1 C 2 A 3 D 4 B 5 D 6 C 7 A

8 9 A 10 D 11 C 12 A 13 D 14 B

15 A 16 B 17 D 18 D 19 B 20 A 21 C

22 A 23 A 24 C 25 D 26 C 27 B 28 A

29 B 30 31 C 32 A 33 A 34 C 35 D

36 A 37 C 38 D 39 B 40 C 41 D 42 B

43 C 44 A 45 C 46 B 47 A 48 D 49 B

50 A 51 B 52 D 53 C 54 D 55 B 56 B

57 A 58 D 59 C 60 D 61 C 62 B 63 D

64 B 65 C 66 67 C 68 A 69 D 70 B

71 C 72 B 73 D 74 75 D 76 D 77 A

78 C 79 C 80 C 81 C,B 82 A,B 83 B,C 84 D,B

85 B,D

Explanations:

1.

2.

( ) 8

( )

100

R 5Ω R 7.5

2

RMS value = 32 ≈ 4 ’∆ ÷« 2 ◊

32 162

17

3.

4.

RNS value = 17

( )

( ) In the steady state C is open

∴ VC = 10V

V Z I11 1Z I1 2

V Z I Z I Z I Z22 2Z I1 21 1 1Z2I21 1

» ÿV » Z Z ÿ» ÿ… Ÿ1…V Z

1 2I1Ÿ… Ÿ

5.

6.

⁄( )

( )

2 1Z2Z1⁄ ⁄I2

Page 1 of 14

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2

T s 2

ss 1

4s 4

0 Ω s j2j

7.

8.

9.

( ) All the root locus branches starts at S = 0

1 also going to ∞ with an angle 18060

3So, answer is (A).

( ) Phase = -180 + tan-1() Phase = -180, = 0

Phase crossover frequency = 0;

H2

gain 2∞gain margin = ∞

( )

10. ()

11. ()

∞x −4 ∞ 0 1 1

−3

S —x dx11

−4 —

1

4 4

S 12. ()

dx

4

, dt −3x t sx s −3x s

x s

xo

s3

3t

x t xoe−13. () 14. ()

15. ()

16. ()

17. ()

18. ()

2

P

19. ()

20. ()

21. ()

XVseries

P;1

P2

≈800 ’∆ ÷«400 ◊

4;P2

P1

4

22. () PMMC voltmeter reads only D∴ V = 2V

Page 2 of 14

23. () D1‰ short

D2 ‰ open

I = 0 mA

24. ()

&()#"

I

&('

1mA

2k

2k

¢¡¤£0¥12)&(43)

25. () For MOSFET, VGS = 2V, Vt = 1V

VGS > Vt, so MOSFET is turn-on

Vab = 0V

10V

26 - 31. ()

32. ()

VRY∠V0

VYBV 120VBR∠V

120

1K 1K

2K

a

b

RIR

IB

R1

I R V

RB

R

V

−∠120R

V

V V ∠−60

R

B

R2

IY

I

YB ∠−120R R

0

Y IY

I

RIYIB

− I

IB Y R

−VR

∠−60 ∠−120

−V− R−

0.5 − j0.866 −0.5 − j0.866≈ ’

3V V∠−90 3 ∆ ÷∠90

« ◊R

IR: :IYIB1 :1 : 333. () Page 3 of 14

V rms =

112T

3 2T

1V rms=

6

16

34. () In the steady state, inductor is drawing 1A, 1 acts as a short.

When t = 0+, inductor acts as a open circuit.

iL=1A

35. ()

36. ()

VPQ = 2V

10ΩP

VTh = 2V

For RTh,

RTh = 5Ω37. ()

20Ω

20Ω

4V

10Ω

10Ω

10Ω

Q

P

Q

Characteristic equation is 1 + G(s) H(s) = 0;

K 1 −s1 s 1 0

K œ Ks + S + 1 = 0; S(1 œ K) + (K + 1) = 0

For stable system, 1 œ K > 0

1 > K; K < +1

K 1

38. ()

-

K + 1 > 0

K > -1

R(s)+

- X1

3 +

X2

2s 2

Y(s)

X1 = R(s) œ y(s)

X2 = 3[R(s) œ y(s)] œ R(s) = 2R(s) œ 3y(s)

y(s) = [2R(s) œ 3y(s)]2

Page 4 of 14

s 2

' 4 y s

R s s 8

R

s

1

given

s

y s

4

s s 8 4s

steady state error = lt sy sS →0

steady state error = 0.5 39. ()

ltS→0 s 8

≈ ’

180 0.25w Phase of G(s)H(s) is œ90 - ∆ ÷

−90 −≈180 ’w − −

« ◊

≈180 ’≈ ’w∆ ÷0.25 90 ∆ ÷∆ ÷

« ◊

« ◊« ◊4

when it passes through negative real axis phase = -180

≈180 ’≈ ’

−90 −∆ ÷∆ ÷ −180

« ◊« ◊4

2−

s

gain at this point is

e 4

s

20.5

it cross œ vertical axis at œ0.5+j0

40. ()

K s0.366

s s

1 »

− − −−1 ≈0.366’ÿ

phase margin (PM)= 180

PM = 60, 1 rad/sec.

Substitutions, K = 1.366

41. ()

42. ()

43. ()

44

. ()

Page

5 0f 1490 −tan1

tan ∆«

df x

2 −x −x

2 −x−

f x x e

2 2x −x e

x 245. ()

−x,

dx0

2xe x e 0

gradient = ∂ui ∂uj

∂x dy2

= xi yj3

@(1,3) point, i+2j

magnitude is 5

46. ()

2d x

23

dx2x t

5dt

s2

dt

x s

3s 2

5 5

t s23s 2→∞, s →0

» s.5 ÿ

Lt ft Lt SF s Lt … Ÿ 0t →∞ S→0 S →0

0Lt ft

s23s 2 ⁄

t →∞

47. () As t ‰ ∞Ω S ‰ 0 » 2

ÿ

H SF s

s 5s 23s 6⁄

3

S →0

2s s

2s 2

H SF sS →0

48. () F z1

3

−11

1

;c

z 1 z − 1

inverse z-transform to 8(k) œ (-1)K; k≥0

49, 50 & 51. ()

Page 7 of 14

E is constant

P = 50 KW

52. () Torque = KaIa

E = Kam

P = EIa = KamIa

53 - 56 ()

57. ()

≈ ’2

Te st

∆Ist÷ sfl

Te f

« Ifl◊

620.004

Te st

1.44

Te f

58. () xs0.4Ω/km

x Ω kmm 0.1 /

x0xs2xm

x00.62Ω/km

xlxs−xm0.3Ω/km

59. () Power factor 0.97 lag is due to capacitor install

∴ Q = P tan = 4 tan [cos-1 (0.97)]

Q = 1 MVAR

P = 4 MW

If capacitor goes out of service,

Q = Ptan‰ P Q

→tan

2 14

0.75

cos = 0.8 lag

Q = 2+1=3

60. ()

tan

Y22 = j0.1 œ j20 + j0.1 = -j19.8

Y22 = -j19.8

dF dF1 2 ;

61. () dP1dP2

b2CP1b 4CP

2

P1

2 ;

P P21P2 150

Page 7 of 14

62. () I

C

I1

2

1

Z2

$

I1

10∠30

Z2

I1

I2

+

V2

-

Z2

+

V2

-

Z2 = 40∠−45

63. () Vt=1p.u

≈Vt ’

(1)

0.12

(2) Xs

Xs

(3)

1p.u

Infinite bus

P

P

∆«

1Vt2

xs

Vt Vt1 2

÷÷sin◊

max

6.25

xs

1 1≈x∆ s

0.12« 2

’÷◊

Page 8 of 14

xs0.08

&('

(3) P

max 5 (1) (2)

64. ()

now 0.12 0.08

(1)

(1)

(1)

fx

~

fy

0.12 0.08Infinite bus

fy No. of intersections of the horizontal line with curve 1fx

No. of intersections of the vertical line with curve 2

4 fx 2 x t P sin 4t 30

fy 2 ; 4 2fx

2

x

y2

y(t) = Qsin(2t+15)

65. () Multiplying power m =

m = 5;

I5005Im 100

m 1

66. ()

67. ()

Rm5Rshint

90Ω

200

0Ω

Ω Rshort = 0.10.025Ω

4

‰ 86.124Ω

0.01

error = 90−86.1244.5%86.124

68. ()

69. () W1 = 10.5

W2 = -2.5

W = WHW2; W = 8 kW

Page 9 of 14

' ≈w’ »10.5 − ÿ

tan

3 ∆ 1

−w2

÷ 3 … Ÿ

70.43«w1w2◊ … 10.5 −2.5 Ÿ

cos 0.3347

70. ()

IE1mA,

100

1

0.9909

≈100 ’ IC IE∆ ÷ mA 990.099 A

«101 ◊I I A

≈I

10A

71. ()

72. ()

73. ()

C

R

,BIB

-+

R

9.9

RL

B

Vout

Here op-amp in saturation.

So,

Vout=

Q X2X1;

Q X X12

74. () Ambiguous

75. ()

X1

Page 10 of 14

X1

t

X2

Q

Q X X Q X X

76. () X2

2 1; 1

X1

Q

2

Q

x1 = x2 =1;l Q is always toggling

so, Q is unstable.

77. () Even though voltage across the device ‰ 0, current is flowing

so device is turn on

Energy lost =

78. ()

1

2

VI t1t2

79. () VS = 100V

RL = 5

L = 200 mH

L

40m sec.RL

f = 1 KHz

D = 0.5

T = 1 msec.

Ripple current in the load current is

»…≈∆1 −e

≈

−T ’∆÷ 1 −e

−−1 T ’ÿTa ÷Ÿ

V …«

…

∆Ta ◊«

÷Ÿ◊Ÿ

S…

=

R ………

»≈

−T

1 −eTa

1 1 ’≈−

ŸŸŸŸ

1 1 ’ÿ− .

…∆1 −e 2 40 ÷∆1 −e 2 40 ÷Ÿ…∆ ÷∆ ÷Ÿ

1005

«……

≈◊«

−1 ’◊ŸŸ

… ∆1 −e40÷ Ÿ… ∆

«÷◊

Ÿ⁄

Ripple current = 0.1249 ≈0.125A

≈∆d’÷Tl Ja80. () Test = Tl + J

« dt ◊15 = 7 + J(2) Ω J = 4Nm2

81. (A) ( ) V

Vorms V max1cos 1 cos

outV

2V V 230

out 230 ,o

Page 11 of 14

90

Position 1:

120V

20Ω

10H

40Ω

coil

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Icoil = 120

40 20

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Inductor acts as a short circuit.

Position 2:

120 V = (20 + R) 2

R = 40Ω R

20Ω

10H

40Ω

120V

(B) ( ) Total resistance = 40 + 40 + 20 = 100 Ω = R

Time constant = R10010 secL 10∴i(t) = 2e-10t

Balance energy left = 100 œ 95 = 5% = 0.05.

Since the stored energy is proportional to the square of the current, the current at this instant is

i 0.05 initial current

2 0.05−1at

Ω 2 0.05 2e

82. (A) ( ) t 0.15 sec.

»0 1

ÿ » ÿ1

A … BŸ … Ÿ

0 −3 0⁄ ⁄

» ÿ−1

X … Ÿ

3 ⁄

−

ÿ

»»≈S

−1ÿ0 ’ ≈0 1 ’ÿ

−1 »L SI −A

1 Ÿ −1 …L …∆

−÷ ∆

÷Ÿ Ÿ

… … 0 S 0 −3 Ÿ

» −1 ÿ»1…

«

1

◊ «ÿ

Ÿ

◊⁄ ⁄

−

L1

…»S −1 ÿ Ÿ −

L1…s s s 3 Ÿ

…… 0

Ÿ

S 3⁄ Ÿ⁄

……0

1 ŸŸ

»1 − − t ÿ

s 3 ⁄

1…

……

3 1 e3

− t

ŸŸŸ

(B) ( )

Page 12 of 14

0 e3

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1 −1 BVCSx t L SI −A »x ⁄Ÿ⁄

»À1…

…1 ¤

Àÿ

≈ ’¤ŸŒ Œ ∆≈−

’

1

Œ ŒŒ ∆ ÷Œ

−1

…s s s 3 1 Ÿ

L Ã…

‹Ã 3 ÷ ∆ ÷s ‹∆ ÷ Ÿ

Œ 1 ŒŒ« ◊ 0 Œ… « ◊ ŸŒ0 ŒÕ ›

…Õ»À1…Œ

s 1

3¤ŒÀ

ÿ1¤Ÿ

Ÿ⁄

Œ ŒŒ−

Œ

− …s s s 3 1 Ÿ

L1 Ã…

‹Ã ‹s Ÿ

…Œ 1 ŒŒ 3 ŒŸŒ0 ŒÕ ›

…Õ s 3 Ÿ⁄

» 1 1 3 ÿ

− Ÿ− ……ss s s 3 Ÿ

L1……

3 ŸŸ

» − t ÿs 3 ⁄

t −e3

…x t

− t Ÿ

…3e 3 Ÿ

83. (A) ( ) S = 1000 KVA, 6.6 KV, , 3-phase

XS = 20Ω

p.f = 1

3 SI

L

3V IL Lcos ,1000 10

L −L87.47

3 6600 IL 1

E

E

I sphVph jph

4192.99Vph

3810.62 j1749.4

Einduced7.262KV L −L (B) ( ) KW = KVA cos

P = 1000 KW 3

3

10000 = 7.2 10 6.6 10sin

20

24, 88

84. (A) ( ) When a 3-, fault occurs, the fault current is limited by the positive sequence reactance only

∴ 4000 106 = 220 103 I 3

I 220 10X

X positive = 12.1Ω

(B) ( )

85. (A) ( ) Im = ∂ids1ms

∂Vgs

(B) ( ) Av = -gm rds

Page 13 of 14

small signal equivalent circuit is

Page 14 of 14

Vin ~

G

rds

D