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Answer Keys
1
62.5
7.5
10
459.25
11
12
13
14
15
5.39
16
17
125
18
90
19
20
18.371
21
22
23
24
25.13
25
26
27
28
29
30
30
31
32
33
34
35
8.8
36
37
38
479
39
2.83
40
41
42
43
44
45
46
47
48
49
50
51
52
53
0.168
54
55
54.75
56
57
58
59
60
61
62
16
63
64
16
65
3354.17
Explanations:A 0 0 is one eigen value of A is the maximum eigen value as options (B), (C), (D) are
1.
ve values.
2
1 i (1 i) 1 2i 1
2 2
2
1 i 1 i
log e i log1 i i
2
2
2.
3.
d 3 y 2d 2 y dy
2y 0
dx 3 dx 2
dx
m3 2m 2 m 2 0
1 2
0 1
1 1
1
1
2
2
2
0
m 1 m 2 m 2 0
m 1 m 1 m 2 0
m 1, 1, 2
y e x , y e x , y e 2x are solutions
m 1, 1, 2are values
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4.
Curl F
x
xz3
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j
k
y
z
2
2x yz 2yz 4
i 2z 4 2x 2 y j 0 3xz 2 k 4xyz 0
curl F at 1, 2,0 4i
1
curl of linear velocity
2
1
4i 2i
2
angular velocity
5.
6.
At unity power factor, the reactive power transfer is zero. With load fixed, if the excitation is
increased, the machine becomes over excited delivering reactive power along with the active
power. When a generator exports a reactive power, it works at lagging power factor and when
motor exports reactive power, it works at leading power factor.
VS 1 0O
VS in phase IS only when
1
C
1
1
1
C 2
0.0625F
L 4 4 16
L
7.
j e
j
2
j e
t
j / 4 t
8.
j t
4
2
8
V0 Voltage across 4k R L
IR
I Current through capa cot or
dV
IC C
dt
dV
V0 RC 1
dt
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Now average DC component in the output =
9.
Here,
0.11 0.5 0.2 0
1.2
'V '
ratio is not constant
f
We V 2
We2 30
10.
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w e2 V22
w e1 V12
1002
7.5W
2002
Generated voltage in d.c shunt generator,
Eg
zN P
A
60
0.07 220 2 900
60
E8 462V
Resistance of conductor =
l
0.002
A
l Z
Resistance of armature R a 2
AA
R a 0.002
440
16
0.005
Terminal voltage, VC = E8 Ia Ra
Vt 462 50 0.055
Vt 459.25 V
11.
At t = 0-
I L 0
10V
VC 0
10
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VC 0 10V,IL 0
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10
1A
10
Inductor doe not allow the sudden changes in current so at t 0 ,IL (0 ) 1A IL (0 )
Capacitor does not allow the sudden changes in voltage so at t 0 ,Vc (0 ) 10V Vc (0 )
13.
XY XY ?
XY
X
Y
X+Y
Clearly we can say
x y x y for xy 0
14.
In TTL GATE
Floating i P logic 1
Y 1 AB
AB .1 0
15.
O.L.T.F
K
k
C.L.T.F 2
S 1 S
s sk
Standard 2nd ordered C.L.T.F =
n 2
s 2 2n s 2n
..(1)
..(2)
Compare (1) and (2)
n k;
1
2 k
Given peak overshoot in 50%
12
50% 0.5
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ln 0.5 Subtitute n and value
1 2
0.21
k 5.39
16.
ht
d
c t 0 10e10t 10e10t 100te t 100te t
dt
Find T s
C s
R s
; Given R s
1
s
s 10 s s 10 10s
1
1
10
C s
2
2
s s 10 s 10
s s 10
2
C s
R s .100
C s
1 100
100
T s
2
2
s s 10
R s s 10 2
s 10
100
s 20s 100
0 10, 20 20 1
T s
17.
TJ TA PT R JC R CS R SA PT R JA
R JA 0.4 0.1 0.5 1.0
PT
18.
150 25 125W
1.0
The storage charge QRR
2
1 di 2
t rr 0.5 30 A 3 106 135c
s
2 dt
The peak reverse current = IRR 2QRR
19.
di
90A
dt
If x n k n Then
1
h n h n 1 k n
2
1
H z 1 z 1 k
2
n
1
h n k
2
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20.
1 450 2 2
Ior
3
2 10
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VO
VS
2R
22.
H e j
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Y e j
X e j
b e j
1 ae j
For all pass system H e j 1
b e j 1 ae j
1 b2 2bcos 1 a 2 2a cos
This is possible if b a
24.
Vs Vr 132 kV,
Qr 0UPF load
Qr
A 0.96 2o A
B 100 80o
BB
VS Vr
AVr 2
sin r
sin
B
B
132 132
0.96 1322
sin 80
sin 80 2
100
100
10.11o
0
VS Vr
AVr 2
cos
cos
B
B
132 132
0.96 1322
cos 80 10.11
cos 80 2
100
100
25.13 MW
Pr
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25.
ax
x
Z ey
ay
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az
z cos ay
T
0
a x cos ay a y 1 a z 0 e y J
In xz plane y 0
J a x a y a z A
26.
m2
f x, y xy, x 0 1, y 0 1, h 0.1
yi 1 yi h.f x i , yi for i 0,1,3.....
y1 y 0 h.f x 0 , y 0 1 0.1 f 1,1 1 0.1 1.1
y 2 y1 h.f x1 , y1 1.1 0.1 f 1,1 1.1 0.11 1.21
y3 y 2 h.f x 2 , y 2 1.21 0.1 f 1.2,1.21
1.21 0.12 1.33 two decimal places
27.
Verify the Wronskian condition for options
w 0 linearly dependent
0 independent
28.
The regression line of y on x
is y (0.7) x c1 . (1)
The regression line of x on y
is x (0.4) y c2 (2)
Slope of (1), m1 0.7
Slope of (2), m2
tan
29.
1
2.5
0.4
m1 m2
0.7 2.5
0.65
1 m1m2 1 (0.7)(2.5)
A Divisible by 6
B Divisible by 8
n A
33
25
P A
, P B
n s
200
200
A number which is divisible by 6 & 8
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From 1st 200 +ve integers
= is divisible by LCM of 6 & 8 = 24
No. of numbers which are divisible by 24
From 1st 200 positive integers = 8
8
P A B
200
P A B P A P B P A B
30.
33
25
8
50
0.25
200 200 200
200
From given data the Auto transformer
100A
Will as shown,
kVA ATFR 300 100
30 kVA
300V
200V
31.
1
Z jL || R
jC
jC
R jL 1 jC
jL R
R
L
jL 1 jRC
1 LC jRC
jL 1 jRC 1 LC jRC
1 LC R C
2
For resonating frequency imaginary part of Z = 0
L 1 2 LC 3LR 2C2 0
L 2 L2C 2 LR 2C2 0
2 LC R 2C2 1
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1
LC R 2 C2
Ya Yb
Yb
Ynetwork Y1
Yb
Yb YC
Yb
0.14 0.04 0.4444 0.1111
0.04 0.09 0.1111 0.2778
Y1
Ya
YC
0.5844 0.1511
0.1511 0.3678
33.
V1 Z11
V Z
2 21
Z12 i1
Z22 i 2
For the symmetric lattice network
Za Zd 4
Z b Zc 2
Za Z b 6
3
2
2
Z Za 2 4
Z12 Z21 b
1
2
2
3 1
So Z
1 3
Z11 Z22
34.
A Resistor in parallel with voltage source is redundant. It can be neglected as per the load
parameter calculation is concerned because same voltage 10V will always appear across R
irrespective of its value.
35.
Ia1 I 10A
E b1 V Ia R a R se
200 10 1
190 V
Rse
10A
Ra
Ra
In series motor, torque, T Ia2
Given that, T N2
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From (1) and (2), we get,
Ia N
Ia 2 Ia1
10
N2
N1
800
8A
1000
Speed of dc motor, N
Eb
Eb N Eb NIa
E b2 E b1
N 2 Ia 2
N1 Ia1
800 8
1000 10
121.6V
190
E b2
From motor equation, Eb2 V Ia 2 R a R Se Re
Where Re = external Resistance to inserted.
121.6 200 8 1 Re
121.6 200 8 8Re
8Re 70.4
70.4
Re
8.8
8
36.
Open circuit voltage across slip rings
Rotor emf, E2 kE1
N2
K
N1
ph
ph
48
0.5
96
Open circuit voltage across slip rings =
3 kE1 3 0.5 400 346
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FET A.C Equivalent and apply source transformation
rd
G
V0
VS
RD
RS
id
The loop equation for KVL
i d R S R D rd Vgs
Vgs VS R s i d
Vo i d R D
Solving A V
38.
VO
R D
VS
rd R S 1 R D
R
The offset due to Vio is Vo 1 F Vio
R1
500
1
4m 404mV
5
Due to Iio .Vo R F Iio 500k 150n 75mV
Total offset voltage VO 404 75 479mV
39.
Here, out of Q1, Q2 and Q3, Q1will be on and Q2 and Q3 will be off. Because if any one of Q2 or
Q3 is ON, let say Q2, then voltage at E = (0-0.7) = -0.7V. As this voltage is much lower than +2V
which is at base of Q1 immediately it will be ON and voltage at E will change immediately to (2
0.7) = 1.3V. Now this voltage is higher than voltage at the base of both Q and Q2, they will
1.3 12
remain off. So now IE
mA 2.83mA
4.7
40.
Z A BC A BC A BC A BC
AC AC A C
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Derating factor, DRF = 1 string efficiency
0.1 1
6000
1000
1
n s 1000
P 200
Number of series connected SCRs
nS
6000
6.6 7
1000 0.9
Number of parallel connected SCRs n P
1000
5.5 6
200 0.9
42.
The value of C should be such that RLC load is under damped.
X XL
tan C
R
For load commutation XC X L & must be at least equal
15
10
15 sec
For circuit turn off time, i-e,
15 106 sec
103
f
104 Hz
0.1
2 104 15 106 0.9424778rad 54O
x L
tan 54O c
2
C 1.248F
43.
The circuit produces XS-3 code at the output as listed in to table.
P
0
0
0
0
0
0
0
0
1
1
BCD
Q R
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
0 0
0 0
S
0
1
0
1
0
1
0
1
0
1
A
0
0
0
0
0
1
1
1
1
1
XS3
B C
0 1
1 0
1 0
1 1
1 1
0 0
0 0
0 1
0 1
1 0
D
1
0
1
0
1
0
1
0
1
0
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1. X(t) is real odd
Cn is purely imaginary
Cn is odd Cn C n
2.
C1 C1 1
2
2
3. 2 C1 1 C1 j
There are two solution = x1 t 2 sin t
x 2 t 2 sin t
45.
C.L.T.F
K P K Ds 100
s s 10 K P K Ds 100
C.E s 2 10 100K D s 100K P 0
K K Ds
K V 1000 lims P
100 K P 100
s 0
s s 10
.Sub KP in equation 1 and compared with standard equation
2n 10 100K D ; n 100; 0.5 given
Solving we get K D 0.9
46.
t S.T.M L1 sI A
s4
1
Now sI A
2
s 2 s 4
1
s4
1 1
s 2 s 4
t L1 sI A
47.
1
s 2
1
s 2
e4t
2t
4t
e e
e2t
Given C.E = s5 s4 2s3 2s2 3s 15 0
Arrange in R H tabular form
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s5 1
s
s 0
3
15
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12
2 12
s2
15
12k 15
s1
k
0
s 15
Now
TarGATE15
2 12
Let k
2 12
12
2 2 Ve
12k 15
k
12 15 0 12 ve
k
k
So two sign changes = 2 right side poles and 3 left side poles and given system in unstable.
48.
For a separately excited DC motor
H Eb
at 1000 rpm Eb1 V Ia R a 220 10 1 210V
E
E
210
At 700rpm b2 b1 E b2
700 147V
N2 N1
1000
For constant torque I1 I2
VT Eb2 Ia R a 147 10 1 157V
VT VDC
D
1
D
1 D
157 157D 220D
D 0.416
157 220
49.
If % error is same, then the test resistance is R T R a .R v
100 10 103
R T 1000
50.
For dual slope ADC, maximum conversion time,
2N1 Tc 2N1 clock period 29 1s 512s
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Since the fuses are removed, system becomes unbalanced.
IR 1200A
IY IB 0A
Positive sequence componenents
I
I
1
IR1 I R I Y 2 I B R 0 0 R 400A
3
3
3
IY1 IR1 1 240 40 o 40 240 40 120
IB1 IR1 1120 40 0 40 120
Neagative sequence components
1
IR 2 IR 2 I Y I B .120 240 40 0A
3
3
IY2 I R 2 1120 40 0 40 120A
IB2 2 I R 2 1 120 40 0 40 120A
Zero sequence components
IR 0 IY0 IB0
52.
1
1
IR I y IB 120 0 0 40A
3
3
Electric field, at any point / P due to uniformly charged plane with change density S is
s
an
2 0
an
f
3a x 4a y
1fl
25
f 3x 4y
f 0 at 'P '
Hence a n
E
3, 4
25
3, 4
2 10
9
2 10 36 25
4
67.85a x 90.48 a y V m
53.
Let Xn be the reactance to be placed in the generator neutral.
Base kv 2
112
2.016
Base mVA 60
V
1
f
j5 P.U
Vx j0.2
z Base
If3
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If L G
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3Vf
3
x 0 x1 x 2 3Xn j 0.057 0.2 0.1 3Xn
3
j 0.35 3Xn
Given If 3 If L G
5
3
0.35 3Xn
x n 0.0833 P.U
x actual 0.0833 2.016 0.168
54.
For machine 1
G1
P1
50
62.5 MVA
cos 1 0.8
NS1
120f1 120 50
1500 rpm
P1
4
2NS1 2 1500
50 rad sec
60
60
1
1
2
2
KE1 J1 S1 40000 50 493.48 MJ
2
2
S1
For machine 2
G2
P2
60
100 MVA
cos 2 0.6
NS2
120f 2 120 50
3000 rpm
P2
2
2NS2 2 3000
100 rad sec
60
60
1
1
2
2
K.E 2 I 2 WS2 20000 100 986.96 MJ
2
2
K.E total K.E1 K.E 2 493.48 986.96 1480.44 MJ
S2
Base MVA 200 MVA
H eq
55.
K.E total 1480.44
7.4022 MJ MVA
Base mVA
200
l 300 km
r 0.12 km
x 0.4 km
VS VR 100kV
Z R jx 39 j120
Z R jx
y 3 1016 mho km
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R rl 300 0.13 39
X xl 120
Y j9 104 mho.
z R jx
VS
Y
2
Y
2
VR
model network
j9 104 126.178 71.99 1 0.1135 161.99
yz
A D 1
1
2
2
2
o
B z 126.178 71.99 B B 1 0.05675 161.99
C 1
y2
4
0.946 j0.0475
0.946 1.06o
A
Steady state stability limit
VS VR
A
cos cos VR2
VV
A
100 100 0.946
PR max S R cos VR2
cos 71.99 1.06 1002
B
B
126.178 126.178
54.75 MW
PR
56.
Each term in the series is one less than the square of the preceding term. Thus 22 1 3,32 1 8
82 1 63
57.
Usher is a person who escorts people to seats in a theater, church etc.
Squire is a personal attendant / a man who accompanies a woman
58.
The given sentence is correct. As they did is used equivalent to remained in the context.
59.
The function of anodyne is to remove pain as the function of detergent is to remove stain.
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Observe the Venn diagrams below:
Heroes
Assets
Vegetarians
Books
Novels
Entrepreneur
s
61.
62.
Assuming that places of worship and education are visited bare-foot does not mean that people do
not use vehicles to reach them. Hence this is the assumption that cannot be logically made by the
Commissioner of Police.
1
In original mixture % of liquid Q 100 20%
5
3
In resultant mixture % of liquid Q 100 60%
5
20%
100%
60%
40%
40%
10 litres of Q is mixed with 10 litres of mixture
Total mixture 10 10 20 litres
(10 litres of mixture is taken out)
20
and liquid P 4 16 litre
5
63.
Since possibilities are being asked, we have to assume that sixth and seventh batsmen have scored
zero. Then only possibility of the first 5 batsman scoring the highest possible average will arise
375
75
So in this case possible average
5
64.
Let the usual time taken be t hrs and speed be x km/hr
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20
20
Dis tan ce xt 6 t 8 t
60
60
3 3t 1 4 3t 1
9t 3 12t 4
7
3t 7 t
3
8
7 1
Dis tan ce 6 6 16 km
3
3 3
65.
Bonus received in Jan
30000
100 50
200
60000
100 200
200
100000
March to Dec 10
100 4000
200
Total bonus 4000 200 50 4250
Total fixed salary 3000 12 36000
Feb
Total annual income 36000 4250 40, 250
Total monthly income
40, 250
Rs. 3354.17
12
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