AIMT - 04
Branch : EE
..General Aptitude Part..
Q.1 to Q.5 Carry One Mark
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: -0.33
Q.1
In the following sets of analogies one word is missing, select that word from the words (a), (b),
(c), (d) exhibit the same analogy as established among the three italicized words.
Porcupine : Rodent : : Mildew : ?
(a) Fungus
(b) Germ
(c) Insect
(d) Pathogen
Sol.
Correct answer (a)
Mildew belongs to fungus family
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: -0.33
Q.2
Ans.
Sol.
Deepa moved a distance of 75 metres towards the north. She then turned to the left and
walking for about 25 metres, turned left again and walked 80 metres. Finally, she turned to
the right at an angle of 45°. In which direction was she moving finally?
(A) North-east
(B) North-west
(C) South
(D) South-west
D
Deepa started from A, moved 75 m up to B, turned left and walked 25 m up to D. Turning
to the right at an angle of 45°, she was finally moving in the direction DE i.e., south-west.
Hence, the correct option is (D).
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: -0.33
Q.3
There are 5 stilettos, 4 wedges and 7 heels in a closet. If you have to select 3 foot wears,
the probability that one is stiletto, one is wedge and one is heel, is
1
(A)
2
1
(B)
4
1
(C)
6
(D)
1
8
Ans.
B
Sol.
Probability of selecting three-foot wears  16C3
Probability that one is stiletto, one is wedge and one is heal, is

5
C1  4C1  7C1 1

16
4
C3
Hence, the correct option is (B).
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: -0.33
Q.4
Ans.
Sol.
Pick the odd one from the following options.
(A) CADBE
(B) JHKIL
(C) XVYWZ
(D) ONPMQ
D
Each group contains five consecutive letters from the English alphabet. The arrangement
within the group is similar in (A), (B), and (C), but different in (D). Thus, (D) is the odd
one.
Hence, the correct option is (D).
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
Q.5
: -0.33
Find the missing character from among the given alternatives-
Ans.
Sol.
(A) 7
(B) 25
(C) 49
(D) 129
B
We have: (16 – 6)2 + (5 – 2)2 = 102 + 32 = 109; (22 – 15)2 + (21 – 19)2 = 72 + 22 = 53.
So, missing number = (17 – 13)2 + (51 – 48) 2 = 42 + 32 = 25.
Hence, the correct option is (B).
Q.6 to Q.10 Carry Two Marks
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: -0.66
Q.6 Rashmi was employed in a company at a salary of Rs. 3000 every month for the first year and an
increment of Rs. 1000 in his monthly salary every succeeding year. How much does the Rashmi
earn from the company in 20 years?
(a) Rs. 7,90,000
(b) Rs. 9,10,000
(c) Rs. 30,00,000
(d) Rs. 27,59,000
Sol.
Correct answer (c)
First year the monthly salary is 3000 and it increases by 1000 per month every year, so in the
second year the monthly salary is 4000 and next year 5000 and so on.
So total earning from the company would be written as 3000 × 12 + 4000×12 + ……. Upto 20 terms.
This will constitute an A.P. and its sum will be calculated asThe series can be re-arranged as = 1000 × 12 [3 + 4 + 5 +…….. upto 20 terms]
Sum = 1000×12
[2  3  (20  1)1]
× 20 = 1000 × 12 × 250 = 30,00,000 Rs.
2
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: -0.66
Q.7
Ans.
Sol.
A painter mixes blue paint with white paint so that the mixture contains 10% blue paint. In
a mixture of 40 litres paint, how many litres blue paint should be added so that the mixture
contains 20% of blue paint?
(A) 2.5 litres
(B) 4 litres
(C) 5 litres
(D) 2 litres
C
Quantity of blue paint in the mixture = 10% of 40 = 4 litres
Quantity of white paint in the mixture = 40 – 4 = 36 litres
Let x litres blue paint be added
 According to the question, in the final mixture,
Quantity of blue paint 4  x 20


Quantity of white paint
36
80
20  36
4 x 
80
 x  9  4  5 liters
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: -0.66
Q.8
Read the argument given below and then answer the questions.
Ans.
Sol.
Consumers are using their mobile phones to download tens of millions of games, songs,
ring tones and video programs. They shell out money for these items, even as they resist
paying for similar digital goodies online using their computers. It is a curious equation: pay
for stuff on a tiny, low-resolution screen while getting some of the very same games and
video free on a fancy wide-screen monitor.
If all of the statements in the passage are true, each of the following must also be true
EXCEPT:
(A) Mobile phones are a status symbol for today’s consumers.
(B) The companies producing mobile content are thriving.
(C) The users enjoy the availability of music download at their fingertips.
(D) Stuff on computer seem to be easily available so why pay money.
A
Option (A) is not mentioned anywhere.
Other options are supporting the statements.
Hence, the correct option is (A).
Topic
: General Aptitude
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: -0.66
Q.9 Sita can finish some work in 12 days working 4 hours a day. Gita can finish the same in 15 days
working 3 hours a day. In how many days can they finish it working together at 4 ½ hours a day?
Sol.
(a)
155
31
(b)
160
31
(c)
191
31
(d)
205
31
Correct answer (b)
Number of hours taken by Sita = 12 × 4 = 48
Number of hours taken by Gita = 15 × 3 = 45
Together they will do
45  48
31
work in one hour.

45  48 720 t h
Together they will take
720
720
hours or
days working 1 hour a day
31
31
Working 4½ hours a day, they can complete the work in 
Topic
: General Aptitude
Concept
:
Sub-concept
:
720
720  2
5

 5 days .
9
31 9
31
31
2
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: -0.66
Q.10 Given below are three statements and four conclusions drawn based on the statements.
Statements 1 : Some engineers are writers.
Statements 2 : No writer is an actor.
Statement 3 : All actors are engineers.
Conclusion I : Some writers are engineers.
Conclusion II : All engineers are actors.
Conclusion III : No actor is a writer.
Conclusion IV : Some actors are writers.
Which one of the following options can be logically inferred?
(A) Only conclusion I is correct
(B) Only conclusion I and conclusion III are correct
(C) Either conclusion III or conclusion IV is correct
(D) Only conclusion II and conclusion III are correct
Ans.
B
Sol.
Given :
Statements 1 : Some engineers are writers.
Statements 2 : No writer is an actor.
Statement 3 : All actors are engineers.
We can draw venn diagram of given statement as shown below
Engineer
Actor
Writer
Conclusion I : Some writers are engineers.
Is definitely true as some part of engineer is intersected with writer.
Conclusion II : All engineer are actor.
Is definitely wrong as some part of engineer is intersected with writer which can never be
part of actor.
Conclusion III : no actor is writer.
Is definitely true as no part of actor is writer.
Conclusion IV : actor are writer.
Is definitely wrong as no part of actor is writer.
Only conclusion I and III follows.
Hence, the correct option is (B).
..Technical Part..
Q.1 to Q.25 Carry One Mark
Topic
: Analog Electronics
Concept
: Operational amplifier
Sub-concept
: Basic operation
Concept-field :
Level
:
Question Type
: MSQ
Marks
: +1,
Negative mark
: 00
Q.1
Ans.
Sol.
Which of the following statement(s) is/are correct for Op-Amp?
(A) Ideal Op-Amp has infinite gain
(B) Ideal Op-Amp has infinite input impedance
(C) Gain bandwidth product is constant for ideal Op-Amp
(D) Input current for inverting and non-inverting terminal is zero for ideal Op-Amp
A, B, C, D
Consider below Op-Amp,
Ideal Op-Amp has below characteristics,
A
Gain BW = Constant
Ri  
I1  I 2  0 Amp
Hence, the correct options are (A), (B), (C) & (D).
Topic
: Signals & systems
Concept
: Z-transform
Sub-concept
: Properties
Concept-field :
Level
:
Question Type
: NAT
Marks
: +1,
Negative mark
: 00
Q.2
For
the
inter
connection
system
given
below,
if
h1 (t )  h2 (t )  8(t )
and
h3 (t )  h4 (t )  h5 (t )  u (t ) . The overall impulse response is Pu (t ) , then the value of ' P '
is_______.
Ans.
Sol.
136
Given :
h1 (t )  h2 (t )  8(t )
h3 (t )  h4 (t )  h5 (t )  u (t )
Overall response is,
h(t )   h2 (t )*[h3 (t )  h4 (t )]  h5 (t )* h1(t )
h(t )  8(t )*[u (t )  u (t )]  u (t )  *8(t )
h(t )  16u (t )  u (t )  *8(t )
h(t )  17u (t ) *8(t )  136u (t )
h(t )  Pu (t )
P  136
Topic
: Signals & systems
Concept
: Systems
Sub-concept
: Properties of LTI systems
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.3
Z-transform of x(n) = 2n u(−n − 3) is
−z3
(a) 4(z−2), with ROC |z| < 2
−z3
(b) 4(z−2), with ROC |z| > 2
(c)
(d)
−z3
1
1
2
−z3
, with ROC |z| < 2
4(z− )
1
2
, with ROC |z| >
4(z− )
1
2
Answer: (a)
Sol:
ZT.
1 n
z
1
( ) u(n) ⟷
, |z| >
1
2
2
z−2
Z.T.
1 −n
z −1
( ) u(−n) ⟷
, |z| < 2
1
2
z −1 − 2
1
Z.T.
n
2 u(−n) ⟶ z , |z| < 2
2−z
2z
ZT.
−2
2n u(−n) ⟷
, |z| < 2
(z − 2)
Z.T. −2z 3
2n+3 u(−n − 3) ⟷
, |z| < 2
(z − 2)
Z.T.
−z 3
n
2 u(−n − 3) ⟷ −
, |z| < 2
4(z − 2)
Hence, Option (A) is correct
Topic
: Digital Electronics
Concept
: Minimization of Boolean function
Sub-concept
: Essential prime implicants
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.4
The K-map for a boolean function is shown below. The number of essential prime
implicants for this function is
(A) 3
(B) 4
(C) 5
(D) 6
Ans.
Sol.
f  ABD  BCD  BD
Hence the number of essential prime implicants for this function is 3.
Topic
: Electrical machine
Concept
: Transformer
Sub-concept
: Transformer operation
Concept-field : Transformer efficiency
Level
:
Question Type
: NAT
Marks
: +1,
Negative mark
:0
Q.5
Ans.
Sol.
Consider a 20 kVA, 2200/220 V, 50 Hz transformer. The O.C./S.C. test results are as
follows :
O.C. test : 220 V, 4.2 A, 148 W(l.v. side)
S.C. test : 86 V, 10.5 A. 360 W(h.v. side)
Determine the regulation at 0.8 p.f. lagging and at full load.
2.9
It may be noted that O.C. data is not required in this question for finding the regulation.
Since during S.C. test instruments have been placed on the h.v. side i.e. primary side.
86
 Z 01 
 8.19 
10.5
360
R01 
 3.26 
10.52
X 01  8.192  3.262  7.5 
F.L. primary current, I1 
20, 000
 9.09 A
2200
Total voltage drop as referred to primary  I1 ( R01 cos   X 01 sin )
Drip  9.09(3.26  0.8  7.5  0.6)  64.6 V
% average regn.  64.6 
100
 2.9%
2200
Topic
: Power Electronics
Concept
: Phase controlled rectifiers
Sub-concept
: Single phase full wave rectifier
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.6
In a single phase full-bridge diode rectifier, the constant DC current is 20 A. The average
power supplied to the load is ______ W, if supply voltage Vs is a pulse waveform as shown
below.
Ans.
Sol.
(A) 666.6
(B) 2000
(C) 1273.23
(D) 2666.6
D
The output voltage of the diode rectifier is,
Average value of output voltage is,
2 / 3
V0  200 

2
V0  200   133.33V
3
Power delivered to load is,
P  V0 I 0  133.33  20  2666.6 W
Hence, the correct option is (D).
Topic
: Analog electronics
Concept
: Diode circuit
Sub-concept
: Diode equivalent circuit
Concept-field :
Level
:
Question Type
: MSQ
Marks
: +1,
Negative mark
Q.7
:0
Consider the diodes are ideal, which of the following statements is/are correct?
(A) For R  1 k , V01  V02  12 V .
(B) For R  2 k , V01  V02  0 V
(C) For R  4 k , V01  V02  4 V
(D) For R  4 k , V01  V02  2 V
Ans.
Sol.
B, C
Assume D1 and D2 are ON,
V02  0 V and V01  0 V
By KCL at V02 ,
I D1  I D2  I
I
I D2
0  ( 6) 6

R
R
60 6

  3 mA
2 k 2
So, from equation (i),
…(i)
I D1   3 
6
R
For D1 to ON, I D1  0
6
3
R
6
R   2 k
3
Case 1 : R  1 k
…(ii)
So, both diodes ON,
V01  V02  0 V
Case 2 : R  2 k
Assume D1  OFF and D2  ON (Check)
I D2 
6  ( 6) 12

 3 mA
22
4
I D2  (  v ) , so D2  ON
V01  V02   6  3  2  0 V
So, VD1  0 V
Assumption are correct.
Case 3 : R  4 k
Assume D1 OFF and D2 ON (check)
( D1  OFF )
12
 2 mA
6
V01  V02  6  2  2  2 V
I
Or V01  V02   6  4  2  2 V
So, VD1  0  2   2 V
So, D1  OFF (R.B.) and D2  ON ( vI )
V01  V02  2 V
From the above analysis,
For R  2 k , V01  V02  0 V
For R  4 k , V01  V02  4 V
Hence, the correct options are (B) & (C).
Topic
: Electromagnetic field
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.8
In a hundred turn coil, if the flux through each turn is  2t 4  t 3  5 m Wb, the magnitude
of the induced emf in the coil at t  3sec , is
(A) 18.9 V
(B) 18.9 mV
(C) 10 mV
(D) 0.18 V
Ans.
A
Sol.
Flux through each turn   2t 4  t 3  5 103 Wb
Flux leakage of the coil.
 (t)  100  2t 4  t 3  5  103 Wb - turns
d  t 
 100 8t 3  3t 2 103 volts .
dt
Induced emf 
At t  3,
e  t   100  216  27  103  18.9 V
Hence, the correct option is (A).
Topic
: Measurement
Concept
: Error analysis
Sub-concept
: Error in measurement
Concept-field :
Level
:
Question Type
: NAT
Marks
: +1,
Negative mark
:0
Q.9
A correctly adjusted 1-, 240 V induction watt hour meter has a meter constant of 600
Ans.
Sol.
revolutions per kWH ; at a load current 10 A with 0.8 power factor lagging. If the lag
adjustments is altered so that the phase angle between voltage flux and applied voltage is
860. Then the error introduced in meter is _____%.
– 5.4 to – 5.6
Given :
  860
  cos (0.8)  36.86
 Key Point
True power  VL I L sin (90  )
Measured power  VL I L sin (  )
% error 
Pm  Pt
100
Pt
 sin (  )  cos  
% error  
 100
cos 


 sin (86  36.86)  cos (36.86) 
% error  
  5.5%
cos (36.86)


Topic
: Digital
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.10 Consider the circuit as shown in figure. Which of the following is the redundant logic gate.
Ans.
Sol.
(A) 1
(B) 2
(C) 3
(D) 4
A
Step 1 : SOP from of the given circuit
F  C  B  AC  A  B
F  BC  AC  A.B
Step 2 : K – MAP
B C is prime implicant
So, B C is redundant because of common terms in other min terms.
Hence, the correct option is (A).
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.11 For the circuit given below, the thevenin voltage across the terminals AB is ______.
(correct upto 2 decimal places).
Ans.
Sol.
0.85 to 0.86
Nodal analysis at super node,
Vx  6 Vx Vth
 
0
1
1 1
…(i)
2Vx  Vth  6
KVL in loop,
Vx  2Vth  Vth  0
…(ii)
Vx  3Vth
From equation (i) and (ii),
2(3Vth )  Vth  6
7Vth  6
Vth 
Topic
6
 0.857 V
7
: Signal
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.12 The Laplace transform of the signal x(t ) is X ( s ). If X ( s ) 
of the signal x(t ) in the region of convergence Re ( s )  0 is
Ans.
Sol.
(A) 0
(B) 2
(C) 0.5
(D) Does not exist
C
Given Laplace transform of signal x(t ) is,
X ( s) 
1
s ( s  2)
X ( s) 
1 s
s2 


2  s ( s  2) s ( s  2) 
1
 1
X ( s )  0.5 
 
s 2 s
As region of convergence is, Re ( s )  0
Inverse Laplace transform of X ( s ) is,
x(t )  0.5  e2t u (t )  u (t ) 
x(t )  0.5[1  e2t ] u(t )
The final value of x(t ) will exist at t  
 Final value of x(t )  x()  0.5
Hence, the correct option is (C).
Topic
: Engg. Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
1
then the final value
s ( s  2)
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
tan x  x
. (rounded upto two decimal places)
x  0 x 2 tan x
Q.13 Evaluate lim
Ans.
0.33
Sol.
Here, f ( x)  tan x  x, g ( x)  x 2 tan x so that lim f ( x)  0 and lim g ( x)  0
x 0
x 0
0
form.
0
tan x  x
x
tan x  x
x
 Given limit  lim

 lim
 lim
3
3
x 0
x

0
x

0
x
tan x
x
tan x
 The given limits is in
 lim
x 0
tan x  x
tan x  x
1  lim
3
x

0
x
x3
0

 form 
0

sec2 x  1
1  tan x  1
 lim 
   0.33
2
x 0
x 0 3 
3x
x  3
2
 G.L.  lim
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.14 Two inductive coils having resistance 3 and 5 and quality factor 10 and 14 respectively are
connected in series. Then the overall quality factor is…….
(a) 12.5
(b) 11.2
(c) 10.5
(d) 14.2
Ans.
Correct answer (a)
Q1 
L 1
R1
Q2 
L 2
R2
QT 
(L 1  L 2 )
R1  R2
L 1
L 2
 R1 
 R2
R1
R2

R1  R2

Q1R1  Q2R2
R1  R2
QT 
10  3  14  5 30  70 100


 12.5
8
8
8
Topic
: Signal
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.15 The signals x (t ) and y (t ) are related to each other as y (t )  4 x (2t  5) . If the energy of
y (t ) is 160, then the energy of x (t ) is_____.
Ans.
Sol.
20 to 20
Given :
y (t )  4 x(2t  5)
Let 2t  5  
5
t
2
  5
y
  4 x ( )
 2 
x ( ) 
1   5
y

4  2 
Since '  ' is a dummy variable, it can be replaced with ' t '
x (t ) 
1  t 5
y

4  2 
Given that energy of y (t )  160 (i.e.) Energy [y(t)] = 160
  t 5 
Energy  y      160  2
  2 2 
 1  t 5   160  2
Energy  y     
16
 4  2 2 
Energy [ x (t )]  20
Topic
: Engg.Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.16 A die is rolled and a coin is tossed. The probability that the die shows an even number and
coin shows a head is ______.
Ans.
0.25 to 0.25
Sol.
We know that, sample space for coin S1  {H , T } and for dice S2  {1, 2, 3, 4, 5, 6}
Complete sample space :
S  {(1, H ), (2, H ), (3, H ), (4, H ) , (5, H ) , (6, H ) , (1, T ), (2, T ), (3, T ) , (4, T ), (5, T ) , (6, T )}
According to question :
E  Even and head
E = {(2, H), (4, H), (6, H)}
Required probability,
P( E ) 
n( E ) 3 1
 
n( S ) 12 4
Topic
: Power System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.17 The figure shown above is a one line diagram of a power system network. The ratings and
reactance of the various component are shown. Consider 100 MVA as base MVA for the
whole system base kV for the circuit of generator-1 is 22 kV. Then the correct statements
is/are
(A) X T2  0.22 pu
(B) X T2  0.11pu
(C) X G2  0.37 pu
(D) X G2  0.74 pu
Ans.
Sol.
B, C
Base kV in generator 1 circuit = 22 kV
Transformer -1 rating  22 /132 kV
132
 22  132 kV
22
Transformer-2 rating  115 /11kV
Base voltage in line 
Base voltage in generator 2 circuit 
11
132  12.62 kV
115
2
X T2  0.06 
100  11 

  0.11pu
40  12.62 
2
100  15 
X G2  0.08 

  0.37 pu
30  12.62 
Hence correct option are (B) & (C).
Topic
: Engg.Mathematics
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.18 The solution of differential equation 2
dy y 2  1

, y(1)  5 is
dx
xy
(A) y 2  26 x  1
(B) y 2  25 x  1
(C) y 2  26 x  1
(D) y 2  25 x  1
Ans.
A
Sol.
Given : 2
dy y 2  1

dx
xy
2 ydy dx

y2 1 x
2y
1
dy  dx
2
y 1
x
Integrating both side,

2y
1
dy   dx
2
y 1
x
ln ( y 2  1)  ln x  ln c
ln ( y 2  1)  ln ( xc)
y 2  1  xc
Now, y (1)  5
(52 )  1  1c
25  1  c
c  26
So, y 2  1  26 x
y 2  26 x  1
Hence, the correct option is (A).
Topic
: EMT
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.19 For vector A  2 xiˆ  5 yjˆ  7 zkˆ the value of
 div ( A) dv , where v is the volume bounded
v
by the cylinder x 2  y 2  9 and planes z  0 and z  3 is
(A) 3
(B) 81
(C) 27
(D) 108
Ans.
D
Sol.
Given :
A  2 xiˆ  5 yjˆ  7 zkˆ
div ( A)    A
 


div ( A)   iˆ  ˆj  kˆ   (2 xiˆ  5 yjˆ  7 zkˆ)
y
z 
 x
div ( A)  (2  5  7)  4
Now,
 4 dv  4 dv  4  Volume of cylinder
v
v
 4 dv  4  R H  4   3
2
2
 3  108
v
Hence, the correct option is (D).
Topic
: Power Electronics
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q20.
A three-phase half wave converter operates from a three phase Y-connected 220 V, 50 Hz supply
and load resistance is R = 10 . The average output voltage required is 50% of maximum possible
output voltage then calculate the value of firing delay angle ‘’ (in degrees)
Sol.
Correct answer (67.7) [67.5-68]
Phase voltage VP 
220
3
And max. phase voltage Vm  2VP

2
.220  179.6
3
3 3Vm
3 3(179.6)
 148.528V
2
2
The average output voltage Vdc = 0.5× Vom = 0.5 × 148.53 = 74.26 V

For resistive load, load current is continuous if  
6
V 


Vdc  om 1 cos    
3 
6



148.528 
74.26 
1 cos    
3 
6



0.866  1 cos    
6

Max. Output voltage Vom 



cos      0.134
6

 = 67.7°
Topic
: Engg. Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.21 The value of
Ans.
0.54 to 0.60


0
1
e y y 2 dy is ________. (correct upto 2 decimal places)
3
Sol.
Given :

1
I   e y y 2 dy
3
0
Put y3  t ,
Differentiating with respect to t,
3y 2 dy  dt
1 23
1 21
y dy  y dt  t dt
3
3
1
2

1 1
I   et t 2 dt
0
3
1
1
 n
Take, n  1 
2
2
Using the property of gamma function,
1
I  n
3
1 1 1
I  
 = 0.591
3 2 3
Topic
: Control System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
s2
. The magnitude (in dB) of the
s 8
transfer function at a frequency where the phase of the compensator system is maximum
is _______. (Assume frequency in rad/sec). (correct upto 2 decimal places)
– 6.04 to – 6
Given transfer function of compensator is,
s2
G ( s) 
s 8
Q.22 The transfer function of a compensator is G ( s ) 
Ans.
Sol.
Let m be the frequency at which phase is maximum.
 m  2  8  16  4 rad/sec
Put s  j in G ( s )
 G( j) 
j  2
j  8
2  4
 G ( j) 
2  64
G ( j) 
2m  4
16  4 1


2
m  64
16  64 2
G( j) dB  20log 2  6.02 dB
Topic
: Machine
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.23 A 1500 kVA, star-connected, 6.6 kV salient pole synchronous motor has X d  25  and
X q  14  / phase respectively, armature resistance being negligible. If the excitation is
Ans.
Sol.
cut off. The maximum load motor can supply without loss of synchronous is ______ kW.
684 to 685
Power delivered by salient pole motor.
E f Vt
V 2  Xd  Xq 
P
 sin   t  
 sin 2
Xd
2  Xd Xq 
In the above equation, the second term indicates reluctance power, which is independent
of field excitation and would be present even if field fails to exist.
This reluctance power is maximum at   450 ,
Pmax when excitation fails 
Pmax
when excitation fails 
Vt 2  X d  X q 


2  Xd Xq 
 6.6 2  25  14 


2
 25 14 
Pmax  684.5kW
Topic
: Power Electronics
Concept
: Power Electronics
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
dv
rating of SCR is 400
dt
V/µs and it’s junction capacitance is 25 pF. The switch ‘S’ is closed at t = 0. Find the value of CS for
dv
which SCR is turned on due to
.
dt
Q.24 A circuit is shown in figure below where VDC = 200 V and RL = 20 . The
(1) 0.054 µF
(2) 0.0249 µF
(3) 0.0569 µF
(4) 0.0258 µF
Sol.
Correct answer (2)
Initially SCR remains forward biased (blocking state) and It modelled with Cj
V
After t = 0, current flowing through circuit i dc 1 e t/RC
RL
Voltage across equivalent capacitance,
 t/R C
VC  Vdc 1 e L 


dVc
V
 dc
dt RLC

1
200  1 


  0.025µF
RL dv /dt 20  400 
As we know:
C = Cs + Cj
CS = C – Cj = 0.025 µF – 20 pF
Cs = 0.0249 µF
C
Vdc
Topic
: Power system
Concept
:

Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +1,
Negative mark
: 0.33
Q.25 A 40 km long, 50 Hz, 100 kV transmission line has a voltage drop of 10% across its series
impedance at full load condition. Of this 6% is due to the reactance of the line. At full load
with 0.8 leading power factor, voltage regulation will be approximately equal to
(A) 4.0
(B) 2.8
(C) 4.5
(D) 5.0
Ans. B
Sol. VZ  VR  jVX
VZ  VR  Vx
2
2
0.10  VR  1(0.06) 2
2
VR  0.08
Approximate voltage regulation,
V .R   VR cos   VX sin  100
  (0.08  0.8)  (0.06  0.6)  100  2.8%
Hence, the correct option is (B).
Q.26 to Q.55 Carry Two Marks
Topic
: Measurement
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.26 An unknown inductance resonates at a frequency of 2 MHz with an external capacitance
of 210 pF and has a Q factor of 100. If the frequency of the source is doubled, it is found
that the tuning capacitor required for resonance is 50 pF. Determine the value of the
unknown inductance in the equivalent circuit is ______ μH . (rounded upto one decimal
Ans.
Sol.
place)
29.5 to 29.8
Given :
f1  2 MHz
C1  210 pF
f 2  4 MHz
C2  50 pF
Q  100
Ratio of resonance frequencies,
f
n 2 2
f1
Measurement of self-capacitance,
Cd 
C1  n 2C2
n2  1
210 1012  22  50 1012
22  1
Cd  3.33 pF
Cd 
Unknown inductance,
1
1
L

2
2
(2f1 ) (C1  Cd ) (2f 2 ) (C2  Cd )
L
1
(2 2 10 ) (210 1012  3.33 1012 )
6 2
L  29.68 H
Topic
: Control system
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.27 The input-output representation of a system is shown below
x ( t)
H ( s)
y (t)
Where x(t ) is the input, y (t ) is the output and H ( s) is the transfer function of a first order
system having a positive forward path gain. If x(t )  2sin t then y (t )  2 sin (t  450 ) .
The value of y (t ) at t  1sec when x (t )  u (t ) is______. (correct upto 2 decimal places)
Ans.
0.5 to 0.7
Sol.
Given that when x (t )  2 sin t we get y (t )  2 sin (t  450 ) .
k
where k is the forward path gain and s  x is the pole
sx
k
H ( j) 
j  x
Let H ( s) 
k
H ( j) 
 x
When x (t )  2 sin t
2
2
and H ( j)   tan 1
y (t )  2 H ( j) sin (t  H ( j))
2 H ( j)  2
H ( j) 
1
2
H ( j)  450
 tan 1

 450
x

 tan 450  1
x
But   1rad/s ec
1
1
x
x 1
and also H ( j) 
k
1 1
2
2

1
2
k
1

2
2
k 1
 Required transfer function is

x
1
s 1
When x (t )  u (t )
H (s) 
1
s
X (s) 
 y(t )  L1[Y (s)]  L1[ X (s) H (s)]
1 
1 1 
1  1
 L1  
L  

 s s 1
 s s 1
y(t )  (1  et ) u(t )
At t  1sec
1
y (t )  1   0.632
e
Topic
: Signal System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.28 A signal x(t )  10sinc(20t ) is sampled of f s  40 Hz by using ideal sampler and is passed
through an ideal LPF with cut-off frequency 20 Hz. Then the output spectrum of the filter
is
 f 
(A) 20 rect  
 40 
 f 
(B) rect  
 40 
 f 
(C) rect  
 20 
Ans.
Sol.
 f 
(D) 20 rect  
 20 
D
x(t )  10sinc(20t )
1
 f 
10sinc(20t ) 
 rect  
2
 20 
X(f ) 
1
 f 
rect  
2
 20 
Important :
t
  
A rect   
 A sinc  

 2 
t
A rect   
 A sinc  f  

Duality :
A
f 
A sinc  t   
 rect  


Given : f s  40 Hz
1
40
Sampled spectrum is,
Ts 
Xs( f ) 
1
Ts


n 
n 
 X ( f  nfs)  40  X ( f  n40)
Low pass-filter frequency = 20 Hz
Topic
: Power System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.29 A power flow problem is solved using Newton-Raphson power flow in polar coordinate.
The Jacobian size is 50  50. There is one slack bus in the system. The number of PV buses
is 4. The number of PQ buses is
(A) 23
(B) 26
(C) 25
(D) 27
Ans. A
Sol. Given :
Slack bus = 1
PV buses or Generator buses = 4
In the Newton-Raphson method,
The size of a Jacobian matrix = (2n – m – 2)  (2n – m – 2)
Size of Jacobian matrix = 50  50
Where n = Total number of buses
m = Number of PV buses or generator buses
2n – m – 2 = 50
2n – 4 – 2 = 50
n = 28
Number of PQ buses = Total no of buses - slack bus - PV bus
Number of PQ buses = 28 - 1 - 4 = 23
Hence, the correct option is (A).
Topic
: Signal System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.30 Which of the following statements is/are true about the signal shown below
(A) Average value of x(t ) is 1
(B) Average value of x(t ) is 0
16
3
(D) Energy of x(t ) is infinity
(C) Energy of x(t ) is
Ans.
Sol.
B, C
Given signal x(t ) is a finite duration signal between
t   2 to t   2 .
 Average value of x(t ) will be zero.
Energy of a triangular signal is
2 2
A  where A is the amplitude and  is the half width.
3
Since, there are 4 triangles
2
16
 Energy of x(t )  4   22  0.5 
3
3
Hence, correct options are (B) & (C).
Topic
: Analog Electronics
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.31 In the circuit shown below, the zener diodes Z1 and Z 2 have their breakdown voltages as
2 V and 3 V respectively. The value of voltage V0 (in volts) is
Ans.
Sol.
(A) 2 V
(B) 4 V
(C) 3 V
(D) 5 V
A
Given circuit can be redrawn as shown below
Assume that the zener diodes are not in breakdown. So , the circuit becomes as shown
below.
Voltage at node ' x ' can be given as
5
Vx  10 
 5Volts
55
 Vx  2 Volts and Vx  3volts
Both the zener diodes are coming to be in breakdown region. This makes voltage at node
' x ' 2 Volts and 3 Volts which is impossible as it will violate kirachoff’s law.
 Any one among the two zener diodes must be in breakdown.
Let Z1 be in breakdown region. Then voltage at node x will be 2 V which is less then
breakdown voltage of Z 2 that is 3 V.
Z 2 will not be in breakdown region when Z1 is in breakdown region. Now let Z 2 be in
breakdown region. Then voltage at node x will be 3 V which also makes Z1 go into
breakdown region making voltage at node x as 2 V simultaneously which is impossible as
it violates Kirchhoff’s Law.
 Z1 is in breakdown region and Z 2 is not in breakdown region and the circuit becomes
as shown below,
V0  2Volts
Hence, the correct option is (A).
Topic
: Control System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.32 The open-loop transfer functions of a unity negative feedback control system is given by
k
G s 
 s  5 3
The root locus diagram for the control system is plotted in figure 1
Figure 1
If the damping ratio    of the system is 0.5, then
(A) The value of k at point P would be equal to 125
(B) The value of     600 .
(C) The value of   600 and   300
(D) The value of the constant C  25 3 rad/sec
Ans.
Sol.
A, B
Given :
G s H s 
k
 s  5 3
Numbers of poles P  3
Number of zeros Z  0
Number of branch terminating at infinite = 3
Angle of asymptotes, is given by
 2  11800
A 
,
(where,   0,1,.... P  Z  1 )
PZ
 2  11800
(  0,1, 2)
A 
3
A  600 ,1800 ,3000
Centroid is given by,
5  5  5  0

 5
3
Intersection with imaginary axis can be calculated using Routh array.
Characteristic equation is given by,
k
1
0
 s  5 3
 s  5 3  k  0
s 3   5   3.s.5  s  5   k  0
3
s 3  15s 2  75s  k  125  0
Routh table :
Rows of zeros can be formed by equation
1125  k  125
0
15
kmarginal  1000
Auxiliary equation can be formed as,
15s 2  125  k  0
15s 2  1125  0
s 2  75
s   j5 3
So, n  C  5 3 rad / sec
So, option (D) is not correct.
Root locus diagram :
We know cos   
Given :   0.5
  cos1  0.5  600
In  OAB,
tan  
5 3
 3
5
  tan 1  3   600
So, option (B) is correct and option (C) is incorrect.
k can be calculated as,
Product of length from poles to origin
k
Products of Length from zero to origin
5 5 5
 125
1
Hence, the correct options are (A) & (B).
k
Topic
: Control System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.33 A state space representation of a system is given by
 0 1
1 
x
x, y  1 2 x, x(0)    .

 2 0
0 
The time response of this system will be
(A) cos 2t  2 2 sin 2t
(B) sin 2t
(C) cos 2t  2 2 sin 2t
(D) 2 cos 2t
Ans. C
 0 1
x  Ax  
Sol.
x
 2 0
Take Laplace transform, SX ( s )  x(0)  AX ( s)
X (s)  [ SI  A]1 x(0)
1
 s 1
1  s 1
[ SI  A]  
 2

s  2  2 s 
2 s 
1
X ( s)  [ SI  A]1 x(0)
1  s 1  1 
1 s
 2




s  2  2 s  0 s  2  2
1 s
1
 2
[ s  4]
 Y ( s )  CX ( s)  [1 2] 2


s  2  2 s  2
s
4
Y (s)  2
 2
s 2 s 2

2
Y (t )  cos 2t  (2) 2 sin 2t
Y (t )  cos 2t  2 2 sin 2t
Hence, the correct option is (C).
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.34 Consider the initially uncharged capacitor shown below
If V  2i  2 , then which of the following statements are correct?
(A) The rate of change of current i will be 2 A/s at i  2 A if the power stored in the
capacitor is 48 W.
(B) The rate of change of current i will be always zero.
(C) The energy stored in the capacitor from i  0 to i  4 A will be 12 J .
Ans.
Sol.
(D) The energy stored in the capacitor from i  0 to i  4 A will be 96 J .
A, D
Given :
…(i)
V  2i  2
Power stored in the capacitor is,
P V I
P V 
2dV
dt
From equation (i),
dV
di
2
dt
dt
di
dt
di
P  (2i  2)  4
dt

P  V  2 2
If P  48 W and i  2 A
48  24 
di
dt
di
 2 A/s
dt
Energy stored in the capacitor is,
E   (V  i ) dt
E  V  2
dv
dt
dt
E   2V dv
E   2V
dv
di
di
From equation (i),
dv
2
di
E   2V  2di
E  4  V di
E  4 (2i  2)di
Energy from i  0 to i  4 A will be
i 4
E  4  (2i  2)di
i 0
i4
E  8  (i  1) di
i 0
4
 i2 
E  8   i
 2 0
16

E  8   4  0
2

E  96 J
Hence, the correct options are (A) & (D).
Topic
: Control Systems
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.35 Consider the closed loop system shown in figure below. For this system peak overshoot is
0.3 and time for second peak overshoot is 3 second then the value of k and T is
(A) k  11.2, T  0.03
(B) k  11.2, T  0.3
(C) k  12.6, T  0.6
(D) k  12.6, T  0.06
Ans.
Sol.
A
Characteristics equation is given by,
1  G (s) H (s)  0
k
(1  Ts)  0
s( s  2)
s ( s  2)  k (1  Ts )  0
1
s 2  2s  k  kTs  0
s 2  s(2  kT )  k  0
Compare with, s 2  2n s  2n  0
2n  2  kT
…(i)
2n  k
n  k
…(ii)
Given :
Peak overshoot  0.3
2
𝑒 −𝜉𝜋/√1−𝜉 = 0.3
 
 ln (0.3)
1  2
   1   2 (1.203)
  1   2 (1.203)
2 2  (1  2 )  (1.203)2
2 2  (1  2 ) 1.447
2 2  1.447  1.4472
2 (9.869)  1.447  1.4472
2 (11.316)  1.447
  0.357
  0.35
Given, time for second peak overshoot,
3
tp 
3
d
3
 3 , d  rad/sec
d
d  n 1   2
  n 1   2
n 
3.14
1  (0.35) 2
n  3.352 rad/sec
From equation (ii),
n  k
2n  k
k  3.352 2
k  11.23
From equation (i),
2n  2  Tk
2  0.35  3.352  2  (11.23)T
T  0.03
Hence, the correct option is (A).
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.36 Consider the series R-L network shown below having an initially un-energized inductor
driven by a pulse input Vin (t )
Ans.
The maximum energy (in joules) that will be stored by the inductor will be
(A) 37.4 J
(B) 74.8 J
(C) 30 J
(D) 70 J
A
Sol.
For 0  t  2 , the circuit looks as shown below
At steady state, the inductor will be short circuit,
10
 10 A
1
As the inductor is initially un-energized i (0)  0 A .
 i ( ) 
1
Time constant    1sec
1


t
i(t )  i()  [i(0)  i()]e  , t  0
i(t )  10  [0  10]et , t  0
i(t )  10[1  et ], t  0
The maximum current that will be stored in the circuit is at t  2sec .

imax  10[1  e 2 ]  8.65 A
Maximum energy stored by the inductor will be,
1 2
Emax  Limax
2
1
Emax  1 (8.65) 2  37.4 J
2
Hence, the correct option is (A).
Topic
: Machine
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.37 A 220 V DC shunt motor with an armature resistance of 1 Ω is drawing 22 A from supply
and its field current is 20 A at speed 100 rad/sec. The torque developed by the motor when
drawing 12 A from the supply with same field current is
(A) 20 N-m
(B) 40 N-m
(C) 24 N-m
(D) 43 N-m
Ans. C
Sol. Given :
V  220 V
Ra  1 Ω
I L  22 A
I a  20 A
Eb  V  I a Ra  220  20  200 V
T1 
Eb I a 200  20

 40 N-m
r
100
When I a  12 A with same field current 20 A.
T1 I a1

T2 I a 2
40 12
 24 N-m
20
Hence, the correct option is (C).
 T2 
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.38 For the two-port network shown in figure below
The value of h11 is _______ 
Ans.
Sol.
0.3 to 0.5
The given circuit is the combination of two-port networks connected in parallel. For twoport networks connected in parallel we can easily find the y parameters
For [YA ] ,
I1 A 
V1 A  V2 A
2
I1A  0.5V1A  0.5V2 A
I 2 A   I1 A
I2 A 
V2 A V1 A

2
2
I 2 A  0.5V2 A  0.5V1A
I 2 A  0.5V1A  0.5V2 A
 0.5 0.5
2 4
[YA ]  
[YB ]  


 0.5 0.5 
1 3 
Y-parameters of network,
[Y ]  [YA ]  [YB ]
 0.5 0.5  2 4  2.5 3.5
[Y ]  



 0.5 0.5  1 3 0.5 3.5
Now Y parameters equations,
I1  Y11V1  Y12V2
I 2  Y21V1  Y22V2
I1  2.5V1  3.5V2
…(i)
I 2  0.5V1  3.5V2
…(ii)
Now h11 
V1
I1 V
2 0
From equation (i),
Keep V2  0
I1  2.5V1
h11 
V1
1

 0.4 
I1 2.5
Topic
: Engg.Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.39 A bag contains (2n + 1) coins. It is known that n of these coins have a head on both the
sides whereas the remaining coins are fair. A coin is picked up at random from the bag and
37
is tossed. If the probability that the toss results in a head is
, then the value of n is
50
(A) 37
(B) 12
(C) 13
(D) 50
Ans. B
Sol. Let E1 denote the event that a fair coin is chosen and E2 denote the event that a coin with
both heads is chosen. Let H denote the event that toss results in a head. Then
n 1
n
P ( E1 ) 
, P ( E2 ) 
2n  1
2n  1
H 1 H 
 P   , P  1
 E1  2  E2 
 P (getting head)  P( E1  H )  P( E2  H )
H
H
 P( E1 ) P    P( E2 ) P  
 E1 
 E2 
n 1 1
n
3n  1

 
1 
2n  1 2 2n  1
4n  2
3n  1 37


4n  2 50
n  12
Hence, the correct option is (B).
Topic
: Digital
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.40 Consider the following sequential circuit implemented by using T-Flip-flop. The state
transition for Q2 Q1 in decimal after 5th clock is ________. (Assuming initially Q2Q1  00 )
Ans.
Sol.
3
𝑇1 = 𝑃 ⊙ 𝑄2 = 1 ⊙ 𝑄2 = 𝑄2
Clk
Q2
Q1
T2
T1
Q2
Q1
Decimal
equation
0
1
0
1
0
0
1
1
0
0
1
1
1
1
1
1
0
1
0
1
1
0
1
0
1
1
1
0
0
1
3
1
2
0
3
Present state
1
2
3
4
5.
Topic
: EMT
Concept
:
Sub-concept
:
Next state
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.41 Consider the integral
Ans.
Sol.
  2 x iˆ  2 yjˆ  5z kˆ  . nˆ.ds
over the surface of a sphere of radius
 3 with center at origin, and surface unit normal n̂ pointing away from the origin. Using
the Gauss divergence theorem, the value of this integral is _________.
564 to 566
Given :
I
2 xiˆ  2 yjˆ  5 zkˆ . nˆ . ds
 

The surface is a sphere of radius r  3 and center  Origin
By Gauss Divergence theorem,
 F . nˆ ds    ( .F ). dv
S



where,   iˆ  ˆj  kˆ
 Del/Nebla operator
x
y
z
 


.F   iˆ  ˆj  kˆ  . 2 xiˆ  2 yjˆ  5 zkˆ
y
z 
 x






   2 x    2 y    5z 
y
z
 x

 225  5
So, I   2 xiˆ  2 yjˆ  5 zkˆ nˆ.ds    .F  .dv

S

V
  5. dv  5 dv  5V
V
V
Here, V  volume of sphere with radius  3
4
3
So, I  5    3
3
4 3

Since, Volume of sphere = 3 R 
 180  = 565.48
Topic
: Network
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.42 A sinusoidal source of voltage V and frequency f is connected to a series RC circuit shown
in figure below
The locus of the tip of the current phasor I and R is varied from 0 to  is
(A)
(B)
(C)
(D)
Ans.
A
Sol.
Consider the given circuit,
I
V
R  jX c
I
X 
  tan 1  c 
 R 
R 2  X c2
I
X 
 tan 1  c 
 R 
R 2  X c2
V
V
For R  0 ,
V
I
900
Xc
For R  X c ,
I
V
450
2Xc
For R   ,
I  000
From the above data the locus plot of I when R is varied from 0 to  is,
Hence, the correct option is (A).
Topic
: Power system
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark : 0.66
Q.43 The main bus-bars in generating station are divided into three sections, each section being
connected to tie bar by a similar reactor. One 20 MVA, 3-phase, 50 Hz, 11 kV generator with a
short circuit reactance of 15% is connected to each section bus-bar. When a short circuit takes
place between the phases of one of the section bus-bar, the voltage of the remaining sections
falls to 55% of the normal value. Calculate the reactance of reactors (in ohm).
Sol.
Correct answer (0.37)
Let A, B and C be the three generators, each of 20 MVA and of 15% (0.15 pu) reactance. A’, B’,
and C’ are the three reactors of reactance X per unit based on 20 MVA.
The circuit diagram is shown is shown in figure assuming that the fault occurs in section 3. A and
A' in series are in parallel with B and B' in series.
 reactance of A, B, A' and B' between the tie-bar and neutral
0.15  X
i.e. (0.5X × 0.075) p.u
2
Reactance from neutral to fault via tie bar = [(0.5 X + 0.075) + X]
= (1.5 X +0.075) p.u
The voltage of sections 1 and 2 drops to 55% i.e., 0.55 times of the normal value, therefore, 0.45
times of the normal value is dropped in the reactance of A and B i.e., reactance of A and B
 0.15 
 i.e.
 is 0.45 times the total reactance from neutral to fault
2 

i.e.,
0.15
 0.45(1.5 X  0.075)
2
Or X = 0.0611 p.u
Hence reactance of each reactor
Per unit reac t an ce  (kV)2  1000
=
Base kVA
Topic
: Signal System
Concept
:
Sub-concept
:

0.0611 (11)2  1000
 0.37
20000
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.44 Which
of
the
following
statements
 x 1
 2
f ( x)  
 x 2  1 e x
 x 1

3
(A) f ( x ) is an odd function
(B) f ( x ) is neither even nor odd function
2
(C)
1
 f ( x)dx  2 [e
2
 1]
0
2
1
(D)  f ( x)dx  [e 4  1]
2
0
Ans.
Sol.
A, D
Given function is,
 x3  1
 2
f ( x)  
 x 2  1 e x
 x 1

 ( x  1)( x 2  x  1)
 2
f ( x)  
 x 2  1 e x
x 1


f ( x)   x2  x  1  x2  1 e x
2
is/are
true
about
the
function
f ( x)  xex
2
f ( x)   xex   f ( x)
2

f ( x ) is an odd function
2

2
f ( x)dx   xe x dx
2
0
0
Let x  t
2xdx  dt
2
xdx 
1
dt
2
t limits 0 to 4
2

4
x
t
 xe dx   e
2
0
0
2
 xe
4
x2
dx 
0
2
 xe
x2
0
2
 xe
1
dt
2
x2
0
1 t
e dt
2 0
1
dx  [et ]04
2
1
dx  [e 4  1]
2
Hence, the correct options are (A) & (D).
Topic
: EMT
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.45 A magnetic circuit has 130 turns-coil, the cross section area 5 104 m 2 and the length of
the magnetic circuit 25  102 m . The value of magnetic field intensity and relative
permeability when the current is 2 A and flux is 0.8 103 Wb , are
(A) 1040 AT/m and 1224.27
(B) 300 AT/m and 500 103
(C) 300 AT/m and 397.9
(D) 1200 AT/m and 500 106
Ans.
Sol.
A
Given :
A  5 104 m 2
l  25 102 m
i  2A
  0.8 103 Wb
 0  4 107 H/ m
N = 130
We know, N   Li
130  0.8 103  L  2
 L = 0.052 H
r 0 N 2 A
l
 Relative permittivity of core is,
Again, L 
0.052  25 10 2
 1224.27
410 7 1302  5 10 4
Again, Ni  H  l
 130  2  H  25 102
 H  1040 AT/m
 Magnetic field intensity is 1040 AT/m.
Hence, the correct option is (A).
r 
Topic
: Power Electronics
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.46 A line commutated inverter feeds power to 400 V, 50 Hz, 3- AC side, using a battery of
500 V. The battery is connected to converter by a large filter choke having resistance of
4  . While delivering 4 kW power to ac side from dc side, which of the following
statement(s) is/are correct?
(A) The efficiency of the energy transfer of inverter is 93.13%.
(B) Firing angle of the 3- rectifier is 129.850 .
(C) Firing angle of the 3- rectifier is 141.590 .
Ans.
Sol.
(D) The efficiency of the energy transfer inverter is 95.62%
A, C
Given :
Line commuted inverter (i.e.   900 )
3- AC side : 400 V, 50 Hz
DC side : 500 V battery large filter choke with R  4 
Power transferred = 4 kW.
Power is transferred from battery to AC side
EI 0  P  I 02 R
P  4000 W
500 I 0  4000  I 02  4
I 0  116.409,8.59 A
Always choose the lesser value of I 0 for better efficiency.
So, I 0  8.59 A
Now,
3VmL
cos    E  I 0 R

3  400 2
cos   500  8.59  4

  141.590
Now,  

Output
100
Output  Losses
Ptransferred
Ptransferred  I 02 R
100

4000
100  93.13%
4000  (8.59) 2  4
Hence, the correct options are (A) & (C).
Topic
: Analog Electronics
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.47 In the circuit given below, assume that VCC  15 V , Z1, Z2, Z3 and Z4 and identical Zener
diodes with breakdown voltage of 5 V, R1 = R4  5 k, R2  R3  10 k . Find V0 when Q1
is OFF.
Ans.
Sol.
(A) 7.5 V
(B) 5 V
(C) 10 V
(D) 8.33 V
C
Given : VCC = 15 V, Z1, Z2, Z3, Z4 are identical Zener diodes Q1 is OFF. ( IC = 0)
Apply KCL at node,
V0  15
V  5 V0  5
0 0

0
5 k
10 k 10 k

2V0  30  V0  5  V0  5
10 k
4V0  40
V0  10 V
Hence, the correct option is (C).
Topic
: Engg. Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.48 Consider the following two normal distributions
f1 ( x)  e x
2
 1

1  ( x2 2 x1)
f 2 ( x)  e  4 
2
If  and  denote the mean and standard deviation respectively, then
(A) 1  2 and 12   22
(B) 1  2 and 12   22
(C) 1  2 and 12   22
(D) 1  2 and 12   22
Ans.
Sol.
C
The density function of the normal distribution is given by :
f ( x) 
1
1  x  
 

2  
2
e
22
Where,   Mean and   Standard deviation
Given, f1 ( x)  ex
2
On comparing with the above equation, we have
1
1  0 and
1
212
12 
1
2
 1

1  4  ( x2 2 x1)
e
Also, f 2 ( x) 
2
1  41 ( x 1)2
1  41 ( x ( 1))2
f 2 ( x) 
e

e
2
2
On comparing, we have
2  1 and
1
2
2
2

1
  22  2
2
So, 1  2 and 12   22
Hence, the correct option is (C).
Topic
: Engg.Maths
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.49 Consider two random variables X and Y where X is uniformly distributed over [–1, 2].
Assuming Y  2 X  3, the probability density function (pdf) of Y i.e. fY ( y) is ________
1
 ; 1  y  4
(A) fY ( y )   5

 0 ; otherwise
2
 ; 1  y  4
(B) fY ( y )   5

 0 ; otherwise
1
 ; 1 y  7
(C) fY ( y )   6

 0 ; otherwise
1
 ; 1  y  6
(D) fY ( y )   6

 0 ; otherwise
Ans.
Sol.
C
Given :
X UDF [1, 2]
So, pdf of X is –
1
 , 1  x  2
f X ( x)   3

 0 , otherwise
Transformation of random variable,
dY
fY ( y )
 f X ( x)
dX
fY ( y ) 2  f X ( x )
1
f X ( x)
2
Hence, the correct option is (C).
fY ( y ) 
Range of Y :1  y  7
Topic
: Power System
Concept
: Power flow through transmission line
Sub-concept
: Voltage profile in transmission line
Concept-field :
Level
:
Question Type
: MSQ
Marks
: +2,
Negative mark
:0
Q.50 The one-line diagram of a three phase power system is shown in the following figure.
Impedances are marked in per unit on a 100 MVA, 400 kV base. The load at bus-3 is
S3  77 MW  j14 MVAR and the load at bus-2 is S2  15.92 MW  j33.4 MVAR . The
voltage at bus-3 is maintained at 40000 kV. The correct statements is/are
(A) V2  1.116.260 pu
(B) V2  1.219.260 pu
(C) V1  1.236.870 pu
(D) V1  480 V
Ans.
Sol.
A, C, D
Given :
S3  77 MW  j14 MVAR
S2  15.92 MW  j33.4 MVAR
S 2 and S 3 in pu
S3  0.77  j 0.14pu
S2  0.1592  j 0.334 pu
I 3* 
S3
 I 3  0.77  j 0.14 pu .
V3
V2  V3  I 3 j 0.4
 100  j 0.308  0.056
 1.056  j 0.308  1.116.260
I 2* 
S2
V2
0.1592  j 0.334
 0.336480.7750
0
1.116.26
I1  I 2  I3  0.824  j 0.192
I 2* 
V1  V2  j 0.5I1
 1.056  j 0.308  j 0.412  0.096
 0.96  j 0.72  1.236.870 pu
V1  1.2  400  480 V
Hence, the correct options are (A), (C), (D).
Topic
: Control System
Concept
:
Sub-concept
:
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.51 Consider the system shown in figure, where r  t  is the input to the system and c  t  is the
response of the system in time domain. Assuming zero initial conditions, if r  t   3t 3u t 
Fig.-1
2d 2c  t 
d c  t  4d c  t  3d c  t 



dt 5
dt 4
dt 3
dt 2
(A)  54  18t  18t 2  3t 3  u  t   36   t 
5
4
3
Then,

3dc t 
 2c  t  is
dt
(B) (54  36t  9t 2  3t 3 )u(t )  18(t )
(C)  36  54t  18t 2  9t 3  u  t   54   t 
(D)  54  18t  9t 2  3t 3  u  t   54   t 
Ans.
Sol.
B
From figure, the transfer function of the system can be written as,
C s
s 4  3s 3  2s 2  s  1
 5
R  s  s  4s 4  3s 3  2s 2  3s  2
Cross multiplying, we get
 s5  4s 4  3s3  2s 2  3s  2  C  s    s 4  3s3  2s 2  s  1 R  s 
Taking the inverse Laplace transform,
d 5c  t  4d 4c  t  3d 3c  t  2d 2c  t  3dc t 




 2c  t 
dt 5
dt 4
dt 3
dt 2
dt
d 4 r  t  3d 3r  t  2d 2 r  t  dr  t 




 r t 
dt 4
dt 3
dt 2
dt
Given, r  t   3t 3u t 
dr  t  d  3t 3 

 9t 2u  t 
dt
dt
d 2r t 
 2 18tu (t )  36tu  t 
dt 2
d 3r  t 
3
 3 18u (t )  54u  t 
dt 3
d 4r t 
 18  t 
dt 4
d 5c  t  4d 4c  t  3d 3c  t  2d 2c  t  3dc t 




 2c  t 
dt 5
dt 4
dt 3
dt 2
dt
 18   t   54u(t )  36tu(t )  9t 2u(t )  3t 3u(t )
2
 (54  36t  9t 2  3t 3 )u(t )  18(t )
Hence, the correct option is (B).
Topic
: Electrical machine
Concept
: Induction generator
Sub-concept
: Single phase synchronous motor
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.52 A cylindrical rotor synchronous motor, with E f  1.5 pu , is connected to an infinite bus.
The machine has a synchronous reactance of 1.2 pu and output power delivered to the load
Ans.
Sol.
is 0.5 pu with all losses ignored, find the magnitude of reactive power delivered or
absorbed by the synchronous motor.
(A) 1.25 pu delivering
(B) 1.25 pu absorbing
(C) 0.312 pu delivering
(D) 0.312 pu absorbing
C
EV
P
sin 
Xs
0.5 
1.5 1
sin 
1.2
sin   0.4
and cos   1  0.42  0.9165
V
V  E cos  
Reactive power, Q 
Xs
1
1  1.5  0.9165  0.312 pu
1.2
Q is negative, motor is delivering reactive power to the infinite bus.
Hence, the correct option is (C).
Q
Topic
: Electrical machine
Concept
: DC machine
Sub-concept
: DC motor
Concept-field : DC shunt motor
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.53 A 220 V dc shunt motor running at 1000 rpm takes an armature current of 17 A. What will
be additional resistance to be added in armature circuit to reduce the speed of the motor to
600 rpm with armature current remaining same and armature resistance is 0.4  ?
(A) 5.016 
(B) 5.6 
(C) 4.6 
(D) 4.016 
Ans.
Sol.
A
Given :
V = 220 V
I a  17 A
Ra  0.4 
N1  1000 rpm and N 2  600 rpm
Eb1  V  I a Ra  220  17  0.4  213.2 V
Eb2  V  I a  Ra  Rc   220 17  0.4  Rc 
N
Eb

[   Constant, armature resistance control method]
N2 Eb2

N1 Eb1
600 220  17  0.4  Rc 

1000
213.2
 220 17  0.4  Rc   127.92
 Rc  5.416  0.4  5.016 
Hence, the correct option is (A).
Topic
: Power Electronics
Concept
: Choppers
Sub-concept
: Buck-Boost convertor
Concept-field :
Level
:
Question Type
: MCQ
Marks
: +2,
Negative mark
: 0.66
Q.54 In the circuit shown below,
Ans.
Sol.
If the frequency of switching is 20 kHz, then the rms value of ripple current through L is
(A) 31.6 mA
(B) 43.5 mA
(C) 37.4 mA
(D) 42.6 mA
B
Given :
Vs  12.6 V
L  1 mH
f  20 kHz
V0  5 V
Ripple current through inductor, I 
Vs (1  )
fL
As it is a Buck converter,
V0  Vs
5  (12.6)
  0.397
12.6  0.397  (1  0.397)
 0.1508 A
I 
20 103 1103
I  0.0754  (0.0754)
iL ripple  0.1508A
RMS value of this wave form is,
A 0.0754

 43.53mA
3
3
Hence, the correct option is (B).
I Lripple (rms) 
Topic
: Power Electronics
Concept
: Invertors
Sub-concept
: Three-phase bridge invertor
Concept-field : Three-phase 1800 mode VSI
Level
:
Question Type
: MSQ
Marks
: +2,
Negative mark
:0
Q.55 A 3- , 180 mode, bridge inverter has star connected load of R  4  and L  20 mH .
0
Ans.
Sol.
The inverter is fed from 220 V dc and output frequency is 50 Hz. If only fundamental
component of current is considered, then which of the following statement(s) is/are correct?
(A) Fundamental power across the load is 2125.87 W.
(B) Fundamental component of load current is 13.31 A
(C) Fundamental power across the load is 4141.1 W.
(D) Fundamental component of load current is 9.66 A.
A, B
Given :
3- , star connected bridge inverter
Operating in 1800 conduction mode
Vs  220 V
f  50 Hz
R4
L  20 mH

Vnph 
2Vs
sin nt
n  6 k 1 n
Vnph 
2Vs
sin nt
n
2Vs
I 01 

2 Z
Z  R  jL
Z  4  j (20 103  2   50)  4  j 6.28 
Z  7.44 
( I 01 )rms 
2  220
 13.31A
2   7.44
Power across the load  3I 021 R
 3 13.312  4  2125.87 W
Hence, the correct options are (A) & (B).
