circuits solutions ulaby chapter 2

Problem 2.1 An AWG-14 copper wire has a resistance of 17.1 at 20C. Howlong is it?

Solution: AWG-14 has a diameter of 1.6 mm (Table 2-2), and at 20C, coppersconductivity is = 5.81107 (S/m) [Table 2-1].

R =

A = RA

= Rpi(d/2)2

= 17.15.81107pi(

1.61032

)2 2 km.

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Problem 2.2 A 3-km long AWG-6 metallic wire has a resistance of approximately6 at 20C. What material is it made of?

Solution:

R =

A, AWG-6 has a diameter of 4.1 mm.

=

RA

=

Rpi(d/2)2 =3103

6pi(4.1103/2)2 = 3.79107 (S/m),

which, according to Table 2-1, is approximately the value of the conductivity ofaluminum.


Problem 2.3 A thin-film resistor made of germanium is 2 mm in length and itsrectangular cross section is 0.2 mm 1 mm, as shown in Fig. P2.3. Determine theresistance that an ohmmeter would measure if connected across its:

(a) Top and bottom surfaces(b) Front and back surfaces(c) Right and left surfaces

0.2 mm 1 mm

2 mm

y

z

x

Figure P2.3: Film resistor of Problem 2.3.

Solution:(a)

R =

A = 0.22 mm, A = 1 mm2 mm = 2106 m2

=2104

2.132106 47 .

(b)

R =

A = 1 mm, A = 2 mm0.2 mm = 4107 m2

=103

2.134107 1,174 .

(c)

R =

A = 2 mm, A = 1 mm0.2 mm = 2107 m2

=2103

2.134107 4,695 .


Problem 2.4 A resistor of length consists of a hollow cylinder of radius asurrounded by a layer of carbon that extends from r = a to r = b, as shown inFig. P2.4.

(a) Develop an expression for the resistance R.(b) Calculate R at 20C for a = 2 cm, b = 3 cm, and = 10 cm.

Carbon

Hollow 2b2a

l

Figure P2.4: Carbon resistor of Problem 2.4.

Solution:(a) R = A .

The area through which current can flow is the cross section consisting of carbon.Hence,

A = pib2pia2.

Thus,R =

pi(b2a2) .

(b)R =

0.17.14104pi(0.0320.022) = 0.89 (m).


Problem 2.5 A standard model used to describe the variation of resistance withtemperature T is given by

R = R0(1+T ),

where R is the resistance at temperature T (measured in C), R0 is the resistance atT = 0C, and is a temperature coefficient. For copper, = 4 103 C1. Atwhat temperature is the resistance greater than R0 by 1%?

Solution: R = 1.01R0. Hence,

RR0

= 1.01 = 1+T

T = 0.01

T =0.01

4103 = 2.5C.


Problem 2.6 A light bulb has a filament whose resistance is characterized by atemperature coefficient = 6 103 C1 (see resistance model given in Problem2.5). The bulb is connected to a 100-V household voltage source via switch. Afterturning on the switch, the temperature of the filament increases rapidly from theinitial room temperature of 20C to an operating temperature of 1800C. When itreaches its operating temperature, it consumes 80 W of power.

(a) Determine the filament resistance at 1800C.(b) Determine the filament resistance at room temperature.(c) Determine the current that the filament draws at room temperature and also at

1800C.(d) If the filament deteriorates when the current through it approaches 10 A, is the

damage done to the filament greater when it is first turned on or later on whenit arrives at its operating temperature?

Solution:(a) R = resistance at 1800C.

p =V 2

R; R =

V 2

p=

(100)2

80 = 125 .

(b) R = R0(1+T ).

R0 =R

1+T=

1251+61031800 = 10.6 @ 0

C.

At T = 20C,

R = R0(1+T ) = 10.6(1+610320) = 11.9 @ 20C.

(c) At T = 20C,I =

VR(20C) =

10011.9 = 8.43 A.

At T = 1800C,I =

pV

=80100 = 0.8 A.

(d) The damage is greater at room temperature because the current is much closer to10 A.


Problem 2.7 A 110-V heating element in a stove can boil a standard-size pot ofwater in 1.2 minutes, consuming a total of 136 kJ of energy. Determine the resistanceof the heating element and the current flowing through it.

Solution:

W = p t = i t

i =W

t =136103

1101.260 = 17.2 A.

R =

i=

11017.2

= 6.41 .


Problem 2.8 A certain copper wire has a resistance R characterized by the modelgiven in Problem 2.5 with = 4 103 C1. If R = 60 at 20C and the wire isused in a circuit that cannot tolerate an increase in the magnitude of R by more than10 percent over its value at 20C, what would be the highest temperature at whichthe circuit can be operated within its tolerance limits?

Solution: The model is given by

R = R0(1+T ).

We are given that R = 60 @ T = 20C, and the maximum R that can be toleratedis 66 .

At 20C, 60 = R0(1+20).At unknown T , 66 = R0(1+T ).The ratio gives

6660 =

1+T1+20 .

Solving for T , we have

T =6+132

60 =6+132+4103

604103 = 27.2C.


Problem 2.9 The circuit shown in Fig. P2.9 includes two identical potentiometerswith per-length resistance of 20 /cm. Determine Ia and Ib.

Ia

80 mA4 cm

6 cm

10 cm 2.5 cm

7.5 cm10 cm

Ib

Figure P2.9: Circuit of Problem 2.9.

Solution: Effectively, the circuit looks as shown in Fig. P2.9(a).Ia

Ib

R1

R2

80 mA

Fig. P2.9 (a)

Resistors R1 and R2 are:

R1 = (4+7.5)20 = 230 ,R2 = (6+2.5)20 = 170 .

Hence, by current division

Ia = (80 mA(

R2R1 +R2

)

= 8102 170400 = 34 mA,

Ib = 8102230400 = 46 mA.


Problem 2.10 Determine VL in the circuit of Fig. P2.10.

+

_

+

_

12 V VL4 6 6 10

5 5

5 5


The parallel combination of the 4 and 6 resistors is

R =464+6 = 2.4 .

By voltage division

VL = 12(

2.45+2.4+5

)= 2.32 V.


Problem 2.11 Select the value of R in the circuit of Fig. P2.11 so that VL = 9 V.

500

I0

+_

500

12 V

6 mA

+_

VL

R

3I0


Solution: The voltage across the 500- resistor in the right-hand segment is

VL = (3I0 +6103)500.

Setting VL = 9 V leads to

I0 =13

(VL500 610

3)

=13

(9

500 6103

)= 4 mA.

The left-hand loop has to satisfy KVL:

12+ I0R+500I0 = 0,

which leads to

R =12500I0

I0=

12I0500

=12

4103 500 = 2500 .


Problem 2.12 A high-voltage direct-current generating station delivers 10 MWof power at 250 kV to a city, as depicted in Fig. P2.12. The city is representedby resistance RL and each of the two wires of the transmission line between thegenerating station and the city is represented by resistance RTL. The distance betweenthe two locations is 2000 km and the transmission lines are made of 10-cmdiametercopper wire. Determine (a) how much power is consumed by the transmission lineand (b) what fraction of the power generated by the generating station is used by thecity.

+_

RLV0

RTL

RTL

2000 km

Station

(city)

Figure P2.12: Diagram for Problem 2.12.

Solution: For copper, = 1.72108 -m.

For each wire of the transmission line, Eq. (2.2) leads to

RTL =

A=

1.721082106

pi(5102)2

= 4.4 .Delivering 10 MW at 250 kV to the city means that

PL =V 2LRL

= 107 W, with VL = 2.5105 V.

Hence,

RL =V 2LPL

=(2.5105)2

107 = 6.25103 .

Also, the current flowing through RL is

I =VLRL

=2.51056.25103 = 40 A.

(a) The power consumed by transmission lines isPTL = 2I2RTL

= 2 (40)24.4 = 14080 W.Total power generated by the source is

Ps = PTL +PL = 107 +14080 = 1.001408 MW,and the fraction used by the city is

Fraction = PLPs

=107

1.001408107

= 0.9986 99.86%.


Problem 2.13 Determine the current I in the circuit of Fig. P2.13 given that I0 = 0.

3 1

2

1

1

1

24 V

I0 = 0

I

+_

+ _

+_

+_

+_

+_

I2

V2

V1

V4

V5V3

I1

I1I2

a

b

+

_

Figure P2.13: Circuit for Problem 2.13.

Solution: Since I0 = 0, the middle branch in the bridge section is of no consequence.The voltage between nodes a and b is the same following either path between them.Hence,

I11+ I11 = I21+ I21,

or I1 = I2, which is obvious considering that all 4 resistors in the bridge section arethe same. For the left loop that includes the 24-V source,

24+V1 +V2 +V3 = 0V1 = 3IV2 = I1V3 = I1

and I = I1 + I2 = 2I1. Hence,

24+3I +2I1 = 024+3I + I = 0

or

I =244

= 6 A.


Problem 2.14 Determine currents I1 to I3 in the circuit of Fig. P2.14.

7 12

4

2

8 18 V

1 A

3 A

I2

a

bI3

I1

+

_


Solution: For the loop containing the 18-V source,

18+32+8I1 = 0.

Hence, I1 = 1.5 A.KCL at node a gives

31 I1 I2 = 0I2 = 2 I1 = 21.5 = 0.5 A.

KCL at node b gives

1+ I2 I3 = 0I3 = 1+ I2 = 1+0.5 = 1.5 A.


Problem 2.15 Determine Ix in the circuit of Fig. P2.15.

Ix

2

5

12 V 1 A+_

I


Solution:KVL gives: 12+5I +2Ix = 0.KCL gives: I +1 Ix = 0.Solution of the two equations yields Ix = 177 = 2.43 A.



I2I1

I4

I3

6 4

1 8

12 V

4 A

5 V

1 V

+_

+_

+_

+

_

+

_

+

_


Solution: Application of KVL to the outer-parameter loop gives

12+41+8I3 +51 = 0,

which givesI3 = 0.5 A.

KVL for the left-most loop is

12+41+4I1 = 0,

which leads toI1 = 2 A.

KCL at the top center node gives

4 I1 I2 I3 = 0

or

I2 = 4 I1 I3 = 420.5 = 1.5 A.

KCL at the bottom left node gives

I4 + I14 = 0,

or

I4 = 4 I1 = 42 = 2 A.



I1 I2 I3 I4

2 6 A

4 2 4


Solution: The same voltage exists across all four resistors. Hence,

2I1 = 4I2 = 2I3 = 4I4.

Also, KCL mandates thatI1 + I2 + I3 + I4 = 6

It follows that I1 = 2 A, I2 = 1 A, I3 = 2 A, and I4 = 1 A.


Problem 2.18 Determine the amount of power dissipated in the 3-k resistor in thecircuit of Fig. P2.18.

103V02 k 3 k10 mA V0+

_


Solution: In the left loop,

V0 = 101032103 = 20 V.

The dependent current source is I0 = 103V0 = 20 mA.The power dissipated in the 3-k resistor is

p = I20 R = (20103)23103 = 1.2 W.


Problem 2.19 Determine Ix and Iy in the circuit of Fig. P2.19.

4IxI

2 6

4 10 V+

_

+

_

Ix Iy


Solution: Application of KVL to the two loops gives

10+2Ix +4I = 04I +6Iy4Ix = 0.

Additionally,I = Ix Iy.

Solution of the three equations yields

Ix = 3.57 A, Iy = 2.86 A.


Problem 2.20 Find Vab in the circuit in Fig. P2.20.

Ia

b

2 2

2

6 V 12 V

+

_

+

_Vab

+

_


Solution: For the lower loop, KVL gives

6+4I +12 = 0,

or

I =1.5 A.

Moving from a to b via the 12-V supply,

Vab = (1.5)2+12 = 9 V.


Problem 2.21 Find I1 to I3 in the circuit of Fig. P2.21.

3 k

4 k2 k16 V

+

_

12 V

8 V

+

_

+

_

I1 I3

I2 +

_

+

_

+

_


Solution: Application of KVL to the outer-perimeter loop gives

16+3103I18+12 = 0,

which leads toI1 = 4 mA.

For the left loop,

16+3103I1 +4103I2 = 0,

I2 =163103I1

4103 =1631034103

4103 = 1 mA.

Consequently,I3 = I1 I2 = 41 = 3 mA.


Problem 2.22 Find I in the circuit of Fig. P2.22.

2I

3 10 V+

_

I+

_

+

_

+

_


Solution:10+2I +3I = 0.

Hence,I =

105 = 2 A.


Problem 2.23 Determine the amount of power supplied by the independent currentsource in the circuit of Fig. P2.23.

V1

I

2

2

0.2 A4

V1

+

_


Solution: KCL at top node gives

0.2+ V14 I = 0

Also I = V1/2 (for the 2- resistor).Hence,

0.2+ V14

V12

= 0

V1 = 0.8 V.

The voltage across the 0.2-A source is 2V1 = 1.6 V.Power dissipated is

P = V I = 1.60.2 = 0.32 W.


Problem 2.24 Given that in the circuit of Fig. P2.24, I1 = 4 A, I2 = 1 A, and I3 =1 A, determine node voltages V1, V2, and V3.

R1 = 18

1 6 6

6 18 40 V

+

_

V1I1 V2

V3

I2

I3


Solution:R1 = 18

1 6 6

6 18 40 V

+

_

V1 V2V3

I4

Fig. P2.24 (a)

V1 = 40 (1 ) I1 = 404 = 36 V.

KCL at node V1 leads to

I4 = I1 I2 = 41 = 3 A.

Hence,V2 = V1 +6I4 = 3663 = 18 V.

The voltage across R1 is 18I2.Hence,

V3 = V118I2 = 36181 = 18 V.


Problem 2.25 After assigning node V4 in the circuit of Fig. P2.25 as the groundnode, determine node voltages V1, V2, and V3.

6 6

3 3

12 V3 A

+

_

1 A

6

1 A

V1 V3V2

V4


Solution:

6 6

3 3

12 V3 A

+

_

I1

1 A

6

1 A

V1 V3V2

V4

Fig. P2.25 (a)

From KCL at node V1, the sum of currents leaving the node is

3+ I11 = 0,

or

I1 =3+1 =2 A.

Node voltages (relative to V4):

V1 =61 =6 V,V2 = V13I1 =63(2) = 0,V3 = 61 = 6 V.


Problem 2.26 After assigning node V1 in the circuit of Fig. P2.25 as the groundnode, determine node voltages V2, V3, and V4.

Solution:

6 6

3 3

12 V3 A

+

_

I1

1 A

6

1 A

V1V3

V2

V4

Fig. P2.26 (a)

From KCL at node V1, the sum of currents leaving the node is

3+ I11 = 0,

or

I1 =3+1 =2 A.

Hence, relative to node V1:

V2 =3I1 =3(2) = 6 V.V3 = 12 V

(because () terminal of voltage source is at V1)V4 = 6(1) = 6 V.


Problem 2.27 In the circuit of Fig. P2.27, I1 = 42/81 A, I2 = 42/81 A, andI3 = 24/81 A. Determine node voltages V2, V3, and V4 after assigning node V1 asthe ground node.

9

V3 V4

V2

V1

6 9

6

9

6

6 V

+

_

6 V

+

_

+

_

+

_

I3

I1I2


Solution: At node V3, KCL gives

I2 + I4 I3 = 0,

or

I4 = I3 I2 =2481

4281 =

1881 A.

I4

9

V3

V1 6

V4

V2

V1

6 9

6

9

6

6 V

+

_

6 V

+

_

+

_

+

_

I3

I1I2

Fig. P2.27 (a)

V3 = V169I2

= 06+9 4281 = 1.33 =43 V.

V2 = V36I4

=43 6

(

1881

)= 0,

V4 = V1 +69I1

= 0+69(

4281

)=

43 V.


Problem 2.28 The independent source in Fig. P2.28 supplies 48 W of power.Determine I2.

I2 0.25I112 V

I3I1 R

R

RR

+

_


Solution: From

P = V0I1,

I1 =PV0

=4812

= 4 A.

Current of dependent current source is the same as I3. Hence,

I3 = 0.25I1 = 0.254 = 1 A.

By KCL,I2 = I1 I3 = 41 = 3 A.


Problem 2.29 Given that I1 = 1 A in the circuit of Fig. P2.29, determine I0.

I0

I1 = 1 AI2

1 2 4 8 16

I3I4I5


Solution: Since the 16- and 8- resistors are connected in parallel, they have thesame voltage across them, namely

V = 16 I1 = 161 = 16 V.

By KCL, I0 equals the sum of the currents flowing in all five resistors:

I0 =161

+162

+164

+168 +

1616

= 31 A.


Problem 2.30 What should R be in the circuit of Fig. P2.30 so that Req = 4 ?

Req

a

b

R2

5

1

6


Solution: The parallel combination of R and 2- resistor is

R1 =2R

2+R.

R1 is in series with 5- resistor. Hence

R2 = R1 +5 =2R

2+R+5.

R2 is in parallel with 6- resistor:

R3 =6

(2R

2+R+5

)

6+ 2R2+R

+5,

and

Req = 1+R3 = 1+6

(2R

2+R+5

)

11+2R

2+R

= 4.

Solving for R leads toR = 2 .


Problem 2.31 Find I0 in the circuit of Fig. P2.31.

3 6

4

12 18 A

I0


Solution: Combining the 3- and 6- resistors in parallel gives

R =363+6 =

189 = 2 .

The new circuit becomes

6 12

18 A

I0

Current division leads to

I0 =(

Req12

)18 = 6126+12 18 = 6 A.


Problem 2.32 For the circuit in Fig. P2.32, find Ix for t < 0 and t > 0.

t = 0

2 3 2

1

4

2 4

4 15 V+

_

Ix

+

_

Figure P2.32: Circuit with SPDT switch for Problem 2.32.

Solution:For t < 0:

2 3 4

2 4

4 15 V+

_

Ix

2 3 4

2 2

15 V

+

_

Ix

2 2 6 || 3 =

2

15 V

+

_

Ix

1

2

15 V

+

_

Ix

= 2 6 3

6 + 3

Ix =15

2+1= 5 A

For t > 0:

2

15 V

+

_

Ix

Ix =152

= 7.5 A.


Problem 2.33 Determine Req at terminals (a,b) in the circuit of Fig. P2.33.

8 16

32 8

4 Req

a

b


Solution:

a

b

8 16

32 8

4 Req

Terminals (a,b) are connected together through a short circuit. Hence,

Req = 0.


Problem 2.34 Select R in the circuit of Fig. P2.34 so that VL = 5 V.

Solution: Multiple application of the source-transformation method leads to the finalcircuit below.

R 1 k

2 k5 k5 mA VL

+

_

R 1 k

2 k

5 k

25 VVL

+

_

+

_

1 k

2 k VLR1Is =

R1 = R + 5k

Is

+

_

1 k

VL

R2

Vs

+

_

123

R3+

_

25

R1

Figure P2.34: Circuit

for Problem 2.34.

R2 = R1 2 k =2R1103

R1 +2103=

2103(R+5103)R+7103

Vs = IsR2 =25R2R1

=50103

R+7103

Since no current flows through R3,

VL = Vs =50103

R+7103 .

Setting VL = 5 V leads toR = 3 k.


Problem 2.35 If R = 12 in the circuit of Fig. P2.35, find I.

Solution:

R

4 20 V

R+

_

I 4 20 V

+

_

I

R/2 R/2

R/2 R/2

20 V

+

_

I4

RR

RR

RR

RR

4 20 V

+

_

I


for Problem 2.35.

R2

=122

= 6

I =20

4+6 = 2 A.


Problem 2.36 Use resistance reduction and source transformation to find Vx in thecircuit of Fig. P2.36. All resistance values are in ohms.

Solution:

16 16

4

12 16 1664

10 A

Vx+ _

8

4

12 864

10 A

Vx+ _

6 864

10 A

Vx+ _

834

10 A

Vx+ _

83 4

Vx+ _

30 V+

_


for Problem 2.36.

Vx =304

3+4+8 = 8 V.


Problem 2.37 Determine A if Vout/Vs = 9 in the circuit of Fig. P2.37.

Solution:

AI1

I1+

_

12 12 3

3

6 VoutVs

+

_

AI1

+

_

6

3

2 Vs

I

Vout

+

_


for Problem 2.37.

I =Vs9

I1 =I2

=Vs18

Vout = AI12 =AVs18 2 =

AVs9

VoutVs

=A9 = 9.

HenceA = 81.


Problem 2.38 For the circuit in Fig. P2.38, find Req at terminals (a,b).

Solution:

a

b

8 16

32 8

4 Req

Req = 4+5 = 9 .


Problem 2.39 Find Req at terminals (c,d) in the circuit of Fig. P2.38.

Solution:

a

b

c

d

5 3 5

5 3 6 6

c

d

5

5 12 6 Req

c

d

5

5 4 Req

Req = 4+5+5 = 14 .


Problem 2.40 Simplify the circuit to the right of terminals (a,b) in Fig. P2.40 tofind Req, and then determine the amount of power supplied by the voltage source. Allresistances are in ohms.

Solution:

25 V

3 5 8

668

412 12

a

b

+

_

Req

25 V

3 5 8

68

412 4

a

b

+

_

Req

25 V

3 5

68

46 12 || 12 = 6

8 || 8 || 4 = 2

5 + 6 || 6 = 5 + 3 = 8

a

b

+

_

Req

25 V

3

84

8

a

b

+

_

Req

25 V

3

2

a

b

+

_

Req


for Problem 2.40.

Req = 3+2 = 5

P =V 2

Req=

(25)25 = 125 W.


Problem 2.41 For the circuit in Fig. P2.41, determine Req at(a) Terminals (a,b)(b) Terminals (a,c)(c) Terminals (a,d)(d) Terminals (a, f )

e f

ba

c

d

2

2

2

2

2

2

2

2

2

2


Solution: All resistances are in ohms.(a)

ba

2

2

2

2

2

2

Req

ba

22

Req

= 1.56 2

6 + 2

Req = 1.5+2+2 = 5.5 .(b)

Req a

c

2

2

2

222

Req a

c

2

82

Req = 2+2 = 4 .(c)

Req

a

2

2

2

22

d2

Req

a

2

d2

= 1.56 2

6 + 2

Req = 2+2+1.5 = 5.5 .(d)

Req

a

2

2

2

2

2

2

f

2

= 24 4

4 + 4Req

a

2

f

Req = 2+2+2 = 6 .


Problem 2.42 Find Req for the circuit in Fig. P2.42. All resistances are in ohms.

Solution:

1010

5

5

10

10

10

Req

Req 10 20 205

5

Req 55

5


for Problem 2.42.

Req = 15 .


Problem 2.43 Apply voltage and current division to determine V0 in the circuit ofFig. P2.43 given that Vout = 0.2 V.

Solution:

Vout = 0.2 V

I5

2

1

4

2

4

8

+

_

+

_

V1

+

_V2

+

_

V3

+

_

V4

+

_

V5

+

_

V0


for Problem 2.43.

I3

I4

I2

I1

I1 =0.21

= 0.2 A

I2 =V22

=I12

(2+1) = 0.3 A

I3 = I1 + I2 = 0.5 A

I4 =V44

=V3 +V2

4=

4I3 +2I24 = 0.65 A

I5 = I3 + I4 = 1.15 AV0 = V4 +V5 = 4I4 +8I5 = 11.8 V.


Problem 2.44 Apply source transformations and resistance reductions to simplifythe circuit to the left of nodes (a,b) in Fig. P2.44 into a single voltage source and aresistor. Then, determine I.

5 A4

3 A

12 2

10 a

b

I


Solution:


5 A

3 A

12 2

10 a

b

30 V

12

10 2 a

b

+_

+

_

10 V

12

12 a

b

+

_

40 V

12 12

a

b

4012

4012

6

a

b

4

6 a

b

+

_

20 V

I

Fig. P2.44 (a)

I =20

6+4 = 2 A.


Problem 2.45 Determine the open-circuit voltage Voc across terminals (a,b) inFig. P2.45.

Solution:

Voc

6

2 A

+

_

3 30 V

5 a

b

+

_

Voc

6

2 A

+

_

3 6 A

a

b

5

Voc

6

+

_

8 A

a

b

R = =3 + 5

3 5 15

8

Voc

6

+

_

15 V

a

b

15

8

Fig. P2.45 (a)

+

_

Hence,Voc = 15 V.


Problem 2.46 Use circuit transformations to determine I in the circuit of Fig. P2.46.

Solution:

4

4 2 A

3

6 3 A

30 V

2

+

_

I

4

8 V

3

6

12 V 10 A

2

4

I+

_

+_

8

20 V

= 2

10 A2

I+

_

6 3

9

8 2

20 V

20 V

2

I+

_

+ _

Fig. P2.46 (a)

I =2020

8+2+2 = 0.



+

_

12 V

I1

I26

12 I3

I46

3

Figure P2.47: Circuit of Problems 2.47 and 2.48.

Solution:

I1 =1212

= 1 A,

I2 =126 = 2 A,

I3 =123 = 4 A,

I4 =126 = 2 A.


Problem 2.48 Replace the 12-V source in the circuit of Fig. P2.47 with a 4-Acurrent source pointing upwards. Then, determine currents I1 to I4.

Solution:

4 A

I1

I26

12 I3

I46

3

Req =(

112

+16 +

13 +

16

)1=

129 .

I1 =Req124 =

49 A,

I2 =Req6 4 =

89 A,

I3 =Req3 4 =

169 A,

I4 =Req6 4 =

89 A.


Problem 2.49 Determine current I in the circuit of Fig. P2.49.

Solution: Resistance combining leads to

Fig. P2.49 (a)

30 60

5

50 V

10 25 40

+

_

10 10

I

= 24

5

50 V

25 40

+

_

5

I

64

5

50 V+

_

30

I

5

50 V+

_

= 20.43

I

40 60

40 + 60

30 64

94

I =50

5+20.43 = 1.97 A.


Problem 2.50 Determine the equivalent resistance Req at terminals (a,b) in thecircuit of Fig. P2.50.

Solution:

5

4

6

4

4

5

Req

a

b

10

4

4

5

5

a

b

R = 4 + 4 + (5 || 5 || 10) = 10

a

b

Fig. P2.50 (a)

R = 4+4+(5 5 10) = 10 .


*2.51 Determine current I in the circuit of Fig. P2.51.

Solution:

1 k

6 mA

2 mA

2 k

2 k

16 V

8 V

2 k 5 mA

+_

+

_

I

1 k

8 mA

2 k

2 k

16 V

8 V

2 k 5 mA

+_

+

_

I

8 mA

= 0.5 k(2k || 2k || 1k)

16 V

8 V

2 k 5 mA

+_

+

_

I

0.5 k

16 V

8 V

2 k 5 mA

+_

+

_

4 V+

_

I

0.5 k2 k 5 mA24 mA

I1

I1

0.5 k2 k 19 mA

I1

Fig. P2.51 (a)

I1 = (19 mA)(

0.5k2.5k

)= 3.8 mA.


Problem 2.52 Determine voltage Va in the circuit of Fig. P2.52.

Solution:

V a

4

2.5 A

2

2 A

2 +

_

2 A

5 A

4

4

V a

4

10 V 10 V

16 V

2

2

+_

2 A

2 A

4

8

V a

+_

4 V

+_

2 A 2 A

4

8

V a

+_

4 A

4

8 V a

+_

+_

+_

+_

I

Fig. P2.52 (a)

By current division,I =

484+8 = 2.67 A


andV = 4I = 10.67 V.


Problem 2.53 Convert the circuit in Fig. P2.53(a) from a to a Y configuration.

Solution:

a

b

c

d

3 1

6 a

b

c

d

R1 R2

R3

Figure P2.53(a)

R1 =63

6+3+1 = 1.8

R2 =61

10 = 0.6

R3 =31

10 = 0.3 .


Problem 2.54 Convert the circuit in Fig. P2.53(b) from a T to a configuration.

Solution:

8 2

4

Ra

Rc

Rb

Figure P2.53(b)

Ra =28+24+48

8 =568 = 7

Rb =28+24+48

4=

564

= 14

Rc =28+24+48

2=

562

= 28 .


Problem 2.55 Find the power supplied by the generator in Fig. P2.55.

Solution:

R1 = 18

Y

1 6 6

6 18 20 V+

_

1 18

18 18 18

18

20 V

+

_

1 9

18 9 20 V+

_

1

9 20 V+

_

I

I =2010 = 2 A

P = V I = 202 = 40 W.


Problem 2.56 Repeat Problem 2.55 after replacing R1 with a short circuit.

Solution:

Y

1 6 6

6 18 20 V+

_

1 18

18 18 18 20 V+

_

1

18 9 20 V+

_

1

20 V

+

_

I

= 6 9 18

9 + 18

I =20

6+1 =207

P = V I = 20 207

= 57.1 W.


Problem 2.57 Find I in the circuit of Fig. P2.57.

Solution:

9 6 9

6 9 6

3 V

+

_

3 V

+

_

I

I

9

18 18

18 9

9

3 V 3 V

+

_+_

I

9 18 18

6 9

1/3 A

3 V

+

_

I

12 18

9

1/6 A

3 V

+

_

+_

I

36/5 9

6/5 V

3 V

+

_

2 V

I

18

6 9 6

3 V

+

_

+_

T Transformation

9 || 8 = 6

18 || 12 =36

5

I =3+ 659+ 365

0.26 A.


Problem 2.58 Find the power supplied by the voltage source in Fig. P2.58.

Solution:

6 R = 6 6

3 3

4 V

+

_

6 Rb = 15 Rc = 15

Ra = 7.5

6

4 V

+

_

Ra =33+36+36

6 = 7.5

Rb =33+36+36

3 = 15

Rc = Rb = 15 .

4.3 4.3 7.5

4.006

4 V

+

_

4 V

+

_

I

R =(4.3+4.3)7.54.3+4.3+7.5 = 4.006

I =4

4.006 = 0.998 A

P = I2R = (0.998)24.006 = 3.99 4 W.


Problem 2.59 Repeat Problem 2.58 after replacing R with a short circuit.

Solution:

6 6

3 3

4 V

+

_

6 3 3 6

4 V

+

_

2

4 V

+

_

I

= 2 6 3

6 + 3

I =4

2+2= 1 A

P = V I = 41 = 4 W.


Problem 2.60 Find I in the circuit of Fig. P2.60. All resistances are in ohms.

Solution:

14

2 2

2 2

12 V

I+_

1

2 2

12 V

I

+_

R1 = 0.5

R3 = 1R2 = 1

Y

R1 =22

2+2+4= 0.5

R2 =24

2+2+4= 1

R3 = R2 = 1 .

1

3 3

12 V

I

+_

0.5

1

12 V

I

+_

0.5

1.5

I =12

1+0.5+1.5 = 4 A.


Problem 2.61 Find Req for the circuit in Fig. P2.61.

Solution:

Req

18

6

6 18

1 6

18

9

18 18

18

1

18

18

18

9

Req

9

18

1

9

18

9

Req

18

1 9

9

Req

9

1

Req

Req = 9+1 = 10 .


Problem 2.62 Find Req at terminals (a,b) in Fig. P2.62 if(a) Terminal c is connected to terminal d by a short circuit(b) Terminal e is connected to terminal f by a short circuit(c) Terminal c is connected to terminal e by a short circuit

All resistance values are in ohms.

e

d

a bf

c

3

3 3

3 3

3Req


Solution:(a)

d

c

a33 3

3 3

b 3

d

c

a9

9 9

9 9

9b a4.5 4.5b

Req = 9

(b)d

c

ae f

3

3

b 3

3 3

3

a3 3b

Req = 6

(c)


dc

ae

3

3

b 3

3 3

3

d

c

a3

3

b 3

3

3

a3 3b

Req = 3 + 3 + 2 = 8

= 26 3

6 + 3


Problem 2.63 For the Wheatstone bridge circuit of Fig. 2-30, solve the followingproblems:

(a) If R1 = 1 , R2 = 2 , and Rx = 3 , to what value should R3 be adjusted soas to achieve a balanced condition?

(b) If V0 = 6 V, Ra = 0.1 , and Rx were then to deviate by a small amount toRx = 3.01 , what would be the reading on the ammeter?

Solution:(a) From Eq. (2.45),

R3R1

=RxR2

, R3 =R1RxR2

=13

2= 1.5 .

(b)

R1

I1 I2

I4I3 R3

RaIa

R2

Rx

V0 V1 V2

V0

+

KCL equations at nodes V1 and V2 are

I1 = I3 + IaI4 = I2 + Ia

KVL for the left loop and outside perimeter loop are

V0 + I1R1 + I3R3 = 0,V0 + I2R2 + I4Rx = 0.

Also, for the upper triangle of the bridge,

I1R1 + IaRa I2R2 = 0.

Simultaneous solution of the five equations leads to

Ia =(R3R2RxR1)V0

R2Rx(R1 +R3)+(R2 +Rx)[R1R3 +Ra(R1 +R3)].

For R1 = 1 , R2 = 2 , R3 = 1.5 , Rx = 3.01 , and V0 = 6 V,

Ia =2.5 mA.


Problem 2.64 If V0 = 10 V in the Wheatstone-bridge circuit of Fig. 2-31 andthe minimum voltage Vout that a voltmeter can read is 1 mV, what is the smallestresistance fraction (R/R) that can be measured by the circuit?

Solution:

Vout =V04

(RR

)

RR

=4VoutV0

=4103

10 = 4104

,

or 4 parts in 10,000.


Problem 2.65 Suppose the cantilever system shown in Fig. 2-38 is used in theWheatstone-bridge sensor of Fig. 2-31 with V0 = 2 V, = 1 109 m2/N,L = 0.5 cm, W = 0.2 cm, and H = 0.2 mm. If the measured voltage is Vout =2 V,what is the force applied to the cantilever?

Solution: From Example 2-16,

Vout =V04

FL

WH2.

Solving for F , we have

F = 4(

VoutV0

)WH2

L

= 4(22

)0.2102 (0.2103)2

(1109)0.5102

= 64 N.


Problem 2.66 A touch sensor based on a piezoresistor built into a micromechanicalcantilever made of silicon is connected in a Wheatstone-bridge configuration with aV0 = 1 V. If L = 1.44 cm and W = 1 cm, what should the thickness H be so thatthe touch sensor registers a voltage magnitude of 10 mV when the touch pressure is10 N?

Solution: From Example 2-16:

Vout =V04

FL

WH2.

Solving for H, we have

H =[(

V04Vout

)

FLW

]1/2

=

[1

4 (10103) (1109)101.44102

102

]1/2= 0.6 mm.


Problem 2.67 Determine I1 and I2 in the circuit of Fig. P2.67. Assume VF = 0.7 Vfor both diodes.

I2I153 53

6 V

+_

0.7 V 0.7 V

+_+

_


Solution: The diode in the left-hand loop is reverse biased, so

I1 = 0.

In the right-hand loop, the diode is forward biased. Hence,

I2 =60.7

53 = 0.1 A.


Problem 2.68 Determine V1 in the circuit of Fig. P2.68. Assume VF = 0.7 V for alldiodes.

50

25

100

9 V

+_

+

_

V1


Solution:

I =93(0.7)

50+100+25 0.04 A,

V1 = 100I = 4 V.


Problem 2.69 If the voltage source in the circuit of Fig. P2.69 generates a singlesquare wave with an amplitude of 2 V, generate a plot for out for the same timeperiod.

100

+

_

out

+

_

s(t)

2 V

-2 V

tT

s(t)

Figure P2.69: Circuit and voltage waveform for Problem 2.69.

Solution: Current will flow through the loop only when s(t) is greater than 0.7 V.Hence,

out =

{20.7 = 1.3 V for 0 t T/20 for T/2 t T.

1.3 V

tTT/2

vout


Problem 2.70 If the voltage source in the circuit of Fig. P2.70(a) generates thesingle square waveform shown in Fig. P2.70(b), generate plots for i1(t) and i2(t).

Solution:

+

_

i1

s(t)

i2

73 146

s(t)

t (s)4

8 V

8 V

2

(b) Square wave

(a)

t (s)

0.1 A

0.05 Ai2

i1

8 0.7= 0.05 A

146

8 + 0.7= 0.1 A

73


Problem 2.71 If the voltage source in the circuit of Fig. P2.70(a) generates thesingle triangular waveform shown in Fig. P2.70(c), generate plots for i1(t) and i2(t).

Solution:

+

_

i1

s(t)

i2

73 146

s(t)

t (s)4

8 V

8 V

2

(c) Triangular wave

(a)

31 2 4t (s)

0.1 A

0.05 Ai2

i1


Problem 2.72 Use the DC Operating Point Analysis in Multisim to solve forvoltage Vout in the circuit of Fig. P2.72. Solve for Vout by hand and compare withthe value generated by Multisim. See the solution for Exercise 2-14 (on CDROM ) for howto incorporate circuit variables into algebraic expressions.

Vout10 10

25 15

2.5 V

+

_

+

_

Circuit for Problem 2.72.

Solution:

A. By-Hand Solution

By voltage division:

Vout =(

1010+25

)2.5 = 0.714 V.

Vout

25

10

25

2.5 V

+

_

+

_


B. By Multisim

Circuit in MultiSIM Schematic Capture

DC Operating Point Solution. The value in the last row, V(1)-V(3), is the specific

solution to the problem.


Problem 2.73 Find the ratio Vout/Vin for the circuit in Fig. P2.73 using DCOperating Point Analysis in Multisim. See the Multisim Tutorial included on theCD on how to reference currents in ABM sources (you should not just type in I(V1)).

Vin

Iin

Vout100Iin10 k

1 k

1 k

+

_

+

_


Solution:

Vout = (100Iin)1000

= 105 Vin104 = 10Vin.

Hence,VoutVin

= 10.

Circuit in MultiSIM Schematic Capture

DC Operating Point solution


Problem 2.74 Use DC Operating Point Analysis in Multisim to solve for all sixlabeled resistor currents in the circuit of Fig. P2.74.

I1

I3

I5

1

1

1

I2

I4

I6

1

1

1

1 A2 V

3 V

+ _

+ _


Solution:

Circuit in MultiSIM Schematic Capture DC Operating Point solution

I1 =V5V3

1= 1.52 =0.5 A,

I2 =V5V4

1= 1.50 = 1.5 A,

I3 =V3V1

1= 22 = 0,

I4 =V4V2

1= 0+1 = 1 A,

I5 =V11

= 2 A,

I6 =V21

=1 A.


Problem 2.75 Find the voltages across R1, R2, and R3 in the circuit of Fig. P2.75using the DC Operating Point Analysis tool in Multisim.

R1

R2V1

10

R3

15

30 15 V 1.5I

I+

_


Solution: When built, the circuit should look like that shown below:

To set the current source up so that it refers to the proper current, double-click onthe current source. This will bring up the window shown below:

In ABM syntax, this becomes: 1.5*V(2)/30. Enter this as shown in the figure above.Using DC Operating Analysis, select the following outputs:

V(1)V(2)V(3)V(1) V(2)V(2) V(3)

The numerical results are shown in the grapher window:


Problem 2.76 Find the equivalent resistance looking into the terminals of thecircuit in Fig. P2.76 using a test voltage source and current probes in the InteractiveSimulation in Multisim. Compare the answer you get to what you obtain from seriesand parallel combining of resistors carried out by hand.


Solution: The two 12-k resistors in parallel are equal to 6 k because for anyparallel combination of identically valued resistors,

Req =RN

,

where Req is the total equivalent resistance of the combination, R is the resistance ofthe elements, and N is the number of resistors in the parallel network.

On the right side there is now a series combination of a 6-k resistor, a 10-kresistor, and a 3.9-k resistor. Together these equal 19.9 k.

This new value is in parallel with a 1-k resistor. Therefore19.9 1 = 952.153 .

This is then in series with a 1-k resistor, so1.952153 k.

This is then in parallel with another 1-k resistor, so1.952 1 = 661.25 .

Finally, this value is in series with a 4.7-k resistor, soReq = 4.7 k+661.25 = 5.361 k.

When you construct the circuit, add a voltage source across the two pins, and add thecurrent probe right above the test source. Run the Interactive Simulation to get thefollowing screen shot:


Note that you can set the test voltage to anything you want. Unity test voltage (andcurrent) sources usually just make things easier to solve.

Take the values from this simulation, to get

Req =1.00

187106 = 5.348 k,

which is pretty close (0.24% off) from what we obtained by hand. The differencescan be explained by rounding.


m2.1 Kirchhoffs Laws: Determine currents I1 to I3 and the voltage V1 in thecircuit of Fig. m2.1 with component values Isrc = 1.8 mA, Vsrc = 9.0 V, R1 = 2.2 k,R2 = 3.3 k, and R3 = 1.0 k.

+

_

+

_

Vsrc

Isrc

R2

R1

I2

R3

V1

I1

I3

Figure m2.1: Circuit for Problem m2.1.

Solution:

Summary Comparison:

Results: I1 (mA) I2 (mA) I3 (mA) V1 (V)

Analysis -0.56 2.36 -1.80 9.59

Simulaon -0.56 2.36 -1.80 9.58

Measurement -0.51 2.43 -1.92 9.80

Rela ve Dierences:

Simulaon -- Analysis 0.0% 0.0% 0.0% -0.1%

Measurement -- Analysis -8.9% 3.0% 6.7% 2.2%


Analytical Solution:


Multisim Results:


myDAQ Results:

Further Exploration:

No change observed in I1, I2, and I3, but V1 varied from 7.88 V to 9.88 V.

Explanaon: The current source maintains a constant current in the lower branch and therefore the remaining

circuit sees no dierence. However, as the voltage changes across R3 the current source voltage adapts its own

voltage V1.


m2.2 Equivalent Resistance: Find the equivalent resistance between the followingterminal pairs in the circuit of Fig. m2.2 under the stated conditions:

(a) a-b with the other terminals unconnected,(b) a-d with the other terminals unconnected,(c) b-c with a wire connecting terminals a and d, and(d) a-d with a wire connecting terminals b and c.

Use these component values: R1 = 10 k, R2 = 33 k, R3 = 15 k, R4 = 47 k,and R5 = 2 k.

R2

R4d

R1 R5R3

c

ba


Solution:

Summary Comparison:

Results: R_AB (kohms) R_AD (kohms) R_BC (kohms) R_AD (kohms)

Analysis 16.72 9.77 21.45 9.025

Simulaon 16.72 9.766 21.45 9.025

Measurement 16.64 9.79 21.4 9.04

Rela ve Dierences:

Simulaon -- Analysis 0.0% 0.0% 0.0% 0.0%

Measurement -- Analysis -0.5% 0.2% -0.2% 0.2%


Multisim Results:


myDAQ Results:

See summary comparison table for DMM ohmmeter measurements.

Further Exploration Results:

Measurement Voltage (V) Current (mA) Calculated resistance (k)

R_AB 4.89 0.29 16.9

R_AD 4.88 0.50 9.76

R_BC with wire 4.91 0.23 21.4

R_AD with wire 4.88 0.54 9.04

These calculated resistance values agree with the myDAQ ohmmeter measurements within 2%.


m2.3 Current and Voltage Dividers: Apply the concepts of voltage dividers,current dividers, and equivalent resistance to find the currents I1 to I3 and the voltagesV1 to V3 in the circuit of Fig. m2.3. Use these component values: Vsrc = 12 V,R1 = 1.0 k, R2 = 10 k, R3 = 1.5 k, R4 = 2.2 k, R5 = 4.7 k, and R6 = 3.3 k.

+

_

V2 R4

+

_

V1Vsrc

R5

R3

R2

R1

R6

I1

I2

I3

+ _V3

+

_


Solution:

Summary Comparison:

Results: I1 (mA) I2 (mA) I3 (mA) V1 (V) V2 (V) V3 (V)

Analysis 1.82 1.40 0.45 4.19 2.09 -5.99

Simulaon 1.82 1.40 0.45 4.19 2.09 -5.99

Measurement 1.83 1.41 0.45 4.21 2.10 -5.96

Rela ve Dierences:

Simulaon -- Analysis 0.2% 0.2% 0.0% 0.0% 0.0% 0.0%

Measurement -- Analysis 0.8% 0.9% 1.1% 0.5% 0.3% -0.6%


Multisim Results:


myDAQ Results:

See summary comparison table for DMM voltmeter and ammeter measurements.


m2.4 Wye-Delta Transformation: Find (a) the currents I1 and I2 in the circuit ofFig. m2.4 and (b) the power delivered by each of the two voltage sources. Use thesecomponent values: V1 = 15 V, V2 = 15 V, R1 = 3.3 k, R2 = 1.5 k, R3 = 4.7 k,R4 = 5.6 k, R5 = 1.0 k, and R6 = 2.2 k.

R4

R3

R1

I1

V1 V2

R2 R5

R6

+_ +_

I2


Solution:

Summary Comparison:

Results: I1 (mA) I2 (mA) P1 (mW) P2 (mW)

Analysis 4.05 4.45 60.7 66.7

Simulaon 4.05 4.45 60.7 66.7

Measurement 4.05 4.45 60.6 66.5

Rela ve Dierences:

Simulaon -- Analysis 0.0% 0.0% 0.0% 0.0%

Measurement -- Analysis 0.0% 0.0% -0.2% -0.3%


Multisim Results:

Interac ve simula on results: I1 = 4.05 mA and I2 = 4.4 5mA


Interac ve simula on results: PL = 60.703 mW delivered, PR = 66.687 mW delivered


myDAQ Results:

DMM ammeter on 20mA range se!ng measures I1 = 4.06 mA and I2 = 4.45 mA.

DMM voltmeter on 20V range se!ng measures 14.91 V for the le" source and 14.85 V for the right

source. Calculate power as product of voltage and current: PL = 60.5 mW and PR = 66.1 mW.


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circuits solutions ulaby chapter 2

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