General Aptitude - eeedrmcet.zohosites.com Questions/2016B.pdf · |EE| GATE-2016-PAPER-02...
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1
General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1. The chairman requested the aggrieved shareholders to ___________him.
(A) bare with (B) bore with (C) bear with (D) bare
Key: (C)
2. Identify the correct spelling out of the given options:
(A) Managable (B) Manageable (C) Manageble (D) Managible
Key: (B)
3. Pick the odd one out in the following
13, 23, 33, 43, 53
(A) 23 (B) 33 (C) 43 (D) 53
Key: (B)
Exp: Given numbers are 13, 23, 33, 43, 53.
All the numbers have second digits as 3.
If We sum of the digits of each number we get 4, 5, 6, 7, 8.
All given numbers are irrational numbers except 33 which is rotation.
So odd one out is 33.
4. R2D2 is a robot. R2D2 can repair aeroplanes. No other robot can repair aeroplanes.
Which of the following can be logically inferred from the above statements?
(A) R2D2 is a robot which can only repair aeroplanes.
(B) R2D2is the only robot which can repair aeroplanes.
(C) R2D2 is a robot which can repair only aeroplanes.
(D) Only R2D2 is a robot.
Key: (B)
5. If 29y 6 3, then y 4y 3 is _________.
(A) 0 (B) 1 3 (C) 1 3 (D) undefined
Key: (C)
Exp: 9Y 6 3
Possibility (A): 9y 9 y 1
Possibility (B): 1
9y 3 y3
When y=1
2 24y 4(1) 3 4 1
y 13 3 3 3
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2
When 1
y3
2
14
1 1 4 3 13
3 3 9 9 9 3
Q. No. 6 – 10 Carry Two Mark Each
6. The following graph represents the installed capacity for cement production (in tones) and the
actual production (in tones) of nine cement plants of a cement company Capacity utilization of a
plant is defined as ratio of actual production of cement to installed capacity. A plant with installed
capacity of at least 200 tonnes is called a large plant and a plant with lesser capacity is called a
small plant. The difference between total production of large plants and small plants, in tones is
Key: 120
Exp: Largent plant
Installed Capacity 220 200 250 200
Actual production 160 190 230 190
Plant number 1 4 8 9
Total production of larger plants = 160+190+230+190=770 tonnes
Smaller Plants
Installed Capacity 180 190 160 150 140
Actual production 150 160 120 100 120
Plant number
Total production of smallest plants = 150+160+120+100+120= 650tonnes
Difference = 770-650 = 120 tonnes
300
250
200
150
100
50
0
1 2 3 4 5 6 7 8 9
Plant Number
Capacity
production
tonnes
220
160
180
150
190
160
200
190
160
120
150
100
140
120
250
230
200190
InstalledCapacity Actual Production
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3
7. A poll of students appearing for masters in engineering indicated that 60% of the students
believed that mechanical engineering is a profession unsuitable for women. A research study on
women with master or higher degrees in mechanical engineering found that 99% of such women
were successful in their professions.
Which of the following can be logically inferred from the above paragraph?
(A) Many students have isconceptions regarding various engineering disciplines
(B) Men with advanced degrees in mechanical engineering believe women are well suited to be
mechanical engineers.
(C) Mechanical engineering is a profession well suited for women with masters or higher degrees
in mechanical engineering.
(D) The number of women pursuing high degrees in mechanical engineering is small.
Key: (C)
8. Sourya committee had proposed the establishment of Sourya Institutes of Technology (SITs) in
line with Indian Institutes of Technology (IITs) to cater to the technological and industrial needs
of a developing country.
Which of the following can be logically inferred from the above sentence?
Based on the proposal,
(i) In the initial years, SIT students will get degrees from IIT.
(ii) SITs will have a distinct national objective
(iii) SIT like institutions can only be established in consultation with IIT.
(iv) SITs will serve technological needs of a developing country.
(A) (iii) and (iv) only (B) (i) and (iv) only
(C) (ii) and (iv) only (D) (ii) and (iii) only
Key: (C)
9. Shaquille O’ Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free
throws out of 100 attempts on average. What is the probability that he will successfully make
exactly 6 free throws in 10 attempts?
(A) 0.2508 (B) 0.2816 (C) 0.2934 (D) 0.6000
Key: (A)
10. The numeral in the units position of 870 127 424211 146 3 is _____.
Key: (7)
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Electrical Engineering
Q. No. 1 –25 Carry One Mark Each
1. The output expression for the Karnaugh map shown below is
(A) A B
(B) A C
(C) A C
(D) A C
Key: (B)
Exp:
2. The circuit shown below is an example of a
(A) low pass filter
(B) band pass filter
(C) high pass filter
(D) notch filter
Key: (A)
Exp: 2
21 i 2
22
s
1R
RcsZ R , Z1 1 R CS
Rc
2 2
1 2 1
Z RTF
Z 1 R CS R
WhenS 0 ; 2
1
RTF non zero value
R
When S ; TF 0
Given circuit is an example of low pass filter
BC
00 01 11 10A
0
1
1 0 0 1
1111
2R
C
15V
15V
1RinV
outV
B C B C B C B C
1 0 0
1
1
111
A
A
F A,B,C A C
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5
3. The following figure shows the connection of an ideal transformer with primary to secondary
turns ratio of 1:100. The applied primary voltage is 100V (rms), 50Hz, AC. The rms value of the
current I, in ampere, is _____.
Key: (10)
Exp: Where
3secondary
L 22
Z 80 j40Z 10 8 4i
n 100 1
L
100 100 100I 10 36.86
j10 Z J10 8 j4 8 j6
Since all the calculation done with respect to RMS, I also in RMS
4. Consider a causal LTI system characterized by differential equation
dy t 1
y t 3x tdt 6
. The
response of the system to input 1
3x t 3e u t ,
where u(t) denotes the unit step function, is
(A) t
39e u t
(B) t
69e u t
(C) t t
3 69e u t 6e u t
(D) t t
6 354e u t 54e u t
Key: (D)
Exp: 1
S y s 3 s6
3
H s1
s6
Similarly 3
X s1
s3
9 54 54
y s1 1 1 1
s s s s3 6 6 3
Thus t
t 36y t 54 e u t 54e u t 0
LX 10
I~100V
I :100R 80k
CX 40k
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6
5. Suppose the maximum frequency in a band-limited signal x(t) is 5kHz. Then, the maximum
frequency in x(t) cos 2000 t , in kHz, is ______.
Key: 6
Exp: Maximum frequency of x t 5kHz
Maximum frequency of cos 2000 t 1kHz
x t cos 2000 t Gives convolution between their respective spectrums in frequency domain
max frequency of x t cos 2000 t 6kHz
6. Consider the function f (z) z z* where z is a complex variable and z* denotes its complex
conjugate. Which one of the following is TRUE?
(A) f(z) is both continuous and analytic
(B) f(z) is both continuous but not analytic
(C) f(z) is not continuous but is analytic
(D) f(z) is neither continuous not analytic
Key: (B)
Exp: f z z z
x iy x iy z is conjugate of z
2x i 0
f z 2x is continues but not analytic, since
C – R equations will not satisfy
7. A 3 3 matrix P is such that, 3P P. Then the eigenvalues of P are
(A) 1, 1, - 1
(B) 1,0.5 j0.866,0.5 j0.866
(C) 1, 0.5 j0.866,0.5 j0.866
(D) 0, 1, - 1
Key: (D)
Exp: If is an Eigen value of p then 3 is an Eigen
value of p3.
3p p
3
3 2 20 1 0 0; 1
0; 1
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7
8. The solution of the differential equation, for "t 0, y t 2y ' t y t 0 with initial
conditions y 0 = 0 and y ' 0 1, is u t denotes the unit step function),
(A) tte u t (B) t te te u t
(C) t te te u t (D) te u t
Key: (A)
Exp: The operator form of the of given D.E is
2D 2D 1 y 0
The A.E is 2D 2D 1 0
2
t
1 2
D 1 0 D 1, 1.
y t e C C t
Given y 0 0 & y' 0 1 i.e, t 0; y 0
from (s); 10 C 1C 0 y' 0 1
From (1), t t
1 2 2
dye C C t e C
dt
dy
At t 0; y ' 1dt
2 21 0 C 0 1 C 2 21 C C 1
From (1) t ty(t) e (t) te u(t)
9. The value of the line integral 2 2
c
2xy dx 2x ydy dz along a path joining the origin (0,0,0) and
the point (1,1,1) is
(A) 0 (B) 2 (C) 4 (D) 6
Key: (B)
Exp: Let 2 2f 2xy i 2x y j k
2 2
i j k
curl f f 0yx z
2xy 2x y 1
f is irrotational
Consider a straight line passing through 0,0,0 & 1,1,1
x y z
i.e, t x y z t dx dy dz dt1 1 1
1
2 2 3 3
t 0C
2xy dx 2x ydy dz 2t dt 2t dt dt
1 1
3 4
0t 04t 1 dt t t 2
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8
10. Let f(x) be a real, periodic function satisfying f x f x . The general form of its Fourier
series representation would be
(A) 0 kk 1f x a a cos kx
(B) kk 1
f x b sin kx
(C) 0 2kk 1f x a a cos kx
(D) 2k 1k 1
f x a sin 2k 1 x
Key: (B)
Exp: We know that a periodic function f(x) defined in (-c, c) can be represented by the poisoins series
on n
n 1 n 1
a n x n xf x a cos b sin
2 c c
If a periodic function f(x) is odd, its Fourier expression contains only sine terms
11. A resistance and a coil are connected in series and supplied from a single phase, 100V, 50Hz ac
source as shown in the figure below. The rms values of plausible voltage across the resistance
RV and coil CV respectively, in volts, are
(A) 65, 35 (B) 50, 50 (C) 60, 90 (D) 60, 80
Key: (C)
12. The voltage (v) and current (A) across a load are as follows.
v t 100 sin t , Oi t 10sin t 60 2sin 3 t sin 5 t
The average power consumed by the load, in W, is ________.
Key: 250
Exp: The instantaneous power of load is
p t V i t
100sin t 10sin( t 60 100sin t 2sin3 t
100sin t 5sin5 t
T
avg
0
since P P t dt , in the above expression
Only 1st term will result non zero answer
Remaining 2 terms will be 0.
so directly consider
P t 100sin t 10 sin t 60
~SV
RV
CV
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avg rms rms V IP V I cos 100 10 1000 1
cos 60 250 watt2 22 2
13. A power system with two generators is shown in the figure below. The system (generators, buses
and transmission lines) is protected by six overcurrent relays 1 6R to R . Assuming a mix of
directional and nondirectional relays at appropriate locations, the remote backup relays for R4 are
(A) 1 2R , R (B) 2 6R , R (C) 2 5R , R (D) 1 6R , R
Key: (D)
14. A power system has 100 buses including 10 generator buses. For the load flow analysis using
Newton-Raphson method in polar coordinates, the size of the Jacobian is
(A) 189 189 (B) 100 100 (C) 90 90 (D) 180 180
Key: (A)
Exp: Total no of Buses = 100
generator buses = 10 – 1 = 9 (n)
load buses = 100 -10 = 90(m)
Jacobeam matrix size = 2m n 2m n 2 90 9 2 90 9 189 189
15. The inductance and capacitance of a 400kV. Three-phase. 50Hz lossless transmission line are 1.6
mH/km/phase and 10nF km phase respectively. The sending end voltage is maintained at 400kV.
To maintain a voltage of 400kV at the receiving end, when the line is delivering 300MW load, the
shunt compensation required is
(A) Capacitive (B) Inductive (C) Resistive (D) Zero
Key: (B)
Exp: 3
LX j L j314 1.6 10 j0.5024
C 9
j jX j31847.3376
C 314 10 10
Since C LX X , the shunt compensation is inductive
16. A parallel plate capacitor field with two dielectrics is shown in the figure below. If the electric
field in the region A is 4kV/cm, the electric field in the region B, in kV/cm, is
1S
~
1R
3R
2R
4R
6R5R 2S
~
r
A
1 r
B
4 2cm
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(A) 1 (B) 2 (C) 4 (D) 16
Key: (C)
Exp: Since the voltage & distance b w the two plates are same for both the regions. The electric field
is same for both the regions V
Ed
Electric field in region B = 4kV cm
17. A 50MVA, 10kV, 50Hz, star-connected, unloaded three-phase alternator has a synchronous
reactance of 1 p.u. and a sub-transient reactance of 0.2 p.u. If a 3-phase short circuit occurs close
to the generator terminals, the ratio of initial and final values of the sinusoidal component of the
short circuit current is _________.
Key: 5
Exp: prefault
SC
VI
X P.U
SC
S
I initial 15
I cfinal 0.2
18. Consider a liner time-invariant system transfer function
1H s
s 1
. If the input is cos(t) and
the steady state output is Acos t , then the value of A is _________.
Key: 0.707
Exp: 1
2
1H tan
1
O1H 45
2
So when input is cost then O/P
O1y t cos t 45
2
So 1
A 0.7072
19. A three-phase diode bridge rectifier is feeding a constant DC current of 100A to a highly inductive
load. If three-phase, 415V, 50Hz AC source is supplying to this bridge rectifier then the rms value
of the current in each diode, in ampere, is _________.
Key: 57.73
Exp: OI 100A
RMS, diode current = 100
57.73A3
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11
20. A buck-bost DC-DC converter shown in the figure
below, is used to convert 24 V battery voltage to 36 V
DC voltage to feed a load of 72 W. It is operated at
20kHz with an inductor of 2 mH and output capacitor
of 1000 F. All devices are considered to be ideal.
The peak voltage across the solid-state switch (S), in
volt, is _________.
Key: 60
21. For the network shown in the figure below, the frequency in rad s at which the maximum
phase lag occurs is.______.
Key: 0.316
Exp: The given circuit is standard lag compensator
Whose Transfer function
11
s 1 1 ssG s1 1 10s 1 s
a 1s
So 1, 10 the frequency at which maximum phase lag happen
m
1 10.316 rad sec
10
22. The direction of rotation of a single-phase capacitor run induction motor is reversed by
(A) interchanging the terminals of the AC supply
(B) interchanging the terminals of the capacitor
(C) interchanging the terminals of the auxiliary winding.
(D) interchanging the terminals of both the windings
Key: (C)
23. In the circuit shown below, the voltage and current are ideal. The voltage (Vout) across the current
source, in volts, is
outsV
2
10V
5A
S
2mH
Load
36V24V
9
1
1F
0in
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12
(A) 0 (B) 5 (C) 10 (D) 20
Key: (D)
Exp: Writing KVL
O
O
V 10 10 0
V 20V
24. The graph associated with an electrical network has 7 branches and 5 nodes. The number of
independent KCL equations and the number of independent KVL equations, respectively, are
(A) 2 and 5 (B) 5 and 2 (C) 3 and 4 (D) 4 and 3
Key: (D)
Exp: No of branches = 7
Nodes = 5
No of KCL equations = No of Modal equations = n – 1 = 5 – 1 = 4
No of KVL equations = No of Mesh equations = b-(n – 1) = 7-4 = 3
Since no information given regarding how many simple & principal node, if we assume all
principal nodes then the answer for nodal is 5 – 1 = 4
25. The electrodes, whose cross-sectional view is shown in the figure below, are at the same potential
The maximum electric field will be at the point
(A) A (B) B (C) C (D) D
Key: (A)
Exp: At A
Fields are additive 1 2F F
At C
Fields are C
subtractive
1 2F F
At D field is due to one electrode 2F
At B field make an angle
2 2
1 2 1 2F F 2FF cos
So maximum electric field is at ‘A’
A D C
B
OV
10V
10V
5A
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13
Q. No. 26 – 50 Carry Two Mark Each
26. The Boolean expression a b c d b c simplifies to
(A) 1 (B) a.b (C) a.b (D) 0
Key: (D)
Exp: F = a b c d b c = a d b b c c = a d 1 1 = 1 = 0
27. For the circuit shown below, taking the opamp is ideal, the output voltage outV in terms of the
input voltages 1 2V , V and 3V is
(A) 1 2 31.8V 7.2V V (B) 1 2 32V 8V 9V
(C) 1 2 37.2V 1.8V V (D) 1 2 38V 2V 9V
Key: (D)
Exp: x 1 x 2 1 2x
V V V V 4V V0; V
1 4 5
x 3 x outV V V V
01 9
1 2
out1 2
3
4V VV4V V 5V 0
5 9
1 2 out
1 2 3
4V V 5V4V V 5V 0
9
1 2 3 1 2 out36V 9V 45V 4V V 5V 0
1 2 3out
40V 10V 45VV
5
out 1 2 3V 8V 2V 9V
9
1
4
2V
1V
3V
outV
1
xV
xV
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14
28. Let 1 1 2 2x t X and x t X be two signals whose Fourier Transforms are as shown
in the figure below. In the figure h(t) = 2 t
e
denotes the impulse response.
For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can
be uniquely reconstructed from its samples, is
(A) 12B (B) 1 22 B B (C) 1 24 B B (D)
Key: (B)
Exp: 1 2y t x t x t * h t
In frequency duration
1 2y X *X H
Max. Frequency of 1 1X B
Max, frequency of 2 2X B
Max frequency of 1 2 1 2X *X B B
Max frequency of H
Thus max, freq of 1 2y B B
Max frequency
Nyquist frequency = 1 22 B B
29. The value of the integral sin 2 t
2 dtt
is equal to
(A) 0 (B) 0.5 (C) 1 (D) 2
Key: (D)
1X
1B1B
2
1B
2
1B2B
2X
2B
1x t
2x t
2 th t e
y t
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15
Exp: 0
sin 2 t sin 2 t sin 2 t2 dt 2 2 dt is an even function
t t t
0.t
0
sin 2 t4 e .dt
t
By the defining of L.T; we have
sin 2 t
4L ; where S 0t
4 sin 2 t
L ; wheres 0t
1 14 2 sin at atan ; where s 0 L tan
s t s
Putting s = 0; than
14 4tan 2
2
sin 2 t
2 dt 2t
30. Let y(x) be the solution of the differential equation 2
2
d y dy4 4y 0
dx dx with initial conditions
x 0
dyy 0 0 and 1.
dx
Then the value of y (1) is _________.
Key: 7.398
Exp: The operate form of given D.E is
2D 4D 4 y 0
The A.E is 2D 4D 4 0
2
D 2 0 D 2,2
2
D 2 0 D 2,2
The solution is 2x
1 2y e C C x 1
Given that y 0 0 & y ' 0 1
i.e x 0 y 0 i.e at x 0, y' 1
from (1); 10 1 C 0 from 1 2x 2x
2 1 21 y e C C C x 2c
1C 0 2 21 C 0 C 1
2xfrom 1 y e 0 1.x 2xy xe
2 1 2 2y 1 1e e y 1 e or y 1 7.389
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16
31. The line integral of the vector field F = 2 2ˆ ˆ ˆ5xzi 3x 2y j x zk along a path from (0,0,0) to
(1,1,1) parametrized by 2t, t , t is _________.
Key: 4.4167
Exp: 2 2F 5xzi 3x 2y j x z k
2x t; y t ; z t
dx dt d 2tdt; dz dt
The line integral of the vector field is
2 2
C
F.dr 5xzdx 3x 2y dy x z dz
1
2 3 3
t 0
5t dt 10t dt t dt
1
2 3
t 0
5t dt 11t dt
1 13 4
0 0
t t5 11
3 4
20 23 535 11 4.4167
3 4 1̀2 12
32. Let P = 3 1
1 3
. Consider the set S of all vector x
y
such than 2 2a x
a b 1 where pb y
.
Then S is
(A) A circle of radius 10 (B) a circle of radius = 1
10
(C) an ellipse with major axis along 1
1
(D) an ellipse with minor axis along 1
1
Key: (D)
Exp: 3 1
P1 3
a x a 3 1 x
Pb y b 1 3 y
3x y a x 3y b
2 2a b 1
2 2
3x y x 3y 1
2 210x 10y 12xy 1
a 10; b 10; h 6
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17
It represents ellipse
The length of semi-axes is 2 4 2ab h r a b r 1 0
4 264r 20r 1 0 2 21 1r or r
4 16
Both 2r values are positive, so it represents ellipse
1 1
r or r2 4
Length of major axis = 2r 1
Length of minor axis = 1 12r 2
24
Equation of the major axis is 2
1
1a x hy 0
r
10 4 x 6y 0 x y 0
Equation of the minor axis is 2
2
1a x hy 0
r
10 16 x 6y 0 y x 0
Major axis exists along y = -x and minor axis exists along y = x.
1
The vector1
Lies on the line y = x
33. Let the probability density function of random variable, X, be given as:
3x 4x
x
3f x e u x ae u x
2
where u(x) is the unit step function. Then the value of ‘a’ and Prob X 0 , respectively, are
(A) 1
2,2
(B) 1
4,2
(C) 1
2,4
(D) 1
4,4
Key: (A)
Exp: we have xf x dx 1
0
x x
0
f x dx f x dx 1
0
4x 3x
0
3a e dx e dx 1
2
u x 1 for x 0
0, other wise
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18
u x 1 for x 0
0,otherwise
0
4x 3x
0
3a e dx e dx 1
2
04x 3x
0
e 3 ea 1
4 2 3
1 1
a 0 0 1 14 2
a 1 a 11 a 2
4 2 4 2
Prob 0 0 0
4x 4x
xx 0 f x dx a.e dx a e dx
04xe
2y
2 11 0
4 2
34. The driving point input impedance seen from the source VS of the circuit shown below, in , is
__________.
Key: 20
Exp: The Driving point impedance is nothing but the ratio of voltage to current from the defined port.
In this case it is S
S
V
I
Writing KCL at node x
x xS 1
V VI 4V 0
3 6
Substituting these in Eq(1)
S s S sS s
V 2I V 2II 8I 0
3 6
S S
1 1 2 2V I 1 8
3 6 3 6
S SV 2 1 I 6 4 48 2
S
S
V 6020
I 3
1V
2SI
3 14V
2
4
SV
1V
6
1V
2SI
3 14V
2
4SV
xV
xV
3
1I
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19
35. The z-parameters of the two port network shown in the figure are 11 12Z 40 , Z 60 ,
21 22Z 80 , Z 100 . The average power delivered to LR 20 , in watts, is ____.
Key: 35.55
Exp: In the given terminated 2 port network the Z matrix is known and for load of 20 we want to
find power on the load.
→ The get it assuming LR as load let first obtain the thevenin equivalent of 2 port
→ Thevenin equivalent means th thV & R
2
th 2 I 0V V i.e., O.C voltage of port 2
SC 2 2I I V 0 i.e., s.c current of port 2,
thin
sc.
VR
I
→ Evaluation of thV .
The Z matrix equation is
1 1 2V 40I 60I
2 1 2V 80I 100I
In the above two equations if 2I 0 then
1 1V 40I (1)
2 1V 80I (2)
From the input side we can say 1 1V 20 10I
1 120 10I 40I
1
2I A5
Then equation 2 becomes 2
3
2 1
2V 80 I 80 32V
5
th 2so V V 32
Evaluation of ISC
In the Z matrix equation if we put 2V 0 then
1 1 2V 40I 60I …(5)
1 20 80I 100I …(4)
101I
1V
20V Z
2V
LR
2I
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20
1 2 1 1 2
10 100I I & V 20 10I 20 I
8 8
Using these 1 1V & I in equation 3
2 2 2
100 40020 I I 60I
8 8
2 2 2160 100I 400I 480I
2 2160 20I I 8A
SC 2I I 8A
inin
sc
V 32R 4
I 8
Now the ckt is from port 2is
2
20 20P I 20
32
204 20
35.55watt
36. In the balanced 3-phase, 50Hz, circuit shown below,
the value of inductance (L) is 10mH. The value of
the capacitance (C) for which all the line current are
zero, in millifarads, is _____.
Key: 3.04
Exp: LI 0 2ph
L Cph
L C
j L j
X .X 3 cZ
j L jX X
3 c
L 1
b c
3
3314
10 10 C
C 3.04 mF
LI
CL
3
L 3 L 3
CC
thV 32
thR 4
LR 20
L
L
C C
L
C
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21
37. In the circuit shown below, the initial capacitor voltage is 4V.
Switch S1 is closed at t = 0. The charge in C lost by the
capacitor from t 25 s to t 100 s is _______.
Key: 6.99
Exp: It is given CV 0 4
1
R 5 , C 54f so 40000RC
→ Since it is a source free network we can say
t
C CV t V 0 e ; t 0 40000 t4e
→ We are asked to find the charge last by capacitor
From t 25 s to 100 s
We know in a capacitor Q CV
or Q C V
C CQ C V 25 sec V 100 sec
→ 40000t
CV 4e 640000 25 10
C
t 25 sec
V 4e 1.47
640000 100 10
C
t 100 sec
V 4e 0.073
Q 5 1.47 0.073 6.99 s
38. The single line diagram of a balanced power system is shown in the figure. The voltage magnitude
at the generator internal bus is constant and 1.0 p.u. the p.u. reactances of different components in
the system are also shown in the figure. The infinite bus voltage magnitude is 1.0p.u. A three
phase fault occurs at the middle of line 2.
The ratio of the maximum real power that can be transferred during the pre-fault condition to the
maximum real power that can be transferred under the faulted condition is ______
1S
4V
5 F
5
Generator int ernal bus j0.1
j0.5
Line 1
inf inite bus
j0.5
Line 2
j0.1
j0.2~
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22
Key: 2.286
Exp:
ac
0.375 0.125 0.125 0.0729 0.0729 0.375X
0.0729
1.143
EV
Pe max1.143
Pe Prefault 1.1432.286
Pe during fault 20.5
39. The open loop transfer function of a unity feedback control system is given by
K s 1G s , K 0, T 0.
s 1 Ts 1 2s
The closed loop system will be stable if
(A) 4 K 1
0 TK 1
(B)
4 T 20 K
T 2
(C) T 2
0 KT 1
(D)
8 K 10 T
K 1
Key: (C)
Exp: To comment closed 100b system stability we need the characteristic equation. Here it is given that
it is a unity feedback system.
Unity feedback system
So the characteristic equation is
S 1 TS 1 2S K S 1 0
During fault
0.2
0.6
0.1 0.25
~
0.25
0.2
0.35 ~
0.6
0.25
b
C
Before fault
0.1
0.2
0.5
0.1 0.5 ~
0.1 0.5Xeq 0.2 0.5 P.U
2
e
eq
EVP max 2EV
X
0.6b c
x cx b
x e
0.35 0.25
b
0.175 0.125
0.0729
C
a0.375 0.125
0.0729
C
Converting Y
acXC
a
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23
2S TS 1 2S KS K 0
2 2 3S 2S TS 2TS KS K 0
3 2S 2T S 2 T S 1 k k 0
3 22 T k 1 ks s s 0
2T 2T 2T
for stability using R t criterion
2 T K 1 K
2T 2T 2T
T 2 K 1 K K 1 1
K T 2
1 1
1̀K T 2
1 T 1 T 2
KK T 2 T 1
40. At no load condition a 3-phase, 50Hz, lossless power transmission line has sending –end and
receiving-end voltage of 400 kV and 420kV respectively. Assuming the velocity of traveling
wave to be the velocity of light, the length of the line, in km, is __________.
Key: 294.84
Exp: 2 2 10
S r
10V V 1
18
2 2 10314 10
400 420 1 294.84km18
41. The power consumption of industry is 500kVA, at 0.8 p.f. lagging. A synchronous motor is added
to raise the power factor of the industry to unity. If the power intake of the motor is 100kW. The
p.f. of the motor is _________.
Key: 0.3162
Exp: 2cos 1
2
1
1
1 2
0
36.86
Pcos
S
P 400; P 100
motor 1 1 1 2 2Q P tan P P tan 400 tan36.86 500 tan 300 kW
motor
m
S 100 j300
cos 0.3162
2 400 100
36.86
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24
42. The flux linkage and current (i) relation for an electromagnetic system is i g . When i
= 2A and g air gap length 10cm, the magnitude of mechanical force on the moving part, in
N, is ________.
Key: 186 to 190
43. The starting line current of a 415V. 3-phase, delta connected induction motor is 120A, when the
rated voltage is applied to its stator winding. The starting line current at a reduced voltage of
110V, in ampere is _________.
Key: 31 to 33
44. A single-phase, 2kVA, 100 200V transformer is reconnected as an auto-transformer such that its
kVA rating is maximum. The new rating in kVA, is _______.
Key: 6
Exp:
`
45. A full-bridge converter supplying in RLE load is shown in figure. The firing angle of the bridge
converter is 120O. The supply voltage m t 200 sin 100 t V, R 20 , E 800V. The
inductor L is large enough to make the output current LI a smooth dc current. Switches are
lossless. The real power fed back to the source. In kW is ________.
`
Key: 6
Exp: 0
2Vm 200V cos 2 cos120 200V
Load
L
R 20
E 800V
LI
3T1T
4T 2T
BridgeinV~
20A
100V
200V
300V
30A
10A
max
KVA rating 300 200
6kVA
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25
00
E V 800 200I 30A
R 20
fedback 0 0P V I 200 30 6kW .
46. A three – phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected
resistive load of 30 phase . If it is fed from a 600V battery, with 180O conduction of solid-state
devices, the power consumed by the load, in kW, is _________.
Key: 24
Exp: ph
2 2V Vdc 600 200 2 V
3 3
ph
R 30R 10
3 3
2
2
Load
ph
200 23VphP 3 24kW
R 10
47. A DC-DC boost converter, as shown in the figure below, is used to boost 360V to 400V, at a
power of kW. All devices are ideal. Considering continuous inductor current, the rms current in
the solid state switch (S), in ampere, is ______.
Key: 3 to 4
48. A single-phase bi-directional voltage source converter (VSC) is shown in the figure below. All
devices are ideal. It is used to charge a battery at 400V with power of 5kW from a source
sV 220V(rms),50Hz sinusoidal AC mains at unity p.f. If its ac side interfacing inductor is 5mH
and the switches are operated at 20kHz, then the phase shift between AC mains voltage SV
and fundamental AC rms VSC voltage C1V , in degree, is________.
600V
30
30
30
10mH
1mF
Load
400 V
S360V
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26
Key: 9.1 to 9.3
49. Consider a linear time invariant system x Ax, with initial condition x(0) at t = 0. Suppose
and are eigenvectors of 2 2 matrix A corresponding to distinct eigenvalues 1 2and
respectively. Then the response x t of the system due to initial condition x (0) = is
(A) 1te (B) 1te (C) 2te (D) 2 2t te e
Key: (A)
Exp: Eigen values are nothing but pole location
Here with respect to the pole is 1
wrt Pole is 2
The section should be of form t2
1te e but we are asking w.r.t
Initial condition x 0 only so the response should be 1te
50. A second-order real system has the following properties:
(a) the damping ratio 0.5 and undamped natural frequency n 10 rad s,
(b) the steady state value of the output, to a unit step input, is 1.02.
The transfer function of the system is
(A) 2
1.02
s 5s 100 (B)
2
102
s 10s 100 (C)
2
100
s 10s 100 (D)
2
102
s 5s 100
Key: (B)
Exp: The standard 2nd
order T/F is 2
n
2 2
n n
Ks 2 s
it is given that n0.5 & 10
2
100G s K
s 10s 100
Now to satisfy the steady state O/P 1.02
2s 0
1 100y t K 1.02 K 1.02
s s 10s 100
2 2
1.02 100 102G s
s 10s 100 s 10s 100
5mH
SXSI
~220VAC
1mH 400V
SI
SV
1 1I XC1V
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27
51. Three single-phase transformers are connected to form a delta-star three-phase transformer of
110kV 11kV. The transformer supplies at 11kV a load of 8MW at 0.8 p.f. lagging to a nearby
plant. Neglect the transformer losses. The ratio of phase current in delta side to star side is
(A) 1:10 3 (B) 10 3 :1 (C) 1:10 (D) 3 :10
Key: (A)
Exp: 1 2
11N : N 110: 10 3 :1
3
1 1 2 2I N I N
i 2I 10 3 I .1
1
2
I 1
I 10 3
52. The gain at the breakaway point of the root locus of a unity feedback system with open loop
transfer function
KsG s
s 1 s 4
is
(A) 1 (B) 2 (C) 5 (D) 9
Key: (A)
Exp:
KsG s
s 1 s 4
To find Break away point
We need to find the root of dk
0ds
where
2s 1 s 4 s 5s 4
Ks s
2 2
2
d ds s 5s 4 s 5s 4 s
dk ds ds
ds s
2S 2S 5 S 5S 4 0
2 22S 5S S 5S 4 0
2S 4 0 S 2
From the pole zero plot it is clean that Break away point must be S 2 as it is in between 2
poles
Now to find gain at this point use magnitude condition
s 2
KS1
s 1 s 4
KS1 K 1
1 2
1 4
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28
53. Two identical unloaded generators are connected in parallel as shown in the figure. Both the
generators are having positive, negative and zero sequence impedance of j0.4p.u, j0.3p.u and
terminals of the generators, the fault current, in p.u., is ________.
Key: 6
Exp: 0Z 0.15 P.U ; 1
0.4Z 0.2 P.U
2 ; 2
0.3Z 0.15P.U
2
f
O 1 2
3Vprefault 3I 6p.u
Z Z Z 0.15 0.2 0.15
54. An energy meter, having meter constant of 1200 revolutions kWh, makes 20 revolutions in 30
seconds for a constant load. The load, in kW is ______
Key: 2
Exp:
20revolutionsK 1200re v kwh
30P kW hr
3600
P 2kW
55. A rotating conductor of 1m length is placed in a radially
outward (about the z-axis) magnetic flux density (B) of 1 Tesla
as shown in figure below. Conductor is parallel to and at 1m
distance from the z-axis. The speed of the conductor in r.p,m.
required to induce a voltage of 1V across it, should be ______.
Key: 9.55
Exp: Voltage = B
Velocity Voltage 1
1m sB 1 1
i.e, 1m takes
2 2 1 2 m Takes 2 r
2 r for 1 rotation
in 1 minute = 60
2
Velocity = 60
9.55 rpm2
~
~
z
B
1m
1m