eee ies conventional solved papers of 2013

7/25/2019 EEE IES Conventional Solved Papers of 2013

1/19

1) (a) The circuit in the question will look like :

Finding thevein equivalent across inductor terminals

th

16kV * 8 6.4V

16k 4k (By potential divider rule)

16 4thR k k

16*43.2k

16 4

The equivalent circuit looks like as shown in adjoining figure

Steady state

Inductor can be short circuited

6.4Vi 2mAR 3.2

Time - constant

1L 0.3125msR 3.2k

(b) Output of a LTI system

y t x t * h t

y(t) x h t d

t

o

u u t d 1d tu t

y(t) tu(t)

(c) LX wL 0.1w

c

1 1X

wC w


2/19

Re-drawing the circuit in phasor form

Equivalent of1 2

Z & Z

1 2

eq

1 2

Z ZZ

Z Z

=

2 2

j1 * j w j 1 jwjw

j w j 1 w 1 w1 w

Total impedance of circuit

in eq 3 2

1 jwZ Z z j0.1w

1 w

For current and voltage to be in phase Zinshould be real

Im Zin 0

2

w0.1w 0

1 w

2

1w 0.1 0

1 w

Either w = 0 or 210.1 1 w

21 w 10

2w 9 w 3rad / sec


3/19

(d) 04.6

y 2

zB 10e a mWb / m

(i) r 0

On comparison

r

4.6

r m

1 , where Xm =magnetic susceptibility

m = 3.6

(ii) B H

y 3y

z7

B 10e *10H 1730e a A / m

4.6 * 4 *10

(iii)o

BM H

y 3

y yz7

10e * 10 1730e 6227.75e a A / m4 * 10

(f)

x

H i 2exp j wt z20

On comparing with

H A exp j wt kz

20k

(i)22 40mk20

(ii)8c 3 *10

f 7.5MHz40


4/19

(iii)

x y zi i i

XBx y z

2exp j wt z 0 020

x y zd d dd d di 0 0 i * 0 2exp j wt z i 2exp j wt z 0

dy dz dx 20 20dz dy dx

y

X B i 2 * jexp j wt z20 20

yji exp j wt z d

2010

By Maxwells Equations

0 0dE

XB dt

0 0

dE 1XB

dt

2

0 0

1c

2 2

Lc XB c i jexp j wt z

10 20

2

y

E

E dt c i * j exp j wt z dt10 20t

2cexp j wt z iy

10w 20


5/19

28

6 6

1 6t sec,z 5m

15

3 *10 * 1

E exp j 2 * * 7.5 *10 * *10 * 5201510 *2 * *7 .5*10

16

y6

9 * 10 *E exp j i

410 * 2 * 7.5 * 1 0

8 y36 *10 exp j i4

(g) By the use of KCL

nm 1 2I I I

mo am nm 1 2I I I 2I I

By KVL

1 am moV 10 I 5 I

1 1 210 * I 5 2I I

1 1 2V 20I 5I

2 nm moV I 10 I 5

1 2 1 2I I 10 2I I 5

2 1 2V 20I 15I

In terms of Z-parameters

1 11 1 12 2V Z I Z I

2 21 1 22 2V Z I Z I


6/19

On Comparing

11 12 21 22Z 20 ; Z 5 ; Z 20 ; Z 15

20Z

20

5

15

(i) G G

1G( jw) cos sin 0

N

Assume G(jw) = x+ iy

For unity feedback system,

C( jw) G( jw) x iyM( jw)

R( jw) 1 G( jw) x iy 1

2 2

2 2

x yM( jw)

(1 x) y

Assume |M(jw)| = M

=> 2 2 2 2M (1 x) y x y

Squaring both sides we get

2 2 2 2 2M (1 x) y x y

Re-arranging this equation we get,

2 222

2 2

M Mx y

1 M 1 M

In our case, C( jw)

M( jw) 6dBR(jw)

=> 20log(|M(jw)|) = 6dB

=>|M(jw)| = 2

The polar plot of the equation (i)

for M=2 looks like


7/19

dB-Magnitude Phase Angle Plot

2) (a) The hollow sphere looks like as shown in figure

By Gausss law

freeD .d A Q

2D.dA Q D.4 r Q

2Q

D r4 r

for all values of r

if r < a

20 0

D QE r

E 4 r

If a < r < b

2

0 r 0 r

D QE r

4 r

If r > b

20 0

D QE r

4 r


8/19

2

0

2

0 v

2

0

Q

4 rQ

E4 r

Q

4 r

r a

a r b

r b

0 0

P E D D E

2 2 2v

v

Q Q Q 11 r4 r 4 r 4 r

(b) Forward Path

1 1 2 3 4P G G G G

2 1 5 8 4P G G G G

Loop

1 1 2 9L G G G

2 3 4 10L G G G

3 5 6L G G

4 7L G

5 1 5 8 4 10 9L G G G G G G

1 71 G (1P does not touch L4)

2 1 ( 2P touches all loops)

1 2 3 4 5 1 4 2 3 2 41 L L L L L L L L L L L


9/19

Gain =>

1 2 3 4 7 1 5 4 8

1 2 9 3 4 10 5 6 7 1 5 8 4 10 9 1 2 9 7 3 4 10 5 6 3 4 10 7

G G G G 1 G G G G G

1 G G G G G G G G G G G G G G G G G G G G G G G G G G G G

3) (a)

(i) This is an image charge problem, as there is an infinite conducting plane beneath the

charges, image charge needs to be symmetrically placed beneath the surface. The

charge configuration then looks like

Electric field at b

b bc bd baE E E E

0 0

bc 2 2

00

Q QE

8 a4 2a

0

ba 2

0

QE

4 a

;

0bd 2

0

QE

4 a

0 0

b2 2 2

0 0 0

Q Q QE x y cos45x sin45y

4 a 4 a 8 a

0

2

0

Q 111 x 1 y4 a 2 2 2 2

2

0b b

0 20

Q 11F Q E 1 x 1 y4 a 2 2 2 2

2 22

0b

2

0

Q 1 1F 1 1

4 a 2 2 2 2


10/19

2

0

2

0

Q 1 11 11 1

8 84 a 2 2

2

0

2

0

3Q

8 a

(ii) If one of charges is reversed, the charge configuration looks like

b bd ba bc

E E E E

0

bd 20

Q

E y4 a

0

ba 2

0

QE x

4 a

0bc 2

0

QE cos 45 x sin45 y

8 a

0b 2

0

Q 1 1E 1 x 1 y

4 a 2 2 2 2

0b 0 b 2

0

Q 1 1E Q E 1 x 1 y

4 a 2 2 2 2

2

0b 2

0

Q 1F 2

24 a

(b) Assure VRNis reference voltage

0RN

415V 0 V

3

0YN

415V 120 V

3

0BN

415V 120 V

3


11/19

For a balanced 3-supply line voltage leads phase voltage by 030

0RYV 415 30 V

0YBV 415 90 V

0BRV 415 150 V

I wattmeter 0415 30

4.15 66.9 A100 36.9

V wattmeter 0BN 415V 120 V3

Swattmeter *

0415* 4.15VI 120 66.93

415* 4.1553.1

3

=597W

(c) Z(s)

2

2 22 2

s s 10 s 1 1

2 s 4 s 16s 4 s 16

2 21 s 1 s

2 2s 4 s 16

Realizing Foster-1 form

n

is02 2

i 1 i

KkZ s K S

s s w

0 s 0K limsZ(s) 0


12/19

2 2s s s

z s 1 1 1 1K lim lim lim 0

s 2 2s 4 s 16

1 1 1 1Z s4 162 2s ss s

= 2 41 3

1 1

8 322s 2ss s

y y y y

11Y sC 2s

1C 2F

21

1 8YssL

;1

1L H8

3 2Y sC 2s

2C 2F

42

1 32Ys

sL

21L H32

4. a) The network given in the question looks like

LZ sL

C

1Z

SC

Equivalent impedance of R & L

eqZ R sL

eq C2

eq C

1R sLZ Z R sLsCZ s1Z Z s LC sRC 1R sLsC


13/19

Pole of Z(s)

2

1 2

RC RC 4 LCZ ,Z

2LC

=

2R R 1

2L 2L LC

Zeroes of Z(s)

0 RZ s L

On comparisonR

1 R LL

..(i)

2R 1 3j

22L LC

2

2

R 1 34LC4L

21 R 1 3

44 L LC

1 314 4LC

11

LC (ii)

Z(j0) = 1

Replace s by j0 in Z(s)

Z(j0) = R = 1

R 1 From (i)

L = 1H

From (ii)

C = 1F

(b) To find thevenin equivalent across terminals AB, we must open circuit A & B

Since secondary is open circuited, voltage induced in primary due to secondary is zero.

1 1100 0 2 j I j4 I

1100 0

I A2 5 j


14/19

AB 2 1V I j8 I j3

since 2I 0

AB Th100 0

V V j32 5j

= 55.7 21.8

For ThZ , assume 2I 0

1 1 2V 2 5j I 3j I

2 1 2V j3I j8I

For ThZ , 1V need to be shorted

1

V 0

2

1

3jII

2 5j

22 2

3j 3j IV 8 jI

2 5j

1

2Th

2 v 0

V 9 2 5j 8 jZ 0.62 6.45j

I 2 5j

For maximum power transfer, Th*

LZ Z = (0.62 6.45j)

(c) Torque expression for moving iron meter is given by

2d

1 dLT I

2 d

dT = deflecting torque

For spring controlled

CT k k = spring constant

21 dLk I2 d

dL

3 H/ rad2d

2 61k I 3 1022

22 6 63IIk 10 104 2


15/19

2 2

26

3I 3 5

25I2 10 k 2 254 4

=75

2 125

4

61.2rad

55

2

1.2rad

5. (a) 22100

AJ a mP

2 3

1 1 100 1 100 100.J A

By continuity equationp

.Jt

Substituting value of .J

3

100 p

t

3

100t

(b) Conversion time of n-bits successive Approximation ADCn

Tf

6C

8T 8 s

1 10

aT 8 s

(c) The s model of capacitor with initial condition is


16/19

Equivalent circuit in s domain looks like

Applying KVL

0V 1v I s Rs sC

00 0V v sV V C sV Vs

I S sRC 11 1R sRsC C

00

VCV sV

I S1RC sRC

=

00 0 0

V 1 1V ssV RC RCV V V

1R R 1s s

RC RC

00

V 1V RCV

1R 1s

RC

t

0 RC

0

V V 1i t 1 e u tV RCR

6) (a) 1C 10 F

2C 5 F

3C 2 F

The maximum voltage across 2 3C & C can be 5 + 2 = 7 V

Break down strength of 1C = 10 V


17/19

Which is not possible, else either 2 3C or C will be damaged

max,C2 2 max,C2Q = C V 5 5 25 C

max,C3 3 max,C3Q = C V 2 2 4 C

In series ; change must be same

C2 C3Q =Q =4 C

22

C2Q 4V 0.8VC 5

;3

3

C3Q 4V 2VC 2

V =2 3V V = 2.8 V

In parallel connection, voltage is same

2 31V V V 2.8V 1 1 1Q C V 10 2.8 28 C

Maximum safe voltage = 2.8 V

Equation capacitance of 2 32 32 3

C C 5 2 10C &C F

C C 5 2 7

1 3eq 2C C C || C

=801010 F7 7

Total charge =eq

80C V 2.8 32 C

7

(c) The bridge looks like as shown in adjoining Owens Bridge under balanced condition


18/19

42 31L R R C

= 10 x 842 x 1 x 106

= 8.42 mH

41

2

3R C 10 1R 74.07C 0.135

Power factor angle of will =

31 11

1

wL 2 1000 8.42 10tan tan

R 74.07

35.53

Power factor = cos = 0.813

7) (b) 0B H M

0

BM H

In first case

B =22Wb m ; H = 1200 A m

61 7

2M 1200 1.59 10 A / m

4 10

In second case2B 1.4Wb m ; H = 400 A/m

62 7

1.4M 400 1.113 10 A / m

4 10

Change in magnetization =6

1 2M M M 0.476 10 A /m


19/19

(c)

2 2

2

s 4 s 9Z S

s s 6

Degree of numerator = 4

Degree of Denominator = 3

Difference = 1

It is a ratio of even to odd polynomial

Lowest power of s in Numerator = 0

Lowest power of s in Denominator = 1

Difference = 1

There is a pole at 0

There is a pole at .

Pole zero plot

Poles and zeroes are alternating on jw axis and hence it satisfies all properties required for

LC impedance function.

eee ies conventional solved papers of 2013

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