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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
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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