA 896.361.26104.62255

1060 oo

o

I

)896.3200cos(1.266 oti

Chapter 9, Solution 37.

5j)1)(5(jLj

j-)2.0)(5(j

1Cj

1

Let Z , j-1 5j210j

5j2)5j)(2(

5j||22Z

Then, s21

2x I

ZZZ

I , where 0s 2I

3212.28j5

20j)2(

5j210j

j-

5j210j

xI

Therefore, )t(i x 2.12 sin(5t + 32 ) A

Page 399: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 38.

(a) 2j-)6/1)(3(j

1Cj

1F

61

43.18-472.4)4510(2j4

2j-I

Hence, i(t) = 4.472 cos(3t �– 18.43 ) A

43.18-89.17)43.18-472.4)(4(4IV Hence, v(t) = 17.89 cos(3t �– 18.43 ) V

(b) 3j-)12/1)(4(j

1Cj

1F

121

12j)3)(4(jLjH3

87.3610j34050

ZV

I

Hence, i(t) = 10 cos(4t + 36.87 ) A

69.336.41)050(j128

12jV

Hence, v(t) = 41.6 cos(4t + 33.69 ) V Chapter 9, Solution 39.

10j810j5j

)10j-)(5j(8)10j-(||5j8Z

34.51-124.334.51403.6

20j108040

ZV

I

34.51-248.6210j5j

10j-1 III

66.1283.124-5j-5j

2 III

Therefore, )t(i1 6.248 cos(120 t �– 51.34 ) A

)t(i2 3.124 cos(120 t + 128.66 ) A

Page 400: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 40.

(a) For , 1j)1)(1(jLjH1

20j-)05.0)(1(j

1Cj

1F05.0

802.0j98.120j2

40j-j)20j-(||2jZ

05.22-872.105.22136.2

04j0.8021.9804

o ZV

I

Hence, i )t(o 1.872 cos(t �– 22.05 ) A

(b) For , 55j)1)(5(jLjH1

4j-)05.0)(5(j

1Cj

1F05.0

2.4j6.12j1

4j-5j)4j-(||25jZ

14.69-89.014.69494.4

04j41.6

04o Z

VI

Hence, i )t(o 0.89 cos(5t �– 69.14 ) A

(c) For , 1010j)1)(10(jLjH1

2j-)05.0)(10(j

1Cj

1F05.0

9j12j2

4j-10j)2j-(||210jZ

66.38-4417.066.839.055

049j1

04o Z

VI

Hence, i )t(o 0.4417 cos(10t �– 83.66 ) A

Page 401: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 41.

, 1j)1)(1(jLjH1

j-)1)(1(j

1Cj

1F1

j21

1j-1)j-(||)j1(1Z

j210s

ZV

I , II )j1(c

18.43-325.6j2

)10)(j1()j1()j1)(j-( IIV

Thus, v(t) = 6.325 cos(t �– 18.43 ) V

Chapter 9, Solution 42.

200

100j-)1050)(200(j

1Cj

1F50 6-

20j)1.0)(200(jLjH1.0

20j40j2-1

j100-j10050

)(50)(-j100-j100||50

9014.17)060(7020j

)060(20j403020j

20joV

Thus, )t(vo 17.14 sin(200t + 90 ) V or )t(vo 17.14 cos(200t) V

Page 402: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 43.

22j)1)(2(jLjH1

5.0j-)1)(2(j

1Cj

1F1

69.33328.3045.1j1

5.1j15.0j2j

5.0j2jo II

Thus, )t(io 3.328 cos(2t + 33.69 ) A

Chapter 9, Solution 44.

2002j)1010)(200(jLjmH10 -3

j-)105)(200(j

1Cj

1mF5 3-

4.0j55.010

j35.0j25.0

j31

2j1

41

Y

865.0j1892.14.0j55.0

11Y

Z

7.956-96.0865.0j1892.6

065

06Z

I

Thus, i(t) = 0.96 cos(200t �– 7.956 ) A

Page 403: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 45. We obtain I by applying the principle of current division twice. o

I I2 I2 Io

Z1 -j2 Z2 2

(a) (b)

2j-1Z , 3j1j2-2

j4-j42||-j2)(4j2Z

j1j10-

)05(3j12j-

2j-

21

12 I

ZZZ

I

1110-

j1j10-

j-1j-

j2-2j2-

2o II -5 A

Chapter 9, Solution 46.

405)40t10cos(5i ss I

j-)1.0)(10(j

1Cj

1F1.0

2j)2.0)(10(jLjH2.0

Let 6.1j8.02j4

8j2j||41Z , j32Z

)405(6.0j8.3

j1.60.8s

21

1o I

ZZZ

I

46.94325.297.8847.3

)405)(43.63789.1(oI

Thus, )t(io 2.325 cos(10t + 94.46 ) A

Page 404: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 47.

First, we convert the circuit into the frequency domain.

-j10

Ix

+

2 j4 5 0 20

63.524607.063.52854.10

5626.8j588.42

5

4j2010j)4j20(10j2

5Ix

is(t) = 0.4607cos(2000t +52.63) A

Chapter 9, Solution 48.

Converting the circuit to the frequency domain, we get: 10 30 V1

+

j20

Ix -j20 20 -40

We can solve this using nodal analysis.

Page 405: Solution Manual for Fundamentals of Electric Circuits

A)4.9t100sin(4338.0i

4.94338.020j30

29.24643.15I

29.24643.1503462.0j12307.0

402V

402)01538.0j02307.005.0j1.0(V

020j300V

20j0V

104020V

x

x

1

1

111

Chapter 9, Solution 49.

4j1

)j1)(2j(2)j1(||2j2TZ

1 I Ix

j2 -j

IIIj1

2jj12j

2jx , where

21

05.x 0I

4jj1

2jj1

xII

45-414.1j1j

j1)4(

4jj1

Ts ZIV

)t(vs 1.414 sin(200t �– 45 ) V Chapter 9, Solution 50. Since = 100, the inductor = j100x0.1 = j10 and the capacitor = 1/(j100x10-3) = -j10 .

j10 Ix

-j10 +

vx

20

5 40

Page 406: Solution Manual for Fundamentals of Electric Circuits

Using the current dividing rule:

V)50t100cos(50v5050I20V

505.2405.2j40510j2010j

10jI

x

xx

x

Chapter 9, Solution 51.

5j-)1.0)(2(j

1Cj

1F1.0

j)5.0)(2(jLjH5.0 The current I through the 2- resistor is

4j32j5j11 s

s

III , where 010I

13.53-50)4j3)(10(sI Therefore,

)t(is 50 cos(2t �– 53.13 ) A Chapter 9, Solution 52.

5.2j5.2j1

5j5j5

25j5j||5

101Z , 5.2j5.25.2j5.25j-2Z

I2

Z1 Z2IS

Page 407: Solution Manual for Fundamentals of Electric Circuits

sss21

12 j5

45.2j5.12

10III

ZZZ

I

)5.2j5.2(2o IV

ss j5)j1(10

)j1)(5.2(j5

4308 II

)j1(10)j5)(308(

sI 2.884 -26.31 A

Chapter 9, Solution 53.

Convert the delta to wye subnetwork as shown below. Z1 Z2 Io 2 Z3 + 10

60 V 8o30 - Z

,3077.24615.0210

46,7692.01532.021042

21 jjxj

Zjjxj

Z

2308.01538.1210

123 j

jZ

6062.0726.4)3077.25385.9//()2308.01538.9()10//()8( 23 jjjZZ

163.0878.66062.0726.42 1 jjZZ

A 64.28721.83575.188.6

3060Z

3060I oo

ooo

Page 408: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 54.

Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. Vs

V1

+

+

V2

+

Z 2 Z

)j1(2)j1(o1 IV )j1(42 12 VV

)j1(621s VVV

sV 8.485 -45 V Chapter 9, Solution 55.

-j4

I I1

+

Vo

I2+

Z 12

-j20 V j8

-j0.58j4

8jo

1

VI

j8j4-

)8j((-j0.5)j4-

)8j(12

ZZZII

5.0j8

j8

-j0.521

ZZIII

)8j(12j20- 1 ZII

)8j(2j-

2j

812j20- Z

Z

Page 409: Solution Manual for Fundamentals of Electric Circuits

21

j23

j26-4- Z

279.6864.1618.43-5811.1

25.26131.26

21

j23

j26-4-Z

Z = 2.798 �– j16.403

Chapter 9, Solution 56.

30H3 jLj

30/13F jCj

15/11.5F jCj

06681.0

1530

1530

)15///(30 jj

j

jxj

jj

m 333606681.02033.0

)06681.02(033.0)06681.02//(30

jjjjj

jj

Z

Chapter 9, Solution 57.

2H2 jLj

jCj

11F

2.1j6.2j22j)j2(2j1)j2//(2j1Z

S 1463.0j3171.0Z

1Y

Page 410: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 58.

(a) 2j-)1010)(50(j

1Cj

1mF10 3-

5.0j)1010)(50(jLjmH10 -3

)2j1(||15.0jinZ

2j22j1

5.0jinZ

)j3(25.05.0jinZ

inZ 0.75 + j0.25 (b) 20j)4.0)(50(jLjH4.0

10j)2.0)(50(jLjH2.0

20j-)101)(50(j

1Cj

1mF1 3-

For the parallel elements,

20j-1

10j1

2011

pZ

10j10pZ Then,

inZ 10 + j20 + = pZ 20 + j30 Chapter 9, Solution 59.

)4j2(||)2j1(6eqZ

)4j2()2j1()4j2)(2j1(

6eqZ

5385.1j308.26eqZ

eqZ 8.308 �– j1.5385

Chapter 9, Solution 60.

878.91.51122.5097.261525)1030//()5020()1525( jjjjjjZ

Page 411: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 61.

All of the impedances are in parallel.

3j11

5j1

2j11

j111

eqZ

4.0j8.0)3.0j1.0()2.0j-()4.0j2.0()5.0j5.0(1

eqZ

4.0j8.01

eqZ 1 + j0.5

Chapter 9, Solution 62.

2 20j)102)(1010(jLjmH -33

100j-)101)(1010(j

1Cj

1F1 6-3

50 j20

+

+ V +

Vin1 0 A 2V

-j100

50)50)(01(V

)50)(2()100j20j50)(01(inV 80j15010080j50inV

01in

in

VZ 150 �– j80

Page 412: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 63. First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta.

5.22j1020

450j200z,333.13j3015j

450j200z

45j2010

300j150j200z

32

1

�–j12 �–j16 8

z2

z3

�–j16

z1

10

ZT 10 Now all we need to do is to combine impedances.

93.6j69.34)821.3j7.21938.8j721.8(z12j8Z

821.3j70.21)16j10(z

938.8j721.833.29j40

)16j10)(333.13j30()16j10(z

1T

3

2

Chapter 9, Solution 64.

A7.104527.14767.1j3866.0Z

9030I

5j192j6

)8j6(10j4Z

T

T

Page 413: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 65.

)4j3(||)6j4(2TZ

2j7)4j3)(6j4(

2TZ

TZ 6.83 + j1.094 = 6.917 9.1

1.9917.610120

TZV

I 17.35 0.9 A

Chapter 9, Solution 66.

)j12(145170

5j60)10j40)(5j20(

)10j40(||)5j20(TZ

TZ 14.069 �– j1.172 = 14.118 -4.76

76.9425.476.4-118.14

9060

TZV

I

I

I1

20

+ Vab

I2

j10

IIIj122j8

5j6010j40

1

IIIj12j4

5j605j20

2

21ab 10j20- IIV

Page 414: Solution Manual for Fundamentals of Electric Circuits

IIVj12

40j10j12

)40j(160-ab

IIV145

j)(150)-12(j12

150-ab

)76.9725.4)(24.175457.12(abV

abV 52.94 273 V

Chapter 9, Solution 67.

(a) 20j)1020)(10(jLjmH20 -33

80j-)105.12)(10(j

1Cj

1F5.12 6-3

)80j60(||20j60inZ

60j60)80j60)(20j(

60inZ

22.20494.6733.23j33.63inZ

inin

1Z

Y 0.0148 -20.22 S

(b) 10j)1010)(10(jLjmH10 -33

50j-)1020)(10(j

1Cj

1F20 6-3

2060||30

)10j40(||2050j-inZ

10j60)10j40)(20(

50j-inZ

56.74-75.5092.48j5.13inZ

inin

1Z

Y 0.0197 74.56 S = 5.24 + j18.99 mS

Page 415: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 68.

4j-

1j3

12j5

1eqY

)25.0j()1.0j3.0()069.0j1724.0(eqY

eqY 0.4724 + j0.219 S

Chapter 9, Solution 69.

)2j1(41

2j-1

411

oY

6.1j8.05

)2j1)(4(2j1

4oY

6.0j8.0joY

)6.0j8.0()333.0j()1(6.0j8.0

13j-

1111

oY

41.27028.2933.0j8.11

oY

2271.0j4378.041.27-4932.0oY

773.4j4378.05joY

97.22773.4j4378.0

5.0773.4j4378.0

1211

eqY

2078.0j5191.01

eqY

3126.02078.0j5191.0

eqY 1.661 + j0.6647 S

Page 416: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 70.

Make a delta-to-wye transformation as shown in the figure below.

cb

n

a

Zan

Zcn Zbn

2

Zeq

8

-j5

9j75j15

)10j15)(10(15j1010j5)15j10)(10j-(

anZ

5.3j5.45j15

)15j10)(5(bnZ

3j1-5j15

)10j-)(5(cnZ

)5j8(||)2( cnbnaneq ZZZZ

)8j7(||)5.3j5.6(9j7eqZ

5.4j5.13)8j7)(5.3j5.6(

9j7eqZ

2.0j511.59j7eqZ

2.9j51.12eqZ 15.53 -36.33

Page 417: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 71.

We apply a wye-to-delta transformation.

j4

Zeq

-j2 Zbc

Zab

Zac

a b

c

1

j12j

2j22j

4j2j2abZ

j12

2j2acZ

j2-2j-2j2

bcZ

8.0j6.13j1

)j1)(4j()j1(||4j||4j abZ

2.0j6.0j2

)j1)(1()j1(||1||1 acZ

6.0j2.2||1||4j acab ZZ

6.0j2.21

2j2-1

2j-11

eqZ

1154.0j4231.025.0j25.05.0j 66.644043.03654.0j173.0

eqZ 2.473 -64.66 = 1.058 �– j2.235

Page 418: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 72.

Transform the delta connections to wye connections as shown below. a

R3

R2R1

-j18 -j9

j2

j2 j2

b

6j-18j-||9j- ,

8102020)20)(20(

R1 , 450

)10)(20(R 2 , 4

50)10)(20(

3R

44)j6(j2||)82j(j2abZ

j4)(4||)2j8(j24abZ

j2-12)4jj2)(4(8

j24abZ

4054.1j567.3j24abZ

abZ 7.567 + j0.5946

Page 419: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 73. Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b).

a

R3

R2R1

-j18 -j9

j2

j2 j2

b

8.4j-10j48

6j8j8j)6j-)(8j(

1Z

-j4.812 ZZ

4.6jj1064-

10j)8j)(8j(

3Z

))(2())(4()4)(2( 313221 ZZZZZZ

6.9j4.46)4.6j)(8.4j2()4.6j)(8.4j4()8.4j4)(8.4j2(

25.7j5.14.6j

6.9j4.46aZ

688.6j574.38.4j4

6.9j4.46bZ

945.8j727.18.4j2

6.9j4.46cZ

3716.3j07407688.12j574.3

)88.61583.7)(906(||6j bZ

602.2j186.025.11j5.1

j7.25)-j4)(1.5(||4j- aZ

Page 420: Solution Manual for Fundamentals of Electric Circuits

1693.5j5634.0945.20j727.1

)07.7911.9)(9012(||12j cZ

)||12j||4j-(||)||6j( cabeq ZZZZ

)5673.2j7494.0(||)3716.3j7407.0(eqZ

eqZ 1.508 75.42 = 0.3796 + j1.46 Chapter 9, Solution 74.

One such RL circuit is shown below.

Z

+

Vi = 1 0 j20

20 20 V

j20 +

Vo

We now want to show that this circuit will produce a 90 phase shift.

)3j1(42j120j20-

40j20)20j20)(20j(

)20j20(||20jZ

)j1(31

3j63j1

)01(12j2412j4

20 iVZ

ZV

903333.03j

)j1(31

j1j

20j2020j

o VV

This shows that the output leads the input by 90 .

Chapter 9, Solution 75.

Since , we need a phase shift circuit that will cause the output to lead the input by 90 .

)90tsin()tcos(This is achieved by the RL circuit shown

below, as explained in the previous problem.

Page 421: Solution Manual for Fundamentals of Electric Circuits

10 10

+

Vi j10 j10 +

Vo

This can also be obtained by an RC circuit.

Chapter 9, Solution 76.

Let Z = R �– jX, where fC2

1C

1X

394.9566116R|Z|XXR|Z| 222222

F81.27394.95x60x2

1fX21C

Chapter 9, Solution 77.

(a) ic

co jXR

jX-VV

where 979.3)1020)(102)(2(

1C1

X 9-6c

))53.979(tan-90(979.35

979.3j3.979-5

j3.979- 1-22

i

o

VV

)51.38-90(83.1525

979.3

i

o

VV

51.49-6227.0i

o

VV

Therefore, the phase shift is 51.49 lagging

Page 422: Solution Manual for Fundamentals of Electric Circuits

(b) )RX(tan-90-45 c-1

C1

XR)RX(tan45 cc1-

RC1

f2

)1020)(5)(2(1

RC21

f 9- 1.5915 MHz

Chapter 9, Solution 78. 8+j6 R Z -jX

5)6(8

)]6(8[)6(8//XjRXjRXjRZ

i.e 8R + j6R �– jXR = 5R + 40 + j30 �–j5X Equating real and imaginary parts:

8R = 5R + 40 which leads to R=13.33 6R-XR =30-5 which leads to X=4.125 .

Chapter 9, Solution 79.

(a) Consider the circuit as shown.

40

Z1

V1V2

j30 +

Vi j60 +

Vo j10

30 20

Z2

Page 423: Solution Manual for Fundamentals of Electric Circuits

21j390j30

)60j30)(30j()60j30(||30j1Z

21.80028.9896.8j535.131j43

)21j43)(10j()40(||10j 12 ZZ

Let V . 01i

896.8j535.21)01)(21.80028.9(

20 i2

22 V

ZZ

V

77.573875.02V

03.2685.47)77.573875.0)(87.81213.21(

21j4321j3

40 221

11 VV

ZZ

V

61.1131718.01V

111o )j2(52

2j12j

60j3060j

VVVV

)6.1131718.0)(56.268944.0(oV 2.1401536.0oV

Therefore, the phase shift is 140.2

(b) The phase shift is leading. (c) If , then V120iV

2.14043.18)2.1401536.0)(120(oV V and the magnitude is 18.43 V.

Chapter 9, Solution 80. 4.75j)10200)(60)(2(jLjmH200 3-

)0120(4.75j50R

4.75j4.75j50R

4.75jio VV

(a) When 100R ,

69.2688.167)0120)(904.75(

)0120(4.75j150

4.75joV

oV 53.89 63.31 V

Page 424: Solution Manual for Fundamentals of Electric Circuits

(b) When 0R ,

45.5647.90)0120)(904.75(

)0120(4.75j50

4.75joV

oV 100 33.55 V

(c) To produce a phase shift of 45 , the phase of = 90 + 0 = 45 . oVHence, = phase of (R + 50 + j75.4) = 45 . For to be 45 , R + 50 = 75.4 Therefore, R = 25.4

Chapter 9, Solution 81.

Let Z , 11 R2

22 Cj1

RZ , 33 RZ , and x

xx Cj1

RZ .

21

3x Z

ZZ

Z

22

1

3

xx Cj

1R

RR

Cj1

R

)600(400

1200R

RR

R 21

3x 1.8 k

)103.0(1200400

CRR

CC1

RR

C1 6-

23

1x

21

3

x 0.1 F

Chapter 9, Solution 82.

)1040(2000100

CRR

C 6-s

2

1x 2 F

Chapter 9, Solution 83.

)10250(1200500

LRR

L 3-s

1

2x 104.17 mH

Page 425: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 84.

Let s

11 Cj1

||RZ , 22 RZ , 33 RZ , and . xxx LjRZ

1CRjR

Cj1

R

CjR

s1

1

s1

s

1

1Z

Since 21

3x Z

ZZ

Z ,

)CRj1(R

RRR

1CRjRRLjR s1

1

32

1

s132xx

Equating the real and imaginary components,

1

32x R

RRR

)CR(R

RRL s1

1

32x implies that

s32x CRRL

Given that , k40R1 k6.1R 2 , k4R 3 , and F45.0Cs

k16.0k40

)4)(6.1(R

RRR

1

32x 160

)45.0)(4)(6.1(CRRL s32x 2.88 H Chapter 9, Solution 85.

Let , 11 RZ2

22 Cj1

RZ , 33 RZ , and 4

44 Cj1

||RZ .

jCRRj-

1CRjR

44

4

44

44Z

Since 324121

34 ZZZZZ

ZZ

Z ,

Page 426: Solution Manual for Fundamentals of Electric Circuits

223

44

14

Cj

RRjCR

RRj-

2

3232

424

24414

CjR

RR1CR

)jCR(RRj-

Equating the real and imaginary components,

3224

24

241 RR

1CRRR

(1)

2

324

24

24

241

CR

1CRCRR

(2) Dividing (1) by (2),

2244

CRCR

1

4422

2

CRCR1

4422 CRCR1

f2

4242 CCRR21

f

Chapter 9, Solution 86.

84j-1

95j1

2401

Y

0119.0j01053.0j101667.4 3-Y

2.183861.41000

37.1j1667.410001

YZ

Z = 228 -18.2

Page 427: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 87.

)102)(102)(2(j-

50Cj

150 6-31Z

79.39j501Z

)1010)(102)(2(j80Lj80 -33

2Z

66.125j802Z

1003Z

321

1111ZZZZ

66.125j801

79.39j501

10011

Z

)663.5j605.3745.9j24.1210(101 3-

Z

3-10)082.4j85.25( 97.81017.26 -3

Z = 38.21 -8.97

Chapter 9, Solution 88.

(a) 20j12030j20j-Z

Z = 120 �– j10

(b) If the frequency were halved, Cf2

1C1

Lf2L

would cause the capacitive

impedance to double, while would cause the inductive impedance to halve. Thus,

40j12015j40j-Z Z = 120 �– j65

Page 428: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 89.

Cj

1R||LjinZ

C1

LjR

RLjCL

Cj1

LjR

Cj1

RLj

inZ

22

in

C1

LR

C1

LjRRLjCL

Z

To have a resistive impedance, 0)Im( inZ . Hence,

0C1

LCL

RL 2

C1

LCR 2

1LCCR 2222

C1CR

L 2

222

(1) Ignoring the +1 in the numerator in (1),

)1050()200(CRL 9-22 2 mH Chapter 9, Solution 90.

Let , 0145sV L377jL)60)(2(jLjX

jXR800145

jXR80sV

I

Page 429: Solution Manual for Fundamentals of Electric Circuits

jXR80)145)(80(

801 IV

jXR80)145)(80(

50 (1)

jXR80)0145)(jXR(

)jXR(o IV

jXR80)145)(jXR(

110 (2)

From (1) and (2),

jXR80

11050

511

)80(jXR

30976XR 22 (3)

From (1),

23250

)145)(80(jXR80

53824XRR1606400 22

47424XRR160 22 (4)

Subtracting (3) from (4),

R16448R160 102.8 From (3),

204081056830976X2

LL37786.142X 0.3789 H

Page 430: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 91.

Lj||RCj

1inZ

LjRLRj

Cj-

inZ

222

222

LRLRjRL

Cj-

To have a resistive impedance, 0)Im( inZ . Hence,

0LR

LRC1-

222

2

222

2

LRLR

C1

22

222

LRLR

C

where 7102f2

)109)(1020)(104()10400)(104(109

C 46142

121424

nF72

169C 2

2

C = 235 pF

Chapter 9, Solution 92.

(a) oo

o

o xYZ

Z 5.134.4711048450

751006

(b) ooo xxZY 5.612121.0104845075100 6

Page 431: Solution Manual for Fundamentals of Electric Circuits

Chapter 9, Solution 93.

Ls 2 ZZZZ )9.186.05.0(j)2.238.01(Z

20j25Z

66.3802.320115S

L ZV

I

LI 3.592 -38.66 A

Page 432: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 1.

1

45-10)45tcos(10 60-5)30tsin(5

jLjH1

j-Cj

1F1

The circuit becomes as shown below.

Vo3

+

10 -45 V +

2 Io

j

5 -60 V

Applying nodal analysis,

j-j)60-5(

3)45-10( ooo VVV

oj60-1545-10j V 9.24715.73150-1545-10oV

Therefore, )t(vo 15.73 cos(t + 247.9 ) V

Chapter 10, Solution 2.

1045-4)4t10cos(4

150-20)3t10sin(20 10jLjH1

5j-2.0j

1Cj

1F02.0

Page 433: Solution Manual for Fundamentals of Electric Circuits

The circuit becomes that shown below.

Io

Vo10

j10 -j5 20 -150 V +

4 -45 A

Applying nodal analysis,

5j-10j45-4

10)150-20( ooo VVV

o)j1(1.045-4150-20 V

98.150816.2)j1(j

45-4150-210j

oo

VI

Therefore, )t(io 2.816 cos(10t + 150.98 ) A Chapter 10, Solution 3.

402)t4cos(2

-j1690-16)t4sin(16 8jLjH2

3j-)121)(4(j

1Cj

1F121

The circuit is shown below.

Vo

+

2 0 A-j16 V

4 -j3 6

1

j8

Page 434: Solution Manual for Fundamentals of Electric Circuits

Applying nodal analysis,

8j612

3j416j- ooo VVV

o8j61

3j41

123j4

16j-V

02.35-835.388.12207.115.33-682.4

04.0j22.156.2j92.3

oV

Therefore, )t(vo 3.835 cos(4t �– 35.02 ) V

Chapter 10, Solution 4.

16 4,10-16)10t4sin(4jLjH1

j-)41)(4(j

1Cj

1F25.0

j4

1

Ix -j

16 -10 V +

V1

0.5 Ix

+

Vo

j121

4j)10-16( 1

x1 V

IV

But

4j)10-16( 1

x

VI

So, j18j

))10-16((3 11 VV

Page 435: Solution Manual for Fundamentals of Electric Circuits

4j1-10-48

1V

Using voltage division,

04.69-232.8)4j1j)(--(1

10-48j1

11o VV

Therefore, )t(vo 8.232 sin(4t �– 69.04 ) V

Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be Vx and we get:

03j

V2j

V4

3010I5.0V xxxx

xx

xxx V5.0j2j

VIbut3030I5.1V)4j6j3(

Combining these equations we get:

A38.97615.425.1j3

30305.0jI

25.1j33030Vor3030V)75.0j2j3(

x

xx

Chapter 10, Solution 6. Let Vo be the voltage across the current source. Using nodal analysis we get:

010j20

V3

20V4V oxo where ox V

10j2020V

Combining these we get:

30j60V)35.0j1(010j20

V3

10j20V4

20V

oooo

5.0j2)3(20Vor

5.0j230j60V xo 29.11 �–166 V.

Page 436: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 7. At the main node,

501

30j

20j401V

3j196.520j40

058.31j91.11550V

30jV306

20j40V15120 o

o

V 15408.1240233.0j04.0

7805.4j1885.3V o

Chapter 10, Solution 8.

,200

20j1.0x200jLjmH100

100j10x50x200j

1Cj

1F506

The frequency-domain version of the circuit is shown below.

0.1 Vo 40 V1 Io V2 + -j100

6 o15 20 Vo j20 -

At node 1,

40VV

100jV

20V

V1.0156 21111

o

or 21 025.0)01.0025.0(5529.17955.5 VVjj (1)

Page 437: Solution Manual for Fundamentals of Electric Circuits

At node 2,

212

121 V)2j1(V30

20jV

V1.040

VV (2)

From (1) and (2),

BAVor0

)5529.1j7955.5(VV

)2j1(3025.0)01.0j025.0(

2

1

Using MATLAB,

V = inv(A)*B leads to V 09.1613.110,23.12763.70 21 jVj

o21

o 17.82276.740

VVI

Thus, A )17.82t200cos(276.7)t(i o

o Chapter 10, Solution 9.

10 33 10,010)t10cos(10jLjmH10

20j-)1050)(10(j

1Cj

1F50 6-3

Consider the circuit shown below.

V120 V2

-j20

20

Io

30

+

Vo 4 Io+

10 0 V

j10 At node 1,

20j-202010 2111 VVVV

21 j)j2(10 VV (1)

Page 438: Solution Manual for Fundamentals of Electric Circuits

At node 2,

10j3020)4(

20j-2121 VVVV

, where 20

1o

VI has been substituted.

21 )8.0j6.0()j4-( VV

21 j4-8.0j6.0

VV (2)

Substituting (2) into (1)

22 jj4-

)8.0j6.0)(j2(10 VV

or 2.26j6.0

1702V

26.70154.62.26j6.0

170j3

310j30

302o VV

Therefore, )t(vo 6.154 cos(103 t + 70.26 ) V Chapter 10, Solution 10.

2000,100j10x50x2000jLj mH 50 3

250j10x2x2000j

1Cj

1F26

Consider the frequency-domain equivalent circuit below. V1 -j250 V2 36<0o 2k j100 0.1V1 4k

Page 439: Solution Manual for Fundamentals of Electric Circuits

At node 1,

212111 V004.0jV)006.0j0005.0(36

250jVV

100jV

2000V

36 (1)

At node 2,

212

121 V)004.0j00025.0(V)004.0j1.0(0

4000V

V1.0250j

VV (2)

Solving (1) and (2) gives

o2o 43.931.89515.893j6.535VV

vo (t) = 8.951 sin(2000t +93.43o) kV

Chapter 10, Solution 11.

cos( 2,01)t260-8)30t2sin(8

2jLjH1 j-)21)(2(j

1Cj

1F21

4jLjH2 2j-)41)(2(j

1Cj

1F41

Consider the circuit below.

-j

-j

2 Io

2 Io

2 Io 2 Io 2 I

2 2

2 Io

2

Page 440: Solution Manual for Fundamentals of Electric Circuits

At node 1,

2jj-2)60-8( 2111 VVVV

21 j)j1(60-8 VV (1)

At node 2,

02j4j)60-8(

2j1 221 VVV

12 5.0j60-4 VV (2)

Substituting (2) into (1), 1)5.1j1(30460-81 V

5.1j130460-81

1V

46.55-024.5j5.1

30460-81j-1

o

VI

Therefore, )t(io 5.024 cos(2t �– 46.55 ) Chapter 10, Solution 12.

20 1000,020)t1000sin(10jLjmH10

20j-)1050)(10(j

1Cj

1F50 6-3

The frequency-domain equivalent circuit is shown below.

2 Io

-j20 20

V2V1

20 0 A

10

Io

j10

Page 441: Solution Manual for Fundamentals of Electric Circuits

At node 1,

1020220 211

o

VVVI ,

where

10j2

o

VI

102010j2

20 2112 VVVV

21 )4j2(3400 VV (1)

At node 2,

10j20j-1010j2 22212 VVVVV

21 )2j3-(2j VV

or (2) 21 )5.1j1( VVSubstituting (2) into (1),

222 )5.0j1()4j2()5.4j3(400 VVV

5.0j1400

2V

6.116-74.35)5.0j1(j

4010j

2o

VI

Therefore, )t(io 35.74 sin(1000t �– 116.6 ) A Chapter 10, Solution 13. Nodal analysis is the best approach to use on this problem. We can make our work easier by doing a source transformation on the right hand side of the circuit. �–j2 18 j6

+

50 0º V

+

+

Vx 3

40 30º V

Page 442: Solution Manual for Fundamentals of Electric Circuits

06j1850V

3V

2j3040V xxx

which leads to Vx = 29.36 62.88 A.

Chapter 10, Solution 14.

At node 1,

30204j10

02j-

0 1211 VVVV

100j2.1735.2j)5.2j1(- 21 VV (1)

At node 2,

30204j5j-2j

1222 VVVV

100j2.1735.2j5.5j- 12 VV (2)

Equations (1) and (2) can be cast into matrix form as

3020030200-

5.5j-5.2j5.2j5.2j1

2

1

VV

38.15-74.205.5j205.5j-5.2j5.2j5.2j1

120600)30200(3j5.5j-302005.2j30200-

1

7.1081020)5j1)(30200(302005.2j30200-5.2j1

2

38.13593.2811V

08.12418.4922V

Page 443: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 15.

We apply nodal analysis to the circuit shown below.

5 A

I

V2V1

-j2 +

2

2 I

j

-j20 V 4

At node 1,

j2j-5

220j- 2111 VVVV

21 j)5.0j5.0(10j5- VV (1)

At node 2,

4j25 221 VVV

I ,

where 2j-1V

I

j25.05

2V V1

(2) Substituting (2) into (1),

1)j1(5.0j25.0

5j10j5- V

4j140j

20j10-)j1( 1V

1740j

17160

20j10-)45-2( 1V

5.31381.151V

Page 444: Solution Manual for Fundamentals of Electric Circuits

)5.31381.15)(905.0(2j-1V

I

I 7.906 43.49 A

Chapter 10, Solution 16.

At node 1,

5j-10202j 21211 VVVVV

21 )4j2()4j3(40j VV At node 2,

10jj1

5j-1022121 VVVVV

21 )j1()2j1(-)j1(10 VV Thus,

2

1

j1j2)(1-j2)(12-4j3

)j1(1040j

VV

11.31-099.5j5j1j2)(1-j2)(12-4j3

96.12062.116100j60j1j)(110j2)(12-40j

1

29.129142.13j110-90j)(110j2)(1-

40j4j32

1

1V 22.87 132.27 V

22V 27.87 140.6 V

Page 445: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 17.

Consider the circuit below.

At node 1,

Io

V2V1

j4

+

100 20 V 2

-j2

1

3

234j20100 2111 VVVV

21 2j)10j3(

320100 V

V (1)

At node 2,

2j-2120100 2212 VVVV

21 )5.0j5.1(5.0-20100 VV (2)

From (1) and (2),

2

1

2j-310j1)j3(5.05.0-

2010020100

VV

5.4j1667.02j-310j1

5.0j5.15.0-

2.286j45.55-2j-20100

5.0j5.1201001

5.364j95.26-20100310j1201005.0-

2

Page 446: Solution Manual for Fundamentals of Electric Circuits

08.13-74.6411V

35.6-17.8122V

9j3333.031.78j5.28-

222121

o

VVI

oI 9.25 -162.12

Chapter 10, Solution 18.

Consider the circuit shown below.

j6

2

+

Vo -j2

4 j5

4 45 A -j

+

Vx 2 Vx

8 V1 V2

At node 1,

6j82454 211 VVV

21 )3j4()3j29(45200 VV (1)

At node 2,

2j5j4j-2

6j822

x21 VV

VVV

, where VV

1x

21 )41j12()3j104( VV

21 3j10441j12

VV (2)

Substituting (2) into (1),

22 )3j4(3j104)41j12(

)3j29(45200 VV

Page 447: Solution Manual for Fundamentals of Electric Circuits

2)17.8921.14(45200 V

17.8921.1445200

2V

222o 258j6-

3j42j-

2j5j42j-

VVVV

17.8921.1445200

2513.23310

oV

oV 5.63 189 V

Chapter 10, Solution 19.

We have a supernode as shown in the circuit below.

j2

-j4

+

Vo

V2V3

V1 4

2 0.2 Vo

Notice that . 1o VVAt the supernode,

2j24j-4311223 VVVVVV

321 )2j1-()j1()2j2(0 VVV (1)

At node 3,

42j2.0 2331

1

VVVVV

0)2j1-()2j8.0( 321 VVV (2)

Subtracting (2) from (1),

Page 448: Solution Manual for Fundamentals of Electric Circuits

21 j2.10 VV (3) But at the supernode,

21 012 VV or (4) 1212 VVSubstituting (4) into (3),

)12(j2.10 11 VV

o1 j2.112j

VV

81.39562.19012

oV

oV 7.682 50.19 V

Chapter 10, Solution 20.

The circuit is converted to its frequency-domain equivalent circuit as shown below.

R

Cj1

+

Voj L+

Vm 0

Let LC1

Lj

Cj1

Lj

CL

Cj1

||Lj 2Z

m2m

2

2

mo VLj)LC1(R

LjV

LC1Lj

R

LC1Lj

VR Z

ZV

)LC1(RL

tan90L)LC1(R

VL2

1-22222

moV

Page 449: Solution Manual for Fundamentals of Electric Circuits

If , then

AoV

A22222

m

L)LC1(R

VL

and )LC1(R

Ltan90 2

1-

Chapter 10, Solution 21.

(a) RCjLC1

1

Cj1

LjR

Cj1

2i

o

VV

At , 011

i

o

VV

1

As , i

o

VV

0

At LC1

,

LC1

jRC

1

i

o

VV

CL

Rj-

(b) RCjLC1

LC

Cj1

LjR

Lj2

2

i

o

VV

At , 0i

o

VV

0

As , 11

i

o

VV

1

At LC1

,

LC1

jRC

1

i

o

VV

CL

Rj

Page 450: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 22.

Consider the circuit in the frequency domain as shown below.

R1

Cj1

R2 +

Vo

j L

+

Vs

Let Cj

1||)LjR( 2Z

LCRj1LjR

Cj1

LjR

)LjR(Cj

1

22

2

2

2

Z

CRjLC1LjR

R

CRjLC1LjR

R2

22

1

22

2

1s

o

ZZ

VV

s

o

VV

)CRRL(jLCRRRLjR

2112

21

2

Chapter 10, Solution 23.

0CVj

Cj1Lj

VR

VV s

s2 VRCVj1LC

RCVjV

s2

232VV

LC1RLCjRCjRCjLC1

Page 451: Solution Manual for Fundamentals of Electric Circuits

)LC2(RCjLC1V)LC1(V 22

s2

Chapter 10, Solution 24.

For mesh 1,

22

121

s Cj1

Cj1

Cj1

IIV (1)

For mesh 2,

22

12 Cj

1LjR

Cj1

0 II (2)

Putting (1) and (2) into matrix form,

2

1

22

221s

Cj1LjR

Cj1

Cj1

Cj1

Cj1

0 IIV

212

221 CC1

Cj1

LjRCj1

Cj1

2s1 Cj

1LjRV and

2

s2 Cj

V

11I

212

221

2s

CC1

Cj1

LjRCj1

Cj1

Cj1

LjRV

22I

212

221

2

s

CC1

Cj1

LjRCj1

Cj1

CjV

Chapter 10, Solution 25.

2

010)t2cos(10

Page 452: Solution Manual for Fundamentals of Electric Circuits

-j690-6)t2sin(6 4jLjH2

2j-)41)(2(j

1Cj

1F25.0

The circuit is shown below.

4 j4

For loop 1,

I2+

10 0 V +

-j2 I1

Io

6 -90 V

02j)2j4(10- 21 II

21 j)j2(5 II (1) For loop 2,

0)6j-()2j4j(2j 21 II 321 II (2)

In matrix form (1) and (2) become

35

11jj2

2

1

II

)j1(2 , 3j51 , 3j12

45414.1j1)j1(2

42121o III

Therefore, )t(io 1.414 cos(2t + 45 ) A Chapter 10, Solution 26.

We apply mesh analysis to the circuit shown below. For mesh 1,

0204010- 21 II

21 241 II (1) For the supermesh,

0)10j30(20)20j20( 312 III 0)j3()2j2(2- 321 III (2)

Page 453: Solution Manual for Fundamentals of Electric Circuits

At node A, 21o III (3)

At node B, o32 4III (4)

Substituting (3) into (4) 2132 44 IIII

123 45 III (5) Substituting (5) into (2) gives

21 )3j17()4j14(-0 II (6) From (1) and (6),

2

1

3j17)4j14(-2-4

01

II

4j40

3j17j3170

2-11 , 4j14

0j4)(14-14

2

4j408j245

45 12123 III

25.70154.6j10

)4j1(1530 3o IV

Therefore, )t(vo 6.154 cos(103 t + 70.25 ) V Chapter 10, Solution 27.

For mesh 1,

020j)20j10j(3040- 21 II

21 2jj-304 II (1) For mesh 2,

020j)20j40(050 12 II

21 )2j4(2j-5 II (2) From (1) and (2),

2

1

)2j4(-2j-2jj-

5304

II

56.116472.4j42-

Page 454: Solution Manual for Fundamentals of Electric Circuits

8.21101.2110j)2j4)(304(-1

27.15444.412085j-2

11I 4.698 95.24 A

2

2I 0.9928 37.71 A

Chapter 10, Solution 28.

25.0j4x1j

1Cj

1F1,4jLjH1

The frequency-domain version of the circuit is shown below, where

o2

o1 3020V,010V .

1 j4 j4 1 -j0.25 + + V1 I1 1 I2 V2 - -

o2

o1 3020V,010V

Applying mesh analysis,

21 I)25.0j1(I)75.3j2(10 (1)

(2) 21

o I)75.3j2(I)025.0j1(3020 From (1) and (2), we obtain

Page 455: Solution Manual for Fundamentals of Electric Circuits

2

1II

75.3j225.0j125.0j175.3j2

10j32.1710

Solving this leads to

o2

o1 1536505.4111.2j1438.4I,69.356747.19769.0j3602.1I

Hence,

A )15346cos(651.4i A, )69.35t4cos(675.1i o2

o1

Chapter 10, Solution 29.

For mesh 1,

02030)j2()5j5( 21 II

21 )j2()5j5(2030 II (1) For mesh 2,

0)j2()6j3j5( 12 II

21 )3j5()j2(-0 II (2) From (1) and (2),

2

1

j3-5j)(2-j)(2-5j5

02030

II

21.948.376j37

96.10-175)96.30-831.5)(2030(1 56.4608.67)56.26356.2)(2030(2

1

1I 4.67 -20.17 A

2

2I 1.79 37.35 A

Page 456: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 30. Consider the circuit shown below.

For mesh 1,

I3

I1

I2

Io

j4

+

10 0 V 2

3

1

-j2

321 34j)4j3(20100 III (1) For mesh 2,

321 2j)4j3(4j-0 III (2) For mesh 3,

321 )2j5(23-0 III (3) Put (1), (2), and (3) into matrix form.

00

20100

j2-52-3-j2-j43j4-3-j4-4j3

3

2

1

III

30j106j2-52-3-

j2-j43j4-3-j4-4j3

)26j8)(20100(j2-503-

j2-0j4-3-201004j3

2

)20j9)(20100(02-3-0j43j4-

20100j4-4j3

3

30j106)6j1)(20100(23

23o III

oI 5.521 -76.34 A

Page 457: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 31.

Consider the network shown below.

80 j60 20 Io

-j40 I1 I3-j40 + 100 120 V +

I2 60 -30 V

For loop 1,

040j)40j80(20100- 21 II

21 4j)j2(42010 II (1) For loop 2,

040j)80j60j(40j 321 III

321 220 III (2)

For loop 3, 040j)40j20(30-60 23 II

32 )2j1(24j30-6- II (3) From (2),

123 22 III Substituting this equation into (3),

21 )2j1()2j1(2-30-6- II (4) From (1) and (4),

2

1

2j1)2j1(2-4j)j2(4

30-6-12010

II

3274.3720j322j14j2-

4j-4j8

44.9325.8211.82j928.4-30-6-4j2-

120104j82

2

2o II 2.179 61.44 A

Page 458: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 32.

Consider the circuit below.

j4

3 Vo -j2

Io

I1 I2 +

+

Vo 2 4 -30 V

For mesh 1,

03)30-4(2)4j2( o1 VI where )30-4(2 1o IV Hence,

0)30-4(630-8)4j2( 11 II

1)j1(30-4 I or 15221I

)30-4)(2(2j-

32j-

31

oo I

VI

)152230-4(3joI

oI 8.485 15 A

32j- o

o

IV 5.657 -75 V

Chapter 10, Solution 33.

Consider the circuit shown below.

Page 459: Solution Manual for Fundamentals of Electric Circuits

5 A

I1 +

2

I

-j2 2 I I2

I4

I3

j

-j20 V 4

For mesh 1,

02j)2j2(20j 21 II 10j-j)j1( 21 II (1)

For the supermesh,

0j42j)2jj( 4312 IIII (2) Also,

)(22 2123 IIIII

213 2 III (3) For mesh 4,

54I (4) Substituting (3) and (4) into (2),

5j)j4-()2j8( 21 II (5) Putting (1) and (5) in matrix form,

5j10j-

j42j8jj1

2

1

II

5j3- , 40j5-1 , 85j15-2

5j3-45j1021

21 III 7.906 43.49 A

Page 460: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 34.

The circuit is shown below.

-j2

j4

8

10

3 A I2

I3

I1+

40 90 V

5

Io

20

j15

For mesh 1,

0)4j10()2j8()2j18(40j- 321 III (1) For the supermesh,

0)2j18()19j30()2j13( 132 III (2) Also,

332 II (3) Adding (1) and (2) and incorporating (3),

0)15j20()3(540j- 33 II

48.38465.13j58j3

3I

3o II 1.465 38.48 A

Page 461: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 35.

Consider the circuit shown below.

4 j2

1

I1

I3

I2+

-j3 8

-j4 A

10

20 V

-j5

For the supermesh,

0)3j9()8j11(820- 321 III (1) Also,

4j21 II (2) For mesh 3,

0)3j1(8)j13( 213 III (3) Substituting (2) into (1),

32j20)3j9()8j19( 32 II (4) Substituting (2) into (3),

32j)j13()3j9(- 32 II (5) From (4) and (5),

32j32j20

j13)3j9(-)3j9(-8j19

3

2

II

69j167 , 148j3242

45.22-69.18055.24-2.356

69j167148j3242

2I

2I 1.971 -2.1 A

Page 462: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 36.

Consider the circuit below.

j4 -j3

I1 I2

I3

+

Vo

2 2

+ 4 90 A 2 12 0 V

2 0 A

Clearly, 4j9041I and 2-3I

For mesh 2, 01222)3j4( 312 III

01248j)3j4( 2I

64.0j52.3-3j48j16-

2I

Thus, 28.9j04.7)64.4j52.3)(2()(2 21o IIV

oV 11.648 52.82 V Chapter 10, Solution 37. I1 + Ix 120 V Z o90 - I2 Z=80-j35 Iz - Iy

Page 463: Solution Manual for Fundamentals of Electric Circuits

o30120 V Z + I3 For mesh x,

120jZIZI zx (1) For mesh y,

60j92.10330120ZIZI ozy (2)

For mesh z,

0ZI3ZIZI zyx (3) Putting (1) to (3) together leads to the following matrix equation:

BAI0

60j92.103120j

III

)105j240()35j80()35j80()35j80()35j80(0)35j80(0)35j80(

z

y

x

Using MATLAB, we obtain

j0.15251.3657-j0.954- 2.1806-j1.4115 1.9165-

B*inv(A)I

A 6.1433802.24115.1j9165.1II ox1

A 37.963802.23655.2j2641.0III oxy2

A 63.233802.2954.0j1806.2II oy3

Chapter 10, Solution 38.

Consider the circuit below.

Page 464: Solution Manual for Fundamentals of Electric Circuits

Io

I1

j2

1

2 -j4

10 90 V

1

I2

I3 I44 0 A

+ 2 0 A

A

Clearly, 21I (1)

For mesh 2, 090104j2)4j2( 412 III (2)

Substitute (1) into (2) to get 5j22j)2j1( 42 II

For the supermesh, 04j)4j1(2j)2j1( 2413 IIII

4j)4j1()2j1(4j 432 III (3) At node A,

443 II (4) Substituting (4) into (3) gives

)3j1(2)j1(2j 42 II (5) From (2) and (5),

6j25j2

j12j2j2j1

4

2

II

3j3 , 11j91

)j10-(31

3j3)11j9(--

- 12o II

oI 3.35 174.3 A

Page 465: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 39.

For mesh 1,

o321 6412I15jI8I)15j28( (1)

For mesh 2, 0I16jI)9j8(I8 321 (2)

For mesh 3, 0I)j10(I16jI15j 321 (3) In matrix form, (1) to (3) can be cast as

BAIor006412

III

)j10(16j15j16j)9j8(8

15j8)15j28( o

3

2

1

Using MATLAB, I = inv(A)*B

A 6.1093814.03593.0j128.0I o1

A 4.1243443.02841.0j1946.0I o2

A 42.601455.01265.0j0718.0I o3

Page 466: Solution Manual for Fundamentals of Electric Circuits

A 5.481005.00752.0j0666.0III o21x

Chapter 10, Solution 40.

Let i , where i is due to the dc source and is due to the ac source. For

, consider the circuit in Fig. (a). 2O1OO ii 1O 2Oi

1Oi

4 2

iO1 +

8 V

(a)Clearly,

A428i 1O For , consider the circuit in Fig. (b). 2Oi

4 2

10 0 V j4

IO2

+

(b)If we transform the voltage source, we have the circuit in Fig. (c), where 342||4 .

2 2.5 0 A 4

IO2

j4

(c) By the current division principle,

)05.2(4j34

342OI

56.71-79.075.0j25.02OI

Page 467: Solution Manual for Fundamentals of Electric Circuits

Thus, A)56.71t4cos(79.0i 2O Therefore,

2O1OO iii 4 + 0.79 cos(4t �– 71.56 ) A

Chapter 10, Solution 41. Let vx = v1 + v2. For v1 we let the DC source equal zero. 5 1

j100V)j55j1(tosimplifieswhich01

Vj

V5

20V1

111

V1 = 2.56 �–39.8 or v1 = 2.56sin(500t �– 39.8) V

Setting the AC signal to zero produces:

+ �– 20 0 �–j

+

V1

1

+ �– 6 V

+

V2

5

The 1-ohm resistor in series with the 5-ohm resistor creating a simple voltage divider yielding: v2 = (5/6)6 = 5 V.

vx = {2.56sin(500t �– 39.8) + 5} V.

Page 468: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 42. Let ix = i1 + i2, where i1 and i2 which are generated by is and vs respectively. For i1 we let is = 6sin2t A becomes Is = 6 0, where =2.

63.41983.431.3j724.32j52j1126

4j22j34j2I1

i1= 4.983sin(2t �– 41.63) A �–j4 2

j2

3

i1 is For i2, we transform vs = 12cos(4t �– 30) into the frequency domain and get Vs = 12 �–30.

Thus, 2.8385.54j32j2

30122I or i2 = 5.385cos(4t + 8.2) A

�–j2 2

+

Vs

j4

3

i2

ix = [5.385cos(4t + 8.2) + 4.983sin(2t �– 41.63)] A.

Page 469: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 43.

Let i , where i is due to the dc source and is due to the ac source. For , consider the circuit in Fig. (a).

2O1OO ii 1O 2Oi

1Oi

4 2

iO1 +

8 V

(a)Clearly,

A428i 1O For , consider the circuit in Fig. (b). 2Oi

4 2

10 0 V j4

IO2

+

(b)If we transform the voltage source, we have the circuit in Fig. (c), where 342||4 .

2 2.5 0 A 4

IO2

j4

(c) By the current division principle,

)05.2(4j34

342OI

56.71-79.075.0j25.02OI Thus, A)56.71t4cos(79.0i 2O Therefore,

2O1OO iii 4 + 0.79 cos (89)(4t �– 71.56 ) A

Page 470: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 44. Let v , where v21x vv 1 and v2 are due to the current source and voltage source respectively.

For v1 , , 6 30jLjH 5 The frequency-domain circuit is shown below. 20 j30 + 16 V1 Is -

Let o5.1631.12497.3j8.1130j36

)30j20(16)30j20//(16Z

V )5.26t6cos(7.147v5.267.147)5.1631.12)(1012(ZIV o1

ooos1

For v2 , , 2 10jLjH 5 The frequency-domain circuit is shown below. 20 j10 + 16 V2 + Vs

- - -

Page 471: Solution Manual for Fundamentals of Electric Circuits

Using voltage division,

V )52.15t2sin(41.21v52.1541.2110j36

)050(16V10j2016

16V o2

oo

s2

Thus,

V )52.15t2sin(41.21)5.26t6cos(7.147v oox

Chapter 10, Solution 45.

Let I , where I is due to the voltage source and is due to the current source. For I , consider the circuit in Fig. (a).

21o II

1

1 2I

10 IT

+ 20 -150 V -j5 j10

I1

(a)

10j-5j-||10j

j1150-2

10j10150-20

TI

Using current division,

150-)j1(-j1

150-25j5j-

5j10j5j-

T1 II

For , consider the circuit in Fig. (b). 2I

-j5 j10

I2

4 -45 A 10

(b)

j210j-

5j-||10

Using current division,

Page 472: Solution Manual for Fundamentals of Electric Circuits

45-)j1(2-)45-4(10j)j2(10j-

)j2(10j-2I

022105-2-21o III 98.150816.2366.1j462.2-oI

Therefore, oi 2.816 cos(10t + 150.98 ) A Chapter 10, Solution 46.

Let v , where , , and are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For consider the circuit in Fig. (a).

321o vvv 1v 2v 3v

1v

2 H 6

1/12 F +

+

v1 10 V

(a) The capacitor is open to dc, while the inductor is a short circuit. Hence,

V10v1 For , consider the circuit in Fig. (b). 2v

2 4jLjH2

6j-)12/1)(2(j

1Cj

1F

121

4 0 A+

V2 -j6 6 j4

(b)

Applying nodal analysis,

2222

4j

6j

61

4j6j-64 V

VVV

56.2645.215.0j1

242V

Page 473: Solution Manual for Fundamentals of Electric Circuits

Hence, V)56.26t2sin(45.21v2 For , consider the circuit in Fig. (c). 3v

3 6jLjH2

4j-)12/1)(3(j

1Cj

1F

121

6 j6

+

V3 -j4 + 12 0 V

(c)

At the non-reference node,

6j4j-612 333 VVV

56.26-73.105.0j1

123V

Hence, V)56.26t3cos(73.10v3 Therefore, ov 10 + 21.45 sin(2t + 26.56 ) + 10.73 cos(3t �– 26.56 ) V Chapter 10, Solution 47.

Let i , where i , i , and are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For , consider the circuit in Fig. (a).

321o iii 1 2 3i

1i

+

2

1 1/6 F 2 H 24 V

i1

4

(a) Since the capacitor is an open circuit to dc,

Page 474: Solution Manual for Fundamentals of Electric Circuits

A424

24i1

For , consider the circuit in Fig. (b). 2i

1 2jLjH2

6j-Cj

1F

61

1 j2 -j6

2 I2I1+

10 -30 V

I2

4

(b)For mesh 1,

02)6j3(30-10- 21 II

21 2)j21(330-10 II (1)

For mesh 2,

21 )2j6(2-0 II

21 )j3( II (2)

Substituting (2) into (1)

215j1330-10 I 1.19504.02I

Hence, A)1.19tsin(504.0i2 For , consider the circuit in Fig. (c). 3i

3 6jLjH2

2j-)6/1)(3(j

1Cj

1F

61

1 j6 -j2

2 0 A2

I3

4

(c)

Page 475: Solution Manual for Fundamentals of Electric Circuits

2j3)2j1(2

)2j1(||2

Using current division,

3j13)2j1(2

2j3)2j1(2

6j4

)02(2j3

)2j1(2

3I

43.76-3352.03I Hence A)43.76t3cos(3352.0i3 Therefore, oi 4 + 0.504 sin(t + 19.1 ) + 0.3352 cos(3t �– 76.43 ) A Chapter 10, Solution 48.

Let i , where i is due to the ac voltage source, is due to the dc voltage source, and is due to the ac current source. For , consider the circuit in Fig. (a).

3O2O1OO iii

3Oi1O 2Oi

1Oi

2000 050)t2000cos(50

80j)1040)(2000(jLjmH40 3-

25j-)1020)(2000(j

1Cj

1F20 6-

I IO1

50 0 V 80 +

-j25

100

j80 60 (a)

3160)10060(||80

33j3230

25j80j316050

I

Using current division,

Page 476: Solution Manual for Fundamentals of Electric Circuits

9.454618010

31-

16080I80-

1O II

1.134217.01OI Hence, A)1.134t2000cos(217.0i 1O For , consider the circuit in Fig. (b). 2Oi

+

100

24 V

iO2

80

60

(b)

A1.01006080

24i 2O

For , consider the circuit in Fig. (c). 3Oi

4000 02)t4000cos(2

160j)1040)(4000(jLjmH40 3-

5.12j-)1020)(4000(j

1Cj

1F20 6-

-j12.5

80

60

I2

I1

I3

2 0 A

j160

IO3

100

(c) For mesh 1,

21I (1) For mesh 2,

Page 477: Solution Manual for Fundamentals of Electric Circuits

080160j)5.12j160j80( 312 III Simplifying and substituting (1) into this equation yields

32j8)75.14j8( 32 II (2) For mesh 3,

08060240 213 III Simplifying and substituting (1) into this equation yields

5.13 32 II (3) Substituting (3) into (2) yields

125.54j12)25.44j16( 3I

38.71782.125.44j16

125.54j123I

38.71782.1-- 33O II

Hence, A)38.7t4000sin(1782.1-i 3O Therefore, Oi 0.1 + 0.217 cos(2000t + 134.1 ) �– 1.1782 sin(4000t + 7.38 ) A Chapter 10, Solution 49.

200,308)30t200sin(8 j)105)(200(jLjmH5 3-

5j-)101)(200(j

1Cj

1mF1 3-

After transforming the current source, the circuit becomes that shown in the figure below.

5 3 I

+ 40 30 V

j

-j5

56.56472.44j8

30405jj35

3040I

i 4.472 sin(200t + 56.56 ) A

Page 478: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 50.

55 10,050)t10cos(50 40j)104.0)(10(jLjmH4.0 3-5

50j-)102.0)(10(j

1Cj

1F2.0 6-5

After transforming the voltage source, we get the circuit in Fig. (a).

j40

20 +

Vo -j50 2.5 0 A 80

(a)

Let 5j2

100j-50j-||20Z

and 5j2

250j-)05.2(s ZV

With these, the current source is transformed to obtain the circuit in Fig.(b).

Z j40

+

Vo +

80 Vs

(b)

By voltage division,

5j2250j-

40j805j2

100j-80

40j8080

so VZ

V

6.40-15.3642j36

)250j-(8oV

Therefore, ov 36.15 cos(105 t �– 40.6 ) V

Page 479: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 51.

The original circuit with mesh currents and a node voltage labeled is shown below.

Io

40 j10 -j20 4 -60 V 1.25 0 A

The following circuit is obtained by transforming the voltage sources.

Io

4 -60 V -j20 j10 40 1.25 0 A

Use nodal analysis to find . xV

x401

20j-1

10j1

025.160-4 V

x)05.0j025.0(464.3j25.3 V

61.1697.8429.24j42.8105.0j025.0

464.3j25.3xV

Thus, from the original circuit,

10j)29.24j42.81()20j64.34(

10j3040 x

1

VI

678.4j429.0-10j

29.4j78.46-1I 4.698 95.24 A

4029.24j42.31

40050x

2

VI

7.379928.06072.0j7855.02I 0.9928 37.7 A Chapter 10, Solution 52.

We transform the voltage source to a current source.

12j64j2060

sI

Page 480: Solution Manual for Fundamentals of Electric Circuits

The new circuit is shown in Fig. (a).

-j2

6

2

Is = 6 �– j12 A

-j3 j4

4

Ix

5 90 A

(a)

Let 8.1j4.24j8

)4j2(6)4j2(||6sZ

)j2(1818j36)8.1j4.2)(12j6(sss ZIV With these, we transform the current source on the left hand side of the circuit to a voltage source. We obtain the circuit in Fig. (b).

Zs -j2

4

-j3

+

Vs

Ix

j5 A

(b) Let )j12(2.02.0j4.22jso ZZ

207.6j517.15)j12(2.0

)j2(18

o

so Z

VI

With these, we transform the voltage source in Fig. (b) to a current source. We obtain the circuit in Fig. (c).

Zo

-j3

4

Ix

Io j5 A

(c)

Page 481: Solution Manual for Fundamentals of Electric Circuits

Using current division,

)207.1j517.15(2.3j4.62.0j4.2

)5j(3j4 o

o

ox I

ZZ

I

5625.1j5xI 5.238 17.35 A Chapter 10, Solution 53.

We transform the voltage source to a current source to obtain the circuit in Fig. (a).

4 j2

+

Vo 2

j4 -j3

5 0 A -j2

(a)

Let 6.1j8.02j4

8j2j||4sZ

j4)6.1j8.0)(5()05( ss ZV 8 With these, the current source is transformed so that the circuit becomes that shown in Fig. (b).

Zs -j3 j4

+

Vo +

Vs 2 -j2

(b)Let 4.1j8.03jsx ZZ

6154.4j0769.34.1j8.0

8j4

s

sx Z

VI

With these, we transform the voltage source in Fig. (b) to obtain the circuit in Fig. (c).

j4

+

Vo Zx -j2 2 Ix

(c)

Page 482: Solution Manual for Fundamentals of Electric Circuits

Let 5714.0j8571.04.1j8.28.2j6.1

||2 xy ZZ

7143.5j)5714.0j8571.0()6154.4j0769.3(yxy ZIV With these, we transform the current source to obtain the circuit in Fig. (d).

Zy j4

+ -j2

+

Vo Vy

(d)

Using current division,

2j4j5714.0j8571.0)7143.5j(2j-

2j4j2j-

yy

o VZ

V (3.529 �– j5.883) V

Chapter 10, Solution 54.

059.2224.133050

)30(50)30//(50 jjjx

j

We convert the current source to voltage source and obtain the circuit below. 13.24 �– j22.059

40 j20

+ + - 115.91 �–j31.06V I V - + -

134.95-j74.912 V

Applying KVL gives -115.91 + j31.058 + (53.24-j2.059)I -134.95 + j74.912 = 0

or 8055.17817.4059.224.53

97.10586.250j

jj

I

Page 483: Solution Manual for Fundamentals of Electric Circuits

But I)20j40(VV0VI)20j40(V s

V 15406.124)8055.1j7817.4)(20j40(05.31j91.115V o

which agrees with the result in Prob. 10.7. Chapter 10, Solution 55.

(a) To find , consider the circuit in Fig. (a). thZ

j20 10

Zth-j10

(a)

10j20j)10j-)(20j(

10)10j-(||20j10thN ZZ

20j10 22.36 -63.43 To find , consider the circuit in Fig. (b). thV

j20 10

+

Vth+

50 30 V -j10

(b)

)3050(10j20j

10j-thV -50 30 V

43.63-36.223050-

th

thN Z

VI 2.236 273.4 A

Page 484: Solution Manual for Fundamentals of Electric Circuits

(b) To find , consider the circuit in Fig. (c). thZ

-j5

j10 Zth

8

(c)

5j810j)5j8)(10j(

)5j8(||10jthN ZZ 10 26

To obtain , consider the circuit in Fig. (d). thV

-j5

Io

4 0 A

+

Vthj10 8

(d) By current division,

5j832

)04(5j10j8

8oI

5j8320j

10j oth IV 33.92 58 V

26105892.33

th

thN Z

VI 3.392 32 A

Page 485: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 56.

(a) To find , consider the circuit in Fig. (a). thZ

j4

-j2 Zth

6

(a)

4j62j4j

)2j-)(4j(6)2j-(||4j6thN ZZ

= 7.211 -33.69 By placing short circuit at terminals a-b, we obtain,

NI 2 0 A

)02()69.33-211.7(ththth IZV 14.422 -33.69 V

(b) To find , consider the circuit in Fig. (b). thZ

j10

-j5 60 Zth

30

(b)

2060||30

5j20)10j20)(5j-(

)10j20(||5j-thN ZZ

= 5.423 -77.47

Page 486: Solution Manual for Fundamentals of Electric Circuits

To find and , we transform the voltage source and combine the 30 and 60 resistors. The result is shown in Fig. (c).

thV NI

a

4 45 A -j5

j10

20 IN

b (c)

)454)(j2(52

)454(10j20

20NI

= 3.578 18.43 A

)43.18578.3()47.77-423.5(Nthth IZV = 19.4 -59 V

Chapter 10, Solution 57.

To find , consider the circuit in Fig. (a). thZ

5 -j10 2

j20 Zth

(a)

10j5)10j5)(20j(

2)10j5(||20j2thN ZZ

12j18 21.633 -33.7 To find , consider the circuit in Fig. (b). thV

5 -j10 2

+

Vth + j20 60 120 V

(b)

Page 487: Solution Manual for Fundamentals of Electric Circuits

)12060(2j1

4j)12060(

20j10j520j

thV

= 107.3 146.56 V

7.33-633.2156.1463.107

th

thN Z

VI 4.961 -179.7 A

Chapter 10, Solution 58.

Consider the circuit in Fig. (a) to find . thZ

Zth

-j6

j10 8

(a)

)j2(54j8

)6j8)(10j()6j8(||10jthZ

= 11.18 26.56 Consider the circuit in Fig. (b) to find . thV

+

Vth

Io

5 45 A

-j6

j10

8

(b)

)455(2j43j4

)455(10j6j8

6j8oI

)j2)(2()455)(3j4)(10j(

10j oth IV 55.9 71.56 V

Page 488: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 59.

The frequency-domain equivalent circuit is shown in Fig. (a). Our goal is to find and

across the terminals of the capacitor as shown in Figs. (b) and (c). thV

thZ

3 j 3 j

Zth

b

a+

Vo -j + 5 -60 A10 -45 V +

(b) (a)

3 j Zth

+

Vth 10 -45 V +

+

+

Vo

b

a

(d)

+

Vth5 -60 A -j

(c) From Fig. (b),

)3j1(103

j33j

j||3thZ

From Fig.(c),

0j

60-53

45-10 thth VV

3j1301545-10

thV

From Fig. (d),

Page 489: Solution Manual for Fundamentals of Electric Circuits

301545-10j

j-th

tho V

ZV

oV 15.73 247.9 V Therefore, ov 15.73 cos(t + 247.9 ) V Chapter 10, Solution 60.

(a) To find , consider the circuit in Fig. (a). thZ

10 -j4

b

a

j5 4

(a)

Zth

)4j24j-(||4)5j||104j-(||4thZ

2||4thZ 1.333 To find , consider the circuit in Fig. (b). thV

+

Vth j5 4

V2V1

+ 4 0 A20 0 V

10 -j4

(b) At node 1,

4j-5j1020 2111 VVVV

205.2j)5.0j1( 21 VV (1)

At node 2,

Page 490: Solution Manual for Fundamentals of Electric Circuits

44j-4 221 VVV

16j)j1( 21 VV (2)

Substituting (2) into (1) leads to

2)3j5.1(16j28 V

333.5j83j5.1

16j282V

Therefore,

2th VV 9.615 33.69 V

(b) To find , consider the circuit in Fig. (c). thZ

c dZth

j5 4

10 -j4

(c)

j210j

4||4j-)5j||104(||4j-thZ

)4j6(64j-

)4j6(||4j-thZ 2.667 �– j4

To find ,we will make use of the result in part (a). thV

)2j3()38(333.5j82V )j5()38(16j16j)j1( 21 VV

8j31621th VVV 9.614 56.31 V

Page 491: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 61. First, we need to find and across the 1 resistor. thV thZ

4 -j3 j8 6

Zth

(a) From Fig. (a),

8.0j4.45j10

)8j6)(3j4()8j6(||)3j4(thZ

thZ 4.472 -10.3

4 -j3 j8 6

+

Vth

+ 2 A -j16 V

(b) From Fig. (b),

8j62

3j416j- thth VV

45.43-93.204.0j22.056.2j92.3

thV

43.8-46.545.43-93.20

1 th

tho Z

VV

02.35-835.3oV Therefore, ov 3.835 cos(4t �– 35.02 ) V

Page 492: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 62.

First, we transform the circuit to the frequency domain. 1,012)tcos(12

2jLjH2

4j-Cj

1F

41

8j-Cj

1F

81

To find , consider the circuit in Fig. (a). thZ

3 Io

21

Io 4

-j8

Vx

-j4 +

1 V

j2 Ix

(a)At node 1,

2j1

34j-4

xo

xx VI

VV, where

4- x

o

VI

Thus, 2j

14

24j-

xxx VVV

8.0j4.0xV At node 2,

2j1

8j-1

3 xox

VII

83

j)5.0j75.0( xx VI

425.0j1.0-xI

24.103-29.2229.2j5246.0-1

xth I

Z

Page 493: Solution Manual for Fundamentals of Electric Circuits

To find , consider the circuit in Fig. (b). thV

3 Io

V2

12 0 V

Io 4

-j8

1 2+

Vth

V1

-j4 +

j2

(b)At node 1,

2j4j-3

412 211

o1 VVV

IV

, where 4

12 1o

VI

21 2j)j2(24 VV (1)

At node 2,

8j-3

2j2

o21 V

IVV

21 3j)4j6(72 VV (2)

From (1) and (2),

2

1

3j-4j62j-j2

7224

VV

6j5- , 24j-2

8.219-073.322th VV

Thus,

229.2j4754.1)8.219-073.3)(2(

22

thth

o VZ

V

3.163-3.25.56-673.28.219-146.6

oV

Therefore, ov 2.3 cos(t �– 163.3 ) V

Page 494: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 63.

Transform the circuit to the frequency domain.

200,304)30t200cos(4 k2j)10)(200(jLjH10

kj-)105)(200(j

1Cj

1F5 6-

NZ is found using the circuit in Fig. (a).

-j k

2 kZN

j2 k

(a)

k1j1j-2j||2j-NZ We find using the circuit in Fig. (b). NI

-j k

j2 k 2 k4 30 A IN

(b)

j12||2j By the current division principle,

75657.5)304(jj1

j1NI

Therefore,

Ni 5.657 cos(200t + 75 ) A

NZ 1 k

Page 495: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 64.

is obtained from the circuit in Fig. (a). NZ

ZN60 40

-j30 j80

(a)

50j100)50j)(100(

50j||100)30j80j(||)4060(NZ

40j20NZ 44.72 63.43

To find , consider the circuit in Fig. (b). NI

IN

I2

I1

Is3 60 A

j80

60 40

-j30

(b)

603sI For mesh 1,

060100 s1 II 608.11I

For mesh 2,

080j)30j80j( s2 II 608.42I

21N III 3 60 A

Page 496: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 65.

2,05)t2cos(5 8j)4)(2(jLjH4

2j-)4/1)(2(j

1Cj

1F

41

j-)2/1)(2(j

1Cj

1F

21

To find , consider the circuit in Fig. (a). NZ

2

ZN

-j2 -j

(a)

)10j2(131

3j2)2j2(j-

)2j2(||j-NZ

To find , consider the circuit in Fig. (b). NI

2 +

IN-j2

5 0 V

-j

(b)

5jj-05

NI

The Norton equivalent of the circuit is shown in Fig. (c).

Io

ZN IN j8

(c)

Page 497: Solution Manual for Fundamentals of Electric Circuits

Using current division,

94j210j50

8j)10j2)(131()5j)(10j2)(131(

8j NN

No I

ZZ

I

47.77-05425294.0j1176.0oI Therefore, oi 0.542 cos(2t �– 77.47 ) A

Chapter 10, Solution 66.

10

5j)5.0)(10(jLjH5.0

10j-)1010)(10(j

1Cj

1mF10 3-

Vx

j5 2 Vo +

Vo

-j10

1 A 10

(a)

To find , consider the circuit in Fig. (a). thZ

10j105j21 xx

o

VVV , where

10j10 x

o

V10

V

2j2110j10-

5j10j1019

1 xxx V

VV

44.5095.21135142.14

1x

thN

VZZ 0.67 129.56

To find and , consider the circuit in Fig. (b). thV NI

2 Vo I

++

Vth

+

Vo

-j10

j5 10

12 0 V

-j2 A

(b)

Page 498: Solution Manual for Fundamentals of Electric Circuits

012)2(5j)2j-)(10()5j10j10( oVI where )2j-)(10(o IV Thus,

20j188-)105j10( I

105j10-20j188

I

200105j)40j21(5j)2(5j oth IIVIV

076.2j802.11-200105j10-

)20j188(105jthV

thV 11.97 170 V

56.12967.017097.11

th

thN Z

VI 17.86 40.44 A

Chapter 10, Solution 67.

079.1j243.116j20

)6j8(125j23

)5j13(10)6j8//(12)5j13//(10ZZ ThN

37.454j93.25)4560(6j20)6j8(V,44.21j78.13)4560(

5j2310V o

bo

a

A 09.9734.38ZV

I V, 599.11.433VVV o

Th

ThN

obaTh

Chapter 10, Solution 68.

10j1x10jLj H1

2j

201x10j

1Cj

1 F201

We obtain VTh using the circuit below.

Page 499: Solution Manual for Fundamentals of Electric Circuits

Io 4 a + + + -j2 j10 Vo 6<0o Vo/3 - 4Io - - b

5.2j2j10j)2j(10j)2j//(10j

ooo I10j)5.2j(xI4V (1)

0V31I46 oo (2)

Combining (1) and (2) gives

oooTho 19.5052.11

3/10j460jI10jVV,

3/10j46I

)19.50t10sin(52.11v o

Th To find RTh, we insert a 1-A source at terminals a-b, as shown below. Io 4 a + + -j2 j10 Vo Vo/3 - 4Io -

1<0o

12V

I0V31I4 o

ooo

10jV

2jV

I41 ooo

Page 500: Solution Manual for Fundamentals of Electric Circuits

Combining the two equations leads to

4766.1j2293.14.0j333.0

1Vo

477.12293.11

VZ o

Th

Chapter 10, Solution 69.

This is an inverting op amp so that

Cj1R--

i

f

s

o

ZZ

VV

-j RC

When and ms VV RC1 ,

90-VVj-VRCRC1

j- mmmoV

Therefore,

)90tsin(V)t(v mo - Vm cos( t) Chapter 10, Solution 70.

This may also be regarded as an inverting amplifier.

44 104,02)t104cos(2

k5.2j-)1010)(104(j

1Cj

1nF10 9-4

i

f

s

o -ZZ

VV

where and k50iZ kj40

100j-)k5.2j-(||k100fZ .

Thus, j40

2j-

s

o

VV

If , 02sV

57.88-1.043.1-01.40

90-4j40

4j-oV

Therefore,

)t(vo 0.1 cos(4x104 t �– 88.57 ) V

Page 501: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 71.

oo 308)30t2cos(8

0. k1j10x5.0x2j

1Cj

1F5

6

At the inverting terminal,

)j6.0(308)j1.0(Vk2308

k10308V

k1j308V

o

ooo

oo

o

o 747.4283.9j1.0

)j6.0)(4j9282.6(V

vo(t) = 9.283cos(2t + 4.75o) V

Chapter 10, Solution 72.

44 10,04)t10cos(4

k100j-)10)(10(j

1Cj

1nF1 9-4

Consider the circuit as shown below.

4 0 V

Vo Vo

-j100 k

50 k

+

+Io

100 k

At the noninverting node,

5.0j14

100j-504

ooo V

VV

A56.26-78.35mA)5.0j1)(100(

4k100

oo

VI

Therefore,

)t(io 35.78 cos(104 t �– 26.56 ) A

Page 502: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 73.

As a voltage follower, o2 VV

k-j20)1010)(105(j

1Cj1

nF10C 9-31

1

k-j10)1020)(105(j

1Cj1

nF20C 9-32

2

Consider the circuit in the frequency domain as shown below.

-j20 k

Zin

Io

V1

V2Is

-j10 k

20 k10 k

+

+ Vo

VS

At node 1,

2020j-10o1o11s VVVVVV

o1s )j1()j3(2 VVV (1) At node 2,

10j-0

20oo1 VVV

o1 )2j1( VV (2) Substituting (2) into (1) gives

os 6j2 VV or so 31

-j VV

so1 31

j32

)2j1( VVV

s1s

s k10)j1)(31(

k10V

VVI

Page 503: Solution Manual for Fundamentals of Electric Circuits

k30j1

s

s

VI

k)j1(15j1

k30

s

sin I

VZ

inZ 21.21 45 k Chapter 10, Solution 74.

11i Cj

1RZ ,

22f Cj

1RZ

11

22

i

f

s

ov

Cj1

R

Cj1

R-ZZ

VV

A11

22

2

1

CRj1CRj1

CC

At , 0 vA2

1

CC

As , vA1

2

RR

At 11CR

1, vA

j1CRCRj1

CC 1122

2

1

At 22CR

1, vA

22112

1

CRCRj1j1

CC

Chapter 10, Solution 75.

3102

k-j500)101)(102(j

1Cj1

nF1CC 9-31

21

Page 504: Solution Manual for Fundamentals of Electric Circuits

Consider the circuit shown below.

100 k

-j500 k

+

Vo

-j500 k

20 k

V1

V2

100 k20 k

+

+

VS

At node 1,

500j-100500j-211o1s VVVVVV

2o1s 5j)5j2( VVVV (1) At node 2,

100500j-221 VVV

21 )5j1( VV (2) But

2RRR o

o43

32

VVV (3)

From (2) and (3),

o1 )5j1(21

VV (4)

Substituting (3) and (4) into (1),

ooos 21

5j)5j1)(5j2(21

VVVV

os )25j26(21

VV

25j262

s

o

VV

0.0554 43.88

Chapter 10, Solution 76.

Page 505: Solution Manual for Fundamentals of Electric Circuits

Let the voltage between the -jk capacitor and the 10k resistor be V1.

o1o

o1o11o

V6.0jV)6.0j1(302

k20VV

k10VV

k4jV302

(1)

Also,

o1oo1 V)5j1(Vk2j

Vk10VV

(2)

Solving (2) into (1) yields

V 34.813123.03088.0j047.0V oo

Chapter 10, Solution 77.

Consider the circuit below.

1

2

+

VS

+

Vo

C1

C2 R2 R1

V1

V1

R3

+

At node 1,

11

1s CjR

VVV

111s )CRj1( VV (1) At node 2,

)(CjRR

0o12

2

o1

3

1 VVVVV

322

31o1 RCj

RR

)( VVV

Page 506: Solution Manual for Fundamentals of Electric Circuits

13223

o RCj)RR(1

1 VV (2)

From (1) and (2),

3223

2

11

so RRCjR

R1

CRj1V

V

s

o

VV

)RRCjR()CRj1(RRCjRR

322311

32232

Chapter 10, Solution 78. 400,02)t400sin(2

k5j-)105.0)(400(j

1Cj

1F5.0 6-

k10j-)1025.0)(400(j

1Cj

1F25.0 6-

Consider the circuit as shown below.

20 k

At node 1,

10 k2 0 V

10 k -j5 k

20 k

V1 V2

-j10 k40 k

+

+ Vo

205j-10j-102 o12111 VVVVVV

o21 4j)6j3(4 VVV (1) At node 2,

105j221 VVV

21 )5.0j1( VV (2) But

oo2 31

402020

VVV (3)

Page 507: Solution Manual for Fundamentals of Electric Circuits

From (2) and (3),

o1 )5.0j1(31

VV (4)

Substituting (3) and (4) into (1) gives

oooo 61

j134

j)5.0j1(31

)6j3(4 VVVV

46.9945.3j6

24oV

Therefore, )t(vo 3.945 sin(400t + 9.46 ) V

Chapter 10, Solution 79.

1000,05)t1000cos(5

k10j-)101.0)(1000(j

1Cj

1F1.0 6-

k5j-)102.0)(1000(j

1Cj

1F2.0 6-

Consider the circuit shown below.

20 k

+

Vs = 5 0 V

V1

-j5 k

-j10 k40 k

+Vo

+

10 k

+

Since each stage is an inverter, we apply ii

fo

-V

ZZ

V to each stage.

Page 508: Solution Manual for Fundamentals of Electric Circuits

1o 15j-40-

VV

(1) and

s1 10)10j-(||20-

VV

(2) From (1) and (2),

0510j20

)10-j)(20(-10

8j-oV

56.2678.35)j2(16oV

Therefore, )t(vo 35.78 cos(1000t + 26.56 ) V Chapter 10, Solution 80. 4

1000,60-4)60t1000cos(

k10j-)101.0)(1000(j

1Cj

1F1.0 6-

k5j-)102.0)(1000(j

1Cj

1F2.0 6-

The two stages are inverters so that

10j5-

5020

)60-4(10j-

20oo VV

o52

2j-

)60-4()2j(2j-

V

60-4)5j1( oV

31.71-922.35j1

60-4oV

Therefore, )t(vo 3.922 cos(1000t �– 71.31 ) V

Page 509: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 81. The schematic is shown below. The pseudocomponent IPRINT is inserted to print the value of Io in the output. We click Analysis/Setup/AC Sweep and set Total Pts. = 1, Start Freq = 0.1592, and End Freq = 0.1592. Since we assume that w = 1. The output file includes: FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E-01 1.465 E+00 7.959 E+01 Thus, Io = 1.465 79.59o A

Chapter 10, Solution 82. The schematic is shown below. We insert PRINT to print Vo in the output file. For AC Sweep, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we print out the output file which includes: FREQ VM($N_0001) VP($N_0001) 1.592 E-01 7.684 E+00 5.019 E+01

Page 510: Solution Manual for Fundamentals of Electric Circuits

which means that Vo = 7.684 50.19o V

hapter 10, Solution 83.

he schematic is shown below. The frequency is

C

T 15.1592

10002/f

Page 511: Solution Manual for Fundamentals of Electric Circuits

When the circuit is saved and simulated, we obtain from the output file

REQ VM(1) VP(1)

hus, vo = 6.611cos(1000t �– 159.2o) V

F1.592E+02 6.611E+00 -1.592E+02 T

hapter 10, Solution 84.

he schematic is shown below. We set PRINT to print Vo in the output file. In AC

FREQ VM($N_0003)

1.592 E-01 1.664 E+00 -1.646

amely, Vo = 1.664 -146.4o V

C TSweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes: VP($N_0003) E+02 N

hapter 10, Solution 85.

C

The schematic is shown below. We let rad/s so that L=1H and C=1F.

When the circuit is saved and simulated, we obtain from the output file

FREQ VM(1) VP(1)

From this, we conclude that

5.167228.2Vo V

1

1.591E-01 2.228E+00 -1.675E+02

Page 512: Solution Manual for Fundamentals of Electric Circuits

Chapter 10, Solution 86.

e insert three pseudocomponent PRINTs at nodes 1, 2, and 3 to print V1, V2, and V3,

FREQ VM($N_0002)

1.592 E-01 6.000 E+01 3.000

FREQ VM($N_0003)

1.592 E-01 2.367 E+02 -8.483

Winto the output file. Assume that w = 1, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After saving and simulating the circuit, we obtain the output file which includes: VP($N_0002) E+01 VP($N_0003) E+01

Page 513: Solution Manual for Fundamentals of Electric Circuits

FREQ VM($N_0001)

1.592 E-01 1.082 E+02 1.254

herefore,

V1 = 60 30o V

VP($N_0001) E+02 T

V2 = 236.7 -84.83o V V3 = 108.2 125.4o V

hapter 10, Solution 87.

he schematic is shown below. We insert three PRINTs at nodes 1, 2, and 3. We set otal Pts = 1, Start Freq = 0.1592, End Freq = 0.1592 in the AC Sweep box. After

VM($N_0004) VP($N_0004)

1.696

FREQ VM($N_0001) VP($N_0001)

-1.386

C TTsimulation, the output file includes: FREQ 1.592 E-01 1.591 E+01 E+02 1.592 E-01 5.172 E+00 E+02

Page 514: Solution Manual for Fundamentals of Electric Circuits

FREQ VM($N_0003) VP($N_0003)

-1.524

ore,

o V

1.592 E-01 2.270 E+00 E+02 Theref

V1 = 15.91 169.6 V2 = 5.172 -138.6o V V3 = 2.27 -152.4o V

hapter 10, Solution 88.

ow. We insert IPRINT and PRINT to print Io and Vo in the utput file. Since w = 4, f = w/2 = 0.6366, we set Total Pts = 1, Start Freq = 0.6366, nd End Freq = 0.6366 in the AC Sweep box. After simulation, the output file includes:

6.366 E-01 3.496 E+01 1.261

FREQ IM(V_PRINT2) IP _PRINT2)

6.366 E-01 8.912 E-01

C The schematic is shown beloa FREQ VM($N_0002) VP($N_0002) E+01 (V -8.870 E+01

Page 515: Solution Manual for Fundamentals of Electric Circuits

Therefore, Vo = 34.96 12.6o V, Io = 0.8912 -88.7o A

vo = 34.96 cos(4t + 12.6o)V, io = 0.8912cos(4t - 88.7o )A

onsider the circuit below.

Chapter 10, Solution 89. C

At node 1,

in

2

2

1

in

RVVV0

in

R

in1

22in R

R- VVV (1)

At node 3,

Cj1R4in

3

in2 VVVV

V

C R1 R2 Vin

+

Iin

in

3

R3 R4

+ +

V1

2 4

Page 516: Solution Manual for Fundamentals of Electric Circuits

3

2in4in CRj

-VV

VV (2)

rom (1) and (2),

F

in13

24in RCRj

R-- VVV

1R

Thus,

in43

2

4

4inin RCRj

RR

VVV

I

eq2

431

in

inin Lj

RRRCRj

IV

Z

where 2

431eq R

CRRRL

Chapter 10, Solution 90.

et

LRCj1

RC

1||RZ

C

j4

CjRCj1

Cj1

R3Z

onsider the circuit shown below.

R2Z4

+ Vo Vi +

Z3 R1

Page 517: Solution Manual for Fundamentals of Electric Circuits

i21

2i

43

4o RR

RVV

ZZZ

V

21

2

i

o

RRR

CjRCj1

Cj1R

Cj1R

VV

21

22 RR

R)RCj1(RCj

RCj

21

2222

i

o

RRR

RC3jCR1RCj

V

V

For and to be in phase, oV iVi

o

VV

must be purely real. This happens when

0CR1 222

f2RC1

or RC21

f

At this freque cy, n

21

2

i

ov RR

R31

VV

A

Chapter 10, Solution 91.

(a) Let

2V voltage at the noninverting terminal of the op amp outputoV voltage of the op amp

Rk10 opZ1Cj

LjRsZ

s in Section 10.9, A

Page 518: Solution Manual for Fundamentals of Electric Circuits

Cj

LjRR

R

o

o

ps

p

o

2

ZZZ

VV

)1LC(j)RR(CCR

2o

o

o

2

VV

For this to be purely real,

LC1

01LC o2o

)102)(104.0(21

LC21

f9-3-o

of 180 kHz

(b) At oscillation,

o

o

oo

oo

o

2

RRR

)RR(CCR

VV

This must be compensated for by

52080

12

ov V

VA

oo

o R4R51

RRR

40 k

Chapter 10, Solution 92.

Let voltage at the noninverting terminal of the op amp 2V

oV output voltage of the op amp

os RZ

)1LC(jRLRL

Lj1

CjR1

1R||

Cj1

||Lj 2pZ

As in Section 10.9,

)1LC(jRLRL

R

)1LC(jRLRL

2o

2

ps

p

o

2

ZZZ

VV

Page 519: Solution Manual for Fundamentals of Electric Circuits

)1LC(RjRLRRLRL

2ooo

2

VV

For this to be purely real,

LC21

f1LC o2o

(a) At , o

oooo

o

o

2

RRR

LRRLRL

VV

This must be compensated for by

11k100k1000

1RR

1o

f

2

ov V

VA

Hence,

R10R111

RRR

oo

100 k

(b) )102)(1010(2

1f

9-6-o

of 1.125 MHz Chapter 10, Solution 93. As shown below, the impedance of the feedback is

j L

1Cj1

2Cj1 ZT

21T Cj

1Lj||

Cj1

Z

Page 520: Solution Manual for Fundamentals of Electric Circuits

)CLCCC(j

LC1

Cj-

LjCj-

Cj-

LjCj-

212

21

2

21

21TZ

In order for to be real, the imaginary term must be zero; i.e. TZ

0CLCCC 212o21

T21

212o LC

1CLCCC

To LC2

1f

Chapter 10, Solution 94. If we select C

nF20C21

nF102

CCC

CCC 1

21

21T

Since T

o LC21

f ,

mH13.10)1010)(102500)(4(

1C)f2(

1L 9-62

T2

)1020)(1050)(2(1

C1

X 9-32

c 159

We may select and , say k20R i if RR k20R f . Thus,

21 CC 20 nF, L 10.13 mH if RR 20 k Chapter 10, Solution 95. First, we find the feedback impedance.

C

L1L2

ZT

Page 521: Solution Manual for Fundamentals of Electric Circuits

Cj1

Lj||Lj 21TZ

)1)LL(C(j)L1(CL

Cj

LjLj

Cj

LjLj

212

212

21

21

TZ

In order for to be real, the imaginary term must be zero; i.e. TZ

01)LL(C 212o

)LL(C1

f221

oo

)LL(C21

f21

o

Chapter 10, Solution 96.

(a) Consider the feedback portion of the circuit, as shown below.

V2V1

+

R

R

j L

Vo j L

21 LjLjR

LjRLj

VVVV 12 (1)

Applying KCL at node 1,

LjRRLj111o VVVV

LjR1

R1

Lj 11o VVV

Page 522: Solution Manual for Fundamentals of Electric Circuits

)LjR(RLRL2j

122

1o VV

(2) From (1) and (2),

2

22

o )LjR(RLRL2j

1Lj

LjRVV

RLjLRL2jRLjR 222

2

o

VV

RLjLR

3

1222

o

2

VV

o

2

VV

)LRRL(j31

(b) Since the ratio o

2

VV

must be real,

0L

RR

L

o

o

LR

Lo

2

o

LR

f2 oo

L2R

fo

(c) When o

31

o

2

VV

This must be compensated for by 3vA . But

3RR

11

2vA

12 R2R

Page 523: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 1.

)t50cos(160)t(v )9018030t50cos(2)30t50sin(20-)t(i

)60t50cos(20)t(i

)60t50cos()t50cos()20)(160()t(i)t(v)t(p W)60cos()60t100cos(1600)t(p

)t(p W)60t100cos(1600800

)60cos()20)(160(21

)cos(IV21

P ivmm

P W800

Chapter 11, Solution 2. First, transform the circuit to the frequency domain.

030)t500cos(30 , 500 150jLjH3.0

100j-)10)(20)(500(

j-Cj

1F20 6-

I

I1

I2

+

30 0 V j150

-j100

200

2.0j-902.0150j

0301I

)t500sin(2.0)90t500cos(2.0)t(i1

06.0j12.056.261342.0j2

3.0100j200030

2I

Page 524: Solution Manual for Fundamentals of Electric Circuits

)56.25t500cos(1342.0)t(i2

49.4-1844.014.0j12.021 III )35t500cos(1844.0)t(i

For the voltage source,

])35t500cos(1844.0[])t500cos(30[)t(i)t(v)t(p At , s2t )351000cos()1000cos(532.5p

)935.0)(5624.0)(532.5(p p W91.2

For the inductor,

])t500sin(2.0[])t500cos(30[)t(i)t(v)t(p At , s2t )1000sin()1000cos(6p

)8269.0)(5624.0)(6(p p W79.2

For the capacitor,

63.44-42.13)100j-(2c IV )56.25t500cos(1342.0[])44.63500cos(42.13[)t(i)t(v)t(p

At , s2t )56.261000cos()44.631000cos(18p

)1329.0)(991.0)(18(p p W37.2

For the resistor,

56.2584.26200 2R IV ])56.26t500cos(1342.0[])56.26t500cos(84.26[)t(i)t(v)t(p

At , s2t )56.251000(cos602.3p 2

21329.0)(602.3(p p W0636.0

Page 525: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 3. 10 , 3010)30t2cos( 2

2jLjH1

-j2Cj

1F25.0

4 2 I I1

I2

+

10 30 V j2 -j2

2j22

)2j2)(2j()2j2(||2j

565.11581.12j24

3010I

565.101581.1j22j

1 III

565.56236.22

2j22 II

For the source,

)565.11-581.1)(3010(21*IVS

5.2j5.718.43905.7S

The average power supplied by the source = W5.7 For the 4- resistor, the average power absorbed is

)4()581.1(21

R21

P 22I W5

For the inductor,

5j)2j()236.2(21

21 2

L

2

2 ZIS

The average power absorbed by the inductor = W0

Page 526: Solution Manual for Fundamentals of Electric Circuits

For the 2- resistor, the average power absorbed is

)2()581.1(21

R21

P 22

1I W5.2

For the capacitor,

5.2j-)2j-()581.1(21

21 2

c

2

1 ZIS

The average power absorbed by the capacitor = W0

Chapter 11, Solution 4.

20 10

I2I1+

-j10 50 V j5

For mesh 1,

21 10j)10j20(50 II

21 j)j2(5 II (1) For mesh 2,

12 10j)10j5j10(0 II

12 2j)j2(0 II (2) In matrix form,

2

1

j22jjj2

05

II

4j5 , )j2(51 , -j102

1.12746.14j5

)j2(511I

66.128562.1j4-5

j10-22I

For the source,

12.1-65.4321 *

1IVS

Page 527: Solution Manual for Fundamentals of Electric Circuits

The average power supplied )1.12cos(65.43 W68.42 For the 20- resistor,

R21

P2

1I W48.30

For the inductor and capacitor, P W0

For the 10- resistor,

R21

P2

2I W2.12

Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 2

+

j6

�–j2 8 �–40

W4159.112

6828.1P

38.256828.1

2j26j)2j2(6j1

408I

21

1

P3H = P0.25F = 0

W097.522

258.2P

258.238.256828.12j26j

6jI

22

2

Page 528: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 6.

20 10

I2I1+

-j10 50 V j5

For mesh 1,

04)604(2j)2j4( o1 VI (1) )604(2 2o IV (2)

For mesh 2, 04)604(2)j2( o2 VI (3)

Substituting (2) into (3), 0)604(8608)j2( 22 II

j106040

2I

Hence,

j10608j-

j106040

6042oV

Substituting this into (1),

j10j14

)608j(j10

6032j608j)2j4( 1I

125.06498.28j21

)14j1)(604(1I

)4()498.2(21

R21

P 22

14 I W48.12

Chapter 11, Solution 7.

20 10

I2I1+

-j10 50 V j5

Page 529: Solution Manual for Fundamentals of Electric Circuits

Applying KVL to the left-hand side of the circuit, oo 1.04208 VI (1)

Applying KCL to the right side of the circuit,

05j105j

8 11o

VVI

But, o11o 105j10

5j1010

VVVV

Hence, 01050j

5j108 o

oo

VVI

oo 025.0j VI (2)

Substituting (2) into (1), )j1(1.0208 oV

j12080

oV

25-2

1010

o1

VI

)10(2

10021

R21

P2

1I W250

Chapter 11, Solution 8. We apply nodal analysis to the following circuit.

At node 1,

Io V2V1

6 0 A 0.5 Io j10

-j20

I2

40

20j-10j6 211 VVV

21 120j VV (1)

At node 2,

Page 530: Solution Manual for Fundamentals of Electric Circuits

405.0 2

oo

VII

But, j20-

21o

VVI

Hence, 40j20-

)(5.1 221 VVV

21 )j3(3 VV (2)

Substituting (1) into (2),

0j33360j 222 VVV

j6)-1(37360

j6360j

2V

j6)-1(379

402

2

VI

)40(379

21

R21

P2

2

2I W78.43

Chapter 11, Solution 9.

rmsV8)2)(4(V26

1V so

mW1064

RV

P2o

10 mW4.6

The current through the 2 -k resistor is

mA1k2

Vs

RIP 2

2 mW2 Similarly,

RIP 26 mW6

Page 531: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 10.

No current flows through each of the resistors. Hence, for each resistor, P W0 .

Chapter 11, Solution 11.

, , 377 410R -910200C754.0)10200)(10)(377(RC -94

02.37)RC(tan -1

k37.02-375.637.02-)754.0(1

k10Z

2ab

mA)68t377cos(2)22t377sin(2)t(i

68-2I

3

2-3

ab2rms 10)37.02-375.6(

2102

ZIS

mVA37.02-751.12S

)02.37cos(SP mW181.10

Chapter 11, Solution 12.

(a) We find using the circuit in Fig. (a). ThZ

Zth

8 -j2

(a)

882.1j471.0)4j1(178

j28(8)(-j2)

-j2||8ThZ

*ThL ZZ 882.1j471.0

Page 532: Solution Manual for Fundamentals of Electric Circuits

We find using the circuit in Fig. (b). ThV

Io +

Vth -j2 4 0 A 8

(b)

)04(2j8

2j-oI

j28j64-

I8 oThV

)471.0)(8(68

64

R8P

2

L

2

Thmax

VW99.15

(b) We obtain from the circuit in Fig. (c). ThZ

5 -j3

Zth

j2

4

(c)

167.1j5.23j9

)3j4)(5(2j)3j4(||52jThZ

*ThL ZZ 167.1j5.2

Page 533: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 13.

(a) We find at the load terminals using the circuit in Fig. (a). ThZ

j100

-j40 Zth

80

(a)

6.1j2.51j6080

j100)(-j40)(80j100)(80||-j40ThZ

*ThL ZZ 6.1j2.51

(b) We find at the load terminals using Fig. (b). ThV

j100 Io

+

Vth 3 20 A -j40 80

(b)

6j8)203)(8(

)203(40j100j80

80oI

6j8)2024)(40j(-

40j- oTh IV

)2.51)(8(

241040

R8P

2

L

2

Thmax

VW5.22

Page 534: Solution Manual for Fundamentals of Electric Circuits

From Fig.(d), we obtain V using the voltage division principle. Th

5 -j3

+

Vth

+

10 30 V

j2

4

(d)

303

10j33j4

)3010(3j93j4

ThV

)5.2)(8(3

10105

R8P

2

L

2

Thmax

VW389.1

Chapter 11, Solution 14.

I

+

VTh

_

16

�–j10

j8

j24

10

ZTh 40 90º A

3.2j245.8ZZ

3.2j245.87.7j245.810j8j1624j10)8j16)(24j10(10jZ

Th

Th

Page 535: Solution Manual for Fundamentals of Electric Circuits

W6.456245.8)245.8x2(

2

V

245.8IP

V12.158j53.7166.6555.173

)8j16(40j8j1624j10

10)8j16(IV

2

2

2Th

2rmsmax

Th

Chapter 11, Solution 15. To find Z , insert a 1-A current source at the load terminals as shown in Fig. (a). Th

+

Vo

2 1 1 -j

2 Vo j 1 A

(a)At node 1,

2oo2oo j

j-j1VV

VVVV (1)

At node 2,

o2o2

o )j2(j1j-

21 VVVV

V (2)

Substituting (1) into (2), 222 )j1()j)(j2(j1 VVV

j11

2V

5.0j5.02

j11

2Th

VV

*ThL ZZ 5.0j5.0

Page 536: Solution Manual for Fundamentals of Electric Circuits

We now obtain from Fig. (b). ThV

1 -j

+

Vth +

12 0 V +

Vo 2 Vo j

(b)

j112

2 ooo

VVV

j112-

oV

0)2j-( Thoo VVV

j1)2j1)(12(

j2)-(1 oTh VV

)5.0)(8(2512

R8P

2

L

2

Thmax

VW90

Chapter 11, Solution 16.

520/14

11F20/1,4H1,4 jxjCj

jLj

We find the Thevenin equivalent at the terminals of ZL. To find VTh, we use the circuit shown below. 0.5Vo 2 V1 4 V2 + + + 10<0o Vo -j5 j4 VTh

- - -

Page 537: Solution Manual for Fundamentals of Electric Circuits

At node 1,

2121

111 25.0)2.01(5

425.0

5210

VjVVV

VjVV

(1)

At node 2,

)25.025.0(5.004

25.04 21

21

21 jVVjV

VVV

(2)

Solving (1) and (2) leads to

2 6.1947 7.0796 9.4072 48.81oThV V j Chapter 11, Solution 17. We find ThR at terminals a-b following Fig. (a).

ba

j20

30

40

-j10 (a)

10j40-j10))(40(

20j30)20j)(30(

)10j-(||4020j||30ThZ

41.9j353.285.13j23.9ThZ

44.4j583.11ThZ *ThL ZZ 44.4j583.11

We obtain from Fig. (b). ThV

I1 I2

+ VTh

j20

30

40

-j10

j5 A

(b)

Page 538: Solution Manual for Fundamentals of Electric Circuits

Using current division,

3.2j1.1-)5j(10j7020j30

1I

7.2j1.1)5j(10j7010j40

2I

70j1010j30 12Th IIV

)583.11)(8(5000

R8P

L

2

Th

max

VW96.53

Chapter 11, Solution 18. We find Z at terminals a-b as shown in the figure below. Th

a

b

Zth

40 80 -j10

40

j20

j1080(80)(-j10)

20j20-j10)(||8040||4020jThZ

154.10j23.21ThZ

*ThL ZZ 15.10j23.21

Chapter 11, Solution 19. At the load terminals,

j9j)(6)(3

-j2)j3(||62j-ThZ

561.1j049.2ThZ

576.2R ThL Z

Page 539: Solution Manual for Fundamentals of Electric Circuits

To get V , let Th 439.0j049.2)j3(||6Z . By transforming the current sources, we obtain

756.1j196.8)04(Th ZV

608.20258.70

R8P

L

2

Thmax

VW409.3

Chapter 11, Solution 20. Combine j20 and -j10 to get

-j20-j10||20j To find , insert a 1-A current source at the terminals of , as shown in Fig. (a). ThZ LR

Io

-j20

4 Io

+ V2V1

-j10

40

1 A

(a)At the supernode,

10j-20j-401 211 VVV

21 4j)2j1(40 VV (1)

Also, , where o21 4IVV40

- 1o

VI

1.11.1 2

121

VVVV (2)

Substituting (2) into (1),

22 4j1.1

)2j1(40 VV

Page 540: Solution Manual for Fundamentals of Electric Circuits

4.6j144

2V

71.6j05.11

2Th

VZ

ThLR Z 792.6 To find , consider the circuit in Fig. (b). ThV

+

120 0 V

+

Vth

Io

-j20

4 Io

+ V2V1

-j10

40

(b)At the supernode,

j10-j20-40120 211 VVV

21 4j)2j1(120 VV (3)

Also, , where o21 4IVV40

120 1o

VI

1.1122

1

VV (4)

Substituting (4) into (3),

2)818.5j9091.0(82.21j09.109 V

92.43-893.18818.5j9091.082.21j09.109

2Th VV

)792.6)(8()893.18(

R8P

2

L

2

Thmax

VW569.6

Page 541: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 21. We find Z at terminals a-b, as shown in the figure below. Th

100 -j10

40

b

a

Zth

j30

50

])30j40(||10010j-[||50ThZ

where 634.14j707.3130j140

)30j40)(100()30j40(||100

634.4j707.81)634.4j707.31)(50(

)634.4j707.31(||50ThZ

73.1j5.19ThZ

ThLR Z 58.19 Chapter 11, Solution 22.

ttti 0,sin4)(

8)02

(16

42sin

216

sin161

00

22 tttdtI rms

A 828.28rmsI

Page 542: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 23.

6t2,52t0,15

)t(v

6550

dt5dt1561

V6

22

2

022

rms

rmsV V574.9

Chapter 11, Solution 24.

, 2T2t15,-1t0,5

)t(v

25]11[225

dt-5)(dt521

V2

12

1

022

rms

rmsV V5

Chapter 11, Solution 25.

266.33

32f

332

]16016[31

dt4dt0dt)4(31

dt)t(fT1

f

rms

32

221

10

2T0

22rms

Page 543: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 26.

, 4T4t2102t05

)t(v

5.62]20050[41

dt)10(dt541

V4

22

2

022

rms

rmsV V906.7

Chapter 11, Solution 27. , 5T 5t0,t)t(i

333.815

1253t

51

dtt51

I 50

35

022

rms

rmsI A887.2

Chapter 11, Solution 28.

5

22

2

022

rms dt0dt)t4(51

V

533.8)8(1516

3t16

51

V 20

32rms

rmsV V92.2

2533.8

RV

P2rms W267.4

Page 544: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 29.

, 20T25t152t40-

15t5t220)t(i

25

152

15

522

eff dt2t)-40(dt)t220(201

I

25

15

215

5

22eff dt400)t40t(dt)tt20100(

51

I

2515

23

155

322

eff t400t203t

3t

t10t10051

I

332.33]33.8333.83[51

I2eff

effI A773.5

RIP 2

eff W400 Chapter 11, Solution 30.

4t21-2t0t

)t(v

1667.1238

41

dt-1)(dtt41

V4

22

2

022

rms

rmsV V08.1

Chapter 11, Solution 31.

6667.81634

21

)4()2(21

)(21 2

0

1

0

2

1

222 dtdttdttvV rms

V 944.2rmsV

Page 545: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 32.

2

1

1

0

222rms dt0dt)t10(

21I

105t

50dtt50I 10

51

042

rms

rmsI A162.3

Chapter 11, Solution 33.

3t202t1t10201t010

)t(i

0dt)t1020(dt1031

I2

121

022

rms

33.133)31)(100(100dt)tt44(100100I32

122

rms

333.133

Irms A667.6

Chapter 11, Solution 34.

472.420f

20363t9

31

dt6dt)t3(31dt)t(f

T1f

rms

2

0

3

32

220

2T0

22rms

Page 546: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 35.

6

52

5

42

4

22

2

12

1

022

rms dt10dt20dt30dt20dt1061

V

67.466]1004001800400100[61

V2rms

rmsV V6.21

Chapter 11, Solution 36.

(a) Irms = 10 A

(b) V 528.42916

234

222

rmsrms VV (checked)

(c) A 055.92

3664rmsI

(d) V 528.42

16225

rmsV

Chapter 11, Solution 37.

)302cos(6)10sin(48321oo ttiiii

A 487.9902

362

166432

22

12

rmsrmsrmsrms IIII

Chapter 11, Solution 38.

08.157j)5.0)(50)(2(jLjH5.0

08.157j30jXR LZ

Page 547: Solution Manual for Fundamentals of Electric Circuits

08.157j30)210( 2

*

2

ZV

S

Apparent power 160

)210( 2

S VA6.275

)19.79cos(36

08.157tancoscospf 1-

pf (lagging)1876.0

Chapter 11, Solution 39.

4j12)8j12)(4j(

)8j12(||4jTZ

74.7456.4)11j3(4.0TZ

)74.74cos(pf 2631.0

Chapter 11, Solution 40. At node 1,

)1(V4.0)6667.0j4.1(V60j92.103

50VV

30jV

20V30120

21

2111o

At node 2,

212221 )25.16(0401050

VjVjVVVV

(2)

Solving (1) and (2) leads to

V1 = 45.045 + j66.935, V2 = 9.423 + j9.193

Page 548: Solution Manual for Fundamentals of Electric Circuits

(a) 4030 0 jj PP

W665.820/3.173||

21 2

22

10 RV

RV

P rms

W6.034100/1.4603||

21 2

2150 R

VVP

W86.8740/3514|30120|

21 2

120 R

VP

o

(b) 6092.10330120,3467.0944.22030120 1 jVj

VI o

s

o

VA 8.177||,3.1065.14221

SSjIVS s

(c ) pf = 142.5/177.8 = 0.8015 (leading). Chapter 11, Solution 41.

(a) -j6j

(-j2)(-j3)-j3||-j2)2j5j(||2j-

56.31-211.76j4TZ

)-56.31cos(pf (leading)5547.0

(b) 52.1j64.03j4

)j4)(2j()j4(||2j

5.214793.044.0j64.144.0j64.0

)j52.1j64.0(||1Z

)5.21cos(pf (lagging)9304.0

Page 549: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 42.

683.30cos86.0pf

kVA798.9)683.30sin(

5sin

QSsinSQ

683.30536.44220

683.3010798.9 3**

VS

IIVS

Peak current 536.442 A98.62 Apparent power = S = kVA798.9

Chapter 11, Solution 43.

(a) V 477.53021

29

2532

22

12

rmsrmsrmsrms VVVV

(b) W310/302

RV

P rms

Chapter 11, Solution 44.

opf 46.49cos65.0

kVA 385.32)7599.065.0(50)sin(cos jjjSS

Thus, Average power = 32.5 kW, Reactive power = 38 kVAR

Page 550: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 45.

(a) V 9.4622002

6020

222

rmsrms VV

AI rms 061.1125.125.0

12

2

(b) W74.49rmsrms IVP

Chapter 11, Solution 46.

(a) 30-110)60-5.0)(30220(*IVSS VA55j26.95 Apparent power =110 VA

Real power = W26.95

Reactive power = VAR55 pf is leading because current leads voltage

(b) *IVS 151550)252.6)(10-250( S VA2.401j2.1497 Apparent power =1550 VA

Real power =1497 W2.

Reactive power = VAR2.401 pf is lagging because current lags voltage

(c) 15288)154.2)(0120(*IVSS VA54.74j2.278 Apparent power = VA288

Real power = W2.278

Reactive power = VAR54.74 pf is lagging because current lags voltage

Page 551: Solution Manual for Fundamentals of Electric Circuits

(d) 135-1360)180-5.8)(45160(*IVSS VA7.961j7.961- Apparent power =1360 VA

Real power = W7.961-

Reactive power = VAR7.961- pf is leading because current leads voltage

Chapter 11, Solution 47.

(a) , 10112V 50-4I

6022421 *IVS VA194j112

Average power =112 W Reactive power =194 VAR

(b) , 0160V 4525I

45-20021 *IVS VA42.141j42.141

Average power =141 W42. Reactive power = VAR42.141-

(c) 3012830-50

)80( 2

*

2

ZV

S V64j51.90 A

Average power = W51.90

Reactive power = VAR64

(d) )45100)(100(2ZIS kVA071.7j071.7

Average power = kW071.7

Reactive power = kVAR071.7

Page 552: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 48.

(a) jQPS V150j269 A

(b) 84.259.0cospf

31.4588)84.25sin(

2000sin

QSsinSQ

48.4129cosSP

S VA2000j4129

(c) 75.0600450

SQ

sinsinSQ

59.48 , 6614.0pf

86.396)6614.0)(600(cosSP S VA450j9.396

(d) 121040

)220(S

22

Z

V

8264.012101000

SP

coscosSP

26.34

25.681sinSQ

S VA2.681j1000

Chapter 11, Solution 49.

(a) kVA))86.0(sin(cos86.04

j4 1-S

S kVA373.2j4

Page 553: Solution Manual for Fundamentals of Electric Circuits

(b) 6.0sincos8.026.1

SP

pf

sin2j6.1S kVA2.1j6.1

(c) VA)505.6)(20208(*

rmsrms IVS70352.1S kVA2705.1j4624.0

(d) 56.31-11.72

1440060j40)120( 2

*

2

ZV

S

56.317.199S VA16.166j77.110 Chapter 11, Solution 50.

(a) ))8.0(sin(cos8.0

1000j1000jQP 1-S

750j1000S

But, *

2

rms

ZV

S

23.23j98.30750j1000

)220( 22

rms*

SV

Z

Z 23.23j98.30

(b) ZIS2

rms

22

rms)12(2000j1500

I

SZ 89.13j42.10

(c) 60-6.1)604500)(2(

)120(2

222

rms*

SV

SV

Z

606.1Z 386.1j8.0

Page 554: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 51.

(a) )6j8(||)5j10(2TZ

j1820j110

2j18

)6j8)(5j10(2TZ

5.382188.8768.0j152.8TZ

)5.382cos(pf (lagging)9956.0

(b) )5.382-188.8)(2(

)16(22

1 2

*

2

*

ZV

IVS

5.38263.15S

cosSP W56.15

(c) sinSQ VAR466.1

(d) SS VA63.15

(e) 382.563.15S VA466.1j56.15

Chapter 11, Solution 52.

749j4200SSSS500j1000S

2749j12009165.0x3000j4.0x3000S

1500j20006.08.0

2000j2000S

CBA

C

B

A

(a) .leading9845.07494200

4200pf

22

(b) 11.5555.3545120749j4200

IIVS rmsrmsrms

Irms = 35.55 �–55.11 A.

Page 555: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 53. S = SA + SB + SC = 4000(0.8�–j0.6) + 2400(0.6+j0.8) + 1000 + j500 = 5640 + j20 = 5640 0.2

(a)

A8.2997.9388.2946.66x2I

8.2946.66

230120

2.05640V

SV

SSVS

Irmsrms

CA

rms

Brms

(b) pf = cos(0.2) 1.0 lagging.

Chapter 11, Solution 54.

(a) ))8.0(sin(cos8.0

1000j1000jQP 1-S

750j1000S

But, *

2

rms

ZV

S

23.23j98.30750j1000

)220( 22

rms*

SV

Z

Z 23.23j98.30

(b) ZIS2

rms

22

rms)12(2000j1500

I

SZ 89.13j42.10

(c) 60-6.1)604500)(2(

)120(2

222

rms*

SV

SV

Z

606.1Z 386.1j8.0

Page 556: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 55. We apply mesh analysis to the following circuit.

40 0 V rms I2+

j10

20

I3

I1+

-j20

50 90 V rms

For mesh 1, 21 I20I)20j20(40

21 II)j1(2 (1) For mesh 2, 12 I20I)10j20(50j-

21 I)j2(I-2j5- (2) Putting (1) and (2) in matrix form,

2

1

II

j22-1-j1

j5-2

j1 , 3j41 , 5j-12

13.8535.3)j7(21

j13j4

I 11

56.31-605.33j2j15j1-

I 22

8.66808.35.3j5.1)3j2()5.0j5.3(III 213

For the 40-V source,

)j7(21

)40(-- *1IVS Vj20140- A

For the capacitor,

c

2

1 ZIS Vj250- A For the resistor,

R2

3IS V290 A For the inductor,

L

2

2 ZIS VA130j For the j50-V source,

)3j2)(50j(*2IVS Vj100150- A

Page 557: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 56.

8.1j6.02j6

)2j-)(6(6||2j-

j2.23.66||-j2)(4j3

The circuit is reduced to that shown below.

+

Vo

Io

2 30 A 5

3.6 + j2.2

47.0895.0)302(2.2j6.82.2j6.3

oI

47.0875.45 oo IV

)30-)(247.0875.4(21

21 *

so IVS

17.0875.4S VA396.1j543.4

Chapter 11, Solution 57. Consider the circuit as shown below.

At node o,

+

V2

V1Vo

j2 +

24 0 V

4 -j1 2

1 2 Vo

j-1424 1ooo VVVV

Page 558: Solution Manual for Fundamentals of Electric Circuits

1o 4j)4j5(24 VV (1)

At node 1, 2j

2j-

1o

1o VV

VV

o1 )4j2( VV (2)

Substituting (2) into (1),

o)168j4j5(24 V

j41124-

oV , j411

j4)-(-24)(21V

The voltage across the dependent source is

o1o12 4)2)(2( VVVVV

4j11)4j6)(24-(

)44j2(j411

24-2V

)2(21

21 *

o2*

2 VVIVS

)4j6(137576

j4-1124-

4j11)4j6)(24-(

S

S VA82.16j23.25

Chapter 11, Solution 58.

4 k

Ix

8 mA

-j3 k j1 k

10 k

From the left portion of the circuit,

mA4.0500

2.0oI

mA820 oI

Page 559: Solution Manual for Fundamentals of Electric Circuits

From the right portion of the circuit,

mAj7

16)mA8(

3jj1044

xI

)1010(50

)1016(R 3

2-32

xIS

S mVA2.51

Chapter 11, Solution 59. Consider the circuit below.

4 k

Ix

8 mA

-j3 k j1 k

10 k

30j40j20-50240

4 ooo VVV

o)38.0j36.0(88 V

46.55-13.16838.0j36.0

88oV

43.4541.8j20-

o1

VI

83.42-363.330j40

o2

VI

Reactive power in the inductor is

)30j()363.3(21

21 2

L

2

2 ZIS VAR65.169j

Reactive power in the capacitor is

)20j-()41.8(21

21 2

c2

1 ZIS VARj707.3-

Page 560: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 60.

15j20))8.0(sin(cos8.0

20j20S 1-1

749.7j16))9.0(sin(cos9.0

16j16S 1-

2

29.32585.42749.22j36SSS 21

But o

*o V6IVS

6S

Vo 29.32098.7

)29.32cos(pf (lagging)8454.0

Chapter 11, Solution 61. Consider the network shown below.

I2

So

I1+

Vo

Io

S2

S3S1

kVA8.0j2.12S

kVA937.1j4))9.0(sin(cos9.0

4j4 1-

3S

Let kVA137.1j2.5324 SSS

But *2o4 2

1IVS

Page 561: Solution Manual for Fundamentals of Electric Circuits

104j74.2290100

10)137.1j2.5)(2(2 3

o

4*2 V

SI

104j74.222I

Similarly, kVA2j2))707.0(sin(cos707.02

j2 1-1S

But *1o1 2

1IVS

40j40-100j

10)4j4(V2 3

o

1*1

SI

j40-401I

83.96145j144-17.2621o III

*ooo IV

21

S

VA)96.83-145)(90100(21

oS

oS kVA862.0j2.7

Chapter 11, Solution 62. Consider the circuit below

0.2 + j0.04

+

V1

+

V2+

I2

I1

I 0.3 + j0.15

Vs

Page 562: Solution Manual for Fundamentals of Electric Circuits

25.11j15))8.0(sin(cos8.0

15j15 1-

2S

But *

222 IVS

12025.11j15

2

2*2 V

SI

09375.0j125.02I

)15.0j3.0(221 IVV )15.0j3.0)(09375.0j125.0(1201V

0469.0j02.1201V

843.4j10))9.0(sin(cos9.0

10j10 1-

1S

But *

111 IVS

02.002.12084.25111.11

1

1*1 V

SI

0405.0j0837.025.82-093.01I

053.0j2087.021 III )04.0j2.0(1s IVV

)04.0j2.0)(053.0j2087.0()0469.0j02.120(sV 0658.0j06.120sV

sV V03.006.120 Chapter 11, Solution 63. Let S . 321 SSS

929.6j12))866.0(sin(cos866.012

j12 1-1S

916.9j16))85.0(sin(cos85.0

16j16 1-

2S

Page 563: Solution Manual for Fundamentals of Electric Circuits

20j1520j)6.0(sin(cos

)6.0)(20(1-3S

*o2

1987.22j43 IVS

11098.22j442*

o VS

I

oI A27.58-4513.0

Chapter 11, Solution 64.

I2 I1

+

j12

Is

8 120 0º V

Is + I2 = I1 or Is = I1 �– I2

But,

333.3j83.20Ior

333.3j83.20120

400j2500

V

SIVI

923.6j615.412j8

120I

2

22

1

S

Is = I1 �– I2 = �–16.22 �– j10.256 = 19.19 �–147.69 A.

Page 564: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 65.

k-j1001010j-

Cj1

nF1C 9-4

At the noninverting terminal,

j14

j100-10004

ooo V

VV

45-2

4oV

)45t10cos(2

4)t(v 4

o

W1050

12

12

4R

VP 3

22rms

P W80

Chapter 11, Solution 66. As an inverter,

)454(j34j4)(2--

si

fo V

ZZ

V

mAj3)j2)(4-(6

)45j4)(4(2-mA

2j6o

o

VI

The power absorbed by the 6-k resistor is

36-

22

o 10610540420

21

R21

P I

P mW96.0

Page 565: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 67.

51.02

11F1.0,6H3,2 jxjCj

jLj

42510

50)5//(10 jj

jj

The frequency-domain version of the circuit is shown below. Z2=2-j4

Z1 =8+j6 I1 - + Io + + V Vo206.0 o - -

123Z

(a) oo

jj

jI 87.1606.0

682052.05638.0

680206.0

1

mVA 86.3618mVA 8.104.14)87.1606.0)(203.0(21

1* ooo

s jIVS

(b) ooooso j

jZV

IVZZ

7.990224.0)206.0()68(12

)42(,31

2V

mW 904.2)12()0224.0(5.0||21 22 RIP o

Page 566: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 68.

Let S cLR SSS

where 0jRI21

jQP 2oRRRS

LI21

j0jQP 2oLLLS

C1

I21

j0jQP 2occcS

Hence, SC1

LjRI21 2

o

Chapter 11, Solution 69.

(a) Given that 12j10Z

19.501012

tan

cospf 6402.0

(b) 09.354j12.295)12j10)(2(

)120(2

2

*

2

ZV

S

The average power absorbed )Re(P S W1.295

(c) For unity power factor, 01

09.354, which implies that the reactive power due

to the capacitor is Qc

But 2VC21

X2V

Qc

2

c

22c

)120)(60)(2()09.354)(2(

VQ2

C F4.130

Page 567: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 70.

6.0sin8.0cospf 528)6.0)(880(sinSQ

If the power factor is to be unity, the reactive power due to the capacitor is

VAR528QQc

But 2c2

c

2rms

VQ2

CVC21

XV

Q

2)220)(50)(2()528)(2(

C F45.69

Chapter 11, Solution 71.

67.666.0

508.08.0,6.0

,50,065.1067071.0150 22

2211 SPQ

SQxQP

5067.66,065.106065.106 21 jSjS

9512.098.17cos,98.176.18106.56735.17221

oo pfjSSS

058.56)098.17(tan735.172)tan(tan 21o

c PQ

F 33.10120602

058.5622 xxV

QC

rms

c

Chapter 11, Solution 72.

(a) 54.40)76.0(cos-11

84.25)9.0(cos-12

)tan(tanPQ 21c

kVAR])84.25tan()54.40tan()[40(Qc kVAR84.14Qc

22rms

c

)120)(60)(2(14840

VQ

C mF734.2

Page 568: Solution Manual for Fundamentals of Electric Circuits

(b) , 54.401 02

kVAR21.34kVAR]0)54.40tan()[40(Qc

22rms

c

)120)(60)(2(34210

VQ

C mF3.6

Chapter 11, Solution 73.

(a) kVA7j1022j15j10S 22 710S S kVA21.12

(b) 240

000,7j000,10**

VS

IIVS

167.29j667.41I A35-86.50

(c) 35107

tan 1-1 , 26.16)96.0(cos-1

2

])tan(16.26-)35tan([10]tantan[PQ 211c

cQ kVAR083.4

22rms

c

)240)(60)(2(4083

VQ

C F03.188

(d) , 222 jQPS kW10PP 12

kVAR917.2083.47QQQ c12

kVA917.2j102S

But *22 IVS

2402917j000,102*

2 VS

I

154.12j667.412I A16.26-4.43

Page 569: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 74.

(a) 87.36)8.0(cos-11

kVA308.0

24cos

PS

1

11

kVAR18)6.0)(30(sinSQ 111

kVA18j241S

19.18)95.0(cos-12

kVA105.4295.0

40cos

PS

2

22

kVAR144.13sinSQ 222

kVA144.13j402S

kVA144.31j6421 SSS

95.2564144.31

tan 1-

cospf 8992.0

(b) , 95.252 01

kVAR144.31]0)95.25tan([64]tantan[PQ 12c

22rms

c

)120)(60)(2(144,31

VQ

C mF74.5

Page 570: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 75.

(a) VA59.323j75.5175j8

576050j80)240( 2

*1

2

1 ZV

S

VA91.208j13.3587j12

576070j120

)240( 2

2S

VA96060

)240( 2

3S

321 SSSS VA68.114j88.1835

(b) 574.388.183568.114

tan 1-

cospf 998.0

(c) ]0)574.3tan([88.35.18]tantan[PQ 12c

VAR68.114Qc

22rms

c

)240)(50)(2(68.114

VQ

C F336.6

Chapter 11, Solution 76.

The wattmeter reads the real power supplied by the current source. Consider the circuit below.

Vo4

8

-j3

j2 12 0 V +

3 30 A

8j23j412

303 ooo VVV

Page 571: Solution Manual for Fundamentals of Electric Circuits

19.86347.11322.11j7547.004.3j28.2

52.23j14.36oV

)30-3)(19.86347.11(21

21 *

oo IVS

19.56021.17S

)Re(P S W471.9

Chapter 11, Solution 77.

The wattmeter measures the power absorbed by the parallel combination of 0.1 F and 150 .

0120)t2cos(120 , 2 8jLjH4

-j5Cj

1F1.0

Consider the following circuit.

j8 6 I

+

120 0 V Z

5.4j5.15j15

(15)(-j5)-j5)(||15Z

25.02-5.14)5.4j5.1()8j6(

120I

)5.4j5.1()5.14(21

21

21 22* ZIIVS

VA06.473j69.157S

The wattmeter reads

)Re(P S W69.157

Page 572: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 78.

4The wattmeter reads the power absorbed by the element to its right side.

02)t4cos(2 ,

4jLjH1

-j3Cj

1F

121

Consider the following circuit.

10 I

+

20 0 V Z

3j4)3j-)(4(

4j53j-||44j5Z

08.2j44.6Z

7.21-207.108.2j44.16

20I

)08.2j44.6()207.1(21

21 22

ZIS

)Re(P S W691.4

Chapter 11, Solution 79. The wattmeter reads the power supplied by the source and partly absorbed by the 40- resistor.

Page 573: Solution Manual for Fundamentals of Electric Circuits

20j10x500x100j

1Cj

1F500,j10x10x100jmH 10

,100

63

The frequency-domain circuit is shown below.

20 Io

I 40 j V1 V2 +1 2 Io 10<0o -j20 - At node 1,

21

21212121o

1

V)40j6(V)40j7(10jVV

20)VV(3

20VV

jVV

I240

V10 (1)

At node 2,

2122121 )19()20(02020

VjVjjVVV

jVV (2)

Solving (1) and (2) yields V1 = 1.5568 �–j4.1405

0703.22216.421,4141.08443.0

4010 1 jVISj

VI

P = Re(S) = 4.222 W.

Chapter 11, Solution 80.

(a) 4.6

110ZV

I A19.17

Page 574: Solution Manual for Fundamentals of Electric Circuits

(b) 625.18904.6)110( 22

ZV

S

41.34825.0pfcos

76.1559cosSP kW6.1

Chapter 11, Solution 81.

kWh consumed kWh77132464017 The electricity bill is calculated as follows :

(a) Fixed charge = $12 (b) First 100 kWh at $0.16 per kWh = $16 (c) Next 200 kWh at $0.10 per kWh = $20 (d) The remaining energy (771 �– 300) = 471 kWh

at $0.06 per kWh = $28.26. Adding (a) to (d) gives $ 26.76

Chapter 11, Solution 82.

(a) 0,000,5 11 QP 171,17)82.0sin(cos000,30,600,2482.0000,30 1

22 QxP 171,17j600,29)QQ(j)PP(SSS 212121

AkV 22.34|S|S

(b) Q = 17.171 kVAR

(c ) 865.0220,34600,29

SP

pf

VAR 2833)9.0tan(cos)865.0tan(cos600,29

)tan(tanPQ11

21c

(d) F 46.130240602

283322 xxV

QC

rms

c

Page 575: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 83.

(a) oooVIS 35840)258)(60210(21

21

W1.68835cos840cos oSP

(b) S = 840 VA (c) VAR 8.48135sin840sin oSQ (d) (lagging) 8191.035cos/ oSPpf

Chapter 11, Solution 84.

(a) Maximum demand charge 000,72$30400,2 Energy cost 000,48$10200,104.0$ 3

Total charge = $ 000,120

(b) To obtain $120,000 from 1,200 MWh will require a flat rate of

kWhper10200,1

000,120$3 kWhper10.0$

Chapter 11, Solution 85.

(a) 15 655.51015602 mH 3 jxxxj

We apply mesh analysis as shown below. I1 + Ix 120<0o V 10 - In 30 Iz + 10 120<0o V Iy - j5.655 I2

Page 576: Solution Manual for Fundamentals of Electric Circuits

For mesh x, 120 = 10 Ix - 10 Iz (1) For mesh y, 120 = (10+j5.655) Iy - (10+j5.655) Iz (2) For mesh z, 0 = -10 Ix �–(10+j5.655) Iy + (50+j5.655) Iz (3) Solving (1) to (3) gives Ix =20, Iy =17.09-j5.142, Iz =8 Thus, I1 =Ix =20 A I2 =-Iy =-17.09+j5.142 = A 26.16385. o17 In =Iy - Ix =-2.091 �–j5.142 = A 5.119907.5 o

(b) 5.3085.1025)120(21,12002060)120(

21

21 jISxIS yx

VA 5.3085.222521 jSSS

(c ) pf = P/S = 2225.5/2246.8 = 0.9905 Chapter 11, Solution 86. For maximum power transfer

*LThi

*ThL ZZZZZ

)104)(1012.4)(2(j75LjR -66LZ

55.103j75LZ

iZ 55.103j75 Chapter 11, Solution 87. jXRZ

k6.11050

80RR 3-

RR I

VIV

22222222

)6.1()3(RXXR ZZ k5377.2X

77.576.1

5377.2tan

RX

tan 1-1-

cospf 5333.0

Page 577: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 88.

(a) 55220)552)(110(S

)55cos(220cosSP W2.126

(b) SS VA220 Chapter 11, Solution 89.

(a) Apparent power S kVA12

kW36.9)78.0)(12(cosSP kVAR51.7))78.0(sin(cos12sinSQ -1

jQPS kVA51.7j36.9

(b) 3

22

**

2

10)51.7j36.9()210(

SV

ZZV

S

Z 6.27j398.34

Chapter 11, Solution 90 Original load :

kW2000P1 , 79.3185.0cos 11

kVA94.2352cos

PS

1

11

kVAR5.1239sinSQ 111

Additional load :

kW300P2 , 87.368.0cos 22

kVA375cos

PS

2

22

kVAR225sinSQ 222

Page 578: Solution Manual for Fundamentals of Electric Circuits

Total load :

jQP)QQ(j)PP( 212121 SSS kW23003002000P

kVAR5.14642255.1239Q The minimum operating pf for a 2300 kW load and not exceeding the kVA rating of the generator is

9775.094.2352

2300SP

cos1

or 177.12

The maximum load kVAR for this condition is

)177.12sin(94.2352sinSQ 1m kVAR313.496Qm

The capacitor must supply the difference between the total load kVAR ( i.e. Q ) and the permissible generator kVAR ( i.e. ). Thus, mQ

mc QQQ kVAR2.968 Chapter 11, Solution 91

cosSP

)15)(220(2700

SP

cospf 8182.0

3.1897)09.35sin()15(220sinSQ

When the power is raised to unity pf, 01 and Q 3.1897Qc

22rms

c

)220)(60)(2(3.1897

VQ

C F104

Page 579: Solution Manual for Fundamentals of Electric Circuits

Chapter 11, Solution 92

(a) Apparent power drawn by the motor is

kVA8075.0

60cos

PSm

kVAR915.52)60()80(PSQ 2222

m Total real power

kW8020060PPPP Lcm Total reactive power

020915.52QQQQ Lcm kVAR91.32 Total apparent power

22 QPS kVA51.86

(b) 51.86

80SP

pf 9248.0

(c) 550

86510VS

I A3.157

Chapter 11, Solution 93

(a) kW7285.3)7457.0)(5(P1

kVA661.48.0

7285.3pfP

S 11

kVAR796.2))8.0(sin(cosSQ -1

11 kVA796.2j7285.31S

kW2.1P2 , VAR0Q2

kVA0j2.12S

kW2.1)120)(10(P3 , VAR0Q3 kVA0j2.13S

Page 580: Solution Manual for Fundamentals of Electric Circuits

kVAR6.1Q4 , 8.0sin6.0cos 44

kVA2sin

QS

4

44

kW2.1)6.0)(2(cosSP 444

kVA6.1j2.14S

4321 SSSSS kVA196.1j3285.7S

Total real power = 7 kW3285. Total reactive power = 1 kVAR196.

(b) 27.93285.7196.1

tan 1-

cospf 987.0

Chapter 11, Solution 94

57.457.0cos 11 kVA1000MVA1S1 kW700cosSP 111

kVAR14.714sinSQ 111 For improved pf,

19.1895.0cos 22 kW700PP 12

kVA84.73695.0

700cos

PS

2

22

kVAR08.230sinSQ 222

Page 581: Solution Manual for Fundamentals of Electric Circuits

P1 = P2 = 700 kW

Q2

Q1

S1

S2

1 2

Qc

(a) Reactive power across the capacitor

kVAR06.48408.23014.714QQQ 21c Cost of installing capacitors 06.48430$ 80.521,14$

(b) Substation capacity released 21 SS kVA16.26384.7361000

Saving in cost of substation and distribution facilities

16.263120$ 20.579,31$

(c) Yes, because (a) is greater than (b). Additional system capacity obtained by using capacitors costs only 46% as much as new substation and distribution facilities.

Chapter 11, Solution 95

(a) Source impedance css XjRZ Load impedance 2LL XjRZ For maximum load transfer

LcLs*sL XX,RRZZ

LC1

XX Lc

or f2LC1

Page 582: Solution Manual for Fundamentals of Electric Circuits

)1040)(1080(21

LC21

f9-3-

kHz814.2

(b) )10)(4(

)6.4(R4

VP

2

L

2s mW529 (since V is in rms) s

Chapter 11, Solution 96

ZTh

+

VTh ZL

(a) Hz300,V146VTh

8j40ZTh

*ThL ZZ 8j40

(b) )40)(8(

)146(R8

VP

2

Th

2

Th W61.66

Chapter 11, Solution 97

22j2.100)20j100()j1.0)(2(ZT

22j2.100240

ZV

IT

s

22

22

L

2

)22()2.100()240)(100(

I100RIP W3.547

Page 583: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 1.

(a) If , then 400abV

30-3

400anV V30-231

bnV V150-231

cnV V270-231

(b) For the acb sequence, 120V0V ppbnanab VVV

30-3V23

j21

1V ppabV

i.e. in the acb sequence, lags by 30 . abV anV Hence, if , then 400abV

303

400anV V30231

bnV V150231

cnV V90-231 Chapter 12, Solution 2.

Since phase c lags phase a by 120 , this is an acb sequence.

)120(30160bnV V150160 Chapter 12, Solution 3. Since V leads by 120 , this is an bn cnV abc sequence.

)120(130208anV V250208

Page 584: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 4.

120cabc VV V140208

120bcab VV V260208

303260208

303ab

an

VV V230120

120-anbn VV V110120

Chapter 12, Solution 5. This is an abc phase sequence.

303anab VV

or 3030420

303ab

an

VV V30-5.242

120-anbn VV V150-5.242

120ancn VV V905.242

Chapter 12, Solution 6.

26.5618.115j10YZ The line currents are

26.5618.110220

Y

ana Z

VI A26.56-68.19

120-ab II A146.56-68.19

120ac II A93.4468.19

Page 585: Solution Manual for Fundamentals of Electric Circuits

The line voltages are

303200abV V30381

bcV V90-381

caV V210-381 The load voltages are

anYaAN VZIV V0220

bnBN VV V120-220

cnCN VV V120220 Chapter 12, Solution 7. This is a balanced Y-Y system.

+

440 0 V ZY = 6 j8

Using the per-phase circuit shown above,

8j60440

aI A53.1344

120-ab II A66.87-44

120ac II A13.73144 Chapter 12, Solution 8. , V220VL 9j16YZ

29.36-918.6)9j16(3

2203VV

Y

L

Y

pan ZZ

I

LI A918.6

Page 586: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 9.

15j200120

YL

ana ZZ

VI A36.87-8.4

120-ab II A156.87-8.4

120ac II A83.138.4

As a balanced system, nI A0

Chapter 12, Solution 10. Since the neutral line is present, we can solve this problem on a per-phase basis.

For phase a,

36.5355.620j270220

2A

ana Z

VI

For phase b,

120-1022

120-2202B

bnb Z

VI

For phase c,

97.3892.165j12

1202202C

cnc Z

VI

The current in the neutral line is

)-( cban IIII or cban- IIII

)78.16j173.2-()66.8j5-()9.3j263.5(- nI

02.12j91.1nI A81-17.12

Page 587: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 11.

90-310220

90-390-3BCbc

an

VVV

anV V100127

120BCAB VV V130220

V110-220120-BCAC VV

If , then 6030bBI

18030aAI , 60-30cCI

21032.1730-3

1803030-3

aAAB

II

9032.17BCI , 30-32.17CAI

CAAC -II A15032.17

BCBC VZI

9032.170220

BC

BC

IV

Z 80-7.12

Chapter 12, Solution 12. Convert the delta-load to a wye-load and apply per-phase analysis.

Ia

110 0 V +

ZY

45203Y

ZZ

Page 588: Solution Manual for Fundamentals of Electric Circuits

45200110

aI A45-5.5

120-ab II A165-5.5

120ac II A755.5

Chapter 12, Solution 13.

First we calculate the wye equivalent of the balanced load.

ZY = (1/3)Z = 6+j5

Now we only need to calculate the line currents using the wye-wye circuits.

A07.58471.615j8120110

I

A07.178471.615j8120110

I

A93.61471.65j610j2

110I

c

b

a

Page 589: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 14. We apply mesh analysis. 2j1 A a + ZL 100 ZV 0o L - I3 n I1 B C - -

100 + - + c I

V 120100 o V 120o 1212 jZ L

2 b 2j1 2j1 For mesh 1,

0)1212()21()1614(120100100 321 IjIjjIo or

6.861506.8650100)1212()21()1614( 321 jjIjIjIj (1) For mesh 2,

0)1614()1212()21(120100120100 231 IjIjjIoo or

2.1736.86506.8650)1212()1614()21( 321 jjjIjIjIj (2) For mesh 3,

0)3636()1212()1212( 321 IjIjIj (3) Solving (1) to (3) gives

016.124197.4,749.16098.10,3.19161.3 321 jIjIjI

A 3.9958.191o

aA II

A 8.159392.712o

bB III

A 91.5856.192o

cC II

Page 590: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 15.

Convert the delta load, , to its equivalent wye load. Z

10j83Ye

ZZ

14.68-076.85j20

)10j8)(5j12(|| YeYp ZZZ

047.2j812.7pZ

047.1j812.8LpT ZZZ

6.78-874.8TZ

We now use the per-phase equivalent circuit.

Lp

pa

VZZ

I , where 3

210pV

78.666.13)6.78-874.8(3

210aI

aL II A66.13 Chapter 12, Solution 16.

(a) 15010)180-30(10- ACCA II This implies that 3010ABI

90-10BCI

30-3ABa II A032.17

bI A120-32.17

cI A12032.17

(b) 3010

0110

AB

AB

IV

Z 30-11

Page 591: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 17. Convert the -connected load to a Y-connected load and use per-phase analysis.

Ia

+

ZL

Van ZY

4j33Y

ZZ

48.37-931.19)5.0j1()4j3(

0120

LY

ana ZZ

VI

But 30-3ABa II

30-348.37-931.19

ABI A18.37-51.11

BCI A138.4-51.11

CAI A101.651.11

)53.1315)(18.37-51.11(ABAB ZIV

ABV V76.436.172

BCV V85.24-6.172

CAV V8.5416.172 Chapter 12, Solution 18.

901.762)303)(60440(303anAB VV

36.87159j12Z

Page 592: Solution Manual for Fundamentals of Electric Circuits

36.8715901.762AB

AB ZV

I A53.1381.50

120-ABBC II A66.87-81.50

120ABCA II A173.1381.50 Chapter 12, Solution 19.

18.4362.3110j30Z The phase currents are

18.4362.310173ab

AB ZV

I A18.43-47.5

120-ABBC II A138.43-47.5

120ABCA II A101.5747.5 The line currents are

30-3ABCAABa IIII

48.43-347.5aI A48.43-474.9

120-ab II A168.43-474.9

120ac II A71.57474.9 Chapter 12, Solution 20.

36.87159j12Z The phase currents are

36.87150210

ABI A36.87-14

120-ABBC II A156.87-14

120ABCA II A83.1314 The line currents are

30-3ABa II A66.87-25.24

120-ab II A186.87-25.24

120ac II A53.1325.24

Page 593: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 21.

(a) )rms(A66.9896.1766.38806.12

1202308j10

120230IAC

(b)

A34.17110.31684.4j75.30220.11j024.14536.6j729.16

66.3896.1766.15896.178j100230

8j10120230

IIIII ABBCBABCbB

Chapter 12, Solution 22.

Convert the -connected source to a Y-connected source.

30-12030-3

20830-

3

VpanV

Convert the -connected load to a Y-connected load.

j8)5j4)(6j4(

)5j4(||)6j4(3

||Y

ZZZ

2153.0j723.5Z

Ia

+

ZL

Van Z

2153.0j723.730120

L

ana ZZ

VI A28.4-53.15

120-ab II A148.4-53.15

120ac II A91.653.15

Page 594: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 23.

(a) oAB

AB ZV

I6025

208

oo

oo

ABa II 90411.146025

303208303

A 41.14|| aL II

(b) kW 596.260cos25

3208)208(3cos321

oLL IVPPP

Chapter 12, Solution 24. Convert both the source and the load to their wye equivalents.

10j32.1730203Y

ZZ

02.24030-3ab

an

VV

We now use per-phase analysis.

Ia

+

1 + j

Van 20 30

3137.212.240

)10j32.17()j1(an

a

VI A31-24.11

120-ab II A151-24.11

120ac II A8924.11

Page 595: Solution Manual for Fundamentals of Electric Circuits

But 30-3ABa II

30-331-24.11

ABI A1-489.6

120-ABBC II A121-489.6

120ABCA II A119489.6

Chapter 12, Solution 25.

Convert the delta-connected source to an equivalent wye-connected source and consider the single-phase equivalent.

Ya 3

)3010(440Z

I

where 78.24-32.146j138j102j3YZ

)24.78-32.14(320-440

aI A4.7874.17

120-ab II A115.22-74.17

120ac II A124.7874.17

Chapter 12, Solution 26. Transform the source to its wye equivalent.

30-17.7230-3

VpanV

Now, use the per-phase equivalent circuit.

ZV

I anaA , 32-3.2815j24Z

Page 596: Solution Manual for Fundamentals of Electric Circuits

32-3.2830-17.72

aAI A255.2

120-aAbB II A118-55.2

120aAcC II A12255.2

Chapter 12, Solution 27.

)15j20(310-220

330-

Y

aba Z

VI

aI A46.87-081.5

120-ab II A166.87-081.5

120ac II A73.13081.5

Chapter 12, Solution 28. Let 0400abV

307.7)60-30(3

30-4003

30-

Y

ana Z

VI

aLI I A7.7

30-94.23030-3an

YaAN

VZIV

ANpV V V9.230

Page 597: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 29.

, cosIV3P pp 3V

V Lp , pL II

cosIV3P LL

pL

L I05.20)6.0(3240

5000cosV3

PI

911.6)05.20(3

240I3

VIV

L

L

p

pYZ

13.536.0cos

(leading)53.13-911.6YZ

YZ 53.5j15.4

83336.0

5000pfP

S

6667sinSQ

S VA6667j5000

Chapter 12, Solution 30.

Since this a balanced system, we can replace it by a per-phase equivalent, as shown below. + ZL Vp

-

Page 598: Solution Manual for Fundamentals of Electric Circuits

3,

33 *

2L

pp

pp

VV

ZV

SS

kVA 454421.14530)208( 2

*

2o

op

L

ZV

S

kW 02.1cosSP

Chapter 12, Solution 31.

(a) kVA 5.78.0/6cos

,8.0cos,000,6 Ppp

PSP

kVAR 5.4sinPp SQ

kVA 5.1318)5.46(33 jjSS p For delta-connected load, Vp = VL= 240 (rms). But

608.4144.6,10)5.1318(

)240(3333

22*

*

2

jZxjS

VZ

ZV

S Pp

pp

p

(b) A 04.188.02403

6000cos3

xxIIVP LLLp

(c ) We find C to bring the power factor to unity

F 2.207240602

4500kVA 5.4 22 xxV

QCQQ

rms

cpc

Chapter 12, Solution 32.

LLIV3S

3LL 1050IV3S S

)440(35000

IL A61.65

Page 599: Solution Manual for Fundamentals of Electric Circuits

For a Y-connected load,

61.65II Lp , 03.2543

4403

VV L

p

872.361.6503.254

IV

p

pZ

ZZ , 13.53)6.0(cos-1

)sinj)(cos872.3(Z

)8.0j6.0)(872.3(Z

Z 098.3j323.2

Chapter 12, Solution 33. LLIV3S

LLIV3S S

For a Y-connected load,

pL II , pL V3V

ppIV3S

)208)(3(4800

V3S

IIp

pL A69.7

2083V3V pL V3.360

Page 600: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 34.

3

2203

VV L

p

5873.6)16j10(3

200V

Y

pa Z

I

pL II A73.6

58-73.62203IV3 LLS

S VA8.2174j1359

Chapter 12, Solution 35.

(a) This is a balanced three-phase system and we can use per phase equivalent circuit. The delta-connected load is converted to its wye-connected equivalent

10203/)3060(31'' jjZZ y

IL

+ 230 V Z�’y Z�’�’y

-

5.55.13)1020//()1040(//' '' jjjZZZ yyy

A 953.561.145.55.13

230j

jI L

(b) kVA 368.1361.3* jIVS Ls (c ) pf = P/S = 0.9261

Page 601: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 36.

(a) S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA

(b) 49.5252.5942003

10)6614.075.0(3

36

** jx

xjVS

IIVp

pppS

kW 19.25)4()36.79(|| 22

lpL RIP (c) kV 2.709-4.443 kV 21.04381.4)4( ojjIV pLsV Chapter 12, Solution 37.

206.0

12pfP

S

kVA16j1220SS

But LLIV3S

20831020

I3

L A51.55

p

2

p3 ZIS

For a Y-connected load, pL II .

2

3

2

L

p )51.55)(3(10)16j12(

I3

SZ

pZ 731.1j298.1

Page 602: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 38.

As a balanced three-phase system, we can use the per-phase equivalent shown below.

14j100110

)12j9()2j1(0110

aI

)12j9()1410(

)110(21

21

22

2

Y

2

ap ZIS

The complex power is

)12j9(296

)110(23

32

pSS

S VA81.735j86.551

Chapter 12, Solution 39. Consider the system shown below.

I2

I1

I3

5

5

-j6

10

j3

B C

A

8

4

b

a

+

+

100 -120

100 120 100 0 +

5

c

For mesh 1,

321 )6j8(5)6j18(100 III (1) For mesh 2,

312 10520120-100 III

321 24-120-20 III (2)

Page 603: Solution Manual for Fundamentals of Electric Circuits

For mesh 3, 321 )3j22(10)6j8(-0 III (3)

To eliminate , start by multiplying (1) by 2, 2I

321 )12j16(10)12j36(200 III (4) Subtracting (3) from (4),

31 )15j38()18j44(200 II (5) Multiplying (2) by 45 ,

321 5.2525.1-120-25 III (6) Adding (1) and (6),

31 )6j5.10()6j75.16(65.21j5.87 II (7) In matrix form, (5) and (7) become

3

1

6j5.10-6j75.1615j38-18j44

65.12j5.87200

II

25.26j5.192 , 2.935j25.9001 , 6.1327j3.1103

144.4j242.538.33-682.67.76-28.194

46.09-1.129811I

694.6j485.177.49-857.67.76-28.194

85.25-2.133233I

We obtain from (6), 2I

312 21

41

120-5 III

)347.3j7425.0()0359.1j3104.1()33.4j-2.5(2I

713.8j4471.0-2I The average power absorbed by the 8- resistor is

W89.164)8(551.2j756.3)8(P22

311 II The average power absorbed by the 4- resistor is

W1.188)4()8571.6()4(P 22

32 I

Page 604: Solution Manual for Fundamentals of Electric Circuits

The average power absorbed by the 10- resistor is

W12.78)10(019.2j1.9321-)10(P 22323 II

Thus, the total real power absorbed by the load is

321 PPPP W1.431 Chapter 12, Solution 40. Transform the delta-connected load to its wye equivalent.

8j73Y

ZZ

Using the per-phase equivalent circuit above,

46.75-567.8)8j7()5.0j1(

0100aI

For a wye-connected load,

567.8II aap I

)8j7()567.8)(3(3 2p

2

p ZIS

)7()567.8)(3()Re(P 2S k541.1 W

Chapter 12, Solution 41.

kVA25.68.0

kW5pfP

S

But LLIV3S

40031025.6

V3S

I3

LL A021.9

Page 605: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 42. The load determines the power factor.

13.53333.13040

tan

(leading)6.0cospf

kVA6.9j2.7)8.0(6.02.7

j2.7S

But p

2

p3 ZIS

80)40j30)(3(

10)6.9j2.7(3

3

p

2

p ZS

I

A944.8Ip

pL II A944.8

)944.8(31012

I3S

V3

LL V6.774

Chapter 12, Solution 43.

p

2

p3 ZIS , Lp II for Y-connected loads

)047.2j812.7()66.13)(3( 2S S kVA145.1j373.4

Page 606: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 44. For a -connected load,

Lp VV , pL I3I

LLIV3S

273.31)240(3

10)512(V3

SI

322

LL

At the source,

LLL'L ZIVV

)3j1)(273.31(0240'LV

819.93j273.271'LV

'LV V04.287

Also, at the source,

*L

'L

' 3 IVS )273.31)(819.93j273.271(3'S

078.19273.271

819.93tan 1-

cospf 9451.0

Chapter 12, Solution 45. LLIV3S

LL V3

-I

S, kVA6.635

708.010450

pfP 3

S

A45-8344403

-)6.635(LI

At the source,

)2j5.0(0440 LL IV

Page 607: Solution Manual for Fundamentals of Electric Circuits

)76062.2)(45-834(440LV 137.1719440LV

7.885j1.1914LV

LV V.8324109.2 Chapter 12, Solution 46. For the wye-connected load,

pL II , pL V3V Zpp VI

*

2

L*

2

p*pp

3333

ZV

Z

VIVS

W121100

)110( 2

*

2

L

ZV

S

For the delta-connected load,

Lp VV , pL I3I , Zpp VI

*

2

L*

2

p*pp

333

ZV

Z

VIVS

W363100

)110)(3( 2

S

This shows that the delta-connected load will deliver three times more average

power than the wye-connected load. This is also evident from 3Y

ZZ .

Chapter 12, Solution 47.

87.36)8.0(cos(lagging)8.0pf -1 kVA150j20087.362501S

19.18-)95.0(cos(leading)95.0pf -1

kVA65.93j28519.81-3002S

Page 608: Solution Manual for Fundamentals of Electric Circuits

0)1(cos0.1pf -1

kVA4503S

kVA45.37.93635.56j935321T SSSS

LLT IV3S

)108.13(3107.936

I3

3

L rmsA19.39

)45.3cos(cospf (lagging)9982.0

Chapter 12, Solution 48.

(a) We first convert the delta load to its equivalent wye load, as shown below. A A ZA 18-j12 40+j15 ZC C B C 60

ZB

B

923.1577.73118

)1218)(1540(j

jjj

Z A

105.752.203118

).1540(60j

jj

Z B

3303.6992.83118

)1218(60j

jj

ZC

The system becomes that shown below.

Page 609: Solution Manual for Fundamentals of Electric Circuits

a 2+j3 A + 240<0o ZA - I1 - - ZB ZC 240<120o 240<-120o + + 2+j3 c I2 b B C 2+j3

We apply KVL to the loops. For mesh 1, 0)()2(120240240 21 lBBAl

o ZZIZZZI or

85.207360)105.1052.22()13.11097.32( 21 jIjIj (1) For mesh 2,

0)2()(120240120240 21 CBllBoo ZZZIZZI

or

69.415)775.651.33()105.1052.22( 21 jIjIj (2) Solving (1) and (2) gives

89.11165.15,328.575.23 21 jIjI

A 6.14281.10,A 64.1234.24 121o

bBo

aA IIIII

A 9.14127.192o

cC II (b) ooo

aS 64.126.5841)64.1234.24)(0240( ooo

bS 6.224.2594)6.14281.10)(120240( ooo

bS 9.218.4624)9.14127.19)(120240( kVA 54.24.12kVA 55.0386.12 o

cba jSSSS

Page 610: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 49.

(a) For the delta-connected load, (rms) 220,1020 Lpp VVjZ ,

kVA 56.26943.629045808)1020(

22033 2

*

2o

p

p jj

xZV

S

(b) For the wye-connected load, 3/,1020 Lpp VVjZ ,

kVA 56.26164.2)1020(3

22033 2

*

2o

p

p

jx

ZV

S

Chapter 12, Solution 50.

kVA 3kVA, 4.68.4)8.06.0(8 121 SjjSSS Hence,

kVA 4.68.112 jSSS

But p

LLp

p

p

ZV

SV

VZV

S *

2

2*

2

2.

3,3

34.8346.210)4.68.1(

2403

2

2

** jZ

xjSV

Z pL

p

Chapter 12, Solution 51. Apply mesh analysis to the circuit as shown below.

Za

i2

i1

Zc

Zb

+

+

+

150 -120

150 0

n 150 120

Page 611: Solution Manual for Fundamentals of Electric Circuits

For mesh 1,

0)(150- 2b1ba IZIZZ

21 )9j12()j18(150 II (1) For mesh 2,

0)(120-150- 1b2cb IZIZZ

12 )9j12()9j27(120-150 II (2) From (1) and (2),

2

1

9j279j12-9j12-j18

120-150150

II

27j414 , 8.3583j9.37801 , 2.1063j9.5792

47.256.123.73-88.414

43.475.520911I

57.66-919.23.73-88.414

61.39-1.121122I

1a II A47.256.12

4647j3201-12

12b III

3.73-88.414235.443.5642

bI A239.176.13

2c -II A122.34919.2

Chapter 12, Solution 52. Since the neutral line is present, we can solve this problem on a per-phase basis.

6066020120120

AN

ana Z

VI

040300120

BN

bnb Z

VI

Page 612: Solution Manual for Fundamentals of Electric Circuits

150-33040120-120

CN

cnc Z

VI

Thus,

cban- IIII 150-304606- nI

)5.1j598.2-()4()196.5j3(- nI 4075.5696.3j405.4- nI

nI A22075.5

Chapter 12, Solution 53.

3

250Vp

Since we have the neutral line, we can use per-phase equivalent circuit for each phase.

60401

30250

aI A60-608.3

45-601

3120-250

bI A75-406.2

0201

3120250

cI A120217.7

cban- IIII

)25.6j609.3-()324.2j6227.0()125.3j804.1(- nI

801.0j1823.1nI A34.12- 428.1

Page 613: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 54. Consider the circuit shown below.

Iaa

B C

A

j50

-j50

50

IAB

c b

Vp 0

+

Vp 120 Vp -120

+

+

303100abAB VV

50303100

AB

ABAB Z

VI A30464.3

90-5090-3100

BC

BCBC Z

VI A0464.3

90501503100

CA

CACA Z

VI A60464.3

Page 614: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 55. Consider the circuit shown below.

Iaa A

Ib

I1

I2

c b

220 0

+

220 120 220 -120

+

+

30 + j40100 �– j120

B C

60 + j80

Ic

For mesh 1, 0)120j100()40j160(0220120-220 21 II

21 )6j5()2j8(120-1111 II (1) For mesh 2,

0)120j100()80j130(120-220120220 12 II

21 )4j5.6()6j5(-12011120-11 II (2) From (1) and (2),

2

1

j4-6.5j65-j65-2j8

j19.053-526.9j5.16

II

15j55 , 35.99j04.311 , 8.203j55.1012

87.91-8257.115.2601.57

72.65-08.10411I

78.77-994.315.2601.5763.51-7.2272

2I

87.91-8257.11a II

Page 615: Solution Manual for Fundamentals of Electric Circuits

71.23-211.215j55

45.104j51.701212b III

23.011994.3- 2c II

7.266j99.199)80j60()8257.1( 2AN

2

aA ZIS

6.586j9.488)120j100()211.2( 2BN

2

bB ZIS

1.638j6.478)40j30()994.3( 2CN

2

cC ZIS

CBA SSSS V2.318j5.1167 A Chapter 12, Solution 56.

(a) Consider the circuit below.

For mesh 1,

0)(10j0440120-440 31 II

60-21.7610j

)866.0j5.1)(440(31 II (1)

For mesh 2,

0)(20120-440120440 32 II

1.38j20

)732.1j)(440(23 II (2)

For mesh 3, 05j)(20)(10j 32313 IIIII

I2

I1

b

+

+

440 -120440 120

440 0

+

BI3

20

j10

a A

-j5

C c

Page 616: Solution Manual for Fundamentals of Electric Circuits

Substituting (1) and (2) into the equation for mesh 3 gives,

6042.152j5

j0.866)-1.5)(440(3I (3)

From (1),

3013266j315.11460-21.7631 II From (2),

50.9493.1209.93j21.761.38j32 II

1a II A30132

j27.9-38.10512b III A143.823.47

2c -II A230.99.120

(b) kVA08.58j)10j(2

31AB IIS

kVA04.29)20(2

32BC IIS

kVA-j116.16-j5)((152.42)-j5)( 22

3CA IS

kVA08.58j04.29CABCAB SSSS Real power absorbed = 29 kW04.

(c) Total complex supplied by the source is S kVA08.58j04.29

Page 617: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 57.

We apply mesh analysis to the circuit shown below. Ia

+ Va 50j80 - I1 - - 3020 j

Vc Vb + + Ib

I2 Ic

4060 j

263.95165)3020()80100( 21 jVVIjIj ba (1) 53.190)1080()3020( 21 jVVIjIj cb (2)

Solving (1) and (2) gives 722.19088.0,6084.08616.1 21 jIjI .

A 55.1304656.11136.1528.0,A 1.189585.1 121o

bo

a jIIIII

A 8.117947.12o

c II Chapter 12, Solution 58. The schematic is shown below. IPRINT is inserted in the neutral line to measure the current through the line. In the AC Sweep box, we select Total Ptss = 1, Start Freq. = 0.1592, and End Freq. = 0.1592. After simulation, the output file includes

FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E�–01 1.078 E+01 �–8.997 E+01

Page 618: Solution Manual for Fundamentals of Electric Circuits

i.e. In = 10.78 �–89.97 A

Chapter 12, Solution 59. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, we obtain an output file which includes

FREQ VM(1) VP(1) 6.000 E+01 2.206 E+02 �–3.456 E+01 FREQ VM(2) VP(2) 6.000 E+01 2.141 E+02 �–8.149 E+01 FREQ VM(3) VP(3) 6.000 E+01 4.991 E+01 �–5.059 E+01

i.e. VAN = 220.6 �–34.56 , VBN = 214.1 �–81.49 , VCN = 49.91 �–50.59 V

Page 619: Solution Manual for Fundamentals of Electric Circuits

hapter 12, Solution 60.

he schematic is shown below. IPRINT is inserted to give Io. We select Total Pts = 1,

FREQ IM(V_PRINT4) IP(V_PRINT4)

.592 E�–01 1.421 E+00 �–1.355 E+02

om which, Io = 1.421 �–135.5 A

C TStart Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. Upon simulation, the output file includes

1

fr

Page 620: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 61. The schematic is shown below. Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. Once the circuit is simulated, we get an output file which includes

FREQ VM(2) VP(2) 1.592 E�–01 2.308 E+02 �–1.334 E+02 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 1.115 E+01 3.699 E+01

from which IaA = 11.15 37 A, VBN = 230.8 �–133.4 V

Page 621: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 62. Because of the delta-connected source involved, we follow Example 12.12. In the AC Sweep box, we type Total Pts = 1, Start Freq = 60, and End Freq = 60. After simulation, the output file includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 6.000 E+01 5.960 E+00 �–9.141 E+01 FREQ IM(V_PRINT1) IP(V_PRINT1) 6.000 E+01 7.333 E+07 1.200 E+02

From which Iab = 7.333x107 120 A, IbB = 5.96 �–91.41 A

Page 622: Solution Manual for Fundamentals of Electric Circuits
Page 623: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 63.

Let F 0333.0X1 C and H, 20X/ L that so 1

The schematic is shown below..

.

When the file is saved and run, we obtain an output file which includes the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 1.592E-01 1.867E+01 1.589E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 1.592E-01 1.238E+01 1.441E+02 From the output file, the required currents are:

A 1.14438.12 A, 9.15867.18 oAC

oaA II

Page 624: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 64. We follow Example 12.12. In the AC Sweep box we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation the output file includes

FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 4.710 E+00 7.138 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 6.781 E+07 �–1.426 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E�–01 3.898 E+00 �–5.076 E+00 FREQ IM(V_PRINT4) IP(V_PRINT4) 1.592 E�–01 3.547 E+00 6.157 E+01 FREQ IM(V_PRINT5) IP(V_PRINT5) 1.592 E�–01 1.357 E+00 9.781 E+01 FREQ IM(V_PRINT6) IP(V_PRINT6) 1.592 E�–01 3.831 E+00 �–1.649 E+02

from this we obtain

IaA = 4.71 71.38 A, IbB = 6.781 �–142.6 A, IcC = 3.898 �–5.08 A

IAB = 3.547 61.57 A, IAC = 1.357 97.81 A, IBC = 3.831 �–164.9 A

Page 625: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 65. Due to the delta-connected source, we follow Example 12.12. We type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. The schematic is shown below. After it is saved and simulated, we obtain an output file which includes

FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 6.581 E+00 9.866 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 1.140 E+01 �–1.113 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E�–01 6.581 E+00 3.866 E+01

Thus, IaA = 6.581 98.66 A, IbB = 11.4 �–111.3 A, IcC = 6.581 38.66 A

Page 626: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 66. Chapter 12, Solution 66.

(a) 3

2083

VV L

p V120 (a)

3

2083

VV L

p V120

(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,

(b) Because the load is unbalanced, we have an unbalanced three-phase system. Assuming an abc sequence,

A05.248

01201I

A120-340

120-1202I

A120260

1201203I

A05.248

01201I

A120-340

120-1202I

A120260

1201203I

23

j5.0-)2(23

j0.5-)3(5.2- 321N IIII23

j5.0-)2(23

j0.5-)3(5.2- 321N IIII

Page 627: Solution Manual for Fundamentals of Electric Circuits

A90866.0866.0j23

jNI

Hence, 1I A5.2 , 2I A3 , 3I A2 , NI A866.0

(c) )48()5.2(RIP 2

1211 W300

)40()3(RIP 22

222 W360

)60()2(RIP 23

233 W240

(d) 321T PPPP W900

Chapter 12, Solution 67.

(a) The power to the motor is kW221)85.0)(260(cosSPT

The motor power per phase is

kW67.73P31

P Tp

Hence, the wattmeter readings are as follows:

2467.73Wa kW67.97

1567.73Wb kW67.88

967.73Wc kW67.83

(b) The motor load is balanced so that 0IN . For the lighting loads,

A200120

000,24Ia

A125120

000,15Ib

A75120000,9

Ic

If we let

Page 628: Solution Manual for Fundamentals of Electric Circuits

A02000IaaI A120-125bI

A12075cI Then,

cbaN- IIII

23

j0.5-)75(23

j5.0-)125(200- NI

A602.86100- NI

NI A3.132 Chapter 12, Solution 68.

(a) )4.8)(330(3IV3S LL VA4801

(b) SP

cospfcosSP

24.48014500

pf 9372.0

(c) For a wye-connected load,

Lp II A4.8

(d) 3

3303

VV L

p V53.190

Chapter 12, Solution 69.

MVA 8.0SMVA, 323.15.1)661.075.0(2SMVA, 72.096.0)6.08.0(2.1 321 jjjS

9833.03153.326.3MVA, 603.026.3321

SP

pfjSSSS

MVA 1379.0)99.0tan(cos)9833.0[tan(cos26.3)tan(tan 11

newoldc PQ

mF 28106.6602

101379.031

62

6

xxx

xxC

Page 629: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 70.

8004001200PPP 21T

1600-1200400-PPQ 12T

-63.43-28001600-

PQ

tanT

T

cospf (leading)4472.0

406

240IV

ZL

Lp

pZ 63.43-40

Chapter 12, Solution 71.

(a) If , 0208abV 120-208bcV , 120208caV ,

04.1020

0208

Ab

abAB Z

VI

75-708.1445-210

120-208

BC

bcBC Z

VI

97.381622.6213

120208

CA

caCA Z

VI

97.381604.10CAABaA III

867.15j055.24.10aAI 51.87-171.20aAI

75-708.1497.8316BCCAcC III

101.0364.30cCI

)cos(PaAab IVaAab1 IV

Page 630: Solution Manual for Fundamentals of Electric Circuits

)87.510cos()171.20)(208(P1 W2590

)cos(PcCcb IVcCcb2 IV

But 60208- bccb VV

)03.10160cos()64.30)(208(P2 W4808

(b) W17.7398PPP 21T

VAR25.3840)PP(3Q 12T

TTT jQPS VA25.3840j17.7398

TTS S V8335 A Chapter 12, Solution 72. From Problem 12.11,

V130220ABV and A18030aAI

)180130cos()30)(220(P1 W4242

190220- BCCB VV 60-30cCI

)60190cos()30)(220(P2 W2257-

Chapter 12, Solution 73. Consider the circuit as shown below.

I1

I2

Z

Z

240 -120 V +

Ia

Ib

+

240 -60 V

Z

Ic

Page 631: Solution Manual for Fundamentals of Electric Circuits

71.5762.3130j10Z

131.57-59.771.5762.31

60-240aI

191.57-59.771.5762.31120-240

bI

0120-24060-240c ZI

108.4359.771.5731.62

240-cI

101.57-146.13ca1 III

138.43146.13cb2 III

).57101146.13)(60-240(ReReP *111 IV W2360

)138.43-146.13)(120-240(ReReP *

222 IV W8.632- Chapter 12, Solution 74. Consider the circuit shown below.

Z = 60 j30

For mesh 1,

+

208 -60 V I2

I1+

208 0 V

Z

Z

212208 IZIZ For mesh 2,

21 2-60-208- IZIZ

Page 632: Solution Manual for Fundamentals of Electric Circuits

In matrix form,

2

1

2--2

60-208-208

II

ZZZZ

23Z , Z)866.0j5.1)(208(1 , Z)732.1j)(208(2

56.56789.1)30j60)(3(

)866.0j5.1)(208(11I

116.5679.1)30j60)(3()732.1j)(208(2

2I

)56.56-789.1)(208(ReReP *

111 IV W98.208

)63.4479.1))(60-208(Re)-(ReP *222 IV W65.371

Chapter 12, Solution 75.

(a) 60012

RV

I mA20

(b) 600120

RV

I mA200

Chapter 12, Solution 76. If both appliances have the same power rating, P,

sVP

I

For the 120-V appliance, 120

PI1 .

For the 240-V appliance, 240P

I2 .

Power loss =appliance -V240 for the

240RP

appliance -V120 for the120

RP

R

2

2

2

2

2I

Since 22 2401

1201

, the losses in the 120-V appliance are higher.

Page 633: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 77. , lineloadTg PPPP 85.0pf

But 3060pf3600cos3600PT

)80)(3(25003060Pg W320 Chapter 12, Solution 78.

79.3185.06051

cos 11

kVAR61.31)5268.0)(60(sinSQ 111

kW51PP 12

19.1895.0cos 22

kVA68.53cos

PS

2

22

kVAR759.16sinSQ 222

kVAR851.14759.1661.3QQQ 21c

For each load,

kVAR95.43

QQ c

1c

221c

)440)(60)(2(4950

VQ

C F82.67

Page 634: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 79. Consider the per-phase equivalent circuit below.

Ia

n

a

Van +

2 A

ZY = 12 + j5

N

5j140255

2Y

ana Z

VI A19.65-15.17

Thus,

120-ab II A139.65-15.17

120ac II A100.3515.17

)22.6213)(19.65-15.17(YaAN ZIV V2.97223

Thus, 120-ANBN VV V117.63-223

120ANCN VV V122.97223 Chapter 12, Solution 80.

)7071.07071.0(8)]83.0sin(cos83.0[6 21

321 jSjSSSS kVA 31.26368.10 2SjS (1)

But

kVA j18.28724.383 VA )6.08.0)(6.84)(208(33 jIVS LL (2) From (1) and (2),

kVA 28.5676.246.20746.132 jS Thus, the unknown load is 24.76 kVA at 0.5551 pf lagging.

Page 635: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 81.

87.36-(leading)8.0pf 1 kVA36.87-1501S

00.1pf 2

kVA01002S

13.53(lagging)6.0pf 3 kVA53.132003S

kVA95j804S

4321 SSSSS

kVA21.452.451165j420S

LLIV3S

kVA18.44j67.17LS

At the source,

2.209j7.437LT SSS kVA25.551.485TS

7.5423101.485

I3S

V3

L

TT V516

Page 636: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 82.

p

p

ZV

jjS *

2

21 3SkVA, 240320)6.08.0(400

For the delta-connected load, V pL V

kVA 93.8427.1053810)2400(3

2

2 jj

xS

MVA 0829.13737.121 jSSS

Let I = I1 + I2 be the total line current. For I1,

3,3 1

*1

Lpp

VVIVS

735.5798.76,)2400(310)240320(

3 1

31

1* jI

xjVS

IL

For I2, convert the load to wye.

76.2891.273303810

24003032 jj

II oop

5.34735021 jIII

kV 372.5||kV 405.1185.5)63(2400 slineLs VjjIVVV

Chapter 12, Solution 83.

kVA 80SkVA, 135.60135.60)707.0707.0(95.0746120 21 jjxxS

kVA 135.60135.14021 jSSS

A 42.1834803

1049.1523

||3||But 3

xx

VS

IIVSL

LLL

Page 637: Solution Manual for Fundamentals of Electric Circuits

Chapter 12, Solution 84. We first find the magnitude of the various currents.

For the motor,

A248.53440

4000V3

SI

LL

For the capacitor,

A091.4440

1800VQ

IL

cC

For the lighting,

V2543

440Vp

A15.3254800

VP

Ip

LiLi

Consider the figure below.

Ia I1

a

-jXC

R In

ILi

I3

I2

Ic

+

Vab Ib

IC

b

c

n

If , 0VpanV 30V3 pabV 120VpcnV

Page 638: Solution Manual for Fundamentals of Electric Circuits

120091.4X-j C

abC

VI

)30(091.4ab1 Z

VI

where 95.43)72.0(cos-1

73.95249.51I

46.05-249.52I

193.95249.53I

12015.3R

cnLi

VI

Thus,

120091.473.95249.5C1a III

aI A93.96608.8

120091.446.05-249.5C2b III

bI A52.16-271.9

12015.3193.95249.5Li3c III

cI A167.6827.6

Lin -II A60-15.3 Chapter 12, Solution 85. Let Z RY

V56.1383

2403

VV L

p

RV

kW92

27IVP

2p

pp

Page 639: Solution Manual for Fundamentals of Electric Circuits

133.29000

)56.138(P

VR

22p

Thus, YZ 133.2

Chapter 12, Solution 86. Consider the circuit shown below.

1

B

N

A

b

n

I1

I2+

120 0 V rms 15 + j4

24 �– j2

1

a

+

120 0 V rms

1

For the two meshes,

21)2j26(120 II (1)

12)4j17(120 II (2)

In matrix form,

2

1

4j171-1-2j26

120120

II

70j449 , )4j18)(120(1 , )2j27)(120(2

3.6787.48.8642.454

12.5344.1812011I

13.1-15.78.8642.454

4.24-07.2712022I

Page 640: Solution Manual for Fundamentals of Electric Circuits

1aA II A3.6787.4

2bB -II A166.915.7

1212nN III

70j449)6j9)(120(

nNI A42.55-856.2

Chapter 12, Solution 87. 85.18j)5010)(60)(2(jLjmH50L 3-

Consider the circuit below.

1

30

20 I1

I2+

2

+

1

115 V

15 + j18.85

115 V

Applying KVl to the three meshes, we obtain

11520223 321 III (1) 11530332- 321 III (2)

0)85.18j65(3020- 321 III (3) In matrix form,

0115115

j18.856530-20-30-332-20-2-23

3

2

1

III

232,14j775,12 , )8.659j1975)(115(1

)3.471j1825)(115(2 , )1450)(115(3

Page 641: Solution Manual for Fundamentals of Electric Circuits

29.62-52.1248.0919214

47.18208211511I

33.61-33.1148.0919124

14.489.188411522I

75.231,14j775,12)5.188j-150)(115(12

12n III A176.6-448.1

)29.6252.12)(115(IVS *

111 VA6.711j1252

)33.6133.1)(115(*222 IVS V2.721j1085 A

Page 642: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 1. For coil 1, L1 �– M12 + M13 = 6 �– 4 + 2 = 4

For coil 2, L2 �– M21 �– M23 = 8 �– 4 �– 5 = �– 1

For coil 3, L3 + M31 �– M32 = 10 + 2 �– 5 = 7

LT = 4 �– 1 + 7 = 10H

or LT = L1 + L2 + L3 �– 2M12 �– 2M23 + 2M12 LT = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 �– 2M23 2M31

= 10 + 12 +8 + 2x6 �– 2x6 �–2x4 = 22H Chapter 13, Solution 3. L1 + L2 + 2M = 250 mH (1) L1 + L2 �– 2M = 150 mH (2)

Adding (1) and (2),

2L1 + 2L2 = 400 mH

But, L1 = 3L2,, or 8L2 + 400, and L2 = 50 mH

L1 = 3L2 = 150 mH From (2), 150 + 50 �– 2M = 150 leads to M = 25 mH

k = M/ 150x50/5.2LL 21 = 0.2887

Page 643: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus,

Leq = L1 + L2 + 2M

L2

L1

L2 L1

I1 I2

Is

Vs+�–

Leq (a) (b) (b) For the parallel coil, consider Figure (b).

Is = I1 + I2 and Zeq = Vs/Is

Applying KVL to each branch gives,

Vs = j L1I1 + j MI2 (1)

Vs = j MI1 + j L2I2 (2)

or 2

1

2

1

s

s

II

LjMjMjLj

VV

= �– 2L1L2 + 2M2, 1 = j Vs(L2 �– M), 2 = j Vs(L1 �– M)

I1 = 1/ , and I2 = 2/

Is = I1 + I2 = ( 1 + 2)/ = j (L1 + L2 �– 2M)Vs/( �– 2(L1L2 �– M))

Zeq = Vs/Is = j (L1L2 �– M)/[j (L1 + L2 �– 2M)] = j Leq

i.e., Leq = (L1L2 �– M)/(L1 + L2 �– 2M)

Page 644: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 5.

(a) If the coils are connected in series,

60x25)5.0(26025M2LLL 21 123.7 mH

(b) If they are connected in parallel,

mH 36.19x26025

36.1960x25M2LL

MLLL

2

21

221 24.31 mH

Chapter 13, Solution 6.

V1 = (R1 + j L1)I1 �– j MI2

V2 = �–j MI1 + (R2 + j L2)I2

Chapter 13, Solution 7. Applying KVL to the loop,

20 30 = I(�–j6 + j8 + j12 + 10 �– j4x2) = I(10 + j6)

where I is the loop current.

I = 20 30 /(10 + j6)

Vo = I(j12 + 10 �– j4) = I(10 + j8)

= 20 30 (10 + j8)/(10 + j6) = 22 37.66 V Chapter 13, Solution 8. Consider the current as shown below.

j 6 j 4

1

-j3

j2

10 + �– I2I1

4

+ Vo �–

Page 645: Solution Manual for Fundamentals of Electric Circuits

For mesh 1, 10 = (1 + j6)I1 + j2I2 (1)

For mesh 2, 0 = (4 + j4 �– j3)I2 + j2I1 0 = j2I1 +(4 + j)I2 (2)

In matrix form,

2

1

II

j42j2j6j1

010

= 2 + j25, and 2 = �–j20

I2 = 2/ = �–j20/(2 + j25)

Vo = �–j3I2 = �–60/(2 + j25) = 2.392 94.57

Chapter 13, Solution 9. Consider the circuit below.

2 -j1 2

For loop 1,

-j2V+ �– j 4 j 4

8 30o + �– I2I1

8 30 = (2 + j4)I1 �– jI2 (1)

For loop 2, ((j4 + 2 �– j)I2 �– jI1 + (�–j2) = 0 or I1 = (3 �– j2)i2 �– 2 (2) Substituting (2) into (1), 8 30 + (2 + j4)2 = (14 + j7)I2

I2 = (10.928 + j12)/(14 + j7) = 1.037 21.12

Vx = 2I2 = 2.074 21.12

Page 646: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 10. Consider the circuit below.

Io

j L j L

1/j C

I2I1

j M

Iin 0o

221 LLLkM = L, I1 = Iin 0 , I2 = Io

Io(j L + R + 1/(j C)) �– j LIin �– (1/(j C))Iin = 0 Io = j Iin( L �– 1/( C)) /(R + j L + 1/(j C))

Chapter 13, Solution 11. Consider the circuit below.

I1

j L1R1

R2

j L2 1/j C

j M

V2

+ �–

V1 + �–

I3

I2

For mesh 1, V1 = I1(R1 + 1/(j C)) �– I2(1/j C)) �–R1I3 For mesh 2,

0 = �–I1(1/(j C)) + (j L1 + j L2 + (1/(j C)) �– j2 M)I2 �– j L1I3 + j MI3

For mesh 3, �–V2 = �–R1I1 �– j (L1 �– M)I2 + (R1 + R2 + j L1)I3

or V2 = R1I1 + j (L1 �– M)I2 �– (R1 + R2 + j L1)I3

Page 647: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 12.

Let .1 j4 j2 + j6 j8 j10 1V - I1 I2

Applying KVL to the loops,

21 481 IjIj (1)

21 1840 IjIj (2) Solving (1) and (2) gives I1 = -j0.1406. Thus

H 111.711

11 jILjL

IZ eqeq

We can also use the equivalent T-section for the transform to find the equivalent inductance. Chapter 13, Solution 13. We replace the coupled inductance with an equivalent T-section and use series and parallel combinations to calculate Z. Assuming that ,1

10,101020,81018 21 MLMLLMLL cba The equivalent circuit is shown below:

Page 648: Solution Manual for Fundamentals of Electric Circuits

12 j8 j10 2 j10 -j6 Z j4 Z=12 +j8 + j14//(2 + j4) = 13.195 + j11.244 Chapter 13, Solution 14. To obtain VTh, convert the current source to a voltage source as shown below.

5 -j3

+ �– 8 V +

VTh �–

2 j8

a

b

j10 V + �–

j6

I

j2 Note that the two coils are connected series aiding.

L = L1 + L2 �– 2 M

j L = j6 + j8 �– j4 = j10 Thus, �–j10 + (5 + j10 �– j3 + 2)I + 8 = 0

I = (�– 8 + j10)/ (7 + j7) But, �–j10 + (5 + j6)I �– j2I + VTh = 0

Page 649: Solution Manual for Fundamentals of Electric Circuits

VTh = j10 �– (5 + j4)I = j10 �– (5 + j4)(�–8 + j10)/(7 + j7)

VTh = 5.349 34.11 To obtain ZTh, we set all the sources to zero and insert a 1-A current source at the terminals a�–b as shown below.

I2I1 1 A

5 j6 j8 -j3 a

j2

2

+ Vo �–

b Clearly, we now have only a super mesh to analyze.

(5 + j6)I1 �– j2I2 + (2 + j8 �– j3)I2 �– j2I1 = 0 (5 + j4)I1 + (2 + j3)I2 = 0 (1) But, I2 �– I1 = 1 or I2 = I1 �– 1 (2) Substituting (2) into (1), (5 + j4)I1 +(2 + j3)(1 + I1) = 0

I1 = �–(2 + j3)/(7 + j7) Now, ((5 + j6)I1 �– j2I1 + Vo = 0

Vo = �–(5 + j4)I1 = (5 + j4)(2 + j3)/(7 + j7) = (�–2 + j23)/(7 + j7) = 2.332 50

ZTh = Vo/1 = 2.332 50 ohms

Page 650: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 15.

To obtain IN, short-circuit a�–b as shown in Figure (a).

I1

I2 I1

20

j10

j20 a

j5

1 + �–

20

j10

j20 a

j5

IN

60 30o

+ �–

I2

b b (a) (b)For mesh 1,

60 30 = (20 + j10)I1 + j5I2 �– j10I2 or 12 30 = (4 + j2)I1 �– jI2 (1) For mesh 2,

0 = (j20 + j10)I2 �– j5I1 �– j10I1 or I1 = 2I2 (2) Substituting (2) into (1), 12 30 = (8 + j3)I2

IN = I2 = 12 30 /(8 + j3) = 1.404 9.44 A To find ZN, we set all the sources to zero and insert a 1-volt voltage source at terminals a�–b as shown in Figure (b). For mesh 1, 1 = I1(j10 + j20 �– j5x2) + j5I2 1 = j20I1 + j5I2 (3) For mesh 2, 0 = (20 + j10)I2 + j5I1 �– j10I1 = (4 + j2)I2 �– jI1 or I2 = jI1/(4 + j2) (4) Substituting (4) into (3), 1 = j20I1 + j(j5)I1/(4 + j2) = (�–1 + j20.5)I1

I1 = 1/(�–1 + j20.5)

ZN = 1/I1 = (�–1 + j20.5) ohms

Page 651: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 16. To find IN, we short-circuit a-b.

j 8 -j2 a

+ j4 j6 I2 IN 80 V Io0 1 - b

80)28(0)428(80 2121 jIIjjIIjj (1)

2112 606 IIjIIj (2) Solving (1) and (2) leads to

A 91.126246.1362.0584.11148

802

oN j

jII

To find ZN, insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below.

j 8 -j2 2 a

+

j4 j6 I2 2V I1 - b

28)28(0 2

121 jjI

IjIIj (3)

0)62(2 12 jIIj (4)

Solving (3) and (4) leads to I2 = -0.1055 +j0.2975, Vab=-j6I2 = 1.7853 +0.6332

oabN 53.19894.1

1V

Z

Page 652: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 17.

Z = -j6 // Zo where

7.15j5213.045j2j30j

14420jZo

7.9j1989.0Z6j

xZ6jZ

o

o

Chapter 13, Solution 18. Let 5105.0,20,5.1 2121 xLLkMLL We replace the transformer by its equivalent T-section.

5,25520,1055)( 11 MLMLLMLL cba We find ZTh using the circuit below. -j4 j10 j25 j2 -j5 ZTh 4+j6

12.29215.274

)4(627)6//()4(27 jj

jjjjjjZTh

We find VTh by looking at the circuit below.

Page 653: Solution Manual for Fundamentals of Electric Circuits

-j4 j10 j25 j2 + -j5 + VTh 120<0o

4+j6 - -

V 22.4637.61)120(64

4 oTh jj

jV

Chapter 13, Solution 19.

Let H 652540)(.1 1 MLLa

25LH, 552530 C2 MMLLb

Thus, the T-section is as shown below. j65 j55 -j25

Page 654: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 20. Transform the current source to a voltage source as shown below.

4

20 0o + �–

8

j12 + �–

I3

j10 j10

-j5 I2I1

k=0.5

k = M/ 21LL or M = k 21LL

M = k 21 LL = 0.5(10) = 5 For mesh 1, j12 = (4 + j10 �– j5)I1 + j5I2 + j5I2 = (4 + j5)I1 + j10I2 (1) For mesh 2, 0 = 20 + (8 + j10 �– j5)I2 + j5I1 + j5I1 �–20 = +j10I1 + (8 + j5)I2 (2)

From (1) and (2), 2

1

II

5j810j10j5j4

2012j

= 107 + j60, 1 = �–60 �–j296, 2 = 40 �– j100

I1 = 1/ = 2.462 72.18 A

I2 = 2/ = 0.878 �–97.48 A

I3 = I1 �– I2 = 3.329 74.89 A

i1 = 2.462 cos(1000t + 72.18 ) A

i2 = 0.878 cos(1000t �– 97.48 ) A

Page 655: Solution Manual for Fundamentals of Electric Circuits

At t = 2 ms, 1000t = 2 rad = 114.6

i1 = 0.9736cos(114.6 + 143.09 ) = �–2.445

i2 = 2.53cos(114.6 + 153.61 ) = �–0.8391 The total energy stored in the coupled coils is

w = 0.5L1i12 + 0.5L2i2

2 �– Mi1i2 Since L1 = 10 and = 1000, L1 = L2 = 10 mH, M = 0.5L1 = 5mH

w = 0.5(10)(�–2.445)2 + 0.5(10)(�–0.8391)2 �– 5(�–2.445)(�–0.8391)

w = 43.67 mJ Chapter 13, Solution 21. For mesh 1, 36 30 = (7 + j6)I1 �– (2 + j)I2 (1)

For mesh 2, 0 = (6 + j3 �– j4)I2 �– 2I1 jI1 = �–(2 + j)I1 + (6 �– j)I2 (2)

Placing (1) and (2) into matrix form, 2

1

II

j6j2j26j7

03036

= 48 + j35 = 59.41 36.1 , 1 = (6 �– j)36 30 = 219 20.54

2 = (2 + j)36 30 = 80.5 56.56 , I1 = 1/ = 3.69 �–15.56 , I2 = 2/ = 1.355 20.46

Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts

Page 656: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes,

Ix

Ib

Ia

-j50

j30Ib

j60

j20Ia

j10Ia

j80

j30Ic

j40 j10Ib

+ + +

+

I3

50 0 V

j20Ic

Io

I2I1

+

+

+

100 Note the following,

Ia = I1 �– I3 Ib = I2 �– I1 Ic = I3 �– I2

and Io = I3

Now all we need to do is to write the mesh equations and to solve for Io. Loop # 1,

-50 + j20(I3 �– I2) j 40(I1 �– I3) + j10(I2 �– I1) �– j30(I3 �– I2) + j80(I1 �– I2) �– j10(I1 �– I3) = 0

j100I1 �– j60I2 �– j40I3 = 50 Multiplying everything by (1/j10) yields 10I1 �– 6I2 �– 4I3 = - j5 (1)

Loop # 2, j10(I1 �– I3) + j80(I2�–I1) + j30(I3�–I2) �– j30(I2 �– I1) + j60(I2 �– I3) �– j20(I1 �– I3) + 100I2 = 0

-j60I1 + (100 + j80)I2 �– j20I3 = 0 (2)

Page 657: Solution Manual for Fundamentals of Electric Circuits

Loop # 3,

-j50I3 +j20(I1 �–I3) +j60(I3 �–I2) +j30(I2 �–I1) �–j10(I2 �–I1) +j40(I3 �–I1) �–j20(I3 �–I2) = 0 -j40I1 �– j20I2 + j10I3 = 0 Multiplying by (1/j10) yields, -4I1 �– 2I2 + I3 = 0 (3) Multiplying (2) by (1/j20) yields -3I1 + (4 �– j5)I2 �– I3 = 0 (4) Multiplying (3) by (1/4) yields -I1 �– 0.5I2 �– 0.25I3 = 0 (5) Multiplying (4) by (-1/3) yields I1 �– ((4/3) �– j(5/3))I2 + (1/3)I3 = -j0.5 (7) Multiplying [(6)+(5)] by 12 yields (-22 + j20)I2 + 7I3 = 0 (8) Multiplying [(5)+(7)] by 20 yields -22I2 �– 3I3 = -j10 (9) (8) leads to I2 = -7I3/(-22 + j20) = 0.2355 42.3o = (0.17418+j0.15849)I3 (10) (9) leads to I3 = (j10 �– 22I2)/3, substituting (1) into this equation produces, I3 = j3.333 + (-1.2273 �– j1.1623)I3 or I3 = Io = 1.3040 63o amp.

Chapter 13, Solution 23. = 10

0.5 H converts to j L1 = j5 ohms

1 H converts to j L2 = j10 ohms

0.2 H converts to j M = j2 ohms

25 mF converts to 1/(j C) = 1/(10x25x10-3) = �–j4 ohms

The frequency-domain equivalent circuit is shown below.

j10

I2I1

+

�–j4

j5 j2

5 12 0

Page 658: Solution Manual for Fundamentals of Electric Circuits

For mesh 1, 12 = (j5 �– j4)I1 + j2I2 �– (�–j4)I2 �–j2 = I1 + 6I2 (1) For mesh 2, 0 = (5 + j10)I2 + j2I1 �–(�–j4)I1 0 = (5 + j10)I2 + j6I1 (2) From (1), I1 = �–j12 �– 6I2 Substituting this into (2) produces,

I2 = 72/(�–5 + j26) = 2.7194 �–100.89

I1 = �–j12 �– 6 I2 = �–j12 �– 163.17 �–100.89 = 5.068 52.54

Hence, i1 = 5.068cos(10t + 52.54 ) A, i2 = 2.719cos(10t �– 100.89 ) A.

At t = 15 ms, 10t = 10x15x10-3 0.15 rad = 8.59

i1 = 5.068cos(61.13 ) = 2.446

i2 = 2.719cos(�–92.3 ) = �–0.1089

w = 0.5(5)(2.446)2 + 0.5(1)(�–0.1089)2 �– (0.2)(2.446)(�–0.1089) = 15.02 J Chapter 13, Solution 24. (a) k = M/ 21LL = 1/ 2x4 = 0.3535 (b) = 4 1/4 F leads to 1/(j C) = �–j/(4x0.25) = �–j 1||(�–j) = �–j/(1 �– j) = 0.5(1 �– j) 1 H produces j M = j4 4 H produces j16 2 H becomes j8

Page 659: Solution Manual for Fundamentals of Electric Circuits

j4

j8

j16

2

I2I1

+

0.5(1�–j) 12 0

12 = (2 + j16)I1 + j4I2 or 6 = (1 + j8)I1 + j2I2 (1) 0 = (j8 + 0.5 �– j0.5)I2 + j4I1 or I1 = (0.5 + j7.5)I2/(�–j4) (2) Substituting (2) into (1),

24 = (�–11.5 �– j51.5)I2 or I2 = �–24/(11.5 + j51.5) = �–0.455 �–77.41

Vo = I2(0.5)(1 �– j) = 0.3217 57.59

vo = 321.7cos(4t + 57.6 ) mV (c) From (2), I1 = (0.5 + j7.5)I2/(�–j4) = 0.855 �–81.21

i1 = 0.885cos(4t �– 81.21 ) A, i2 = �–0.455cos(4t �– 77.41 ) A At t = 2s,

4t = 8 rad = 98.37

i1 = 0.885cos(98.37 �– 81.21 ) = 0.8169

i2 = �–0.455cos(98.37 �– 77.41 ) = �–0.4249

w = 0.5L1i12 + 0.5L2i2

2 + Mi1i2

= 0.5(4)(0.8169)2 + 0.5(2)(�–.4249)2 + (1)(0.1869)(�–0.4249) = 1.168 J

Page 660: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 25. m = k 21LL = 0.5 H

We transform the circuit to frequency domain as shown below.

12sin2t converts to 12 0 , = 2

0.5 F converts to 1/(j C) = �–j

2 H becomes j L = j4 j1

�–j1

b

a

10

2 Io 4

j2j2

1

+

j4 12 0 Applying the concept of reflected impedance,

Zab = (2 �– j)||(1 + j2 + (1)2/(j2 + 3 + j4))

= (2 �– j)||(1 + j2 + (3/45) �– j6/45)

= (2 �– j)||(1 + j2 + (3/45) �– j6/45)

= (2 �– j)||(1.0667 + j1.8667)

=(2 �– j)(1.0667 + j1.8667)/(3.0667 + j0.8667) = 1.5085 17.9 ohms

Io = 12 0 /(Zab + 4) = 12/(5.4355 + j0.4636) = 2.2 �–4.88

io = 2.2sin(2t �– 4.88 ) A

Page 661: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 26.

M = k 21LL

M = k 21 LL = 0.6 40x20 = 17 The frequency-domain equivalent circuit is shown below.

j17

Io

�–j30

I2I1

50

j40j20

+

10 200 60

For mesh 1, 200 60 = (50 �– j30 + j20)I1 + j17I2 = (50 �– j10)I1 + j17I2 (1)

For mesh 2, 0 = (10 + j40)I2 + j17I1 (2)

In matrix form, 2

1

II

40j1017j17j10j50

060200

= 900 + j100, 1 = 2000 60 (1 + j4) = 8246.2 136 , 2 = 3400 �–30

I2 = 2/ = 3.755 �–36.34 Io = I2 = 3.755 �–36.34 A Switching the dot on the winding on the right only reverses the direction of Io. This can be seen by looking at the resulting value of 2 which now becomes 3400 150 . Thus, Io = 3.755 143.66 A

Page 662: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 27.

Zin = �–j4 + j5 + 9/(12 + j6) = 0.6 + j.07 = 0.922 49.4 I1 = 12 0 /0.922 49.4 = 13 �–49.4 A

Chapter 13, Solution 28.

We find ZTh by replacing the 20-ohm load with a unit source as shown below.

j10 8 -jX

+ j12 j15 I2 1V - I1 For mesh 1, 0 21 10)128( IjIjjX (1) For mesh 2, (2) jIIIjIj 1.05.1010151 2112

Substituting (2) into (1) leads to

XjjXjI

5.18121.08.02.1

2

XjXjj

IZTh 1.08.02.1

5.18121

2

6247275.108.0)1.02.1(

)5.18(1220|| 2

22

22

XXX

XZTh

Solving the quadratic equation yields X = 6.425

Page 663: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 29.

30 mH becomes j L = j30x10-3x103 = j30 50 mH becomes j50 Let X = M Using the concept of reflected impedance,

Zin = 10 + j30 + X2/(20 + j50) I1 = V/Zin = 165/(10 + j30 + X2/(20 + j50)) p = 0.5|I1|2(10) = 320 leads to |I1|2 = 64 or |I1| = 8 8 = |165(20 + j50)/(X2 + (10 + j30)(20 + j50))|

= |165(20 + j50)/(X2 �– 1300 + j1100)|

or 64 = 27225(400 + 2500)/((X2 �– 1300)2 + 1,210,000) (X2 �– 1300)2 + 1,210,000 = 1,233,633 X = 33.86 or 38.13

If X = 38.127 = M

M = 38.127 mH k = M/ 21LL = 38.127/ 50x30 = 0.984

j38.127

I2I1

10

j50j30

+

20 165 0

Page 664: Solution Manual for Fundamentals of Electric Circuits

165 = (10 + j30)I1 �– j38.127I2 (1) 0 = (20 + j50)I2 �– j38.127I1 (2)

In matrix form, 2

1

II

50j20127.38j127.38j30j10

0165

= 154 + j1100 = 1110.73 82.03 , 1 = 888.5 68.2 , 2 = j6291

I1 = 1/ = 8 �–13.81 , I2 = 2/ = 5.664 7.97 i1 = 8cos(1000t �– 13.83 ), i2 = 5.664cos(1000t + 7.97 )

At t = 1.5 ms, 1000t = 1.5 rad = 85.94

i1 = 8cos(85.94 �– 13.83 ) = 2.457 i2 = 5.664cos(85.94 + 7.97 ) = �–0.3862 w = 0.5L1i1

2 + 0.5L2i22 + Mi1i2

= 0.5(30)(2.547)2 + 0.5(50)(�–0.3862)2 �– 38.127(2.547)(�–0.3862)

= 130.51 mJ

Chapter 13, Solution 30.

(a) Zin = j40 + 25 + j30 + (10)2/(8 + j20 �– j6) = 25 + j70 + 100/(8 + j14) = (28.08 + j64.62) ohms (b) j La = j30 �– j10 = j20, j Lb = j20 �– j10 = j10, j Lc = j10 Thus the Thevenin Equivalent of the linear transformer is shown below.

j40 j20 j10 25 8

Zin

j10

�–j6

Zin = j40 + 25 + j20 + j10||(8 + j4) = 25 + j60 + j10(8 + j4)/(8 + j14) = (28.08 + j64.62) ohms

Page 665: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 31.

(a) La = L1 �– M = 10 H Lb = L2 �– M = 15 H Lc = M = 5 H (b) L1L2 �– M2 = 300 �– 25 = 275

LA = (L1L2 �– M2)/(L1 �– M) = 275/15 = 18.33 H

LB = (L1L2 �– M2)/(L1 �– M) = 275/10 = 27.5 H

LC = (L1L2 �– M2)/M = 275/5 = 55 H Chapter 13, Solution 32. We first find Zin for the second stage using the concept of reflected impedance.

Zin�’

LBLb

R Zin�’ = j Lb + 2Mb

2/(R + j Lb) = (j LbR - 2Lb2 + 2Mb

2)/(R + j Lb) (1) For the first stage, we have the circuit below.

Zin

LALa Zin�’

Zin = j La + 2Ma2/(j La + Zin)

= (�– 2La

2 + 2Ma2 + j LaZin)/( j La + Zin) (2)

Page 666: Solution Manual for Fundamentals of Electric Circuits

Substituting (1) into (2) gives,

=

b

2b

22b

2b

a

b

2b

22b

2b

a2a

22a

2

LjRMLRLj

Lj

LjR)MLRLj(

LjML

�–R 2La

2 + 2Ma2R �– j 3LbLa + j 3LbMa

2 + j La(j LbR �– 2Lb2 + 2Mb

2) =

j RLa �– 2LaLb + j LbR �– 2La2 + 2Mb

2

2R(La2 + LaLb �– Ma

2) + j 3(La2Lb + LaLb

2 �– LaMb2 �– LbMa

2) Zin =

2(LaLb +Lb2 �– Mb

2) �– j R(La +Lb) Chapter 13, Solution 33.

Zin = 10 + j12 + (15)2/(20 + j 40 �– j5) = 10 + j12 + 225/(20 + j35)

= 10 + j12 + 225(20 �– j35)/(400 + 1225)

= (12.769 + j7.154) ohms Chapter 13, Solution 34. Insert a 1-V voltage source at the input as shown below. j6 1 8

+ j12 j10 j4 1<0o V I1 I2 - -j2

Page 667: Solution Manual for Fundamentals of Electric Circuits

For loop 1, 21 4)101(1 IjIj (1)

For loop 2,

21112 )32(062)21048(0 IjjIIjIjIjjj (2) Solving (1) and (2) leads to I1=0.019 �–j0.1068

ojI

Z 91.79219.9077.96154.11

1

Alternatively, an easier way to obtain Z is to replace the transformer with its equivalent T circuit and use series/parallel impedance combinations. This leads to exactly the same result. Chapter 13, Solution 35.

For mesh 1, 21 2)410(16 IjIj (1) For mesh 2, 321 12)2630(20 IjIjIj (2) For mesh 3, 32 )115(120 IjIj (3) We may use MATLAB to solve (1) to (3) and obtain

A 41.214754.15385.03736.11

ojI A 85.1340775.00549.00547.02

ojI A 41.110077.00721.00268.03

ojI Chapter 13, Solution 36.

Following the two rules in section 13.5, we obtain the following: (a) V2/V1 = �–n, I2/I1 = �–1/n (n = V2/V1) (b) V2/V1 = �–n, I2/I1 = �–1/n (c) V2/V1 = n, I2/I1 = 1/n (d) V2/V1 = n, I2/I1 = �–1/n

Page 668: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 37.

(a) 5480

2400

1

2

VV

n

(b) A 17.104480

000,50000,50 1222111 IVISVIS

(c ) A 83.202400

000,502I

Chapter 13, Solution 38.

Zin = Zp + ZL/n2, n = v2/v1 = 230/2300 = 0.1

v2 = 230 V, s2 = v2I2*

I2

* = s2/v2 = 17.391 �–53.13 or I2 = 17.391 53.13 A

ZL = v2/I2 = 230 0 /17.391 53.13 = 13.235 �–53.13

Zin = 2 10 + 1323.5 �–53.13

= 1.97 + j0.3473 + 794.1 �– j1058.8

Zin = 1.324 �–53.05 kohms Chapter 13, Solution 39. Referred to the high-voltage side,

ZL = (1200/240)2(0.8 10 ) = 20 10

Zin = 60 �–30 + 20 10 = 76.4122 �–20.31

I1 = 1200/Zin = 1200/76.4122 �–20.31 = 15.7 20.31 A

Since S = I1v1 = I2v2, I2 = I1v1/v2

= (1200/240)( 15.7 20.31 ) = 78.5 20.31 A

Page 669: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 40.

V 60)240(41nVV

VV

n,41

2000500

NN

n 121

2

1

2

W30012

60R

VP22

Chapter 13, Solution 41. We reflect the 2-ohm resistor to the primary side.

Zin = 10 + 2/n2, n = �–1/3

Since both I1 and I2 enter the dotted terminals, Zin = 10 + 18 = 28 ohms

I1 = 14 0 /28 = 0.5 A and I2 = I1/n = 0.5/(�–1/3) = �–1.5 A Chapter 13, Solution 42. 10 1:4 -j50

+ + + + V1 V2 20 Vo 120<0o V I1 - - - - I2

Applying mesh analysis,

11 VI10120 (1)

22 VI)50j20(0 (2)

Page 670: Solution Manual for Fundamentals of Electric Circuits

At the terminals of the transformer,

121

2 V4V4nVV

(3)

211

2 I4I41

n1

II

(4)

Substituting (3) and (4) into (1) gives 120 22 V25.0I40 (5)

Solving (2) and (5) yields 6877.0j4756.2I2

V 52.1539.51I20V o

2o Chapter 13, Solution 43. Transform the two current sources to voltage sources, as shown below.

+

v1

+

v2

1 : 4

12V+ �–

10

20 V + �– I2I1

12 Using mesh analysis, �–20 + 10I1 + v1 = 0 20 = v1 + 10I1 (1) 12 + 12I2 �– v2 = 0 or 12 = v2 �– 12I2 (2) At the transformer terminal, v2 = nv1 = 4v1 (3) I1 = nI2 = 4I2 (4) Substituting (3) and (4) into (1) and (2), we get, 20 = v1 + 40I2 (5) 12 = 4v1 �– 12I2 (6) Solving (5) and (6) gives v1 = 4.186 V and v2 = 4v = 16.744 V

Page 671: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 44.

We can apply the superposition theorem. Let i1 = i1�’ + i1�” and i2 = i2�’ + i2�” where the single prime is due to the DC source and the double prime is due to the AC source. Since we are looking for the steady-state values of i1 and i2,

i1�’ = i2�’ = 0.

For the AC source, consider the circuit below.

+�–i2�”

+

v1

1 : n

+

v2

R

i1�”

Vn 0

v2/v1 = �–n, I2�”/I1�” = �–1/n

But v2 = vm, v1 = �–vm/n or I1�” = vm/(Rn)

I2�” = �–I1�”/n = �–vm/(Rn2)

Hence, i1(t) = (vm/Rn)cos t A, and i2(t) = (�–vm/(n2R))cos t A Chapter 13, Solution 45. 48

+

Z4 �–90

4j8Cj8ZL , n = 1/3

Page 672: Solution Manual for Fundamentals of Electric Circuits

3.7303193.07.1628.125

90436j7248

904I

36j72Z9n

ZZ L2

L

We now have some choices, we can go ahead and calculate the current in the second loop and calculate the power delivered to the 8-ohm resistor directly or we can merely say that the power delivered to the equivalent resistor in the primary side must be the same as the power delivered to the 8-ohm resistor. Therefore,

7210x5098.0722IP 3

28 36.71 mW

The student is encouraged to calculate the current in the secondary and calculate the power delivered to the 8-ohm resistor to verify that the above is correct. Chapter 13, Solution 46. (a) Reflecting the secondary circuit to the primary, we have the circuit shown below.

Zin

+

+

16 60 I1

10 30 /(�–n) = �–5 30

Zin = 10 + j16 + (1/4)(12 �– j8) = 13 + j14 �–16 60 + ZinI1 �– 5 30 = 0 or I1 = (16 60 + 5 30 )/(13 + j14) Hence, I1 = 1.072 5.88 A, and I2 = �–0.5I1 = 0.536 185.88 A

(b) Switching a dot will not effect Zin but will effect I1 and I2.

I1 = (16 60 �– 5 30 )/(13 + j14) = 0.625 25 A and I2 = 0.5I1 = 0.3125 25 A

Page 673: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 47.

0.02 F becomes 1/(j C) = 1/(j5x0.02) = �–j10 We apply mesh analysis to the circuit shown below.

�–j10

+

vo

I3

+

v1

+

v2

3 : 1 10

10 0 + �– I2I1

2 For mesh 1, 10 = 10I1 �– 10I3 + v1 (1) For mesh 2, v2 = 2I2 = vo (2) For mesh 3, 0 = (10 �– j10)I3 �– 10I1 + v2 �– v1 (3) At the terminals, v2 = nv1 = v1/3 (4) I1 = nI2 = I2/3 (5) From (2) and (4), v1 = 6I2 (6) Substituting this into (1), 10 = 10I1 �– 10I3 (7) Substituting (4) and (6) into (3) yields 0 = �–10I1 �– 4I2 + 10(1 �– j)I3 (8) From (5), (7), and (8)

0100

III

10j10410106100333.01

3

2

1

Page 674: Solution Manual for Fundamentals of Electric Circuits

I2 = 33.93j20

100j1002 = 1.482 32.9

vo = 2I2 = 2.963 32.9 V (a) Switching the dot on the secondary side effects only equations (4) and (5). v2 = �–v1/3 (9) I1 = �–I2/3 (10) From (2) and (9), v1 = �–6I2 Substituting this into (1),

10 = 10I1 �– 10I3 �– 6I2 = (23 �– j5)I1 (11) Substituting (9) and (10) into (3),

0 = �–10I1 + 4I2 + 10(1 �– j)I3 (12)

From (10) to (12), we get

0100

III

10j10410106100333.01

3

2

1

I2 = 33.93j20

100j1002 = 1.482 �–147.1

vo = 2I2 = 2.963 �–147.1 V Chapter 13, Solution 48. We apply mesh analysis. 8 2:1 10 + + + V1 V2 I1 - j6 100 0o V - I2 - Ix -j4

Page 675: Solution Manual for Fundamentals of Electric Circuits

121 4)48(100 VIjIj (1)

212 4)210(0 VIjIj (2)

But

211

2 221

VVnVV

(3)

211

2 5.021II

nII

(4)

Substituting (3) and (4) into (1) and (2), we obtain

22 2)24(100 VIj (1)a

22)410(0 VIj (2)a

Solving (1)a and (2)a leads to I2 = -3.5503 +j1.4793

A 4.157923.15.0 221o

x IIII Chapter 13, Solution 49.

101F 201,2 j

Cj

Ix -j10 2 I1 1:3 I2 1 2 + + + V1 V2 6 12<0o V - - -

Page 676: Solution Manual for Fundamentals of Electric Circuits

At node 1,

2111211 2.0)2.01(212

10212

VjjVIIjVVV

(1)

At node 2,

212221

2 )6.01(6.060610

VjVjIV

jVV

I (2)

At the terminals of the transformer, 1212 31,3 IIVV

Substituting these in (1) and (2),

)4.23(60),8.01(612 1212 jVIjVI Adding these gives V1=1.829 �–j1.463 and

ox j

VjVV

I 34.51937.010

410

121

A )34.512cos(937.0 o

x ti Chapter 13, Solution 50.

The value of Zin is not effected by the location of the dots since n2 is involved.

Zin�’ = (6 �– j10)/(n�’)2, n�’ = 1/4 Zin�’ = 16(6 �– j10) = 96 �– j160 Zin = 8 + j12 + (Zin�’ + 24)/n2, n = 5 Zin = 8 + j12 + (120 �– j160)/25 = 8 + j12 + 4.8 �– j6.4 Zin = (12.8 + j5.6) ohms

Chapter 13, Solution 51. Let Z3 = 36 +j18, where Z3 is reflected to the middle circuit.

ZR�’ = ZL/n2 = (12 + j2)/4 = 3 + j0.5 Zin = 5 �– j2 + ZR�’ = (8 �– j1.5) ohms I1 = 24 0 /ZTh = 24 0 /(8 �– j1.5) = 24 0 /8.14 �–10.62 = 8.95 10.62 A

Page 677: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 52. For maximum power transfer,

40 = ZL/n2 = 10/n2 or n2 = 10/40 which yields n = 1/2 = 0.5 I = 120/(40 + 40) = 3/2 p = I2R = (9/4)x40 = 90 watts.

Chapter 13, Solution 53.

(a) The Thevenin equivalent to the left of the transformer is shown below. 8

20 V

+

The reflected load impedance is ZL�’ = ZL/n2 = 200/n2. For maximum power transfer, 8 = 200/n2 produces n = 5.

(b) If n = 10, ZL�’ = 200/10 = 2 and I = 20/(8 + 2) = 2

p = I2ZL�’ = (2)2(2) = 8 watts. Chapter 13, Solution 54. (a)

I1

+

ZTh

ZL/n2 VS

Page 678: Solution Manual for Fundamentals of Electric Circuits

For maximum power transfer,

ZTh = ZL/n2, or n2 = ZL/ZTh = 8/128

n = 0.25 (b) I1 = VTh/(ZTh + ZL/n2) = 10/(128 + 128) = 39.06 mA (c) v2 = I2ZL = 156.24x8 mV = 1.25 V But v2 = nv1 therefore v1 = v2/n = 4(1.25) = 5 V Chapter 13, Solution 55. We reflect Zs to the primary side.

ZR = (500 �– j200)/n2 = 5 �– j2, Zin = Zp + ZR = 3 + j4 + 5 �– j2 = 8 + j2

I1 = 120 0 /(8 + j2) = 14.552 �–14.04

I2I1

+

1 : n Zp Vp Zs

Since both currents enter the dotted terminals as shown above,

I2 = �–(1/n)I1 = �–1.4552 �–14.04 = 1.4552 166 S2 = |I2|2Zs = (1.4552)(500 �– j200) P2 = Re(S2) = (1.4552)2(500) = 1054 watts

Page 679: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 56. We apply mesh analysis to the circuit as shown below.

+

v1

5

+

v2

I2I1

+

1 : 2 2 10 46V For mesh 1, 46 = 7I1 �– 5I2 + v1 (1) For mesh 2, v2 = 15I2 �– 5I1 (2) At the terminals of the transformer, v2 = nv1 = 2v1 (3) I1 = nI2 = 2I2 (4) Substituting (3) and (4) into (1) and (2), 46 = 9I2 + v1 (5) v1 = 2.5I2 (6) Combining (5) and (6), 46 = 11.5I2 or I2 = 4 P10 = 0.5I2

2(10) = 80 watts.

Page 680: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 57.

(a) ZL = j3||(12 �– j6) = j3(12 �– j6)/(12 �– j3) = (12 + j54)/17

Reflecting this to the primary side gives Zin = 2 + ZL/n2 = 2 + (3 + j13.5)/17 = 2.3168 20.04 I1 = vs/Zin = 60 90 /2.3168 20.04 = 25.9 69.96 A(rms) I2 = I1/n = 12.95 69.96 A(rms)

(b) 60 90 = 2I1 + v1 or v1 = j60 �–2I1 = j60 �– 51.8 69.96

v1 = 21.06 147.44 V(rms) v2 = nv1 = 42.12 147.44 V(rms) vo = v2 = 42.12 147.44 V(rms)

(c) S = vsI1

* = (60 90 )(25.9 �–69.96 ) = 1554 20.04 VA Chapter 13, Solution 58. Consider the circuit below. 20

+

vo

I3

+

v1

+

v2

1 : 5 20

80 0 + �– I2I1

100

For mesh1, 80 = 20I1 �– 20I3 + v1 (1) For mesh 2, v2 = 100I2 (2)

Page 681: Solution Manual for Fundamentals of Electric Circuits

For mesh 3, 0 = 40I3 �– 20I1 which leads to I1 = 2I3 (3) At the transformer terminals, v2 = �–nv1 = �–5v1 (4) I1 = �–nI2 = �–5I2 (5) From (2) and (4), �–5v1 = 100I2 or v1 = �–20I2 (6) Substituting (3), (5), and (6) into (1),

4 = I1 �– I2 �– I 3 = I1 �– (I1/(�–5)) �– I1/2 = (7/10)I1 I1 = 40/7, I2 = �–8/7, I3 = 20/7

p20(the one between 1 and 3) = 0.5(20)(I1 �– I3)2 = 10(20/7)2 = 81.63 watts p20(at the top of the circuit) = 0.5(20)I3

2 = 81.63 watts p100 = 0.5(100)I2

2 = 65.31 watts Chapter 13, Solution 59. We apply nodal analysis to the circuit below. 2

I3

v2 v1 I1

4

+

v1

+

v2

2 : 1 8

20 0 + �–

I2

Page 682: Solution Manual for Fundamentals of Electric Circuits

20 = 8I1 + V1 (1) V1 = 2I3 + V2 (2)

V2 = 4I2 (3) At the transformer terminals, v2 = 0.5v1 (4) I1 = 0.5I2 (5) Solving (1) to (5) gives I1 = 0.833 A, I2 = 1.667 A, I3 = 3.333 A V1 = 13.33 V, V2 = 6.667 V.

P8 = 0.5(8)|(20 �– V1)/8|2 = 2.778 W P2 = 0.5(2)I3

2 = 11.11 W, P4 = 0.5V22/4 = 5.556 W

Chapter 13, Solution 60. (a) Transferring the 40-ohm load to the middle circuit,

ZL�’ = 40/(n�’)2 = 10 ohms where n�’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Zin = 4 + 6/n2 = 4 + 96 = 100 ohms, where n = 0.25 I1 = 120/100 = 1.2 A and I2 = I1/n = 4.8 A

I2�’

120 0 + �–

5

10

I1

+

v1

+

v2

1 : 4 4 I2 10

Using current division, I2�’ = (10/25)I2 = 1.92 and I3 = I2�’/n�’ = 0.96 A (b) p = 0.5(I3)2(40) = 18.432 watts

Page 683: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 61. We reflect the 160-ohm load to the middle circuit.

ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3

Io�’

24 0 + �–

14

60

I1

+

v1

+

vo

1 : 5 2 Io

90

14 + 60||90 = 14 + 36 = 50 ohms

We reflect this to the primary side.

ZR�’ = ZL�’/(n�’)2 = 50/52 = 2 ohms when n�’ = 5 I1 = 24/(2 + 2) = 6A 24 = 2I1 + v1 or v1 = 24 �– 2I1 = 12 V vo = �–nv1 = �–60 V, Io = �–I1 /n1 = �–6/5 = �–1.2 Io�‘ = [60/(60 + 90)]Io = �–0.48A I2 = �–Io�’/n = 0.48/(4/3) = 0.36 A

Chapter 13, Solution 62. (a) Reflect the load to the middle circuit.

ZL�’ = 8 �– j20 + (18 + j45)/32 = 10 �– j15 We now reflect this to the primary circuit so that Zin = 6 + j4 + (10 �– j15)/n2 = 7.6 + j1.6 = 7.767 11.89 , where n = 5/2 = 2.5 I1 = 40/Zin = 40/7.767 11.89 = 5.15 �–11.89 S = 0.5vsI1

* = (20 0 )(5.15 11.89 ) = 103 11.89 VA

Page 684: Solution Manual for Fundamentals of Electric Circuits

(b) I2 = �–I1/n, n = 2.5

I3 = �–I2/n�’, n = 3

I3 = I1/(nn�’) = 5.15 �–11.89 /(2.5x3) = 0.6867 �–11.89

p = 0.5|I2|2(18) = 9(0.6867)2 = 4.244 watts Chapter 13, Solution 63. Reflecting the (9 + j18)-ohm load to the middle circuit gives,

Zin�’ = 7 �– j6 + (9 + j18)/(n�’)2 = 7 �– j6 + 1 + j12 = 8 + j4 when n�’ = 3 Reflecting this to the primary side, Zin = 1 + Zin�’/n2 = 1 + 2 �– j = 3 �– j, where n = 2 I1 = 12 0 /(3 �– j) = 12/3.162 �–18.43 = 3.795 18.43A I2 = I1/n = 1.8975 18.43 A I3 = �–I2/n2 = 632.5 161.57 mA

Chapter 13, Solution 64. We find ZTh at the terminals of Z by considering the circuit below. 10 1:n + + V1 V2 + �– �– I1 I2 1 V �– 20

Page 685: Solution Manual for Fundamentals of Electric Circuits

For mesh 1, 02030 121 VII (1) For mesh 2, 12020 221 VII (2)

At the terminals, nI

InVV 1212 ,

Substituting these in (1) and (2) leads to

1)1(20,0)3020( 1212 nVInVIn Solving these gives

5.72040301204030

1 2

222 nn

IZ

nnI Th

Solving the quadratic equation yields n=0.5 or 0.8333 Chapter 13, Solution 65.

40 10 I1 1:2 I2 50 I2 1:3 I3 1 + 2 + + + + 200 V V3 V4 (rms) V1 V2 - - 20 - - - At node 1,

1411411 1025.025.1200

4010200

IVVIVVV

(1)

Page 686: Solution Manual for Fundamentals of Electric Circuits

At node 2,

3413441 403

2040IVVI

VVV (2)

At the terminals of the first transformer,

121

2 22 VVVV

(3)

211

2 22/1 IIII

(4)

For the middle loop,

223322 50050 IVVVIV (5) At the terminals of the second transformer,

343

4 33 VVVV

(6)

322

3 33/1 IIII

(7)

We have seven equations and seven unknowns. Combining (1) and (2) leads to

314 50105.3200 IIV But from (4) and (7), 3321 6)3(22 IIII . Hence

34 1105.3200 IV (8) From (5), (6), (3), and (7),

3122224 45061503)50(3 IVIVIVV Substituting for V1 in (2) gives

433344 21019450)403(6 VIIIVV (9)

Page 687: Solution Manual for Fundamentals of Electric Circuits

Substituting (9) into (8) yields

87.14452.13200 44 VV

W05.1120

42V

P

Chapter 13, Solution 66.

v1 = 420 V (1) v2 = 120I2 (2) v1/v2 = 1/4 or v2 = 4v1 (3) I1/I2 = 4 or I1 = 4 I2 (4)

Combining (2) and (4),

v2 = 120[(1/4)I1] = 30 I1 4v1 = 30I1 4(420) = 1680 = 30I1 or I1 = 56 A

Chapter 13, Solution 67.

(a) V 1604004.04.04.0

112

2

21

2

1 xVVN

NNVV

(b) A 25.311605000000,5 2222 IVIS

(c ) A 5.12400

5000000,5 21112 IVISS

Page 688: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 68. This is a step-up transformer.

I1

I2

N1

2 �– j6

+

v1

+

v2

N2 +

10 + j40 20 30

For the primary circuit, 20 30 = (2 �– j6)I1 + v1 (1) For the secondary circuit, v2 = (10 + j40)I2 (2)

At the autotransformer terminals,

v1/v2 = N1/(N1 + N2) = 200/280 = 5/7,

thus v2 = 7v1/5 (3) Also, I1/I2 = 7/5 or I2 = 5I1/7 (4)

Substituting (3) and (4) into (2), v1 = (10 + j40)25I1/49 Substituting that into (1) gives 20 30 = (7.102 + j14.408)I1 I1 = 20 30 /16.063 63.76 = 1.245 �–33.76 A I2 = 5I1/7 = 0.8893 �–33.76 A Io = I1 �– I2 = [(5/7) �– 1]I1 = �–2I1/7 = 0.3557 146.2 A p = |I2|2R = (0.8893)2(10) = 7.51 watts

Page 689: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 69. We can find the Thevenin equivalent.

I1

I2

75

+

VTh

N2

j125

+

v1

+

v2

N1 +

120 0

I1 = I2 = 0

As a step up transformer, v1/v2 = N1/(N1 + N2) = 600/800 = 3/4

v2 = 4v1/3 = 4(120)/3 = 160 0 rms = VTh. To find ZTh, connect a 1-V source at the secondary terminals. We now have a step-down transformer.

I2

I1

75 j125

+

v2

+

v1

+

1 0 V

v1 = 1V, v2 =I2(75 + j125)

But v1/v2 = (N1 + N2)/N1 = 800/200 which leads to v1 = 4v2 = 1

and v2 = 0.25

I1/I2 = 200/800 = 1/4 which leads to I2 = 4I1

Page 690: Solution Manual for Fundamentals of Electric Circuits

Hence 0.25 = 4I1(75 + j125) or I1 = 1/[16(75 + j125)

ZTh = 1/I1 = 16(75 + j125) Therefore, ZL = ZTh

* = (1.2 �– j2) k Since VTh is rms, p = (|VTh|/2)2/RL = (80)2/1200 = 5.333 watts

Chapter 13, Solution 70. This is a step-down transformer.

I1

I2

+

v2

+

v1

+

30 + j12 120 0 20 �– j40

I1/I2 = N2/(N1 + N2) = 200/1200 = 1/6, or I1 = I2/6 (1) v1/v2 = (N2 + N2)/N2 = 6, or v1 = 6v2 (2)

For the primary loop, 120 = (30 + j12)I1 + v1 (3) For the secondary loop, v2 = (20 �– j40)I2 (4)

Substituting (1) and (2) into (3),

120 = (30 + j12)( I2/6) + 6v2 and substituting (4) into this yields

120 = (49 �– j38)I2 or I2 = 1.935 37.79 p = |I2|2(20) = 74.9 watts.

Page 691: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 71.

Zin = V1/I1 But V1I1 = V2I2, or V2 = I2ZL and I1/I2 = N2/(N1 + N2) Hence V1 = V2I2/I1 = ZL(I2/I1)I2 = ZL(I2/I1)2I1

V1/I1 = ZL[(N1 + N2)/N2] 2 Zin = [1 + (N1/N2)] 2ZL

Chapter 13, Solution 72.

(a) Consider just one phase at a time.

1:n A

B 20MVA

Load

C

a b c

n = VL/ )312470/(7200V3 Lp = 1/3

(b) The load carried by each transformer is 60/3 = 20 MVA.

Hence ILp = 20 MVA/12.47 k = 1604 A ILs = 20 MVA/7.2 k = 2778 A

(c) The current in incoming line a, b, c is

85.1603x3I3 Lp = 2778 A

Current in each outgoing line A, B, C is 2778/(n 3 ) = 4812 A

Page 692: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 73.

(a) This is a three-phase -Y transformer. (b) VLs = nvLp/ 3 = 450/(3 3 ) = 86.6 V, where n = 1/3

As a Y-Y system, we can use per phase equivalent circuit. Ia = Van/ZY = 86.6 0 /(8 �– j6) = 8.66 36.87 Ic = Ia 120 = 8.66 156.87 A ILp = n 3 ILs I1 = (1/3) 3 (8.66 36.87 ) = 5 36.87 I2 = I1 �–120 = 5 �–83.13 A

(c) p = 3|Ia|2(8) = 3(8.66)2(8) = 1.8 kw.

Chapter 13, Solution 74.

(a) This is a - connection. (b) The easy way is to consider just one phase.

1:n = 4:1 or n = 1/4

n = V2/V1 which leads to V2 = nV1 = 0.25(2400) = 600

i.e. VLp = 2400 V and VLs = 600 V

S = p/cos = 120/0.8 kVA = 150 kVA

pL = p/3 = 120/3 = 40 kw 4:1

Ipp

IL

Ips

ILs VLp VLs

Page 693: Solution Manual for Fundamentals of Electric Circuits

But pLs = VpsIps

For the -load, IL = 3 Ip and VL = Vp Hence, Ips = 40,000/600 = 66.67 A

ILs = 3 Ips = 3 x66.67 = 115.48 A

(c) Similarly, for the primary side

ppp = VppIpp = pps or Ipp = 40,000/2400 = 16.667 A

and ILp = 3 Ip = 28.87 A (d) Since S = 150 kVA therefore Sp = S/3 = 50 kVA

Chapter 13, Solution 75.

(a) n = VLs/( 3 VLp) 4500/(900 3 ) = 2.887 (b) S = 3 VLsILs or ILs = 120,000/(900 3 ) = 76.98 A ILs = ILp/(n 3 ) = 76.98/(2.887 3 ) = 15.395 A

Chapter 13, Solution 76. (a) At the load, VL = 240 V = VAB

VAN = VL/ 3 = 138.56 V

Since S = 3 VLIL then IL = 60,000/(240 3 ) = 144.34 A

Page 694: Solution Manual for Fundamentals of Electric Circuits

1:n

0.05 j0.1

0.05 j0.1

0.05

A

B

Balanced

Load 60kVA 0.85pf leading C

240V j0.1 2640V

(b) Let VAN = |VAN| 0 = 138.56 0 cos = pf = 0.85 or = 31.79 IAA�’ = IL = 144.34 31.79 VA�’N�’ = ZIAA�’ + VAN = 138.56 0 + (0.05 + j0.1)(144.34 31.79 ) = 138.03 6.69

VLs = VA�’N�’ 3 = 137.8 3 = 238.7 V (c) For Y- connections,

n = 3 VLs/Vps = 3 x238.7/2640 = 0.1569

fLp = nILs/ 3 = 0.1569x144.34/ 3 = 13.05 A Chapter 13, Solution 77.

(a) This is a single phase transformer. V1 = 13.2 kV, V2 = 120 V

n = V2/V1 = 120/13,200 = 1/110, therefore n = 110 (b) P = VI or I = P/V = 100/120 = 0.8333 A

I1 = nI2 = 0.8333/110 = 7.576 mA

Page 695: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 78. The schematic is shown below.

k = 21LL/M = 3x6/1 = 0.2357 In the AC Sweep box, set Total Pts = 1, Start Freq = 0.1592 and End Freq = 0.1592. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 4.253 E+00 �–8.526 E+00 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 1.564 E+00 2.749 E+01 From this, I1 = 4.253 �–8.53 A, I2 = 1.564 27.49 A The power absorbed by the 4-ohm resistor = 0.5|I|2R = 0.5(1.564)2x4

= 4.892 watts

Page 696: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 79. The schematic is shown below.

k1 = 5000/15 = 0.2121, k2 = 8000/10 = 0.1118 In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the circuit is saved and simulated, the output includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 4.068 E�–01 �–7.786 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 1.306 E+00 �–6.801 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E�–01 1.336 E+00 �–5.492 E+01 Thus, I1 = 1.306 �–68.01 A, I2 = 406.8 �–77.86 mA, I3 = 1.336 �–54.92 A

Page 697: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 80. The schematic is shown below.

k1 = 80x40/10 = 0.1768, k2 = 60x40/20 = 0.482

k3 = 60x80/30 = 0.433 In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After the simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 1.304 E+00 6.292 E+01 i.e. Io = 1.304 62.92 A

Page 698: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 81. The schematic is shown below.

k1 = 8x4/2 = 0.3535, k2 = 8x2/1 = 0.25 In the AC Sweep box, we let Total Pts = 1, Start Freq = 100, and End Freq = 100. After simulation, the output file includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.000 E+02 1.0448 E�–01 1.396 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.000 E+02 2.954 E�–02 �–1.438 E+02 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.000 E+02 2.088 E�–01 2.440 E+01

i.e. I1 = 104.5 13.96 mA, I2 = 29.54 �–143.8 mA,

I3 = 208.8 24.4 mA.

Page 699: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 82. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain the output file which includes FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592 E�–01 1.955 E+01 8.332 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 6.847 E+01 4.640 E+01 FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E�–01 4.434 E�–01 �–9.260 E+01

i.e. V1 = 19.55 83.32 V, V2 = 68.47 46.4 V,

Io = 443.4 �–92.6 mA.

hapter 13, Solution 83.

he schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq

FREQ IM(V_PRINT1) IP(V_PRINT1)

FREQ VM($N_0001) VP($N_0001)

i.e. iX = 1.08 33.91 A

C T= 0.1592, and End Freq = 0.1592. After simulation, the output file includes 1.592 E�–01 1.080 E+00 3.391 E+01 1.592 E�–01 1.514 E+01 �–3.421 E+01

, Vx = 15.14 �–34.21 V.

Page 700: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 84.

he schematic is shown below. We set Total Pts = 1, Start Freq = 0.1592, and End

FREQ IM(V_PRINT1) IP(V_PRINT1)

FREQ IM(V_PRINT2) IP(V_PRINT2)

FREQ IM(V_PRINT3) IP(V_PRINT3)

i.e. I1 = 4.028 �–52.38 A

TFreq = 0.1592. After simulation, the output file includes 1.592 E�–01 4.028 E+00 �–5.238 E+01 1.592 E�–01 2.019 E+00 �–5.211 E+01 1.592 E�–01 1.338 E+00 �–5.220 E+01

, I2 = 2.019 �–52.11 A,

I3 = 1.338 �–52.2 A.

Page 701: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 85.

For maximum power transfer,

Z1 = ZL/n2 or n2 = ZL/Z1 = 8/7200 = 1/900

n = 1/30 = N2/N1. Thus N2 = N1/30 = 3000/30 = 100 turns.

Z1

+

ZL/n2VS

Chapter 13, Solution 86.

n = N2/N1 = 48/2400 = 1/50 ZTh = ZL/n2 = 3/(1/50)2 = 7.5 k

Page 702: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 87. ZTh = ZL/n2 or n = 300/75Z/Z ThL = 0.5 Chapter 13, Solution 88. n = V2/V1 = I1/I2 or I2 = I1/n = 2.5/0.1 = 25 A

p = IV = 25x12.6 = 315 watts Chapter 13, Solution 89. n = V2/V1 = 120/240 = 0.5

S = I1V1 or I1 = S/V1 = 10x103/240 = 41.67 A S = I2V2 or I2 = S/V2 = 104/120 = 83.33 A

Chapter 13, Solution 90.

(a) n = V2/V1 = 240/2400 = 0.1 (b) n = N2/N1 or N2 = nN1 = 0.1(250) = 25 turns (c) S = I1V1 or I1 = S/V1 = 4x103/2400 = 1.6667 A S = I2V2 or I2 = S/V2 = 4x104/240 = 16.667 A

Chapter 13, Solution 91.

(a) The kVA rating is S = VI = 25,000x75 = 1875 kVA (b) Since S1 = S2 = V2I2 and I2 = 1875x103/240 = 7812 A

Page 703: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 92.

(a) V2/V1 = N2/N1 = n, V2 = (N2/N1)V1 = (28/1200)4800 = 112 V (b) I2 = V2/R = 112/10 = 11.2 A and I1 = nI2, n = 28/1200

I1 = (28/1200)11.2 = 261.3 mA (c) p = |I2|2R = (11.2)2(10) = 1254 watts.

Chapter 13, Solution 93. (a) For an input of 110 V, the primary winding must be connected in parallel, with series-aiding on the secondary. The coils must be series-opposing to give 12 V. Thus the connections are shown below.

12 V 110 V

(b) To get 220 V on the primary side, the coils are connected in series, with series-aiding on the secondary side. The coils must be connected series-aiding to give 50 V. Thus, the connections are shown below.

50 V

220 V

Page 704: Solution Manual for Fundamentals of Electric Circuits

Chapter 13, Solution 94. V2/V1 = 110/440 = 1/4 = I1/I2 There are four ways of hooking up the transformer as an auto-transformer. However it is clear that there are only two outcomes.

V2

V1

V2

V1

V2

V1

V2

V1

(1) (2) (3) (4) (1) and (2) produce the same results and (3) and (4) also produce the same results. Therefore, we will only consider Figure (1) and (3). (a) For Figure (3), V1/V2 = 550/V2 = (440 �– 110)/440 = 330/440 Thus, V2 = 550x440/330 = 733.4 V (not the desired result) (b) For Figure (1), V1/V2 = 550/V2 = (440 + 110)/440 = 550/440 Thus, V2 = 550x440/550 = 440 V (the desired result) Chapter 13, Solution 95.

(a) n = Vs/Vp = 120/7200 = 1/60 (b) Is = 10x120/144 = 1200/144

S = VpIp = VsIs

Ip = VsIs/Vp = (1/60)x1200/144 = 139 mA

Page 705: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 1.

RCj1RCj

Cj1RR

)(i

o

VV

H

)(H0

0

j1j

, where RC1

0

20

0

)(1)(H H

0

1-tan2

)(H

This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC10 . Thus, the sketches of H and are shown below.

H

0 = 1/RC

1 0.7071

0

0

90

0 = 1/RC

45

Page 706: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 2.

RLj11

LjRR

)(H0j1

1, where

LR

0

20 )(1

1)(H H

0

1-tan-)(H

The frequency response is identical to the response in Example 14.1 except that

LR0 . Hence the response is shown below.

H

0 = R/L

0.7071 1

0

0 = R/L

-45

-90

0

Chapter 14, Solution 3.

(a) The Thevenin impedance across the second capacitor where V is taken is o

sRC1R

RsC1||RRThZ

sRC1sC1RsC1 i

iTh

VVV

ZTh

sC1

+

Vo +

VTh

Page 707: Solution Manual for Fundamentals of Electric Circuits

)sC1)(sRC1(sC1sC1

Th

iTh

Tho Z

VV

ZV

))sRC1(sRCsRC1)(sRC1(1

)sRC1)(sC1(1

)sThi

o

ZVV

H(

)s(H1sRC3CRs

1222

(b) 08.01080)102)(1040(RC -3-63

There are no zeros and the poles are at

RC0.383-

s1 4.787-

RC2.617-

s2 32.712-

Chapter 14, Solution 4.

(a) RCj1

RCj

1||R

)RCj1(LjRR

RCj1R

Lj

RCj1R

)(i

o

VV

H

)(HLjRRLC-

R2

(b) )LjR(Cj1

)LjR(CjCj1LjR

LjR)(H

)(HRCjLC1

RCjLC-2

2

Page 708: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 5.

(a) Cj1LjR

Cj1)(

i

o

VV

H

)(HLCRCj1

12

(b) RCj1

RCj

1||R

)RCj1(LjR)RCj1(Lj

)RCj1(RLjLj

)(i

o

VV

H

)(HRLCLjR

RLCLj2

2

Chapter 14, Solution 6.

(a) Using current division,

Cj1LjRR

)(i

o

II

H

)25.0)(10()25.0)(20(j1)25.0)(20(j

LCRCj1RCj

)( 22H

)(H 25.25j15j

(b) We apply nodal analysis to the circuit below.

1/j C+

IoVx

0.5 Vx

R Is j L

Page 709: Solution Manual for Fundamentals of Electric Circuits

Cj1Lj5.0

Rxxx

s

VVVI

But )Cj1Lj(2Cj1Lj

5.0ox

xo IV

VI

Cj1Lj5.0

R1

x

s

VI

)Cj1Lj(21

R1

)Cj1Lj(2 o

s

II

1R

)Cj1Lj(2

o

s

II

)LC1(2RCjRCj

R)Cj1Lj(211)( 2

s

o

II

H

)25.01(2jj

)( 2H

)(H 25.0j2j

Chapter 14, Solution 7.

(a) Hlog2005.0 10

Hlog105.2 10-3

-3105.210H 005773.1

(b) Hlog206.2- 10

Hlog0.31- 10 -0.3110H 4898.0

(c) Hlog207.104 10

Hlog235.5 10 235.510H 510718.1

Page 710: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 8.

(a) 05.0H05.0log20H 10dB 26.02- , 0

(b) 125H

125log20H 10dB 41.94 , 0

(c) 43.63472.4j2

10j)1(H

472.4log20H 10dB 01.13 , 43.63

(d) 23.55-254.47.1j9.3j2

6j1

3)1(H

254.4log20H 10dB 577.12 , 23.55-

Chapter 14, Solution 9.

)10j1)(j1(1

)(H

10/j1log20j1log20-H 1010dB )10/(tan)(tan- -1-1

The magnitude and phase plots are shown below.

HdB

0.1

-40

10/j11

log20 10

j11

log20 10

-20

10 1 100

1

j11

arg

-135

10/j1arg

-90

-180

0.1 10 1 100-45

Page 711: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 10.

5j1j1

10)j5(j

50)j(H

j1log20

5j1

1log20

20 log1

-40

0.1

10

1 100

20

-20

HdB 40

-135

5/j11arg

j1arg

-90

-180

0.1 10

1

100 -45

Chapter 14, Solution 11.

)2j1(j)10j1(5

)(H

2j1log20jlog2010j1log205log20H 10101010dB

2tan10tan90- -1-1

Page 712: Solution Manual for Fundamentals of Electric Circuits

The magnitude and phase plots are shown below.

-40

0.1 10 1 100

20 14

-20

HdB 40 34

-45

45

1001 10 0.1

-90

90

Page 713: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 12.

201.0log20,)10/1(

)1(1.0)(jjj

wT

The plots are shown below. |T| (db) 20 0 0.1 1 10 100 -20 -40 arg T 90o

0 0.1 1 10 100 -90o

Page 714: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 13.

)10j1()j()j1)(101(

)j10()j(j1

)( 22G

10j1log20jlog40j1log2020-G 101010dB

10tantan-180 -1-1 The magnitude and phase plots are shown below.

GdB 40

20

0.1 1 10 100-20

-40

90

0.1 1 10 100-90

-180

Page 715: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 14.

2

5j

2510j

1j

j12550

)(H

jlog20j1log202log20H 101010dB

2

10 )5j(52j1log20

512510

tantan90- 21-1-

The magnitude and phase plots are shown below.

-90

-40

0.1 10 1 100

20 6

40 26

-20

HdB

-180

0.1 10 1 100

90

Page 716: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 15.

)10j1)(2j1()j1(2

)j10)(j2()j1(40

)(H

10j1log202j1log20j1log202log20H 10101010dB

10tan2tantan -1-1-1

The magnitude and phase plots are shown below.

-45

6

40

-40

0.1 10 1 100

20

-20

HdB

90

-90

0.1 10 1 100

45

Chapter 14, Solution 16.

2

10j1)j1(100

j)(G

Page 717: Solution Manual for Fundamentals of Electric Circuits

GdB

-20

jlog20

10jlog40

-90 2

10j1

1arg j11arg

arg(j )

-180

0.1 10 1 100

90

20 log(1/100) -60

-40

0.1 10 1 100

20

Chapter 14, Solution 17.

2)2j1)(j1(j)41(

)(G

2j1log40j1log20jlog204-20logG 10101010dB

2tan2tan--90 -1-1

The magnitude and phase plots are shown below.

GdB

-20

20

-12

1001 10 0.1

-40

Page 718: Solution Manual for Fundamentals of Electric Circuits

-90

-180

0.1 10 1 100

90

Chapter 14, Solution 18.

)10j1)(5j1(j50)2j1(4

)(2

G

jlog202j1log40504log20G 101010dB

10j1log205j1log20 1010 where 94.21-504log1020

10tan5tan2tan290- -1-1-1

The magnitude and phase plots are shown below.

GdB

-20

180

0.1 10 1 100

90

-90

-60

-40

0.1 10 1 100

20

Page 719: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 19.

)10010j1(100j

)( 2H

10010j1log20100log20jlog20H 2

101010dB

100110

tan90 21-

The magnitude and phase plots are shown below.

-90

20

-60

-40

0.1 10 1 100

40

-20

HdB

-180

0.1 10 1 100

90

Page 720: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 20.

)10j1)(j1()j1(10

)(2

N

2

101010dB j1log2010j1log20j1log2020N

10tantan1

tan 1-1-2

1-

The magnitude and phase plots are shown below.

40

180

0.1 10 1 100

90

-90

20

-40

0.1 10 1 100-20

NdB

Chapter 14, Solution 21.

)10010j1)(10j1(100)j1(j

)( 2T

100log20j1log20jlog20T 101010dB

Page 721: Solution Manual for Fundamentals of Electric Circuits

10010j1log2010j1log20 21010

100110

tan10tantan90 21-1-1-

The magnitude and phase plots are shown below.

180

-180

0.1 10 1 100

90

-90

-40

-60

0.1 10 1 100

20

-20

TdB

Chapter 14, Solution 22. 10kklog2020 10

A zero of slope at dec/dB20 2j12

A pole of slope t dec/dB20- a20j1

120

Page 722: Solution Manual for Fundamentals of Electric Circuits

A pole of slope t dec/dB20- a100j1

1100

Hence,

)100j1)(20j1()2j1(10

)(H

)(H)j100)(j20(

)j2(104

Chapter 14, Solution 23.

A zero of slope at the origindec/dB20 j

A pole of slope t dec/dB20- a1j1

11

A pole of slope at dec/dB40- 2)10j1(1

10

Hence,

2)10j1)(j1(j

)(H

)(H 2)j10)(j1(j100

Chapter 14, Solution 24. The phase plot is decomposed as shown below.

-45 100/j1

1arg

)10/j1(arg

)j(arg

100

90

-90

0.1 10 1 1000

45

Page 723: Solution Manual for Fundamentals of Electric Circuits

)j100(j)j10)(10(k

)100j1(j)10j1(k

)(G

where k is a constant since arg 0k .

Hence, )(G)j100(j

)j10(k, where 10k constant isk

Chapter 14, Solution 25.

s/krad5)101)(1040(

1LC1

6-3-0

R)( 0Z k2

C4

L4

jR)4(0

00Z

)101)(105(4

10404105

j2000)4( 6-33-

3

0Z

)5400050(j2000)4( 0Z

)4( 0Z k75.0j2

C2

L2

jR)2(0

00Z

)101)(105(2

)1040(2

)105(j2000)2( 6-3

3-3

0Z

)52000100(j2000)4( 0Z

)2( 0Z k3.0j2

C21

L2jR)2(0

00Z

Page 724: Solution Manual for Fundamentals of Electric Circuits

)101)(105)(2(1

)1040)(105)(2(j2000)2( 6-33-3

0Z

)2( 0Z k3.0j2

C41

L4jR)4(0

00Z

)101)(105)(4(1

)1040)(105)(4(j2000)4( 6-33-3

0Z

)4( 0Z k75.0j2

Chapter 14, Solution 26.

(a) kHz 51.2210101052

12

139 xxxLC

fo

(b) krad/s 101010

1003xL

RB

(c ) 142.14101.0

101050

1013

36

xx

RL

LCRL

Q o

Chapter 14, Solution 27. At resonance,

10RZ , LC1

0

LR

B and R

LB

Q 00

Hence,

H1650

)80)(10(RQL

0

Page 725: Solution Manual for Fundamentals of Electric Circuits

F25)16()50(

1L

1C 22

0

s/rad625.01610

LR

B

Therefore, R 10 , L H16 , C F25 , B s/rad625.0

Chapter 14, Solution 28. Let 10R .

H5.02010

BR

L

F2)5.0()1000(

1L

1C 22

0

5020

1000B

Q 0

Therefore, if then 10R L H5.0 , C F2 , Q 50

Chapter 14, Solution 29.

j 1/j

1 Z

j

j1j

j1

jZ

Page 726: Solution Manual for Fundamentals of Electric Circuits

2

2

1j1

jZ

Since and are in phase, )t(v )t(i

211

0)Im(Z

0124

618.02

411-2

s/rad7861.0

Chapter 14, Solution 30. Select 10R .

mH5H05.0)20)(10(

10Q

RL

0

F2.0)05.0)(100(

1L

1C 2

0

s/rad5.0)2.0)(10(

1RC1

B

Therefore, if then 10R L mH5 , C F2.0 , B s/rad5.0

Chapter 14, Solution 31.

LL

XLLX

rad/s 10x796.810x40

10x6.5x10x10x2X

RLRB 6

3

36

L

Page 727: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 32. Since 10Q ,

2B

01 , 2B

02

120106

QB

60 s/krad50

025.061 s/rad10975.5 6

025.062 s/rad10025.6 6

Chapter 14, Solution 33.

pF 84.5610x40x10x6.5x2

80Rf2

QCRCQ36o

o

H 21.1480x10x6.5x2

10x40Qf2

RLL

RQ6

3

oo

Chapter 14, Solution 34.

(a) krad/s 443.110x60x10x8

1LC1

63o

(b) rad/s 33.310x60x10x5

1RC1B

63

(c) 9.43210x60x10x5x10x443.1RCQ 633

o

Page 728: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 35. At resonance,

3-10251

Y1

RR1

Y 40

)40)(10200(80

RQ

CRCQ 30

0 F10

)1010)(104(1

C1

LLC1

6-1020

0 H5.2

8010200

QB

30 s/krad5.2

5.22002B

01 s/krad5.197

5.22002B

01 s/krad5.202

Chapter 14, Solution 36.

s/rad5000LC1

0

R)(R1

)( 00 ZY k2

kS75.18j5.0L

4C

4j

R1

)4(0

00Y

01875.0j0005.01

)4( 0Z 3.53j4212.1

kS5.7j5.0L

2C

2j

R1

)2(0

00Y

Page 729: Solution Manual for Fundamentals of Electric Circuits

0075.0j0005.01

)2( 0Z 74.132j85.8

kS5.7j5.0C2

1L2j

R1

)2(0

00Y

)2( 0Z 74.132j85.8

kS75.18j5.0C4

1L4j

R1

)4(0

00Y

)4( 0Z 3.53j4212.1

Chapter 14, Solution 37.

22 )C

1L(R

)C

1L(jRLRjCL

LjCj

1R

)Cj

1R(Lj)

Cj1R//(LjZ

1)LCCR(0)

C1L(R

C1L

CLLR

)ZIm( 22222

2

Thus,

22CRLC

1

Chapter 14, Solution 38.

222 LRLjR

CjCjLjR

1Y

At resonance, , i.e. 0)Im(Y

Page 730: Solution Manual for Fundamentals of Electric Circuits

0LR

LC 22

02

00

CL

LR 220

2

2

3-6-3-2

2

0 104050

)1010)(1040(1

LR

LC1

0 s/rad4841

Chapter 14, Solution 39.

(a) s/krad810x)8690(2)ff(2B 31212

17610x)88(2)(21 3

21o

nF89.1910x2x10x8

1BR1C

RC1B

33

(b) H4.16410x89.19x)176(

1

C

1LLC1

92o

2o

(c ) s/krad9.552176o (d) s/krad13.258B

(e) 228

176B

Q o

Chapter 14, Solution 40.

(a) L = 5 + 10 = 15 mH

63010x20x10x15

1LC1 1.8257 k rad/sec

Page 731: Solution Manual for Fundamentals of Electric Circuits

633

0 10x20x10x25x10x8257.1RCQ 912.8

63 10x2010x25

1RC1B 2 rad

(b) To increase B by 100% means that B�’ = 4.

4x10x25

1BR1C

3 10 µF

Since F10CC

C

21

21CC and C1 = 20 µF, we then obtain C2 = 20 µF.

Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one.

Chapter 14, Solution 41.

(a) This is a series RLC circuit.

862R , H1L , F4.0C

4.01

LC1

0 s/rad5811.1

85811.1

RL

Q 0 1976.0

LR

B s/rad8

(b) This is a parallel RLC circuit.

F263)6)(3(

F6andF3

F2C , k2R , mH20L

Page 732: Solution Manual for Fundamentals of Electric Circuits

)1020)(102(1

LC1

3-6-0 s/krad5

)1020)(105(102

LR

Q 3-3

3

020

)102)(102(1

RC1

B 6-3 s/krad250

Chapter 14, Solution 42.

(a) )LjR(||)Cj1(inZ

RCjLC1LjR

Cj1

LjR

CjLjR

2inZ

22222

2

in CR)LC1()RCjLC1)(LjR(

Z

At resonance, Im( 0)inZ , i.e.

CR)LC1(L0 22 CRLLC 22

LCCRL 2

0 LR

C1 2

(b) )Cj1R(||LjinZ

RCj)LC1()RCj1(Lj

Cj1LjR)Cj1R(Lj

2inZ

22222

22

in CR)LC1(]RCj)LC1[()LjRLC-(

Z

At resonance, Im( 0)inZ , i.e.

LCR)LC1(L0 2232

Page 733: Solution Manual for Fundamentals of Electric Circuits

1)CRLC( 222

0 22CRLC1

(c) )Cj1Lj(||RinZ

RCj)LC1()LC1(R

Cj1LjR)Cj1Lj(R

2

2

inZ

22222

22

in CR)LC1(]RCj)LC1)[(LC1(R

Z

At resonance, Im( 0)inZ , i.e.

RC)LC1(R0 2 0LC1 2

0 LC1

Chapter 14, Solution 43. Consider the circuit below.

1/j C

R1 Zin

j L R2

(a) )Cj1R(||)Lj||R( 21inZ

Cj1

R||LjR

LjR2

1

1inZ

Page 734: Solution Manual for Fundamentals of Electric Circuits

LjRLjR

Cj1

R

Cj1

RLjR

LRj

1

12

21

1

inZ

12

21

21in LCR)CRj1)(LjR(

)CRj1(LRjZ

)CRRL(jLCRLCRRLRjLCRR-

2122

12

1

1212

inZ

221

222

21

21

2122

12

11212

in )CRRL()LCRLCRR()]CRRL(jLCRLCRR)[LRjLCRR-(

Z

At resonance, Im( 0)inZ , i.e.

)LCRLCRR(LR)CRRL(LCRR0 22

12

1121213

CLRLRLCRR0 221

321

222

21

3 LC1CR0 222

22

1)CRLC( 222

2

222

0 CRLC1

26-26-0 )109()1.0()109)(02.0(1

0 s/krad357.2 (b) At , s/krad357.20

14.47j)1020)(10357.2(jLj -33

0212.0j9996.014.47j1

14.47jLj||R1

14.47j1.0)109)(10357.2(j

11.0

Cj1

R 6-32

)Cj1R(||)Lj||R()( 210inZ

Page 735: Solution Manual for Fundamentals of Electric Circuits

)14.47j1.0()0212.0j9996.0()14.47j1.0)(0212.0j9996.0(

)( 0inZ

)( 0inZ 1

Chapter 14, Solution 44. We find the input impedance of the circuit shown below.

j (2/3) 1/jZ

1

1/j C

Cj111

j2j

31Z

, 1

2

2

C1jCC

-j0.5jC1

jCj-j1.5

1Z

)t(v and i are in phase when Z is purely real, i.e. )t(

1)1C(C1

C-0.50 2

2 or C F1

221

C1C1

2

2

ZZ

20)10)(2(IZV

V)tsin(20)t(v , i.e. oV V20

Page 736: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 45.

(a) j1

jj||1 ,

j11

j11j1

j1

||1

Transform the current source gives the circuit below.

j11

j1j

1

Ij1

j +

+

Vo

IVj1

j

j1j

j11

1

j11

o

IV

H o)( 2)j1(2j

(b) 2)j1(21

)1(H

2)2(21

)1(H 25.0

Chapter 14, Solution 46.

(a) This is an RLC series circuit.

nF26.1110x10x)10x15x2(

1

L

1CLC1

323o

2o

(b) Z = R, I = V/Z = 120/20 = 6 A

(c ) 12.471520

10x10x10x15x2R

LQ

33o

Page 737: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 47.

RLj1

1LjR

R)(

i

o

VV

H

1)0(H and 0)(H showing that this circuit is a lowpass filter.

At the corner frequency, 2

1)(H c , i.e.

RL

1

RL

1

12

1 c

2c

or LR

c

Hence,

cc f2LR

3-

3

c 1021010

21

LR

21

f kHz796

Chapter 14, Solution 48.

Cj1

||RLj

Cj1

||R)(H

Cj1RCjR

Lj

Cj1RCjR

)(H

)(HRLCLjR

R2

1)0(H and showing that 0)(H this circuit is a lowpass filter.

Page 738: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 49.

At dc, 224

)0(H .

Hence, 2

2)0(H

21

)(H

2c1004

42

2

2.081004 c

2c

10j12

20j24

)2(H

199.01012

)2(H

In dB, )2(Hlog1020 14.023-

10-tan)2(Harg -1 84.3- Chapter 14, Solution 50.

LjR

Lj)(

i

o

VV

H

0)0(H and showing that 1)(H this circuit is a highpass filter.

LR

1

LR

1

12

1)(

c2

c

cH

or cc f2LR

1.0200

21

LR

21

fc Hz3.318

Page 739: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 51.

RC1j

jRCj1

RCj)(H (from Eq. 14.52)

This has a unity passband gain, i.e. 1)(H .

50RC1

c

j5010j

)(10)( HH ^

)(Hj50

10j

Chapter 14, Problem 52.

Design an RL lowpass filter that uses a 40-mH coil and has a cut-off frequency of 5 kHz.

Chapter 14, Solution 53.

cc f2LR

)1040)(10)(2(Lf2R -35

c k13.25 Chapter 14, Solution 54.

311 1020f2

322 1022f2

3

12 102B

3120 1021

2

Page 740: Solution Manual for Fundamentals of Electric Circuits

221

BQ 0 5.11

C1

LLC1

20

0

)1080()1021(1

L 12-23 H872.2

BLRLR

B

)872.2)(102(R 3 k045.18

Chapter 14, Solution 55.

k3.26510x300x10x2x2

1Cf2

1RRC1f2

123ccc

Chapter 14, Solution 56.

s/krad10)104.0)(1025(

1LC1

63o

s/krad4.01025

10LR

B 3-

4.010

Q 25

s/krad8.92.0102Bo1 or kHz56.12

8.91f

s/krad2.102.0102Bo2 or kHz62.12

2.10f2

Therefore,

kHz62.1fkHz56.1

Page 741: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 57.

(a) From Eq 14.54,

LC1

LRss

LRs

LCssRC1sRC

sC1sLR

R)s(2

2H

Since LR

B and LC1

0 ,

)s(H 20

2 sBssB

(b) From Eq. 14.56,

LC1

LR

ss

LC1

s

sC1

sLR

sC1

sL)s(

2

2

H

)s(H 20

2

20

2

sBss

Chapter 14, Solution 58.

(a) Consider the circuit below.

I I1

R1/sC+

Vo +

R 1/sC

Vs

sC2

R

sC1

RsC1

RsC1

R||sC1

R)s(Z

Page 742: Solution Manual for Fundamentals of Electric Circuits

)sRC2(sCsRC1

R)s(Z

)sRC2(sCCRssRC31

)s(222

Z

ZV

I s

)sRC2(RsC2sC1 s

1 ZV

II

222s

1o CRssRC31)sRC2(sC

sRC2R

RV

IV

222s

o

CRssRC31sRC

)s(VV

H

222

CR1

sRC3

s

sRC3

31

)s(H

Thus, 2220 CR

1 or

RC1

0 s/rad1

RC3

B s/rad3

(b) Similarly,

sLR2)sLR(R

sL)sLR(||RsL)s(Z

sLR2LssRL3R

)s(222

Z

ZV

I s , )sLR2(

RsLR2

R s1 Z

VII

222s

1o LssRL3RsLR2

sLR2sLR

sLV

IV

Page 743: Solution Manual for Fundamentals of Electric Circuits

2

22

222s

o

LR

sLR3

s

sLR3

31

LssRL3RsRL

)s(VV

H

Thus, LR

0 s/rad1

LR3

B s/rad3

Chapter 14, Solution 59.

(a) )1040)(1.0(

1LC1

12-0 s/rad105.0 6

(b) 43

1021.0

102LR

B

250102105.0

BQ 4

60

As a high Q circuit,

)150(102B 4

01 s/krad490

)150(102B 4

02 s/krad510

(c) As seen in part (b), Q 250

Page 744: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 60. Consider the circuit below.

Ro

+

Vo

+ R

1/sC

Vi

sL

sC1sLR)sC1sL(R

sC1

sL||R)s(Z

LCssRC1)LCs1(R

)s( 2

2

Z

LCRsRLCRsCsRRR)LCs1(R

R 2o

2oo

2

oi

o

ZZ

VV

H

LCssRC1)LCs1(R

RR 2

2

ooin ZZ

LCssRC1LCRsRLCRsCsRRR

2

2o

2oo

inZ

js

RCjLC1LCRRLCRCRRjR

2

2o

2oo

inZ

222

2o

2o

2o

in )RC()LC1()RCjLC1)(CRRjLCRLCRRR(

Z

0)Im( inZ implies that

0)LC1(CRR]LCRLCRRR[RC- 2

o2

o2

o

Page 745: Solution Manual for Fundamentals of Electric Circuits

0LCRRLCRLCRRR o2

o2

o2

o RLCR2

)104)(101(1

LC1

6-3-0 s/krad811.15

LCRLCRRCRRjR)LC1(R

2o

2oo

2

H

RRR

)0(HHo

max

or o

oo

2o

2

max RRR

)RR(LCCRRjRR

LC1Rlim)(HH

At and , 1 2 mzxH2

1H

CRRj)RR(LCRR)LC1(R

)RR(2R

oo2

o

2

o

2o

2o

2o

2o

))RR(LCRR()CRR(

)LC1)(RR(2

1

28-226-

-92

)10410()10(96)1041(10

21

21

)10410()10(96)1041(10

028-226-

-92

0)10410()1096()2)(10410( 2-822-6-82

2-822-62-82 )10410()1096()10410)(2(

0)10410()1096( 2-822-6

010010092.8106.1 2-74-15

Page 746: Solution Manual for Fundamentals of Electric Circuits

01025.610058.5 1684

8

82

101471.2109109.2

Hence,

s/krad653.141

s/krad061.172

653.14061.17B 12 s/krad408.2 Chapter 14, Solution 61.

(a) iCj1RCj1

VV , oVV

Since , VV

oiRCj11

VV

i

o)(VV

HRCj1

1

(b) iCj1RR

VV , oVV

Since , VV

oiRCj1RCj

VV

i

o)(VV

HRCj1

RCj

Page 747: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 62. This is a highpass filter.

RCj11

RCj1RCj

)(H

cj11

)(H , )1000(2RC1

c

f1000j11

ffj11

)(c

H

(a) i

o

5j11

)Hz200f(VV

H

5j1mV120

oV mV53.23

(b) i

o

5.0j11

)kHz2f(VV

H

5.0j1mV120

oV m3.107 V

(c) i

o

1.0j11

)kHz10f(VV

H

1.0j1mV120

oV m4.119 V

Chapter 14, Solution 63.

For an active highpass filter,

ii

fiRsC1

RsC)s(H (1)

Page 748: Solution Manual for Fundamentals of Electric Circuits

But

10/s1s10)s(H (2)

Comparing (1) and (2) leads to:

M10C10R10RC

iffi

k100C

1.0R1.0RCi

iii

Chapter 14, Solution 64.

ff

f

fff CRj1

RCj1

||RZ

i

ii

iii Cj

CRj1Cj

1RZ

Hence,

i

f

i

o -)(

ZZ

VV

H)CRj1)(CRj1(

CRj-

iiff

if

This is a bandpass filter. )(H is similar to the product of the transfer function of a lowpass filter and a highpass filter.

Chapter 14, Solution 65.

ii RCj1RCj

Cj1RR

VVV

ofi

i

RRR

VV

Since , VV

Page 749: Solution Manual for Fundamentals of Electric Circuits

iofi

i

RCj1RCj

RRR

VV

i

o)(VV

HRCj1

RCjRR

1i

f

It is evident that as , the gain is i

f

RR

1 and that the corner frequency is RC1

.

Chapter 14, Solution 66.

(a) Proof

(b) When 3241 RRRR ,

CR1ss

RRR

)s(243

4H

(c) When 3R ,

CR1sCR1-

)s(2

1H

Chapter 14, Solution 67.

fii

f R4R41

RR

gain DC

s/rad)500(2CR

1frequencyCorner

ffc

If we select , then k20R f k80R i and

nF915.15)1020)(500)(2(

1C 3

Therefore, if fR k20 , then iR k80 and C nF915.15

Page 750: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 68.

ifi

f R5RRR

5gainfrequency High

s/rad)200(2CR

1frequencyCorner

iic

If we select , then k20R i k100fR and

nF8.39)1020)(200)(2(

1C 3

Therefore, if iR k20 , then fR k100 and C nF8.39

Chapter 14, Solution 69. This is a highpass filter with f 2c kHz.

RC1

f2 cc

3c 104

1f2

1RC

Hz108 may be regarded as high frequency. Hence the high-frequency gain is

410

RRf or R5.2Rf

If we let R k10 , then fR k25 , and 41040001

C nF96.7 .

Page 751: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 70.

(a) )YYY(YYY

YY)s()s(

)s(321421

21

i

o

VV

H

where 11

1 GR1

Y , 22

2 GR1

Y , 13 sCY , 24 sCY .

)s(H)sCGG(sCGG

GG

121221

21

(b) 1GGGG

)0(H21

21 , 0)(H

showing that this circuit is a lowpass filter. Chapter 14, Solution 71. L , 50R , mH F1C40

)1040(KK

1LKK

L 3-

f

m

f

m

mf KK25 (1)

fm

-6

fm KK10

1KK

CC

mf

6

K1

K10 (2)

Substituting (1) into (2),

ff

6

K251

K10

fK -3102.0

fm K25K -3105

Page 752: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 72.

CLLC

KKLC

CL 2f2

f

8-

-6-32f 104

)2)(1()1020)(104(

K

fK -4102

LC

CL

KKCL

CL 2

m2m

3-

3-

-62m 105.2

)104)(2()1020)(1(

K

mK -2105

Chapter 14, Solution 73.

)10800)(12(RKR 3m M6.9

)1040(1000800

LKK

L 6-

f

m F32

)1000)(800(10300

KKC

C-9

fmpF375.0

Chapter 14, Solution 74.

300100x3RK'R 1m1

k 1100x10RK'R 2m2

H 200)2(10

10LKK

'L6

2

f

m

nF 110101

KKC'C

8fm

Page 753: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 75.

20010x20RK'R m

H 400)4(10

10LKK

'L5f

m

F 110x10

1KK

C'C5fm

Chapter 14, Solution 76.

5010

10x50R10x50RK'R3

33

m

mH 1010

10x10x10LH 10LKK

'L3

66

f

m

mF 4010x10x10x40CKK

CFp 40'C 6312

fm

Chapter 14, Solution 77. L and C are needed before scaling.

H25

10BR

LLR

B

F5.312)2)(1600(

1L

1C

LC1

20

0

(a) )2)(600(LKL m H1200

60010125.3

KC

C-4

m

F5208.0

Page 754: Solution Manual for Fundamentals of Electric Circuits

(b) 3f 10

2KL

L mH2

3

-4

f 1010125.3

KC

C nF5.312

(c) 5f

m

10)2)(400(

LKK

L mH8

)10)(400(10125.3

KKC

C 5

-4

fmpF81.7

Chapter 14, Solution 78.

k1)1)(1000(RKR m

H1.0)1(1010

LKK

L 4

3

f

m

F1.0)10)(10(

1KK

CC 43

fm

The new circuit is shown below.

1 k

1 k 0.1 F0.1 H1 k +

Vx I

Page 755: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 79.

(a) Insert a 1-V source at the input terminals.

Ro

+

3Vo

+

Vo

Io V2V1

sL +

R 1/sC

1 V

There is a supernode.

sC1sLR1 21 VV

(1)

But (2) o12o21 33 VVVVVV

Also, sC1sLsLsC1sL

sL 2o2o

VVVV (3)

Combining (2) and (3)

oo12 sLsC1sL

3 VVVV

12

2

o LCs41LCs

VV (4)

Substituting (3) and (4) into (1) gives

12o1

LCs41sC

sLR1

VVV

12

2

121 LCs41sRCLCs41

LCs41sRC

1 VVV

sRCLCs41LCs41

2

2

1V

Page 756: Solution Manual for Fundamentals of Electric Circuits

)sRCLCs41(RsRC

R1

21

o

VI

sCLCs4sRC11 2

oin I

Z

sC1

RsL4inZ (5)

When , , 5R 2L 1.0C ,

)s(inZs

105s8

At resonance,

C1

L40)Im( inZ

or )2)(1.0(2

1LC21

0 s/rad118.1

(b) After scaling,

RKR m 404 505

H2.0)2(10010

LKK

Lf

m

4-

fm10

)100)(10(1.0

KKC

C

From (5),

)s(inZs

1050s8.0

4

)10)(2.0(2

1LC21

4-0 s/rad8.111

Page 757: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 80.

(a) 400)2)(200(RKR m

mH2010

)1)(200(K

LKL 4

f

m

F25.0)10)(200(

5.0KK

CC 4

fm

The new circuit is shown below.

20 mHa

400

Ix

0.25 F 0.5 Ix

b

(b) Insert a 1-A source at the terminals a-b.

V2V1

b

a sL

R

Ix

1/(sC) 1 A 0.5 Ix

At node 1,

sL

sC1 211

VVV (1)

At node 2,

R

5.0sL

2x

21 VI

VV

But, I . 1x sC V

Page 758: Solution Manual for Fundamentals of Electric Circuits

RsC5.0

sL2

121 V

VVV

(2)

Solving (1) and (2),

1sCR5.0LCs

RsL21V

1sCR5.0LCsRsL

1 21

Th

VZ

At , 410

1)400)(1025.0)(10j(5.0)1025.0)(1020()10j(400)1020)(10j(

6-46-3-24

3-4

ThZ

200j6005.0j5.0

200j400ThZ

ThZ ohms435.18-5.632

Chapter 14, Solution 81. (a)

1GR)LGRC(jLC

RLjZtoleadswhich

LjR1)LjR)(CjG(

LjR1CjG

Z1

2

LC1GR

CG

LRj

LCR

Cj

)(Z2

(1)

We compare this with the given impedance:

25001j2

)1j(1000)(Z2

(2)

Page 759: Solution Manual for Fundamentals of Electric Circuits

Comparing (1) and (2) shows that

LR1 R/LmF, 1C1000C1

1CG2CG

LR mS

L4.0RR10

1R10LC

1GR25013

3

Thus, R = 0.4 , L = 0.4 H, C = 1 mF, G = 1 mS

(b) By frequency-scaling, Kf =1000. R�’ = 0.4 , G�’ = 1 mS

mH4.010

4.0KL'L

3f, F1

10

10KC'C

3

3

f

Chapter 14, Solution 82.

fmKK

CC

2001

200K c

f

50002001

101

K1

CC

K 6-f

m

RKR m 5 k , thus, if R2R k10

Page 760: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 83.

pF 1.010x100

10CKK

1'CF15

6

fm

pF 5.0'CF5

M 1k 10x100RK'Rk 10 m

M 2'Rk 20 Chapter 14, Solution 84.

The schematic is shown below. A voltage marker is inserted to measure vo. In the AC sweep box, we select Total Points = 50, Start Frequency = 1, and End Frequency = 1000. After saving and simulation, we obtain the magnitude and phase plots in the probe menu as shown below.

Page 761: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 85. We let A so that Vo

s 01I .VI/ oso The schematic is shown below. The circuit is simulated for 100 < f < 10 kHz.

Page 762: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 86.

The schematic is shown below. A current marker is inserted to measure I. We set Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the AC sweep box. After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below.

Page 763: Solution Manual for Fundamentals of Electric Circuits
Page 764: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 87.

The schematic is shown below. In the AC Sweep box, we set Total Points = 50, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude response as shown below. It is evident from the response that the circuit represents a high-pass filter.

Page 765: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 88. Chapter 14, Solution 88.

The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.

The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the magnitude and phase plots of Vo as shown below.

Page 766: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 89.

The schematic is shown below. In the AC Sweep box, we type Total Points = 101, Start Frequency = 100, and End Frequency = 1 k. After simulation, the magnitude plot of the response Vo is obtained as shown below.

Page 767: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 90.

The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.

Page 768: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 91.

The schematic is shown below. In the AC Sweep box, we set Total Points = 1001, Start Frequency = 1, and End Frequency = 100k. After simulation, we obtain the magnitude plot of the response as shown below. The response shows that the circuit is a high-pass filter.

Page 769: Solution Manual for Fundamentals of Electric Circuits

Chapter 14, Solution 92.

The schematic is shown below. We type Total Points = 101, Start Frequency = 1, and End Frequency = 100 in the AC Sweep box. After simulating the circuit, the magnitude plot of the frequency response is shown below.

Page 770: Solution Manual for Fundamentals of Electric Circuits

hapter 14, Solution 93.

C

R

L

C

2

2

0 LR

LC1

21

f

610

10240400

L

7

6- 28810

)10120)(10240(1

LC1 16

12-6- R

,

Since LC1

L

R

Page 771: Solution Manual for Fundamentals of Electric Circuits

22410

LC21

f8

0 kHz938

If R is reduced to 40 , LC1

LR

.

The result remains the same. Chapter 14, Solution 94.

RC1

c

We make R and C as small as possible. To achieve this, we connect 1.8 k and 3.3 k in parallel so that

k 164.13.38.13.3x8.1R

We place the 10-pF and 30-pF capacitors in series so that

C = (10x30)/40 = 7.5 pF Hence,

rad/s 10x55.11410x5.7x10x164.1

1RC1 6

123c

Chapter 14, Solution 95.

(a) LC2

1f0

When , pF360C

MHz541.0)10360)(10240(2

1f

12-6-0

When , pF40C

MHz624.1)1040)(10240(2

1f

12-6-0

Therefore, the frequency range is MHz624.1fMHz541.0 0

Page 772: Solution Manual for Fundamentals of Electric Circuits

(b) RfL2

Q

At , MHz541.0f0

12

)10240)(10541.0)(2(Q

-66

98.67

At , MHz624.1f0

12

)10240)(10624.1)(2(Q

-66

1.204

Chapter 14, Solution 96.

C2 C1

+

Vo

VoV1

+

Ri

RL

L

Vi

Z1Z2

22

L

2L1 CsR1

RsC

1||RZ

2L

2L2

L

11

12 CsR1

LCRsRsL||

sC1

)sL(||sC1

ZZ

2L

2L2

L

1

2L

2L2

L

12

CsR1LCRsRsL

sC1

CsR1LCRsRsL

sC1

Z

21L3

1L12

2L

2L2

L2 CLCRsCsRLCsCsR1

LCRsRsLZ

Page 773: Solution Manual for Fundamentals of Electric Circuits

ii2

21 R

VZ

ZV

i1

1

22

21

1

1o sLRsL

VZ

ZZ

ZV

ZZ

V

sLR 1

1

22

2

i

o

ZZ

ZZ

VV

where

22

2

RZZ

21Li3

1Li1i2

2Lii2L2

L

2L2

L

CLCRRsCRsRLCRsCRsRRLCRsRsLLCRsRsL

and 2L

2L

L

1

1

LCRssLRR

sLZZ

Therefore,

i

o

VV

)LCRssLR)(CLCRRsCRsRLCRsCRsRRLCRsRsL(

)LCRsRsL(R

2L2

L21Li3

1Li1i2

2Lii2L2

L

2L2

LL

where s . j

Chapter 14, Solution 97.

C2 C1

+

Vo

VoV1

+

Ri

RL

L

Vi

Z2 Z1

Page 774: Solution Manual for Fundamentals of Electric Circuits

2L

2L

2L sC1sLR

)sC1R(sLsC

1R||sLZ , js

i1i

1 sC1RV

ZZ

V

i1i2L

L1

2L

Lo sC1RsC1R

RsC1R

RV

ZZ

VV

)sC1sLR)(sC1R()sC1R(sL)sC1R(sL

sC1RR

)(2L1i2L

2L

2L

L

i

o

VV

H

)(H)1CsR(LCs)1CsRLCs)(1CsR(

CCLRs

2L12

2L22

1i

21L3

where s . j

Chapter 14, Solution 98.

44)432454(2)ff(2B 1212

)44)(20(QBf2 00

)22)(20(2

)44)(20(f0 Hz440

Chapter 14, Solution 99.

Cf21

C1

Xc

2010

)105)(102)(2(1

Xf21

C-9

36c

Lf2LXL

Page 775: Solution Manual for Fundamentals of Electric Circuits

4103

)102)(2(300

f2X

L-4

6L

2010

4103

2

1LC2

1f

9-4-0 MHz826.1

4-1034

)100(LR

B s/rad10188.4 6

Chapter 14, Solution 100.

RC1

f2 cc

)105.0)(1020)(2(1

Cf21

R 6-3c

91.15

Chapter 14, Solution 101.

RC1

f2 cc

)1010)(15)(2(1

Cf21

R 6-c

k061.1

Chapter 14, Solution 102.

(a) When and 0R s LR , we have a low-pass filter.

RC1

f2 cc

)1040)(104)(2(1

RC21

f 9-3c Hz7.994

Page 776: Solution Manual for Fundamentals of Electric Circuits

(b) We obtain across the capacitor. ThR)RR(||RR sLTh

k5.2)14(||5R Th

)1040)(105.2)(2(1

CR21

f 9-3Th

c

cf kHz59.1

Chapter 14, Solution 103.

sC1||RRR

)(12

2

i

o

VV

H , js

)sC1(RR)sC1R(R

sC1R)sC1(R

R

R)(

12

12

1

12

2H

)(H21

12

sCRR)sCR1(R

Chapter 14, Solution 104.

The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total Points = 1001, Start Frequency = 100, and End Frequency = 100 k. After simulation, we obtain the magnitude plot of the response as shown.

Page 777: Solution Manual for Fundamentals of Electric Circuits
Page 778: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 1.

(a) 2

ee)atcosh(

at-at

as1

as1

21

)atcosh(L 22 ass

(b) 2

ee)atsinh(

at-at

as1

as1

21

)atsinh(L 22 asa

Chapter 15, Solution 2.

(a) )sin()tsin()cos()tcos()t(f )tsin()sin()tcos()cos()s(F LL

)s(F 22s)sin()cos(s

(b) )sin()tcos()cos()tsin()t(f

)tsin()cos()tcos()sin()s(F LL

)s(F 22s)cos()sin(s

Chapter 15, Solution 3.

(a) )t(u)t3cos(e-2tL9)2s(

2s2

(b) )t(u)t4sin(e-2tL16)2s(

42

Page 779: Solution Manual for Fundamentals of Electric Circuits

(c) Since 22 ass

)atcosh(L

)t(u)t2cosh(e-3tL4)3s(

3s2

(d) Since 22 asa

)atsinh(L

)t(u)tsinh(e-4tL1)4s(

12

(e) 4)1s(

2)t2sin(e 2

t-L

If )s(F)t(f

)s(Fdsd-

)t(ft

Thus, 1-2t- 4)1s(2dsd-

)t2sin(etL

)1s(2)4)1s((

222

)t2sin(et -tL 22 )4)1s(()1s(4

Chapter 15, Solution 4.

(a) 16s

se6e4s

s6)s(G2

ss

22

(b) 3s

e5s

2)s(Fs2

2

Page 780: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 5.

(a) 4s

)30sin(2)30cos(s)30t2cos( 2L

4s1)30cos(s

dsd

)30t2cos(t 22

22L

1-2 4s1s

23

dsd

dsd

2-21-2 4s1s

23s24s

23

dsd

32

2

222222 4s

1s23)s8(

4s

23s2

4s

1s232

4s

2s-23

32

2

22 4s

1s23)s8(

4ss32s3s3-

32

23

32

2

4ss8s34

4s)42)(ss3(-3

)30t2cos(t 2L 32

32

4ss3s6s3128

(b) 5t-4

)2s(!4

30et30L 5)2s(720

(c) )01s(4s2

)t(dtd

4)t(ut2 2L s4s22

Page 781: Solution Manual for Fundamentals of Electric Circuits

(d) )t(ue2)t(ue2 -t1)--(t

)t(ue2 1)--(tL1s

e2

(e) Using the scaling property,

s21

25)21(s

121

15)2t(u5L

s5

(f) 31s

6)t(ue6 3t-L

1s318

(g) Let f . Then, )t()t( 1)s(F .

)0(fs)0(fs)s(Fs)t(fdtd

)t(dtd 2n1nn

n

n

n

n

LL

0s0s1s)t(fdtd

)t(dtd 2n1nn

n

n

n

n

LL

)t(dtd

n

n

L ns

Chapter 15, Solution 6.

(a) )1t(2L -se2

(b) )2t(u10L 2s-es

10

(c) )t(u)4t(Ls4

s12

(d) )4t(uee2)4t(ue2 4)--(t-4-t LL)1s(e

e24

-4s

Page 782: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 7.

(a) Since 22 4ss

)t4cos(L , we use the linearity and shift properties to

obtain )1t(u))1t(4cos(10L16ses10

2

s-

(b) Since 32

s2

tL , s1

)t(uL ,

3t2-2

)2s(2

etL , and s

e)3t(u

s3-

L

)3t(u)t(uet t2-2Ls

e)2s(

2 -3s

3

Chapter 15, Solution 8.

(a) t3-t2- e10e4)t2(u6)t3(2L

3s10

2s4

2s1

21

631

2

3s10

2s4

s6

32

(b) )1t(ue)1t(ue)1t()1t(uet -t-t-t

)1t(uee)1t(uee)1t()1t(uet -11)--(t-11)--(t-t

1see

)1s(ee

)1t(uet-s-1

2

-s-1t-L

1se

)1s(e 1)-(s

2

1)-(s

(c) )1t(u))1t(2cos(L4s

es2

-s

Page 783: Solution Manual for Fundamentals of Electric Circuits

(d) Since )t4sin()t4cos()4sin()4cos()t4sin())t(4sin( )t(u))t(4sin()t(u)t4sin(

)t(u)t(u)t4sin(L

)t(u))t(4sin()t(u)t4sin( LL

16se4

16s4

2

s-

2 )e1(16s

4 s-2

Chapter 15, Solution 9.

(a) )2t(u2)2t(u)2t()2t(u)4t()t(f

)s(F 2

-2s

2

-2s

se2

se

(b) )1t(uee2)1t(ue2)t(g 1)--4(t-4-4t

)s(G)4s(e

e24

-s

(c) )t(u)1t2cos(5)t(h

)Bsin()Asin()Bcos()Acos()BAcos(

)1sin()t2sin()1cos()t2cos()1t2cos(

)t(u)t2sin()1sin(5)t(u)t2cos()1cos(5)t(h

4s2

)1sin(54s

s)1cos(5)s(H 22

)s(H4s

415.84ss702.2

22

(d) )4t(u6)2t(u6)t(p

)s(P 4s-2s- es6

es6

Page 784: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 10.

(a) By taking the derivative in the time domain, )tsin(et)tcos()ee-t()t(g -t-t-t

)tsin(et)tcos(et)tcos(e)t(g -t-t-t

1)1s(1

dsd

1)1s(1s

dsd

11)(s1s

)s(G 222

2222

2

2 )2s2(s22s

)2s2(s2ss

2s2s1s

)s(G 22

2

)2s2(s2)(ss

(b) By applying the time differentiation property,

)0(f)s(sF)s(G where , f)tcos(et)t(f -t 0)0(

22

2

2 )2s2(s2s)(s)(s

1)1s(1s

dsd-

)s()s(G 22

2

)2s2(s2)(ss

Chapter 15, Solution 11.

(a) Since 22 ass

)atcosh(L

4)1s()1s(6

)s(F 2 3s2s)1s(6

2

(b) Since 22 asa

)atsinh(L

12s4s12

16)2s()4)(3(

)t4sinh(e3 222t-L

1-22t- )12s4s(12

dsd-

)t4sinh(e3t)s(F L

2-2 )12s4s)(4s2)(12()s(F 22 )12s4s(

)2s(24

Page 785: Solution Manual for Fundamentals of Electric Circuits

(c) )ee(21

)tcosh( t-t

)2t(u)ee(21

e8)t(f t-t3t-

)2t(ue4)2t(ue4 -4t-2t

)2t(uee4)2t(uee4 2)--4(t-82)--2(t-4

)t(ueee4)2t(uee4 -2-2s-42)--2(t-4 LL

2se4

)2t(uee44)-(2s

2)-2(t-4-L

Similarly, 4s

e4)2t(uee4

8)-(2s2)-4(t-8-L

Therefore,

4se4

2se4

)s(F8)-(2s4)-(2s

8s6s)e8e16(s)e4e4(e

2

-22-226)-(2s

Chapter 15, Solution 12.

)2s(2

)2s(s2

2

2s

)1t(22)1t(222)1t(2

e)2s(

3s2s

112s

e2s

ee)2s(

ee)s(f

)1t(uee)1t(uee)1t()1t(uete)t(f

Chapter 15, Solution 13.

(a) )()( sFdsdttf

If f(t) = cost, then 22

2

2 )1()12()1)(1()( and

1)(

sssssF

dsd

sssF

22

2

)1(1)cos(

sssttL

Page 786: Solution Manual for Fundamentals of Electric Circuits

(b) Let f(t) = e-t sin t.

221

1)1(1)( 22 sss

sF

22

2

)22()22)(1()0)(22(

sssss

dsdF

22 )22()1(2)sin(

sss

dsdFtte tL

(c ) s

dssFttf )()(

Let 22)( then ,sin)(s

sFttf

sssds

stt

ss

11122 tantan

2tan1sinL

Chapter 15, Solution 14.

2t1t5101t0t5

)t(f

We may write f as )t(

)2t(u)1t(u)t510()1t(u)t(ut5)t(f )2t(u)2t(5)1t(u)1t(10)t(ut5

s2-2

s-22 e

s5

es10

s5

)s(F

)s(F )ee21(s5 s2-s-2

Page 787: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 15. )2t(u)1t(u)1t(u)t(u10)t(f

se

es2

s1

10)s(Fs2-

s- 2s- )e1(s

10

Chapter 15, Solution 16.

)4t(u5)3t(u3)1t(u3)t(u5)t(f

)s(F s4-s3-s- e5e3e35s1

Chapter 15, Solution 17.

3t03t111t0t

0t0

)t(f2

)3t(u)1t(u1)1t(u)t(ut)t(f 2

)3t(u)1t(u)1t(u)1t2-()1t(u)1t()t(ut 22

)3t(u)1t(u)1t(2)1t(u)1t()t(ut 22

)s(Fs

ee

s2

)e1(s2 s-3

s-2

s-3

Chapter 15, Solution 18.

(a) )3t(u)2t(u3)2t(u)1t(u2)1t(u)t(u)t(g )3t(u3)2t(u)1t(u)t(u

)s(G )e3ee1(s1 s3-s2-s-

Page 788: Solution Manual for Fundamentals of Electric Circuits

(b) )3t(u)1t(u2)1t(u)t(ut2)t(h

)4t(u)3t(u)t28( )3t(u2)1t(u2)1t(u2)1t(u)1t(2)t(ut2

)4t(u)4t(2)3t(u2)3t(u)3t(2 )4t(u)4t(2)3t(u)3t(2)1t(u)1t(2)t(ut2

s4-2

s3-2

s-2 e

s2

es2

)e1(s2

)s(H )eee1(s2 s4-s3-s-2

Chapter 15, Solution 19.

Since 1)t(L and , 2T )s(F s2-e11

Chapter 15, Solution 20.

Let 1t0),tsin()t(g1 )1t(u)t(u)tsin( )1t(u)tsin()t(u)tsin(

Note that sin( )tsin(-)tsin())1t( . So, 1)-u(t1))-t(sin(u(t)t)sin()t(g1

)e1(s

)s(G s-221

2s-1

e1)s(G

)s(G)e1)(s(

)e1(2s-22

-s

Page 789: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 21.

2T

Let )1t(u)t(u2t

1)t(f1

)1t(u21

1)1t(u)1t(21

)t(u2t

)t(u)t(f1

2

-s-ss-

2

s-

21 s2e1-se)12-(2

s1

e21

1-s2

es2

1s1

)s(F

Ts-1

e1)s(F

)s(F)e1(s2

e1-se)12-(2s2-2

-s-s

Chapter 15, Solution 22.

(a) Let 1t0,t2)t(g1 )1t(u)t(ut2 )1t(u2)1t(u)1t(2)t(ut2

s-

2

-s

21 es2

se2

s2

)s(G

1T,e1

)s(G)s(G sT-

1

)s(G)e1(s

)ese1(2s-2

-s-s

(b) Let h , where is the periodic triangular wave. )t(uh 0 0h

Let h be within its first period, i.e. 1 0h

2t1t241t0t2

)t(h1

)2t(u)2t(2)1t(ut2)1t(u4)1t(ut2)t(ut2)t(h1

)2t(u)2t(2)1t(u)1t(4)t(ut2)t(h1

Page 790: Solution Manual for Fundamentals of Electric Circuits

2s-

22

-2ss-

221 )e1(s2

se2

es4

s2

)s(H

)e1()e1(

s2

)s(H 2s-

2-s

20

)s(H)e1(

)e1(s2

s1

2s-

2-s

2

Chapter 15, Solution 23.

(a) Let 2t11-1t01

)t(f1

)2t(u)1t(u)1t(u)t(u)t(f1

)2t(u)1t(u2)t(u)t(f1

2s-2s-s-1 )e1(

s1

)ee21(s1

)s(F

2T,)e1(

)s(F)s(F sT-

1

)s(F)e1(s

)e1(2s-

2-s

(b) Let )2t(ut)t(ut)2t(u)t(ut)t(h 222

1 )2t(u4)2t(u)2t(4)2t(u)2t()t(ut)t(h 22

1

2s-2s-2

2s-31 e

s4

es4

)e1(s2

)s(H

2T,)e1(

)s(H)s(H Ts-

1

)s(H)e1(s

)ss(es4)e1(22s-3

2-2s-2s

Page 791: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 24.

(a) 5s6s

ss10lim)s(sFlim)0(f 2

4

ss

0

10

s5

s6

s1

s110

lim432

3

s

5s6sss10

lim)s(sFlim)(f 2

4

0s0s0

(b) 6s4s

sslim)s(sFlim)0(f 2

2

ss1

The complex poles are not in the left-half plane.

)(f existnotdoes

(c) )5s2s)(2s)(1s(

s7s2lim)s(sFlim)0(f 2

3

ss

10

s5

s21

s21

s11

s7

s2

lim

2

3

s0

100

)5s2s)(2s)(1s(s7s2

lim)s(sFlim)(f 2

3

0s0s0

Page 792: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 25.

(a) )4s)(2s()3s)(1s)(8(

lim)s(sFlim)0(fss

s41

s21

s31

s11)8(

lims

8

)4)(2()3)(1)(8(

lim)s(sFlim)(f0s0s

3

(b) 1s

)1s(s6lim)s(sFlim)0(f 4ss

10

s1

1

s1

s1

6lim)0(f

4

42

s0

All poles are not in the left-half plane.

)(f existnotdoes Chapter 15, Solution 26.

(a) 6s4s

s3slim)s(sFlim)0(f 23

3

ss1

Two poles are not in the left-half plane.

)(f existnotdoes

Page 793: Solution Manual for Fundamentals of Electric Circuits

(b) )4s2s)(2s(

ss2slim)s(sFlim)0(f 2

23

ss

2

2

s

s4

s21

s21

s1

s21

lim 1

One pole is not in the left-half plane.

)(f existnotdoes Chapter 15, Solution 27.

(a) )t(f -te2)t(u

(b) 4s

113

4s11)4s(3

)s(G

)t(g -4te11)t(3

(c) 3s

B1s

A)3s)(1s(

4)s(H

2A , -2B

3s2

1s2

)s(H

)t(h -3t-t e2e2

Page 794: Solution Manual for Fundamentals of Electric Circuits

(d) 4s

C)2s(

B2s

A)4s()2s(

12)s(J 22

62

12B , 3

)-2(12

C 2

2)2s(C)4s(B)4s()2s(A12

Equating coefficients :

2s : -3-CACA01s : 6-2ABBA2C4BA600s : 121224-24C4B4A812

4s3

)2s(6

2s3-

)s(J 2

)t(j -2t-2t-4t et6e3e3

Chapter 15, Solution 28. (a)

)t(ut)e4e2()t(f

5s4

3s2

5s2

)4(2

3s2

)2(2

)s(F

t5t3

(b)

)t(ut2sine5.1t2cose2e2)t(h5s2s

1s21s

2)s(H

1C;2B;2A3CB,3CB;BA;11CA5

CA5s)CBA2(s)BA()1s)(CBs()5s2s(A11s3

5s2sCBs

1sA

)5s2s)(1s(11s3)s(H

ttt2

22

22

Page 795: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 29.

2222

2222

3)2s(3

32

3)2s()2s(2

s2)s(V

6Band2A26s2BsAs26s8s2;3)2s(

BAss2)s(V

v(t) = 0t,t3sine32t3cose2)t(u t2t22

Chapter 15, Solution 30.

(a) 22222213)2s(

332

3)2s()2s(2

3)2s(2)2s(2)s(H

t3sine32t3cose2)t(h t2t2

1

(b) )5s2s(

DCs)1s(

B)1s(

A)5s2s()1s(

4s)s( 2222

22H

22222 )1s(D)1s(Cs)5s2s(B)5s2s)(1s(A4s

or

)1s2s(D)ss2s(C)5s2s(B)5s7s3s(A4s 2232232 Equating coefficients:

ACCA0:s3

DBADC2BA31:s2

2/1C,2/1A2A4D2B2A6D2CB2A70:s

4/1D,4/5B1B4A4DB5A54:constant

Page 796: Solution Manual for Fundamentals of Electric Circuits

)2)1s(1)1s(2

)1s(5

)1s(2

41

)5s2s(1s2

)1s(5

)1s(2

41)s(H 222222

Hence,

)t(ut2sine5.0t2cose2te5e241)t(h tttt

2

(c ) )3s(

1)1s(

1e21

)3s(B

)1s(Ae

)3s)(1s(e)2s()s(H ss

s3

)1t(uee21)t(h )1t(3)1t(

3

Chapter 15, Solution 31.

(a) 3s

C2s

B1s

A)3s)(2s)(1s(

s10)s(F

-5210-

)1s()s(FA 1-s

201-20-

)2s()s(FB -2s

-15230-

)3s()s(FC 3-s

3s15

2s20

1s5-

)s(F

)t(f -3t-2t-t e15e20e5-

(b) 323

2

)2s(D

)2s(C

2sB

1sA

)2s)(1s(1s4s2

)s(F

-1)1s()s(FA 1-s

-1)2s()s(FD -2s3

)4s4s)(1s(B)4s4s)(2s(A1s4s2 222

Page 797: Solution Manual for Fundamentals of Electric Circuits

)1s(D)2s)(1s(C Equating coefficients :

3s : 1 -ABBA02s : 3 A2CCACB5A621s : DC3A4DC3B8A124

-1A-2DDA640s : 116-4D2C4ADC2B4A81

32 )2s(1

)2s(3

2s1

1s1-

)s(F

2t-

22t-2t-t- e

2t

et3ee-f(t)

)t(f 2t-2

t- e2t

t31e-

(c) 5s2s

CBs2s

A)5s2s)(2s(

1s)s(F 22

51-

)2s()s(FA -2s

)2s(C)s2s(B)5s2s(A1s 22

Equating coefficients :

2s : 51

-ABBA0

1s : 1CC0CB2A210s : 121-C2A51

222222 2)1s(54

2)1s()1s(51

2s51-

2)1s(1s51

2s51-

)s(F

)t(f )t2sin(e4.0)t2cos(e2.0e0.2- -t-t-2t

Page 798: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 32.

(a) 4s

C2s

BsA

)4s)(2s(s)3s)(1s(8

)s(F

3)4)(2(

(8)(3)s)s(FA 0s

2)-4(

(8)(-1))2s()s(FB -2s

3(-2))-4(

)(8)(-1)(-3)4s()s(FC -4s

4s3

2s2

s3

)s(F

)t(f -4t-2t e3e2)t(u3

(b) 22

2

)2s(C

2sB

1sA

)2s)(1s(4s2s

)s(F

)1s(C)2s3s(B)4s4s(A4s2s 222

Equating coefficients :

2s : A1BBA11s : CA3CB3A42- 0s : -6B2B-CB2A44

7B1A -12A--5C

2)2s(12

2s6

1s7

)s(F

)t(f -2t-t e)t21(6e7

(c) 5s4s

CBs3s

A)5s4s)(3s(

1s)s(F 22

2

)3s(C)s3s(B)5s4s(A1s 222

Page 799: Solution Manual for Fundamentals of Electric Circuits

Equating coefficients :

2s : A1BBA11s : -3CACA3CB3A400s : 5A2A-9C3A51

-4A1B -83A-C

1)2s()2s(4

3s5

1)2s(8s4

3s5

)s(F 22

)t(f )tcos(e4e5 -2t-3t

Chapter 15, Solution 33.

(a) 1s

C1sBAs

)1s)(1s(6

1s)1s(6

)s(F 224

)1s(C)1s(B)ss(A6 22

Equating coefficients :

2s : -CACA01s : C -ABBA00s : 3B2BCB6

-3A , 3B , 3C

1s3

1s3s-

1s3

1s33s-

1s3

)s(F 222

)t(f )tcos(3)tsin(3e3 -t

(b) 1s

es)s(F 2

s-

)t(f )t(u)tcos(

Page 800: Solution Manual for Fundamentals of Electric Circuits

(c) 323 )1s(D

)1s(C

1sB

sA

)1s(s8

)s(F

8A , -8D

sD)ss(C)ss2s(B)1s3s3s(A8 22323

Equating coefficients :

3s : -ABBA02s : B-ACCACB2A301s : -ADDADCBA300s : 8D,8C,8B,8A

32 )1s(8

)1s(8

1s8

s8

)s(F

)t(f )t(uet5.0ete18 -t2-t-t

Chapter 15, Solution 34.

(a) 4s

311

4s34s

10)s(F 22

2

)t(f )t2sin(5.1)t(11

(b) )4s)(2s(

e4e)s(G

-2s-s

Let 4s

B2s

A)4s)(2s(

1

21A 21B

4s1

2s1

e24s

12s

12

e)s(G 2s-

-s

)t(g )2t(uee2)1t(uee5.0 2)--4(t2)--2(t-1)-4(t-1)-2(t

Page 801: Solution Manual for Fundamentals of Electric Circuits

(c) Let 4s

C3s

BsA

)4s)(3s(s1s

121A , 32B , 43-C

2s-e

4s43

3s32

s1

121

)s(H

)t(h )2t(ue43

e32

121 2)-4(t-2)-3(t-

Chapter 15, Solution 35.

(a) Let 2s

B1s

A)2s)(1s(

3s)s(G

2A , -1B

2t-t- ee2)t(g

2s1

1s2

)s(G

)6t(u)6t(g)t(f)s(Ge)s(F -6s

)t(f )6t(uee2 6)--2(t6)--(t

(b) Let 4s

B1s

A)4s)(1s(

1)s(G

31A , 31-B

)4s(31

)1s(31

)s(G

4t-t- ee

31

)t(g

)s(Ge)s(G4)s(F -2t

)2t(u)2t(g)t(u)t(g4)t(f

)t(f )2t(uee31

)t(uee34 2)-4(t-2)-(t-4t-t-

Page 802: Solution Manual for Fundamentals of Electric Circuits

(c) Let 4sCBs

3sA

)4s)(3s(s

)s(G 22

133-A

)3s(C)s3s(B)4s(As 22

Equating coefficients :

2s : -ABBA01s : CB310s : C3A40

133-A , 133B , 134C

4s4s3

3s3-

)s(G13 2

)t2sin(2)t2cos(3e-3)t(g13 -3t

)s(Ge)s(F -s

)1t(u)1t(g)t(f

)t(f )1t(u))1t(2sin(2))1t(2cos(3e3-131 1)-3(t-

Chapter 15, Solution 36.

(a) 3s

D2s

CsB

sA

)3s)(2s(s1

)s(X 22

61B , 41C , 91-D

)s2s(D)s3s(C)6s5s(B)s6s5s(A1 2323223

Equating coefficients :

3s : DCA0 2s : CBA3D2C3BA50 1s : B5A600s : 61BB61

Page 803: Solution Manual for Fundamentals of Electric Circuits

365-B65-A

3s91

2s41

s61

s365-

)s(X 2

)t(x 3t-2t- e91

e41

t61

)t(u36

5-

(b) 22 )1s(C

1sB

sA

)1s(s1

)s(Y

1A , 1-C

sC)ss(B)1s2s(A1 22

Equating coefficients :

2s : -ABBA01s : -ACCACBA200s : -1C-1,B,A1

2)1s(1

1s1

s1

)s(Y

)t(y -t-t ete)t(u

(c) 10s6s

DCs1s

BsA

)s(Z 2

101A , 51-B

)ss(D)ss(C)s10s6s(B)10s16s7s(A1 2232323

Equating coefficients :

3s : CBA0 2s : DB5A6DCB6A70 1s : A 2-BB5A10DB10A1600s : 101AA101

101A , 51--2AB , 101AC , 104

A4D

Page 804: Solution Manual for Fundamentals of Electric Circuits

10s6s4s

1s2

s1

)s(Z10 2

1)3s(1

1)3s(3s

1s2

s1

)s(Z10 22

)t(z )t(u)tsin(e)tcos(ee211.0 -3t-3t-t

Chapter 15, Solution 37.

(a) Let 4sCBs

sA

)4s(s12

)s(P 22

3412s)s(PA 0s

sCsB)4s(A12 22

Equating coefficients :

0s : 3AA4121s : C02s : -3-ABBA0

4ss3

s3

)s(P 2

)t2cos(3)t(u3)t(p

)s(Pe)s(F -2s

)t(f )2t(u))2t(2cos(13

(b) Let 9sDCs

1sBAs

)9s)(1s(1s2

)s(G 2222

)1s(D)ss(C)9s(B)s9s(A1s2 2323

Equating coefficients :

3s : -ACCA0

Page 805: Solution Manual for Fundamentals of Electric Circuits

2s : -BDDB01s : 82-C,82AA8CA92 0s : 81-D,81BB8DB91

9s1s2

81

1s1s2

81

)s(G 22

9s1

81

9ss

41

1s1

81

1ss

41

)s(G 2222

)t(g )t3sin(241

)t3cos(41

)tsin(81

)tcos(41

(c) Let 13s4s

117s369

13s4ss9

)s(H 22

2

2222 3)2s(3

153)2s(

2s369)s(H

)t(h )t3sin(e15)t3cos(e36)t(9 t2-t2-

Chapter 15, Solution 38.

(a) 26s10s

26s626s10s26s10s

s4s)s(F 2

2

2

2

26s10s26s6

1)s(F 2

2222 1)5s(4

1)5s()5s(6

1)s(F

)t(f )t5sin(e4)t5cos(e6)t( -t-t

Page 806: Solution Manual for Fundamentals of Electric Circuits

(b) 29s4s

CBssA

)29s4s(s29s7s5

)s(F 22

2

sCsB)29s4s(A29s7s5 222

Equating coefficients :

0s : 1AA29291s : 3A47CCA472s : 4 A5BBA5

1A , 4B , 3C

22222 5)2s(5

5)2s()2s(4

s1

29s4s3s4

s1

)s(F

)t(f )t5sin(e)t5cos(e4)t(u -2t-2t

Chapter 15, Solution 39.

(a) 20s4s

DCs17s2s

BAs)20s4s)(17s2s(

1s4s2)s(F 2222

23

)20s4s(B)s20s4s(A1s4s 22323

)17s2s(D)s17s2s(C 223

Equating coefficients : 3s : CA22s : DC2BA44 1s : D2C17B4A200 0s : D17B201

Solving these equations (Matlab works well with 4 unknowns),

-1.6A , -17.8B , 6.3C , 21D

20s4s21s6.3

17s2s17.81.6s-

)s(F 22

22222222 4)2s()4((3.45)

4)2s()2(3.6)(s

4)1s()4((-4.05)

4)1s(1)(-1.6)(s

)s(F

)t(f )t4sin(e45.3)t4cos(e6.3)t4sin(e05.4)t4cos(e6.1- -2t-2t-t-t

Page 807: Solution Manual for Fundamentals of Electric Circuits

(b) 3s6s

DCs9sBAs

)3s6s)(9s(4s

)s(F 2222

2

)9s(D)s9s(C)3s6s(B)s3s6s(A4s 232232

Equating coefficients :

3s : -ACCA02s : DBA61 1s : A -CBC6B6C9B6A300s : D9B34

Solving these equations,

121A , 121B , 121-C , 125D

3s6s5s-

9s1s

)s(F12 22

5.449--0.551,2

12-366-03s6s2

Let 449.5s

F551.0s

E3s6s

5s-)s(G 2

133.1449.5s5s-

E -0.551s

2.133- 551.0s5s-

F -5.449s

449.5s133.2

551.0s133.1

)s(G

449.5s133.2

551.0s133.1

3s3

31

3ss

)s(F12 2222

)t(f -5.449t-0.551t e1778.0e0944.0)t3sin(02778.0)t3cos(08333.0

Page 808: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 40.

Let 5s2s

CBs2s

A

)5s2s)(2s(

13s7s4)s(22

2H

)2s(C)s2s(B)5s2s(A13s7s4 222

Equating coefficients gives:

BA4:s2

1CCB2A27:s

1B 3,Aor 15A5C2A513:constant

222 2)1s(2)1s(

2s3

5s2s1s

2s3)s(H

Hence,

)t2sinsinAt2coscosA(ee3t2sinet2cosee3)t(h tt2ttt2

where o45,2A1sinA,1cosA Thus,

)t(ue3)45t2cos(e2)t(h t2ot Chapter 15, Solution 41. Let y(t) = f(t)*h(t). For 0 < t < 1, f )t( )(h 0 t 1 2

2t0

t

0

2 t22d4)1()t(y

Page 809: Solution Manual for Fundamentals of Electric Circuits

For 1 <t<3, )t(f )(h 0 1 t 2

4t2t8)28(2d)48)(1(d4)1()t(y 2t1

2t0

2t

1

1

0

For 3 < t < 4 h )( )t(f 0 1 t-2 2 3 t 4

222t

2

2t

2 t2t163228d)48()t(y

Thus,

otherwise 0,4 t 3 ,2t16t-32

3 t 1 ,42t-8t1 t 0 ,t2

)t(y 2

2

2

Page 810: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 42.

(a) For , 1t0 )t(f1 and f )(2 overlap from 0 to t, as shown in Fig. (a).

2t

2d))(1()t(f)t(f)t(y

2t0

2t

021

1

t-1 t 1 0

1

t-1 t 1

f2( )f1(t - )

0

(a) (b) For 1 , 2t )t(f1 and f )(2 overlap as shown in Fig. (b).

2t

t2

d))(1()t(y2

11t

21

1t

For , there is no overlap. 2t Therefore,

)t(yotherwise,0

2t1,2tt1t0,2t

2

2

(b) For , the two functions overlap as shown in Fig. (c). 1t0

td)1)(1()t(f)t(f)t(yt

021

1

t-1 t 1 0

1

t-1 t 1

f2( )f1(t - )

0

(c) (d)

Page 811: Solution Manual for Fundamentals of Electric Circuits

For 1 , the functions overlap as shown in Fig. (d). 2t

t2d)1)(1()t(y 11t

1

1t

For , there is no overlap. 2t Therefore,

)t(yotherwise,0

2t1,t21t0,t

(c) For , there is no overlap. For -1-t 0t1 , f )t(1 and overlap

as shown in Fig. (e). )(f2

t1-

2t

1-21 2d)1)(1()t(f)t(f)t(y

22 )1t(

21

)1t2t(21

)t(y

-1

1

t 1 0-1

1

t 1

f1( )f2(t - )

0

(e) (f) For , the functions overlap as shown in Fig. (f). 1t0

t

0

0

1-d)1)(1(d)1)(1()t(y

t0

201-

2

22)t(y

)tt21(21

)t(y 2

For , the two functions overlap. 1t

Page 812: Solution Manual for Fundamentals of Electric Circuits

1

0

0

1-d)1)(1(d)1)(1()t(y

121

121

221

)t(y 10

2

Therefore,

)t(y

1t1,1t0),1t2t-(5.00t1-),1t2t(5.0

-1t,0

2

2

Chapter 15, Solution 43.

(a) For , 1t0 )t(x and )(h overlap as shown in Fig. (a).

2t

2d))(1()t(h)t(x)t(y

2t0

2t

0

1

t-1 t 1

h( )

x(t - )

0

1

t-1 t 1 0

(a) (b) For 1 , 2t )t(x and )(h overlap as shown in Fig. (b).

1t2t21-

2d)1)(1(d))(1()t(y 2t

11

1t

2t

1

1

1t

For , there is a complete overlap so that 2t

1)1t(td)1)(1()t(y t1t

t

1t

Page 813: Solution Manual for Fundamentals of Electric Circuits

Therefore,

)t(y

otherwise,02t,1

2t1,1t2)2t(-1t0,2t

2

2

(b) For , the two functions overlap as shown in Fig. (c). 0t

t0

-t

0- e-2de2)1()t(h)t(x)t(y

2

1

h( ) = 2e-

0 t

x(t- )

(c) Therefore,

)t(y 0t),e1(2 -t

(c) For , 0t1- )t(x and )(h overlap as shown in Fig. (d).

21t0

21t

0)1t(

21

2d))(1()t(h)t(x)t(y

2-1 t+1

1

t-1 t 1

h( )x(t - )

0

(d) For , 1t0 )t(x and )(h overlap as shown in Fig. (e).

1t

1

1

0d)2)(1(d))(1()t(y

Page 814: Solution Manual for Fundamentals of Electric Circuits

21

tt21-

22

2)t(y 21t

1

210

2

t 2t+1

1

t-1-1 10

(e) For 1 , 2t )t(x and )(h overlap as shown in Fig. (f).

2

1

1

1td)2)(1(d))(1()t(y

21

tt21-

22

2)t(y 22

1

21

1t

2

1

t 2t-1 10 t+1

(f) For , 3t2 )t(x and )(h overlap as shown in Fig. (g).

221t

22

1tt

21

t329

22d)2)(1()t(y

1

t2 t-110 t+1

(g)

Page 815: Solution Manual for Fundamentals of Electric Circuits

Therefore,

)t(y

otherwise,03t2,29t3)2t(2t0,21t)2t(-0t1-,21t)2t(

2

2

2

Chapter 15, Solution 44.

(a) For , 1t0 )t(x and )(h overlap as shown in Fig. (a).

td)1)(1()t(h)t(x)t(yt

0

0

h( )1x(t - )

-1

t 2t-1 1

(a) For 1 , 2t )t(x and )(h overlap as shown in Fig. (b).

t23d)1)(1-(d)1)(1()t(y t1

11t

t

1

1

1t

For , 3t2 )t(x and )(h overlap as shown in Fig. (c).

3t-d)-1)(1()t(y 21t

2

1t

Page 816: Solution Manual for Fundamentals of Electric Circuits

0 1 t-1 2 t

-1

1

0 21 t-1 t

-1

1

(b) (c) Therefore,

)t(y

otherwise,03t2,3t2t1,t231t0,t

(b) For , there is no overlap. For 2t 3t2 , f )t(1 and overlap,

as shown in Fig. (d). )(f2

t

221 d)t)(1()t(f)t(f)t(y

2t22t

2t

2t2

2

t-1 t

1

1 5432

f2( ) f1(t - )

0

(d)

0 1

1

2 3 4 5tt-1 (e)

Page 817: Solution Manual for Fundamentals of Electric Circuits

For , 5t3 )t(f1 and )(f2 overlap as shown in Fig. (e).

21

2td)t)(1()t(y t

1t

2t

1t

For , the functions overlap as shown in Fig. (f). 6t5

12t5t21-

2td)t)(1()t(y 25

1t

25

1t

0 1

1

2 3 4 5 t t-1

(f) Therefore,

)t(y

otherwise,06t5,12t5)2t(-5t3,213t2,2t2)2t(

2

2

Chapter 15, Solution 45.

(a) t

t

0)(fd)t()(f)t()t(f

)t(f)t()t(f

(b) t

0d)t(u)(f)t(u)t(f

Since t0t1

)t(u

t

od)(f)t(u)t(f

Page 818: Solution Manual for Fundamentals of Electric Circuits

Alternatively,

s)s(F

)t(u)t(fL

t

o1 d)(f)t(u)t(f

s)s(F

L

Chapter 15, Solution 46.

(a) Let t

0 1221 d)t(x)t(x)t(x)t(x)t(y For , 3t0 )t(x1 and )(x 2 overlap as shown in Fig. (a).

)ee(4dee4dee4)t(y t2-t-t

0-t-t

0)-(t-2-

x1(t- )

4

3

x2( )

t0

(a) For , the two functions overlap as shown in Fig. (b). 3t

)e1(e4e-e4dee4)t(y 3-t-30

-t-3

0)-(t-2-

4

3 t0

(b)

Page 819: Solution Manual for Fundamentals of Electric Circuits

Therefore,

)t(y3t),e1(e4

3t0),ee(43-t-

t-2t-

(b) For 1 , 2t )(x1 and )t(x 2 overlap as shown in Fig. (c).

1td)1)(1()t(x)t(x)t(y t1

t

121

1 3

1

t-1 t 2

x1( )x2(t - )

0

(c) For , the two functions overlap completely. 3t2

1)1t(td)1)(1()t(y t1t

t

1t

For , the two functions overlap as shown in Fig. (d). 4t3

t4d)1)(1()t(y 31t

3

1t

1 3

1

t-1 t20

(d)Therefore,

)t(y

otherwise,04t3,t43t2,12t1,1t

(c) For , 0t1- )t(x1 and )(x 2 overlap as shown in Fig. (e).

t

1-)-(t-

21 de4)1()t(x)t(x)t(y

Page 820: Solution Manual for Fundamentals of Electric Circuits

1)(t-t

1-t- e14dee4)t(y

1-1 t-1

1

x1(t - )

x2( )

(e) For , 1t0

t

0)-(t-0

1-)-(t- de4)1-(de4)1()t(y

4e4e8ee4ee4)t(y 1)-(t-tt0

-t01-

-t For , the two functions overlap completely. 1t

1

0)-(t-0

1-)-(t- de4)1-(de4)1()t(y

1)-(t1)-(t-t10

-t01-

-t e4e4e8ee4ee4)t(y Therefore,

)t(y1t,e4e4e8

1t0,4e4e80t1-,e14

1)(t-1)(t-t-

1)(t-t-

1)-(t

Chapter 15, Solution 47.

)tcos()t(f)t(f 21

t

0211- d)tcos()cos()s(F)s(FL

)BAcos()BAcos(21

)Bcos()Acos(

Page 821: Solution Manual for Fundamentals of Electric Circuits

t

0211- d)]2tcos()t[cos(

21)s(F)s(FL

t0

t021

1-

2-)2tsin(

21

)tcos(21

)s(F)s(FL

)s(F)s(F 21

1-L )tsin(5.0)tcos(t5.0 Chapter 15, Solution 48.

(a) Let 222 2)1s(2

5s2s2

)s(G

)t2sin(e)t(g -t

)s(G)s(G)s(F

d)t(g)(g)s(G)s(G)t(ft

01-L

t

0)t(-- d))t(2sin(e)2sin(e)t(f

)BAcos()BAcos(21

)Bsin()Asin(

t

0-t- d))2t(2cos()t2cos(ee

21

)t(f

t

02-

-tt

02-

-t

d)4t2cos(e2e

de)t2cos(2e

)t(f

t

02-

-tt0

-2-t

d)4sin()t2sin()4cos()t2cos(e2e

2-e

)t2cos(2

e)t(f

t

02-

-tt2-t- d)4cos(e)t2cos(

2e

)1e-()t2cos(e41

)t(f

t

02-

-t

d)4sin(e)t2sin(2e

Page 822: Solution Manual for Fundamentals of Electric Circuits

)e1()t2cos(e41

)t(f t2-t-

t0

-2-t

)4sin(4)2cos(4-164

e)t2cos(

2e

t0

-2-t

)4cos(4)2sin(4-164

e)t2sin(

2e

)t(f )t4cos()t2cos(20e

)t2cos(20e

)t2cos(4

e)t2cos(

2e -3t-t-3t-t

)t2sin(10e

)t4sin()t2cos(10e -t-3t

)t4cos()t2sin(10e

)t4sin()t2sin(20e -t-t

(b) Let 1s

2)s(X ,

4ss

)s(Y

)t(ue2)t(x -t , )t(u)t2cos()t(y

)s(Y)s(X)s(F

01- d)t(x)(y)s(Y)s(X)t(f L

t

0)(t- de2)2cos()t(f

t0

t- )2sin(2)2cos(41

ee2)t(f

1)t2sin(2)t2cos(ee52

)t(f tt-

)t(f t-e52

)t2sin(54

)t2cos(52

Page 823: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 49.

Let x(t) = u(t) �– u(t-1) and y(t) = h(t)*x(t).

)2s(s)e1(4)

se

s1()s(X)s(H)t(y

s1

s11 L

2s4LL

But

2s1

s1

21

2sB

sA

)2s(s1

2se

se

2s1

s12)s(Y

ss

)1t(u]e1[4)t(u]e1[2)t(y )1t(2t2

Chapter 15, Solution 50.

Take the Laplace transform of each term.

4ss3

)s(V10)0(v)s(Vs2)0(v)0(vs)s(Vs 22

4ss3

)s(V102)s(Vs22s)s(Vs 22

4ss7s

4ss3

s)s(V)10s2s( 2

3

22

10s2sDCs

4sBAs

)10s2s)(4s(s7s

)s(V 2222

3

)4s(D)s4s(C)10s2s(B)s10s2s(As7s 232233

Equating coefficients :

3s : A1CCA12s : DBA20 1s : 4B2A6C4B2A107

Page 824: Solution Manual for Fundamentals of Electric Circuits

0s : B-2.5DD4B100

Solving these equations yields

269

A , 2612

B , 2617

C , 2630-

D

10s2s30s17

4s12s9

261

)s(V 22

222222 3)1s(47

3)1s(1s

174s

26

4ss9

261

)s(V

)t(v )t3sin(e7847

)t3cos(e2617

)t2sin(266

)t2cos(269 t-t-

Chapter 15, Solution 51.

Taking the Laplace transform of the differential equation yields

1s10)s(V6)]0(v)s(sV5)0('v)0(sv)s(Vs2

or )3s)(2s)(1s(

24s16s2)s(V1s

10104s2)s(V6s5s2

2

Let 3C,0B,5A,3s

C2s

B1s

A)s(V

Hence,

)t(u)e3e5()t(v t3t

Page 825: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 52.

Take the Laplace transform of each term.

01)s(I2)0(i)s(Is3)0(i)0(is)s(Is2

0133s)s(I)2s3s( 2

2sB

1sA

)2s)(1s(5s

)s(I

4A , 3-B

2s3

1s4

)s(I

)t(i )t(u)e3e4( -2t-t

Chapter 15, Solution 53.

Take the Laplace transform of each term.

4ss

)s(V6)0(y)s(Ys5)0(y)0(ys)s(Ys 22

4ss

54s)s(Y)6s5s( 22

4s)4s)(9s(s

4ss

9s)s(Y)6s5s( 2

2

22

4sDCs

3sB

2sA

)4s)(3s)(2s(36s5s9s

)s(Y 22

23

427

)s(Y)2s(A -2s , 1375-

)s(Y)3s(B -3s

Page 826: Solution Manual for Fundamentals of Electric Circuits

When , 0s

265

D4D

3B

2A

)4)(3)(2(36

When , 1s

521C

5D

5C

4B

3A

)5)(12(546

Thus,4s

265s5213s

13752s427

)s(Y 2

)t(y )t2sin(525

)t2cos(521

e1375

e4

27 3t-2t-

Chapter 15, Solution 54.

Taking the Laplace transform of the differential equation gives

3s5

)s(V2)0(v)s(Vs3)0(v)0(vs)s(Vs2

3ss2

13s

5)s(V)2s3s( 2

)3s)(2s)(1s(s2

)2s3s)(3s(s2

)s(V 2

3sC

2sB

1sA

)s(V

23A , -4B , 25C

3s25

2s4

1s23

)s(V

)t(v )t(u)e5.2e4e5.1( -3t-2t-t

Page 827: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 55.

Take the Laplace transform of each term.

)0(y)0(ys)s(Ys6)0(y)0(ys)0(ys)s(Ys 223

22 2)1s(1s)0(y)s(Ys8

Setting the initial conditions to zero gives

5s2s1s

)s(Y)s8s6s( 223

5s2sEDs

4sC

2sB

sA

)5s2s)(4s)(2s(s)1s(

)s(Y 22

401

A , 201

B , 104

3-C ,

653-

D , 657-

E

22 2)1s(7s3

651

4s1

1043

2s1

201

s1

401

)s(Y

2222 2)1s(4

651

2)1s()1s(3

651

4s1

1043

2s1

201

s1

401

)s(Y

)t(y )t2sin(e652

)t2cos(e653

e104

3e

201

)t(u401 t-t-4t-2t-

Chapter 15, Solution 56.

Taking the Laplace transform of each term we get:

0)s(Vs

12)0(v)s(Vs4

8)s(Vs

12s4

3ss2

12s4s8

)s(V 22

)t(v t3cos2

Page 828: Solution Manual for Fundamentals of Electric Circuits

Chapter 15, Solution 57.

Take the Laplace transform of each term.

4ss

)s(Ys9

)0(y)s(Ys 2

4s4ss

4ss

1)s(Ys

9s2

2

2

2

9sDCs

4sBAs

)9s)(4s(s4ss

)s(Y 2222

23

)4s(D)s4s(C)9s(B)s9s(As4ss 232323

Equating coefficients :

0s : D4B901s : C4A942s : DB13s : CA1

Solving these equations gives

0A , 54-B , 1C , 59D

9s59

9ss

4s54-

9s59s

4s54-

)s(Y 22222

)t(y )t3sin(6.0)t3cos()t2sin(0.4-

Chapter 15, Solution 58.

We take the Laplace transform of each term and obtain

10s6s

se)s(Ve)s(Vs

10)]0(v)s(sV[)s(V62

s2s2

1)3s(

e3e)3s()s(V2

s2s2

Page 829: Solution Manual for Fundamentals of Electric Circuits

Hence, 3( 2) 3( 2)( ) cos( 2) 3 sin( 2) ( 2)t tv t e t e t u t

Chapter 15, Solution 59.

Take the Laplace transform of each term of the integrodifferential equation.

2s6

)s(Ys3

)s(Y4)0(y)s(Ys

12s

6s)s(Y)3s4s( 2

)3s)(2s)(1s(s)s4(

)3s4s)(2s()s4(s

)s(Y 2

3sC

2sB

1sA

)s(Y

5.2A , 6B , -10.5C

3s5.10

2s6

1s5.2

)s(Y

)t(y -3t-2t-t e5.10e6e5.2

Chapter 15, Solution 60.

Take the Laplace transform of each term of the integrodifferential equation.

16s4

s4

)s(Xs3

)s(X5)0(x)s(Xs2 2

16s64s36s4s2

16ss4

4s2)s(X)3s5s2( 2

23

22

)16s)(5.1s)(1s(32s18s2s

)16s)(3s5s2(64s36s4s2

)s(X 2

23

22

23

Page 830: Solution Manual for Fundamentals of Electric Circuits

16sDCs

5.1sB

1sA

)s(X 2

-6.235)s(X)1s(A -1s

329.7)s(X)5.1s(B -1.5s

When , 0s

2579.0D16D

5.1B

A(1.5)(16)

32-

)16s16ss(B)24s16s5.1s(A32s18s2s 232323

)5.1s5.2s(D)s5.1s5.2s(C 223

Equating coefficients of the s terms, 3

0935.0-CCBA1

16s0.25790.0935s-

5.1s329.7

1s6.235-

)s(X 2

)t(x )t4sin(0645.0)t4cos(0935.0e329.7e6.235- -1.5t-t

Page 831: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 1.

Consider the s-domain form of the circuit which is shown below.

I(s)

+

1

1/s 1/s

s

222 )23()21s(1

1ss1

s1s1s1

)s(I

t23sine

32)t(i 2t-

)t(i A)t866.0(sine155.1 -0.5t

Chapter 16, Solution 2.

8/s s

s4 2

+

+

Vx

4

Page 832: Solution Manual for Fundamentals of Electric Circuits

V)t(u)e2e24(v

38j

34s

125.0

38j

34s

125.0s25.016

)8s8s3(s

2s16V

s32s16)8s8s3(V

0VsV)s4s2(s

)32s16()8s4(V

0

s84

0V2

0Vs

s4V

t)9428.0j3333.1(t)9428.0j3333.1(x

2x

2x

x2

x2

x

xxx

vx = Vt3

22sine2

6t3

22cose)t(u4 3/t43/t4

Chapter 16, Solution 3. s

5/s 1/2 +

Vo

1/8

Current division leads to:

)625.0s(165

s16105

s81

21

21

s5

81Vo

vo(t) = V)t(ue13125. t625.00

Page 833: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 4.

The s-domain form of the circuit is shown below.

6 s

10/s1/(s + 1) +

+

Vo(s)

Using voltage division,

1s1

10s6s10

1s1

s106ss10

)s(V 2o

10s6sCBs

1sA

)10s6s)(1s(10

)s(V 22o

)1s(C)ss(B)10s6s(A10 22

Equating coefficients :

2s : -ABBA01s : A5-CCA5CBA600s : -10C-2,B,2AA5CA1010

22222o 1)3s(4

1)3s()3s(2

1s2

10s6s10s2

1s2

)s(V

)t(vo V)tsin(e4)tcos(e2e2 -3t-3t-t

Chapter 16, Solution 5.

s2

2s1 2

Io

s

Page 834: Solution Manual for Fundamentals of Electric Circuits

A)t(ut3229.1sin7559.0e

or

A)t(ueee3779.0eee3779.0e)t(i

3229.1j5.0s)646.2j)(3229.1j5.1(

)3229.1j5.0(

3229.1j5.0s)646.2j)(3229.1j5.1(

)3229.1j5.0(

2s1

)3229.1j5.0s)(3229.1j5.0s)(2s(s

2VsI

)3229.1j5.0s)(3229.1j5.0s)(2s(s2

2ss

s22s

1

2s

21

s1

12s

1V

t2

t3229.1j2/t90t3229.1j2/t90t2o

22

2o

2

Chapter 16, Solution 6.

2

2s5

Io

10/s

s

Use current division.

t3sine35t3cose5)t(i

3)1s(

5

3)1s(

)1s(5

10s2s

s52s

5

s102s

2sI

tto

22222o

Page 835: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 7. The s-domain version of the circuit is shown below. 1/s 1 Ix + 2s

1

2s

�– Z

2

2

2 s21

1s2s2

s21

s21s2

s1

)s2(s1

1s2//s11Z

)5.0ss(

CBs)1s(

A

)5.0ss)(1s(

1s2

1s2s2

s21x1s

2ZVI

22

2

2

2x

)1s(C)ss(B)5.0ss(A1s2 222

BA2:s2

2CC2CBA0:s

-4B ,6A3 0.5A or CA5.01:constant

222x866.0)5.0s(

)5.0s(41s

6

75.0)5.0s(

2s41s

6I

A)t(ut866.0cose46)t(i t5.0x

Page 836: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 8.

(a) )1s(s

1s5.1ss22)s21(

s1)s21//(1

s1Z

2

(b) )1s(s2

2s3s3

s11

1s1

21

Z1 2

2s3s3

)1s(s2Z2

Chapter 16, Solution 9.

(a) The s-domain form of the circuit is shown in Fig. (a).

s1s2)s1s(2

)s1s(||2Zin 1s2s)1s(2

2

2

1

1

2 s 2/s 1/s

s

2

(a) (b)(b) The s-domain equivalent circuit is shown in Fig. (b).

2s3)2s(2

s23)s21(2

)s21(||2

2s36s5

)s21(||21

2s36s5

s

2s36s5

s

2s36s5

||sZin 6s7s3)6s5(s

2

Page 837: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 10. To find ZTh, consider the circuit below. 1/s Vx + 1V 2 Vo 2Vo - Applying KCL gives

s/12VV21 x

o

But xo Vs/12

2V . Hence

s3)1s2(V

s/12V

s/12V41 x

xx

s3)1s2(

1VZ x

Th

To find VTh, consider the circuit below. 1/s Vy +

1

2s

2 Vo 2Vo

- Applying KCL gives

)1s(34V

2VV2

1s2

oo

o

Page 838: Solution Manual for Fundamentals of Electric Circuits

But 0Vs1V2V ooy

)1s(s3)2s(4

s2s

)1s(34)

s21(VVV oyTh

Chapter 16, Solution 11. The s-domain form of the circuit is shown below.

4/s s

I2I1+

+

2 4/(s + 2) 1/s

Write the mesh equations.

21 I2Is4

2s1

(1)

21 I)2s(I-22s

4- (2)

Put equations (1) and (2) into matrix form.

2

1

II

2s2-2-s42

2)(s4-s1

)4s2s(s2 2 ,

)2s(s4s4s2

1 , s6-

2

4s2sCBs

2sA

)4s2s)(2s()4s4s(21

I 22

21

1

)2s(C)s2s(B)4s2s(A)4s4s(21 222

Equating coefficients :

2s : BA21 1s : CB2A22-

Page 839: Solution Manual for Fundamentals of Electric Circuits

0s : C2A42 Solving these equations leads to A 2 , 23-B , -3C

221 )3()1s(3s23-

2s2

I

22221 )3()1s(3

323-

)3()1s()1s(

23-

2s2

I

)t(i1 A)t(u)t732.1sin(866.0)t732.1cos(e5.1e2 -t-2t

2222

2 )3()1s(3-

)4s2s(2s

s6-I

)t3sin(e33-

)t(i t-2 A)t(u)t732.1sin(e1.732- -t

Chapter 16, Solution 12. We apply nodal analysis to the s-domain form of the circuit below.

Vo

10/(s + 1) +

1/(2s) 4

s

3/s

oo

o

sV24

Vs3

s

V1s

10

1s15s1510

151s

10V)ss25.01( o

2

1s25.0sCBs

1sA

)1s25.0s)(1s(25s15

V 22o

Page 840: Solution Manual for Fundamentals of Electric Circuits

740

V)1s(A 1-so

)1s(C)ss(B)1s25.0s(A25s15 22

Equating coefficients :

2s : -ABBA01s : C-0.75ACBA25.015 0s : CA25

740A , 740-B , 7135C

43

21

s

23

32

7155

43

21

s

21

s

740

1s1

740

43

21

s

7135

s740-

1s740

V 222o

t23

sine)3)(7()2)(155(

t23

cose740

e740

)t(v 2t-2t-t-o

)t(vo V)t866.0sin(e57.25)t866.0cos(e714.5e714.5 2-t2-t-t

Chapter 16, Solution 13.

Consider the following circuit.

Vo

1/(s + 2)

1/s 2s

Io

2 1

Applying KCL at node o,

ooo V

1s21s

s12V

1s2V

2s1

Page 841: Solution Manual for Fundamentals of Electric Circuits

)2s)(1s(1s2

Vo

2sB

1sA

)2s)(1s(1

1s2V

I oo

1A , -1B

2s1

1s1

Io

)t(io A)t(uee -2t-t

Chapter 16, Solution 14. We first find the initial conditions from the circuit in Fig. (a).

1 4

io

+

vc(0)+

5 V

(a)

A5)0(io , V0)0(vc

We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).

2s 5/s

Io

Vo

+

1 4

15/s 4/s

(b)At node o,

0s440V

s5

s2V

1s15V ooo

Page 842: Solution Manual for Fundamentals of Electric Circuits

oV)1s(4

ss2

11

s5

s15

o

2

o

22

V)1s(s42s6s5

V)1s(s4

s2s2s4s4s

10

2s6s5)1s(40

V 2o

s5

)4.0s2.1s(s)1s(4

s5

s2V

I 2o

o

4.0s2.1sCBs

sA

s5

I 2o

sCsB)4.0s2.1s(A)1s(4 s2

Equating coefficients :

0s : 10AA4.041s : -84-1.2ACCA2.142s : -10-ABBA0

4.0s2.1s8s10

s10

s5

I 2o

2222o 2.0)6.0s()2.0(10

2.0)6.0s()6.0s(10

s15

I

)t(io A)t(u)t2.0sin()t2.0cos(e1015 0.6t-

Chapter 16, Solution 15.

First we need to transform the circuit into the s-domain.

2s5

Vo

+

+

5/s

s/4

+ Vx

10 3Vx

Page 843: Solution Manual for Fundamentals of Electric Circuits

2s5VV

2s5VV,But

2ss5V120V)40ss2(0

2ss5sVVs2V120V40

010

2s5V

s/50V

4/sV3V

xoox

xo2

oo2

xo

ooxo

We can now solve for Vx.

)40s5.0s)(2s(

)20s(5V

2s)20s(10V)40s5.0s(2

02s

s5V1202s

5V)40ss2(

2

2x

2x

2

xx2

Chapter 16, Solution 16. We first need to find the initial conditions. For 0t , the circuit is shown in Fig. (a). To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit.

1

+ Vo

1 F

Vo/2 + 1 H io

+

2

3 V

(a)

Page 844: Solution Manual for Fundamentals of Electric Circuits

Hence,

A1-33-

i)0(i oL , V1-vo

V5.221-

)1-)(2(-)0(vc

We now incorporate the initial conditions for as shown in Fig. (b). 0t

I2I1 +

s

2.5/s +

1

+ Vo

1/s

Vo/2 +

Io

+

2

-1 V

5/(s + 2)

(b)For mesh 1,

02

Vs5.2

Is1

Is1

22s

5- o21

But, 2oo IIV

s5.2

2s5

Is1

21

Is1

2 21 (1)

For mesh 2,

0s5.2

2V

1Is1

Is1

s1 o12

1s5.2

Is1

s21

Is1

- 21 (2)

Page 845: Solution Manual for Fundamentals of Electric Circuits

Put (1) and (2) in matrix form.

1s5.2

s5.2

2s5

I

I

s1

s21

s1-

s1

21

s1

2

2

1

s3

2s2 , )2s(s

5s4

2-2

3s2s2CBs

2sA

)3s2s2)(2s(132s-

II 22

22

2o

)2s(C)s2s(B)3s2s2(A132s- 222

Equating coefficients :

2s : B A22-1s : CB2A200s : C2A313

Solving these equations leads to

7143.0A , , -3.429B 429.5C

5.1ss714.2s7145.1

2s7143.0

3s2s2429.5s429.3

2s7143.0

I 22o

25.1)5.0s()25.1)(194.3(

25.1)5.0s()5.0s(7145.1

2s7143.0

I 22o

)t(io A)t(u)t25.1sin(e194.3)t25.1cos(e7145.1e7143.0 -0.5t-0.5t-2t

Page 846: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 17. We apply mesh analysis to the s-domain form of the circuit as shown below.

2/(s+1)

I3

+

1/s

I2I1

4

s

1 1

For mesh 3,

0IsIs1

Is1

s1s

2213 (1)

For the supermesh,

0Iss1

I)s1(Is1

1 321 (2)

But (3) 4II 21

Substituting (3) into (1) and (2) leads to

s1

14Is1

sIs1

s2 32 (4)

1s2

s4-

Is1

sIs1

s- 32 (5)

Adding (4) and (5) gives

1s2

4I2 2

1s1

2I2

Page 847: Solution Manual for Fundamentals of Electric Circuits

)t(i)t(i 2o A)t(u)e2( -t Chapter 16, Solution 18.

vs(t) = 3u(t) �– 3u(t�–1) or Vs = )e1(s3

se

s3 s

s

1

+

Vo

+

1/s

Vs 2

V)]1t(u)e22()t(u)e22[()t(v

)e1(5.1s

2s2

)e1()5.1s(s

3V

VV)5.1s(02

VsV

1VV

)1t(5.1t5.1o

sso

soo

oso

Chapter 16, Solution 19. We incorporate the initial conditions in the s-domain circuit as shown below.

I

1/s

+

2 I V1 Vo

1/s

s

+

2

2 4/(s + 2)

Page 848: Solution Manual for Fundamentals of Electric Circuits

At the supernode,

o11 sV

s1

sV

22

V)2s(4

o1 Vss1

Vs1

21

22s

2 (1)

But and I2VV 1o s1V

I 1

2s2Vs

s)2s(s2V

Vs

)1V(2VV oo

11

1o (2)

Substituting (2) into (1)

oo Vs2s

2V

2ss

s1s2

s1

22s

2

oVs2s1s2

)2s(s)1s2(2

s1

22s

2

o

22

V2s

1s4s2s9s2

)2s(ss9s2

732.3sB

2679.0sA

1s4s9s2

V 2o

443.2A , 4434.0-B

732.3s4434.0

2679.0s443.2

Vo

Therefore,

)t(vo V)t(u)e4434.0e443.2( -3.732t-0.2679t

Page 849: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 20. We incorporate the initial conditions and transform the current source to a voltage source as shown.

1 s

+ 1/s 2/s

Vo

+

1

1/(s + 1) 1/s

At the main non-reference node, KCL gives

s1

sV

1V

s11Vs2)1s(1 ooo

s1s

V)s1s)(1s(Vs21s

soo

oV)s12s2(2s

1s1s

s

)1s2s2)(1s(1s4s2-

V 2

2

o

5.0ssCBs

1sA

)5.0ss)(1s(5.0s2s-

V 22o

1V)1s(A 1-so

)1s(C)ss(B)5.0ss(A5.0s2s- 222

Equating coefficients : 2s : -2BBA1-1s : -1CCBA2-0s : -0.515.0CA5.00.5-

222o )5.0()5.0s()5.0s(2

1s1

5.0ss1s2

1s1

V

)t(vo V)t(u)2tcos(e2e 2-t-t

Page 850: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 21. The s-domain version of the circuit is shown below. 1 s

V1 Vo + 2/s 2 1/s

- At node 1,

10/s

ooo V

sVsV

ssVVV

s )12

()1(1021

102

11

1

(1)

At node 2,

)12

(2

21

1 ss

VVsVV

sVV

oooo (2)

Substituting (2) into (1) gives

ooo VsssVs

Vsss )5.12()12

()12/)(1(10 22

2

5.12)5.12(10

22 ssCBs

sA

sssVo

CsBsssA 22 )5.12(10

BAs 0:2 CAs 20:

-40/3C -20/3,B ,3/205.110:constant AA

22222 7071.0)1(7071.0414.1

7071.0)1(11

320

5.1221

320

sss

ssss

sVo

Taking the inverse Laplace tranform finally yields

V)t(ut7071.0sine414.1t7071.0cose1320)t(v tt

o

Page 851: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 22. The s-domain version of the circuit is shown below. 4s V1 V2

1s

12 1 2 3/s

At node 1,

s4V

s411V

1s12

s4VV

1V

1s12 2

1211 (1)

At node 2,

1s2s34VVV

3s

2V

s4VV 2

212221 (2)

Substituting (2) into (1),

222

2 V23s

37s

34

s41

s4111s2s

34V

1s12

)89s

47s(

CBs)1s(

A

)89s

47s)(1s(

9V22

2

)1s(C)ss(B)89s

47s(A9 22

Equating coefficients:

BA0:s2

A43CCA

43CBA

470:s

-18C -24,B ,24AA83CA

899:constant

Page 852: Solution Manual for Fundamentals of Electric Circuits

6423)

87s(

3

6423)

87s(

)8/7s(24)1s(

24

)89s

47s(

18s24)1s(

24V222

2

Taking the inverse of this produces:

)t(u)t5995.0sin(e004.5)t5995.0cos(e24e24)t(v t875.0t875.0t2

Similarly,

)89s

47s(

FEs)1s(

D

)89s

47s)(1s(

1s2s349

V22

2

1

)1s(F)ss(E)89s

47s(D1s2s

349 222

Equating coefficients:

ED12:s2

D436FFD

436or FED

4718:s

0F 4,E ,8DD833or FD

899:constant

6423)

87s(

2/7

6423)

87s(

)8/7s(4)1s(

8

)89s

47s(

s4)1s(

8V222

1

Thus, )t(u)t5995.0sin(e838.5)t5995.0cos(e4e8)t(v t875.0t875.0t

1 Chapter 16, Solution 23.

The s-domain form of the circuit with the initial conditions is shown below. V

I

1/sC sL R -2/s

4/s 5C

Page 853: Solution Manual for Fundamentals of Electric Circuits

At the non-reference node,

sCVsLV

RV

C5s2

s4

LC1

RCs

ss

CVs

sC56 2

LC1RCssC6s5

V 2

But 88010

1RC1

, 208041

LC1

22222 2)4s()2)(230(

2)4s()4s(5

20s8s480s5

V

)t(v V)t2sin(e230)t2cos(e5 -4t-4t

)20s8s(s4480s5

sLV

I 2

20s8sCBs

sA

)20s8s(s120s25.1

I 22

6A , -6B , -46.75C

22222 2)4s()2)(375.11(

2)4s()4s(6

s6

20s8s75.46s6

s6

I

)t(i 0t),t2sin(e375.11)t2cos(e6)t(u6 -4t-4t

Page 854: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 24. At t = 0-, the circuit is equivalent to that shown below. + 9A 4 5 vo -

20)9(54

4x5)0(vo

For t > 0, we have the Laplace transform of the circuit as shown below after transforming the current source to a voltage source. 4 16 Vo + 36V 10A 2/s 5 - Applying KCL gives

8.12B,2.7A,5.0s

BsA

)5.0s(ss206.3V

5V

2sV

1020

V36o

ooo

Thus,

)t(ue8.122.7)t(v t5.0o

Page 855: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 25. For , the circuit in the s-domain is shown below. 0t

s 6 I

Applying KVL,

+

V

+

(2s)/(s2 + 16)

+

9/s

2/s

0s2

Is9

s616ss2

2

)16s)(9s6s(32s4

I 22

2

)16s()3s(s288s36

s2

s2I

s9V 22

2

16sEDs

)3s(C

3sB

sA

s2

22

)s48s16s3s(B)144s96s25s6s(A288s36 2342342

)s9s6s(E)s9s6s(D)s16s(C 232343

Equating coefficients :

0s : A144288 (1) 1s : E9C16B48A960 (2) 2s : E6D9B16A2536 (3) 3s : ED6CB3A60 (4) 4s : (5) DBA0

Solving equations (1), (2), (3), (4) and (5) gives

2A , B , C7984.1- 16.8- , D 2016.0- , 765.2E

Page 856: Solution Manual for Fundamentals of Electric Circuits

16s)4)(6912.0(

16ss2016.0

)3s(16.8

3s7984.1

s4

)s(V 222

)t(v V)t4sin(6912.0)t4cos(2016.0et16.8e7984.1)t(u4 -3t-3t

Chapter 16, Solution 26.

Consider the op-amp circuit below.

R2

+Vo

0

+

R1

1/sC

+Vs

At node 0,

sC)V0(R

V0R

0Vo

2

o

1

s

o2

1s V-sCR1

RV

211s

o

RRCsR1-

VV

But 21020

RR

2

1 , 1)1050)(1020(CR 6-31

So, 2s

1-VV

s

o

)5s(3Ve3V s

-5ts

Page 857: Solution Manual for Fundamentals of Electric Circuits

5)2)(s(s3-

Vo

5sB

2sA

5)2)(s(s3

V- o

1A , -1B

2s1

5s1

Vo

)t(vo )t(uee -2t-5t

Chapter 16, Solution 27.

Consider the following circuit.

For mesh 1,

I2

2s

I1+

10/(s + 3) 1

2s s

1

221 IsII)s21(3s

10

21 I)s1(I)s21(3s

10 (1)

For mesh 2,

112 IsII)s22(0

21 I)1s(2I)s1(-0 (2) (1) and (2) in matrix form,

2

1

II

)1s(2)1s(-)1s(-1s2

0)3s(10

1s4s3 2

Page 858: Solution Manual for Fundamentals of Electric Circuits

3s)1s(20

1

3s)1s(10

2

Thus 1

1I )1s4s3)(3s()1s(20

2

2

2I)1s4s3)(3s(

)1s(102 2

I1

Chapter 16, Solution 28.

Consider the circuit shown below.

+

Vo I1 +

1

2s s I2

s 6/s 2

For mesh 1,

21 IsI)s21(s6

(1)

For mesh 2,

21 I)s2(Is0

21 Is2

1-I (2)

Substituting (2) into (1) gives

2

2

22 Is

2)5s(s-IsI

s2

12s)-(1s6

or 2s5s

6-I 22

Page 859: Solution Manual for Fundamentals of Electric Circuits

4.561)0.438)(s(s12-

2s5s12-

I2V 22o

Since the roots of s are -0.438 and -4.561, 02s52

561.4sB

438.0sA

Vo

-2.914.123

12-A , 91.2

4.123-12-

B

561.4s91.2

0.438s2.91-

)s(Vo

)t(vo V)t(uee91.2 t438.0-4.561t

Chapter 16, Solution 29.

Consider the following circuit.

1 : 2Io

+

10/(s + 1)

1

4/s 8

Let 1s2

8s48)s4)(8(

s4

||8ZL

When this is reflected to the primary side,

2n,nZ

1Z 2L

in

1s23s2

1s22

1Zin

3s21s2

1s10

Z1

1s10

Iin

o

Page 860: Solution Manual for Fundamentals of Electric Circuits

5.1sB

1sA

)5.1s)(1s(5s10

Io

-10A , 20B

5.1s20

1s10-

)s(Io

)t(io A)t(uee210 t-1.5t

Chapter 16, Solution 30.

)s(X)s(H)s(Y , 1s3

1231s

4)s(X

22

2

)1s3(34s8

34

)1s3(s12

)s(Y

22 )31s(1

274

)31s(s

98

34

)s(Y

Let 2)31s(s

98-

)s(G

Using the time differentiation property,

3t-3t-3t- eet31-

98-

)et(dtd

98-

)t(g

3t-3t- e

98

et278

)t(g

Hence,

3t-3t-3t- et274

e98

et278

)t(u34

)t(y

)t(y 3t-3t- et274

e98

)t(u34

Page 861: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 31.

s1

)s(X)t(u)t(x

4ss10

)s(Y)t2cos(10)t(y 2

)s(X)s(Y

)s(H4s

s102

2

Chapter 16, Solution 32.

(a) )s(X)s(H)s(Y

s1

5s4s3s

2

5s4s

CBssA

)5s4s(s3s

22

CsBs)5s4s(A3s 22

Equating coefficients :

0s : 53AA53 1s : 57- A41CCA41 2s : 53--ABBA0

5s4s7s3

51

s53

)s(Y 2

1)2s(1)2s(3

51

s6.0

)s(Y 2

)t(y )t(u)tsin(e2.0)tcos(e6.06.0 -2t-2t

Page 862: Solution Manual for Fundamentals of Electric Circuits

(b) 22t-

)2s(6

)s(Xet6)t(x

22 )2s(6

5s4s3s

)s(X)s(H)s(Y

5s4sDCs

)2s(B

2sA

)5s4s()2s()3s(6

)s(Y 2222

Equating coefficients :

3s : (1) -ACCA02s : DBA2DC4BA60 (2) 1s : D4B4A9D4C4B4A136 (3) 0s : BA2D4B5A1018 (4)

Solving (1), (2), (3), and (4) gives

6A , 6B , 6-C , -18D

1)2s(18s6

)2s(6

2s6

)s(Y 22

1)2s(6

1)2s()2s(6

)2s(6

2s6

)s(Y 222

)t(y )t(u)tsin(e6)tcos(e6et6e6 -2t-2t-2t-2t

Chapter 16, Solution 33.

)s(X)s(Y

)s(H , s1

)s(X

16)2s()4)(3(

16)2s(s2

)3s(21

s4

)s(Y 22

)s(Ys)s(H20s4s

s1220s4s

s2)3s(2

s4 22

2

Page 863: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 34.

Consider the following circuit.

+

Vo(s)

Vo

10/s

2

4+

Vs

s

Using nodal analysis,

s10V

4V

2sVV ooos

o2

os V)s2s(101

)2s(41

1V10s

41

2s1

)2s(V

o2

s V30s9s2201

V

s

o

VV

30s9s220

2

Chapter 16, Solution 35.

Consider the following circuit.

+

Vo

I V1

+

2/s

2I

s

Vs 3

At node 1,

3sV

II2 1 , where s2VV

I 1s

Page 864: Solution Manual for Fundamentals of Electric Circuits

3sV

s2VV

3 11s

1s1 V

2s3

V2s3

3sV

s1 V2s3

V2s3

3s1

s21 V2s9s3

)3s(s3V

s21o V2s9s3

s9V

3s3

V

s

o

VV

)s(H2s9s3

s92

Chapter 16, Solution 36.

From the previous problem,

s21 V

2s9s3s3

3sV

I3

s2 V2s9s3

sI

But o

2

s Vs9

2s9s3V

2

2

3 9 23 9 2 9 9

ooVs s s

I Vs s s

IV

)s(H o 9

Page 865: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 37.

(a) Consider the circuit shown below.

3 2s

+

Vx 2/s +

I1 I2+

Vs 4Vx

For loop 1,

21s Is2

Is2

3V (1)

For loop 2,

0Is2

Is2

s2V4 12x

But, s2

)II(V 21x

So, 0Is2

Is2

s2)II(s8

1221

21 Is2s6

Is6-

0 (2)

In matrix form, (1) and (2) become

2

1s

II

s2s6s6-s2-s23

0V

s2

s6

s2s6

s2

3

4s6s

18

s1 Vs2s6

, s2 Vs6

Page 866: Solution Manual for Fundamentals of Electric Circuits

s1

1 Vs64s18

)s2s6(I

32s9ss3

VI

s

1

9s2s33s

2

2

(b) 22I

21

21x s2

)II(s2

V

ss

x

V4-)s6s2s6(Vs2V

s

s

x

2

4V-Vs6

VI

2s3-

Chapter 16, Solution 38.

(a) Consider the following circuit.

1

+

Vo

s Is

1/s

VoV1

1/s+

Vs

1 Io

At node 1,

sVV

Vs1

VV o11

1s

o1s Vs1

Vs1

s1V (1)

Page 867: Solution Manual for Fundamentals of Electric Circuits

At node o,

oooo1 V)1s(VVs

sVV

o

21 V)1ss(V (2)

Substituting (2) into (1)

oo2

s Vs1V)1ss)(s11s(V

o23

s V)2s3s2s(V

s

o1 V

V)s(H

2s3s2s1

23

(b) o

2o

231ss V)1ss(V)2s3s2s(VVI

o23

s V)1s2ss(I

s

o2 I

V)s(H

1s2ss1

23

(c) 1

VI o

o

)s(HIV

II

)s(H 2s

o

s

o3 1s2ss

123

(d). )s(HVV

VI

)s( 1s

o

s

o4H

2s3s2s1

23

Chapter 16, Solution 39.

Consider the circuit below.

+

Vo

Io

+

1/sC

R Vb

Va

+

Vs

Page 868: Solution Manual for Fundamentals of Electric Circuits

Since no current enters the op amp, flows through both R and C. oI

sC1

RI-V oo

sCI-

VVV osba

sC1sC1R

VV

)s(Hs

o 1sRC

Chapter 16, Solution 40.

(a) LRs

LRsLR

RVV

)s(Hs

o

)t(h )t(ueLR LRt-

(b) s1)s(V)t(u)t(v ss

LRsB

sA

)LRs(sLR

VLRs

LRV so

1A , -1B

LRs1

s1

Vo

)t(ue)t(u)t(v L-Rt

o )t(u)e1( L-Rt Chapter 16, Solution 41.

)s(X)s(H)s(Y

1s2

)s(H)t(ue2)t(h t-

Page 869: Solution Manual for Fundamentals of Electric Circuits

s5)s(X)s(V)t(u5)t(v ii

1sB

sA

)1s(s10

)s(Y

10A , -10B

1s10

s10

)s(Y

)t(y )t(u)e1(10 -t

Chapter 16, Solution 42.

)s(X)s(Y)s(Ys2

)s(X)s(Y)1s2(

)21s(21

1s21

)s(X)s(Y

)s(H

)t(h )t(ue5.0 2-t

Chapter 16, Solution 43.

i(t)

+

1

1F u(t)

1H

First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KVL we get: '

CC vi;0'ivi)t(u

Page 870: Solution Manual for Fundamentals of Electric Circuits

Thus,

)t(uivi

iv

C'

'C

Finally we get,

)t(u0i

v10)t(i;)t(u

10

iv

1110

iv CCC

Chapter 16, Solution 44. 1/8 F 1H

)t(u4 2

+

+

vx

4

First select the inductor current iL and the capacitor voltage vC to be the state variables. Applying KCL we get:

LCxxLC

'C

C

'C

Cx

x'L

xL'C

'Cx

L

i3333.1v3333.0vor;v2i4v2

vv

8v

4vv

v)t(u4i

v4i8vor;08

v2

vi

LC

'L

LCLCL'C

i3333.1v3333.0)t(u4i

i666.2v3333.1i333.5v3333.1i8v

Now we can write the state equations.

L

Cx

L

C'L

'C

iv

3333.13333.0

v;)t(u40

iv

3333.13333.0666.23333.1

iv

Page 871: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 45.

First select the inductor current iL (current flowing left to right) and the capacitor voltage vC (voltage positive on the left and negative on the right) to be the state variables. Applying KCL we get:

2o'L

oL'CL

o'C

vvi

v2i4vor0i2

v4

v

1Co vvv

21C

'L

1CL'C

vvvi

v2v2i4v

)t(v)t(v

01vi

10)t(v;)t(v)t(v

0211

vi

2410

vi

2

1

C

Lo

2

1

C

L

C

L

Chapter 16, Solution 46.

Page 872: Solution Manual for Fundamentals of Electric Circuits

First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get:

sC'L

sLC'Cs

C'CL

vvi

iiv25.0vor0i4

vvi

Thus,

s

s

L

Co

s

s'L

'C

'L

'C

iv

0000

iv

01

)t(v;iv

0110

iv

01125.0

iv

Chapter 16, Solution 47.

First select the inductor current iL (left to right) and the capacitor voltage vC (+ on the left) to be the state variables.

Letting i1 = 4

v'C and i2 = iL and applying KVL we get:

Loop 1:

1CL'CL

'C

C1 v2v2i4vor0i4

v2vv

Loop 2:

21C21CL

L'L

2'L

'C

L

vvvv2

v2v2i4i2i

or0vi4

vi2

Page 873: Solution Manual for Fundamentals of Electric Circuits

1CL1CL

1 v5.0v5.0i4

v2v2i4i

)t(v)t(v

0005.0

vi

015.01

)t(i)t(i

;)t(v)t(v

0211

vi

2410

vi

2

1

C

L

2

1

2

1

C

L

C

L

Chapter 16, Solution 48.

Let x1 = y(t). Thus, )t(zx4x3yxandxyx 21'22

''1

This gives our state equations.

)t(z0xx

01)t(y;)t(z10

xx

4310

xx

2

1

2

1'2

'1

Chapter 16, Solution 49.

zxyorzyzxxand)t(yxLet 2'''

121 Thus,

z3x5x6zz2z)zx(5x6zyx 21''

21''

2 This now leads to our state equations,

)t(z0xx

01)t(y;)t(z3

1xx

5610

xx

2

1

2

1'2

'1

Chapter 16, Solution 50.

Let x1 = y(t), x2 = .xxand,x '23

'1

Thus, )t(zx6x11x6x 321

"3

We can now write our state equations.

)t(z0xxx

001)t(y;)t(z100

xxx

6116100010

xxx

3

2

1

3

2

1

'3

'2

'1

Page 874: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 51.

We transform the state equations into the s-domain and solve using Laplace transforms.

s1B)s(AX)0(x)s(sX

Assume the initial conditions are zero.

)s/2(0

4s24s

8s4s1

s1

20

s244s

)s(X

s1B)s(X)AsI(

2

1

222222

221

2)2s(2

2)2s()2s(

s1

2)2s(4s

s1

8s4s4s

s1

)8s4s(s8)s(X)s(Y

y(t) = )t(ut2sint2cose1 t2

Chapter 16, Solution 52.

Assume that the initial conditions are zero. Using Laplace transforms we get,

s/4s/3

2s214s

10s6s1

s/2s/1

0411

4s212s

)s(X 2

1

222211)3s(8.1s8.0

s8.0

)1)3s((s8s3X

2222 1)3s(16.

1)3s(3s8.0

s8.0

)t(u)tsine6.0tcose8.08.0()t(x t3t3

1

Page 875: Solution Manual for Fundamentals of Electric Circuits

222221)3s(4.4s4.1

s4.1

1)3s((s14s4X

2222 1)3s(12.0

1)3s(3s4.1

s4.1

)t(u)tsine2.0tcose4.14.1()t(x t3t3

2

)t(u)tsine8.0tcose4.44.2(

)t(u2)t(x2)t(x2)t(yt3t3

211

)t(u)tsine6.0tcose8.02.1()t(u2)t(x)t(y t3t3

12 Chapter 16, Solution 53.

If is the voltage across R, applying KCL at the non-reference node gives oV

oo

oo

s VsL1

sCR1

sLV

VsCRV

I

RLCsRsLIsRL

sL1

sCR1

IV 2

sso

RsLRLCsIsL

RV

I 2so

o

LC1RCssRCs

RsLRLCssL

II

)s(H 22s

o

The roots

LC1

)RC2(1

RC21-

s 22,1

both lie in the left half plane since R, L, and C are positive quantities. Thus, the circuit is stable.

Page 876: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 54.

(a) 1s

3)s(H1 ,

4s1

)s(H2

)4s)(1s(3

)s(H)s(H)s(H 21

4sB

1sA

)s(H)t(h 1-1- LL

1A , 1-B

)t(h )t(u)ee( -4t-t

(b) Since the poles of H(s) all lie in the left half s-plane, the system is stable. Chapter 16, Solution 55.

Let be the voltage at the output of the first op amp. 1oV

sRC1

RsC1

VV

s

1o , sRC

1VV

1o

o

222s

o

CRs1

VV

)s(H

22CRt

)t(h

)t(hlim

t, i.e. the output is unbounded.

Hence, the circuit is unstable.

Page 877: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 56.

LCs1sL

sC1

sL

sC1

sL

sC1

||sL 2

RsLRLCssL

LCs1sL

R

LCs1sL

VV

2

2

2

1

2

LC1

RC1

ss

RC1

s

VV

21

2

Comparing this with the given transfer function,

RC1

2 , LC1

6

If , k1RR21

C F500

C61

L H3.333

Chapter 16, Solution 57. The circuit in the s-domain is shown below.

+

Vx

V1

+

R1

C R2

L

Vi

Z

CsRLCs1sLR

sC1sLR)sLR()sC1(

)sLR(||sC1

Z2

22

2

22

Page 878: Solution Manual for Fundamentals of Electric Circuits

i1

1 VZR

ZV

i12

21

2

2o V

ZRZ

sLRR

VsLR

RV

CsRLCs1sLR

R

CsRLCs1sLR

sLRR

ZRZ

sLRR

VV

22

21

22

2

2

2

12

2

i

o

sLRRCRsRLCRsR

VV

212112

2

i

o

LCRRR

CR1

LR

ss

LCRR

VV

1

21

1

22

1

2

i

o

Comparing this with the given transfer function,

LCRR

51

2 CR

1L

R6

1

2 LCR

RR25

1

21

Since and , 4R1 1R 2

201

LCLC41

5 (1)

C41

L1

6 (2)

201

LCLC45

25

Substituting (1) into (2),

01C24C80C41

C206 2

Thus, 201

,41

C

Page 879: Solution Manual for Fundamentals of Electric Circuits

When 41

C , 51

C201

L .

When 201

C , 1C20

1L .

Therefore, there are two possible solutions. C F25.0 L H2.0 or C F05.0 L H1 Chapter 16, Solution 58.

We apply KCL at the noninverting terminal at the op amp. )YY)(V0(Y)0V( 21o3s

o21s3 V)YY(-VY

21

3

s

o

YYY-

VV

Let , 11 sCY 12 R1Y , 23 sCY

11

12

11

2

s

o

CR1sCsC-

R1sCsC-

VV

Comparing this with the given transfer function,

1CC

1

2 , 10CR

1

11

If , k1R1

421 101

CC F100

Page 880: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 59. Consider the circuit shown below. We notice that o3 VV and o32 VVV .

Y4

Y3

Y1

V1

V2Y2

+

+ Vo

Vin

At node 1, 4o12o111in Y)VV(Y)VV(Y)VV(

)YY(V)YYY(VYV 42o42111in (1) At node 2,

3o2o1 Y)0V(Y)VV(

o3221 V)YY(YV

o2

321 V

YYY

V (2)

Substituting (2) into (1),

)YY(VV)YYY(Y

YYYV 42oo421

2

321in

)YYYYYYYYYYYYYY(VYYV 42

2243323142

2221o21in

43323121

21

in

o

YYYYYYYYYY

VV

1Y and must be resistive, while and must be capacitive. 2Y 3Y 4Y

Let 1

1 R1

Y , 2

2 R1

Y , 13 sCY , 24 sCY

Page 881: Solution Manual for Fundamentals of Electric Circuits

212

2

1

1

1

21

21

in

o

CCsRsC

RsC

RR1

RR1

VV

2121221

212

2121

in

o

CCRR1

CRRRR

ss

CCRR1

VV

Choose R , then k11

6

212110

CCRR1

and 100CRRR

221

21

R

We have three equations and four unknowns. Thus, there is a family of solutions. One such solution is

2R k1 , C1 nF50 , 2C F20

Page 882: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 60. With the following MATLAB codes, the Bode plots are generated as shown below. num=[1 1]; den= [1 5 6]; bode(num,den);

Page 883: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 61. We use the following codes to obtain the Bode plots below. num=[1 4]; den= [1 6 11 6]; bode(num,den);

Page 884: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 62. The following codes are used to obtain the Bode plots below. num=[1 1]; den= [1 0.5 1]; bode(num,den);

Page 885: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 63. We use the following commands to obtain the unit step as shown below. num=[1 2]; den= [1 4 3]; step(num,den);

Page 886: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 64. With the following commands, we obtain the response as shown below. t=0:0.01:5; x=10*exp(-t); num=4; den= [1 5 6]; y=lsim(num,den,x,t); plot(t,y)

Page 887: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 65. We obtain the response below using the following commands. t=0:0.01:5; x=1 + 3*exp(-2*t); num=[1 0]; den= [1 6 11 6]; y=lsim(num,den,x,t); plot(t,y)

Page 888: Solution Manual for Fundamentals of Electric Circuits

Chapter 16, Solution 66. We obtain the response below using the following MATLAB commands. t=0:0.01:5; x=5*exp(-3*t); num=1; den= [1 1 4]; y=lsim(num,den,x,t); plot(t,y)

Chapter 16, Solution 67.

Using the result of Practice Problem 16.14,

)YYY(YYYYY-

VV

321432

21

i

o

Page 889: Solution Manual for Fundamentals of Electric Circuits

When , 11 sCY F5.0C1

12 R

1Y , k10R1

23 YY , 24 sCY , F1C2

)RsC2(RsC1RsC-

)R2sC(sCR1RsC-

VV

1112

11

11221

11

i

o

1RC2sRCCsRsC-

VV

122121

211

i

o

1)10)(1010(1)2(s)10)(1010)(110(0.5s)10)(1010(0.5s-

VV

36-236-6-2

3-6

i

o

42i

o

102s400ss100-

VV

Therefore,

a 100- , b 400 , c 4102 Chapter 16, Solution 68.

(a) Let 3s

)1s(K)s(Y

Ks31

)s11(Klim

3s)1s(K

lim)(Yss

i.e. . K25.0

Hence, Y )s()3s(4

1s

(b) Consider the circuit shown below.

I

Vs = 8 V +

t = 0

YS

Page 890: Solution Manual for Fundamentals of Electric Circuits

s8V)t(u8V ss

)3s(s)1s(2

3s1s

s48

)s(V)s(YZV

I ss

3sB

sA

I

32A , 34-B

)t(i A)t(ue4231 3t-

Chapter 16, Solution 69.

The gyrator is equivalent to two cascaded inverting amplifiers. Let be the voltage at the output of the first op amp.

1V

ii1 -VVRR-

V

i1o VsCR

1V

RsC1-

V

CsRV

RV

I 2oo

o

CsRIV 2

o

o

CRLwhen,sLIV 2

o

o

Page 891: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 1.

(a) This is periodic with = which leads to T = 2 / = 2. (b) y(t) is not periodic although sin t and 4 cos 2 t are independently periodic. (c) Since sin A cos B = 0.5[sin(A + B) + sin(A �– B)],

g(t) = sin 3t cos 4t = 0.5[sin 7t + sin(�–t)] = �–0.5 sin t + 0.5 sin7t which is harmonic or periodic with the fundamental frequency

= 1 or T = 2 / = 2 . (d) h(t) = cos 2 t = 0.5(1 + cos 2t). Since the sum of a periodic function and a constant is also periodic, h(t) is periodic. = 2 or T = 2 / = . (e) The frequency ratio 0.6|0.4 = 1.5 makes z(t) periodic.

= 0.2 or T = 2 / = 10. (f) p(t) = 10 is not periodic. (g) g(t) is not periodic.

Chapter 17, Solution 2.

(a) The frequency ratio is 6/5 = 1.2. The highest common factor is 1. = 1 = 2 /T or T = 2 .

(b) = 2 or T = 2 / = . (c) f3(t) = 4 sin2 600 t = (4/2)(1 �– cos 1200 t)

= 1200 or T = 2 / = 2 /(1200 ) = 1/600. (d) f4(t) = ej10t = cos 10t + jsin 10t. = 10 or T = 2 / = 0.2 .

Page 892: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 3.

T = 4, o = 2 /T = /2 g(t) = 5, 0 < t < 1 10, 1 < t < 2 0, 2 < t < 4

ao = (1/T) = 0.25[ + ] = T

0dt)t(g

1

0dt5

2

1dt10 3.75

an = (2/T) = (2/4)[ T

0 o dt)tncos()t(g1

0dt)t

2ncos(5 +

2

1dt)t

2ncos(10 ]

= 0.5[1

0

t2

nsinn25 +

2

1

t2

nsinn210 ] = (�–1/(n ))5 sin(n /2)

an = (5/(n ))(�–1)(n+1)/2, n = odd 0, n = even

bn = (2/T) = (2/4)[ T

0 o dt)tnsin()t(g1

0dt)t

2nsin(5 +

2

1dt)t

2nsin(10 ]

= 0.5[1

0

t2

ncosn

5x2 �– 2

1

t2

ncosn

10x2 ] = (5/(n ))[3 �– 2 cos n + cos(n /2)]

Chapter 17, Solution 4.

f(t) = 10 �– 5t, 0 < t < 2, T = 2, o = 2 /T =

ao = (1/T) = (1/2) = T

0dt)t(f

2

0dt)t510(

2

0

2 )]2/t5(t10[5.0 = 5

an = (2/T) = (2/2) T

0 o dt)tncos()t(f2

0dt)tncos()t510(

= �– 2

0dt)tncos()10(

2

0dt)tncos()t5(

= 2

022 tncos

n5 +

2

0

tnsinn

t5 = [�–5/(n2 2)](cos 2n �– 1) = 0

Page 893: Solution Manual for Fundamentals of Electric Circuits

bn = (2/2) 2

0dt)tnsin()t510(

= �– 2

0dt)tnsin()10(

2

0dt)tnsin()t5(

= 2

022 tnsin

n5 +

2

0

tncosn

t5 = 0 + [10/(n )](cos 2n ) = 10/(n )

Hence f(t) = )tnsin(n110

1n

5

Chapter 17, Solution 5.

1T/2,2T

5.0]x2x1[21dt)t(z

T1a

T

0o

2

20

0

T

0on 0ntsin

n2nt..sin

n1ntdtcos21ntdtcos11dtncos)t(z

T2a

22

00

T

0on

evenn 0,

oddn,n6

ntcosn2ntcos

n1ntdtsin21ntdtsin11dtncos)t(z

T2b

Thus,

ntsinn65.0)t(z

oddn1n

Page 894: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 6.

.0a,functionoddanisthisSince

326)1x21x4(

21dt)t(y

21a

22,2T

n

20o

o

oddn1n

evenn,0

oddn,n4

21

10

21

10

20 on

)tnsin(n143)t(y

))ncos(1(n2))ncos(1(

n2))ncos(1(

n4

))ncos()n2(cos(n2)1)n(cos(

n4)tncos(

n2)tncos(

n4

dt)tnsin(2dt)tnsin(4dt)tnsin()t(y22b

Chapter 17, Solution 7.

0a,6

T/2,12T 0

10

4

4

2

T

0on ]dt6/tncos)10(dt6/tncos10[

61dtncos)t(f

T2a

3/n5sin3/nsin3/n2sin2n106/tnsin

n106/tnsin

n10 10

44

2

10

4

4

2

T

0on ]dt6/tnsin)10(dt6/tnsin10[

61dtnsin)t(f

T2b

Page 895: Solution Manual for Fundamentals of Electric Circuits

3/n2sin23/ncos3/n5cosn106/ntncos

n106/tncos

n10 10

44

2

1nnn 6/tnsinb6/tncosa)t(f

where an and bn are defined above.

Chapter 17, Solution 8.

T/22,T1, t 1- ),t1(2)t(f o

2ttdt)1t(221dt)t(f

T1a

1

1

1

1

2T

0o

0tnsinn1tnsin

nttncos

n

12tdtncos)1t(222dtncos)t(f

T2a

1

122

1

1

T

0on

ncosn4tncos

n1tncos

nttnsin

n12tdtnsin)1t(2

22dtnsin)t(f

T2b

1

122

1

1

T

0on

1n

ntncos

n)1(42)t(f

Chapter 17, Solution 9. f(t) is an even function, bn=0.

4//2,8 TT

183.3104/sin)4(4

1004/cos1082)(1 2

0

2

00

tdttdttfT

aT

o

Page 896: Solution Manual for Fundamentals of Electric Circuits

dtntntdttntdtntfT

aT

on

2

0

2

0

2/

0

4/)1(cos4/)1(cos5]04/cos4/cos10[840cos)(4

For n = 1,

102/sin25]12/[cos52

0

2

01 tdttdtta

For n>1,

2)1(sin

)1(20

2)1(sin

)1(20

4)1(sin

)1(20

4)1(sin

)1(20 2

0

nn

nn

nn

tnn

an

0sin102sin420,3662.62/sin20sin10

32 aa

Thus,

3213210 0,0,362.6,10,183.3 bbbaaaa Chapter 17, Solution 10.

T/2,2T o

21

tjn10

tjnT

0

10

21

tjntjntojnn jn

e2jn

e421dte)2(dte4

21dte)t(h

T1c

,even n ,0

oddn ,n

j6]6ncos6[

n2je2e24e4

n2jc jnn2jnj

n

Thus,

tjn

oddnn

en

6j)t(f

Page 897: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 11.

2/T/2,4T o

T

0

01

10

2/tjn2/tjntojnn dte)1(dte)1t(

41dte)t(y

T1c

10

2/tjn01

2/tjn22

2/tjnn e

jn2e

jn2)12/tjn(

4/n

e41c

jn2e

jn2e

jn2)12/jn(e

n

4jn2

n

441 2/jn2/jn2/jn

2222

But

2/nsinj2/nsinj2/ncose,2/nsinj2/nsinj2/ncose 2/jn2/jn

2/nsinn2/nsin)12/jn(j1n

1c22n

2/tjn22

ne2/nsinn2/nsin)12/jn(j1

n

1)t(y

Chapter 17, Solution 12.

A voltage source has a periodic waveform defined over its period as v(t) = t(2 - t) V, for all 0 < t < 2 Find the Fourier series for this voltage. v(t) = 2 t �– t2, 0 < t < 2 , T = 2 , o = 2 /T = 1 ao =

(1/T) dt)tt2(21dt)t(f

2

0

2T

0 32)3/21(

24)3/tt(

21 23

2

032

Page 898: Solution Manual for Fundamentals of Electric Circuits

an = 2

02

T

0

2 )ntsin(n

t2)ntcos(n21dt)ntcos()tt2(

T2

2

0

223 )ntsin(tn)ntsin(2)ntcos(nt2

n1

232 n4)n2cos(n4

n1)11(

n2

bn = dt)ntsin()tnt2(1dt)ntsin()tnt2(T2 2T

0

2

2

0

22302 ))ntcos(tn)ntcos(2)ntsin(nt2(

n1))ntcos(nt)nt(sin(

n1n2

0n

4n4

Hence, f(t) = 1n

2

2

)ntcos(n4

32

Chapter 17, Solution 13. T = 2 , o = 1

ao = (1/T) dttsin10[21dt)t(h

0

T

0+ ]dt)tsin(20

2

30)tcos(20tcos10

21 2

0

an = (2/T) T

0 o dt)tncos()t(h

= [2/(2 )] 0

2dt)ntcos()tsin(20dt)ntcos(tsin10

Since sin A cos B = 0.5[sin(A + B) + sin(A �– B)] sin t cos nt = 0.5[sin((n + 1)t) + sin((1 �– n))t] sin(t �– ) = sin t cos �– cost sin = �–sin t sin(t �– )cos(nt) = �–sin(t)cos(nt)

Page 899: Solution Manual for Fundamentals of Electric Circuits

an = 0

2dt)]t]n1sin([)t]n1[sin([20dt)]t]n1sin([)t]n1[sin([10

21

= 2

0 n1)t]n1cos([2

n1)t]n1cos([2

n1)t]n1cos([

n1)t]n1cos([5

an = n1

)]n1cos([3n1

)]n1cos([3n1

3n1

35

But, [1/(1+n)] + [1/(1-n)] = 1/(1�–n2)

cos([n�–1] ) = cos([n+1] ) = cos cos n �– sin sin n = �–cos n

an = (5/ )[(6/(1�–n2)) + (6 cos(n )/(1�–n2))]

= [30/( (1�–n2))](1 + cos n ) = [�–60/( (n�–1))], n = even = 0, n = odd

bn = (2/T) T

0 o dttnsin)t(h

= [2/(2 )][ + 0

dtntsintsin102

dtntsin)tsin(20

But, sin A sin B = 0.5[cos(A�–B) �– cos(A+B)] sin t sin nt = 0.5[cos([1�–n]t) �– cos([1+n]t)] bn = (5/ ){[(sin([1�–n]t)/(1�–n)) �– (sin([1+n]t)/

0)]n1(

+ [(2sin([1-n]t)/(1-n)) �– (2sin([1+n]t)/ 2)]n1( }

= n1

)]n1sin([n1

)]n1sin([5 = 0

Thus, h(t) = 1k

2 )1k4()kt2cos(6030

Page 900: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 14. Since cos(A + B) = cos A cos B �– sin A sin B.

f(t) = 1n

33 )nt2sin()4/nsin(1n

10)nt2cos()4/ncos(

1n10

2

Chapter 17, Solution 15.

(a) Dcos t + Esin t = A cos( t - ) where A = 22 ED , = tan-1(E/D)

A = 622 n1

)1n(16 , = tan-1((n2+1)/(4n3))

f(t) = 101n

3

21

622 n41n

tannt10cosn1

)1n(16

(b) Dcos t + Esin t = A sin( t + )

where A = 22 ED , = tan-1(D/E)

f(t) = 101n

2

31

622 1nn4

tannt10sinn1

)1n(16

Chapter 17, Solution 16.

If v2(t) is shifted by 1 along the vertical axis, we obtain v2*(t) shown below, i.e.

v2*(t) = v2(t) + 1.

1

v2*(t)

2

t

-2 -1 0 1 2 3 4 5

Page 901: Solution Manual for Fundamentals of Electric Circuits

Comparing v2*(t) with v1(t) shows that

v2

*(t) = 2v1((t + to)/2) where (t + to)/2 = 0 at t = -1 or to = 1 Hence v2

*(t) = 2v1((t + 1)/2) But v2

*(t) = v2(t) + 1 v2(t) + 1 = 2v1((t+1)/2) v2(t) = -1 + 2v1((t+1)/2)

= -1 + 1 2

1t5cos251

21t3cos

91

21tcos8

2

v2(t) = 2

52

t5cos251

23

2t3cos

91

22tcos8

2

v2(t) = 2

t5sin251

2t3sin

91

2tsin8

2

Chapter 17, Solution 17. We replace t by �–t in each case and see if the function remains unchanged.

(a) 1 �– t, neither odd nor even.

(b) t2 �– 1, even

(c) cos n (-t) sin n (-t) = - cos n t sin n t, odd

(d) sin2 n(-t) = (-sin t)2 = sin2 t, even

(e) et, neither odd nor even.

Page 902: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 18.

(a) T = 2 leads to o = 2 /T =

f1(-t) = -f1(t), showing that f1(t) is odd and half-wave symmetric.

(b) T = 3 leads to o = 2 /3 f2(t) = f2(-t), showing that f2(t) is even.

(c) T = 4 leads to o = /2

f3(t) is even and half-wave symmetric. Chapter 17, Solution 19. This is a half-wave even symmetric function.

ao = 0 = bn, o = 2 /T /2

an = 2/T

0 o dt)tncos(Tt41

T4

= [4/(n )2](1 cos n ) = 8/(n2 2), n = odd

= 0, n = even

f (t) = oddn

22 2tncos

n18

Chapter 17, Solution 20. This is an even function.

bn = 0, T = 6, = 2 /6 = /3

ao = 3

2

2

1

2/T

0dt4dt)4t4(

62dt)t(f

T2

Page 903: Solution Manual for Fundamentals of Electric Circuits

= )23(4)t4t2(31 2

1

2 = 2

an = 4/T

0dt)3/tncos()t(f

T4

= (4/6)[ + ] 2

1dt)3/tncos()4t4(

3

2dt)3/tncos(4

= 3

2

2

122 3

tnsinn3

616

3tnsin

n3

3tnsin

nt3

3tncos

n9

616

= [24/(n2 2)][cos(2n /3) cos(n /3)]

Thus f(t) = 1n

22 3tncos

3ncos

3n2cos

n1242

At t = 2,

f(2) = 2 + (24/ 2)[(cos(2 /3) cos( /3))cos(2 /3)

+ (1/4)(cos(4 /3) cos(2 /3))cos(4 /3) + (1/9)(cos(2 ) cos( ))cos(2 ) + -----]

= 2 + 2.432(0.5 + 0 + 0.2222 + -----)

f(2) = 3.756 Chapter 17, Solution 21. This is an even function.

bn = 0, T = 4, o = 2 /T = /2. f(t) = 2 2t, 0 < t < 1 = 0, 1 < t < 2

ao = 1

0

1

0

2

2ttdt)t1(2

42 = 0.5

an = 1

0

2/T

0 o dt2

tncos)t1(244dt)tncos()t(f

T4

Page 904: Solution Manual for Fundamentals of Electric Circuits

= [8/( 2n2)][1 cos(n /2)]

f(t) = 1n

22 2tncos

2ncos1

n8

21

Chapter 17, Solution 22.

Calculate the Fourier coefficients for the function in Fig. 16.54. f(t)

4

t

-5 -4 -3 -2 -1 0 1 2 3 4 5

Figure 17.61 For Prob. 17.22

This is an even function, therefore bn = 0. In addition, T=4 and o = /2.

ao = 1

0

21

0

2T

0ttdt4

42dt)t(f

T2 1

an = 1

0

2T

0 o dt)2/tncos(t444dt)ntcos()t(f

T4

1

022 )2/tnsin(

nt2)2/tncos(

n44

an = )2/nsin(n8)1)2/n(cos(

n16

22

Chapter 17, Solution 23. f(t) is an odd function.

f(t) = t, 1< t < 1 ao = 0 = an, T = 2, o = 2 /T =

Page 905: Solution Manual for Fundamentals of Electric Circuits

bn = 1

0

2/T

0 o dt)tnsin(t24dt)tnsin()t(f

T4

= 1022 )tncos(tn)tnsin(

n2

= [2/(n )]cos(n ) = 2( 1)n+1/(n )

f(t) = 1n

1n

)tnsin(n)1(2

Chapter 17, Solution 24.

(a) This is an odd function.

ao = 0 = an, T = 2 , o = 2 /T = 1

bn = 2/T

0 o dt)ntsin()t(fT4

f(t) = 1 + t/ , 0 < t <

bn = 0

dt)ntsin()/t1(24

= 0

2 )ntcos(nt)ntsin(

n1)ntcos(

n12

= [2/(n )][1 2cos(n )] = [2/(n )][1 + 2( 1)n+1]

a2 = 0, b2 = [2/(2 )][1 + 2( 1)] = 1/ = 0.3183

(b) n = n o = 10 or n = 10

a10 = 0, b10 = [2/(10 )][1 cos(10 )] = 1/(5 )

Thus the magnitude is A10 = 2

102 ba10

= 1/(5 ) = 0.06366

and the phase is 10 = tan 1(bn/an) = 90

Page 906: Solution Manual for Fundamentals of Electric Circuits

(c) f(t) = 1n

)ntsin()]ncos(21[n2

f( /2) = 1n

)2/nsin()]ncos(21[n2

For n = 1, f1 = (2/ )(1 + 2) = 6/ For n = 2, f2 = 0 For n = 3, f3 = [2/(3 )][1 2cos(3 )]sin(3 /2) = 6/(3 ) For n = 4, f4 = 0 For n = 5, f5 = 6/(5 ), ----

Thus, f( /2) = 6/ 6/(3 ) + 6/(5 ) 6/(7 ) ---------

= (6/ )[1 1/3 + 1/5 1/7 + --------] f( /2) 1.3824 which is within 8% of the exact value of 1.5.

(d) From part (c) f( /2) = 1.5 = (6/ )[1 1/3 + 1/5 1/7 + - - -] (3/2)( /6) = [1 1/3 + 1/5 1/7 + - - -] or /4 = 1 1/3 + 1/5 1/7 + - - - Chapter 17, Solution 25. This is an odd function since f( t) = f(t).

ao = 0 = an, T = 3, o = 2 /3.

b n = 1

0

2/T

0 o dt)3/nt2sin(t34dt)tnsin()t(f

T4

Page 907: Solution Manual for Fundamentals of Electric Circuits

= 1

022 3

nt2cosn2t3

3nt2sin

n49

34

= 3

n2cosn23

3n2sin

n49

34

22

f(t) = 1n

22 3t2sin

3n2cos

n2

3n2sin

n3

Chapter 17, Solution 26. T = 4, o = 2 /T = /2

ao = 4

3

3

1

1

0

T

0dt1dt2dt1

41dt)t(f

T1 = 1

an = dt)tncos()t(fT

T

0 o2

an = 4

3

3

2

2

1dt)2/tncos(1dt)2/tncos(2dt)2/tncos(1

42

= 4

3

3

2

2

1 2tnsin

n2

2tnsin

n4

2tnsin

n22

= 2

nsin2n3sin

n4

bn = T

0 o dt)tnsin()t(fT2

= 4

3

3

2

2

1dt

2tnsin1dt

2tnsin2dt

2tnsin1

42

= 4

3

3

2

2

1 2tncos

n2

2tncos

n4

2tncos

n22

Page 908: Solution Manual for Fundamentals of Electric Circuits

= 1)ncos(n4

Hence f(t) =

1n

)2/tnsin()1)n(cos()2/tncos())2/nsin()2/n3(sin(n41

Chapter 17, Solution 27. (a) odd symmetry. (b) ao = 0 = an, T = 4, o = 2 /T = /2 f(t) = t, 0 < t < 1 = 0, 1 < t < 2

bn = 1

022

1

0 2tncos

nt2

2tnsin

n4dt

2tnsint

44

= 02

ncosn2

2nsin

n4

22

= 4( 1)(n 1)/2/(n2 2), n = odd

2( 1)n/2/(n ), n = even

a3 = 0, b3 = 4( 1)/(9 2) = 0.045 (c) b1 = 4/ 2, b2 = 1/ , b3 = 4/(9 2), b4 = 1/(2 ), b5 = (25 2)

Frms = 2n

2n

2o ba

21a

Frms

2 = 0.5 bn2 = [1/(2 2)][(16/ 2) + 1 + (16/(8 2)) + (1/4) + (16/(625 2))]

= (1/19.729)(2.6211 + 0.27 + 0.00259) Frms = 14659.0 = 0.3829

Compare this with the exact value of Frms = 6/1dttT2 1

0

2 = 0.4082

Page 909: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 28. This is half-wave symmetric since f(t T/2) = f(t).

ao = 0, T = 2, o = 2 /2 =

an = 1

0

2/T

0 o dt)tncos()t22(24dt)tncos()t(f

T4

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n14

= [4/(n2 2)][1 cos(n )] = 8/(n2 2), n = odd

0, n = even

bn = 4 1

0dt)tnsin()t1(

= 1

022 )tncos(

nt)tnsin(

n1)tncos(

n14

= 4/(n ), n = odd

f(t) = 1k

22)tnsin(

n4

)tncos(n

8 , n = 2k 1

Chapter 17, Solution 29. This function is half-wave symmetric.

T = 2 , o = 2 /T = 1, f(t) = t, 0 < t <

For odd n, an = 0

dt)ntcos()t(T2 =

02 )ntsin(nt)ntcos(n2 = 4/(n2 )

bn = 020

)ntcos(nt)ntsin(n2dt)ntsin()t(2 = 2/n

Page 910: Solution Manual for Fundamentals of Electric Circuits

Thus,

f(t) = 1k

2)ntsin(

n1

)ntcos(n

22 , n = 2k 1

Chapter 17, Solution 30.

2/T

2/T

2/T2/T

2/T2/T oo

tojnn tdtnsin)t(fjtdtncos)t(f

T1dte)t(f

T1c (1)

(a) The second term on the right hand side vanishes if f(t) is even. Hence

2/T

0on tdtncos)t(f

T2c

(b) The first term on the right hand side of (1) vanishes if f(t) is odd. Hence,

2/T

0on tdtnsin)t(f

T2jc

Chapter 17, Solution 31.

If oo /T2

'T2'/T'T),t(f)t(h

'T

0o

'T

0on tdt'ncos)t(f

'T2tdt'ncos)t(h

'T2'a

Let T'T,/dtd,,t

n

T

0on a/dncos)(f

T2'a

Similarly, nn b'b

Page 911: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 32. When is = 1 (DC component)

i = 1/(1 + 2) = 1/3

For n 1, n = 3n, Is = 1/n2 0

I = [1/(1 + 2 + j n2)]Is = Is/(3 + j6n)

= )n2tan(n41n3

1

)3/n6(tann413

0n1

2212

2

Thus,

i(t) = 1n

1

22))n2(tann3cos(

n41n3

131

Chapter 17, Solution 33.

For the DC case, the inductor acts like a short, Vo = 0. For the AC case, we obtain the following:

so

ooso

VVn5n5.2j1

04Vjn

n2jV

10VV

)5n5.2(jn

4

n5n5.2j1

1n4A

n5n5.2j1

VV

22nn

so

n5n5.2tan;

)5n5.2(n

4A22

1n22222n

Page 912: Solution Manual for Fundamentals of Electric Circuits

V)tnsin(A)t(v1n

nno

Chapter 17, Solution 34. For any n, V = [10/n2] (n /4), = n. 1 H becomes j nL = jn and 0.5 F becomes 1/(j nC) = j2/n

2 jn

j2/n

+

+

Vo

V Vo = { j(2/n)/[2 + jn j(2/n)]}V = { j2/[2n + j(n2 2)]}[(10/n2) (n /4)]

)]n2/)2n((tan)2/()4/n[(

4nn

20)n2/)2n((tan)2n(n4n

)2/)4/n((20

21

22

212222

vo(t) = 1n

21

22 n22ntan

24nntcos

4nn

20

Chapter 17, Solution 35.

If vs in the circuit of Fig. 17.72 is the same as function f2(t) in Fig. 17.57(b), determine the dc component and the first three nonzero harmonics of vo(t).

1 1 H

1 F 1

+

+

vo

vS

Figure 17.72 For Prob. 17.35

Page 913: Solution Manual for Fundamentals of Electric Circuits

1

f2(t)

2

t

-2 -1 0 1 2 3 4 5

Figure 17.57(b) For Prob. 17.35

The signal is even, hence, bn = 0. In addition, T = 3, o = 2 /3. vs(t) = 1 for all 0 < t < 1 = 2 for all 1 < t < 1.5

ao = 34dt2dt1

32 5.1

1

1

0

an = 5.1

1

1

0dt)3/tn2cos(2dt)3/tn2cos(

34

= )3/n2sin(n2)3/tn2sin(

n26)3/tn2sin(

n23

34 5.1

1

1

0

vs(t) = 1n

)3/tn2cos()3/n2sin(n12

34

Now consider this circuit,

j2n /3 1

-j3/(2n ) 1

+

+

vo

vS

Let Z = [-j3/(2n )](1)/(1 �– j3/(2n )) = -j3/(2n - j3) Therefore, vo = Zvs/(Z + 1 + j2n /3). Simplifying, we get

vo = )18n4(jn12

v9j22

s

Page 914: Solution Manual for Fundamentals of Electric Circuits

For the dc case, n = 0 and vs = ¾ V and vo = vs/2 = 3/8 V. We can now solve for vo(t)

vo(t) = volts3

tn2cosA83

1nnn

where n23

3ntan90and

63

n4n16

)3/n2sin(n6

A 1on222

22

n

where we can further simplify An to this, 81n4n

)3/n2sin(944nA

Chapter 17, Solution 36.

vs(t) = oddn1n

nn )ntcos(A

where n = tan 1[(3/(n ))/( 1/(n ))] = tan 1( 3) = 100.5

An = 2

nsin9n1

2nsin

n1

n9 22

2222

n = n and 2 H becomes j nL = j2n

Let Z = 1||j2n = j2n/(1 + j2n) If Vo is the voltage at the non-reference node or across the 2-H inductor. Vo = ZVs/(1 + Z) = [j2n/(1 + j2n)]Vs/{1 + [j2n/(1 + j2n)]} = j2nVs/(1 + j4n) But Vs = An n Vo = j2n An n/(1 + j4n)

Page 915: Solution Manual for Fundamentals of Electric Circuits

Io = Vo/j = [2n An n]/ 2n161 tan 14n

= 2

2

n161

n22

nsin9n1

100.5 tan 14n

Since sin(n /2) = ( 1)(n 1)/2 for n = odd, sin2(n /2) = 1

Io = 2

1

n161

n4tan5.100102

io(t) = )n4tan5.100ntcos(n161

1102

oddn1n

1

2

Chapter 17, Solution 37. From Example 15.1,

vs(t) = 1k

),tnsin(n1205 n = 2k 1

For the DC component, the capacitor acts like an open circuit.

Vo = 5

For the nth harmonic, Vs = [20/(n )] 0

10 mF becomes 1/(j nC) = j/(n x10x10 3) = j100/(n )

vo = 22

1s

n25nn5tan90100

n20

100jn20100j

20n

100j

Vn

100j

vo(t) = )n5tan90tnsin(

n25n

1100 1

22

Page 916: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 38.

1ks 1k2n,tnsin

n12

21)t(v

n,Vj1

jV ns

n

no

For dc, 0V,5.0V,0 osn

For nth harmonic, os 90

n2V

22

1o

122

oo

n1

ntan290n2

ntann1

90nV

1k2n),ntantncos(n1

2)t(v 1

1k 22o

Chapter 17, Solution 39.

Comparing vs(t) with f(t) in Figure 15.1, vs is shifted by 2.5 and the magnitude is 5 times that of f(t). Hence

vs(t) = 1k

),tnsin(n1105 n = 2k 1

T = 2, o = 2 //T = , n = n o = n

For the DC component, io = 5/(20 + 40) = 1/12 For the kth harmonic, Vs = (10/(n )) 0

100 mH becomes j nL = jn x0.1 = j0.1n

50 mF becomes 1/(j nC) = j20/(n ) 40

j20/(n )

+

I Io

Z

20

VS j0.1n

Page 917: Solution Manual for Fundamentals of Electric Circuits

Let Z = j20/(n )||(40 + j0.1n ) = n1.0j40

n20j

)n1.0j40(n20j

= )20n1.0(jn40

800jn2n1.0jn4020j

n1.0j40(20j2222

Zin = 20 + Z = )20n1.0(jn40)1200n2(jn802

22

22

I = )]1200n2(jn802[n

)200n(jn400ZV

22

22

in

s

Io = )20n1.0(jn40

I20j

)n1.0j40(n20j

In20j

22

= )]1200n2(jn802[n

200j22

= 2222

221

)1200n2()802(n)}n802/()1200n2{(tan90200

Thus

io(t) = 1k

nn )tnsin(I200201 , n = 2k 1

where n8021200n2tan90

221

n

In = )1200n2()n804(n

1222

Page 918: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 40. T = 2, o = 2 /T =

ao = 2/12ttdt)t22(

21dt)t(v

T

1

0

21

0

T

0

1

an = 1

0

T

0dt)tncos()t1(2dt)tncos()t(v

T2

= 1

022 )tnsin(

nt)tncos(

n1)tnsin(

n12

= 2222

22

)1n2(4oddn,

n4

evenn,0)ncos1(

n2

bn = 1

0

T

0dt)tnsin()t1(2dt)tnsin()t(v

T2

= n2)tncos(

nt)tnsin(

n1)tncos(

n12

1

022

vs(t) = )tncos(A21

nn

where 4422n

21

n )1n2(16

n4A,

n2)1n2(tan

For the DC component, vs = 1/2. As shown in Figure (a), the capacitor acts like an open circuit.

+

Vx

i

Vo2Vx

+

Vo

Vx

+

+

3

1 0.5V

(a)

Page 919: Solution Manual for Fundamentals of Electric Circuits

Vo

(1/4)F

2Vx

+

Vo

Vx

+

+

3

1

VS (b)

Applying KVL to the circuit in Figure (a) gives �–0.5 �– 2Vx + 4i = 0 (1) But �–0.5 + i + Vx = 0 or �–1 + 2Vx + 2i = 0 (2) Adding (1) and (2), �–1.5 + 6i = 0 or i = 0.25 Vo = 3i = 0.75 For the nth harmonic, we consider the circuit in Figure (b).

n = n , Vs = An �– , 1/(j nC) = �–j4/(n ) At the supernode, (Vs �– Vx)/1 = �–[n /(j4)]Vx + Vo/3 Vs = [1 + jn /4]Vx + Vo/3 (3) But �–Vx �– 2Vx + Vo = 0 or Vo = 3Vx Substituting this into (3), Vs = [1 + jn /4]Vx + Vx = [2 + jn /4]Vx = (1/3)[2 + jn /4]Vo = (1/12)[8 + jn ]Vo

Vo = 12Vs/(8 + jn ) = )8/n(tann64

A12122

n

Vo = ))]n2/()1n2((tan)8/n([tan)1n2(

16n

4

n64

12 11442222

Page 920: Solution Manual for Fundamentals of Electric Circuits

Thus

vo(t) = 1n

nn )tncos(V43

where Vn = 442222 )1n2(

16n

4

n64

12

n = tan�–1(n /8) �– tan�–1( (2n �– 1)/(2n)) Chapter 17, Solution 41. For the full wave rectifier,

T = , o = 2 /T = 2, n = n o = 2n

Hence

vin(t) = 1n

2 nt2cos1n4

142

For the DC component,

Vin = 2/

The inductor acts like a short-circuit, while the capacitor acts like an open circuit.

Vo = Vin = 2/

For the nth harmonic, Vin = [�–4/( (4n2 �– 1))] 0

2 H becomes j nL = j4n

0.1 F becomes 1/(j nC) = �–j5/n

Z = 10||(�–j5/n) = �–j10/(2n �– j)

Vo = [Z/(Z + j4n)]Vin = �–j10Vin/(4 + j(8n �– 10))

= )1n4(

04)10n8(j4

10j2

Page 921: Solution Manual for Fundamentals of Electric Circuits

= 22

1

)10n8(16)1n4(

)}5.2n2(tan90{40

Hence vo(t) = 1n

nn )nt2cos(A2

where

An = 29n40n16)1n4(

2022

n = 90 �– tan�–1(2n �– 2.5)

Chapter 17, Solution 42.

1ks 1-2kn ,tnsin

n1205v

nn ,VRCjV)V0(Cj

R0V

onsn

oons

For n = 0 (dc component), Vo=0. For the nth harmonic,

22

5

9422o

oo

n2

10

10x40x10xn

2090n20

RCn901V

Hence,

1k22

5o 1-2kn ,tncos

n

1

2

10)t(v

Alternatively, we notice that this is an integrator so that

1k22

5so 1-2kn ,tncos

n

1

2

10dtvRC1)t(v

Page 922: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 43.

(a) Vrms = )1020(2130)ba(

21a 222

1n

2n

2n

20 = 33.91 V

(b) Irms = )24(216 222 = 6.782 A

(c) P = VdcIdc + )cos(IV21

nnnn

= 30x6 + 0.5[20x4cos(45o-10o) �– 10x2cos(-45o+60o)]

= 180 + 32.76 �– 9.659 = 203.1 W Chapter 17, Solution 44.

(a) W73.300535.319.27045cos1025cos6021vip oo

(b) The power spectrum is shown below.

p 27.19 3.535 0 1 2 3 Chapter 17, Solution 45. n = 1000n j nL = j1000nx2x10�–3 = j2n 1/(j nC) = �–j/(1000nx40x10�–6) = �–j25/n

Page 923: Solution Manual for Fundamentals of Electric Circuits

Z = R + j nL + 1/(j nC) = 10 + j2n �– j25/n

I = V/Z

For n = 1, V1 = 100, Z = 10 + j2 �– j25 = 10 �– j23 I1 = 100/(10 �– j23) = 3.987 73.89

For n = 2, V2 = 50, Z = 10 + j4 �– j12.5 = 10 �– j8.5 I2 = 50/(10 �– j8.5) = 3.81 40.36

For n = 3, V3 = 25, Z = 10 + j6 �– j25/3 = 10 �– j2.333 I3 = 25/(10 �– j2.333) = 2.435 13.13 Irms = 0.5 222 435.281.3987.3 = 3.014 A

p = VDCIDC + 3

1nnnnn )cos(IV

21

= 0 + 0.5[100x3.987cos(73.89 ) + 50x3.81cos(40.36 ) + 25x2.435cos(13.13 )]

= 0.5[110.632 + 145.16 + 59.28] = 157.54 watts Chapter 17, Solution 46. (a) This is an even function

Irms = 2/T

0

2T

0

2 dt)t(fT2dt)t(f

T1

f(t) = 2t1,01t0,t22

T = 4, o = 2 /T = /2

Irms2 =

1

0

321

0

2 )3/ttt(2dt)t1(442

Page 924: Solution Manual for Fundamentals of Electric Circuits

= 2(1 �– 1 + 1/3) = 2/3 or Irms = 0.8165 A (b) From Problem 16.14, an = [8/(n2 2)][1 �– cos(n /2)], ao = 0.5

a1 = 8/ 2, a2 = 4/ 2, a3 = 8/(9 2), a4 = 0, a5 = 9/(25 2), a6 = 4/(9 2)

Irms = 66623.08116

62564

81641664

21

41A

21a 4

1n

2no

Irms = 0.8162 A Chapter 17, Solution 47. Let I = IDC + I1 + I2

For the DC component

IDC = [5/(5 + 10)](3) = 1 A

Is

I

5

j8 10

For AC, = 100 j L = j100x80x10�–3 = j8

In = 5Is/(5 + 10 + j8)

For Is = 0.5 �–60 I1 = 10 �–60 /(15 + j8) or |I1| = 10/ 22 815

For Is = 0.5 �–120

I2 = 2.5 �–120 /(15 + j8) or |I2| = 2.5/ 22 815 p10 = (IDC

2 + |I1|2/2 + |I2|2/2)10 = (1 + [100/(2x289)] + [6.25/(2x289)])x10

p10 = 11.838 watts

Page 925: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 48. (a) For the DC component, i(t) = 20 mA. The capacitor acts like an open circuit so that v = Ri(t) = 2x103x20x10�–3 = 40

For the AC component, n = 10n, n = 1,2

1/(j nC) = �–j/(10nx100x10�–6) = (�–j/n) k Z = 2||(�–j/n) = 2(�–j/n)/(2 �– j/n) = �–j2/(2n �– j)

V = ZI = [�–j2/(2n �– j)]I

For n = 1, V1 = [�–j2/(2 �– j)]16 45 = 14.311 �–18.43 mV

For n = 2, V2 = [�–j2/(4 �– j)]12 �–60 = 5.821 �–135.96 mV

v(t) = 40 + 0.014311cos(10t �– 18.43 ) + 0.005821cos(20t �– 135.96 ) V

(b) p = VDCIDC + 1n

nnnn )cos(IV21

= 20x40 + 0.5x10x0.014311cos(45 + 18.43 ) +0.5x12x0.005821cos(�–60 + 135.96 )

= 800.1 mW Chapter 17, Solution 49.

(a) 5.2)5(21dt4dt1

21dt)t(z

T1Z

0

2T

0

2rms

2

581.1Zrms

(b)

9193.2...251

161

91

411

18

141

n

3621

41)ba(

21aZ

21n 1n

22n2

n2

o2

rms2

7086.1rmsZ

(c ) 071.8100x11.581

1.7086 %error

Page 926: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 50.

cn = T

0 ontj

1n2,dte)t(f

To

1

= 1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e�–jn tdt which leads to v = �–[1/(2jn )]e�–jn t

cn = 1

1

tjn1

1

tjn dtejn21e

jn2t

= 1

1

tjn222

tjnjn e)j(n2

1eenj

= [j/(n )]cos(n ) + [1/(2n2 2)](e�–jn �– ejn )

cn = n

)1(j)nsin(n2

j2n

)1( n

22

nj

Thus

f(t) = n

tjnn

oec =n

tjnn enj

)1(

Chapter 17, Solution 51.

T/2,2T o

20

T

0

2222

03

tjntjn2tojn

n 2tjn2tn)jn(

e21dtet

21dte)t(f

T1c

)jn1(n

2)n4jn4(n2j1c 22

2233n

n

tjn22

e)jn1(n

2)t(f

Page 927: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 52.

cn = T

0 ontj

1n2,dte)t(f

To

1

= 1 tjn dtte

21

Using integration by parts,

u = t and du = dt dv = e�–jn tdt which leads to v = �–[1/(2jn )]e�–jn t

cn = 1

1

tjn1

1

tjn dtejn21e

jn2t

= 1

1

tjn222

tjnjn e)j(n2

1eenj

= [j/(n )]cos(n ) + [1/(2n2 2)](e�–jn �– ejn )

cn = n

)1(j)nsin(n2

j2n

)1( n

22

nj

Thus

f(t) = n

tjnn

oec =n

tjnn enj

)1(

Chapter 17, Solution 53. o = 2 /T = 2 cn =

1

0

t)jn1(T

0

tjnt dtedtee oo

= 1en2j1

1en2j1

1 )n2j1(1

0

t)n2j1(

= [1/(j2n )][1 �– e�–1(cos(2 n) �– jsin(2n ))]

= (1 �– e�–1)/(1 + j2n ) = 0.6321/(1 + j2n

f(t) = n

tn2j

n2j1e6321.0

Page 928: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 54. T = 4, o = 2 /T = /2

cn = T

0

ntj dte)t(fT

o1

= 4

2

2/tjn2

1

2/tjn1

0

2/tjn dte1dte1dte241

= jnn2j2/jnjn2/jn eeee2e2n2j

= jn2/jn e23e3n2j

f(t) = n

tjnn

oec

Chapter 17, Solution 55. T = 2 , o = 2 /T = 1

cn = T

0

tjn dte)t(iT

o1

But i(t) = 2t,0

t0),tsin(

cn = 0

jntjtjt

0

tjn dte)ee(j2

121dte)tsin(

21

n11e

n11e

41

)n1(je

)n1(je

j41

)1n(j)n1(j

0

)n1(jt)n1(jt

Page 929: Solution Manual for Fundamentals of Electric Circuits

nne1enne1e)1n(4

1 )n1(j)n1(j)n1(j)n1(j2

But ej = cos( ) + jsin( ) = �–1 = e�–j

cn = )n1(2

e12neneee)1n(4

12

jnjnjnjnjn

2

Thus

i(t) = n

tjn2

jn

e)n1(2

e1

Chapter 17, Solution 56. co = ao = 10, o = co = (an �– jbn)/2 = (1 �– jn)/[2(n2 + 1)]

f(t) = 0n

n

tjn2

e)1n(2

)jn1(10

Chapter 17, Solution 57. ao = (6/�–2) = �–3 = co cn = 0.5(an �–jbn) = an/2 = 3/(n3 �– 2)

f(t) = 0n

n

nt50j3

e2n

33

Chapter 17, Solution 58. cn = (an �– jbn)/2, (�–1) = cos(n ), o = 2 /T = 1

cn = [(cos(n ) �– 1)/(2 n2)] �– j cos(n )/(2n)

Thus

f(t) = jnt2 e

n2)ncos(

jn2

1)ncos(4

Page 930: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 59. For f(t), T = 2 , o = 2 /T = 1. ao = DC component = (1x + 0)/2 = 0.5

For h(t), T = 2, o = 2 /T = . ao = (3x1 �– 2x1)/2 = 0.5

Thus by replacing o = 1 with o = and multiplying the magnitude by five, we obtain

h(t) = 1

0nn

t)1n2(j

)1n2(e5j

2

Chapter 17, Solution 60. From Problem 16.17, ao = 0 = an, bn = [2/(n )][1 �– 2 cos(n )], co = 0 cn = (an �– jbn)/2 = [j/(n )][2 cos(n ) �– 1], n 0. Chapter 17, Solution 61.

(a) o = 1.

f(t) = ao + )tncos(A non

= 6 + 4cos(t + 50 ) + 2cos(2t + 35 )

+ cos(3t + 25 ) + 0.5cos(4t + 20 ) = 6 + 4cos(t)cos(50 ) �– 4sin(t)sin(50 ) + 2cos(2t)cos(35 )

�– 2sin(2t)sin(35 ) + cos(3t)cos(25 ) �– sin(3t)sin(25 ) + 0.5cos(4t)cos(20 ) �– 0.5sin(4t)sin(20 )

= 6 + 2.571cos(t) �– 3.73sin(t) + 1.635cos(2t)

�– 1.147sin(2t) + 0.906cos(3t) �– 0.423sin(3t) + 0.47cos(4t) �– 0.171sin(4t)

Page 931: Solution Manual for Fundamentals of Electric Circuits

(b) frms = 1n

2n

2o A

21a

frms

2 = 62 + 0.5[42 + 22 + 12 + (0.5)2] = 46.625 frms = 6.828 Chapter 17, Solution 62.

(a) s3141.0202TT/220o

(b) f ...)90t40cos(1.5)90t20cos(43)tncos(Aa)t( o

1n

onono

....t80sin8.1t60sin7.2t40sin1.5t20sin43)t(f

Chapter 17, Solution 63.

This is an even function.

T = 3, o = 2 /3, bn = 0.

f(t) = 5.1t1,2

1t0,1

ao = 5.1

1

1

0

2/T

0dt2dt1

32dt)t(f

T2 = (2/3)[1 + 1] = 4/3

an = 5.1

1

1

0

2/T

0 o dt)3/tn2cos(2dt)3/tn2cos(134dt)tncos()t(f

T4

= 5.1

1

1

0 3tn2sin

n26

3tn2sin

n23

34

= [�–2/(n )]sin(2n /3)

Page 932: Solution Manual for Fundamentals of Electric Circuits

f2(t) = 1n 3

tn2cos3n3sin

n12

34

ao = 4/3 = 1.3333, o = 2 /3, an = �–[2/(n )]sin(2n t/3)

An = 3n2sin

n2b2

n2na

A1 = 0.5513, A2 = 0.2757, A3 = 0, A4 = 0.1375, A5 = 0.1103

The amplitude spectra are shown below. 1.333

0.1103 0.1378

0

0.275

0.551

4321n

5 0

An

Chapter 17, Solution 64.

The amplitude and phase spectra are shown below.

Page 933: Solution Manual for Fundamentals of Electric Circuits

An 3.183 2.122 1.591 0.4244 0 2 4 6 n 0 2 4 6

-180o

Chapter 17, Solution 65. an = 20/(n2 2), bn = �–3/(n ), n = 2n

An = 22442n n

9n400b2

na

= 22n44.441

n3 , n = 1, 3, 5, 7, 9, etc.

n An 1 2.24 3 0.39 5 0.208 7 0.143 9 0.109

Page 934: Solution Manual for Fundamentals of Electric Circuits

n = tan�–1(bn/an) = tan�–1{[�–3/(n )][n2 2/20]} = tan�–1(�–nx0.4712)

n n 1 �–25.23 3 �–54.73 5 �–67 7 �–73.14 9 �–76.74

�–90

0 10

n

n

14 62 18

�–25.23

�–54.73

�–67

�–76.74 �–73.14

�–60

�–30

�–90

2.24

An

n

0.0.143 0.109 0.208

0.39

18 14 10 6 2 0

Chapter 17, Solution 66. The schematic is shown below. The waveform is inputted using the attributes of VPULSE. In the Transient dialog box, we enter Print Step = 0.05, Final Time = 12, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, the output plot is shown below. The output file includes the following Fourier components.

Page 935: Solution Manual for Fundamentals of Electric Circuits

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)

DC COMPONENT = 5.099510E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 3.184E+00 1.000E+00 1.782E+00 0.000E+00 2 1.000E+00 1.593E+00 5.002E-01 3.564E+00 1.782E+00 3 1.500E+00 1.063E+00 3.338E-01 5.347E+00 3.564E+00 4 2.000E+00 7.978E-01 2.506E-01 7.129E+00 5.347E+00 5 2.500E+00 6.392E-01 2.008E-01 8.911E+00 7.129E+00 6 3.000E+00 5.336E-01 1.676E-01 1.069E+01 8.911E+00 7 3.500E+00 4.583E-01 1.440E-01 1.248E+01 1.069E+01 8 4.000E+00 4.020E-01 1.263E-01 1.426E+01 1.248E+01 9 4.500E+00 3.583E-01 1.126E-01 1.604E+01 1.426E+01 TOTAL HARMONIC DISTORTION = 7.363360E+01 PERCENT

hapter 17, Solution 67.

he Schematic is shown below. In the Transient dialog box, we type �“Print step = 0.01s, Final time = 36s, Center frequency = 0.1667, Output vars = v(1),�” and click Enable Fourier. After simulation, the output file includes the following Fourier components,

C T

Page 936: Solution Manual for Fundamentals of Electric Circuits

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 2.000396E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED

PHASE (DEG)

.000E+00 -8.996E+01 0.000E+00

NO (HZ) COMPONENT COMPONENT (DEG)

1 1.667E-01 2.432E+00 12 3.334E-01 6.576E-04 2.705E-04 -8.932E+01 6.467E-01 3 5.001E-01 5.403E-01 2.222E-01 9.011E+01 1.801E+02 4 6.668E-01 3.343E-04 1.375E-04 9.134E+01 1.813E+02 5 8.335E-01 9.716E-02 3.996E-02 -8.982E+01 1.433E-01 6 1.000E+00 7.481E-06 3.076E-06 -9.000E+01 -3.581E-02 7 1.167E+00 4.968E-02 2.043E-02 -8.975E+01 2.173E-01 8 1.334E+00 1.613E-04 6.634E-05 -8.722E+01 2.748E+00 9 1.500E+00 6.002E-02 2.468E-02 9.032E+01 1.803E+02

TAL HARMON DISTORT N = 2.280 E+01 PERC T TO IC IO 065 EN

Page 937: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 68. The schematic is shown below. We set the final time = 6T=12s and the center frequency = 1/T = 0.5. When the schematic is saved and run, we obtain the Fourier series from the output file as shown below.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 1.990000E+00 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.273E+00 1.000E+00 9.000E-01 0.000E+00 2 1.000E+00 6.367E-01 5.001E-01 -1.782E+02 1.791E+02 3 1.500E+00 4.246E-01 3.334E-01 2.700E+00 1.800E+00 4 2.000E+00 3.185E-01 2.502E-01 -1.764E+02 -1.773E+02 5 2.500E+00 2.549E-01 2.002E-01 4.500E+00 3.600E+00 6 3.000E+00 2.125E-01 1.669E-01 -1.746E+0 -1.755E+02 7 3.500E+00 1.823E-01 1.431E-01 6.300E+00 5.400E+00 8 4.000E+00 1.596E-01 1.253E-01 -1.728E+02 -1.737E+02 9 4.500E+00 1.419E-01 1.115E-01 8.100E+00 7.200E+00 Chapter 17, Solution 69.

Page 938: Solution Manual for Fundamentals of Electric Circuits

The schematic is shown below. In the Transient dialog box, set Print Step = 0.05 s, Final Time = 120, Center Frequency = 0.5, Output Vars = V(1) and click enable Fourier. After simulation, we obtain V(1) as shown below. We also obtain an output file which includes the following Fourier components.

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 5.048510E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 4.056E-01 1.000E+00 -9.090E+01 0.000E+00 2 1.000E+00 2.977E-04 7.341E-04 -8.707E+01 3.833E+00 3 1.500E+00 4.531E-02 1.117E-01 -9.266E+01 -1.761E+00 4 2.000E+00 2.969E-04 7.320E-04 -8.414E+01 6.757E+00 5 2.500E+00 1.648E-02 4.064E-02 -9.432E+01 -3.417E+00 6 3.000E+00 2.955E-04 7.285E-04 -8.124E+01 9.659E+00 7 3.500E+00 8.535E-03 2.104E-02 -9.581E+01 -4.911E+00 8 4.000E+00 2.935E-04 7.238E-04 -7.836E+01 1.254E+01 9 4.500E+00 5.258E-03 1.296E-02 -9.710E+01 -6.197E+00 TOTAL HARMONIC DISTORTION = 1.214285E+01 PERCENT

Page 939: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 70. The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s, Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components.

Page 940: Solution Manual for Fundamentals of Electric Circuits

FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) DC COMPONENT = 7.658051E-01 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00 2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 -4.928E+01 3 1.500E+00 2.111E-01 1.973E-01 -3.985E+01 -4.990E+01 4 2.000E+00 1.247E-01 1.166E-01 -5.870E+01 -6.874E+01 5 2.500E+00 8.538E-02 7.980E-02 -5.680E+01 -6.685E+01 6 3.000E+00 6.139E-02 5.738E-02 -6.563E+01 -7.567E+01 7 3.500E+00 4.743E-02 4.433E-02 -6.520E+01 -7.524E+01 8 4.000E+00 3.711E-02 3.469E-02 -7.222E+01 -8.226E+01 9 4.500E+00 2.997E-02 2.802E-02 -7.088E+01 -8.092E+01 TOTAL HARMONIC DISTORTION = 4.352895E+01 PERCENT

Page 941: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 71. The schematic is shown below. We set Print Step = 0.05, Final Time = 12 s, Center Frequency = 0.5, Output Vars = I(1), and click enable Fourier in the Transient dialog box. After simulation, the output waveform is as shown. The output file includes the following Fourier components.

Page 942: Solution Manual for Fundamentals of Electric Circuits

FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L1) DC COMPONENT = 8.374999E-02 HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 5.000E-01 2.287E-02 1.000E+00 -6.749E+01 0.000E+00 2 1.000E+00 1.891E-04 8.268E-03 8.174E+00 7.566E+01 3 1.500E+00 2.748E-03 1.201E-01 -8.770E+01 -2.021E+01 4 2.000E+00 9.583E-05 4.190E-03 -1.844E+00 6.565E+01 5 2.500E+00 1.017E-03 4.446E-02 -9.455E+01 -2.706E+01 6 3.000E+00 6.366E-05 2.783E-03 -7.308E+00 6.018E+01 7 3.500E+00 5.937E-04 2.596E-02 -9.572E+01 -2.823E+01 8 4.000E+00 6.059E-05 2.649E-03 -2.808E+01 3.941E+01 9 4.500E+00 2.113E-04 9.240E-03 -1.214E+02 -5.387E+01 TOTAL HARMONIC DISTORTION = 1.314238E+01 PERCENT Chapter 17, Solution 72. T = 5, o = 2 /T = 2 /5 f(t) is an odd function. ao = 0 = an

bn = 2/T

0

10

0o dt)tn4.0sin(1054dt)tnsin()t(f

T4

= 1

0

)nt4.0cos(n2

5x8 = )]n4.0cos(1[n20

f(t) = 1n

)tn4.0sin()]n4.0cos(1[n120

Chapter 17, Solution 73.

p = R

V21

RV 2

n2DC

= 0 + 0.5[(22 + 12 + 12)/10] = 300 mW

Page 943: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 74.

(a) An = 2n

2n ba , = tan�–1(bn/an)

A1 = 22 86 = 10, 1 = tan�–1(6/8) = 36.87 A2 = 22 43 = 5, 2 = tan�–1(3/4) = 36.87

i(t) = {4 + 10cos(100 t �– 36.87 ) �– 5cos(200 t �– 36.87 )} A

(b) p = RI5.0RI 2n

2DC

= 2[42 +0.5(102 + 52)] = 157 W Chapter 17, Solution 75. The lowpass filter is shown below. R + + vs C vo - -

1nos tncos

Tnsin

n1

TA2

TAv

T/n2n,VRCj1

1V

Cj1R

Cj1

V onsn

s

n

no

For n=0, (dc component), TAVsoV (1)

Page 944: Solution Manual for Fundamentals of Electric Circuits

For the nth harmonic,

o

n122

n2o 90

Tnsin

nTA2

RCtanCR1

1V

When n=1, 22

2o

CRT

41

1T

nsinTA2|V| (2)

From (1) and (2),

4222

222

10x09.39.30CRT

41

CRT

41

110

sinTA2x50

TA

mF 59.2410x4

10x09.3x1010R2

TC10CRT

413

4251022

2

Chapter 17, Solution 76.

vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10. Hence

vo(t) = 1k

)tnsin(n1205 , n = 2k �– 1

T = 2, o = 2 /T = 2 , n = n o = 2n

j nL = j2n ; Z = R||10 = 10R/(10 + R) Vo = ZVs/(Z + j2n ) = [10R/(10R + j2n (10 + R))]Vs

Vo = s2222

1

V)R10(n4R100

)}R10)(R5/n{(tanR10

Vs = [20/(n )] 0

The source current Is is

Is = )R10(n2jR10

n20)R10(

n2jR10

R10V

n2jZV ss

Page 945: Solution Manual for Fundamentals of Electric Circuits

= 2222

1

)R10(n4R100

)}R10)(3/n{(tann20)R10(

ps = VDCIDC + )cos(IV21

nnsnsn

For the DC case, L acts like a short-circuit.

Is = R10

)R10(5

R10R

510

, Vs = 5 = Vo

222

122

222

12

s

)R10(16R100

)R10(5

2tancos)R10(10

)R10(4R100

)R10(5

tancos)R10(20

21

R10)R10(25p

ps = 1n

onDC

RV

21

RV

= 222222 )R10(10R100R100

)R10(4R100R100

21

R25

We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we consider only the DC component as an approximation. In fact the DC component has the latgest share of the power for both input and output signals.

R10

)R10(25x107

R25

100 = 70 + 7R which leads to R = 30/7 = 4.286

Page 946: Solution Manual for Fundamentals of Electric Circuits

Chapter 17, Solution 77.

(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest common factor is 2. Hence o = 2.

T = 2 / o = 2 /2 =

(b) The average value is the DC component = �–2

(c) Vrms = 1n

2n

2no )ba(

21a

)136810(21)2( 2222222

rmsV = 121.5

Vrms = 11.02 V

Chapter 17, Solution 78.

(a) p = R

VR

VR

V21

RV 2

rms,n2DC

2n

2DC

= 0 + (402/5) + (202/5) + (102/5) = 420 W

(b) 5% increase = (5/100)420 = 21

pDC = 21 W = 105R21VtoleadswhichR

V 2DC

2DC

VDC = 10.25 V

Chapter 17, Solution 79. From Table 17.3, it is evident that an = 0,

bn = 4A/[ (2n �– 1)], A = 10.

A Fortran program to calculate bn is shown below. The result is also shown.

Page 947: Solution Manual for Fundamentals of Electric Circuits

C FOR PROBLEM 17.79 DIMENSION B(20) A = 10 PIE = 3.142 C = 4.*A/PIE DO 10 N = 1, 10 B(N) = C/(2.*FLOAT(N) �– 1.) PRINT *, N, B(N) 10 CONTINUE STOP END

n bn 1 12.731 2 4.243 3 2.546 4 1.8187 5 1.414 6 1.1573 7 0.9793 8 0.8487 9 0.7498 10 0.67

Chapter 17, Solution 80. From Problem 17.55, cn = [1 + e�–jn ]/[2 (1 �– n2)]

This is calculated using the Fortran program shown below. The results are also shown.

C FOR PROBLEM 17.80 COMPLEX X, C(0:20) PIE = 3.1415927

A = 2.0*PIE DO 10 N = 0, 10 IF(N.EQ.1) GO TO 10 X = CMPLX(0, PIE*FLOAT(N)) C(N) = (1.0 + CEXP(�–X))/(A*(1 �– FLOAT(N*N))) PRINT *, N, C(N)

10 CONTINUE STOP END

Page 948: Solution Manual for Fundamentals of Electric Circuits

n cn 0 0.3188 + j0 1 0 2 �–0.1061 + j0 3 0 4 �–0.2121x10�–1 + j0 5 0 6 �–0.9095x10�–2 + j0 7 0 8 �–0.5052x10�–2 + j0 9 0 10 �–0.3215x10�–2 + j0

Chapter 17, Solution 81. (a)

A

3T 2TT0

f(t) = 1n

o2 )tncos(1n4

1A4A2

The total average power is pavg = Frms

2R = Frms2 since R = 1 ohm.

Pavg = Frms2 =

T

0

2 dt)t(fT1 = 0.5A2

(b) From the Fourier series above |co| = 2A/2, |cn| = 4A/[ (4n2 �– 1)]

Page 949: Solution Manual for Fundamentals of Electric Circuits

n o |cn| 2|cn|2 % power 0 0 2A/ 4A2/( 2) 81.1% 1 2 o 2A/(3 ) 8A2/(9 2) 18.01% 2 4 o 2A/(15 ) 2A2/(225 2) 0.72% 3 6 o 2A/(35 ) 8A2/(1225 2) 0.13% 4 8 o 2A/(63 ) 8A2/(3969 2) 0.04%

(c) 81.1% (d) 0.72% Chapter 17, Solution 82.

P = 1n

2n

2DC

RV

21

RV

Assuming V is an amplitude-phase form of Fourier series. But

|An| = 2|Cn|, co = ao |An|2 = 4|Cn|2

Hence,

P = 1n

2n

2o

Rc

R2

c

Alternatively,

P = R

V

2rms

where

n

2n

1n

2n

2o

1n

2n

2o

2rms cc2cA

21aV

= 102 + 2(8.52 + 4.22 + 2.12 + 0.52 + 0.22)

= 100 + 2x94.57 = 289.14

P = 289.14/4 = 72.3 W

Page 950: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 1. )2t()1t()1t()2t()t('f

2jjj2j eeee)(Fj cos22cos2

F( ) = j

]cos2[cos2

Chapter 18, Solution 2.

otherwise,0

1t0,t)t(f

- (t-1)

(t)

f �”(t)

t �– �’(t-1)

1

- (t-1)

0

1

f �‘(t)

t

f"(t) = (t) - (t - 1) - '(t - 1) Taking the Fourier transform gives

- 2F( ) = 1 - e-j - j e-j

F( ) = 2

j 1e)j1(

or F 1

0

tj dtet)(

But c)1ax(aedxex 2

axax

102

j

)1tj(j

e)(F 1ej11 j

2

Page 951: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 3.

2t2,21)t('f,2t2,t

21)t(f

2

2

222

tjtj )1tj(

)j(2edtet

21)(F

)12j(e)12j(e2

1 2j2j2

2j2j2j2j2 eeee2j

21

2sin2j2cos4j2

12

F( ) = )2cos22(sinj2

Chapter 18, Solution 4.

1

�–2 (t�–1)

2 (t+1)

�–1

g�’ 2

0 �–2

t

�–2 �’(t�–1)

�–2 (t+1)1

�–2 (t�–1)

2 �’(t+1)

�–1

g�”

4 (t)

0 �–2

t

Page 952: Solution Manual for Fundamentals of Electric Circuits

4sin4cos4ej2e24ej2e2)(G)j(

)1t(2)1t(2)t(4)1t(2)1t(2g

jjjj2

)1sin(cos4)(G2

Chapter 18, Solution 5.

1

0

h�’(t)

�–1

�–2 (t)

1 t

(t+1)

�– (t�–1)

1

0

h�”(t)

�–1

�–2 �’(t)

1 t

j2sinj2j2ee)(H)j(

)t(2)1t()1t()t(h

jj2

H( ) = sinj2j22

Page 953: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 6.

dtetdte)1()(F tj0

1

1

0

tj

)1(cos1tsinttcos1tsin1

tdtcosttdtcos)(F Re

2102

01

0

1

1

0

Chapter 18, Solution 7. (a) f1 is similar to the function f(t) in Fig. 17.6. )1t(f)t(f1

Since 2F j

)1(cos)(

)(Fe)(F j1

j)1(cose2 j

Alternatively, )

)2t()1t(2)t()t(f '

1

e2e(eee21)(Fj jjj2jj1

)2cos2(e j

F1( ) = j

)1e j (cos2

(b) f2 is similar to f(t) in Fig. 17.14.

f2(t) = 2f(t)

F2( ) = 2

)cos1(4

Page 954: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 8.

(a) 21

tj2

21

tj10

tj

2

1

tj1

0

tj

)1tj(e2ej4e

j2

dte)t24(dte2)(F

2j

22jj

2e)2j1(2e

j4

j2e

j22)(F

(b) g(t) = 2[ u(t+2) �– u(t-2) ] - [ u(t+1) �– u(t-1) ]

sin22sin4)(G

Chapter 18, Solution 9.

(a) y(t) = u(t+2) �– u(t-2) + 2[ u(t+1) �– u(t-1) ]

sin42sin2)(Y

(b) )j1(e22)1tj(e2dte)t2()(2

j

210

1

02

tjtjZ

Chapter 18, Solution 10.

(a) x(t) = e2tu(t) X( ) = 1/(2 + j )

(b) 0t,e

0t,ee

t

t)t(

1

1

0

1

1

0

tjttjttj dteedteedte)t(y)(Y

Page 955: Solution Manual for Fundamentals of Electric Circuits

10

t)j1(0

1

t)j1(

)j1(e

j1e

j1sinjcos

j1sinjcose

12 1

2

Y( ) = )sin(cose11

2 12

Chapter 18, Solution 11. f(t) = sin t [u(t) - u(t - 2)]

2

0

2

0

tjtjtjtj dteeej2

1dtetsin)(F

2

0

t)(jt)(j dt)ee(j2

1

20

t)(j20

t)(j

)(jee

)(j1

j21

2j2j e1e1

21

2j22 e22)(2

1

F( ) = 1e 2j22

Chapter 18, Solution 12.

(a) F dtedtee)(0

2

0

t)j1(tjt

20

t)j1(ej1

1 j1

1e 2j2

Page 956: Solution Manual for Fundamentals of Electric Circuits

(b) 0

1

1

0

tjtj dte)1(dte)(H

)cos22(j11e

j1e1

j1 jj

j

2/sin4 2 2

2/2/sin

j

Chapter 18, Solution 13. (a) We know that )]a()a([]at[cosF .

Using the time shifting property,

)a(e)a(e)]a()a([e)]a3/t(a[cos 3/j3/ja3/jF

(b) sin tsinsintcoscostsin)1t( g(t) = -u(t+1) sin (t+1)

Let x(t) = u(t)sin t, then 22 1

1

1)j(

1)(X

Using the time shifting property,

1

ee1

1)(G2

jj

2

(c ) Let y(t) = 1 + Asin at, then Y )]a()a([Aj)(2)(

h(t) = y(t) cos bt

Using the modulation property,

)]b(Y)b(Y[21)(H

)ba()ba()ba()ba(2Aj)b()b()(H

Page 957: Solution Manual for Fundamentals of Electric Circuits

(d) )14j(ej

e1)1tj(ej

edte)t1()(2

4j4j

2402

tj4

0

tjtjI

Chapter 18, Solution 14.

(a) )t3cos()0(t3sin)1(t3cossint3sincost3cos)t3cos( (f )t(ut3cose)t t

F( ) = 9j1

j12

(b)

)1t(u)1t(utcos)t('g

g(t)

t 1 -1

-1

1

-

-1 1

g�’(t)

t

)1t()1t()t(g)t("g 2 jj22 ee)(G)(G

sinj2)ee()(G jj22

G( ) = 22

sinj2

Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F( ) = G( )e-j

Page 958: Solution Manual for Fundamentals of Electric Circuits

jj

22j eee)(FG

22

sin2j

G( ) = 22

sinj2

(c) tcos)0(tsin)1(tcossintsincostcos)1t(cos Let ex )t(he)1t(u)1t(cos)t( 2)1t(2

and )t(u)tcos(e)t(y t2

22)j2(j2)(Y

)1t(x)t(y je)(X)(Y

22

j

j2ej2)(X

)(He)(X 2

)(Xe)(H 2

= 22

2j

j2ej2

(d) Let x )t(y)t(u)t4sin(e)t( t2

)t(x)t(pwhere )t(ut4sine)t(y t2

22 4j2

j2)(Y

16j2

j2)(Y)(X 2

Page 959: Solution Manual for Fundamentals of Electric Circuits

)(X)(p162j

2j2

(e) 2j2j ej1)(23e

j8)(Q

Q( ) = 2j2j e)(23ej6

Chapter 18, Solution 15. (a) F 3j3j ee)( 3sinj2

(b) Let g je2)(G),1t(2)t(

)(F F t

dt)t(g

)()0(Fj

)(G

)()1(2je2 j

= je j2

(c) F 121)t2(

j211

31)(F

2j

31

Page 960: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 16.

(a) Using duality properly

2

2t

2t

22

or 4t42

F( ) = F 2t4 4

(b) tae 22aa2

22 taa2 ae2

22 ta8 2e4

G( ) = F 2t48 2e4

Chapter 18, Solution 17.

(a) Since H( ) = F 000 FF21)t(ftcos

where F( ) = F 2,j1tu 0

2j12

2j12

21H

Page 961: Solution Manual for Fundamentals of Electric Circuits

2222

2j22

2

H( ) = 4

j22

2 2

(b) G( ) = F 000 FF2j)t(ftsin

where F( ) = F j1tu

10j110

10j110

2jG

10

j10

j2j1010

2j

= 100

101010

2 2

j

Chapter 18, Solution 18.

Let f tuet t

jj1F

tcostf 1F1F21

Hence 1j1

11j1

121Y

1j11j1jj1jj1

21

1jjjj1

j12

= 2j2

j12

Page 962: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 19.

dteee21dte)t(fF tj1

0

t2jt2jtj

1

0

t2jt2j dtee21F

1

0

t2jt2j e2j

1e2j

121

2j

1e2j

1e21 2j2j

But 2j2j e12sinj2cose

2

12

1j

1e21F

j

= 1e4

j j22

Chapter 18, Solution 20.

(a) F (cn) = cn ( ) F on

tjnn ncec o

F n

tjnn

oec n

on nc

(b) 2T 1T2

o

T

0 0

jnttjnn 0dte1

21dtetf

T1c o

Page 963: Solution Manual for Fundamentals of Electric Circuits

1en2

jejn1

21 jn

0jnt

But e njn )1(ncosnsinjncos

evenn,0

0n,oddn,n

jn

n 11n2jc

for n = 0

0n 2

1dt121c

Hence

oddn0n

n

jntenj

21)t(f

F( ) =

oddn0n

n

nnj

21

Chapter 18, Solution 21.

Using Parseval�’s theorem,

d|)(F|21dt)t(f 22

If f(t) = u(t+a) �– u(t+a), then

da

asina421a2dt)1(dt)t(f

22a

a22

or

aa4

a4da

asin2

2 as required.

Page 964: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 22.

F dtej2ee)t(ftsin)t(f tj

tjtj

o

oo

dtedte)t(fj2

1 tjtj oo

= oo FFj2

1

Chapter 18, Solution 23.

(a) f(3t) leads to j15j6

303/j53/j2

1031

F t3fj15j6

30

(b) f(2t) j10j4

202/j152/j2

1021

f(2t-1) = f [2(t-1/2)] j10j4

e20 2/j

(c) f(t) cos 2t 2F212F

21

= 2j52j2

52j52j2

5

(d) F j5j2

10jFjt'f

(e) f t

dtt 0FjF

Page 965: Solution Manual for Fundamentals of Electric Circuits

5x2

10xj5j2j

10

= j5j2j

10

Chapter 18, Solution 24. (a) FX + F [3]

= 1e jj6

(b) 2tfty

FeY 2j 1eje j2j

(c) If h(t) = f '(t)

1ejjFjH j je1

(d) 53F

53x10

23F

23x4)(G,t

35f

3ft 10t24g

1e

53

j61e

23

j6 5/3j2/3j

= 1e10j1e4j 5/3j2/3j

Page 966: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 25.

(a) js,2s

Bs2ss

s A10F

52

10B,52

10A

2j

5j5F

f(t) = tue5t2

t2sgn5

(b) 2j

B1j

A2j1j

4jF

js,2s

B1s

A2s1s

4ssF

A = 5, B = 6

j2

6j15F

f(t) = tue6e5 t2t

Chapter 18, Solution 26.

(a) )t(ue)t( )2t(f (b) )t(ute)t( t4h

(c) If sin2)(X)1t(u)1t(u)t(x

By using duality property,

Page 967: Solution Manual for Fundamentals of Electric Circuits

ttsin2)t(g)1(u2)1(u2)(G

Chapter 18, Solution 27.

(a) Let js,10s

Bs10ss

s A100F

1010

100B,1010100A

10j

10j10F

f(t) = tue10tsgn t105

(b) js,3s

Bs2

As3s2

s10sG

6530B,4

520A

3j

62j

4G

g(t) = tue6tue t3t24

(c) 90020j

60130040jj

60H 22

h(t) = tu)t30sin(e t202

21

21

1jj2de

21ty

tj

41

Page 968: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 28.

(a) d)j2)(j5(

e)(21de)(F

21)t(f

tjtj

201

)2)(5(1

21 0.05

(b) )12j)(2j(

e210de

)1j(j)2(10

21)t(f

t2jtj

2j1e

25j t2j

2

e)j2( t2j

(c) )j3)(j2(

e220d

)53)(j2(e)1(20

21)t(f

jttj

)j55(2e20 jt

jte)j1(

(d) Let )(F)(F)j5(j

5)j5()(5)( 21F

5.051

25de

j5)(5

21)t(f tj

1

1B,1A,5s

BsA

)s5(s5)s(2F

5j

1j1)(F2

t5t52 e)t(u

21e)tsgn(

21)t(f

)t(f)t(f)t(f 21 u t5e)t(

Page 969: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 29.

(a) f(t) = F -1[ )]( F -1 )3(4)3(4

t3cos421 = t3cos81

21

(b) If )2t(u)2t(u)t(h

2sin2)(H

)(H4)(G t

t2sin821)t(g

g(t) = t

t2sin4

(c) Since

cos(at) ↔ )a()a( Using the reversal property,

)2t()2t(2cos2 or F -1 2cos6 )2t(3)2t(3

Chapter 18, Solution 30.

(a) ja

1)(X,j2)(Y)tsgn()t(y

)]t(u)t(u[a)t(2)t(hja22

j)ja(2

)(X)(Y)(H

(b) j2

1)(Y,j1

1)(X

)t(ue)t()t(hj2

11j2j1)(H t2

(c ) In this case, by definition, )t(u btsine)t(y)t(h at

Page 970: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 31.

(a) ja

1)(H,)ja(

1)(Y2

)t(ue)t(xja

1)(H)(Y)(X at

(b) By definition, )1t(u)1t(u )t(y)t(x

(c ) j2)(H,

)ja(

1)(Y

)t(ue2a)t(

21)t(x

)ja(2a

21

)ja(2j

)(H)(Y)(X at

Chapter 18, Solution 32.

(a) Since 1j

je )1t(ue 1t

and F f(-t)

1j

e j

1F 1tuetf 1t1

f1(t) = e 1tu1t

(b) From Section 17.3,

1t

22 e2

If ,e2F2 then

f2(t) = 1t

22

Page 971: Solution Manual for Fundamentals of Electric Circuits

(b) By partial fractions

1j

41

1j41

1j41

1j41

1j1j1F 22223

Hence tueteete41t tttt

3f

= tue1t41

tue1t4

tt1

(d) 2j1e

21deF

21tf

tjtj

14 = 21

Chapter 18, Solution 33. (a) Let 1tu1tutsin2tx

From Problem 17.9(b),

22

sinj4X

Applying duality property,

22 ttsinj2tX

21tf

f(t) = 22ttsinj2

(b) sinjcosj2sinj2cosjF

j

ej

eeej 2jjj2j

2tsgn211tsgn

21tf

But sgn( 1)t(u2)t

Page 972: Solution Manual for Fundamentals of Electric Circuits

212tu

211tutf

= u 2tu1t Chapter 18, Solution 34. First, we find G( ) for g(t) shown below.

1tu1tu102tu2tu10tg 1t1t102t2t10t'g The Fourier transform of each term gives

20

t

g(t)

10

0

�–2 �–1 1 2

10 (t+2) 10 (t+1)

�–1 �–2 1 2t

g �‘(t)

0

�–10 (t-2) �–10 (t-1)

jj2j2j ee10ee10Gj sinj202sinj20

sin202sin20G 40 sinc(2 ) + 20 sinc( )

Note that G( ) = G(- ).

Page 973: Solution Manual for Fundamentals of Electric Circuits

G2F

tG21tf

= (20/ )sinc(2t) + (10/ )sinc(t) Chapter 18, Solution 35. (a) x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties,�’�’

)j6(ee

3/j21

31)(X

3/j3/j

(b) Using the modulation property,

3j1

7j1

21

)5(j21

)5(j21

21)]5(F)5(F[

21)(Y

(c ) j2

j)(Fj)(Z

(d) 2)j2(

1)(F)(F)(H

(e) 22 )j2(1

)j2()j0(j)(F

ddj)(I

Chapter 18, Solution 36.

i

o

VV

H

j2V10

VHV iio

Page 974: Solution Manual for Fundamentals of Electric Circuits

(a) vi = 4 (t) Vi( ) = 4

j2

40Vo

tue40)t(v t2

o

vo(2) = 40e�–4 = 0.7326 V

(b) tue6v ti j1

6Vi

j1j2

60Vo

js,2s

B1s

A1s2s

60sVo

601

60B,60160A

j2

60j1

60Vo

)t(uee60)t(v t2t

o

01831.013533.060ee60)2(v 42o

= 7.021 V (c) vi(t) = 3 cos 2t

Vi( ) = [ ( + 2) + ( - 2)]

j2

2210Vo

deV21)t(v tj

oo

dj2

e25dej225

tjtj

Page 975: Solution Manual for Fundamentals of Electric Circuits

45t2j45t2jt2jt2j

ee22

52j2

e52j2

e5

45t2cos2

5

4518.229cos2

5454cos2

52vo

vo(2) = �–3.526 V Chapter 18, Solution 37.

j2

2jj2

By current division,

4j82j

2j

j22j4

j22j

II

Hs

o

H( ) = 3j4

j

Chapter 18, Solution 38.

j1Vi

j1

j55V

2j1010V io

Let j5j

5j5

521o VVV

Page 976: Solution Manual for Fundamentals of Electric Circuits

5s

BsA

5ss5V2 A = 1, B = -1, s = j

j5

1j1V2 t5

2 e)tsgn(21tv

j5

5V1 dej5

521tv tj

1

v1(t) 5.051

25

t5

210 etsgn5.05.0tvtvtv But tu21tsgn

tuetu5.05.0t t5

ov tuetu t5 Chapter 18, Solution 39.

j22

tjs e11

j1dte)t1()(V

j226

3

33s e11

j1

j10

10

10xj10

)(V)(I

Chapter 18, Solution 40. )2t()1t(2)t()t(v 2jj2 ee21)(V

2

2jj ee21V

Now j

2j1j12Z

Page 977: Solution Manual for Fundamentals of Electric Circuits

2j1

j1ee2ZVI 2

2jj

j2j ee5.05.0j5.0j

1

But 5.0s

BsA

)5.0s(s1 A = 2, B = -2

j2jj2j ee5.05.0j5.0

2ee5.05.0j2I

i(t) = )1t(ue2)2t(ue)t(ue)1tsgn()2tsgn(21

)tsgn(21 )1t(5.0)2t(5.0t5.0

Chapter 18, Solution 41. 2

j21

+

1/s 0.5s

+

V

2

j29j4

j2j4jV42j

02jV2Vj

2j2

1V

22

V( ) = )j24)(j2(

)2j5.4(j22

Page 978: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 42.

By current division, Ij2

2Io

(a) For i(t) = 5 sgn (t),

)j2(j20

j10

j22I

j10I

o

Let 10B,10A,2s

BsA

)2s(s20

oI

j2

10j10

oI

io(t) = 5 A)t(ue10)tsgn( t2 (b)

)1t(4)t(4)t('i

4 i(t)

t 1

�–4 (t�–1)

4 (t)

i�’(t)

t

1

je44Ij

je14I

j

j

j

o e1j2

1j14

)j2(je18I

j2

e4je4

j24

j4 jj

Page 979: Solution Manual for Fundamentals of Electric Circuits

io(t) = A1tue4)t(ue4)1tsgn(2)tsgn(2 )1t(2t2 Chapter 18, Solution 43.

j51Ie5i,

j50

10x20j

1Cj

1 mF 20 st

s3

js,)5s)(25.1s(

50j50I

j5040

40V so

5j1

25.1j1

340

5sB

25.1sAVo

)t(u)ee(340)t(v t5t25.1

o

Chapter 18, Solution 44. 1H j

We transform the voltage source to a current source as shown in Fig. (a) and then combine the two parallel 2 resistors, as shown in Fig. (b).

(a)

+

Vo

1 Vs/2

2

Io

+

Vo j2 Vs/2

(b)

Io

j

,122 2

Vj1

1 soI

)j1(2Vj

IjV soo

Page 980: Solution Manual for Fundamentals of Electric Circuits

)2t(10t10)t(vs 2j

s e1010Vj

sVj

e110 2j

Hence 2j2j

o ej1

5j1

5j1

e15V

)2t(ue5)t(ue5)t(v )2t(t

o

01e5)1(v 1o 1.839 V

Chapter 18, Solution 45.

)t(ute2)t(v)1j(

2)2(

j1j2

j1

V to2o

Chapter 18, Solution 46.

F41 4j

41j

1

2H j 2 3 )t(3

)t(ue t

j11

The circuit in the frequency domain is shown below: 2 Vo

Io( )

+

3

+

1/(1+j )

�–j4/

j2

Page 981: Solution Manual for Fundamentals of Electric Circuits

At node Vo, KCL gives

2jV

4jV3

2

Vj1

1oo

o

ooo

V2jVj3jV2j1

2

2jj2

3jj1

2

Vo

2jj22j

j133j2

2jV

I

2

oo

Io( ) = )28(j64

3j32

222

Chapter 18, Solution 47.

Transferring the current source to a voltage source gives the circuit below:

1/(j )2

+ Vo

+

1

8 V j /2

Page 982: Solution Manual for Fundamentals of Electric Circuits

Let j23j4

2j1

2j

22j12inZ

By voltage division,

j23j4j1

8Zj1

88Z

j1

j1

Vin

in

o

= 234jj2j28

= 235j2j28

Chapter 18, Solution 48.

0.2F 5jCj

1

As an integrator,

4.010x20x10x20RC 63

t

o io dtvRC1v

)0(VjV

RC1V i

io

j2j

24.0

1

j2j2125.0mA

20VI o

o

125.0j2

125.0j125.0

Page 983: Solution Manual for Fundamentals of Electric Circuits

dte2125.0tue125.0)tsgn(125.0)t(i tjt2

o

2125.0)t(ue125.0)t(u25.0125.0 t2

io(t) = mA)t(ue125.0)t(u25.0625.0 t2

Chapter 18, Solution 49.

Consider the circuit shown below:

2

j

i2i1

+

j2

j

1

+

vo

VS

21Vs For mesh 1, 0IjI2I2j2V 221s 21s Ij2Ij12V (1) For mesh 2, 112 IjI2Ij30

2

I3I 2

1 (2)

Substituting (2) into (1) gives

22

s Ij2j2

Ij3j122V

2

22s I4j44j322V

22 4j2I

Page 984: Solution Manual for Fundamentals of Electric Circuits

js,2s4s

V2sI 2

s2

2o IV24jj

112j2

dev21)t(v tj

oo

24jj

d1e2j21

24jj

d1e2j21

2

tj

2

tj

24j1

e2j21

24j1

e2j21 jtjt

17

e4j1j221

e17

4j1j221

)t(v

jt

jto

jtjt e7j6341e7j6

341

64.13tj64.13tj e271.0e271.0

vo(t) = 0 V)64.13tcos(542. Chapter 18, Solution 50.

Consider the circuit shown below:

ji2i1

+

+

vo

j1

1 j0.5

VS

Page 985: Solution Manual for Fundamentals of Electric Circuits

For loop 1, 0I5.0jIj12 21 (1) For loop 2, 0I5.0jIj1 12 (2) From (2),

j

Ij12

5.0jIj1

I 221

Substituting this into (1),

22 I

2j

jIj12

2

22 I

234j4j2

22 5.14j4j2I

22o j5.14j4j2IV

2

o

j3

8j38

j34

V

2222

38j

34

316

38j

34

j344

)t(Vo V)t(ut38

sine657.5)t(ut38

cose4 3/t43/t4

Page 986: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 51.

j11

j11

j1

j1//1Z

js,)5.1s)(1s(

1

j12

j231

j12

j112

j11

V2Z

ZV oo

)t(u)ee(2)t(v5.1s

21s

25.1s

B1s

AV t5.1too

00

t3t5.2t2t3t5.2t2

0

2t5.1t2

3e

5.2e2

2e4dt)ee2e(4

dt)ee(4dt)t(fW

J 1332.0)31

5.22

21(4W

Chapter 18, Solution 52.

J = 0

2

0

2 d)(F1dt)t(f2

= 23

1)3/(tan31d

911

0

1

0 22 = (1/6)

Page 987: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 53.

J = dt)t(f2d)(F 2

0

2

f(t) = 0t,e

0t,e2

t2

t

J = 0

t40t4

0

t40 t4

4e

4e2dtedte2 = 2 [(1/4) + (1/4)] =

Chapter 18, Solution 54.

W1 = 0

t2

0

t22 e8dte16dt)t(f = 8 J

Chapter 18, Solution 55.

f(t) = 5e2e�–tu(t) F( ) = 5e2/(1 + j ), |F( )|2 = 25e4/(1 + 2)

W1 = 0

14

0 2

4

0

2 )(tane25d1

1e25d)(F1

= 12.5e4 = 682.5 J

or W1 = 12.5e0

t242 dtee25dt)t(f 4 = 682.5 J

Chapter 18, Solution 56.

W1 = 0

2t22 dt)t2(sinedt)t(f

But, sin2(A) = 0.5(1 �– cos(2A))

Page 988: Solution Manual for Fundamentals of Electric Circuits

W1 = 0

t2

0

t2

0

t2 )]t4sin(4)t4cos(2[164

e2

e21dt)]t4cos(1[5.0e

= (1/4) + (1/20)(�–2) = 0.15 J

Chapter 18, Solution 57.

W1 = 0t20 t22 e2dte4dt)t(i = 2 J

or I( ) = 2/(1 �– j ), |I( )|2 = 4/(1 + 2)

W1 = 2

4)(tan4d)1(

124d)(I

21

0

12

2 = 2 J

In the frequency range, �–5 < < 5,

W = )373.1(4)5(tan4tan4 15

0

1 = 1.7487

W/ W1 = 1.7487/2 = 0.8743 or 87.43% Chapter 18, Solution 58.

m = 200 = 2 fm which leads to fm = 100 Hz (a) c = x104 = 2 fc which leads to fc = 104/2 = 5 kHz (b) lsb = fc �– fm = 5,000 �– 100 = 4,900 Hz (c) usb = fc + fm = 5,000 + 100 = 5,100 Hz

Chapter 18, Solution 59.

j43

j25

2j4

6j2

10

)(V)(V)(H

i

o

Page 989: Solution Manual for Fundamentals of Electric Circuits

js,)4s)(1s(

12)2s)(1s(

20j1

4j4

3j2

5)(V)(H)(V io

Using partial fraction,

j44

j220

j116

4sD

1sC

2sB

1sA)(Vo

Thus, V)t(ue4e20e16)t(v t4t2t

o Chapter 18, Solution 60. 2

1/j j +

V

Is

2j1

Ij

j2j1

j1

IjV2

ss

92.712418.1566.12j48.38

9027.50

4j41

8902V

.componentDCnoisthere,inductortheacrossappearsvoltagetheSince

21

9.803954.013.25j91.156

9083.62

8j161

5904V22

mV)1.129t4cos(3954.0)92.41t2cos(2418.1)t(v

Page 990: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 61.

lsb = 8,000,000 �– 5,000 = 7,995,000 Hz usb = 8,000,000 + 5,000 = 8,005,000 Hz Chapter 18, Solution 62.

For the lower sideband, the frequencies range from 10,000 �– 3,500 Hz = 6,500 Hz to 10,000 �– 400 Hz = 9,600 Hz

For the upper sideband, the frequencies range from 10,000 + 400 Hz = 10,400 Hz to 10,000 + 3,500 Hz = 13,500 Hz Chapter 18, Solution 63.

Since fn = 5 kHz, 2fn = 10 kHz

i.e. the stations must be spaced 10 kHz apart to avoid interference. f = 1600 �– 540 = 1060 kHz The number of stations = f /10 kHz = 106 stations Chapter 18, Solution 64.

f = 108 �– 88 MHz = 20 MHz The number of stations = 20 MHz/0.2 MHz = 100 stations Chapter 18, Solution 65.

= 3.4 kHz fs = 2 = 6.8 kHz

Page 991: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 66.

= 4.5 MHz fc = 2 = 9 MHz Ts = 1/fc = 1/(9x106) = 1.11x10�–7 = 111 ns Chapter 18, Solution 67.

We first find the Fourier transform of g(t). We use the results of Example 17.2 in conjunction with the duality property. Let Arect(t) be a rectangular pulse of height A and width T as shown below.

Arect(t) transforms to Atsinc( 2/2)

�– m/2

G( )

m/2

F( )

�–T/2

A f(t)

t

T/2

According to the duality property, A sinc( t/2) becomes 2 Arect( ) g(t) = sinc(200 t) becomes 2 Arect( ) where A = 1 and /2 = 200 or T = 400 i.e. the upper frequency u = 400 = 2 fu or fu = 200 Hz

The Nyquist rate = fs = 200 Hz

The Nyquist interval = 1/fs = 1/200 = 5 ms

Page 992: Solution Manual for Fundamentals of Electric Circuits

Chapter 18, Solution 68.

The total energy is

WT = dt)t(v2

Since v(t) is an even function,

WT = 0

t4

0

t4

4e5000dte2500 = 1250 J

V( ) = 50x4/(4 + 2)

W = 5

1 22

25

1

2 d)4(

)200(21d|)(V

2|1

But C)a/x(tana1

axx

a21dx

)xa(1 1

222222

W = 5

1

12

4

)2/(tan21

48110x2

= (2500/ )[(5/29) + 0.5tan-1(5/2) �– (1/5) �– 0.5tan�–1(1/2) = 267.19

W/WT = 267.19/1250 = 0.2137 or 21.37%

Chapter 18, Solution 69.

The total energy is

WT = d4

40021d)(F

2 22

21

= 2

100)4/(tan)4/1(4000

1 = 50

W = 2

0

12

0

2 )4/(tan)4/1(2400d)(F

21

= [100/(2 )]tan�–1(2) = (50/ )(1.107) = 17.6187

W/WT = 17.6187/50 = 0.3524 or 35.24%

Page 993: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 1. To get z and , consider the circuit in Fig. (a). 11 21z

1 4 I2 = 0

+

V2

Io+

V1 2 6 I1

(a)

4)24(||611

111 I

Vz

1o 21

II , 1o2 2 IIV

11

221 I

Vz

To get z and , consider the circuit in Fig. (b). 22 12z

1 4 I1 = 0

Io' +

V2

+

V1 2 6 I2

(b)

667.1)64(||22

222 I

Vz

22'

o 61

1022

III , 2o1 '6 IIV

12

112 I

Vz

Page 994: Solution Manual for Fundamentals of Electric Circuits

Hence,

][z667.1114

Chapter 19, Solution 2.

Consider the circuit in Fig. (a) to get and . 11z 21z

+

V2

I2 = 0

1

1

+

V1 1

Io

1

1

Io'1 1

I1

1 1 1 1 (a)

])12(||12[||121

111 I

Vz

733.21511

24111)411)(1(

243

2||1211z

'

o'

oo 41

311

III

11'

o 154

41111

III

11o 151

154

41

III

1o2 151

IIV

06667.0151

121

221 z

IV

z

Page 995: Solution Manual for Fundamentals of Electric Circuits

To get z , consider the circuit in Fig. (b). 22

1 1 1 1 I1 = 0

+

V2

+

V1 1 1 1 I2

1 1 1 1 (b)

733.2)3||12(||12 112

222 z

IV

z

Thus,

][z733.206667.0

06667.0733.2

Chapter 19, Solution 3.

(a) To find and , consider the circuit in Fig. (a). 11z 21z

Io +

V2

+

V1 1 j

I2 = 0-j

I1

(a)

j1j1j)j1(j

)j1(||j1

111 I

Vz

By current division,

11o jj1j

jIII

Page 996: Solution Manual for Fundamentals of Electric Circuits

1o2 jIIV

j1

221 I

Vz

To get z and , consider the circuit in Fig. (b). 22 12z

I1 = 0

+

V2

+

V1 1

-j

j I2

(b)

0)jj(||12

222 I

Vz

21 jIV

j2

112 I

Vz

Thus,

][z0jjj1

(b) To find and , consider the circuit in Fig. (c). 11z 21z

+ V1 -j 1

+ V2

1

I2 = 0j -j

I1

(c)

Page 997: Solution Manual for Fundamentals of Electric Circuits

5.0j5.1j1

j-j1-j)(||11j

1

111 I

Vz

12 )5.0j5.1( IV

5.0j5.11

221 I

Vz

To get z and , consider the circuit in Fig. (d). 22 12z

+ V2 -j 1

+ V1

1

I1 = 0 j -j

I2

(d)

j1.5-1.5(-j)||11-j2

222 I

Vz

21 )5.0j5.1( IV

5.0j5.12

112 I

Vz

Thus,

][z5.1j5.15.0j5.15.0j5.15.0j5.1

Chapter 19, Solution 4. Transform the network to a T network.

Z1 Z3

Z2

Page 998: Solution Manual for Fundamentals of Electric Circuits

5j12120j

5j10j12)10j)(12(

1Z

j512j60-

2Z

5j1250

3Z

The z parameters are

j4.26--1.77525144

j5)--j60)(12(22112 Zzz

26.4j775.1169

)5j12)(120j(1212111 zzZz

739.5j7758.1169

)5j12)(50(2121322 zzZz

Thus,

][z739.5j775.126.4j775.1-

26.4j775.1-26.4j775.1

Chapter 19, Solution 5. Consider the circuit in Fig. (a).

s 1 I2 = 0

Io

1/s1/s

+

V2

+

V1 1 I1

(a)

s1

s11s

1s1

s11s

1

s1

s1||

s1

1

s1

s1

s1||s1

||111z

Page 999: Solution Manual for Fundamentals of Electric Circuits

1s3s2s1ss

23

2

11z

12

11o

1ss1s

s1s

s

s1

s11s

11s

1

s1

s1s1

||1

s1

||1IIII

123o 1s3s2ss II

1s3s2ss1

231

o2I

IV

1s3s2s123

1

221 I

Vz

Consider the circuit in Fig. (b).

s 1 I1 = 0

1/s1/s 1

+

V1

+

V2 I2

(b)

1s1

s1||s1

s1

||1s1||s1

2

222 I

Vz

1ss

ss1

1s1

s1

1s1

s1s1

1s1

s1s1

222z

1s3s2s2s2s

23

2

22z

2112 zz

Hence,

Page 1000: Solution Manual for Fundamentals of Electric Circuits

][z

1s3s2s2s2s

1s3s2s1

1s3s2s1

1s3s2s1ss

23

2

23

2323

2

Chapter 19, Solution 6.

To find and , connect a voltage source to the input and leave the output open as in Fig. (a).

11z 21z 1V

Vo

+

V2

20 I1

30 0.5 V2+

10

V1

(a)

505.0

10o

2o1 V

VVV

, where oo2 53

302030

VVV

oo

oo1 2.455

35 VV

VVV

ooo1

1 32.010

2.310

VVVV

I

125.1332.02.4

o

o

1

111 V

VIV

z

875.132.06.0

o

o

1

221 V

VIV

z

To obtain and , use the circuit in Fig. (b). 22z 12z

I2

0.5 V2 +

10

+

V1 30

20

V2

(b)

Page 1001: Solution Manual for Fundamentals of Electric Circuits

22

22 5333.030

5.0 VV

VI

875.15333.01

2

222 I

Vz

2221 -9)5.0)(20( VVVV

-16.8755333.0

9-

2

2

2

112 V

VIV

z

Thus,

][z1.8751.87516.875-125.13

Chapter 19, Solution 7. To get z11 and z21, we consider the circuit below. I2=0 I1 20 100 + + + vx 50 60 V1

- V2 - -

12vx - +

1xxxxx1 V

12140

V160

V12V50V

20VV

88.29IVz)

20V(

12181

20VVI

1

111

1x11

Page 1002: Solution Manual for Fundamentals of Electric Circuits

37.70IVzI37.70

I81

121x20)

12140

(8

57V)

12140

(8

57V

857

V12)160

V13(60V

1

2211

11xxx

2

To get z12 and z22, we consider the circuit below. I2 I1=0 20 100 + + + vx 50 60 V1

- V2 - -

12vx - +

2x22

222x V09.060

V12V150V

I,V31

V50100

50V

11.1109.0/1IVz

2

222

704.3IVzI704.3I

311.11V

31VV

2

112222x1

Thus,

11.1137.70704.388.29

]z[

Page 1003: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 8. To get z11 and z21, consider the circuit below.

j4 I1 -j2 5 I2 =0 + j6 j8 + V2 V1 10 - -

4j10IVzI)6j2j10(V

1

11111

)4j10(IVzI4jI10V

1

221112

To get z22 and z12, consider the circuit below.

j4 I1=0 -j2 5 I2 + j6 j8 + V2 V1 10 - -

Page 1004: Solution Manual for Fundamentals of Electric Circuits

8j15IVzI)8j105(V

2

22222

)4j10(IVzI)4j10(V

2

11221

Thus,

)8j15()4j10()4j10()4j10(

]z[

Chapter 19, Solution 9.

It is evident from Fig. 19.5 that a T network is appropriate for realizing the z parameters.

R2 R1 2 6

R3 4

410R 12111 zz 6

46R 12222 zz 2

21123R zz 4 Chapter 19, Solution 10.

(a) This is a non-reciprocal circuit so that the two-port looks like the one shown in Figs. (a) and (b).

+

+

V1

I1

+

+

V2 z21 I1 z12 I2

z22 I2 z11

(a)

Page 1005: Solution Manual for Fundamentals of Electric Circuits

(b) This is a reciprocal network and the two-port look like the one shown in

Figs. (c) and (d).

+

V1

I2I1

+

V2z12

z22 �– z12z11 �– z12

(c)

+

+

V1

I2 I1

+

+

V2 5 I1 20 I2

10 25

(b)

s5.01

1s2

11211 zz

s21222 zz

s1

12z

1 F

+

V1

I2I1

+

V2

0.5 F 2 H 1

(d)

Page 1006: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 11. This is a reciprocal network, as shown below. 1+j5 3+j 1 j5 3 5-j2 5 -j2 Chapter 19, Solution 12. This is a reciprocal two-port so that it can be represented by the circuit in Figs. (a) and (b).

2

2 8 I1

+

V1

+

V2

I2

4

+

V1

I2I1

+

V2

(a)

z12

z22 �– z12z11 �– z12

j1

Io

2

(b) From Fig. (b),

111 10)4||48( IIV

Page 1007: Solution Manual for Fundamentals of Electric Circuits

By current division,

1o 21

II , 1o2 2 IIV

1

1

1

2

10II

VV

1.0

Chapter 19, Solution 13. This is a reciprocal two-port so that the circuit can be represented by the circuit below.

40 50 20

10 I2 30 I1 +

+ 120 0 V

rms I1 I2 +

100

We apply mesh analysis. For mesh 1,

2121 91201090120- IIII (1) For mesh 2,

2121 -4012030 IIII (2) Substituting (2) into (1),

3512-

-35-3612 2222 IIII

)100(3512

21

R21

P2

2

2I W877.5

Page 1008: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 14. To find , consider the circuit in Fig. (a). ThZ

I2I1

+

V1 +

ZS Vo = 1

(a)

2121111 IzIzV (1)

2221212 IzIzV (2) But

12V , 1s1 - IZV

Hence, 2s11

1212121s11

-)(0 I

Zzz

IIzIZz

222s11

1221-1 Iz

Zzzz

22

2Th

1II

VZ

s11

122122 Zz

zzz

To find , consider the circuit in Fig. (b). ThV

+

I2 = 0ZS

+

V2 = VTh

I1

+

V1VS

(b)

02I , V s1s1 ZIV

Page 1009: Solution Manual for Fundamentals of Electric Circuits

Substituting these into (1) and (2),

s11

s1111s1s Zz

VIIzZIV

s11

s211212 Zz

VzIzV

2Th VVs11

s21

ZzVz

Chapter 19, Solution 15.

(a) From Prob. 18.12,

24104060x80120

ZzzzzZ

s11

211222Th

24ZZ ThL

(b) 192)120(1040

80VZz

zs

s11

21ThV

W19224x8

192R8

VP2

Th

Th2

max

Chapter 19, Solution 16. As a reciprocal two-port, the given circuit can be represented as shown in Fig. (a).

+ j6

b

a

(a)

15 0 V

5 4 �– j6 10 �– j6

j4

Page 1010: Solution Manual for Fundamentals of Electric Circuits

At terminals a-b, )6j105(||6j)6j4(ThZ

6j4.26j415

)6j15(6j6j4ThZ

ThZ 4.6

6j)015(6j1056j

6jThV V906

The Thevenin equivalent circuit is shown in Fig. (b).

6.4

+

Vo+

6 90 V j4

(b)From this,

14818.3)6j(4j4.6

4joV

)t(vo V)148t2cos(18.3

Chapter 19, Solution 17. To obtain z and , consider the circuit in Fig. (a). 11 21z

4 Io I2 = 0

Io' +

V2

+

V1

2

8 I1

(a)6

Page 1011: Solution Manual for Fundamentals of Electric Circuits

In this case, the 4- and 8- resistors are in series, since the same current, , passes through them. Similarly, the 2- and 6- resistors are in series, since the same current,

, passes through them.

oI

'oI

8.420

)8)(12(8||12)62(||)84(

1

111 I

Vz

11o 52

1288

III 1'

o 53

II

But 024- '

oo2 IIV

111'

oo2 52-

56

58-

2-4 IIIIIV

-0.452-

1

221 I

Vz

To get z and , consider the circuit in Fig. (b). 22 12z

+

V1

I1 = 0

8

2 +

V2

4

I2

(b)6

2.420

)14)(6(14||6)68(||)24(2

222I

Vz

-0.42112 zz

Thus,

][z4.20.4-0.4-8.4

We may take advantage of Table 18.1 to get [y] from [z].

20)4.0()2.4)(8.4( 2z

Page 1012: Solution Manual for Fundamentals of Electric Circuits

21.020

2.4

z

2211

zy 02.0

204.0-

z

1212

zy

02.020

4.0-

z

2121

zy 24.0

208.4

z

1122

zy

Thus,

][y S24.002.002.021.0

Chapter 19, Solution 18. To get y and , consider the circuit in Fig.(a). 11 21y

I2I1

+

+

V2 = 0

(a)

6

3

3

6 V1

111 8)3||66( IIV

81

1

111 V

Iy

12-

832-

366- 11

12

VVII

121-

1

221 V

Iy

To get y and , consider the circuit in Fig.(b). 22 12y

6 3 IoI1 I2

+

+

V1 = 0 3 6 V2

(b)

Page 1013: Solution Manual for Fundamentals of Electric Circuits

21

6||31

)6||63(||31

2

222 V

Iy

2- o

1

II , 22o 3

163

3III

12-

21

61-

6- 2

22

1

VV

II

212

112 12

1-y

VI

y

Thus,

][y S

21

121-

121-

81

Chapter 19, Solution 19. Consider the circuit in Fig.(a) for calculating and y . 11y 21

1/s

I2I1

+

+

V2 = 0 s

1

1

V1

(a)

1111 1s22

)s1(2s2

2||s1

IIIV

5.0s2

1s2

1

111 V

Iy

2-

1s2-

2)s1()s1-( 11

12

VIII

Page 1014: Solution Manual for Fundamentals of Electric Circuits

-0.51

221 V

Iy

To get y and , refer to the circuit in Fig.(b). 22 12y

1 I1 I2

1/s s

1

+

+

V1 = 0 V2

(b)

222 2ss2

)2||s( IIV

s1

5.0s22s

2

222 V

Iy

2-

s22s

2ss-

2ss- 2

221

VVII

-0.52

112 V

Iy

Thus,

][y Ss10.50.5-

0.5-5.0s

Chapter 19, Solution 20. To get y11 and y21, consider the circuit below.

Page 1015: Solution Manual for Fundamentals of Electric Circuits

3ix I1 2 I2 + ix + I1 V1 4 6 V2 =0

- -

Since 6-ohm resistor is short-circuited, ix = 0

75.0VIyI

68)2//4(IV

1

111111

5.0VIyV

21)V

86(

32I

244I

1

2211112

To get y22 and y12, consider the circuit below.

3ix I1 2 + ix + V1=0 4 6 V2 I2

-

-

Page 1016: Solution Manual for Fundamentals of Electric Circuits

1667.061

VIy

6V

2Vi3iI,

6Vi

2

222

22xx2

2x

0VIy0

2Vi3I

2

112

2x1

Thus,

S1667.05.0075.0

]y[

Chapter 19, Solution 21. To get and , refer to Fig. (a). 11y 21y

At node 1,

10

I2I1

+

+

V2 = 0

(a)

5

0.2 V1V1

V1

4.04.02.05 1

11111

11 V

IyVV

VI

-0.2-0.21

22112 V

IyV

and 12y , refer to the circuit in Fig. (b).

I

o get

ince , the dependent current source can be replaced with an open circuit.

22y

0.2 V1

T

+

V1 0 +

I1 I2

V2 5

(b)

10

V1

=

01VS

Page 1017: Solution Manual for Fundamentals of Electric Circuits

1.0101

102

22222 V

IyIV

02

112 V

I y

Thus,

onsequently, the y parameter equivalent circuit is shown in Fig. (c)

][y S1.02.0-

04.0

C .

hapter 19, Solution 22.

and refer to the circuit in Fig. (a).

At node 1,

I I1 2

0.1 S 0.2 V1

+ +

(c)

0.4 S V1 V2

C

(a) To get y11 21y

2

1

(a)

+

V2 0 V1 = +

I1 I2 V1

+

Vx

3

Vx/2

Vo

o11o11

1 5.05.121

VVIVVV

(1)

At node 2,

o1o1o 2.1

322VV

VVV (2)

ubstituting (2) into (1) gives,

I

1V

S

Page 1018: Solution Manual for Fundamentals of Electric Circuits

9.09.06.05.11

1111111 V

IyVVVI

-0.4-0.43

-

1

2211

o2 V

IyV

VI

o get and refer to the circuit in Fig. (b).

so that the dependent current source can be replaced by an

open circuit.

22y 12y

2 3

T

1

(b)

+

V1 0

I1 I2V1

+

x Vx/2 + V2 V =

01x V

V

2.051

5)023(2

222222 V

IyIIV

-0.2-0.2-2

112221 V

IyVII

Thus,

][y S0.20.4-0.2-9.0

(b) o get and refer to Fig. (c). 11y 21y

j

T

)5.0j5.1(||jj1

1||j-j)||11(||jinZ j-

-j

1 1

V1

(c)

+

V2 0

I1 I2

+

Io

Io'

Io''

Zin

=

Page 1019: Solution Manual for Fundamentals of Electric Circuits

8.0j6.05.0j5.1

)5.0j5.1(j

8.0j6.08.0j6.0

11

in1

1111in1 ZV

IyIZV

1o 5.0j5.1j

II , 1'

o 5.0j5.15.0j5.1

II

1I

j2)5.0j5.1)(j1(j1j- 1

o''

o

III

111

2''

o'

o2 )4.0j2.1(5

j25.2

)5.0j5.1(- IIIIII

112 )2.1j4.0()8.0j6.0)(4.0j2.1(- VVI

121

221 j1.2-0.4 y

VI

y

To get refer to the circuit in Fig.(d).

22y

8.0j.0)j-||11(||joutZ

-j

1 1

(d)

+

V1 = 0

I1 I2

Zout

+

V2

j

6

8.0j6.01

out22 Z

y

Thus,

][y S8.0j6.0j1.20.4-

j1.20.4-8.0j6.0

Page 1020: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 23. (a)

1s1

y1s

1s1

//1y 1212

1s2s

1s1

1y1y1yy 12111211

1s

1ss1s

1sysysyy

2

12221222

1s1ss

1s1

1s1

1s2s

]y[ 2

(b) Consider the network below. I1 I2 1 + + + Vs V1 V2 2 - - -

[y]

11s VIV (1)

22 I2V (2)

2121111 VyVyI (3)

2221212 VyVyI (4) From (1) and (3)

Page 1021: Solution Manual for Fundamentals of Electric Circuits

212111s2121111s VyV)y1(VVyVyVV (5) From (2) and (4),

22221

12221212 V)y5.0(y1

VVyVyV5.0 (6)

Substituting (6) into (5),

)y5.0)(y1(y1

y

s/2V

s2

VyVy

)y5.0)(y1(V

221121

12

2

212221

2211s

)5.3s5.7s6s2(s

)1s(2

1s1ss

21

1s3s2

)1s(1s

1

s/2V 2322

Chapter 19, Solution 24. Since this is a reciprocal network, a network is appropriate, as shown below.

Y2

4

8 4

1/4 S

1/8 S 1/4 S

(a)

Y3 Y1

(b) (c)

Page 1022: Solution Manual for Fundamentals of Electric Circuits

S41

41

21

12111 yyY , 1Z 4

S41

- 122 yY , 2Z 4

S81

41

83

21223 yyY , 3Z 8

Chapter 19, Solution 25. This is a reciprocal network and is shown below. 0.5 S 0.5S 1S Chapter 19, Solution 26. To get y and , consider the circuit in Fig. (a). 11 21y

4

+

Vx

2

2 Vx

1 I2

+

+

V2 = 0 1

2

V1

(a)

Page 1023: Solution Manual for Fundamentals of Electric Circuits

At node 1,

x1xx

xx1 -2

412

2VV

VVV

VV (1)

But 5.15.122

2 1

1111

11x11 V

IyV

VVVVI

Also, 1x2xx

2 -3.575.124

VVIVV

I

-3.51

221 V

Iy

To get y and , consider the circuit in Fig.(b). 22 12y

4

+

+

Vx

2

2 Vx

1 I2

1

2

I1 V2

(b) At node 2,

42 x2

x2

VVVI (2)

At node 1,

x2xxxx2

x -23

1242 VVV

VVVVV (3)

Substituting (3) into (2) gives

2xxx2 -1.55.121

2 VVVVI

-1.52

222 V

Iy

5.022

-

2

112

2x1 V

Iy

VVI

Page 1024: Solution Manual for Fundamentals of Electric Circuits

Thus,

][y S5.1-5.3-5.05.1

Chapter 19, Solution 27. Consider the circuit in Fig. (a).

+

V2 = 0

(a)

20 I1 10 0.1 V2

+

I1

+

4 I2

V1

25.04

41

1

1

11111 I

IVI

yIV

55201

221112 V

IyVII

Consider the circuit in Fig. (b).

4 I1 I2

+

V1 = 0 +

+

20 I1 10 0.1 V2 V2

(b)

025.041.0

1.042

11221 V

IyVI

6.06.01.05.010

202

222222

212 V

IyVVV

VII

Page 1025: Solution Manual for Fundamentals of Electric Circuits

Thus,

][y S6.05

025.025.0

Alternatively, from the given circuit,

211 1.04 VIV

212 1.020 VII Comparing these with the equations for the h parameters show that

411h , -0.112h , 2021h , 1.022h Using Table 18.1,

25.0411

1111 h

y , 025.041.0-

11

1212 h

hy

54

20

11

2121 h

hy , 6.0

424.0

11

h22 h

y

as above. Chapter 19, Solution 28. We obtain and by considering the circuit in Fig.(a). 11y 21y

I2

+

V2 = 0

(a)

1

2

+

V1

4

6 I1

4.34||61inZ

2941.01

in1

111 ZV

Iy

11

12 346-

4.3106-

106-

VV

II

Page 1026: Solution Manual for Fundamentals of Electric Circuits

-0.176534

6-

1

221 V

Iy

To get y and , consider the circuit in Fig. (b). 22 12y

1 4 I1 Io

+

V2

+

V1 = 0 2 6 I2

(b)

2

2

22 IV

2434

)734(2)734)(2(

76

4||2)1||64(||21

y

7059.03424

22y

o1 76-

II 222o 347

4814

)734(22

VIII

-0.176534

6-34

6-

2

11221 V

IyVI

Thus,

][y S0.70590.1765-0.1765-2941.0

The equivalent circuit is shown in Fig. (c). After transforming the current source to a voltage source, we have the circuit in Fig. (d).

6/34 S

18/34 S 4/34 S 1 A 2

(c)

Page 1027: Solution Manual for Fundamentals of Electric Circuits

8.5 5.667

1.889 +

(d)

+

V8.5 V 2

5454.0167.149714.0

)5.8)(9714.0(667.55.8889.1||2)5.8)(889.1||2(

V

2)5454.0(

RV

P22

W1487.0

Chapter 19, Solution 29.

(a) Transforming the subnetwork to Y gives the circuit in Fig. (a).

+

V2

+

V1

Vo1 1

2 10 A -4 A

(a) It is easy to get the z parameters

22112 zz , 32111z , 322z

54921122211z zzzz

22z

2211 5

3y

zy ,

52--

z

122112

zyy

Thus, the equivalent circuit is as shown in Fig. (b).

2/5 S I2I1

1/5 S

+

V2

+

V1 1/5 S 10 A -4 A

(b)

Page 1028: Solution Manual for Fundamentals of Electric Circuits

21211 235052

53

10 VVVVI (1)

21212 3-220-53

52-

-4 VVVVI

2121 5.1105.110 VVVV (2) Substituting (2) into (1),

222 25.43050 VVV V8

21 5.110 VV V22

(b) For direct circuit analysis, consider the circuit in Fig. (a). For the main non-reference node,

122

410 oo V

V

o1o1 10

110 VV

VVV22

41

4- o2o2 VV

VVV8

Chapter 19, Solution 30.

(a) Convert to z parameters; then, convert to h parameters using Table 18.1. 60211211 zzz , 10022z

24003600600021122211z zzzz

241002400

22

z11 z

h , 6.010060

22

1212 z

zh

-0.6-

22

2121 z

zh , 01.0

1

2222 z

h

Page 1029: Solution Manual for Fundamentals of Electric Circuits

Thus,

][hS0.010.6-

6.024

(b) Similarly,

3011z 20222112 zzz

200400600z

1020200

11h 12020

12h

-121h 05.0201

22h

Thus,

][hS0.051-

110

Chapter 19, Solution 31. We get h and by considering the circuit in Fig. (a). 11 21h

2 4 I1

1 1

+

V1

V3 V4

(a)

2 I2

I1

At node 1,

431433

1 2222

VVIVVV

I (1)

At node 2,

14

24

143 V

IVV

431431 6-2163-8 VVIVVI (2)

Page 1030: Solution Manual for Fundamentals of Electric Circuits

Adding (1) and (2),

1441 6.3518 IVVI

1143 8.283 IIVV

1131 8.3 IIVV

8.31

111 I

Vh

-3.6-3.61

-

1

2211

42 I

IhI

VI

To get h and h , refer to the circuit in Fig. (b). The dependent current source can be replaced by an open circuit since

22 12

04 1I .

1 1 2 I1 I2

+

+

V1 4 I1 = 0 2 V2

(b)

4.052

1222

2

112221 V

VhVVV

S2.051

5122 2

222

222 V

Ih

VVI

Thus,

][hS0.23.6-

4.038

Chapter 19, Solution 32.

(a) We obtain and by referring to the circuit in Fig. (a). 11h 21h

1 s s I2

+

V1

+

V2 = 0 1/s I1

(a)

Page 1031: Solution Manual for Fundamentals of Electric Circuits

1211 1ss

s1s1

||ss1 IIV

1ss

1s 21

111 I

Vh

By current division,

1s1-

1s-

s1ss1-

21

221

112 I

Ih

III

To get h and h , refer to Fig. (b). 22 12

1 s s I1 = 0 I2

+

+

V1 1/s V2

(b)

1s1

1ss1ss1

22

1122

221 V

Vh

VVV

1ss

s1s1

s1

s 22

22222 V

IhIV

Thus,

][h

1ss

1s1-

1s1

1ss

1s

22

22

(b) To get g and g , refer to Fig. (c). 11 21

1 s s I1 I2 = 0

+

+

V2 1/s V1

(c)

Page 1032: Solution Manual for Fundamentals of Electric Circuits

1sss

s1s11

s1

s1 21

11111 V

IgIV

1ss1

1sss1s1s1

21

2212

112 V

Vg

VVV

To get g and g , refer to Fig. (d). 22 12

I1

+

V2

+

V1 = 0

I2s 1

1/s

s

I2

(d)

222 s1s1s)1s(

s)1s(||s1

s IIV

1ss1s

s 22

222 I

Vg

1ss1-

1ss-

s1s1s1-

22

1122

221 I

Ig

III

Thus,

][g

1ss1s

s1ss

11ss

1-1ss

s

22

22

Chapter 19, Solution 33. To get h11 and h21, consider the circuit below.

Page 1033: Solution Manual for Fundamentals of Electric Circuits

4 j6 I2 + -j3 + I1 V1 5 V2=0 - -

2821.1j0769.3IV

h6j9

I)6j4(5I)6j4//(5V

1

111

111

Also, 2564.0j3846.0II

hI6j9

5I

1

22112

To get h22 and h12, consider the circuit below. 4 j6 I2 I1 + + -j3 + V1 5 V2

- -

2564.0j3846.06j9

5VV

hV6j9

5V

2

11221

2821.0j0769.0)6j9(3j

3j9)6j9//(3j

1VI

hI)6j9//(3jV 2

222

22

Thus,

2821.0j0769.02564.0j3846.02564.0j3846.02821.1j0769.3

]h[

Page 1034: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 34. Refer to Fig. (a) to get h and . 11 21h

300

100

10 1

+

Vx

+

V1

(a)

10 Vx

50 2

+

V2 = 0 +

I2

I1

At node 1,

x1xx

1 4300300

0100

VIVV

I (1)

11x 754

300IIV

But 8585101

1111x11 I

VhIVIV

At node 2,

111xxxx

2 75.1430075

575

300530050100

IIIVVVV

I

75.141

221 I

Ih

To get h and h , refer to Fig. (b). 22 12

300

I2

+

I1 = 0 2 50

10 Vx

+

+

V1

+

Vx

1 10

100 V2

(b)

Page 1035: Solution Manual for Fundamentals of Electric Circuits

At node 2,

x22x22

2 8094005010

400VVI

VVVI

But 4400

100 22x

VVV

Hence, 2222 29209400 VVVI

S0725.040029

2

222 V

Ih

25.041

4 2

112

2x1 V

Vh

VVV

][hS0725.075.14

25.085

To get g and g , refer to Fig. (c). 11 21

300

I1 I2 = 0

+

2 50

10 Vx

+

+

V2

+

Vx

1 10

100 V1

(c) At node 1,

x1xxx

1 5.14350350

10100

VIVVV

I (2)

But x11x1

1 1010

VVIVV

I

or (3) 11x 10IVV

Page 1036: Solution Manual for Fundamentals of Electric Circuits

Substituting (3) into (2) gives

11111 5.144951455.14350 VIIVI

S02929.0495

5.14

1

111 V

Ig

At node 2,

xxx2 -8.42861035011

)50( VVVV

1111 4955.14)286.84(-8.4286286.84-8.4286 VVIV

-5.96-5.961

22112 V

VgVV

To get g and g , refer to Fig. (d). 22 12

300

I1

+

V1 = 0

Io Io

+

V2

50

10 Vx

+

+

Vx

10

100 I2

(d)

091.9100||10

091.93005010 2x2

2

VVVI

x22 818.611818.7091.309 VVI (4)

But 22x 02941.0091.309

091.9VVV (5)

Substituting (5) into (4) gives

22 9091.309 VI

Page 1037: Solution Manual for Fundamentals of Electric Circuits

34.342

222 I

Vg

091.30934.34

091.30922

o

IVI

)091.309)(1.1(34.34-

110100- 2

o1

III

-0.1012

112 I

Ig

Thus,

][g34.345.96-

0.101-S02929.0

Chapter 19, Solution 35. To get h and h consider the circuit in Fig. (a). 11 21

I2

+

V2 = 0

4 1

(a)

+

V1

1 : 2

I1

144

n4

Z 2R

22)11(1

111111 I

VhIIV

-0.521-

2-NN-

1

221

1

2

2

1

II

hII

Page 1038: Solution Manual for Fundamentals of Electric Circuits

To get h and h , refer to Fig. (b). 22 12

+

1 : 2

+

V1

I21 4 I1 = 0

V2

(b) Since , . 01I 02IHence, . 022h At the terminals of the transformer, we have and which are related as 1V 2V

5.021

2nNN

2

112

1

2

1

2

VV

hVV

Thus,

][h05.0-5.02

Chapter 19, Solution 36. We replace the two-port by its equivalent circuit as shown below.

+

V2 100

2 I1

-2 I1 +

+

+

V1

I1 4

3 V2

16 I2

10 V 25

2025||100

112 40)2)(20( IIV (1)

032010- 21 VI

111 140)40)(3(2010 III

Page 1039: Solution Manual for Fundamentals of Electric Circuits

141

1I , 1440

2V

14136

316 211 VIV

708-)2(

125100

12 II

(a) 13640

1

2

VV

2941.0

(b) 1

2

II

1.6-

(c) 136

1

1

1

VI

S10353.7 -3

(d) 140

1

2

IV

40

Chapter 19, Solution 37.

(a) We first obtain the h parameters. To get h and h refer to Fig. (a). 11 21

6 3 I2

+

V2 = 0

+

V1 3 6 I1

(a)

26||3

88)26(1

111111 I

VhIIV

Page 1040: Solution Manual for Fundamentals of Electric Circuits

32-

32-

636-

1

221112 I

IhIII

To get h and h , refer to the circuit in Fig. (b). 22 12

6 3 I1 = 0 I2

+

+

V1 3 6 V2

(b)

49

9||3

94

49

2

22222 V

IhIV

32

32

366

2

112221 V

VhVVV

][hS

94

32-

32

8

The equivalent circuit of the given circuit is shown in Fig. (c).

8 I1 I2

-2/3 I1

+

V2 2/3 V2 9/4 +

+ 10 V 5

(c)

1032

8 21 VI (1)

Page 1041: Solution Manual for Fundamentals of Electric Circuits

1112 2930

2945

32

49

||532

IIIV

21 3029

VI (2)

Substituting (2) into (1),

1032

3029

)8( 22 VV

252300

2V V19.1

(b) By direct analysis, refer to Fig.(d).

6 3

3

+

V2+

6 10 V 5

(d)

Transform the 10-V voltage source to a 6

10-A current source. Since

, we combine the two 6- resistors in parallel and transform

the current source back to

36||6

V536

10 voltage source shown in Fig. (e).

3 3

+

V2+

5 V 3 || 5

(e)

815

8)5)(3(

5||3

Page 1042: Solution Manual for Fundamentals of Electric Circuits

6375

)5(8156

8152V V19.1

Chapter 19, Solution 38. We replace the two-port by its equivalent circuit as shown below.

50 I1

+

V2 200 k +

+

800

10-4 V2

+

V1

I1 200 I2

10 V 50 k

1

sin I

VZ , k4050||200

1

6312 )10-2()1040(-50 IIV

For the left loop,

12

-4s

100010

IVV

11

6-4s 1000)10-2(10 IIV

111s 8002001000 IIIV

1

sin I

VZ 800

Alternatively,

L22

L211211sin 1 Zh

ZhhhZZ

)1050)(105.0(1)1050)(50)(10(

800200 35-

3-4

inZ 800

Page 1043: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 39. To get g11 and g21, consider the circuit below which is partly obtained by converting the delta to wye subnetwork. I1 R1 R2 I2 + + R3 V2 V1 10 - -

2.320

8x8R,R6.1488

8x4R 321

8919.0VVgV8919.0V

6.12.132.13V

1

221112

06757.08.14

1VIgI8.14)102.36.1(IV1

111111

To get g22 and g12, consider the circuit below. 1.6 1.6 I1 + V2 V1=0 I2 13.2 -

Page 1044: Solution Manual for Fundamentals of Electric Circuits

8919.0IIgI8919.0I

6.12.132.13I

2

112221

027.3IVgI027.3)6.1//2.136.1(IV

2

222222

027.38919.08919.006757.0

]g[

Chapter 19, Solution 40. To get g and g , consider the circuit in Fig. (a). 11 21

I2 = 0j10

12

-j6 I1

+

+

V2

(a)

V1

S0333.0j0667.06j12

1)6j12(

1

11111 V

IgIV

4.0j8.0j2

2)6j12(

12

1

1

1

221 I

IVV

g

To get g and g , consider the circuit in Fig. (b). 12 22

I1 -j6

12

j10

+

V1 = 0

I2

I2

(b)

Page 1045: Solution Manual for Fundamentals of Electric Circuits

4.0j8.0--j6-12

12-j6-12

12-21

2

11221 g

II

gII

22 -j6)||1210j( IV

2.5j4.2j6-12-j6))(12(

10j2

222 I

Vg

][g2.5j4.24.0j8.0

4.0j8.0-S0333.0j0667.0

Chapter 19, Solution 41.

For the g parameters 2121111 IgVgI (1)

2221212 IgVgV (2) But and s1s1 ZIVV

222121L22 - IgVgZIV

2L22121 )(0 IZgVg

or 221

L221

)(-I

gZg

V

Substituting this into (1),

221

122111L11221 -

)(I

ggggZgg

I

or gL11

21

1

2

Zgg-

II

Also, 222s1s212 )( IgZIVgV

2221s21s21 IgIZgVg 2222gL11ss21 )( IgIZgZVg

But L

22

-ZV

I

Page 1046: Solution Manual for Fundamentals of Electric Circuits

L

222sgLs11s212 ][

ZV

gZZZgVgV

s21L

22sgLs11L2 ][Vg

ZgZZZgZV

22sgLs11L

L21

s

2

gZZZgZZg

VV

22s1221s2211Ls11L

L21

s

2

gZggZggZZgZZg

VV

s

2

VV

s2112L22s11

L21

Zgg)Zg)(Zg1(Zg

Chapter 19, Solution 42.

(a) The network is shown in Fig. (a).

20 I1 I2

100 0.5 I1

+

V2

+

V1 -0.5 I2+

(a)

(b) The network is shown in Fig. (b).

s 2 I1 I2

10 12 V1

+

V2

+

V1 +

(b)

Page 1047: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 43.

(a) To find and , consider the network in Fig. (a). A C

+

V2

I2I1

+

Z

V1

(a)

12

121 V

VAVV

002

11 V

ICI

To get B and , consider the circuit in Fig. (b). D

+

V2 = 0

I2I1

+

Z

V1

(b)

11 IZV , 12 -II

ZIIZ

IV

B1

1

2

1

---

1-

2

1

II

D

Hence,

][T10Z1

Page 1048: Solution Manual for Fundamentals of Electric Circuits

(b) To find and , consider the circuit in Fig. (c). A C

I1 I2

Z

+

V2+

V1

(c)

12

121 V

VAVV

YZV

ICVIZV

1

2

1211

To get B and , refer to the circuit in Fig.(d). D

I2

+

V1 Y

+

V2 = 0I1

(d)

021 VV 12 -II

0-

2

1

IV

B , 1-

2

1

II

D

Thus,

][T1Y01

Chapter 19, Solution 44. To determine and , consider the circuit in Fig.(a). A C

Page 1049: Solution Manual for Fundamentals of Electric Circuits

j15

I1

+

V2

(a)

+ 20

-j20

Io'

I2 = 0

Io

-j10

Io

V1

11 ])20j15j(||)10j-(20[ IV

111 310

j20j15-

-j10)(-j5)(20 IIV

1'

o II

11o 32

5j10j-10j-

III

'20-j20)( oo2 IIV = 11 I20I340j = 1I

340j20

1

1

2

1

I340j20

)310j20( IVV

A 0.7692 + j0.3461

340j20

1

2

1

VI

C 0.03461 + j0.023

Page 1050: Solution Manual for Fundamentals of Electric Circuits

To find and , consider the circuit in Fig. (b). B D

j15

+

V2 = 0

I2I1 -j10

+

-j20

20 V1

(b) We may transform the subnetwork to a T as shown in Fig. (c).

10j20j10j15j

-j10))(15j(1Z

340

-jj15-

)-j10)(-j20(2Z

20jj15--j20))(15j(

3Z

j10 j20 I1 I2

+

V2 = 0 + 20 �– j40/3 V1

(c)

112 j32j3

20j340j20340j20

- III

6923.0j5385.02j3j3-

2

1

II

D

11 20j340j20)340j20)(20j(

10j IV

Page 1051: Solution Manual for Fundamentals of Electric Circuits

)18j24(j])7j9(210j[ 111 IIV

j55)-15(136

j3j2)-(3-

)18j24(j--

1

1

2

1

I

IIV

B

j25.385-6.923B

][Tj0.69230.5385Sj0.0230.03461

j25.3856.923-3461.0j7692.0

Chapter 19, Solution 45. To obtain A and C, consider the circuit below. I1 sL 1/sC I2 =0 + + V1 R1 V2 - - R2

1

21

2

11

21

12 R

sLRRVV

AVsLRR

RV

12

1112 R

1VICRIV

To obtain B and D, consider the circuit below.

Page 1052: Solution Manual for Fundamentals of Electric Circuits

I1 sL 1/sC I2 + + V1 R1 V2=0 - - R2

CsRCsR1

II

DICsR1

CsRI

sC1R

RI

1

1

2

11

1

11

1

12

2C1

1

1

1211

1

1

21 IsR

)CsR1(CsR1

R)sLR)(CsR1(I

sC1R

sCR

sLRV

)sLR)(CsR1(RCsR

1IVB 211

12

1

Chapter 19, Solution 46. To get and C , refer to the circuit in Fig.(a). A

+

V2

I1

+

1

2 4 Ix

1

Ix

2 I2 = 01

+

Vo V1

(a)At node 1,

2o12oo

1 23212

VVIVVV

I (1)

Page 1053: Solution Manual for Fundamentals of Electric Circuits

At node 2,

2ooo

x2o -2

24

41

VVVV

IVV

(2)

From (1) and (2),

S-2.525-

-522

121 V

ICVI

But 21o1

1 1VV

VVI

21212 -3.52.5- VVVVV

-3.52

1

VV

A

To get B and , consider the circuit in Fig. (b). D

+

V2 = 0

+

Vo

Ix

+

1 1 2 I2

4 Ix 2

1 I1

V1 (b) At node 1,

o1oo

1 3212

VIVV

I (3)

At node 2,

041 x

o2 I

VI

�– I (4) o2oo2 -302 VIVV

Adding (3) and (4),

2121 -0.502 IIII (5)

5.0-

2

1

II

D

Page 1054: Solution Manual for Fundamentals of Electric Circuits

But o11o1

1 1VIV

VVI (6)

Substituting (5) and (4) into (6),

2221 65-

31-

21-

IIIV

8333.065-

2

1

IV

B

Thus,

][T0.5-S2.5-

0.83333.5-

Chapter 19, Solution 47. To get A and C, consider the circuit below. 6 I1 1 4 I2=0 + + + + Vx 2 5Vx V2

V1 - - - -

x1xxxx1 V1.1V

10V5V

2V

1VV

3235.04.3/1.1VVAV4.3V5)V4.0(4V

2

1xxx2

02941.04.3/1.0VICV1.0VV1.1

1VVI

2

1xxx

x11

Page 1055: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 48.

(a) Refer to the circuit below.

I2I1

+

[ T ]+

V2V1 ZL

221 304 IVV (1)

221 1.0 IVI (2) When the output terminals are shorted, 02V . So, (1) and (2) become

21 -30IV and 21 -II Hence,

1

1in I

VZ 30

(b) When the output terminals are open-circuited, 02I .

So, (1) and (2) become 21 4 VV

21 1.0 VI or 12 10IV

11 40IV

1

1in I

VZ 40

(c) When the output port is terminated by a 10- load, . 22 I-10V

So, (1) and (2) become 2221 -7030-40 IIIV

2221 -2- IIII

11 35IV

1

1in I

VZ 35

Page 1056: Solution Manual for Fundamentals of Electric Circuits

Alternatively, we may use DZCBZA

ZL

Lin

Chapter 19, Solution 49. To get and C , refer to the circuit in Fig.(a). A

1/s

+ 1 1/s

(a)

1 1/s

+

V2

I1 I2 = 0

V1

1s1

s11s1

s1

||1

12 s1||1s1s1||1

VV

1s2s

1s1

s1

1s1

1

2

VV

A

)1s(s1s2

||1s

11s

1s1

||1s

1111 IIV

)1s3)(1s(1s2

)1s(s1s2

1s1

)1s(s1s2

1s1

1

1

IV

But s

1s221 VV

Page 1057: Solution Manual for Fundamentals of Electric Circuits

Hence, )1s3)(1s(

1s2s

1s2

1

2

IV

s)1s3)(1s(

1

2

IV

C

To get B and , consider the circuit in Fig. (b). D

+

V2 = 0

I2

1 1/s

I1

+

1/s

1/s 1 V1

(b)

1s2s21

||1s1

||s1

||1 1111

IIIV

1

1

2 12ss-

s1

1s1

1s1-

II

I

s1

2s

1s2-

2

1

II

D

s1-

s-s-1s2

1s21

2

1221 I

VB

IIV

Thus,

][T

s1

2s

)1s3)(1s(s1

1s22

Chapter 19, Solution 50. To get a and c, consider the circuit below.

Page 1058: Solution Manual for Fundamentals of Electric Circuits

I1=0 2 s I2

+ + V1 4/s V2

- -

21

22221 s25.01VVaV

4s4V

s/4ss/4V

4ss25.0s

VIc

s/4sV)s25.01(

s/4sVI

or I)s/4s(V

2

3

1

212

22

22

To get b and d, consider the circuit below. I1 2 s I2

+ +

V1=0 4/s V2

- -

s5.01IId

2sI2I

s/42s/4I

1

2221

2ss5.0I

VbI2

)2s2s

)(4s2s(

I2s

)4s2s(I)s4//2s(V

2

1

21

2

22

22

1s5.04s

ss25.02ss5.01s25.0

]t[2

222

Page 1059: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 51. To get a and c , consider the circuit in Fig. (a).

j2 j

I1 = 0 I2

+

V1

(a)

+

j 1 -j3

V2

222 -j2)3jj( IIV

21 -jIV

2j-

j2-

2

2

1

2

II

VV

a

jj-

1

1

2

VI

c

To get b and d , consider the circuit in Fig. (b).

+

V1 = 0

I1

j2 j

I2

+

1 -j3 j

V2

(b) For mesh 1,

21 j)2j1(0 II

or j2j

2j1

1

2

II

Page 1060: Solution Manual for Fundamentals of Electric Circuits

j-2-

1

2

II

d

For mesh 2, 122 j)3jj( IIV

1112 )5j-2(j)2j-)(j2( IIIV

5j2-

1

2

IV

b

Thus,

][tj2-j5j22

Chapter 19, Solution 52. It is easy to find the z parameters and then transform these to h parameters and T parameters.

322

221

RRRRRR

][z

223221z R)RR)(RR(

133221 RRRRRR

(a)

3232

2

32

2

32

133221

2222

21

22

12

22

z

RR1

RRR-

RRR

RRRRRRRR

1-][

zzz

zz

zh

Thus,

32

32111 RR

RRRh , 21

32

212 h-

RRR

h , 32

22 RR1

h

as required.

(b)

2

32

2

2

133221

2

21

21

22

21

21

z

21

11

RRR

R1

RRRRRRR

RRR

1][

zz

z

zzz

T

Page 1061: Solution Manual for Fundamentals of Electric Circuits

Hence,

2

1

RR

1A , )RR(RR

RB 322

13 ,

2R1

C , 2

3

RR

1D

as required. Chapter 19, Solution 53.

For the z parameters, 2121111 IzIzV (1)

2221122 IzIzV (2) For ABCD parameters,

221 IBVAV (3)

221 IDVCI (4) From (4),

21

2 ICD

CI

V (5)

Comparing (2) and (5),

Cz

121 ,

CD

z 22

Substituting (5) into (3),

211 IBC

ADI

CA

V

21 IC

BCADI

CA (6)

Comparing (6) and (1),

CA

z11 CC

BCADz T

12

Thus,

[Z] =

CD

C1

CCA T

Page 1062: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 54.

For the y parameters 2121111 VyVyI (1)

2221212 VyVyI (2) From (2),

221

22

21

21 V

yy

yI

V

or 221

212

221

1-I

yV

yy

V (3)

Substituting (3) into (1) gives

221

112122

21

22111

-I

yy

VyVy

yyI

or 221

112

21

y1

-I

yy

Vy

I (4)

Comparing (3) and (4) with the following equations

221 IBVAV

221 IDVCI clearly shows that

21

22

yy-

A , 21y1-

B , 21

y

y-

C , 21

11

yy-

D

as required. Chapter 19, Solution 55.

For the z parameters 2121111 IzIzV (1)

2221212 IzIzV (2) From (1),

211

121

111

1I

zz

Vz

I (3)

Substituting (3) into (2) gives

Page 1063: Solution Manual for Fundamentals of Electric Circuits

211

1221221

11

212 I

zzz

zVzz

V

or 211

z1

11

212 I

zV

zz

V (4)

Comparing (3) and (4) with the following equations

2121111 IgVgI

2221212 IgVgV indicates that

1111 z

1g ,

11

1212 z

z-g ,

11

2121 z

zg ,

11

z22 z

g

as required. Chapter 19, Solution 56.

(a) 5j74j)j3)(j2(y

0405.0j2568.00676.0j0946.03784.0j2703.02973.0j2162.0

/y/y/y/y

]z[y11y21

y12y22

(b) 6.0j8.32.0j4.06.1j8.02.0j4.0

y/y/yy/yy/1

]h11y1121

111211[

(c ) 75.0j25.075.1j25.1

25.0j5.0j25.0y/yy/

y/1y/y]t

122212y

121211[

Chapter 19, Solution 57.

1)1)(20()7)(3(T

CD

C

CCA

z 1][T

7113

Page 1064: Solution Manual for Fundamentals of Electric Circuits

BA

B

BBD

y 1-

-

][T

S

203

201-

201-

207

DC

D

DDB

h 1-][T

S71

71-

71

720

AB

A

AAC

g 1

-

][T

320

31

31-

S31

TT

TT][ AC

BD

t3S1

207

Chapter 19, Solution 58.

The given set of equations is for the h parameters.

S0.42-21

][h 4.4-2))(2()4.0)(1(h

(a)

11

h

11

21

11

12

11

-1

][

hhh

hh

hy S

4.42-2-1

(b)

2121

22

21

11

21

h

1--

--

]

hhh

hh

hT[

5.0S2.05.02.2

Page 1065: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 59.

2.00.080.12-0.4)(2)()2)(06.0(g

(a)

11

g

11

21

11

12

11

-1

[

ggg

gg

gz]

333.3333.3667.6667.16

(b)

2222

21

22

12

22

g

1-][

ggg

gg

gy S0.50.1-0.2-1.0

(c)

g

11

g

21

g

12

g

22

-

-

][ gg

gg

hS0.31-

210

(d)

21

g

21

11

21

22

21

1

][

ggg

gg

gT

1S3.0105

Chapter 19, Solution 60.

28.002.03.021122211y yyyy

(a)

y

11

y

21

y

12

y

22

-

-

][ yy

yy

z143.23571.0

7143.0786.1

Page 1066: Solution Manual for Fundamentals of Electric Circuits

(b)

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

S0.46670.1667-3333.0667.1

(c)

12

22

12

y

1212

11

--

1--

][

yy

y

yyy

t5.2S4.1

53

Chapter 19, Solution 61.

(a) To obtain and , consider the circuit in Fig. (a). 11z 21z

1 Io

+

V1

1 1 I2 = 0

+

V2 1 I1

(a)

1111 35

32

1])11(||11[ IIIV

35

1

111 I

Vz

11o 31

211

III

0- 1o2 IIV

1112 34

31

IIIV

Page 1067: Solution Manual for Fundamentals of Electric Circuits

34

1

221 I

Vz

To obtain and , consider the circuit in Fig. (b). 22z 12z

1

+

V2

I1

+

V1

1 1

1 I2

(b) Due to symmetry, this is similar to the circuit in Fig. (a).

35

1122 zz , 34

1221 zz

][z

35

34

34

35

(b)

2222

21

22

12

22

z

1-][

zzz

zz

zh

S53

54-

54

53

(c)

21

22

21

21

z

21

11

1][

zz

z

zzz

T

45

S43

43

45

Chapter 19, Solution 62. Consider the circuit shown below.

Page 1068: Solution Manual for Fundamentals of Electric Circuits

20 k

a 40 k10 k

Ib

b

I1

50 k

30 k

+

V1

+

+

V2

I2

Since no current enters the input terminals of the op amp,

13

1 10)3010( IV (1)

But 11ba 43

4030

VVVV

133b

b 10803

1020V

VI

which is the same current that flows through the 50-k resistor. Thus, b

32

32 10)2050(1040 IIV

133

23

2 10803

10701040 VIV

23

12 1040821

IVV

2

31

32 104010105 IIV (2)

From (1) and (2),

][z k40105040

8

21122211z 1016zzzz

Page 1069: Solution Manual for Fundamentals of Electric Circuits

21

22

21

21

z

21

11

1][

zz

z

zzz

DCBA

T381.0S52.9

k24.15381.0

Chapter 19, Solution 63. To get z11 and z21, consider the circuit below. I1 1:3 I2=0 + + + + V�’1 V�’2 9 V2 V1 4 - - - -

3n,1n9Z 2R

8.0IVzI

54I)Z//4(V

1

11111R1

4.2IVzI)5/4(3nV'nV'VV

1

22111122

To get z21 and z22, consider the circuit below. I1=0 1:3 I2 + + + + V�’1 V�’2 9 V2 V1 4 - - - -

3n,36)4(n'Z 2R

Page 1070: Solution Manual for Fundamentals of Electric Circuits

2.7IVzI

4536x9I)'Z//9(V

2

22222R2

4.2IVzI4.2

3V

nVV

2

1212

221

Thus,

2.74.24.28.0

]z[

Chapter 19, Solution 64.

kj-)10)(10(

j-Cj

1F 6-31

Consider the op amp circuit below.

40 k

+

V1

1 +

V2

I2 I1

2

Vx20 k 10 k

-j k

+

At node 1,

100

j-20xxx1 VVVV

x1 )20j3( VV (1)

At node 2,

2x2x

41-

400

100

VVVV

(2)

But 3x1

1 1020VV

I (3)

Substituting (2) into (3) gives

Page 1071: Solution Manual for Fundamentals of Electric Circuits

26-

16-

321

1 105.121050102025.0

VVVV

I (4)

Substituting (2) into (1) yields

21 )20j3(41-

VV

or (5) 21 )5j75.0(0 VV Comparing (4) and (5) with the following equations

2121111 VyVyI

2221212 VyVyI indicates that and that 02I

][y S5j75.01

105.121050 -6-6

-6-6

y 10)250j65(10)5.12.25j5.77(

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

S5j3.11020.25-102

4

4

Chapter 19, Solution 65. The network consists of two two-ports in series. It is better to work with z parameters and then convert to y parameters.

For aN ,2224

][ az

For bN ,1112

][ bz

3336

][][][ ba zzz

Page 1072: Solution Manual for Fundamentals of Electric Circuits

9918z

z

11

z

21

z

12

z

22

-

-

][ zz

zz

y S

32

31-

31-

31

Chapter 19, Solution 66. Since we have two two-ports in series, it is better to convert the given y parameters to z parameters.

-6-3-321122211y 10200)1010)(102(yyyy

10000500

-

-

][

y

11

y

21

y

12

y

22

a yy

yy

z

200100100600

100100100100

10000500

][z

i.e. 2121111 IzIzV

2221212 IzIzV or (1) 211 100600 IIV

212 200100 IIV (2) But, at the input port,

11s 60IVV (3) and at the output port,

2o2 -300IVV (4) From (2) and (4),

221 -300200100 III

21 -5II (5) Substituting (1) and (5) into (3),

121s 60100600 IIIV

Page 1073: Solution Manual for Fundamentals of Electric Circuits

22 100-5))(660( II (6) 2-3200I

From (4) and (6),

2

2

2

o

3200-300-

II

VV

09375.0

Chapter 19, Solution 67. The y parameters for the upper network is

21-1-2

][y , 314y

32

31

31

32

-

-

][

y

11

y

21

y

12

y

22

a yy

yy

z

1111

][ bz

35343435

][][][ ba zzz

19

16925

z

21

22

21

21

z

21

11

1][

zz

z

zzz

T25.1S75.0

75.025.1

Chapter 19, Solution 68.

For the upper network , [ aN42-2-4

]ay

Page 1074: Solution Manual for Fundamentals of Electric Circuits

and for the lower network , [ bN211-2

]by

For the overall network,

63-3-6

][][][ ba yyy

27936y

11

y

11

21

11

12

11

-1

][

yyy

yy

yh

S29

21-

21-

61

Chapter 19, Solution 69. We first determine the y parameters for the upper network . aNTo get y and , consider the circuit in Fig. (a). 11 21y

21

n , s4

ns12RZ

111R1 s4s2

s4

2)2( IIIZV

)2s(2s

1

111 V

Iy

2ss-

-2n

- 11

12

VI

II

2ss-

1

221 V

Iy

To get y and , consider the circuit in Fig. (b). 22 12y

Page 1075: Solution Manual for Fundamentals of Electric Circuits

2 : 1

+

V1 =0

+

V2

I1 2 1/s I2

I2

(b)

21

)2(41

)2)(n( 2'RZ

222'

R2 s22s

21

s1

s1

IIIZV

2ss2

2

222 V

Iy

2221 2ss-

2ss2

21-

n- VVII

2ss-

2

112 V

Iy

2ss2

2ss-

2ss-

)2s(2s

][ ay

For the lower network , we obtain y and by referring to the network in Fig. (c). bN 11 21y

2 I1 I2

+

V2 = 0+

s V1

(c)

21

21

11111 V

IyIV

Page 1076: Solution Manual for Fundamentals of Electric Circuits

21-

2-

-1

221

112 V

Iy

VII

To get y and , refer to the circuit in Fig. (d). 22 12y

2 I1 I2

+

V2

+

V1 = 0 s I2

(d)

s22s

2ss2

)2||s(2

222222 V

IyIIV

2-

s22s

2ss-

2ss-

- 2221

VVII

21-

2

112 V

Iy

s2)2s(21-21-21

][ by

][][][ ba yyy

)2s(s24s4s5

2)(s22)(3s-

2)(s22)(3s-

2s1s

2

Chapter 19, Solution 70.

We may obtain the g parameters from the given z parameters.

1052025

][ az , 150100250az

30252550

][ bz , 8756251500bz

Page 1077: Solution Manual for Fundamentals of Electric Circuits

11

z

11

21

11

12

11

zzz

zz-

z1

][g

60.20.8-04.0

][ ag , 17.50.5

0.5-02.0][ bg

][][][ ba ggg23.50.7

1.3-S06.0

Chapter 19, Solution 71. This is a parallel-series connection of two two-ports. We need to add their g parameters together and obtain z parameters from there. For the transformer,

2121 I2I,V21V

Comparing this with

221221 DICVI,BIAVV

shows that

2005.0

]T[ 1b

To get A and C for Tb2 , consider the circuit below. I1 4 I2 =0 + + 5 V1 V2 - 2 -

Page 1078: Solution Manual for Fundamentals of Electric Circuits

1211 I5V,I9V

2.05/1VIC,8.15/9

VVA

2

1

2

1

Chapter 19, Solution 72. Consider the network shown below.

+

V1

Ia1 Ia2

+

V2

I2

+ Va2

+ Vb2

Ib2

+ Va1

Nb

Na

Ib1

+ Vb1

I1

2a1a1a 425 VIV (1)

2aa12a 4- VII (2)

2b1b1b 16 VIV (3)

2bb12b 5.0- VII (4)

1b1a1 VVV

2b2a2 VVV

2b2a2 III

1a1 II Now, rewrite (1) to (4) in terms of and 1I 2V

211a 425 VIV (5)

212a 4 - VII (6)

21b1b 16 VIV (7)

2b12b 5.0- VII (8) Adding (5) and (7),

Page 1079: Solution Manual for Fundamentals of Electric Circuits

21b11 51625 VIIV (9) Adding (6) and (8),

21b12 5.14- VIII (10)

11a1b III (11) Because the two networks and are independent, aN bN

212 5.15- VII or (12) 212 6667.0333.3 IIV Substituting (11) and (12) into (9),

2111 5.15

5.125

41 IIIV

211 333.367.57 IIV (13)

Comparing (12) and (13) with the following equations

2121111 IzIzV

2221212 IzIzV indicates that

][z6667.0333.3333.367.57

Alternatively,

14-425

][ ah , 0.51-116

][ bh

1.55-541

][][][ ba hhh 5.86255.61h

2222

21

22

12

22

h

1-][

hhh

hh

hz

6667.0333.3333.367.57

as obtained previously.

Page 1080: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 73. From Example 18.14 and the cascade two-ports,

2132

][][ ba TT

2132

2132

]][[][ ba TTT7S4

127

When the output is short-circuited, 02V and by definition

21 - IBV , 21 - IDI Hence,

DB

IV

Z1

1in 7

12

Chapter 19, Solution 74. From Prob. 18.35, the transmission parameters for the circuit in Figs. (a) and (b) are

101

][ a

ZT ,

1101

][ b ZT

Z

We partition the given circuit into six subcircuits similar to those in Figs. (a) and (b) as shown in Fig. (c) and obtain [ for each. ]T

(b)

Z

(a)

s

T5 T6 T3 T4T1 T2

1 1/s 1 1/s

s

Page 1081: Solution Manual for Fundamentals of Electric Circuits

1101

][ 1T , , 10s1

][ 2T1s01

][ 3T

][][ 24 TT , ][][ 15 TT , ][][ 36 TT

1s01

1101

][][][][][][][][][][][ 4321654321 TTTTTTTTTTT

11s01

10s1

][][][11s01

][][][][ 3214321 TTTTTTT

11ss1ss

1s01

][][2

21 TT

1s1s2ss

s1ss10s1

][223

2

1T

1s1s2sss2s1s2s3ss

1101

223

3234

][T1s2ss2s4s4s2s

s2s1s2s3ss23234

3234

Note that as expected. 1CDAB Chapter 19, Solution 75. (a) We convert [za] and [zb] to T-parameters. For Na, 162440z .

25.125.042

z/zz/1z/z/z

]T[212221

21z2111a

For Nb, . 88880y

4445.05

y/yy/y/1y/y

]T[211121y

212122b

Page 1082: Solution Manual for Fundamentals of Electric Circuits

125.525.5617186

]T][T[]T[ ba

We convert this to y-parameters. .3BCADT

94.100588.01765.03015.0

B/AB/1B/B/D

]y[ T

(b) The equivalent z-parameters are

0911.00178.00533.03067.3

C/DC/1C/C/A

]z[ T

Consider the equivalent circuit below. I1 z11 z22 I2 + + + + ZL Vi z12 I2 z21 I1 Vo - - - -

212111i IzIzV (1)

222121o IzIzV (2)

But (3) Lo2L2o Z/VIZIV From (2) and (3) ,

21L

22

21o1

L

o22121o zZ

zz1

VIZV

zIzV (4)

Substituting (3) and (4) into (1) gives

0051.0VV3.194

Zz

Zzzz

zz

VV

i

.o

L

12

L21

2211

21

11

o

i

Page 1083: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 76. To get z11 and z21, we open circuit the output port and let I1 = 1A so that

21

2211

1

111 V

IV

z,VIV

z

The schematic is shown below. After it is saved and run, we obtain

122.1Vz,849.3Vz 221111 Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that

22

2221

2

112 V

IV

z,VIV

z

The schematic is shown below. After it is saved and run, we obtain

849.3Vz,122.1Vz 222112 Thus,

849.3122.1122.1949.3

]z[

Page 1084: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 77. We follow Example 19.15 except that this is an AC circuit. (a) We set V2 = 0 and I1 = 1 A. The schematic is shown below. In the AC Sweep Box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, the output file includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 3.163 E�–.01 �–1.616 E+02 FREQ VM($N_0001) VP($N_0001) 1.592 E�–01 9.488 E�–01 �–1.616 E+02

From this we obtain

h11 = V1/1 = 0.9488 �–161.6

h21 = I2/1 = 0.3163 �–161.6 .

Page 1085: Solution Manual for Fundamentals of Electric Circuits

(b) In this case, we set I1 = 0 and V2 = 1V. The schematic is shown below. In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtain an output file which includes

FREQ VM($N_0001) VP($N_0001) 1.592 E�–01 3.163 E�–.01 1.842 E+01 FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 9.488 E�–01 �–1.616 E+02

From this,

h12 = V1/1 = 0.3163 18.42

h21 = I2/1 = 0.9488 �–161.6 .

Thus, [h] = 6.1619488.06.1613163.0

42.183163.06.1619488.0

Page 1086: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 78 For h11 and h21, short-circuit the output port and let I1 = 1A. 6366.02/f . The schematic is shown below. When it is saved and run, the output file contains the following: FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 1.202E+00 1.463E+02 FREQ VM($N_0003) VP($N_0003) 6.366E-01 3.771E+00 -1.350E+02 From the output file, we obtain

o1

o2 135771.3V,3.146202.1I

so that o2

21o1

11 3.146202.11Ih,135771.3

1Vh

Page 1087: Solution Manual for Fundamentals of Electric Circuits

For h12 and h22, open-circuit the input port and let V2 = 1V. The schematic is shown below. When it is saved and run, the output file includes: FREQ VM($N_0003) VP($N_0003) 6.366E-01 1.202E+00 -3.369E+01 FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 3.727E-01 -1.534E+02 From the output file, we obtain

o1

o2 69.33202.1V,4.1533727.0I

so that o2

22o1

12 4.1533727.01Ih,69.33202.1

1Vh

Thus,

o

oo

4.1533727.03.146202.169.33202.1135771.3]h[

Page 1088: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 79 We follow Example 19.16. (a) We set I1 = 1 A and open-circuit the output-port so that I2 = 0. The schematic is shown below with two VPRINT1s to measure V1 and V2. In the AC Sweep box, we enter Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ VM(1) VP(1) 3.183 E�–01 4.669 E+00 �–1.367 E+02 FREQ VM(4) VP(4) 3.183 E�–01 2.530 E+00 �–1.084 E+02

From this, z11 = V1/I1 = 4.669 �–136.7 /1 = 4.669 �–136.7 z21 = V2/I1 = 2.53 �–108.4 /1 = 2.53 �–108.4 .

Page 1089: Solution Manual for Fundamentals of Electric Circuits

(b) In this case, we let I2 = 1 A and open-circuit the input port. The schematic is shown below. In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.3183, and End Freq = 0.3183. After simulation, the output file includes

FREQ VM(1) VP(1) 3.183 E�–01 2.530 E+00 �–1.084 E+02 FREQ VM(2) VP(2) 3.183 E�–01 1.789 E+00 �–1.534 E+02

From this, z12 = V1/I2 = 2.53 �–108.4 /1 = 2.53 �–108..4 z22 = V2/I2 = 1.789 �–153.4 /1 = 1.789 �–153.4 .

Thus,

[z] = 2 4.153789.14.10853.

4.10853.27.136669.4

Page 1090: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 80 To get z11 and z21, we open circuit the output port and let I1 = 1A so that

21

2211

1

111 V

IV

z,VIV

z

The schematic is shown below. After it is saved and run, we obtain

37.70Vz,88.29Vz 221111 Similarly, to get z22 and z12, we open circuit the input port and let I2 = 1A so that

22

2221

2

112 V

IV

z,VIV

z

The schematic is shown below. After it is saved and run, we obtain

11.11Vz,704.3Vz 222112 Thus,

11.1137.70704.388.29

]z[

Page 1091: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 81 (a) We set V1 = 1 and short circuit the output port. The schematic is shown below. After simulation we obtain

y11 = I1 = 1.5, y21 = I2 = 3.5

Page 1092: Solution Manual for Fundamentals of Electric Circuits

(b) We set V2 = 1 and short-circuit the input port. The schematic is shown below. Upon simulating the circuit, we obtain

y12 = I1 = �–0.5, y22 = I2 = 1.5

[Y] = 5.15.35.05.1

Page 1093: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 82 We follow Example 19.15. (a) Set V2 = 0 and I1 = 1A. The schematic is shown below. After simulation, we obtain

h11 = V1/1 = 3.8, h21 = I2/1 = 3.6

(b) Set V1 = 1 V and I1 = 0. The schematic is shown below. After simulation, we obtain

h12 = V1/1 = 0.4, h22 = I2/1 = 0.25

Hence, [h] = 25.06.34.08.3

Page 1094: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 83

To get A and C, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 11 and V2 = 34.

02941.0341

VIC,3235.0

VVA

2

1

2

1

Similarly, to get B and D, we open-circuit the output and let I1 = 1A. The schematic is shown below. When the circuit is saved and simulated, we obtain V1 = 2.5 and I2 = -2.125.

4706.0125.21

IID,1765.1

125.25.2

IVB

2

1

2

1

Thus,

4706.002941.01765.13235.0

]T[

Page 1095: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 84

(a) Since A = 0I2

1

2VV

and C = 0I2

1

2VI

, we open-circuit the output port and let V1

= 1 V. The schematic is as shown below. After simulation, we obtain A = 1/V2 = 1/0.7143 = 1.4 C = I2/V2 = 1.0/0.7143 = 1.4

Page 1096: Solution Manual for Fundamentals of Electric Circuits

(b) To get B and D, we short-circuit the output port and let V1 = 1. The schematic is shown below. After simulating the circuit, we obtain B = �–V1/I2 = �–1/1.25 = �–0.8

8

Thus =

D = �–I1/I2 = �–2.25/1.25 = �–1.

DCBA

8.14.18.04.1

lution 85

(a) Since A =

Chapter 19, So

0I2

1

2VV

and C = 0I2

1

2VI

, we let V1 = 1 V and open-

circuit the output port. The schematic is shown below. In the AC Sweep box, we set n

FREQ IM(V_PRINT1) IP(V_PRINT1) 01

REQ VM($N_0002) VP($N_0002)

From this, we obtain

A

Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After simulation, we obtaian output file which includes

1.592 E�– 6.325 E�–01 1.843 E+01 F1.592 E�–01 6.325 E�–01 �–7.159 E+01

= 59.71581.159.716325.0

11 V2

Page 1097: Solution Manual for Fundamentals of Electric Circuits

C = 90159.716325.0

43.186325.0I1

V2

= j

(b) Similarly, since B =

0V2

1

2IV

0V2

1

2II

, we let V1 = 1 V and short-

circuit the output port. The schematic is shown below. Again, we set Total Pts = 1, Start

FREQ IM(V_PRINT1) IP(V_PRINT1) 01

REQ IM(V_PRINT3) IP(V_PRINT3) 01

From this,

B

Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box. After simulation, we get an output file which includes the following results:

and D =

1.592 E�– 5.661 E�–04 8.997 E+01 F1.592 E�– 9.997 E�–01 �–9.003 E+01

= j901909997.0

11

2

I

D =909997.0

97.8910x661.5I 4

2

1

I = 5.561x10�–4

= DCBA

410x661.5jj59.71581.1

Page 1098: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 86

(a) By definition, g11 = 0I1

1

2VI

, g21 = 0I2

1

2VV

.

We let V1 = 1 V and open-circuit the output port. The schematic is shown below. After simulation, we obtain g11 = I1 = 2.7 g21 = V2 = 0.0

(b) Similarly,

g12 = I

, g22 = 0V2

1

1

I

0V2

2

1IV

g12 = I1 = 0

We let I2 = 1 A and short-circuit the input port. The schematic is shown below. After simulation,

Page 1099: Solution Manual for Fundamentals of Electric Circuits

g22 = V2 = 0

Thus [g] =

000S727.2

hapter 19, Solution 87

a =

C

(a) Since 0I1

2

1VV

and c = 0I1

2

1VI

,

we open-circuit the input port and let V2 = 1 V. The schematic is shown below. In the C Sweep box, set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592. After

IP(V_PRINT2) 1.592 E�–01 5.000 E�–01 1.800 E+02

1) ) .592 E�–01 5.664 E�–04 8.997 E+01

From this,

Asimulation, we obtain an output file which includes

FREQ IM(V_PRINT2)

FREQ VM($N_000 VP($N_00011

a = 97.891 176597.8910x664.5 4

c = 97.8928.88297.8910x664.5

1805.04

Page 1100: Solution Manual for Fundamentals of Electric Circuits

(b) Similarly,

b = 0V1

2

1IV

and d = 0V1

2

1II

We short-circuit the input port and let V2 = 1 V. The schematic is shown below. After simulation, we obtain an output file which includes

FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592 E�–01 5.000 E�–01 1.800 E+02

FREQ IM(V_PRINT3) IP(V_PRINT3) 1.592 E�–01 5.664 E�–04 �–9.010 E+01

From this, we get

b = = �–j1765 1.9010x664.5

14

d = = j888.28 1.9010x664.5

1805.04

Thus [t] = 2.888j2.888j

1765j1765j

Chapter 19, Solution 88 To get Z , consider the network in Fig. (a). in

Page 1101: Solution Manual for Fundamentals of Electric Circuits

I2I1

+

(a)Zin

+

V1Vs Two-Port

+

V2 RL

Rs

2121111 VyVyI (1)

2221212 VyVyI (2)

But 222121L

22 R

-VyVy

VI

L22

1212 R1

-y

VyV (3)

Substituting (3) into (1) yields

L22

121121111 R1

-y

VyyVyI ,

LL R

1Y

1L22

L11y1 V

Yy

YyI , 21122211y yyyy

or 1

1inZ

IV

L11y

L22

YyYy

L22

121

1

22in21

1

222121

1

2i

-ZA

YyVy

Iy

yI

VyVyII

L22

212221

L11y

L22

L22

in2122in21

ZZ

Yyyy

yYy

YyYy

yyy

iAL11y

L21

YyYy

From (3),

1

2vA

VV

L22

21

Yyy-

Page 1102: Solution Manual for Fundamentals of Electric Circuits

To get Z , consider the circuit in Fig. (b). out

Zout

I2I1

(b)

+

+

V1 Rs Two-Port V2

222121

2

2

2outZ

VyVyV

IV

(4)

But 1s1 R- IV Substituting this into (1) yields

2121s111 R- VyIyI

2121s11 )R1( VyIy

s

1

s11

2121 R

-R1

Vy

VyI

or s11

s12

2

1

R1R-

yy

VV

Substituting this into (4) gives

s11

s211222

out

R1R

1Z

yyy

y

s2221s221122

s11

RRR1

yyyyyy

outZs22y

s11

YyYy

Page 1103: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 89

Lfereoeieie

Lfev R)hhhh(h

Rh-A

54-6-

5

v 10)72106.210162640(26401072-

A

182426401072-

A5

v 1613-

)1613(log20Alog20gain dc v 15.64

Chapter 19, Solution 90

(a) Loe

Lfereiein Rh1

RhhhZ

L6-

L-4

R10201R12010

20001500

L5-

-3

R10211012

500

L

-3L

-2 R1012R10500

L2 R2.010500

LR k250

(b) Lfereoeieie

Lfev R)hhhh(h

Rh-A

34-6-

3

v 10250)1012010202000(200010250120-

A

33

6

v 1071021030-

A 3333-

Page 1104: Solution Manual for Fundamentals of Electric Circuits

36-Loe

fei 1025010201

120Rh1

hA 20

120101020)2000600(2000600

hhh)hR(hR

Z 4-6-fereoeies

iesout

k40

2600Zout k65

(c) 3-svc

s

c

b

cv 104-3333AA VV

VV

VV

V13.33-

Chapter 19, Solution 91 k2.1R s , k4R L

(a) Lfereoeieie

Lfev R)hhhh(h

Rh-A

34-6-

3

v 104)80105.110201200(120010480-

A

124832000-

A v 25.64-

(b) 36-Loe

fei 10410201

80Rh1

hA 074.74

(c) ireiein AhhZ

074.74105.11200Z -4in k2.1

(d) fereoeies

iesout hhh)hR(

hRZ

0468.02400

80105.11020240012001200

Z 4-6-out k282.51

Page 1105: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 92 Due to the resistor , we cannot use the formulas in section 18.9.1. We will need to derive our own. Consider the circuit in Fig. (a).

240R E

hoe hfe Ib

hie

+

hre Vc

Zin

IE

+

Ib Ic

+

Vc

RE

(a)

+

Vb

Rs

RL Vs

cbE III (1)

Ecbcrebieb R)(hh IIVIV (2)

oeE

cbfec

h1R

hV

II (3)

But (4) Lcc R-IV Substituting (4) into (3),

c

oeE

Lbfec

h1R

Rh III

or Loe

oeEfe

b

ci R(h1

)hR1(hA

II

(5)

)240000,4(10301)10x30x2401(100A 6-

6

i

iA 79.18

From (3) and (5),

Page 1106: Solution Manual for Fundamentals of Electric Circuits

oeE

cbfeb

ELoe

oeEfec

h1R

h)RR(h1

h)R1(h VIII (6)

Substituting (4) and (6) into (2),

EccrebEieb Rh)Rh( IVIV

EL

ccre

feELoe

oeEfe

oeE

Eiecb R

Rh

h)RR(h1

)hR1(hh1R

)Rh( VV

VV

L

Ere

feELoe

oeEfe

oeE

Eie

c

b

v RR

hh

)RR(h1)hR1(h

h1R

)Rh(A1

VV

(7)

400024010

100424010301

)10x30x2401(10010x301240

)2404000(A1 4-

6-

6

6v

-0.06606.01010x06.6A1 4-3

v

vA �–15.15

From (5),

bLoe

fec Rh1

hII

We substitute this with (4) into (2) to get

cLreEbEieb )RhR()Rh( IIV

bELoe

oeEfeLreEbEieb )RR(h1

)hR1(h)RhR()Rh( IIV

)RR(h1)hR1)(RhR(h

RhZELoe

oeELreEfeEie

b

bin I

V (8)

424010301)10x30x2401)(10410240)(100(2404000Z 6-

63-4

in

inZ 12.818 k

Page 1107: Solution Manual for Fundamentals of Electric Circuits

To obtain , which is the same as the Thevenin impedance at the output, we introduce a 1-V source as shown in Fig. (b).

outZ

hie Rs IcIb

(b)

+

Vb

RE

+

Vc

+

IE

Zout

hre Vc

+

hfe Ib hoe

1 V

From the input loop, 0)(Rh)hR( cbEcreiesb IIVI

But 1cVSo,

0Rh)RhR( cEreEiesb II (9) From the output loop,

bfeoeE

oebfe

oeE

cc h

1hRh

h

h1R

IIV

I

or oeE

fe

oe

fe

cb hR1

hh

hI

I (10)

Substituting (10) into (9) gives

0hR1

hh)hRR(

Rhh

)hRR(oeE

fe

oeieEs

cErefe

cieEs I

I

refe

oe

oeE

ieEscEc

fe

ieEs hhh

hR1hRR

Rh

hRRII

Page 1108: Solution Manual for Fundamentals of Electric Circuits

feieEsE

reoeE

ieEsfeoe

c h)hRR(R

hhR1

hRR)hh(

I

fereoeoeE

ieEs

ieEsfeE

cout

hhhhR1

hRRhRRhR1Z

I

10010103010x30x240140002401200

)40002401200(100240Z4-6-

6

out

152.0544024000Zout 193.7 k

Chapter 19, Solution 93

We apply the same formulas derived in the previous problem.

L

Ere

feELoe

oeEfe

oeE

Eie

v RR

hh

)RR(h1)hR1(h

h1R

)Rh(A1

3800200105.2

15004.01

)002.01(150)10200(

)2002000(A1 4-

5v

-0.0563805263.0105.2004.0A1 4-

v

vA �–17.74

)3800200(101)10x2001(150

)RR(h1)hR1(h

A 5-

5

ELoe

oeEfei 5.144

)RR(h1)hR1)(RhR(h

RhZELoe

oeELreEfeEiein

Page 1109: Solution Manual for Fundamentals of Electric Circuits

04.1)002.1)(108.3105.2200)(150(2002000Z

3-4

in

289662200Zin

inZ 31.17 k

fereoeoeE

ieEs

ieEsfeEout

hhhhR1

hRRhRRhR

Z

0.0055-33200

150105.2002.1

10320020002001000150200Z

4-5-out

outZ �–6.148 M

Chapter 19, Solution 94 We first obtain the ABCD parameters.

Given , 6-101000200

][h -421122211h 102hhhh

2-8-

6-

2121

22

21

11

21

h

10-10-2-102-

1--

-

][

hhh

hh

hT

The overall ABCD parameters for the amplifier are

4-10-

-2-8

2-8-

-6

2-8-

-6

1010102102

10-10-2-102-

10-10-2-102-

][T

0102102 -12-12

T

6-4-

T

1010-0200

1-][

DC

D

DDB

h

Page 1110: Solution Manual for Fundamentals of Electric Circuits

Thus, , h200h ie 0re , h , h -4fe -10 -6

oe 10

34-

34

v 104)0102(200)104)(10(

A 5102

0200Rh1Rhh

hZLoe

Lfereiein 200

Chapter 19, Solution 95

Let s5s

8s10s13

24

22A y

Z

Using long division,

B13

2

A Lss5s8s5

s ZZ

i.e. and H1L1 s5s8s5

3

2

BZ

as shown in Fig (a).

y22 = 1/ZA

L1

ZB

(a)

8s5s5s1

2

3

BB Z

Y

Using long division,

C22B sC8s5

s4.3s2.0 YY

where and F2.0C2 8s5s4.3

2CY

Page 1111: Solution Manual for Fundamentals of Electric Circuits

as shown in Fig. (b).

43

2

CC Cs

1Ls

s4.38

4.3s5

s4.38s51

YZ

i.e. an inductor in series with a capacitor

H471.14.3

5L3 and F425.0

84.3

C4

Thus, the LC network is shown in Fig. (c).

Chapter 19, Solution 96 This is a fourth order network which can be realized with the network shown in Fig. (a).

1.471 H0.425 F

L1

1

L3

C4C2

(c)

0.2 F

1 H

C2

(b)

Yc = 1/ZC

L1

(a)

)s613.2s613.2()1s414.3s()s( 324

Page 1112: Solution Manual for Fundamentals of Electric Circuits

s613.2s613.21s414.3s

1

s613.2s613.21

)s(H

3

24

3

which indicates that

s613.22.613s1-

321y

s613.2s613.21s414.3s

3

4

22y

We seek to realize . By long division, 22y

A43

2

22 Css613.2s613.2

1s414.2s383.0 Yy

i.e. and F383.0C4 s613.2s613.21s414.2

3

2

AY

as shown in Fig. (b).

L1

C4C2

L3

YA

y22(b)

1s414.2s613.2s613.21

2

3

AA Y

Z

By long division,

B32A Ls1s414.2

s531.1s082.1 ZZ

i.e. and H082.1L3 1s414.2s531.1

2BZ

Page 1113: Solution Manual for Fundamentals of Electric Circuits

as shown in Fig.(c).

L1

C4C2

L3

ZB

(c)

12

BB Ls

1Cs

s531.11

s577.11

ZY

i.e. and F577.1C2 H531.1L1 Thus, the network is shown in Fig. (d).

1.531 H 1.082 H

1.577 F 0.383 F 1

(d) Chapter 19, Solution 97

s12s24s6

1

s12ss

)24s6()s12s(s

)s(H

3

2

3

3

23

3

Hence,

A3

3

2

22 Cs1

s12s24s6

Zy (1)

where is shown in the figure below. AZ

Page 1114: Solution Manual for Fundamentals of Electric Circuits

C1 C3

L2

ZA y22 We now obtain C and using partial fraction expansion. 3 AZ

Let 12sCBs

sA

)12s(s24s6

22

2

CsBs)12s(A24s6 222

Equating coefficients : 2 0s : AA1224 : 1s C0 2s : 4BBA6Thus,

12ss4

s2

)12s(s24s6

22

2

(2)

Comparing (1) and (2),

F21

A1

C3

s3

s41

s412s1 2

AZ (3)

But 2

1A Ls

1sC

1Z

(4)

Comparing (3) and (4),

F41

C1 and H31

2L

Therefore, 1C F25.0 , 2L H3333.0 , 3C F5.0

Page 1115: Solution Manual for Fundamentals of Electric Circuits

Chapter 19, Solution 98

2.08.01h

005.010x5.210001.0

h/1h/hh/hh/

]T[]T[ 6212122

211121hba

58

5ba

10x510x5.106.010x6.2]T][T[]T[

We now convert this to z-parameters

37

3T

10x33.310x667.60267.010x733.1

C/DC/1C/C/A

]z[

1000 I1 z11 z22 I2 + + + + ZL Vs z12 I2 z21 I1 Vo - - - -

212111s IzI)z1000(V (1)

121222o IzIzV (2)

But (3) Lo2L2o Z/VIZIV Substituting (3) into (2) gives

L21

22

21o1 Zz

zz1VI (4)

We substitute (3) and (4) into (1)

Page 1116: Solution Manual for Fundamentals of Electric Circuits

V74410x136.210x653.7

VZzV

Zzz

z1)z1000(V

54

oL

12o

L21

22

1111s

Chapter 19, Solution 99

)(|| abc31ab ZZZZZZ

cba

bac31

)(ZZZ

ZZZZZ (1)

)(|| cba32cd ZZZZZZ

cba

cba32

)(ZZZ

ZZZZZ (2)

)(|| cab21ac ZZZZZZ

cba

cab21

)(ZZZ

ZZZZZ (3)

Subtracting (2) from (1),

cba

acb21

)(ZZZ

ZZZZZ (4)

Adding (3) and (4),

1Zcba

cb

ZZZZZ

(5)

Subtracting (5) from (3),

2Zcba

ba

ZZZZZ

(6)

Subtracting (5) from (1),

3Zcba

ac

ZZZZZ

(7)

Page 1117: Solution Manual for Fundamentals of Electric Circuits

Using (5) to (7)

2cba

cbacba133221 )(

)(ZZZ

ZZZZZZZZZZZZ

cba

cba133221 ZZZ

ZZZZZZZZZ (8)

Dividing (8) by each of (5), (6), and (7),

aZ1

133221

ZZZZZZZ

bZ3

133221

ZZZZZZZ

cZ2

133221

ZZZZZZZ

as required. Note that the formulas above are not exactly the same as those in Chapter 9 because the locations of and are interchanged in Fig. 18.122. bZ cZ