EE gate-2016-set-1

|EE| GATE-2016-PAPER-01 www.gateforum.com
 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series
Leaders in GATE Preparation  65+ Centers across India
© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.
1
General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1. The man who is now Municipal Commissioner worked as _____________.
(A) the security guard at a university
(B) a security guard at the university
(C) a security guard at university
(D) the security guard at the university
Key: (B)
2. Nobody knows how the Indian cricket team is going to cope with the difficult and seamer-friendly
wickets in Australia.
Choose the option which is closest in meaning to the underlined phase in the above sentence.
(A) put up with (B) put in with (C) put down to (D) put up against
Key: (D)
3. Find the odd one in the following group of words.
Mock, deride, praise, jeer
(A) mock (B) deride (C) praise (D) jeer
Key: (C)
4. Pick the odd one from the following options.
(A) CADBE (B) JHKIL (C) XVYWZ (D) ONPMQ
Key: (D)
5. In a quadratic function, the value of the product of the roots  ,  is 4. Find the value of
n n
n n 
  
  
(A) 4
n (B) n
4 (C) 2n 1
2 
(D) n 1
4 
Key: (B)
Exp: Given 4 
 
 
n n n n
n n
n n
n n n n
n n
n n
1 1
( ) 4
 
     

   
 
    

  
  

2
Q. No. 6 – 10 Carry Two Mark Each
6. Among 150 faculty members in an institute, 55 are connected with each other through Facebook
and 85 are connected through WhatsApp. 30 faculty members do not have Facebook or WhatsApp
accounts. The number of faculty members connected only through Facebook accounts is ______.
(A) 35 (B) 45 (C) 65 (D) 90
Key: (A)
Exp: F  Facebook, W WhatsApp, E Total faculties
given
 
   
 
n(E) 150, n F W 30
n F W n(E) N F W 150 30
n F W 120
  
     
 
     
 
 
n f w n f n(w) n(F w
120 n(F) 85
n(F) 120 85 35
55 n(F) n F W
n F w 55 n(F) 55 35 20
n(w) 85 20 65
    
 
  
  
     
  
7. Computers were invented for performing only high-end useful computations. However, it is no
understatement that they have taken over our world today. The internet, for example, is
ubiquitous. Many believe that the internet itself is an unintended consequence of the original
invention with the advent of mobile computing on our phones, a whole new dimension is now
enabled. One is left wondering if all these developments are good or more importantly, required.
Which of the statement(s) below is/are logically valid and can be inferred from the above paragraph?
(i) The author believes that computers are not good for us
(ii) Mobile computers and the internet are both intended inventions
(A) (i) (B) (ii) only
(C) both (i) and (ii) (D) neither (i) nor (ii)
Key: (D)
8. All hill-stations have a lake. Ooty has two lakes.
Which of the statement(s) below is/are logically valid and can be inferred from the above
sentences?
(i) Ooty is not a hill-station
(ii) No hill-station can have more than one lake.
(A) (i) Only (B) (ii) Only (C) both (i) and (ii) (D) neither (i) nor (ii)
Key: (D)
E, n(E) 150
55 85
F
n(F) 35
W
n(w) 60 
F w
n F w
20



 
30
F W



3
9. In a 2 4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be
observed in the grid?
(A) 21 (B) 27
(C) 30 (D) 36
Key: (C)
Exp: 1: (AEOK)
2: (AEJF), (FJOK)
4: (ABLK), (BCML), (CDNM), (DEON)
2: ACMK, ADNK 2:ECMD,EBLO 2:ACHF,ADIF
2: ECHJ, EBGJ 2:FHMK,FINK 2:JHMD,JGLO
1:BDNL 2:BDIG,GINL
8: ABGF, BCHJ, CDIH, EDI, FGLK, GHML, HINM
Total = 1+2+4+2+2+2+2+2+2+1+2+8=30
10.
Chose the correct expression for f(x) given in the graph.
(A)  f x 1 x 1   (B)  f x 1 x 1  
(C)  f x 2 x 1   (D)  f x 2 x 1  
Key: (C)
Exp: Substituting the coordinates of the straight lines and checking all the four options given, we get
the correct option as C which is f(x)= 2 x 1 
25
2
1.5
1
0.5
0
0.5
1
1.5
2
25
4 3 2 1
0 1 2 3 4X
 f x
A B C D E
F G H I J
K L M N O

4
Electrical Engineering
Q. No. 1 –25 Carry One Mark Each
1. The maximum value attained by the function    f x x x 1   x 2 in the interval  1, 2 is ____.
Key: 0
Exp: f( ) ( 1)( 2)  x x x x 3 2
f( ) 3 2   x x x x 1 2
f ( ) 0 3 6 2 0     x x x
1
1
3
  x
1
But 1 only lies on the interval [1,2]
3
 x
111
At 1 ;f ( ) 6 6
3
   x x x
1
6 1 6 0
3
 
    
 
1
1 is a point of minimum
3
  x
f(x) = x(x-1)(x-2) = 0 at either ends x =1 & x =2
Max value = 0
2. Consider a 3 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
Key: 3
Exp: Consider 3 3
1 1 1
A 1 1 1
1 1 1

 
   
  
It’s only non-zero Eigen value λ = 1 order of the matrix = 1 3 3 
3. The Laplace Transform of      2t
f t e sin 5t u t is
(A) 2
5
s 4s 29 
(B) 2
5
s 5
(C) 2
s 2
s 4s 29

 
(D)
5
s 5
Key: (A)
Exp: Consider
By frequency shifting property,   0S t1
0X(S-S ) (t)e
 x
0thus at S 2
2
5
F(s)
s 4s 29
 
 

x(t) = sin5t u(t) 2
5
X(S)
s +25

2t
f(t) = e sin5tu(t)

2
5
F(s)
(s-2) +25


5
4. A function    y t , such that y 0 1 and   1
y 1 3e ,
 is a solution of the differential equation
2
2
d y dy
2 y 0
dtdt
   . Then  y 2 is
(A) 1
5e
(B) 2
5e
(C) 1
7e
(D) 2
7e
Key: (B)
Exp: The operator function of given D.E is
2
(D +2D+1)y = 0
2
A.E is D +2D+1 = 0 D = -1,-1 
 -x
1 2y = e C + C (1) 
-1
Given y(0)=1 & y(1) =3e
From ; y(0) = 1 i.e , y = 1 at x = 0 11 = C
-1
y(1) = 3e i.e, y = 3/e at x=1
 -1
2
3
e 1 C (1)
e
  2 23 1 C C 2    
-x -2 -2
y = e [1+2x] y(2) = e [1+4] = 5e 
5. The value of the integral
 2C
2z 5
dz
1
z z 4z 5
2

 
   
 
 over the contour z 1. taken in the anti-
clockwise direction, would be
(A)
24 i
13

(B)
48 i
13

(C)
24
13
(D)
12
13
Key: (B)
Exp: Singular points are obtained by
 21
z z -4z+5 0
2
 
  
 
1
z = &z =2 i
2
 
Out of these
1
z =
2
only lies inside C. z =1
By Cauchy’s integrated formula,
 
2
2C C
2z+5 2z+5
dz = dz
1 z -4z+5
z - z -4z+5 0
12
z -
2
 
 
 
   2
1
2 5
48πi2
= 2πi
131 1
4 5
2 2
  
  
   
    
     
    

6
6. The transfer function of a system is
 
 
Y s s
,
R s s 2


The steady state output x(t) is A cos 2t   for
the input  cos 2t . The value of A and , respectively are
(A) O1
, 45
2
 (B) O1
, 45
2
 (C) O
2, 45 (D) O
2, 45
Key: (B)
Exp:
S
H(s) =
S+2
0 1
2
H( ) = 90 tan
2+4
  
  
 
 


If input is COS 2t i.e.,  = 2
01
H( ) = 45
2

01
y(t) = cos(2t 45 )
2
 
0
1
by comparision A= & 45
2

7. The phase cross-over frequency of the transfer function G(s) =
 
3
100
s 1
in rad/s is
(A) 3 (B)
1
3
(C) 3 (D) 3 3
Key: (A)
Exp: Phase Crossover frequency PC
0
PC =: GH 180  
-1 0 -1 0
PC PCGH 3tan 180 3tan w W tan60 3        
8. Consider a continuous-time system with input x(t) and output y(t) given by
     y t x t cos t . This system is
(A) linear and time-invariant (B) Non-linear and time-invariant
(C) linear and time-varying (D) Non-linear and time-varying
Key: (C)
Exp: Linearity
1 1 2 2y (t)= x (t)cost; y (t)= x (t)cost
1 2 1 2y (t) + y (t) = x (t)cost + x (t)cost
1 2 1 2y (t) + y (t) = [x (t) + x (t)]cost
1 2 1 2If x (t) + x (t) = x(t) then y (t) + y (t) = y(t)

7
Thus the system is linear
Time-invariance
consider 1 1 1y (t) x (t) cost ; If x (t) x(t - τ) 
1y (t) x(t - τ)cost 
Define y(t - τ) x(t - τ)cos(t - τ)
1y (t) y(t - τ) system is time-varient 
9. The value of  t
e 2t 2



  dt. where  t is the Dirac delta function, is
 
1
A
2e
 
2
B
e
  2
1
C
e
  2
1
D
2e
Key: (A)
10. A temperature in the range of -40O
C to 55O
C is to be measured with a resolution of 0.1O
C. The
minimum number of ADC bits required to get a matching dynamic range of the temperature
sensor is
(A) 8 (B) 10 (C) 12 (D) 14
Key: (B)
Exp: Usually when voltage information is given we use the formula max min
n
V -V
R =
2
Here based on the given information we can think the systems as following block diagram
 Here the maxV and minV will be the equivalent of maxT and minT respectively
So we can still use the above relation
max min
n
T -T
Resolution =
2
 n n5-(-40)
2 = 2 = 950
0.1
 
2n log (950) 9.89 10   
So minimum requirement is 10 bits
11. Consider the following circuit which uses a 2-to-1 multiplexer
as shown in the figure below. The Boolean expression for
output F in terms of A and B is
(A) A B (B) A B
(C) A B (D) A B
A B
S1
Y
0
F
Temp
Temperature
to
voltage
converter
analog
voltage
nbit
ADC .
.
.

8
Key: (D)
Exp: We can redraw the max circuit as follows
So the Boolean expression of F(A, B)=BA BA=A B=A B 
12. A transistor circuit is given below. The Zener diode breakdown voltage is 53 V as shown The
base to emitter voltage droop to be 0.6V. The value of the current gain  is ________.
Key: 19
Exp: E5.3 = 0.6 + 470 I
EI 0.01A
X 3
10 - 5.3
I = 1mA
4.7 10


BI =1 0.5 0.5mA 
E BI (1 β)I 
3
0.01 (1 β) 0.5 10
     3
0.01
(1 β) 20
0.5 10
   

β 19 
13. In cylindrical coordinate system, the potential produced by a uniform ring charge is given
 f r,z ,  where f is a continuous function of r and z. Let E

be the resulting electric field. Then
the magnitude of E

(A) increases with r (B) is 0
(C) is 3 (D) decreases with z
Key: (B)
Exp: Since the charge is not varying with time the field E is static So E 0 
4.7k
10V
220
470
0.5mA
5.3V
0
F
A
1A
B

9
14. A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the
relative permeability r of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent
points located just inside and just outside the toroid, is ________.
Key: 100
Exp: The field inside and outside the toroid is due to long straight conductor only
Let the two points almost at same distance
The flux density inside toroid = 0 rμ μ I
2πr
The flux density outside toroid = 0μ I
T
2πr
Ratio =
0 r
r
0
μ μ I
2πr μ 100
μ I
2πr
 
15. A BR and R are the input resistances of circuits as shown below. The circuits extend infinitely in
the direction shown. Which one of the following statements is TRUE?
(A) A BR R (B) A BR R 0  (C) A BR R (D)  B A AR R 1 R 
Key: (D)
Exp: By comparing 2 network on the input side
we can say that A
B A B
A
R
R =1//R R =
1 R


16. In a constant V/f induction motor drive, the slip at the maximum torque
(A) is directly proportional to the synchronous speed
(B) remains constant with respect to the synchronous speed
(C) has an inverse relation with the synchronous speed
(D) has no relation with the synchronous speed
2 2 2
1
1 1
BR
AR
2 2 2
1
1 1 1

10
Key: (C)
Exp: 2 2 2
2 2 2
r r r
SmT
x j L j2 L .f
  
 
1
SmT
f

S
120f
N
P

1
SMT
NS
 
17. In the portion of a circuit shown, if the heat generated in 5 resistance is 10 calories per second
then heat generated by the 4 resistance, the calories per second, is ______.
Key: 2
Exp: Here the power information regarding the resistor is given because
E Calorie
P = =watt
t sec

5P = 10
5ΩV
= 10
5


5ΩV = 50
4ΩP is asked
 
2 2
4Ω
4Ω
V 1 4
P = 50
4 4 4 6
1 16 calorie
50 2
4 100 sec
 
   
   
18. In the given circuit, the current supplied by the battery, in ampere, is ________.
`
Key: 0.5
Exp: If we write KCL at node × then
4 6
5
1V
2l1
1
11l
2l


11
1
1 2 2
I
I = 2 I
2
 I
Write KVL in the outer boundary of network
1 21 (1 I ) (2 ) 0    I 1 21 I 2   I
1
1 1 1
I 1
1 =I 2 1 2J I 0.5A
2 2
 
      
 
19. In a 100bus power system, there are 10 generators. In a particular iteration of Newton Raphson
load flow technique (in polar coordinates). Two of the PV buses are converted to PQ type. In this
iteration.
(A) The number of unknown voltage angles increases by two and the number of unknown voltage
magnitudes increases by two.
(B) The number of unknown voltage angles remain unchanged and the number of unknown
voltage magnitudes increases by two
(C) The number of unknown voltage angles increases by two and the number of unknown voltage
magnitudes decreases by two
(D) The number of unknown voltage angles remains unchanged and the number of unknown
voltage magnitudes decreases by two
Key: (B)
Exp: Total no of Buses = 100
Generator Buses = 10
 Load Buses = 100-10=90
Slack Bus = 1
If 2 of the PV buses are converted to PQ type the no of on voltage magnitudes increases by 2 with
constant unknown voltage angles
20. The magnitude of three-phase fault current at buses A and B of a power system are 10 pu and 8
pu, respectively. Neglect all resistance in the system and consider the pre-fault system to be
unloaded. The pre-fault voltage at all buses in the system is 1.0 pu. The voltage magnitude at bus
B during a three-phase fault as but. A is 0.8pu. The voltage magnitude at bus A during a three-
phase fault at bus B, in pu, is _____.
Key: 0.84
Exp: Voltage at ith
bus when fault is at kth
bus is
 
productik
i f
kk f p.u
Vz
V E 1 I
z z X
 
   
 
n
1
Ax
10 X 0.1P.U  
B
1
X
8 B 0.125P.U   
AB
B
AA
Z
V E 1
Z
 
  
 
1V
1I
2I
2I
1
1
1
X

12
AB
AB
Z
0.8 1 1 Z 0.02
0.1
 
    
 
A
BB
2AB 0.02
V 1 1 1
2 0.125
   
       
  
0.84 P.U
21. Consider a system consisting of a synchronous generator working at a lagging power factor, a
synchronous motor working at an overexcited condition and a directly grid-connected induction
generator. Consider capacitive VAr to be a source and inductive VAr to be a sink of reactive
power. Which one of the following statements is TRUE?
(A) Synchronous motor and synchronous generator are sources and induction generator is a sink
of reactive power.
(B) Synchronous motor and induction generator are sources and synchronous generator is a sink
of reactive power.
(C) Synchronous motor is a source and induction generator and synchronous generator are sinks
of reactive power
(D) All are sources of reactive power
Key: (A)
22. A buck converter, as shown in figure (a) below, is working in steady state. The output voltage and
the inductor current can be assumed to be ripple free. Figure (b) shows the inductor voltage LV
during a complete switching interval. Assuming all devices are ideal, the duty cycle of the buck
converter is ______
Key: 0.4
Exp: When M is ON, S L 0V V V 
L S 0V 30 V V   
When M is OFF, L 0 0 0V V 20 V V 20      
S s30 V 20 V 50V    
0
S
V 2
D 0.4
V 5
  
M
 LV
D C 0V
R
 a
gV



LV
0
20V
aT
ONT OFFT
t
 b
30V

13
23. A steady dc current of 100A is flowing through a power module (S.D) as shown in Figure (a). The
V-I characteristics of the IGBT (S) and the diode (D) are shown in Figures (b) and (c),
respectively. The conduction power loss in the power module (S, D), in watts, is _____
Key: 169 to 171
24. A 4pole, lap-connected, separately excited dc motor is drawing a steady current of 40 A while
running at 600 rpm. A good approximation for the waveshape of the current in armature
conductor of the motor is given by
(A) (B)
(C) (D)
Key: (C)
Exp: no of parallel paths = 4
Armature current/conductor =
40
10A
4

For linear commutation, the change from 10A to 10A  is straight line
N = 600rpm
Time for 1 revolution =- 0.1 sec.
For 1 pole-pitch,
0.1
t 25msec
4poles
 
 sI A
0V 1V
 SV Volt
dV dl 0.02 
V l characteristic of 1GBT
 b
100 A
D
S
 a
 0I A
0V 0.7V
 SV Volt
dV dl 0.01 
V l characteristic of diode
 c
10A
t
I
I
40A
t
I
10A
T 25ms
10AT 25ms
I
10A
T 25ms
T 25ms
10A

14
25. If an ideal transformer has an inductive load element at port 2 as shown in the figure below, the
equivalent inductance at port 1 is
`
(A) nL (B) n2
L (C)
n
L
(D)
2
n
L
Key: (B)
Exp: The property of an ideal transformer is of port2 is terminated by an impedance LZ then the
impedance seen from port is
 
L
2
Z L
Z
N / N
In the given problem
 
2
in 2
L
Z n L
1/n
 
Q. No. 26 – 50 Carry Two Mark Each
26. Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were
the permitted pen colours that the candidate could bring. The probability that a candidate comes
with all 3 pens having the same colour is ______.
Key: 0.06
Exp: Total possible options = 43
Favorable choices {BBB, BlBlBl, GGG, RRR} = 4
Probability = 3
4 1
0.06
4 16
 
27. Let n
n 0
S n where 1.


    The value of  in the range 0 1   . Such that S 2  is ______.
Key: 0.293
Exp: n
n = 0
S = n α


2 3
2α α 2α + 3α ]      
2
2α α[1 2α 3α ]       
2
2α α[1 α] if α 1
   
2 1
(1 α) 2 α 1 α 0.293
2

       
Port 1 Port 2
n :1
L

15
28. Let the eigenvalues of a 2 x 2 matrix A be 1, -2 with eigenvectors 1 2x and x respectively. Then
the eigenvalues and eigenvectors of the matrix 2
A 3A 4l  would, respectively, be
(A) 1 22,14; x ,x (B) 1 2 1 22,14; x x ,x x 
(C) 1 22,0; x ,x (D) 1 2 1 22,0; x x , x x 
Key: (A)
Exp: Matrix A
Eigen values 1 , -2
Matrix 2
A -3A+4I
2 2
1 3(1) 4,( 2) 3( 2) 4     
Eigen values 2 , 14 respectively
0 1 2 2A & P(A) = a I + a A + a A have same eigen vectors
29. Let A be a 4 3 real matrix with rank 2. Which one of the following statement is TRUE?
(A) Rank of AT
A is less than 2
(B) Rank of AT
A is equal 2
(C) Rank of AT
A is greater than 2
(D) Rank of AT
A can be any number between 1 and 3
Key: (B)
Exp: Given that  4 3ρ A 2 
From the properties of Rank; we have  T
ρ AA = ρ(A)
T T
Rank of AA = Rank of A A = Rank of A 2 
30. Consider the following asymptotic Bode magnitude plot  is in rad s .
Which one of the following transfer function is best represented by the above Bode magnitude plot?
(A)
  
2
2s
1 0.5s 1 0.25s 
(B)
 
 
4 1 0.5s
s 1 0.25s


(C)
  
2s
1 2s 1 4s 
(D)
  
2
4s
1 2s 1 4s 
 
magnitude
dB
0.dB
20dB dec
12dB
0.5 8

16
Key: (A)
Exp: By looking to the plot we can say that since the initial slope is +20 there must be a zero on the
origin
If we find 2 we can get the
answer by eliminating options
2 1
2 1
M - M
Slope =
log - log 
2
0 12
40
log 8 - log

  

2
12
log 8- log
40
 
2
12
log = log 8-
40
  2 =4 
So one of the corner frequency is 2 = 4s at this frequency 2 poles should exist because the
change in slope is -40db
From this we can say option A satisfies the condition
(i) A zero at origin
(ii) one of corner frequency 4H term will be
s
1+
4
 
 
 
having 2 poles
31. Consider the following state-space representation of a linear time-invariant system
         T1 0 1 1
x t x t , y t c x t , c and x 0
0 2 1 1
     
        
     

The value of y(t) for et log 2 is _________.
Key: 6
32. Loop transfer function of a feedback system is    
 2
s 3
G s H s
s s 3



.Take the Nyquist contour in
the clockwise direction. Then the Nyquist plot of G(s) H(s) encircles – 1 + j0
(A) Once in clockwise direction (B) Twice in clockwise direction
(C) Once in anticlockwise direction (D) Twice in anticlockwise direction
Key: (A)
Exp:
S+3
GH =
S(S-3)
2 1/2
2 2 1/2 2
( +a) 1
GH
( +a)
 

  
1 0 0 1 1
GH tan 180 180 tan 2tan
3 3 3
     
          
  
1 2
 
magnitude
dB
0.dB
20dB dec
12dB
0.5 8

17
1
2
1
GH = 2tan
3




at 0, GH = 0 
at
0
, GH = 0180 
at 01
3, GH = 90
9

So the plot start at 0
0 and goes to 0
180 through 0
90 Since there are 2 poles on origin we will get
2  radius semicircle those will start where the mirror image ends and will terminate where the
actual plot started in clockwise direction. So the plot will be
So the Nyquist plot of G(s) H(s)
encircles – 1 + j0
once in clockwise direction
33. Given the following polynomial equation 3 2
s 5.5s 8.5s 3 0    , the number of roots of the
polynomial, which have real parts strictly less than – 1, is _______
Key: 2
Exp: The polynomial is 3 2
S +5.5S +8.5S+3=0, since we are interested to see the roots wrt S = -1 so in
the above equation replace S by z-1 then the equation is
3 2
(Z-1) +5.5(Z-1) +8.5(Z-1)+3=0
3 2 2
Z -3Z +3Z-1+5.5(Z +1-2Z)+8.5Z-8.5+3=0
3 2
Z +Z ( 3 5.5)+Z(3+8.5-11)+(-1+5.5-8.5+3)=0  
3 2
Z +2.5Z 0.52 1 0   
Using RH table
3
2
1
0
Z 1 0.5
Z 2.5 1
Z 0.9
Z 1


The single sign change in 1st
column indicate that out of 3 roots 1 root lie on the right half of S = -
1 plane if memory remaining 2 lies on left half of S = -1 plane.
w = 
-1+io
M= -1
w = 0

18
34. Suppose    1 2x t and x t have the Fourier transforms as shown below.
Which one of the following statements is TRUE?
(A)  1x t and  2x t are complex and    1 2x t x t is also complex with nonzero imaginary part
(B)        1 2 1 2x t and x t are real and x t x t is also real
(C)        1 2 1 2x t and x t are complex but x t x t is real
(D)        1 2 1 2x t and x t are imaginary but x t x t is real
Key: (C)
Exp: 1 2x (t) & x (t) are complex functions
2 1 2 1x ( ) = x (- ) , x (t) = x (-t) 
1 2x (t) x (t)will be real
35. The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where  x t is
the input signal. A signal z(t) is called eigen-signal of the system T, when     T z t z t ,  ,
where  is a complex number, in general, and is called an eigen value of T. suppose the impulse
response of the system T is real and even. Which of the following statements is TRUE?
(A) cos(t) is and eigen-signal but sin(t) is not
(B) cos(t) is and sin(t) are both eigen-signal but with different eigenvalues
(C) sin(t) is an eigen-signal but cos(t) is not
(D) cos(t) and sin(t) are both eigen-signal with identical eigenvalues
Key: (D)
Exp: Consider the Eigen signal Z(t) = Cos t
For Cost, y(t) cos(t-τ)h(τ)dτ


  (costcosτ +sin tsin τ)h(τ)dτ


 
=cost cosτh(τ)dτ sin t sin τh(τ)dτ
 
 
 
h(τ)is an even signal sinτh(τ)dτ=0



 1X 
1
0.3
0.5
1
0 1 2 
 2X 
0.5
1
2 1 0 1
0.3


19
y(t) =cost cosτh(τ)dτ


 
Thus the integration value will be an Eigen value γ .
Similarly consider the Eigen signal Z(t) = sin t
For sint, y(t) sin(t-τ)h(τ)dτ


  =sin t cosτh(τ)dτ cost sin τh(τ)dτ
 
 
 
Thus y(t)=sin t cosτh(τ)dτ



Eigen value ‘ γ ’is same for both the Eigen functions sint & cost
36. The current state QA QB of a two JK flip-flop system is 00. Assume that the clock rise-time is
much smaller than the delay of the JK flip-flop. The next state of the system is
(A) 00 (B) 01 (C) 11 (D) 10
Key: (C)
Exp:
A A
B B A
J = K =1
J = K =Q
It is given initially A BQ Q = 0
Since it is a synchronous counter, when clock is applied both flipflop will change there state
simultaneously based on JK FF state table
   A A A AJ =1, K =1 , Q =0 Q = 1
 
   B B B BJ =1, K =1 , Q =0 Q = 1

So next state A BQ Q is 11 
5V
J
K
CLK
VQ
AQ
J
K
OQ
Logic 1
AJ
AK BK
BJ
AQ BQ
AQ BQ

20
37. A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is IN0 V 3  Volts.
The expression for the LSB of the output 0B as a Boolean function of 2 1 0X ,X , and X is
(A) 0 2 1X X X   (B) 0 2 1X X X  
(C)  0 2 1X X X (D)  0 2 1X X X
Key: (A)
38. Two electric charges q and -2q are placed at (0,0) and (6,0) on the x-y plane. The equation of the
zero equipotential curve in the x-y plane is
(A) x 2  (B) y 2 (C) 2 2
x y 2  (D)  
2 2
x 2 y 16  
Key: (D)
Exp: The potential due to Q 4-2Q at (x, y) is
2 2 2 2
q 2q
k x + y k (x-6) + y
 
If potential at (x, y) = 0 where
1
k=
4πε
2 2 2 2
q 2q
0
k x + y k (x-6) + y
  2 2 2 2
(x-6) + y 2 x + y 0  
2 2 2 2
(x-6) + y 4(x + y )  2 2 2 2
x +36 -12x + y = 4x 4y 
2 2
x y - 4x-12 = 0 
Option:D
2 2
(x+2) + y 16 2 2
x +4x +4+ y = 16 2 2
x y + 4x-12 = 0 
3V
100
200
200
100


2X


1X


0X
Digital
circuit
1B
0B
INV

21
39. In the circuit shown, switch S2 has been closed for a long time. A time t = 0 switch S1 is closed At
t 0
 , the rate of change of current through the inductor, in amperes per second, is _______.
Key: 2
Exp: At -
t = 0 Network is in steady state with 1S opens 2S (closed) So we can say -
L
3
i (0 )= 1.5A
2

At +
t = 0 indicator behaves as ideal current source of 1.5A if we draw the network at +
t = 0 ,
both switch closed
Writing Nodes equation at LV (0 )
node
L
1 1 3 3
V (0 )= 1.5
1 2 1 2
  
    
 
LV (0 ) = 2

+
di(0 )
L 2
dt
 
+
Ldi (0 )
2A/sec
dt
 
40. A three-phase cable is supplying 800kW and 600kVAr to an inductive load, It is intended to
supply an additional resistive load of 100kW through the same cable without increasing the heat
dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the
load. Given the line voltage is 3.3kV, 50Hz, the capacitance per phase of the bank, expressed in
microfarads, is _________.
Key: 47 to 49
41. A 30MVA, 3-phase, 50Hz 13.8kV, star-connected synchronous generator has positive, negative
and zero sequence reactance, 15%,15% and 5% respectively. A reactance  nX is connected
between the neutral of the generator and ground. A double line to ground fault takes place
involving phases ‘b’ and ‘c’, with a fault impedance of j0.1 p.u. The value of  nX in p.u. that will
limit the positive sequence generator current to 4270 A is ________.
Key: 1.108
Exp: 1X 0.15 P.U
1S
1 2S
2
3V
3V 1H
+
at t = 0
1.5
+
LV (0 )
21
3V
3V

22
6
base 3
30 10
I 1255.109
3 13.8 10

 
 
2X 0.15P.U
0X 0.05P.U
fX 0.1P.U
   actual
base
I 4270
I P.U 3.4 P.U
I 1255.109
 
 a1
1 2 O f n
Ea
I
Z Z || Z 3z 3z

  
 2 0 f n
1
3.4
0.15 Z ||Z 3z 3z

  
 2 f nZ || 20 3z 3z 0.144  
2 f
1 1 1
Z 20 3z 3zln 0.144
 
 
f n20 3z 3z 3.675  
n3z 3.675 0.05 3 0.1   
nZ 1.108P.U
42. If the star side of the star-delta transformer shown in the figure is excited by a negative sequence
voltage, then
(A) VAB leads Vab by 60O
(B) VAB lags Vab by 60O
(C) VAB leads Vab by 30O
(D) VAB lags Vab by 30O
Key: (D)
43. A single-phase thyristor-bridge rectifier is fed from a 230V, 50Hz, single-phase, AC mains. If it is
delivering a constant DC current 10A, at firing angle of 30O
, then value of the power factor at AC
mains is
(A) 0.87 (B) 0.9 (C) 0.78 (D) 0.45
Key: (C)
Exp: IPF = 0.9cos 0.9 cos30 0.78  
A
C
B N
a
c
b

23
44. The switches T1 and T2 are shown in Figure (a).
They are switched in a complementary fashion with sinusoidal pulse width modulation technique.
The modulating voltage    m t 0.8sin 200 t   V and the triangular carrier voltage  c are as
shown in Figure (b). The carrier frequency is 5kHz. The peak value of the 100Hz component of
the load current  Li , in ampere, is ______.
Key: 10
Exp: m
a
V
m 0.8
Vcarrier
 
 0l
Vds
V max ma. 0.8 250 200V
2
   
2 2
2L 16 12 20   
 L
200
I max 10A
20
 
45. The voltage  s across and the current  gl through a semiconductor switch during a turn-ON
transition are shown in figure. The energy dissipated during the turn – ON transition, in mJ, is
_________.
0.8
C
1
t
mV
 b

dcV 2 250V
Li
T1
LX 16 at
100Hz
  R 12 
dcV 2 250V


T2
 a

sv
0
si
0
T1 1 s  T2 1 s 
100A
50A
t
t
600V

24
Key: 75
Exp:   61
E Pdt V t tdt 600 150 10
2
 
     
 
   + 61
100 600 10
2
 
  
 
75mJ
46. A single-phase 400V, 50Hz transformer has an iron loss of 5000W at the condition. When
operated at 200V, 25Hz, the iron loss is 2000W. When operated at 416V, 52Hz, the value of the
hysteresis loss divided by the eddy current loss is
Key: 1.4423
Exp: Since
V 400 200
8
f 50 2.5
 
  
 
is constant, Bm is constant
2
h e 1 2Pcore P P k f k f   
h 1P k f 2
1 2 1 25000 50k 50 . k 50k 2500k   
2
e 2P K f 2
1 1 2 1 22000 25k 25k 25 k 25k 625k    
1 2 2
1 2 1
k 50k 100 k 0.8
k 25k 80 k 60
  

  
∴ Pcore =   2
60f 0.8 f
hP 60f
2
eP 0.8f
When hf 52Hz, P 60 52 3120   
2
eP 0.8 52 2163.2  
h
e
P 3120
1.4423
P 2163.2
 
47. ADC shunt generator delivers 45 A at a terminal voltage of 220V. The armature and the shunt
field resistance are 0.01 and 44 respectively. The stray losses are 375W. The percentage
efficiency of the DC generator is ______.
Key: 86.84
Exp: outP 220 45 990W  
f
220
I 5A
44
 
a L fI I I 45 5 50A    
Armature losses = 2 2
a aI r 50 0.01 25W  
Field losses = 2 2
f fI R 5 44 1100W  
Stray losses = 375W
Total losses = 1500W
out
1
losses 1 P
 
  
 
fI
LI 45A

ar 0.01  220V

44

25
1
1500 1 9900 86.84%
 
    
 
48. A three-phase, 50Hz salient-pole synchronous motor has a per-phase direct-axis reactance  dX
of 0.8pu and a per-phase quadrature-axis reactance  qX of 0.6pu . Resistance of the machine is
negligible. It is drawing full-load current at 0.8pf (leading)j. When the terminal voltage is 1pu,
per-phase induced voltage, in pu, is ______.
Key: 1.608
Exp: dX 0.8P.U P.f 0.8 leading Ra 0  
qX 0.6PU Vt 1P.U
cos 0.8 36..86    
aI 136.86
a a a
t a a a
I qcos I R sin
tan
V I q sin I R cos
   
 
    
a
t
I qcos 1 0.6 0.8
0.3529
V Ia qsin 1 1 0.6 0.6
   
  
     
O
19.44 
36.86 19.44 56.3     
d aI I sin 1 sin56.4 0.832    
q aI I cos 1 0.5547    0.5547
q aEf Vcos Id d I R 0    dVcos I d 1.cos19.44 0.832 0.8 1.608pu      
FE 1.608PU
49. A single-phase, 22kVA, 2200V/220, 50Hz, distribution transformer is to be connected as an auto-
transformer to get an output voltage of 2420 V. Its maximum kVA rating as an auto-transformer is
(A) 22 (B) 24.2 (C) 242 (D) 2420
Key: (C)
Exp: Rated LV current =
22kVA
100A
220

max
2200 10
KVA 242kVA
1000

 
Or
2420 100
242kVA
1000

 
50. A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of
120O
duration is used to trigger the appropriate devices in each half-cycle. The rms value of the
fundamental component of the output voltage, in volts, is
(A) 234 (B) 245 (C) 300 (D) 331
Key: (A)
100A

220V
10A 



2420V

110A
2200V
2200VLOV

26
Exp:    O
n 1,3,s
4Vs n
V t cos sin n t
n 6


 
  
 

S
01
4V1
V cos 6 233.9 234V
2
 
     
51. A single-phase transmission line has two conductors each of 10mm radius. These are fixed a
center-to-center distance of 1m in a horizontal plane. This is now converted to a three-phase
transmission line by introducing a third conductor of the same radius. This conductor is fixed at an
equal distance D from the two single-phase conductors. The three-phase line is fully transposed.
The positive sequence inductance per phase of the three-phase system is to be 5% more than that
of the inductance per conductor of the single-phase system. The distance D, in meters, is _______.
Key: 1.439
Exp: For single phase
m
1
D
L 0.2 n
Ds

 
   
  s
1
0.2 n
D
 
  
 

For three phase
2
3
DM D
L3 0.2 n 0.2 n
DS Ds
 
    
 
 
 
3L 1.05L1  
 
2/3
S
D 1
0.2 1.05 0.2 n
D DS
   
   
  
 
D 1.439 mts 
52. In the circuit shown below, the supply voltage is 10sin  1000t volts. The peak value of the steady
state current through the 1 resistor, in amperes, is _________.
Key: 1
Exp: W =1000, the various impedance at this frequency are
4
240 F
1
4mH
5
500mH
2 F
~
 10sin 1000t
1m
D D
1m
2
3
mD 3 D.D.1 D 

27
3
250μf 4mH 6
-j
Z Z (j 1000 4 10 )
1000 250 10


 
      
( j4) (j4) open circuit    
3
24f 500mH 6
-j
Z Z (j 1000 500 10 )
1000 250 10


 
      
( j500) (j500) open circuit    
Since both LC pair parallel combination becomes
open then the circuit can be redrawn as
1Ω
10sin1000t
I sin1000t
4+1+5
  
So peak value of 1ΩI 1A
53. A dc voltage with ripple is given by  t    100 sin t 5sin 3 t       volts. Measurements of
this voltage  t , made by moving-coil and moving-iron voltmeters, show readings of 1V and 2V
respectively. The value of 2 1V V , in volts, is _______.
Key: 0.312
Exp: 1V 100V
2 2
2
2
10 5
V 100 100.312V
2 2
   
      
   
2 1V V 0.312V 
54. The circuit below is excited by a sinusoidal source. The value of R, in  , for which the
admittance of the circuit becomes a pure conductance at all frequencies is ________.
Key: 14.14
Exp: Admittance becomes pure conductance means the
imaginary part of Y must be zero which imply
resonance condition.
Let first get Y expression interms of L,C
then by equalising imaginary part we will
100 F R
0.02H R
~

C
R
R
L
4
51
10sin1000t
V

28
get the answer.
1 1
1R+j L R-
 Y
C


eq22 2
2
j
R+
R-j L C m [ ] 0
R +( L) 1
R +
C
    
 
 
 
g Y
 


22 2
2
1
L C
R +( L) 1
R +
C
 
 
 
 
 


 Cross multiplying
2
2 2 21 1 1
( L)R L R +( L)
   
    
   C C C
  
  
2 1 1 1
R L- L- 0
C C
    
         C
 
  
2 L 1
R L- 0
C
  
      C


Now by looking into above equation we can say that if
2 L
R
C
 then it will have no depending on frequency
for resonance 2 L
R
C
 
So 6
L 0.02
R 10 2 14.14
C 100 10
    

55. In the circuit shown below, the node voltage VA is _________V.
Key: 11.42
Exp: All the branch currents are expressed interval of AV now writing KCL at node A
A A 1 AV V 10I V 10
5 0
5 5 5
 
    
5 5A 
A 1I 5
5
110I


5
10V 


29
A 1
1 1 1
V 2I 5 1
5 5 10
 
      
 
A
A
V 102 1
V 2 6
5 10 10
  
      
   
A
2 1 2
V 6 2
5 10 10
 
     
 
A
7
V 8
10
 
  
 
A
80
V 11.42V
7
  
AV -10
10
A 1V 10I
5

AV
5
1I
10110I
5A 5
5
AV 5
5



EE gate-2016-set-1

Recomendados

Recomendados

Mais conteúdo relacionado

Mais procurados

Mais procurados (20)

Semelhante a EE gate-2016-set-1

Semelhante a EE gate-2016-set-1 (20)

Último

Último (20)

EE gate-2016-set-1