GATE ESE PSU’s 2019-20
EEE ENGINEERING
GATE EEE OBJ. PAPER SOL.(2011-2019)
GATE (2011-2019) EEE 19 SET PAPER SOLUTION
CONTENT COVERED:
1.GATE EEE PAPER 2011 SOLUTION
2.GATE EEE PAPER 2012 SOLUTION
3.GATE EEE PAPER 2013 SOLUTION
4.GATE EEE PAPER 2014 (1 SET) SOLUTION
5.GATE EEE PAPER 2015 (2 SET) SOLUTION
6. GATE EEE PAPER 2016
(2 SET)
SOLUTION
7. GATE EEE PAPER 2017
(2 SET)
SOLUTION
8. GATE EEE PAPER 2018
(1 SET)
(1 SET)
SOLUTION
Page
1
9. GATE EEE PAPER 2019
SOLUTION
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EE-Paper Code-A GATE 2011
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Q. No. 1 – 25 Carry One Mark Each
1.
Roots of the algebraic equation x3 + x2 + x + 1 = 0 are
(A) ( +1, + j, − j)
(B) ( +1, −1, +1)
(C) (0,0,0)
(D) ( −1, + j, − j)
Answer: - (D)
(
)
Exp: - x3 + x2 + x + 1 = 0 ⇒ x2 + 1 ( x + 1) = 0
2.
⇒ x + 1 = 0; x2 + 1 = 0
⇒ x = −1
x = ±j
With K as a constant, the possible solution for the first order differential equation
dy
= e−3x is
dx
(A) −
1 −3x
e +K
3
(B) −
1 3x
e +K
3
(C) −
1 −3x
e +K
3
(D) −3e− x + K
Answer: - (A)
Exp: -
dy
= e−3x ⇒ dy = e−3x dx
dx
Integrate on both sides
y=
3.
e−3x
1
+ K = − e−3x + K
−3
3
The r.m.s value of the current i(t) in the circuit shown below is
(A)
(B)
1
A
2
1
1F
1Ω
A
2
i ( t)
1Ω
~
+
(1.0 sin t ) V
(C) 1A
(D)
1H
2A
Answer: - (B)
Exp: - ω = 1 rad / sec
XL = 1Ω; XC = 1Ω
1Ω
+
1 Sin t
−
I ( t) =
Sin t
1
= Sin t; Irms =
A
1Ω
2
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∞
4.
The fourier series expansion f ( t ) = a0 + ∑ an cos nωt + bn sin nωt of the periodic
n =1
signal shown below will contain the following nonzero terms
( A ) a0 and bn , n = 1,3,5,...∞
(B ) a0 and an , n = 1,2,3,...∞
( C ) a0 an and bn , n = 1,2, 3,...∞
(D ) a0 and an n = 1,3,5,...∞
f (t )
t
0
Answer: - (D)
Exp: - ⇒ it satisfies the half wave symmetry, so that it contains only odd harmonics.
⇒ It satisfies the even symmetry. So bn = 0
5.
A 4 – point starter is used to start and control the speed of a
(A) dc shunt motor with armature resistance control
(B) dc shunt motor with field weakening control
(C) dc series motor
(D) dc compound motor
Answer: - (A)
6.
A three-phase, salient pole synchronous motor is connected to an infinite bus. Ig
is operated at no load a normal excitation. The field excitation of the motor is
first reduced to zero and then increased in reverse direction gradually. Then the
armature current
(A) Increases continuously
(B) First increases and then decreases steeply
(C) First decreases and then increases steeply
(D) Remains constant
Answer: - (B)
7.
A nuclear power station of 500 MW capacity is located at 300 km away from a
load center. Select the most suitable power evacuation transmission configuration
among the following options
(A)
~
Load center
132kV, 300km double circuit
(B)
~
Load center
132kv,300 km sin gle circuit with 40% series capacitor compensation
(C)
~
Load center
400kV, 300km sin gle circuit
~
(D)
Answer: - (A)
Load center
400kV, 300km double circuit
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The frequency response of a linear system G ( jω ) is provided in the tubular form
below
G ( jω )
1.3
1.2
1.0
0.8
0.5
0.3
∠ G ( jω )
−130O
−140O
−150O
−160O
−180O
−200O
(A)
6 dB and 30O
(B )
6 dB and − 30O
(C)
− 6 dB and 30O
(D )
− 6 dB and − 30O
Answer: - (A)
Exp: - At ∠G ( jw ) = −180 magnitude M=0.5
 1 
So G.M = 20 log 
 = 6dB
 0.5 
At G ( jw ) = 1
phase angle ∠G ( jw ) = −150
So P.M = 180 + ( −150 ) = 300
9.
The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r(t) having a
magnitude of 10 and a duration of one second, as shown in the figure is
r (t)
10
1s
(A) 0
(B) 0.1
t
(C) 1
(D) 10
Answer: - (A)
Exp: - For step input ess = 0.1 =
G (S ) =
1
⇒k =9
1+k
9
S +1
Now the input is pulse r ( t ) = 10 u ( t ) − u ( t − 1) 
1 − e −s 
r ( s ) = 10 

 s 
ess
10.
S 10 1 − e−s 
S.R ( s )
S
= Lt
= Lt
S →0 1 + G ( S ) H ( S )
S →0
S + 10
S +1
=
0
=0
10
Consider the following statement
(i) The compensating coil of a low power factor wattmeter compensates the
effect of the impedance of the current coil.
(ii) The compensating coil of a low power factor wattmeter compensates the
effect 0of the impedance of the voltage coil circuit.
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(A) (i) is true but (ii) is false
(B) (i) is false but (ii) is true
(C) both (i) and (ii) are true
(D) both (i) and (ii) are false
Answer: - (B)
11.
A low – pass filter with a cut-off frequency of 30Hz is cascaded with a high-pass
filter with a cut-off frequency of 20Hz. The resultant system of filters will function
as
(A) an all-pass filter
(B) an all-stop filter
(B) an band stop (band-reject) filter
(D) a band – pass filter
Answer: - (D)
Exp: -
20 30
So it is a band pass filter
12.
+12V
R
vi
−
R
+12V
−
+
−12V
VO
+
−12V
R
R
R
The CORRECT transfer characteristic is
Vo
+12v
+12V
(A)
VO
(B)
+6v
−6v
−6v
Vi
Vi
−12v
−12v
+12v
(C)
−6v
+12v
Vo
+6v
(D)
Vi
−6v
Vo
+6v
Vi
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Answer: - (D)
Exp: - It is a Schmitt trigger and phase shift is zero.
13.
A three-phase current source inverter used for the speed control of an induction
motor is to be realized using MOSFET switches as shown below. Switches S1 to S6
are identical switches.
Id
S1
A
S3
S2
B
S4
A
l.M.
S2
S6
B
The proper configuration for realizing switches S1 to S6 is
A
(A)
(B)
B
A
A
A
(C)
(D)
B
B
B
Answer: - (C)
14.
A point Z has been plotted in the complex plane, as shown in figure below.
Im unit circle
Zo
Re
The plot of the complex number y =
(A)
Im unit circle
y•
(C)
1
is
z
(B )
Re
Im unit circle
y•
Im unit circle
(D )
Re
Im unit circle
y•
Re
Re
y•
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Answer: - (D)
Exp: - Z < 1, so Y > 1
Z is having +ve real part and positive imaginary part (∴ from the characteristics)
So Y should have +ve real part and negative imaginary part.
15.
The voltage applied to a circuit is 100 2 cos (100πt ) volts and the circuit draws
a current of 10 2 sin (100πt + π / 4 ) amperes. Taking the voltage as the
reference phasor, the phasor representation of the current in amperes is
( A ) 10
2 ∠− π/4
(B ) 10
( C ) 10
∠+ π/4
(D ) 10
∠− π/4
2 ∠+ π/4
Answer: - (B)
Exp: - V ( t ) = 100 2 cos (100πt )
π π π
π


i ( t ) = 10 2 sin  100πt + + −  = 10 2 cos 100πt − 
4
2
2
4




So I =
16.
10 2
2
π
π
= 10∠ −
4
4
∠−
In the circuit given below, the value of R required for the transfer of maximum
power to the load having a resistance of 3Ω is
R
10 V
(A)
+
6Ω
(B ) 3 Ω
zero
3Ω
(C)
Load
6Ω
(D ) inf inity
Answer: - (A)
Exp: -
17.
R = 0 : Pmax =
102
3
(∴ RL
= cons tan t )
Given two continuous time signals x ( t ) = e −t and y ( t ) = e−2t which exist for t > 0,
the convolution z(t) = x(t)* y(t) is
(A)
e − t − e −2t
(B )
e−3t
(C)
e+ t
(D )
e− t + e −2t
Answer: - (A)
Exp: - z ( t ) = x ( t ) ∗ y ( t )
z ( s ) = x ( s ) .y ( s ) =
1
1
1
1
.
=
−
s
+
1
s
+
2
s
+
1
s
(
) (
) (
) ( + 2)
L−1 {z ( s )} = z ( t ) = e − t − e −2t
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A single phase air core transformer, fed from a rated sinusoidal supply, is
operating at no load. The steady state magnetizing current drawn by the
transformer from the supply will have the waveform
(A)
(B )
i
i
t
t
(C)
(D )
i
i
t
t
Answer: - (C)
Exp: - It is an air core transformer. So, there is no saturation effect.
19.
A negative sequence relay is commonly used to protect
(A) an alternator
(B) an transformer
(C) a transmission line
(D) a bus bar
Answer: - (A)
20.
For enhancing the power transmission in along EHV transmission line, the most
preferred method is to connect a
(A) Series inductive compensator in the line
(B) Shunt inductive compensator at the receiving end
(C) Series capacitive compensator in the line
(D) Shunt capacitive compensator at the sending end
Answer: - (C)
1
Exp: - P α
Where, X = ( XL − Xc )
X
21.
An open loop system represented by the transfer function G(s) =
( s − 1)
( s + 2 ) ( s + 3)
is
(A) Stable and of the minimum phase type
(B) Stable and of the non - minimum phase type
(C) Unstable and of the minimum phase type
(D) Unstable and of non-minimum phase type
Answer: - (B)
Exp: - Open loop system stability is depends only on pole locations ⇒ system is stable
There is one zero on right half of s-plane so system is non – minimum phase
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The bridge circuit shown in the figure below is used for the measurement of an
unknown element ZX. The bridge circuit is best suited when ZX is a
VS ~
+
R2
C1
D
R4
ZX
(A) low resistance
(C) high resistance
(A) low Q inductor
(B) lossy capacitor
Answer: - (C)
23.
A dual trace oscilloscope is set to operate in the ALTernate mode. The control
input of the multiplexer used in the y-circuit is fed with a signal having a
frequency equal to
(A) the highest frequency that the multiplexer can operate properly
(B) twice the frequency of the time base (sweep) oscillator
(C) the frequency of the time base (sweep) oscillator
(D) haif the frequency of the time base (sweep) oscillator
Answer: - (C)
24.
The output Y of the logic circuit given below is
X
(A) 1
(B) 0
Y
(C) X
(D) X
Answer: - (A)
Exp: - y = x.x + x.x = x + x = 1
Q. No. 26 – 51 Carry Two Marks Each
25.
Circuit turn-off time of an SCR is defined as the time
(A) taken by the SCR turn of
(B) required for the SCR current to become zero
(C) for which the SCR is reverse biased by the commutation circuit
(D) for which the SCR is reverse biased to reduce its current below the holding
current
Answer: - (C)
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Solution of the variables x1 and x2 for the following equations is to be obtained by
employing the Newton-Raphson iterative method.
equation (i)
10x2 sin x1 – 0.8 = 0
equation (ii)
10x22 -10x2 Cos x1 – 0.6 = 0
Assuming the initial valued x1 = 0.0 and x2 = 1.0, the jacobian matrix is
10
(A) 
0
0
(C) 
10
− 0.8

− 0.6 
− 0.8 

− 0.6 
10
(B) 
0
0 

10
10
(D) 
10
0 

− 10 
Answer: - (B)
Exp: - 10x2 sin x1 − 0.5 = 0 ….(i)
10x22 − 10x2 cos x1 − 0.6 = 0
 ∂ (i )

∂x1
J= 
 ∂ (ii)

 ∂x1
27.
∂ (i ) 

∂x2 
at x1 = 0 and x2 = 1
∂ (ii) 

∂x2 
…(ii)
10 0 
J= 

 0 10
The function f(x) = 2x-x2 – x3+3 has
(A) a maxima at x = 1 and minimum at x = 5
(B) a maxima at x = 1 and minimum at x = -5
(C) only maxima at x = 1 and
(D) only a minimum at x = 5
Answer: - (C)
Exp: - f ( x ) = 2x − x2 + 3
f ' ( x ) = 0 ⇒ 2 − 2x = 0 ⇒ x = 1
f '' ( x ) = −2 ⇒ f '' ( x ) < 0
So the equation f(x) having only maxima at x =1
28.
A lossy capacitor Cx, rated for operation at 5 kV, 50 Hz is represented by an
equivalent circuit with an ideal capacitor Cp in parallel with a resistor RP. The
value Cp is found to be 0.102 µ F and the value of Rp = 1.25 M Ω . Then the power
loss and tan ∂ of the lossy capacitor operating at the rated voltage, respectively,
are
(A) 10 W and 0.0002
(B) 10 W and 0.0025
(C) 20 W and 0.025
(D) 20 W and 0.04
Answer: - (C)
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Let the Laplace transform of a function F(t) which exists for t > 0 be F1(s) and
the Laplace transform of its delayed version f(1- τ ) be F2(s). Let F1*(s) be the
complex conjugate of F1(s) with the Laplace variable set as s= σ + jw . If G(s) =
F2 (s).F1 * (s)
, then the inverse Laplace transform of G(s) is
2
F1 (s)
(A) An ideal impulse δ ( t )
(B) An ideal delayed impulse δ ( t − τ )
(C) An ideal step function u ( t )
(D) An ideal delayed step function u ( t − τ )
Answer: - (B)
Exp: - F2 ( t ) = L {f ( t − τ )} = e−STF1 ( S )
G (S) =
e− sτF1 ( s ) .F1* ( s )
F1 ( s )
2
= e− sτ
G ( t ) = L−1 {G ( S )} = δ ( t − τ )
30.
A zero mean random signal is uniformly distributed between limits –a and +a
and its mean square value is equal to its variance. Then the r.m.s value of the
signal is
(A)
a
(B)
3
a
2
(C) a 2
(D) a 3
Answer: - (A)
( a − ( − a) )
2
Exp: - Variance =
31.
12
=
4a2
a2
=
; R.M.S value =
12
3
var iance =
a
3
A 220 V, DC shunt motor is operating at a speed of 1440 rpm. The armature
resistance is 1.0 Ω and armature current is 10A. of the excitation of the machine
is reduced by 10%, the extra resistance to be put in the armature circuit to
maintain the same speed and torque will be
(A) 1.79 Ω
(B) 2.1 Ω
(C) 18.9 Ω
(D) 3.1 Ω
Answer: - (A)
Exp: - Ia1 = 10
Now flux is decreased by 10%, so φ2 = 0.9φ1
Torque is constant so Ia1φ1 = Ia2 φ2 ⇒ Ia2 =
Nα
1=
Eb
φ
⇒
10
= 11.11A
0.9
E
220 − Ia1r1
N1
φ
0.9φ1
= b1 × 2 =
×
N2 Eb2 φ1 220 − Ia2 (r1 + R )
φ1
210 × 0.9
⇒ 1 + R = 2.79 ⇒ R = 1.79Ω
220 − 11.11 (1 + R )
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A load center of 120MW derives power from two power stations connected by
220kV transmission lines of 25km and 75km as shown in the figure below. The
three generators G1,G2 and G3 are of 100MW capacity each and have identical
fuel cost characteristics. The minimum loss generation schedule for supplying the
120 MW load is
~
25 km
~
75 km
~
75 km
P1 = 80MW + losses
(A) P2 = 20MW
P1 = 60MW
(B) P2 = 30MW + losses
P3 = 20MW
P3 = 30MW
P1 = 40MW
(C) P2 = 40MW
P1 = 30MW + losses
(D) P2 = 45MW
P3 = 40MW + losses
P3 = 45MW
Answer: - (A)
Exp: -
Loss α p2 ; Loss α length
For checking all options only option A gives less losses.
33.
The open loop transfer function G(s) of a unity feedback control system is given
as
2

k s + 
3
G ( s ) = 2
s (s + 2)
From the root locus, it can be inferred that when k tends to positive infinity,
(A) Three roots with nearly equal real parts exist on the left half of the s-plane
(B) One real root is found on the right half of the s-plane
(C) The root loci cross the jω axis for a finite value of k ; k ≠ 0
(D) Three real roots are found on the right half of the s-plane
Answer: - (A)
 2
−2 −  − 
 3  = −6 + 2 = − 4 = − 2
Exp: - Centroid σ =
3 −1
6
6
3
Asymptotes =
(29 ± 1) 180
θ1 =
180
= 90
2
θ2 =
18 × 3
= 2700
2
p−z
x
O
x
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A portion of the main program to call a subroutine SUB in an 8085 environment
is given below.
:
:
LXI D,DISP
LP :
CALL SUB
:
It is desired that control be returned to LP+DISP+3 when the RET instruction is
executed in the subroutine. The set of instructions that precede the RET
instruction in the subroutine are
POP H
XTHL
DAD
D
POP D
POP H
INX D
INX H
(A) DAD H
(B)
(C) DAD D
(D) INX D
INX H
PUSH D
PUSH
H
INX D
INX H
XTHL
PUSH H
Answer: - (C)
35.
The transistor used in the circuit shown below has a β of 30 and ICBO is negligible
2.2k
15k
1k
D
VBE = 0.7V
VCE(sat ) = 0.2V
Vz = 5V
−12V
If the forward voltage drop of diode is 0.7V, then the current through collector
will be
(A) 168 mA
(B) 108 mA
Answer: - (D)
Exp: - Transistor is in Saturation region
IC =
36.
(C) 20.54mA
(D) 5.36 mA
12 − 0.2
= 5.36 mA
2.2K
A voltage commutated chopper circuit, operated at 500Hz, is shown below.
M
+
0.1µF
A
iM
iL = 10A
iA
LOAD
200 V
1 mH
−
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If the maximum value of load current is 10A, then the maximum current through
the main (M) and auxiliary (A) thyristors will be
(A) iM max = 12 A and iA max = 10 A
(B) iM max = 12 A and iA max = 2 A
(C) iM max = 10 A and iA max = 12 A
(D) iM max = 10 A and iA max = 8 A
Answer: - (A)
iM max = Io + ICpeak = Io + VS
Exp: -
C
0.1µ
= 10 + 200
= 12A
L
1m
iA max = Io = 10A
37.
2 1 
The matrix A  = 
 is decomposed into a product of a lower triangular
 4 −1
matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and
[U] matrices respectively are
1 0 
1 1 
(A) 
 and 

 4 −1
0 − 2 
2 0 
1 1
(B) 
 and 

 4 −1
0 1
 1 0
2 1 
2 0 
1 1.5
(C) 
 and 
 (D) 
 and 

 4 1
0 −1
 4 −3
0 1 
Answer: - (D)
Exp: - A  = L 
U ⇒ Option D is correct
38.
 1
3
The two vectors [1,1,1] and 1, a, a2  , where a =  − + j
 , are
 2
2 

(A) Orthonormal
(B) Orthogonal
(C) Parallel
(D) Collinear
Answer: - (B)
Exp: - Dot product of two vectors = 1 + a + a2 = 0
So orthogonal
39.
Exp:
A three –phase 440V, 6 pole, 50Hz, squirrel cage induction motor is running at a
slip of 5%. The speed of stator magnetic field to rotor magnetic field and speed
of rotor with respect to stator magnetic field are
(A) zero, - 5 rpm
(B) zero, 955 rpm
(C) 1000rpm, -5rpm
(D) 1000rpm, 955rpm
NS =
120 × f 120 × 50
=
= 1000 rpm ; Rotor speed = NS − SNS = 950 r.p.m
P
6
Stator magnetic field speed = NS = 1000r.p.m
Rotor magnetic field speed = NS = 1000r.p.m
Relative speed between stator and rotor magnetic fields is zero
Rotor speed with respect to stator magnetic field is = 950 − 1000 = −50 r.p.m
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40.
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A capacitor is made with a polymeric dielectric having an εr of 2.26 and a
dielectric breakdown strength of 50kV/cm. The permittivity of free space is
8.85pF/m. If the rectangular plates of the capacitor have a width of 20cm and a
length of 40cm, then the maximum electric charge in the capacitor is
(A) 2µC
(B) 4µC
(C) 8µC
(D) 10µC
Answer: - (C)
Exp:- q = CV =
41.
ε.A
V
× V = ε.A   = εr ε0 A × E = 2.26 × 8.85 × 10−14 × 50 × 103 × 20 × 40 = 8µc
d
 d
The response h(t) of a linear time invariant system to an impulse δ ( t ) , under
initially relaxed condition is h ( t ) = e− t + e−2t . The response of this system for a
unit step input u(t) is
(
)
(
)
(A) u ( t ) + e− t + e−2t (B) e− t + e−2t u ( t ) (C) 1.5 − e− t − 0.5e−2t u ( t )
(D) e−1 δ ( t ) + e−2tu ( t )
Answer: - (C)
Exp: - L (Impulse response) = T.F =
1
1
+
( S + 1) ( S + 2 )



1
1
1   0.5
0.5  
−1  1
+
+
−
Step response = L−1 

 = L  −


(S + 2)  
 S S + 1   S
 S ( S + 1) S ( S + 2 ) 
(
= (1.5 − e
)
= 1 − e− t + 0.5 − 0.5e−2t u ( t )
42.
−t
)
− 0.5e−2t u ( t )
The direct axis and quadrature axis reactance’s of a salient pole alternator are
1.2p.u and 1.0p.u respectively. The armature resistance is negligible. If this
alternator is delivering rated kVA at upf and at rated voltage then its power angle
is
(A) 300
(B) 450
(C) 600
(D) 900
Answer: - (B)
Exp: - Tan δ =
Ia ( x q cos θ + ra sin θ )
Vt + Ia ( x q sin θ − ra cos θ )
Ia = 1 p.u; Vt = 1 p.u
θ = Power factor angle = 00
x d = 1.2p.u; x q = 1.p.u ;ra = 0
Tan δ = 1 ⇒ δ = 450
43.
A 4 1 2 digit DMM has the error specification as : 0.2% of reading + 10 counts. If
a dc voltage of 100V is read on its 200V full scale, the maximum error that can
be expected in the reading is
(A) ±0.1%
(B) ±0.2%
(C) ±0.3%
(D) ±0.4%
Answer: - (C)
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44.
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A three – bus network is shown in the figure below indicating the p.u.
impedances of each element.
− j0.08
j 0.2
j0.1
3
2
1
j0.1
The bus admittance matrix, Y – bus, of the network is
0 
 0.3 −0.2

(A) j  −0.2 0.12 0.08 
 0
0.08 0.02 
5
0 
 −15


(B) j  5
7.5
−12.5
 0
−12.5 2.5 
0 
0.1 0.2

(C) j 0.2 0.12 −0.08 
 0 −0.08 0.10 
5
0 
 −10


(D) j  5
7.5 12.5
 0 12.5 −10 
Answer: - (B)
EXP:Y11 =
45.
1
1
+
= − j15
j0.1 j0.2
A two loop position control system is shown below
R ( s)
+
-
+
-
1
s ( s+1)
Y (s)
ks
The gain k of the Tacho-generator influences mainly the
(A) Peak overshoot
(B) Natural frequency of oscillation
(C) Phase shift of the closed loop transfer function at very low frequencies
( ω → 0)
(D) Phase shift of the closed loop transfer function at very low frequencies
(ω → ∞)
Answer: - (A)
EXP:Y (S)
R ( S)
=
1
S + (k + 1) S + 1
2
−πξ / (1−ξ2 )
k + 1
2ξwn = k + 1 ⇒ ξ = 
;
Peak
over
shoot
=
e

 2 
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46.
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A two – bit counter circuit is shown below
Q
J
>
k
T
>
QB
Q
Q
Q
CLK
It the state QA QB of the counter at the clock time tn is ‘10’ then the state QA QB
of the counter at tn + 3 (after three clock cycles ) will be
(A) 00
Answer: - (C)
EXP:-
(B) 01
(C) 10
Clock
Input
JA =QB
K A =QB
Output
TB= QA
Initial state
47.
(D) 11
QA
QB
1
0
1
1
0
1
1
1
2
0
1
1
0
0
3
1
0
0
1
0
A clipper circuit is shown below.
1k
Vi
D
vz = 10V
~
VO
5V
Assuming forward voltage drops of the diodes to be 0.7V, the input-output
transfer characteristics of the circuit is
10
Vo
Vo
(A)
4.3
4.3
(B)
4.3
4.3
Vi
10
Vi
10
(C)
Vo
Vo
(D)
5.7
−5.7
10
Vi
−0.7 5.7
Vi
−5.7
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Answer: - (C)
Exp:
When -0.7V<Vi < 5.7V outputwillfollowinput,because zener diode andnormal diodes are off
When Vi ≤ −0.7V Zener diode forwardbias and V0 = −0.7 V
When Vi ≥ 5.7V Diode is forwardbias and V0 = 5.7 V
Common Data Questions: 48 & 49
The input voltage given to a converter is
vi = 100 2 sin (100πt ) V
The current drawn by the converter is
ii = 10 2 sin (100πt − π / 3 ) + 5 2 sin ( 300πt + π / 4 ) + 2 2 sin (500πt − π / 6 ) A
48.
The input power factor of the converter is
(A)
(B )
0.31
(C)
0.44
0.5
(D )
0.71
(D )
887 W
Answer: - (C)
Exp: - Input power factor is depends on fundamental components
π

V ( t ) = 100 2 sin (100πt ) V; I ( t ) = 10 2 sin  100πt − 
3


cos φ = cos
49.
π
= 0.5
3
The active power drawn by the converter is
( A ) 181 W
(B ) 500 W
(C)
707 W
Answer: - (B)
Exp: - P = V1(r.m.s)I1(r.m.s) cos θ1 + V3,rms I3,rms cos ( θ3 ) +V5,rms I5,rms cos ( θ5 )
π
P = V1,rms I1,rms cos ( θ1 ) = 100 × 10 cos   = 500 Watts
3
V3,rms = 0 ; V5,rms = 0;
Common Data Questions: 50 & 51
An RLC circuit with relevant data is given below.
IS
VS ~
IRL
R
L
IC
VS = 1 ∠ 0V
C (D ) 94 A
IRL =
2 ∠ − π/4A
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50.
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The power dissipated in the resistor R is
(A)
(B ) 1 W
0.5 W
(C)
(D ) 2 W
2W
Answer: - (B)
Exp: - Total power delivered by the source = power dissipated in ‘R’
 π
P = VI cos θ = 1 × 2 cos   = 1W
 4
51.
The current Ic in the figure above is
(D )
(B )
− j2 A
−j
1
2
(C)
A
+j
1
2
(D )
A
+ j2A
Answer: - (D)
Exp: - IC = IS − IRL = 2∠
π
π
− 2∠ −
= j2A
4
4
Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each
Statement for Linked Answer Questions: 52 & 53
Two generator units G1 and G2 are connected by 15 kV line with a bus at the
mid-point as shown below.
1
~
G1 15kV
L1
2
3
L2
10km
10km
~
15kV
G1 = 250 MVA, 15kV, positive sequence reactance X = 25% on its own base
G2 = 100 MVA, 15kV, positive sequence reactance X = 10 % on its own base
L1 and L2 = 10km, positive sequence reactance X = 0.225 Ω / km
(A)
j0.10
1
j1.0
3
j1.0
~
(B)
j0.25
~
1
j1.0
3
j1.0
j0.10
~
1
~
j0.10
2
~
(C)
j0.10
2
j2.25
3
j2.25 2
j0.10
~
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(D) j0.25
j2.25
1
3
j2.25 2
~
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j0.10
~
Answer: - (A)
Exp: - XL1 = 0.225 × 10 = 2.25Ω
XL2 = 0.225 × 10 = 2.25Ω
Take 100MVA s 15 kV as base
For generator (G1)
X g1 = 0.25 ×
100
= 0.1p.u
25
For Transmission Line (L1 and L2)
ZBase =
(15)
2
100
XTL1 (p.u) =
= 2.25Ω
2.25
= 1p.u
2.25
XTL2 (p.u) = 1p.u
For generator (G2)
 100 
X g2 (p.u) = 0.1 × 
 = 0.1 p.u
 100 
53.
In the above system, the three-phase fault MVA at the bus 3 is
(A)
82.55 MVA
(B )
85.11MVA
( C ) 170.91MVA
(D ) 181.82 MVA
Answer: - (A)
Exp: - X Th1 = 1.1|| 1.1 = 0.55
Fault (MVA) =
Base MVA
100
=
= 181.82 MVA
fault Thevenin' s Impedance 0.55
Statement for Linked Answer Questions: 54 & 55
A solar energy installation utilize a three – phase bridge converter to feed energy
into power system through a transformer of 400V/400 V, as shown below.
Filter Choke
Battery
The energy is collected in a bank of 400 V battery and is connected to converter
through a large filter choke of resistance 10Ω .
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54.
The maximum current through the battery will be
( A ) 14 A
(B ) 40 A
( C ) 80 A
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(D )
94 A
Answer: - (A)
Exp: - V0 = Iara + E
(V0 )max =
3 6Vphase
(∵ cos α = 1)
π
= Iara + E
540.2 = Ia (10 ) + 400
Ia = 14A
55.
The kVA rating of the input transformer is
( A ) 53.2 kVA
(B ) 46.0 kVA
( C ) 22.6 kVA
3
Answer: Exp: - KVA rating = 3VL IL = 3VL
6
6
Io = 3 × 400 ×
× 14 = 7562VA
π
π
Q. No. 56 – 60 Carry One Mark Each
56.
There are two candidates P and Q in an election. During the campaign, 40% of
the voters promised to vote for P, and rest for Q. However, on the day of election
15% of the voters went back on their promise to vote for P and instead voted for
Q. 25% of the voters went back on their promise to vote for Q and instead voted
for P. Suppose, P lost by 2 votes, then what was the total number of voters?
(A) 100
Answer: - (A)
Exp: - P
40%
−6%
+15%
49%
(B) 110
(C) 90
(D) 95
Q
60%
+ 6%
− 15%
51%
∴ 2% = 2
100% = 100
57.
Choose the most appropriate word from the options given below to complete the
following sentence:
It was her view that the country's problems had been_________ by
foreign technocrats, so that to invite them to come back would be
counter-productive.
(A) Identified
(B) ascertained
(C) Texacerbated
(D) Analysed
Answer: - (C)
Exp: -The clues in the question are ---foreign technocrats did something negatively to
the problems – so it is counter-productive to invite them. All other options are
non-negative. The best choice is exacerbated which means aggravated or
worsened.
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58.
Choose the word from the options given below that is most nearly opposite in
meaning to the given word:
Frequency
(A) periodicity
(B) rarity
(C) gradualness
(D) persistency
Answer: - (B)
Exp: - The best antonym here is rarity which means shortage or scarcity.
59.
Choose the most appropriate word from the options given below to complete the
following sentence: Under ethical guidelines recently adopted by the
Indian Medical Association, human genes are to be manipulated only to
correct
diseases
for
which______________
treatments
are
unsatisfactory.
(A) Similar
(B) Most
(C) Uncommon
(D) Available
Answer: - (D)
Exp: - The context seeks to take a deviation only when the existing/present/current/
alternative treatments are unsatisfactory. So the word for the blank should be a
close synonym of existing/present/current/alternative. Available is the closest of
all.
60.
The question below consists of a pair of related words followed by four pairs of
words. Select the pair that best expresses the relation in the original pair:
Gladiator : Arena
(A) dancer : stage
(B) commuter: train
(D) lawyer : courtroom
(C) teacher : classroom
Answer: - (D)
Exp: - The given relationship is worker: workplace. A gladiator is (i) a person, usually a
professional combatant trained to entertain the public by engaging in mortal
combat with another person or a wild.(ii) A person engaged in a controversy or
debate, especially in public.
Q. No. 61 – 65 Carry Two Marks Each
61
The fuel consumed by a motorcycle during a journey while traveling at various
speeds is indicated in the graph below.
Fuel consumption
(kilometers per litre)
120
90
60
30
0
0
15
30
45
60
Speed
(kilometers per hour)
75
90
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The distances covered during four laps of the journey are listed in the table below
Lap
Distance (kilometers)
Average speed
(kilometers per hour)
P
15
15
Q
75
45
R
40
75
S
10
10
From the given data, we can conclude that the fuel consumed per kilometre was
least during the lap
(A) P
(B) Q
(C) R
(D) S
Answer: - (A)
Fuel consumption
Exp: -
62.
P
60 km / l
Q
90 km / l
R
75 km / l
S
30 km / l
Actual
15 1
= l
60 4
75 5
= l
90 6
40
8
=
l
75 15
10 1
= l
30 3
Three friends, R, S and T shared toffee from a bowl. R took 1/3rd of the toffees,
but returned four to the bowl. S took 1/4th of what was left but returned three
toffees to the bowl. T took half of the remainder but returned two back into the
bowl. If the bowl had 17 toffees left, how many toffees-were originally there in
the bowl?
(A) 38
(B) 31
(C) 48
(D) 41
Answer: - (C)
Exp: - Let the total number of toffees is bowl e x
R took
∴
1
3
of toffees and returned 4 to the bowl
Number of toffees with
R =
1
x−4
3
Remaining of toffees in bowl =
1 2

x + 4 − 3
4  3

Number of toffees with S =
Remaining toffees in bowl =
Number of toffees with T =
2
x+ 4
3
3 2

x + 4 + 4
4  3


1  3 2

x + 4 + 4 + 2

2  4  3


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Remaining toffees in bowl =
Given,
63.
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
1 3  2

  x + 4 + 4  + 2
2 4  3



1 3  2
3 2


  x + 4  + 4  + 2 = 17 ⇒
 3 x + 4  = 27 ⇒ x = 48
2 4  3
4




Given that f(y) = | y | / y, and q is any non-zero real number, the value of
| f(q) - f(-q) | is
(A) 0
(B) -1
(C) 1
(D) 2
Answer: - (D)
Exp: - Given, f(y) =
f ( q) − f ( q) =
64.
y
y
q
q
⇒ f ( q) =
+
q
q
=
q
q
2q
q
; f ( −q ) =
−q
=
−q
−q
q
=2
The sum of n terms of the series 4+44+444+.... is
(A) ( 4 / 81) 10n+1 − 9n − 1
(B) ( 4 / 81) 10n−1 − 9n − 1
(C) ( 4 / 81) 10n+1 − 9n − 10 
(D) ( 4 / 81) 10n − 9n − 10
Answer: - (C)
Exp: - Let S=4 (1 + 11 + 111 + ……..) =
{
(
) (
4
(9 + 99 + 999 + .......)
9
}
)
4
(10 − 1) + 102 − 1 + 103 − 1 + .........
9

10n − 1
4
4 
4
=
10 + 102 + ......10n − n = 10
− n =
10n+1 − 9n − 10
9
9
9
81


=
{(
65.
) }
(
)
{
}
The horse has played a little known but very important role in the field of
medicine. Horses were injected with toxins of diseases until their blood built up
immunities. Then a serum was made from their blood. Serums to fight with
diphtheria and tetanus were developed this way.
It can be inferred from the passage that horses were
(A) given immunity to diseases
(B) generally quite immune to diseases
(C) given medicines to fight toxins
(D) given diphtheria and tetanus serums
Answer: - (B)
Exp: - From the passage it cannot be inferred that horses are given immunity as in (A),
since the aim is to develop medicine and in turn immunize humans. (B) is correct
since it is given that horses develop immunity after some time. Refer “until their
blood built up immunities”. Even (C) is invalid since medicine is not built till
immunity is developed in the horses. (D) is incorrect since specific examples are
cited to illustrate and this cannot capture the essence.
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!
" #$
%
"
%
"
( )=&
(ω
"
#
− φ% )
φ
−φ%
*+ #$ &
+
(ω
-&
( )=&
(ω
−σ
)=&
./ − ( ω −
)
(ω
=&
− φ )'
( π )% + φ )
%
" #$
(ω
(
( π )% − φ )
φ%
( )=&
%
− φ − ./,
)
− φ% )
"
−φ + φ% − ./ = − 0/,
− 0/, (
"
)
0/,
φ = ./ + φ%
1
2
(
$
" #$
-
*+ #$
3 −3
3
→
3
→
3
→ &
(
3
→ & "
4
"
"
566
()
/7
87
/
/
%
%/
−87
47
87
07
( )
/7
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Noted-: Single Source Follow, Revise
Multiple Time Best key of Success
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© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the
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|EE-GATE-2013 PAPER|
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Q. No. 1 – 25 Carry One Mark Each
1.
Given a vector field F = y 2 xax − yzay = x 2 az , the line integral
∫ F.dl evaluated along
a segment on the x-axis from x=1 to x=2 is
(A) -2.33
(B) 0
(C) 2.33
(D) 7
Answer: (B)
2.
2 − 2   x1   0 
The equation 
   =   has
1 − 1   x 2   0 
(A) no solution
 x  0 
(B) only one solution  1  =  
 x 2  0 
(C) non-zero unique solution
(D) multiple solutions
Answer: (D)
3.
Square roots of −i, where i =
− 1, are
(A) i, − i
 π
 π
 3π 
 3π 
(B) cos  −  + i sin  −  , cos 
+ i sin 


 4
 4
 4 
 4 
π
 3π 
 3π 
π
(C) cos   + i sin 
 , cos  4  + i sin  4 
4
 4 


 
 3π 
 3π 
 3π 
 3π 
(D) cos 
 + i sin  − 4  , cos  − 4  + i sin  4 
4








Answer: (B)
4.
Three moving iron type voltmeters are connected as shown below. Voltmeter
readings are V, V1 and V2 as indicated. The correct relation among the voltmeter
readings is
− j1Ω
V
(A) V =
V1
2
+
V2
2
V1
(B) V = V1 + V2
j2Ω
V2
I
(C) V = V1 V2
(D) V = V2 − V1
Answer: (B)
5.
Leakage flux in an induction motor is
(A) flux that leaks through the machine
(B) flux that links both stator and rotor windings
(C) flux that links none of the windings
(D) flux that links the stator winding or the rotor winding but not both
Answer: (D)
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6.
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The angle δ in the swing equation of a synchronous generator is the
(A) angle between stator voltage and current
(B) angular displacement of the rotor with respect to the stator
(C) angular displacement of the stator mmf with respect to a synchronously
rotating axis.
(D) angular displacement of an axis fixed to the rotor with respect to a
synchronously rotating axis.
Answer: (D)
7.
Consider a delta connection of resistors and its equivalent star connection as
shown below. If all elements of the delta connection are scaled by a factor k,
k>0, the elements of the corresponding star equivalent will be scaled by a factor
of
RC
Ra
Rb
(B) k
Answer
8.
(C)
1
k
(D) k
(B)
A band-limited signal with a maximum frequency of 5 kHz is to be sampled.
According to the sampling theorem, the sampling frequency in kHz which is not
valid is
(A) 5
(B) 12
Answer
9.
RA
Rc
(A) k 2
RB
For
(C) 15
(D) 20
(A)
a
periodic
signal
v(t) = 30 sin 100 t + 10 cos 300 t + 6 sin(500 t +
π
),
4
the
fundamental frequency in radians/s is
(A) 100
Answer
10.
(B) 300
(C) 500
(D) 1500
(A)
A bulb in a staircase has two switches, one switch being at the ground floor and
the other one at the first floor. The bulb can be turned ON and also can be
turned OFF by any one of the switches irrespective of the state of the other
switch. The logic of switching of the bulb resembles
(A) an AND gate
Answer
(B) an OR gate
(C) an XOR gate
(D)a NAND gate
(C)
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11.
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The Bode plot of a transfer function G(s) is shown in the figure below.
40
32
20
Gain(dB)
0
1
10
−8
100
ω(rad / s)
The gain (20 log G(s) ) is 32 dB and -8 dB at 1 radians/s and 10 radians/s
respectively. The phase is negative for all ω. Then G(s) is
(A)
39.8
s
Answer
12.
(B)
39.8
s2
(C)
32
s
(D)
32
s2
(B)
In the feedback network shown below, if the feedback factor k is increased, then
the
+
−
Vin
V1
Vf = kv out
+
−
A0
+
k
+
−
Vout
+
−
−
(A) input impedance increases and other output impedance decreases
(B) input impedance increases and output impedance also increases.
(C) input impedance decreases and output impedance also decreases.
(D) input impedance decreases and output impedance increases.
Answer: (A)
13.
The input impedance of the permanent magnet moving coil (PMMC) voltmeter is
infinite. Assuming that the diode shown in the figure below is ideal, the reading
of the voltmeter in Volts is
1 kΩ
−
14.14 sin(314 t)V
Voltmeter
100 kΩ
+
(A) 4.46
(B) 3.15
(C) 2.23
(D) 0
Answer: (A)
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14.
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The curl of the gradient of the scalar field defined by V = 2x 2 y + 3y 2 z + 4z2 x is
(A) 4xyax + 6yzay + 8zxaz
(B) 4ax + 6ay + 8az
(C)
( 4xy + 4z ) a
2
x
(
)
(
)
+ 2x 2 + 6yz ay + 3y 2 + 8zx az
(D) 0
Answer: (D)
15.
A continuous random variable X
f(x) = e − x , 0 < x < ∞. Then P{X > 1} is
(A) 0.368
(B) 0.5
has
a
probability
(C) 0.632
density
function
(D) 1.0
Answer: (A)
16.
The flux density at a point in space is given by B = 4xax + 2kyay + 8az Wb / m2 .
The value of constant k must be equal to
(A) -2
(B) -0.5
(C) +0.5
(D) +2
Answer: (A)
17.
A single-phase transformer has no-load loss of 64 W, as obtained from an opencircuit test. When a short-circuit test is performed on it with 90% of the rated
currents flowing in its both LV and HV windings, he measured loss is 81 W. The
transformer has maximum efficiency when operated at
(A) 50.0% of the rated current.
(B) 64.0% of the rated current.
(C) 80.0% of the rated current.
(D) 88.8% of the rated current.
Answer: (C)
18.
A single-phase load is supplied by a single-phase voltage source. If the current
flowing from the load to the source is 10∠ − 150 ° A and if the voltage at the load
terminals is 100∠60 ° V , then the
(A) load absorbs real power and delivers reactive power.
(B) load absorbs real power and absorbs reactive power.
(C) load delivers real power and delivers reactive power.
(D) load delivers real power and absorbs reactive power.
Answer: (B)
19.
A source v s (t) = V cos 100 πt has an internal impedance of
( 4 + j3 ) Ω. If
a purely
resistive load connected to this source has to extract the maximum power out of
the source, its value in Ω should be
(A) 3
Answer
(B) 4
(C) 5
(D) 7
(C)
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20.
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Two systems with impulse responses h1 (t) and h2 (t) are connected in cascade.
Then the overall impulse response of the cascaded system is given by
(A) product of h1 (t) and h2 (t)
(B) Sum of h1 (t) and h2 (t)
(C) Convolution of h1 (t) and h2 (t)
(D) subtraction of h2 (t) and h1 (t)
Answer
21.
(C)
Which one of the following statements is NOT TRUE for a continuous time causal
and stable LTI system?
(A) All the poles of the system must lie on the left side of the jω axis
(B) Zeros of the system can lie anywhere in the s-plane
(C) All the poles must lie within s = 1
(D) All the roots of the characteristic equation must be located on the left side of
the jω axis
Answer
22.
(C)
The impulse response of a system is h(t) = tu(t). For an input u(t − 1), the output
is
(A)
t2
u(t)
2
(B)
t(t − 1)
u(t − 1)
2
(C)
(t − 1)2
u(t − 1)
2
(D)
(t 2 − 1)
u(t − 1)
2
Answer
23.
(C)
Assuming zero initial condition, the response y(t) of the system given below to a
unit step input u(t) is
U(s)
(A) u(t)
Answer
24.
1
s
(B) tu(t)
Y(s)
(C)
t2
u(t)
2
(D) e − tu(t)
(B)
The transfer function
V2 (s)
of the circuit shown below is
V1 (s)
100 µF
+
+
10 kΩ
V1 (s)
V2 (s)
100 µF
−
−
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(A)
0.5s + 1
s +1
Answer
25.
(B)
3s + 6
s+2
(C)
www.gateforum.com
s+2
s +1
(D)
s +1
s+2
(D)
In the circuit shown below what is the output voltage (Vout ) in Volts if a silicon
transistor Q and an ideal op-amp are used?
Q
1 kΩ
5V
(A) -15
Answer
+
−
+15 V
−
+
VBE
Vout
−15 V
(B) -0.7
(C) +0.7
(D) +15
(B)
Q. No. 26 – 55 Carry Two Marks Each
26.
When the Newton-Raphson method is applied to solve the equation
f ( x ) = x3 + 2x − 1 = 0 , the solution at the end of the first iteration with the initial
guess value as x0 = 1.2 is
(A) -0.82
(B) 0.49
(C) 0.705
(D) 1.69
Answer: (C)
27.
A function y = 5x2 + 10x is defined over an open interval x = (1,2). Atleast at one
point in this interval, dy/dx is exactly
(A) 20
(B) 25
(C) 30
(D) 35
Answer: (B)
28.
A 4-pole induction motor, supplied by a slightly unbalanced three-phase 50Hz
source, is rotating at 1440 rpm. The electrical frequency in Hz of the induced
negative sequence current in the rotor is
(A) 100
(B) 98
(C) 52
(D) 48
Answer: (B)
29.
Thyristor T in the figure below is initially off and is triggered with a single pulse of
 100 
 100 
width 10 µs . It is given that L = 
 µH and C =  π  µF . Assuming latching
 π 


and holding currents of the thyristor are both zero and the initial charge on C is
zero, T conducts for
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+
L
T
C
15V
−
(A) 10 µs
(B) 50 µs
(C) 100 µs
(D) 200 µs
Answer: (C)
30.
The following arrangement consists of an ideal transformer and an attenuator
which attenuates by a factor of 0.8. An ac voltage VWX1 = 100V is applied across
WX to get an open circuit voltage across YZ. Next, an ac voltage VYZ2 =100V is
applied across YZ to get an open circuit voltage VWX2
VYZ1
V
, WX2 are respectively,
VWX1
VYZ2
W
across WX. Then,
1:1.25
Y
Z
X
(A) 125/100 and 80/100
(B) 100/100 and 80/100
(C) 100/100 and 100/100
(D) 80/100 and 80/100
Answer
31.
(C)
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the
chosen operating frequency. Their respective resistances are R1 and R 2 . When
connected in series, their effective Q factor at the same operating frequency is
(A) q1R1 + q2R 2
(C)
( q1R1 + q2R 2 ) / (R1 + R2 )
Answer
32.
(B) q1 / R1 + q2 / R 2
(C)
The impulse response of a continuous time system is
h ( t ) = δ ( t − 1) + δ ( t − 3 ) . The value of the step response at t = 2 is
(A) 0
Answer
33.
(D) q1R 2 + q2R1
(B) 1
(C) 2
given
by
(D) 3
(B)
The signal flow graph for a system is given below. The Transfer function,
for the system is
Y (s)
U (s)
1
U (s)
1
s−1
s−1
1
Y (s)
−4
−2
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(A)
s +1
5s + 6s + 2
Answer
34.
(B)
2
s +1
s + 6s + 2
s +1
s + 4s + 2
(C)
2
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(D)
2
1
5s + 6s + 2
2
(A)
In the circuit shown below the op-amps are ideal. Then Vout in Volts is
−2V
1kΩ
1kΩ
−
+
+15V
+15V
+
−15V
Vout
−
−15V
1kΩ
1kΩ
+1V
(A) 4
(B) 6
Answer
35.
1kΩ
(C) 8
(D) 10
(C)
In the circuit shown below, Q1 has negligible collector-to-emitter saturation
voltage and the diode drops negligible voltage across it under forward bias. If Vcc
is +5V, X and Y are digital signals with 0V as logic 0 and Voc as logic 1, then the
Boolean expression for Z is
+ Vcc
R1
Z
R2
X
Diode
Q1
Y
(A) XY
(B) XY
(C) XY
(D) XY
Answer: (B)
36.
The clock frequency applied to the digital circuit shown in the figure below is
1kHz. If the initial state of the output of the flip-flop is 0, then the frequency of
the output waveform Q in kHz is
(A) 0.25
X
(B) 0.5
(C) 1
Clk
T
Q
Q
Q
Q
(D) 2
Answer: (B)
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37.
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z2 − 4
∫ z2 + 4 dz evaluated anticlockwise around the circle z − i = 2, where i =
(A) −4π
(B) 0
(C) 2+ π
−1 , is
(D) 2+2i
Answer: (A)
38.
1
A Matrix has eigenvalues -1 and -2. The corresponding eigenvectors are   and
 −1
1
  respectively. The matrix is
 −2 
1 1
(A) 

 −1 −2 
1 2
(B) 

 −2 −4 
 −1 0 
(C) 

 0 −2 
0 1
(D) 

 −2 −3
Answer: (D)
39.
A dielectric slab with 500mm x 500mm cross-section is 0.4m long. The slab is
subjected to a uniform electric field of E = 6ax + 8aykV / mm . The relative
permittivity of the dielectric material is equal to 2. The value of constant ε0 is
8.85 × 10−12 F / m . The energy stored in the dielectric in Joules is
(A) 8.85 × 10−11
(B) 8.85 × 10−5
(C) 88.5
(D) 885
Answer: (B)
40.
For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5
per unit. The voltage at node 3 is 1.3 −10º per unit. If a capacitor having
reactance of –j3.5 per unit is now added to the network between node 3 and the
reference node, the current drawn by the capacitor per unit is
(A) 0.325 −100º
(B) 0.325 80º
(C) 0.371 −100º
(D) 0.433 80º
Answer: (D)
41.
The separately excited dc motor in the figure below has a rated armature current
of 20A and a rated armature voltage of 150V. An ideal chopper switching at 5
kHz is used to control the armature voltage. If L a = 0.1mH, R a = 1Ω , neglecting
armature reaction, the duty ratio of the chopper to obtain 50% of the rated
torque at the rated speed and the rated field current is
200 V
+
L a ,R a
−
(A) 0.4
(B) 0.5
(C) 0.6
(D) 0.7
Answer: (D)
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A voltage 1000 sin ωt Volts is applied across YZ. Assuming ideal diodes, the
voltage measured across WX in Volts is
1kΩ
W
Y
X
Z
1kΩ
( sin ωt + sin ωt ) / 2
(A) A sinωt
(B)
(C) ( sin ωt − sin ωt ) / 2
(D) 0 for all t
Answer: (D)
43.
Three
capacitors
C1 , C2 , and C3
whose
values
are
10µF, 5µF and 2µF
respectively, have breakdown voltages of 10V, 5V and 2V respectively. For the
interconnection shown, the maximum safe voltage in Volts that can be applied
across the combination and the corresponding total charge in µC stored in the
effective capacitance across the terminals are respectively
C2
C3
C1
(A) 2.8 and 36
Answer
44.
(B) 7 and 119
(C) 2.8 and 32
(D) 7 and 80
(C)
In the circuit shown below, if the source voltage Vs = 100 53.13º V, then the
Thevenin’s equivalent voltage in volts as seen by the load resistance RL is
3Ω
j4Ω
+
I1
(A) 100 90º
Answer
45.
−
VL1
Vs
5Ω
j6Ω
+
j40I2 −
(B) 800 0º
+
− 10V
L1
I2
(C) 800 90º
R L = 10Ω
(D) 100 60º
(C)
The open loop transfer function of a dc motor is given as
ω (s)
Va ( s )
=
10
. When
1 + 10s
connected in feedback as shown below, the approximate value of Ka that will
reduce the time constant of the closed loop system by one hundred times as
compared to that of the open loop system is
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R (s)
(A) 1
Ka
−
10
1 + 10s
(B) 5
Answer
46.
Va ( s )
+
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ω (s)
(C) 10
(D) 100
(C)
In the circuit shown below, the knee current of the ideal Zener diode is 10mA. To
maintain 5V across RL, the minimum value of RL in Ω and the minimum power
rating of the Zener diode in mW respectively are
100Ω
ILoad
10V
Vz = 5V
RL
(A) 125 and 125
(B) 125 and 250
(C) 250 and 125
(D) 250 and 250
Answer: (B)
47.
A strain gauge forms one arm of the bridge shown in the figure below and has a
nominal resistance without any load as R s = 300Ω . Other bridge resistances are
R1 = R 2 = R 3 = 300Ω . The maximum permissible current through the strain gauge
is 20mA. During certain measurement when the bridge is excited by maximum
permissible voltage and the strain gauge resistance is increased by 1% over the
nominal value, the output voltage Vo in mV is
Rs
R1
Vo
+
Vi −
+
−
R3
(A) 56.02
(B) 40.83
R2
(C) 29.85
(D) 10.02
Answer: (C)
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Common Data Questions: 48 & 49
The state variable formulation of a system is given as
 . 
 x1  =  −2 0   x1  + 1 u , x ( 0 ) = 0, x ( 0 ) = 0 and y= 1 0  x1 
1
2

 x 
 .   0 −1 x2  1
 2
x2 
48.
The response y ( t ) to the unit step input is
(A)
1 1 −2t
− e
2 2
(B) 1 −
1 −2t 1 − t
e − e
2
2
(C) e−2t − e − t
(D) 1 − e − t
Answer: (A)
49.
The system is
(A) controllable but not observable
(B) not controllable but observable
(C) both controllable and observable
(D) both not controllable and not observable
Answer: (A)
Common Data Questions: 50 & 51
In the figure shown below, the chopper feeds a resistive load from a battery
source. MOSFET Q is switched at 250 kHz, with a duty ratio of 0.4. All elements
of the circuit are assumed to be ideal
100µH
12V
+
−
50.
Q
20Ω
470µF
The Peak to Peak source current ripple in amps is
(A) 0.96
(B) 0.144
(C) 0.192
(D) 0.288
Answer: (C)
51.
The average source current in Amps in steady-state is
(A) 3/2
(B) 5/3
(C) 5/2
(D) 15/4
Answer: (B)
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Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each
Statement for Linked Answer Questions: 52 & 53
In the following network, the voltage magnitudes at all buses are equal to 1 pu,
the voltage phase angles are very small, and the line resistances are negligible.
All the line reactances are equal to j1 Ω
Bus 1 ( slack )
Bus 2
j1Ω
P2 = 0.1 pu
j1Ω
j1Ω
Bus 3 P = 0.2 pu
3
52.
The voltage phase angles in rad at buses 2 and 3 are
(A) θ2 = −0.1, θ3 = −0.2
(B) θ2 = 0, θ3 = −0.1
(C) θ2 = 0.1, θ3 = 0.1
(D) θ2 = 0.1, θ3 = 0.2
Answer: (C)
53.
If the base impedance and the line-to line base voltage are 100 ohms and 100kV
respectively, then the real power in MW delivered by the generator connected at
the slack bus is
(A) -10
(B) 0
(C) 10
(D) 20
Answer: (C)
Statement for Linked Answer Questions: 54 & 55
The Voltage Source Inverter (VSI) shown in the figure below is switched to
provide a 50Hz, square wave ac output voltage vo across an RL load. Reference
polarity of vo and reference direction of the output current io are indicated in the
figure. It is given that R = 3 ohms, L = 9.55mH.
Q1
Q3
D1
Vdc
D3
L
+
−
io
+
vo
−
Q4
R
Q2
D4
D2
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In the interval when vo < 0 and io > 0 the pair of devices which conducts the load
current is
(A) Q1 , Q2
(B) Q3 , Q4
(C) D1 , D2
(D) D3 , D4
Answer: (D)
55.
Appropriate transition i.e., Zero Voltage Switching (ZVS) / Zero Current
Switching (ZCS) of the IGBTs during turn-on / turn-off is
(A) ZVS during turn off
(B) ZVS during turn-on
(C) ZCS during turn off
(D) ZCS during turn-on
Answer: (D)
Q. No. 56 – 60 Carry One Mark Each
56.
Choose the grammatically CORRECT sentence:
(A) Two and two add four
(B) Two and two become four
(C) Two and two are four
(D) Two and two make four
Answer: (D)
57.
Statement: You can always give me a ring whenever you need.
Which one of the following is the best inference from the above statement?
(A) Because I have a nice caller tune
(B) Because I have a better telephone facility
(C) Because a friend in need in a friend indeed
(D) Because you need not pay towards the telephone bills when you give me a
ring
Answer: (C)
58.
In the summer of 2012, in New Delhi, the mean temperature of Monday to
Wednesday was 41ºC and of Tuesday to Thursday was 43ºC. If the temperature
on Thursday was 15% higher than that of Monday, then the temperature in ºC on
Thursday was
(A) 40
(B) 43
(C) 46
(D) 49
Answer: (C)
Explanations:- Let the temperature of Monday be TM
Sum of temperatures of Tuesday and Wednesday = T and
Temperature of Thursday =TTh
Now, Tm + T = 41 × 3 = 123
& Tth + T = 43 × 3 = 129
∴ TTh − Tm = 6, Also TTh = 1.15Tm
∴0.15Tm = 6 ⇒ Tm = 40
∴ Temperature of thursday = 40 + 6 = 46O C
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59.
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Complete the sentence:
Dare ____________ mistakes.
(A) commit
(B) to commit
(C) committed
(D) committing
Answer: (B)
60.
They were requested not to quarrel with others.
Which one of the following options is the closest in meaning to the word quarrel?
(A) make out
(B) call out
(C) dig out
(D) fall out
Answer: (D)
Q. No. 61 – 65 Carry Two Marks Each
61.
A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and
16km in the third quarter. The average speed of the car in km per hour over the
entire journey is
(A) 30
(B) 36
(C) 40
(D) 24
Answer: (C)
Explanations:-Average speed =
=
62.
Total dis tan ce
Total time
8 + 6 + 16
= 40 km / hr
1 1 1
+ +
4 4 4
Find the sum to n terms of the series 10 + 84 + 734 + …
(A)
(
9 9n + 1
10
) +1
(B)
(
9 9n − 1
8
) +1
(C)
(
9 9n − 1
8
) +n
(D)
(
) +n
9 9n − 1
8
2
Answer: (D)
Explanations:-Using the answer options, substitute n = 2. The sum should add up to 94
63.
Statement: There were different streams of freedom movements in colonial India
carried out by the moderates, liberals, radicals, socialists, and so on.
Which one of the following is the best inference from the above statement?
(A) The emergence of nationalism in colonial India led to our Independence
(B) Nationalism in India emerged in the context of colonialism
(C) Nationalism in India is homogeneous
(D) Nationalism in India is heterogeneous
Answer: (D)
64.
The set of values of p for which the roots of the equation 3x2 + 2x + p (p − 1) = 0
are of opposite sign is
(A)
( −∞, 0 )
(B)
(0,1)
(C)
(1, ∞ )
(D) ( 0, ∞ )
Answer: (B)
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What is the chance that a leap year, selected at random, will contain 53
Sundays?
(A) 2/7
(B) 3/7
(C) 1/7
(D) 5/7
Answer: (A)
Explanations:-There are 52 complete weeks in a calendar year 852 × 7 = 364 days
Number of days in a leap year = 366
∴ Probability of 53 Saturdays =
2
7
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Q. No. 1 – 5 Carry One Mark Each
1.
Choose the most appropriate phrase from the options given below to complete the following
sentence.
India is a post-colonial country because
(A) it was a former British colony
(B) Indian Information Technology professionals have colonized the world
(C) India does not follow any colonial practices
(D) India has helped other countries gain freedom
Answer: (A)
2.
Who ___________ was coming to see us this evening?
(A) you said
(B) did you say
(C) did you say that
Answer: (B)
3.
(D) had you said
Match the columns.
Column 1
Column 2
(1) eradicate
(P) misrepresent
(2) distort
(Q) soak completely
(3) saturate
(R) use
(4) utilize
(S) destroy utterly
(A) 1:S, 2:P, 3:Q, 4:R
(C) 1:Q, 2:R, 3:S, 4:P
Answer: (A)
(B) 1:P, 2:Q, 3:R, 4:S
(D) 1:S, 2:P, 3:R, 4:Q
4.
What is the average of all multiples of 10 from 2 to 198?
(A) 90
(B) 100
(C) 110
Answer: (B)
Exp:
10 + 190 →  200
20 − 180 → 
 9
:

:

90 − 110 

100

5.
⇒
(D) 120
[(200) × 9 + 100] = 1900 = 100
19
19
The value of 12 + 12 + 12 + .... is
(A) 3.464
(B) 3.932
(C) 4.000
(D) 4.444
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Answer: (C)
Exp:
let = 12 + 12 + 12 + .... = y
⇒ 12 + y = y
⇒ 12 + y = y 2
⇒ (y − 4)(y + 3) = 0
⇒ y = 4, y = −3
Q.No. 6 – 10 Carry Two Marks Each
6.
The old city of Koenigsberg, which had a German majority population before World War 2,
is now called Kaliningrad. After the events of the war, Kaliningrad is now a Russian territory
and has a predominantly Russian population. It is bordered by the Baltic Sea on the north and
the countries of Poland to the south and west and Lithuania to the east respectively. Which of
the statements below can be inferred from this passage?
(A) Kaliningrad was historically Russian in its ethnic make up
(B) Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia
(C) Koenigsberg was renamed Kaliningrad, as that was its original Russian name
(D) Poland and Lithuania are on the route from Kaliningrad to the rest of Russia
Answer: (B)
7.
The number of people diagnosed with dengue fever (contracted from the bite of a mosquito)
in north India is twice the number diagnosed last year. Municipal authorities have concluded
that measures to control the mosquito population have failed in this region.
Which one of the following statements, if true, does not contradict this conclusion?
(A) A high proportion of the affected population has returned from neighbouring countries
where dengue is prevalent
(B) More cases of dengue are now reported because of an increase in the Municipal Office’s
administrative efficiency
(C) Many more cases of dengue are being diagnosed this year since the introduction of a new
and effective diagnostic test
(D) The number of people with malarial fever (also contracted from mosquito bites) has
increased this year
Answer: (D)
8.
If x is real and x 2 − 2x + 3 = 11 , then possible values of − x 3 + x 2 − x include
(A) 2, 4
(B) 2, 14
(C) 4, 52
(D) 14, 52
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Answer: (D)
x 2 − 2x + 3 = 11
Exp:
⇒ (x − 4)(x + 2) = 0 ⇒ x = 4, x = −2
Values of − x 3 + x 2 − x
For x = 4
Value = 52
for x = −2
Value = 14
∴ Option D = 14,52
9.
The ratio of male to female students in a college for five years is plotted in the following line
graph. If the number of female students doubled in 2009, by what percent did the number of
male students increase in 2009?
Ra
tio
of
m
al
e
to
fe
m
al
Answer:
Exp:
3.5
3
2.5
2
1.5
1
0.5
0
2008
2009
2010
2011
2012
140%
m
m
=3
= 2.5 m=2.5f
f
f
m'
=3
2f
m ' = 6f
m '− m
=
m
3.5f
%↑ =
× 100
2.5f
7
= = 1.4
8
% ↑= 140%
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10.
At what time between 6 a.m. and 7 a.m will the minute hand and hour hand of a clock make
an angle closest to 60°?
(A) 6: 22 a. m.
(B) 6:27 a.m.
(C) 6: 38 a.m.
(D) 6:45 a.m.
Answer: (A)
Exp:
Angle by minute’s hand
60 min → 360 ο
1min →
360
= 6ο
60
8min → 48o
Angle → 48o with number ‘6’
Angle by hours hand
60 min = 30o
30
× 22
60
= 11
22 min →
Total Angle=48+11=59o.
6.22am
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Q.No. 1 – 25 Carry One Mark Each
1.
Which one of the following statements is true for all real symmetric matrices?
(A) All the eigenvalues are real.
(B) All the eigenvalues are positive
(C) All the eigenvalues are distinct
(D) Sum of all the eigenvalues is zero.
Answer: (A)
Exp:
Eigen values of a real symmetric matrix are all real
2.
Consider a dice with the property that the probability of a face with n dots showing up is
proportional to n. The probability of the face with three dots showing up is_________.
Answer: 1/7
P ( n ) = k.n where n = 1 to 6
Exp:
we know
∑ P ( x ) = 1 ⇒ K [1 + 2 + 3 + 4 + 5 + 6] = 1
x
⇒K=
1
21
∴ required probability is P ( 3) = 3K =
3.
Maximum of the real valued function f ( x ) = ( x − 1)
( A) − ∞
Answer:
Exp:
1
7
( B) 0
2
3
occurs at x equal to
( C) 1
( D) ∞
(C)
f1 (x) =
2
3 ( x − 1)
1
3
is negative, ∀x < 1 or ∀x in (1 − h,1)
is positive, ∀x > 1 or ∀x in (1,1 + h )
h is positive & small
∴ f has local min ima at x = 1 and the min imum value is '0'
4.
All the values of the multi-valued complex function 1i, where i = −1 , are
(A) purely imaginary
(B) real and non-negative
(C) on the unit circle.
(D) equal in real and imaginary parts
Answer: (B)
Exp:
1 = cos ( 2kπ ) + i sin ( 2kπ ) where k is int eger
= ei( 2kπ)
∴1i = e −( 2kπ)
∴ All values are real and non negative
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d2y
dy
+x
− y = 0 . Which of the following is a solution
2
dx
dx
to this differential equation for x > 0?
Consider the differential equation x 2
(A) e x
(B) x 2
(C) 1/x
(D) ln x
Answer: (C)
Exp:
d2 y
dy
+x
− y = 0 is cauchy − Euler equation
2
dx
dx
d
⇒ ( θ2 − 1) .y = 0 where θ =
and z = log x, x = e z
dz
A.E : m 2 − 1 = 0 ⇒ m = −1,1
x2
∴ Solution is y = C1e − Z + C 2 e Z =
C1
+ C2 x
x
1
∴ is a solution
x
6.
Two identical coupled inductors are connected in series. The measured inductances for the
two possible series connections are 380 µH and 240 µH . Their mutual inductance in µH is
________
Answer: 35µH
Exp:
Two possible series connections are
1. Aiding then L equation = L1 + L2 + 2M.
2. Opposing then L equation = L1 + L2 –2M
L1 + L2 + 2M = 380 H
…(1)
L2 + L2 –2M = 240 H
…(2)
From 1 & 2, M = 35µH
7.
The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+, the
switch is moved to position ‘2’ Assuming V02 > V01 , the voltage VC ( t ) across capacitor is
'2'
R
SW
R
V02
V01
C
VC
(A)
vc ( t ) = −V02 (1 − e− t /RC ) − V01
( B)
v c ( t ) = V02 (1 − e − t /RC ) + V01
(C)
v c ( t ) = ( −V02 + V01 ) (1 − e − t /RC ) − V01
( D)
v c ( t ) = ( V02 + V01 ) (1 − e − t /RC ) + V01
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•
Answer: (D)
Exp:
When switching is in position 1
VC ( t ) = ( Initial − final )e
−t
VC ( t ) = V01 1 − e

−t
RC
z
ϑ01
R
+ final value
•
C


•
When switch is in position 2
2
• •
Initial value is
R
−t
VC ( t ) = V01 1 − e RC 


R
V02
•
Final value is –V02
VC ( t ) = V01 [ V02 − V01 ] 1 − e

8.
ϑc
−t
2RC
C


ϑc
•
A parallel plate capacitor consisting two dielectric materials is shown in the figure. The
middle dielectric slab is place symmetrically with respect to the plates.
10 Volt
ε1
ε1
ε2
d/2
d
If the potential difference between one of the plates and the nearest surface of dielectric
interface is 2Volts, then the ratio ε1 : ε 2 is
(A) 1 : 4
Answer:
Exp:
(B) 2 : 3
(C) 3: 2
(D) 4 : 1
(C)
Q = CV C1 = V2
↓
C 2 V1
cons tan t
C=
Aε
d
ε1 V2
=
ε 2 V1
ε1 V2
ε1
ε
=
⇒ V2 =
( V ) ⇒ 6 = 1 (10 )
ε 2 V1
ε1 + ε 2
ε1 + ε 2
ε 3
⇒ 3ε1 + 3ε 2 = 5ε1 ⇒ 1 =
ε1 2
10V
ε1
8V
ε2
2V
0V
ε1
ε1 : ε 2 = 3 : 2
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Consider an LTI system with transfer function
H`( s) =
1
s ( s + 4)
If the input to the system is cos(3t) and the steady state output is Asin ( 3t + α ) , then the value
of A is
(A) 1/30
(B) 1/15
(C) 3/4
(D) 4/3
Answer:
(B)
Exp:
Given H ( s ) =
H ( jω) =
1
s (s + 4)
1
ω ω2 + 16
cos ( ω0 t )
y ( t ) = H ( jω) ω=ω cos ( ω0 t + θ )
H ( jω )
0
where θ = H ( jω)
ω= 0
⇒ A = H ( jω) ω=ω
0
ω0 = 3
⇒A=
10.
1
3 9 + 16
=
1
15
1
Consider an LTI system with impulse response h(t) = e −5t u ( t ) . If the output of the system is
y ( t ) = e −2 t u ( t ) − e −5t u ( t ) then the input, x(t), is given by
( A ) e−3t u ( t )
Answer: (B)
Exp:
x (t )
( B) 2e −3t u ( t )
1
s+5
y ( t ) = e −3t − e−5t u ( t ) ↔ Y ( s ) =
⇒ X (s) =
( D ) 2e −5t u ( t )
y (t )
h (t )
h ( t ) = e−5t u ( t ) ↔ H ( s ) =
⇒ H (s) =
( C) e−5t u ( t )
Y (s)
1
1
−
s+3 s+5
X (s)
Y (s)
( 5 + 5 ) − ( s + 3) = 2
H (s)
( 5 + 3)( s + 5) s + 3
x ( t ) = 2e −3t u ( t )
=
1
s+5
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Assuming an ideal transformer,. The Thevenin’s equivalent voltage and impedance as seen
from the terminals x and y for the circuit in figure are
1Ω
x
sin ( ωt )
y
1: 2
( A ) 2sin ( ωt ) , 4Ω
( C) 1sin ( ωt ) , 2Ω
( B) 1sin ( ωt ) , 1Ω
( D ) 2sin ( ωt ) , 0.5Ω
Answer: A
Exp:
ϑxy = Voc
ϑin ϑxy
=
⇒ ϑxy = ϑoc = 2 sin ωt
1
2
2
2
R xy = 100 ×   ⇒ 4
1
ϑth = 2sin ωt
R th = 4Ω
12.
A single phase, 50kVA, 1000V/100V two winding transformer is connected as an
autotransformer as shown in the figure.
100 V
1100 V
1000 V
The kVA rating of the autotransformer is _____________.
Answer: 550kVA
Exp:
Given,
1000V
100V
3
50 ×10
I2 =
= 500
100
∴ ( kVA ) A.TFr = 1100 × 500 = 550 kVA
I 2 = 500A
50 kVA,
1100 V
1000 V
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A three-phase, 4pole, self excited induction generator is feeding power to a load at a
frequency f1. If the load is partially removed, the frequency becomes f2. If the speed of the
generator is maintained at 1500 rpm in both the cases, then
(A) f1 f 2 > 50 Hz and f1 > f 2
(B) f1 < 50 Hz and f 2 > 50Hz
(C) f1 f 2 < 50 Hz and f 2 > f 2
(D) f1 > 50 Hz and f 2 < 50Hz
Answer: (C)
Exp:
Initially self excited generator supply power to a load at f1 . If load is partially removed then
slightly speed increase, also frequency f 2
∴ f 2 > f1
But both cases f1f 2 < 50 Hz
14.
A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor is
connected in parallel to the motor to improve the power factor of the combination of motor
and capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motor
remains same as before, the reactive power delivered by the capacitor in MVAR is
____________.
Answer: 7MVAR
Exp:
Given, 1 − φ Induction motor draws 12mW at 0.6pf, lag
Let P1 =12mW
cos φ1 = 0.6 pf
To improve pf, cos φ2 = 0.8
( Qc ) del by capacitor = ?
cos φ1 =
P1
12 ×106
⇒ S1 =
S1
0.6
⇒ S1 = 20 MVA
4 Reactive power, Q1 = S12 − P12 =16 MVAR
When capacitor is connected then
cos φ2 =
0.8 =
P1
(∵ Re al power drawn is same )
S2
12 ×106
S2
S2 =15MVA
∴ Re active power ,Q 2 = S22 − P12 = 9MVAR
But motor should draw the same reactive power.
∴ ( Qc )del by capacitor = 16 − 9 = 7 MVAR
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A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power
factor lagging. The three phase base power and base current are 100MVA and 437.38A
respectively. The line-to line load voltage in kV is ___________.
Answer: 117-120
Exp:
Given, 100 mVA, 437.38 A
VL − L (kV) = ?
We know that, S = 3 VL .I L
100 × 106 = 3.VL .I L
VL =
100 × 106
3 × 437.38
VL = 132.001kV
But it is drawing power at a voltage of 0.9 pu
∴ Vpu =
Vactual
VBase
⇒ Vactual = VL − L = Vpu × VB
= 0.9 × 132 = 118.8kV
16.
Shunt reactors are sometimes used in high voltage transmission system to
(A) limit the short circuit current through the line.
(B) compensate for the series reactance of the line under heavily loaded condition.
(C) limit over-voltages at the load side under lightly loaded condition.
(D) compensate for the voltage drop in the line under heavily loaded condition.
Answer: (C)
17.
The closed-loop transfer function of a system is T ( s) =
(
4
. The steady state error
s + 0.4s + 4
2
)
due to unit step input is ________.
Answer: 0
Exp:
Steady state error for Type-1 for unit step input is 0.
18.
The state transition matrix for the system
 x 1 
 x  =
 2
e t
A
( ) t
e
1 0  x1  1
1 1   x  + 1 u is

  2  
0

et 
 et
B
( ) 2t
t e
0

et 
 et
C
( )  t
 − te
0

et 
e t
D
( ) 
0
te t 

et 
Answer: (C)
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1 0 
1
Given A = 
B= 

1 1 
1
[SI − A ]
−1
 s 0   1 0  
= 
−

 0 s   1 1  
 1
 ( s − 1)
−1
[SI − A ] =  1

2
 ( s − 1)
−1



1 
( s − 1) 
0
The state transition matrix
−1
e At = L−1 ( SI − A ) 


 et
e At =  t
 te
19.
0

et 
The saw-tooth voltage wave form shown in the figure is fed to a moving iron voltmeter. Its
reading would be close to ______________
100 V
20 ms
40 ms
Answer: 57.73
Exp:
100 V
20 m sec
t
40 m sec
Moving iron meter reads RMS value only RMS value of saw-tooth waveform is
Meter reads =
ϑmax
3
100
3
= 57.73 volts
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While measuring power of a three-phase balanced load by the two-wattmeter method, the
readings are 100W and 250 W. The power factor of the load is ______________.
Answer: 0.802
Exp:
In two-wattmeter method,
The readings are 100 W & 250 W
Power factor = cos φ

 3 ω1 − ω2
= cos  tan −1 

 ω1 + ω2


 

 3 (150 )  
= cos  tan −1 


 350  
= 0.8029
21.
Which of the following is an invalid state in an 8-4-2-1. Binary Coded Decimal counter
(A) 1 0 0 0
(B) 1 0 0 1
(C) 0 0 1 1
(D) 1 1 0 0
Answer: (D)
Exp:
In binary coded decimal (BCD) counter the valid states are from 0 to 9 only in binary system
0000 to 1001 only. So, 1100 in decimal it is 12 which is invalid state in BCD counter.
22.
The transistor in the given circuit should always be in active region. Take VCE(sat ) = 0.2 V .
VEE = 0.7 V. The maximum value of RC in Ω which can be used is __________.
RC
+
5V
Rs = 2kΩ
β = 100
5V
+
Answer: 22.32Ω
Exp:
IB =
5 − 0.7
= 2.15mA
2k
IC = 0.215A
∴RC =
5 − 0.2
= 22.32Ω
0.215
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20
V. Considering all possible
2
values of RL, the minimum value of RS in Ω to avoid burnout of the Zener diode is
________.
A sinusoidal ac source in the figure has an rms value of
RS
20
V ~
2
5V
1/ 4W
RL
Answer: 300Ω
Exp:
Vm = 20V
Pz = Vz Iz ⇒ I z =
R S ( min ) =
24.
Pz
= 50mA
Vz
20 − 5
= 300Ω
50mA
A step-up chopper is used to feed a load at 400 V dc from a 250 V dc source. The inductor
current is continuous. If the ‘off’ time of the switch is 20 µs, the switching frequency of the
chopper is kHz is __________.
Answer: 31.25 kHz
Exp:
V0 = 400v, Vs = 250 v, Toff = 20 µ sec, F = ?
Given chopper in step up chopper
Vs
1− D
250
250
400 =
⇒ 1− D =
1− D
400
3
D=
= 0.375
8
but Toff = (1 − D ) T
∴ Vo =
(
20 ×10−6 = 1 − 3
8
)T
∴ T = 32 µ sec
1
1
=
= 31.25 Hz
T
32 ×10−6
∴ f = 31.25kHz
Then f =
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In a constant V/f control of induction motor, the ratio V/f is maintained constant from 0 to
base frequency, where V is the voltage applied to the motor at fundamental frequency f.
Which of the following statements relating to low frequency operation of the motor is TRUE?
(A) At low frequency, the stator flux increases from its rated value.
(B) At low frequency, the stator flux decreases from its rated value.
(C) At low frequency, the motor saturates.
(D) At low frequency, the stator flux remains unchanged at its rated value.
Answer: (B)
Exp:
V
control, at low frequency, the voltage also applied to the induction motor
f
is low. Hence the stator flux also decreases from its rated value.
During constant
Q.No. 26 – 55 Carry Two Marks Each
26.
8  ( y/2 ) +1  2x − y 

To evaluate the double integral ∫  ∫
dx  dy , we make the substitution


0
 y/ 2  2  
y
 2x − y 
u=
and v = . The integral will reduce to
 2 
2
(
( A ) ∫0 ∫0 2 u du
4
(
2
( C) ∫0 ∫0
4
1
(
( D) ∫ ( ∫
) dv
) dv
u du ) dv
( B) ∫0 ∫0 2 u du
)
u du dv
4
1
4
2
0
0
Answer: (B)
Exp:
u=
2x − y
y
........(1) and V = ........ ( 2 )
2
2
y
y
⇒ u = 0 ; x = +1⇒ u =1
2
2
y =0⇒ v =0 ; y =8⇒ v = 4
x=
from (1) and ( 2 ) , x = u + v ... ( 3) and y = 2v ... ( 4 )
∂x
∂u
Jacobian transformation; J =
∂y
∂u
∂x
∂v 1
=
∂y 0
∂v
1
=2
2
 ( y 2 ) +1

4
 1

 2x − y  

∴∫  ∫ 
dx
dy
=
 ∫ ( u ) J du  dv
 
∫
0
v=0  u =0

 y2  2  


8
=∫
4
0
( ∫ 2u du ) dv
1
0
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Let X be a random variable with probability density function
 0.2,

f ( x ) = 0.1,
 0,

for x ≤ 1
for 1 < x ≤ 4
otherwise
The probability p ( 0.5 < x < 5) is __________
Answer: 0.4
Exp:
P ( 0.5 < x < 5 ) = ∫ f ( x ) dx
5
0.5
= ∫ f ( x ) dx + ∫ f ( x ) dx + ∫ f ( x ) dx
1
4
0.5
5
1
4
Opposite
Hypotenuse
= ( 0.2 )( x )0.5 + ( 0.1)( x )1 + 0
1
4
= 0.1 + 0.3 = 0.4
28.
The minimum value of the function f ( x ) = x 3 − 3x 2 − 24x + 100 in the interval [-3. 3] is
(A) 20
(B) 28
(C) 16
(D) 32
Answer: (B)
Exp:
f 1 ( x ) = 0 ⇒ x 2 − 2x − 8 = 0
⇒ x = −2, 4 ∈ [ −3,3]
Now f ( −3) = 118 ; f ( 3) = 28
and f ( −2 ) = 128 ; f ( 4 ) = 44
∴ f ( x ) is min imum at x = 3 and the min imum value is f ( 3) = 28
29.
Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage vi
must be outside the range
10 k
Vi ~
1V
(A) −1V to − 2V
(B) −2V to − 4V
10k
2V
Vo
(C) +1V to − 2V
(D) +2V to − 4V
Answer: (B)
Exp:
When both diodes are 0FF, vo =
vi
(Not clipped).
2
∴ For the clipped, vi must bt ouside the range − 2V to − 4V
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30.
The
voltage
across
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the capacitor, as sown in the figure,
v t ( t ) = A1 sin ( ω1t − θ1 ) + A 2 sin ( ω 2 t − θ 2 )
1Ω
expressed
1H
VC ( t )
20sin10t ~
is
↑ 10sin 5t
1F
The value of A1 and A2 respectively, are
(A) 2.0 and 1.98
(B) 2.0 and 4.20
(C) 2.5 and 3.50
Answer: (A)
Exp: By using super position theorem,
1. ϑC1 ( t ) − When 20 sin 10t voltage source is acting,
(D) 5.0 and 6.40
1
1
j ωc
Network function H ( jω) =
⇒
1
10
( j + 1)
R+
j ωc
ϑc1 ( t ) =
2.
1
20 sin (10t − tan −1 (10 ) )
101
ϑc2 ( t ) − When 10 sin 5t current source is acting
ϑc2 =
10 0 × 1
× −0.2 j
1 − 0.2 j
ϑc2 =
−2 j
1 − 0.2 j
ϑc2 ( t ) =
2
1 + ( 0.2 )
2
.sin ( 5t − θ2 )
ϑc2 ( t ) = 1.98 sin ( 5t − θ2 )
VC ( t ) = 2sin (10t − θ1 ) + 1.98 ( 5t − θ2 )
By comparing with given expression,
31.
A1 = 2.0
A 2 = 1.98
The total power dissipated in the circuit, show in the figure, is 1kW.
10 A
2A
1Ω
X c1
XL
R
Load
ac source
~
X c2
V
200 V
The voltmeter, across the load, reads 200 V. The value of XL is _________.
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Answer: 17.34 Ω
Exp: Total power dissipated in the circuit is 1kW.
P = 1kW
1000 = I2 .1 + I 2 .R.
1000 = ( 2 ) .1 + (10 ) .R.
2
2
⇒ R = 9.96 Ω
Z=
V
200
⇒
= 20
I
10
Z = R 2 + XL2
⇒ X 2L = ( Z ) − R 2
2
X 2L = ( 20 ) − ( 9.96 )
2
2
⇒ X L = 17.34 Ω
32.
( )
The magnitude of magnetic flux density B at a point having normal distance d meters from
µ0I
(in SI units). An infinitely
2nd
extended wire is laid along the x-axis and is carrying current of 4 A in the +ve x direction.
Another infinitely extended wire is laid along the y-axis and is carrying 2 A current in the +ve
ˆ J,
ˆ K
ˆ to be unit vectors along x, y and
y direction µ 0 is permeability of free space Assume I,
z axes respectively.
an infinitely extended wire carrying current of l A is
y
( 2,1,0)
2A
1
I amps
d
B=
µ0I
2 πd
4A
1
2
z
Assuming right handed coordinate system, magnetic field intensity, H at coordinate (2,1,0)
will be
(A)
3 ˆ
k weber / m 2
2π
( B)
4 ˆ
iA/m
3π
( C)
3 ˆ
kA/m
2π
( D)
0 A/m
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Answer: (C)
Exp:
H = Hx + Hy
Hx =
I
4
2
aφ =
ax × ay ) = az
(
2πρ
2π (1)
π
Hy =
I
2
−1
aφ =
ay × ax ) =
az
(
2πρ
2π ( 2 )
2π
H=
33.
1
1 3
az
2−  =
π
2  2π
A discrete system is represented by the difference equation
 X1 ( k + 1)   a
a − 1  X1 ( k ) 

=

 
 X 2 ( k + 1)  a + 1 a   X 2 ( k ) 
It has initial condition X1 ( 0 ) = 1; X 2 ( 0 ) = 0 . The pole location of the system for a = 1,
are
(A) 1 ± j0
(B) −1 ± j0
(C) ±1 + j0
(D) 0 ± j1
Answer: (A)
Exp: from the given difference equation,
a − 1
 a
A=

a + 1 a 
The pole locations of the system for a = 1.
1 0 
Then A = 

2 1
SI − A . ⇒ ( s − 1) = 0
2
S = 1 ± j0
34.
An input signal x ( t ) = 2 + 5sin (100πt ) is sampled with a sampling frequency of 400 Hz and
applied to the system whose transfer function is represented by
Y (z)
X (z)
=
1
N
 1 − z− N 

−1 
 1− z 
where, N represents the number of samples per cycle. The output y(n) of the system under
steady state is
(A) 0
(B) 1
(C) 2
(D) 5
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Answer: (C)
Exp:
x ( t ) = 2 + 5sin (100πt )
1 

x ( n Ts ) = 2 + 5sin  100π n.

400 

π 
= 2 + 5sin  n  , N = 8
4 
Y ( e jΩ )
X ( e jΩ )
=
1  1 − e − jΩN 
= H ( e jΩ )

− jΩ 
N  1− e

x [ n ] = x1 [ n ] + x 2 [ n ]
due to x1 [ n ]
y1 [ n ] = H ( e jΩ )
r =0
= x1 [ n ]
y1 [ n ] = 2
y 2 [ n ] = H ( e jΩ )
H ( e jΩ )
Ω=
π
4
π
Ω=
4
π

sin  n + H ( e jΩ ) π 
4
Ω= 
4 

=0
y [ n ] = y1 [ n ] + y 2 [ n ]
y[n] = 2
Thus at steadystate y[n] = 2
35.
A 10 kHz even-symmetric square wave is passed through a bandpass filter with centre
frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is
(A) a highly attenuated square wave at 10kHz
(B) nearly zero.
(C) a nearly perfect cosine wave at 30kHz.
(D) a nearly perfect sine wave at 30kHz.
Answer: (C)
Exp: 10 KHz even symmetric square wave have frequency component present
10KHz, 30KHz, 50KHz, 70KHz
[only odd harmonics due to half wave symmetry]
Since bandpass filter is contered at 30KHz, 30KHz component will pass through
⇒ filter output is nearly perfect cosine wave at 10 KHz
Cosine in due to reason that signal in even signal.
36.
A 250 V dc shunt machine has armature circuit resistance of 0.6Ω and field circuit resistance
of125 Ω . The machine is connected to 250 V supply mains. The motor is operated as a
generator and then as a motor separately. The line current of the machine in both the cases is
50 A. The ratio of the speed as a generator to the speed as a motor is ____________.
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Answer: 1.27
Exp: Given:
2A
50A
2A
50A
48A
52A
250V
125Ω
250V
125Ω
Generator
Motor
E b = V − Ia R a
E b = V + Ia R a
= 250 − 48× 0.6
=
∴
Ng
Nm
= 250 + 52 × 0.6
= 281.2V
221.2V
= ?
We know that
Ng
Nm
=
37.
www.gateforum.com
=
Eg
Eb
(∵ flux is constant )
Ng
281.2
⇒
= 1.27
221.2
Nm
A three-phase slip-ring induction motor, provided with a commutator winding, is shown in
the figure. The motor rotates in clockwise direction when the rotor windings are closed.
3 − phase ac,fHz
f2
Pr ime
mover
Slip Ring Induction Motor
fr
f1
If the rotor winding is open circuited and the system is made to run at rotational speed fr with
the help of prime-mover in anti-clockwise direction, then the frequency of voltage across slip
rings is f1 and frequency of voltage across commutator brushes is f2. The values of f1 and f2
respectively are
(A) f + fr and f
(B) f - fr and f
(A) f - fr and f+ fr
(D) f - fr and f
Answer: (A)
Exp: Whenever the Rotor winding is open circuited and rotating in anti-clockwise direction then the
( Ns + N r ) P
frequency of voltage across slip rings is f1 =
120
f1 = f + f r
At the same time frequency of voltage across commutator brushes if
NP
f2 = s = f
120
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A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All the
coils of a phase are in series. If the coils are connected to realize single-phase winding, the
generated voltage is V1 . If the coils are reconnected to realize three-phase star-connected
winding, the generated phase voltage is V2 . Assuming full pitch, single-layer winding, the
ratio V1 / V2 is
1
3
( A)
( B)
1
2
( C)
( D)
3
2
Answer: (D)
Exp:
Given poles, P=20
Total slots = 180
4 Total no. of conductor = 180 × 6 =1080
4 the ratio of voltage generated when the coils are connected in 1 − φ to when the coils are
connected in 3 − φ, Y-connection.
i.e.,
39.
( V1 ) 1−φ
( V2 ) 3−φ
= 2
For a single phase, two winding transformer, the supply frequency and voltage are both
increased by 10%. The percentage changes in the hysteresis loss and eddy current loss,
respectively, are
(A) 10 and 21
(B) -10 and 21
(C) 21 and 10
(D) -21 and 10
Answer: (A)
Exp:
Given1 − φ Transformer
V and f are increased by 10%
∴ % ∆ Wn = ?
%∆We = ?
Here
40.
∴ Wn ∝ f
V
is constant
f
We ∞ f 2
Wn n ∝ f
We ∝ f 2
as 'f ' increased by 10%
We ∝1.21f 2
⇒ Wn also ↑ 10%
⇒ We ↑ by 21%
A synchronous generator is connected to an infinite bus with excitation voltage Ef = 1.3 pu.
The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6
pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The
reactive power (Q) in pu supplied by the generator to the bus under this condition is
_________.
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Answer: 0.109
Exp:
Given, E f = 1.3 P.u
X s = 1.1P.u
P
= 0.6 pu
V =1.0 pu
Q =?
We know that, P =
EV
sin δ
Xs
1.3×1
× sin δ
1.1
⇒ δ = 30.5°
V
∴Q =
[ E cos δ − V ] = 0.109
Xs
⇒ 0.6 =
41.
There are two generators in a power system. No-load frequencies of the generators are 51.5
Hz and 51Hz, respectively, and both are having droop constant of 1 Hz/MW. Total load in the
system is 2.5 MW. Assuming that the generators are operating under their respective droop
characteristics, the frequency of the power system in Hz in the steady state is __________.
Answer: 50
Exp:
Given, two generators in a power system has no load frequency of 51.5 & 51 Hz.
∴ drop constant=1Hz/mW
Total load=2.5 mW
for generator '1',
generator '2 '
f = − x1 + 51.5
f = − x 2 + 51
∴ − x1 + 51.5 = − x 2 + 51
⇒ x1 − x 2 = 0.5
...(1)
Total load ⇒ x1 + x 2 = 2.5 ...(2)
By solving (1) & (2) ⇒ x1 =
3
= 1.5
2
∴ f = −1.5 + 51.5 = 50 Hz
42.
The horizontally placed conductors of a single phase line operating at 50 Hz are having
outside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. The
permittivity of free space is 8.854 × 10−12 . The capacitance to ground per kilometer of each
line is
(A) 4.2 × 10-9F
(B) 8.4 × 10-9F
(C) 4.2 × 10-12F
(D) 8.4 × 10-12F
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Answer: (B)
Exp: Given, diameters of conductor =1.61m
∴ radius, r = 0.8cm
Spacing between conductors, d=6m
Permitivity ∈0 = 8.85×10-12
∴ capacitance to ground per km = ?
2π∈o 2π × 8.85 × 10−12
=
= 8.4 × 10−12
6
d


ln   ln 
−2 
r
 0.8 × 10 
−19
C / km = 8.4 × 10 F
∴C =
43.
A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The positive,
negative, and the zero sequence reactances of the generator are 0.2 pu, 0.2 pu, and 0.05 pu,
respectively, at the machine base quantities. If a bolted single phase to ground fault occurs at
the terminal of the unloaded generator, the fault current in amperes immediately after the
fault is _________.
Answer: 15500
Exp: Single line to ground fault,
Fault current in If = 3Ia1
positive sub transient circuit,
Ea
∴ Ia1 =
z1 + z 2 + z 0
1 + jo
j0.2 + j0.2 + j0.05
1
=
= − j2.2223pu
j0.45
=
Fault current ( If
Base current =
) = 3 × Ia1
= ( 3×− j2.222 )
100 ×106
3 × 25×103
= − j6.666 pu
= 2309.4 pu
Fault circuit = pu fault circuit in pu x Base circuit in Amp
If = 15396A
44.
A system with the open loop transfer function:
G ( s) =
K
s ( s + 2 ) s2 + 2s + 2
(
)
is connected in a negative feedback configuration with a feedback gain of unity. For the
closed loop system to be marginally stable, the value of K is ______
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Answer: 5
Exp: The characteristic equation 1 + G(s) = 0
1+
k
=0
s ( s + 2 ) ( s 2 + 2s + 2 )
⇒ s 4 + 4s3 + 6s 2 + 4s + k = 0
R−H Arry:
S4
S3
S2
1
4
5
6 k
4 0
k 0
20 − 4k
5
S0
k
S1
0
For marginally stable, 20−4k = 0
20 = 4 k ⇒ k = 5
45.
For the transfer function
G ( s) =
5 (S + 4)
(
s ( s + 0.25) s2 + 4s + 25
)
The values of the constant gain term and the highest corner frequency of the Bode plot
respectively are
(A) 3.2, 5.0
(B) 16.0, 4.0
(C) 3.2, 4.0
(D) 16.0, 5.0
Answer: (A)
Exp: G ( s ) =
5(s + 4)
s ( s + 0.25 ) ( s 2 + 4s + 25 )
If we convert it into time constants,
 s
5 × 4 1 + 
 4
G (s) =
2
s  
4

s  
s [ 0.25] 1 +
25 1 + .s +   
 0.25   25
 5  
 s
3.2 1 + 
 4
G (s) =
s 
4
s2 

s 1 +
1
+
.s
+

25 
 0.25   25
Constant gain term is 3.2
ωn = 5 → highest corner frequency
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46.
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The second order dynamic system
dX
= PX + Qu
dt
y = RX
has the matrices P, Q and R as follows:
 −1 1 
0 
P=
Q =   R = [ 0 1]

 0 −3
1 
The system has the following controllability and observability properties:
(A) Controllable and observable
(B) Not controllable but observable
(C) Controllable but not observable
(D) Not controllable and not observable
Answer: (C)
 −1 1 
0 
Exp: Given P = 
Q= 

 0 −3
1 
For controllability:
0 1 
Q C = [ Q PQ ] ⇒ 

1 −3
Q C ≠ 0 ∴ controllable
For observability:
Q 0 =  R T
0 0 
P T .R T  ⇒ 

1 −3
Q 0 = 0 ∴ Not observable.
47.
Suppose that resistors R1 and R2 are connected in parallel to give an equivalent resistor R. If
resistors R1 and R2 have tolerance of 1% each., the equivalent resistor R for resistors R1 =
300Ω and R 2 = 200 Ω will have tolerance of
(A) 0.5%
Answer: (B)
Exp:
(B) 1%
R 1 = 250 ± 1%
RT =
(D) 2%
R1R 2
R1 + R 2
R 2 = 300 ± 1%
% E RT =
(C) 1.2%
R T = 136.36Ω
∆R T
× 100
RT
 R ∆R R ∆R 2 
= ± T . 1 + T .
 × 100
R2 R2 
 R1 R1
∆R 1 =
R1. ∈ R1
R .∈ R 2
= 2.5; ∆R 2 = 2
100
100
136.36 2.5 136.36 3 
= ±
.
+
.
= ±1%
300 300 
 250 250
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Two ammeters X and Y have resistances of 1.2 Ω and 1.5 Ω respectively and they give full
scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by
connecting shunts so as to give full scale deflection with 15 A. The ammeters along with
shunts are connected in parallel and then placed in a circuit in which the total current flowing
is 15A. The current in amperes indicated in ammeter X is __________.
Answer: 10.157
Exp: X and Y ammeters are connected in parallel
Shunt Registration of X and Y meters:
1.2
 15 × 103

− 1

150


R shx =
•
15A
R shx = 0.01212 Ω
R shy
1.5 Ω
1.2 Ω
R shy
1.5
=
 15 × 103

− 1

 250

R shx
↑
I mx = 150 mA
↑
I my = 250 mA
•
R shy = 0.02542Ω
Current through X ammeter is
=
0.02542
× 15
0.01212
+ 0.02542 )
(
= 10.157 ampers
49.
An oscillator circuit using ideal op-amp and diodes is shown in the figure
R
+ 5V
−
+
C
3kΩ
1kΩ
Vo
−5V
1kΩ
The time duration for +ve part of the cycle is ∆t1 and for-ve part is ∆t 2 .The value of
e
∆t1 −∆t 2
RC
will be___________.
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Answer: 1.3
Exp:
VC ( t ) = Vmax + ( Vinitial − Vmax ) e − t τ
UTP = + Vsat + ( LTP − Vsat ) e − t1
5
 −5

=5+
− 5  e − t1
4
 2

5
 −15  − t1
−5=
e
4
 2 
τ
τ
where
1 5
UTP = 5 × =
4 4
LTP = 5 ×
1 −5
=
2 2
τ
−3.75 = −7.5 e− t τ
0.5 = e− t1
τ
t1 = 0.69 τ
LTP = − Vsat + ( LTP + Vsat ) e− t 2
−5
 −5

= −5 + 
+ 5  e− t 2 τ
2
 2

5
5 − ⇒ 2.5 = −5 + ( 2.5 ) e − t 2
2
7.5 = 2.5e − t 2 τ ⇒ e − t 2 τ = 3
t 2 = −1.098τ
τ
τ
e( 0.69 τ+1.098τ ) τ = 5.98.
50.
The SOP (sum of products) form of a Boolean function is Σ(0,1,3,7,11), where inputs are
A,B,C,D (A is MSB, and D is LSB). The equivalent minimized expression of the function is
( A)
( C)
( B + C)( A + C) ( A + B)( C + D)
( B + C)( A + C) ( A + C)( C + D)
Answer: (A)
Exp:
(B + C)
( B + C)( A + C) ( A + C)( C + D)
( D ) ( B + C)( A + B) ( A + B)( C + D)
( B)
CD
AB
00
01
11
10
00
1
1
1
0
01
0
0
1
0
11
0
0
0
0
10
0
0
1
0
(C + D )
( A + B)
(A + C)
The equivalent minimized expression of this function is = ( B + C ) ( A + C )( A + B )( C + D )
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A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.
J
( A)
T
Clk
K
T flip −
flop
Qn
J
T
( B)
Clk
K
Qn
T flip −
flop
Qn
J
( C)
T
Clk
K
T flip −
flop
Qn
Qn
J
T
( D)
Clk
K
T flip −
flop
Qn
Qn
Answer: (B)
Exp:
Qn
J
K
Q n +1
T
0
0
0
0
0
1
0
0
0
0
0
1
1
0
1
1
1
0
0
0
1
1
1
0
0
1
1
1
1
0
0
1
1
1
1
1
0
1
0
1
JK
00
01
11
10
0
0
0
1
1
1
0
1
1
0
Qn
T = Q n J + Qn k
Analysis:
If you will observe the combinational circuit output expression which is the input for T flip
flop is not matching directly, so you should go through the option. If you will solve the
combinational circuit of option (B) then
(
T = ( J + Qn ) . K + Q n
)
Qn
(
= J.K + JQ n + K.Q n + Q n Q n = J.K + JQ n + K.Q n + 0 ∵ Q n .Q n = 0
)
= J.K + J Q n + K.Q n
Now, according to consensus theorem J-K will become redundant term, so it should be
eliminated.
Hence, T = JQ n + K.Q n , which in matching with our desired result and option-(B) is correct
answer.
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In an 8085 microprocessor, the following program is executed
Address location – Instruction
2000H
XRA A
2001H
MVI B,04H
2003H
MVI A, 03H
2005H
RAR
2006H
DCR B
2007H
JNZ 2005
200AH
HLT
At the end of program, register A contains
(A) 60H
(B) 30H
Answer: (A)
Exp:
(C) 06H
(D) 03H
Address location
Instruction
Operation
2000H
XRA A
[ A ] = 00H, CY = 0, Z = 1
2001H
MVI B, 04H
[ B] = 04H
2003H
MVI A, 03H
[ A ] = 03H
2005H
RAR
Rotate accumulator right with carry
2006H
DCR B
Decrement content of B register by
one
2007H
JNZ 2005H
Jump on no zero to location 2005H
200AH
HLT
Accumulator
0
0
0
0
0
CY
0
1
1
0
Initial :
value
RAR
0
0
0
0
0
0
0
1
1
[B] = 03H
RAR
1
0
0
0
0
0
0
0
1
[B] = 02H
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RAR
1
1
0
0
0
0
0
0
0
0
0
0
B  = 01H
RAR
0
1
1
0
0
0
Now loop will be over
[B] = 00H
53.
and HLT will execute and
program will be over
A fully controlled converter bridge feeds a highly inductive load with ripple free load current.
The input supply ( v s ) to the bridge is a sinusoidal source. Triggering angle of the bridge
converter is α = 30O . The input power factor of the bridge is_________.
is
+
VS ~
−
Load
Answer: 0.78
Exp: For fully controlled converter bridge
The input power factor (PF) = 0.9 × cos α
∴ IPF = 0.9 × cos30
⇒ IPF = 0.78
A single-phase SCR based ac regulator is feeding power to a load consisting of 5 Ω
resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing
angle at which the voltage across the device becomes zero all throughout and the rms value of
current through SCR, under this operating condition, are
(A) 300 and 46 A
(B) 300 and 23 A
(C) 450 and 23 A
(D) 450 and 32 A
Answer: (C)
54.
Exp:
Vs = 230V, 50 Hz
R = 5Ω, L =16mH
The maximum firing angle at which the volt across device becomes zero is the angle at
which device trigger i.e. minimum firing angle to converter.
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 XL 
−1  ωL 
α = φ = tan −1 
 = tan 

 R 
 R 
 2π× 50 ×16 ×10−3 
α = φ = tan −1 
 = 45.1°
5


The
current
α = φ, γ = π
ITrms
 1
= 
 2π

π+α
∫
α
flowing
SCR
is
2

 Vm

 sin ( ωt − α )  .dωt 
 2


1
max
at
their
angle
ie.
when
2
Vm
2 × 230
=
2z
2 × 52 + 5.042
ITrms =
∴ ITrms = 22.9 ≈ 23A
55.
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 µs is
applied to the SCR. The maximum value of R in Ω to ensure successful firing of the SCR is
_________.
SCR
100V
500 Ω
+
200 mH
R
Answer: 6060Ω
Exp:
I L = 40 mt
Width of gate pulse
t = 50 µ sec
When SCR in ON with given pulse width of gate
I = I1 + I2
I =
τ =
)
(
−t
V
V
1− e τ +
R1
R2
L
200 × 10−3
=
R
500
Time constant of RL circuit, τ = 0.4 ×10−3
−6
40 ×10−3
100
=
500
−50 ×10

−3
 1 − e 0.4×10



 + 100
 R2

∴ R = 6060 Ω
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GATE Previous Year
Solved Paper
Electrical Engineering
(Fully Solved)
ORBITMENTOR.COM
GATExplore.com
2015,2016,2017,2018,2019
(2015, 2016, 2017)
Free Download
EE-GATE-2015 PAPER-01|
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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
Which of the following combinations is incorrect?
(A) Acquiescence – Submission
(B) Wheedle – Roundabout
(C) Flippancy – Lightness
(D) Profligate – Extravagant
Answer:
2.
(B)
Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected,
one from each set. What is probability that the sum of the two numbers equals 16?
(A) 0.20
Answer:
Exp:
(B) 0.25
(C) 0.30
(D) 0.33
(A)
4  5  20 Total mass
5,11 
4,12 
 4 favorable
3,13 
2,14 

3.
4 1
  0.2
20 5
Which of the following options is the closest in meaning to the sentence below?
She enjoyed herself immensely at the party.
(A) She had a terrible time at the party.
(B) She had a horrible time at the party.
(C) She had a terrific time at the party
(D) She had a terrifying time at the party
Answer:
4.
(C)
Based on the given statements, select the most appropriate option to solve the given
question.
If two floors in a certain building are 9 feet apart, how many steps are there in a set of
stairs that extends from the first floor to the second floor of the building?
Statements:
(I) Each step is ¾ foot high.
(II) Each step is 1 foot wide.
(A) Statement I alone is sufficient, but statement II alone is not sufficient.
(B) Statement II alone is sufficient, but statement I alone is not sufficient.
(C) Both statements together are sufficient, but neither statement alone is sufficient.
(D) Statement I and II together are not sufficient.
Answer: (D)
Exp:
Though we know height of each step, and of strains is not mentioned.
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Didn’t you buy _________ when you went shopping?
(A) any paper
Answer:
(B) much paper
(C) no paper
(D) a few paper
(A)
Q. No. 6 – 10 Carry Two Marks Each
6.
The given statement is followed by some courses of action. Assuming the statement to be
true, decide the correct option.
Statement:
There has been a significant drop in the water level in the lakes supplying water to the
city.
Course of action:
(I) The water supply authority should impose a partial cut in supply to tackle the
situation.
(II) The government should appeal to all the residents through mass media for minimal
use of water.
(III) The government should ban the water supply in lower areas.
(A) Statements I and II follow.
(B) Statements I and III follow
(C) Statements II and III follow.
(D) All statements follow.
Answer: (A)
7.
The number of students in a class who have answered correctly, wrongly, or not attempted
each question in an exam, are listed in the table below. The marks for each question are
also listed. There is no negative or partial marking.
Q No
Marks
Answered
Correctly
Answered
Wrongly
Not
Attempted
1
2
21
17
6
2
3
15
27
2
3
1
11
29
4
4
2
23
18
3
5
5
31
12
1
What is the average of the marks obtained by the class in the examination?
(A) 2.290
Answer:
Exp:
(B) 2.970
(C) 6.795
(D) 8.795
(B)
21  2  15  3  11  11  1  23  2  31  5
 2.970
21  15  11  23  31
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The pie chart below has the breakup of the number of students from different departments
in an engineering college for the year 2012. The proportion of male to female students in
each department is 5:4. There are 40 males in Electrical Engineering. What is the
difference between numbers of female students in the Civil department and the female
students in the Mechanical department?
Electrical
20%
Computer
Mechanical
science
10%
20%
Civil
30%
Answer:
16
Electrical malestudents  40
Exp:
 Electrical Femalestudents 
4
 40  32
5
Total no.of Student  72.
% Female
20 32
30
48
 Difference is 16.
9.
Select the alternative meaning of the underlined part of the sentence.
The chain snatchers took to their heels when the police party arrived.
(A) took shelter in a thick jungle
(B) open indiscriminate fire
(C) took to flight
(D) unconditionally surrendered
Answer:
(C)
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The probabilities that a student passes in Mathmatics, Physics and Chemistry are m,p, and
c respectively. Of these subjects, the student has 75% chance of passing in at least one, a
50% chance of passing in at least two and a 40% chance of passing in exactly two.
Following relations are drawn in m, p, c:
(I) p + m + c = 27/20
(II) p + m + c = 13/20
(III) (p)  (m)  (c) = 1/10
(A) Only relation I is true
(B) Only relation II is true
(C) Relations II and III are true.
(D) Relations I and III are true.
Answer:
Exp:
(A)
P(atleast two)  p(exat 2)
 0.5  0.4  0.1
0.75  p  m  c  0.1  (0.5  0.11 2)
 p  mc  0.65  0.7
 1.35
27

20
Electrical Engineering
Q. No. 1 – 25 Carry One Mark Each
1.
A moving average function is given by y  t  
signal of frequency
1 t
u    dt. If the input u is a sinusoidal
T t T
1
Hz, then in steady state, the output y will lag u (in degree) by
2T
________.
Answer:
Exp:
90
u(τ) = sin (ωτ)
  2f  2.
1


2T T
T  
t T
cos   
1
y  t    sin    d 
T t T
T t
t

1
cos   t  T   cos t 

1
cos t cos T  sin t sin T  cos t 

2
2
y  t    cos t  sin  90  t 


x  t   sin t

  90
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Consider a one-turn rectangular loop of wire place in a uniform magnetic field as shown
in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the
loop is 0.4, and its inductance is negligible. The magnetic flux density (in Tesla) is a
function of time, and is given by B  t   0.25sin t, where   2 50 radian/second.
The power absorbed (in Watt) by the loop from the magnetic field is __________.
10 cm
5 cm
Answer:
Exp:
0.192
2
Vemf
R
 d
Vemf 
dt
P

1
 B.dS  B.S.  800 sin t
S
d 1
  cos t
dt
8
2

1
p  cos 2 t 
64
R
2

1  cos 2t 
p

0.4  64 
2

Vemf 
2
2

cos 2t
20  0.4  64 0.4  64  2
2

 0.192W
20  0.4  64
pavg 
pavg
3.
If the sum of the diagonal elements of a 2 × 2 matrix is 6, then the maximum possible
value of determinant of the matrix is ________.
Answer:
Exp:
9
Sum of the diagonals elements is -6 for 2×2 matrix
The possible eigen value are
 1, 5 5, 1, 8, 2
2, 3 4, 2 9,3     
3, 1 3, 3 10, 4
Maximum possible value of determinant is -3×-3 = 9.
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When the Wheatstone bridge shown is used to find value of resistance Rx, the
Galvanometer G indicates zero current when R1  50 , R 2  65  & R 3  100 . If R3 is
known with 5% tolerance on its nominal value of 100 , what is range of Rx in ohms?
R2
R1
G
R3
Rx
V


(A) [123.5, 136.5]
(C) [117, 143]
Answer:
Exp:
(B) [125.898, 134.12]
(D) [120.25, 139.75]
(A)
Weinbridge is balanced, R1, Rx = R2R3
50×Rx = 65×100
Rx = 130
Now R3 = 100±100×0.05 = 100±5 = 95/105
Rx 
R 2 R 3 65  105

 136.5 
R1
50
65  95
 123.5
50
Rangeof R x is123.5 to136.5 
Rx 
5.
For the given circuit the Thevenin equivalent is to be determined. The Thevenin voltage,
VTh (in volt), seen from terminal AB is _________.
20i
1


2V
1
i

A
2
B
Answer:
Exp:
3.36
Vth = 2i1
2 = 1[i+i1]+i = 2i+i1
i(1) = -20i + 2i1
∴ 21i = 2i1
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 2
i    i1
 21 
25
 2
 4

2  2i  i1  2   i1  i1    1 i1  i1
21
 21 
 21 
42
 1.68
25
Vth  2i1  3.36V
i1 
6.
The impulse response g(t) of a system, G, is as shown in Figure (a). What is the maximum
value attained by the impulse response of two cascaded blocks of G as shown in Figure
(b)?
gt
1
G
0
(A)
Answer:
Exp:
t
1
a 
2
3
G
b
(B)
3
4
(C)
4
5
(D) 1
(D)
Overall impulse response = g(f)*g(t)
h(f) = g(f)*g(f)
ht
1
m 1
m  1
1
7.
Base load power plants are
P: wind farms.
Q: run-of-river plants.
R: nuclear power plants.
S: diesel power plants.
(A) P, Q and S only (B) P, R and S only
Answer:
(C) P, Q and R only (D) Q and R only
(D)
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Of the four characteristic given below, which are the major requirements for an
instrumentation amplifier?
P: High common mode rejection ratio
Q: High input impedance
R: High linearity
S: High output impedance
(A) P, Q and R only (B) P and R only
Answer:
Exp:
(C) P, Q and S only (D) Q, R and S only
(A)
Additional characteristics include very low DC offset, low drift, low noise, very high
open-loop gain, very high common-mode rejection ratio, and very high input impedances.
Instrumentation amplifiers are used where great accuracy and stability of the circuit both
short and long-term are required.
9.
A random variable X has probability density function f(x) as given below:
a  bx for 0  x  1
f x  
otherwise
 0
If the expected value E  X  2 3, then Pr  X  0.5 is __________.
Answer:
Exp:
0.25

 f  x  dx  1
so   a  bx  dx  1

1
0
b
1
2
2a  b  2 ____ 1
a
given E  X   2 3 
 x a  bx  dx
1
0
2 a b
 
3 2 3
3a  2b  4 ____  2 
from 1 and  2 
a0
b2
p r  X  0.5   f  x  dx  2  x dx  0.25
0.5
0.5
0
0
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10.
1
ˆ where r is the distance from the origin and r̂ is the unit
r,
r2
vector in the radial direction. The divergence of the function over a sphere of radius R,
which includes the origin, is
Consider a function f 
(A) 0
Answer:
Exp:
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F
(B) 2
(C) 4
(D) R
(A)
1
ar
r2
.F 
1  2
1 
1 F
r Fr  
 sin F  

2
r r
r sin  
r sin  
1  2 1
r  2  0 0
r 2 r 
r 
.F  0
.F 
11.
A separately excited DC generator has an armature resistance of 0.1 and negligible
armature inductance. At rated field current and rated rotor speed, its open-circuit voltage
is 200 V. When this generator is operated at half the rated speed, with half the rated field
current, an un-charged 1000 F capacitor is suddenly connected across the armature
terminals. Assume that the speed remains unchanged during the transient. At what time
(in microsecond) after the capacitor is connected will the voltage across it reach 25V?
(A) 62.25
Answer:
Exp:
(B) 69.3
(C) 73.25
(D) 77.3
(B)
E b2 N2 2 0.5N1  0.51


 E b2  0.25  E b1  0.25  200  50
E b1 N12
N1  51
  R  C  0.1 1000
6
50  2000e t 10010  t  69.3 sec
12.
In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are
sufficiently large to ensure continuous inductor current and ripple free capacitor voltage,
the charging current (in Ampere) of the 5 V battery, under steady-state, is ________.
S
20V
Answer:
Exp:


L
3
C

5V

1
V0  DVS  0.4  20  8V
I0 
V0  E 8  5 3

  1A
R
3
3
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If a continuous function f(x) does not have a root in the interval [a, b], then which one of
the following statements is TRUE?
(A) f  a  .f  b   0
(B) f  a  .f  b   0
(C) f  a  .f  b   0
(D) f  a  f  b   0
Answer: (C)
Exp: We know that, (Intermediate value theorem)
If f  a  f  b   0 then f  x  has at least one root in (a, b)
f(x) does not have root is (a, b) means f  a  f  b   0
14.
The primary mmf is least affected by the secondary terminal conditions in a
(A) power transformer
(B) potential transformer
(C) current transformer
Answer: (B)
Q15.
(D) distribution transformer
Consider a HVDC link which uses thyristor based line-commutated converters as shown in the
figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends,
and the current, are given by: V1 =500 kV, V2 =485 kV and I=1.5 kA. If the direction of power
flow is to be reversed (that is, from System 2 to System 1) without changing the electrical
connections, then which one of the following combinations id feasible?
System1
System 2
I


V1
V2


If power is to be reversed
(A) V1  500kV, V2  485kV and I  1.5kA
(B) V1  485kV, V2  500kV and I  1.5 kA
(C) V1  500kV, V2  485kV and O  1.5kA
(D) V1  500kV, V2  485kV, I  1.5kA
Answer:
Exp:
(A)
V  V2
I 1
R
For power to be reversed
I
V2  V1
 Ve 
R
I

V1


V2

V1  500kV; V2  485kV; I  1.5kA
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An inductor is connected in parallel with a capacitor as shown in the figure.
i
L
C
Z
As the frequency of current i is increased, the impedance (Z) of the network varies as
Inductive
(A)
Inductive
(B)
z
z
f
f
Capacitive
Capacitive
(C)
Inductive
(D)
Capacitive
z
z
Inductive
f
f
Answer:
Exp:
(B)
Z = ZL//ZC
jL 
Z
Z
Capacitive
Inductive
1
jC

1 
 jL 

jC 

jL
1   1  LC 

z
f
Capacitive
 L 
Z  j
2

1   LC 
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For the signal-flow graph shown in the figure, which one of the following expressions is
Y(s)
equal to the transfer function
?
X 2 (s) X (s) 0
1
X 2 (s)
X1 (s)
1
G2
G1
1
1
(A)
Answer:
Exp:
G1
1  G 2 (1  G1 )
(B)
Y(s)
G2
1  G1 (1  G 2 )
(C)
G1
1  G1G 2
(D)
G2
1  G1G 2
(A)
P1  G 2
  1   G1G 2  G1   1  G1 1  G 2 
TF 
18.
P11
G2


1  G1 1  G 2 
The voltages developed across the 3 and 2 resistors shown in the figure are 6V and
2V respectively, with the polarity as marked. What is the power (in Watt) delivered by the
5V voltage source?

6v

3

2v

Network 1
Network 2
5

(A) 5
Answer: 5
6V
 2A
Exp: I 
3
(B) 7
5v

(C) 10
(D) 14
2V
 1A
2
I 1  2
I  1A
I
P  5  1  5W
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The self inductance of the primary winding of a single phase, 50 Hz, transformer is 800
mH, and that of the secondary winding is 600 mH. The mutual inductance between these
two windings is 480 mH. The secondary winding of this transformer is short circuited and
the primary winding is connected to a 50 Hz, single phase, sinusoidal voltage source. The
current flowing in both the winding is less than their respective rated currents. The
resistance of both windings can be neglected. In this connection, what is the effective
inductance (in mH) seen by the source?
(A) 416
Answer:
(A)
Exp:
I1
(B) 440
R1
(C) 200
R2
(D) 920
I2
M

x1
V1
x2
ZL

V1
2 M 2
  R1  jX1  
I1
R 2  jX 2  ZL
Given, L1 = 800 mH
L2 = 600 mH
M = 480 mH
W = 314 rad/sec
ZL = 0
R1 R2 neglected

2 M 2
2 M 2 
Zin  jX1 
 j  X1 

jX 2
X2 

Zin 

3142  0.482 
 j 314  0.8 
  j 251.32  120.576
0.6  314 

 j130.744  jw Leff  j314.Leff
Leff  0.416  416 mH
20.
A Bode magnitude plot for the transfer function G(s) of a plant is shown in the figure.
Which one of the following transfer functions best describes the plant?
20log G(j2f )
20
0
20
0.1 1 10 100 1k 10k 100k
f (Hz)
Capacitive
(A)
1000(s  10)
s  1000
(B)
10(s  10)
s(s  1000)
(C)
s  1000
10s(s  10)
(D)
s  1000
10(s  10)
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(D)
Exp:
1


K.1 
.S 
1000

G  S  
1 

1  S 
 10 
0dB
20dB
0dB
MdB = 20dB @ initial frequency
20 log M = 20
10
1k
1p  1z 
20 log K = 20
K = 10
G  S 
21.
10 S  1000  10
1000 S  10   1

S  1000 
10 S  10 
In the 4×1 multiplexer, the output F is given by F  A  B. Find the required input
'I3I2 I1I0 '.
I0
I1
4 1
MUX
I2
F
I3
S1
A
(A) 1010
Answer:
Exp:
(B) 0110
S0
B
(C) 1000
(D) 1110
(B)
F  A  B  AB' A'B
00
01
10
11
AB
S1S0
A 'B' I0 0
A 'B I1 1
AB' I 2 1
AB I3 0
I0  0
I1  1
I2  1
I3  0
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22.
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In the given circuit, the silicon transistor has   75 and collector voltage Vc  9V. Then
the ratio of RB and RC is _______.
15V
RB
RC
VC
Answer:
Exp:
105.1
IC  I B 
6
RC
8.3
,
RB
IB 
   75, IC  I B
 76  IB 
76 
6
8.3
, IB 
RC
RB
8.3
6

RB RC
R B 76  8.3

 105.1
RC
6
23.
A (0-50A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full
scale deflection. The external shunt resistance (in milliohms) needed to extend its range to
(0-500A) is ________.
Answer:
Exp:
0.22
I2  500, I1  500
I2  I1  450
450  R sh  0.1 R sh  0.1 / 450  0.22m
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24.
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Consider the circuit shown in the figure,. In this circuit R  1k, and C  1F. The input
voltage is sinusoidal with a frequency of 50 Hz, represented as phasor with magnitude Vi
and phase angle 0 radian as shown in the figure. The output voltage is represented as a
phasor with magnitude V0 and phase angle  radian. What is the value of output phase
angle (in radian) relative to the phase angle of the input voltage?
R
C

vo  Vo 
vi  Vi 0

C
R
(A)
Answer:
Exp:
(B) 
0
(C)

2
(D) 

2
0
Vt
SCR
 SCR 

 Vt 
 Vin
Vn 1  SCR
 1  SCR 
VCn  V V  Vo

1
R
SC
SCR

 SCR Vin
SCR  Vin 
Vin  
 Vo
1  SCR

 1  SCR
Vo  0;   0
25.
A steady current I is flowing in the –x direction through each of two infinitely long wires
L
at y   as shown in the figure.
z
2
The permeability of the medium

is  0 . The B  field at (0,L,0) is
(A) 
40 I
ẑ
3L
4 I
(B)  0 ẑ
3L
(C) 0
30 I
ẑ
4L
(A)
(D) 
Answer:
Exp:
y  L / 2
Current  I
yL/2
0
Current  I
x
H = H1+H2
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
I
 a z  
2  L 2 
z
I
 a z 
 3L 
2  
 2 
I
2 2
 a z    
2
 L 3L 
4I

 a z 
3L
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
y
x
 L 
,0 
 0,
 2

 L 
 0, ,0 
 2 
Q. No. 26 – 55 Carry Two Marks Each
26.
Consider a discrete time signal given by
x[n]  (0.25)n u[n]  (0.5)n u[n  1]
The region of convergence of its Z-transform would be
(A) the region inside the circle of radius 0.5 and centered at origin.
(B) the region outside the circle of radius 0.25 and centered at origin.
(C) the annular region between the two circles, both centered at origin and having radii
0.25 and 0.5.
(D) the entire Z plane.
Answer: (C)
Exp:
x  n    0.25  u  n    0.5  u  n  1


 
n

ROC1 : Z  0.25
n

ROC2 : Z  0.5
ROC  ROC1  ROC2
0.25  Z  0.5
27.
Two players, A and B, alternately keep rolling a fair dice. The person to get a six first
wins the game. Given that player A starts the game, the probability that A wins the game
is
(A) 5/11
(B) 1/2
(C) 7/13
(D) 6/11
Answer: (D)
6 1
Exp: Probability of getting 6 is 

36 6
1
i.e,. Probability of A wins the game 
6
Probability of A not wins the game  1 
Probability of B wins the game 
1 5

6 6
1
6
5
6
If a starts the game, Probability A win the game
Probability of B not win the game 
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 P  A   P  A  P  B  P  A   P  A  P  B  P  A  P  B  P  A   .....
1 551 55551


 ..........
6 666 66666
1 55 5555

 1 

 .....
6 66 6666


2
4

1 5 5
 1        .....
6   6   6 





1
1
  13  6
 
6   5 2 
6 6 11
1    
 6 
28.
The circuit shown in meant to supply a resistive load R L from two separate DC voltage
sources. The switches S1 and S2 are controlled so that only one of them is ON at any
instant. S1 is turned on for 0.2 ms and S2 is turned on for 0.3 ms in a 0.5 ms switching
cycle time period. Assuming continuous conduction of the inductor current and negligible
ripple on the capacitor voltage, the output voltage V0 (in Volt) across RL is ________.
S1
L
RL
S2
10V 

Answer:

Vo
C
 5V


(7)
Exp:
V0
10V
5V
t  msec 
0.2
V0 
0.5
10  0.2  5  0.3
 7V
0.5
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29.
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Determine the correctness or otherwise of the following Assertion [a] and the Reason p[r].
Assertion: Fast decoupled load flow method gives approximate load flow solution
because it uses several assumptions.
Reason: Accuracy depends on the power mismatch vector tolerance.
(A) Both [a] and [r] are true and [r] is the correct reason for [a].
(B) Both [a] and [r] are true and [r] is not the correct reason for [a].
(C) Both [a] and [r] are false.
(D) [a] is false and [r] is true.
Answer:
30.
(A)
In the given circuit, the parameter k is
positive, and the power dissipated in the
2 resistor is 12.5 W. The value of k is
4V
________.
5
2

Vo 
10


5A
kV0
Answer:
Exp:
0.5
P2  12.5 W
12.5
 2.5
2
V0  2  2.5  5V
i 2 
2.5  KV0  5
KV0  2.5
K
31.
2.5 1
  0.5
5 2
5
In the signal flow diagram given in the
figure, u1 and u2 are possible inputs whereas
y1 and y2 are possible outputs. When would
the SISO system derived from this diagram
u1
be controllable and observable?
(A) When u1 is the only input and y1 is the
only output.
(B) When u2 is the only input and y1 is the
only output.
(C) When u1 is the only input and y2 is the
only output.
(D) When u2 is the only input and y2 is the
only output.
Answer:
(B)
y1
x1
1/ s
1
1
1
2
u2
1/ s
1
x2
1
1
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y2
Noted-: Single Source Follow, Revise
Multiple Time Best key of Success
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32.
In a linear two-port network, when 10 V is applied to Port 1, a current of 4 A flows
through Port 2 when it is short-circuited. When 5V is applied to Port1, a current of 1.25 A
flows through a 1 resistance connected across Port 2. When 3V is applied to Port 1,
then current (in Ampere) through a 2 resistance connected across Port 2 is _________.
Answer:
Exp:
www.gateforum.com
0.545
I2  0.4  3  0.6  2I 2 
I1  y11v1  y12 v1
I2  y 21v1  y 22 v 2
 1.2  1.2I 2
4  10y 21  y 21  0.4 I2  0.545A.
1.25  0.4v1  1.25y22  0.4
y 22  0.6
33.
A self commutating switch SW, operated at duty cycle  is used to control the load
voltage as shown in the figure.
D
VL
L

SW
Vdc
VC
C
RL
Under steady state operating conditions, the average voltage across the inductor and the
capacitor respectively, are
(A) VL  0 and VC 
1
Vdc
1 

1
(B) VL  Vdc and Vc 
Vdc
2
1 
(C) VL  0 and VC 

Vdc
1 
(D) VL 
Answer:
34.
(A)
The figure shown a digital circuit constructed using negative edge triggered J-K flip flops.
Assume a starting state of Q2Q1Q0  000. This state Q2Q1Q0  000 will repeat after
________ number of cycles of the clock CLK.
J0
1
CLK
1
Answer:
Exp:


Vdc and VC 
Vdc
2
1 
J1
Q0
Q0
1
K1
Q2
Clock
Clock
Clock
K0
J2
Q1
Q0
1
K2
Q2
6
First flip flop acts as mod-2 counter
Second 2 flip flops from mod (2n-1) Johnson counter = mod counter
∴ overall modulus = mod – 6 counter
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35.
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The signum function is given by
x
 ;x  0
sgn(x)   x
0; x  0

The Fourier series expansion of sgn(cos(t)) has
(A) only sine terms with all harmonics.
(B) only cosine terms with all harmonics
(C) only sine terms with even numbered harmonics.
(D) only cosine terms with odd numbered harmonics.
Answer: (D)
Exp: sgn(cos t)  1; cos t  0
 1;cos t  0
cos t
t
sign(cos t)
t
1
it represents square wave, which is even and half wave symmetry function, it contains
cosine terms for all odd harmonics.
36.
A DC motor has the following specifications: 10 hp, 37.5 A, 230V; flux/pole = 0.01 Wb,
number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature
resistance = 0.267. The armature reaction is negligible and rotational losses are 600W.
The motor operates from a 230V DC supply. If the motor runs at 1000 rpm, the output
torque produced in (in Nm) is __________.
Answer:
Exp:
E
14.14
2Np 0.01 666  4  1000

 55.5
60A
60  2
Internal power =EI=55.5×37.5=2081.25
Pout=2081.25-600=1481.25
T
Pout 1481.25

 14.14 Nm
1000
w
2 
60
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37.
Find the transfer function
Y s
X s
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of the system given below.

G1


X  S
H

Y  S

Y  S

G2
(A)
G1
G2

1  HG1 1  HG 2
(B)
G1
G2

1  HG1 1  HG 2
(C)
G1  G 2
1  H  G1  G 2 
(D)
G1  G 2
1  H  G1  G 2 
Answer:
Exp:
(C)
From the block diagram
Y  G1  X  HY   G 2  X  HY 
Y  X  G1  G 2   HY  G1  G 2 
Y 1  H  G1  G 2    X  G1  G 2 

38.
G1  G 2
Y

X 1  H  G1  G 2 
The transfer function of a second order real system with a perfectly flat magnitude
response of unity has a pole at (2-j3). List all the poles and zeroes.
(A) Poles at (2±j3), no zeroes
(B) Poles at (±2-j3), one zero at origin
(C) Poles at (2-j3), (-2+j3), zeroes at (-2-j3), (2+j3)
(D) Poles at (2±j3), zeroes at (-2±j3)
Answer:
Exp:
(D)
This is an APF 2  3j
Im
2  3j
0
Re
2  3j
2  3j
39.
Two single-phase transformers T1 and T2 each rated at 500 kVA are operated in parallel.
Percentage impedances of T1 and T2 are (1+j6) and (0.8+j4.8), respectively. To share a
load of 1000 kVA at 0.8 lagging power factor, the contribution of T 2 (in kVA) is
_________.
Answer: 555
Exp:
ST2  S 
6.0880.53
z1
 1000 
 555KVA
10.9480.53
z1  z 2
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40.
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A parallel plate capacitor is partially filled with glass of dielectric constant 4.0 as shown
below. The dielectric strengths of air and glass are 30 kV/cm and 300 kV/cm,
respectively. The maximum voltage (in kilovolts), which can be applied across the
capacitor without any breakdown, is _______.
Air, r  1.0
5mm
10mm
Glass, εr = 4.0
Answer:
Exp:
18.75
A 0
C1 
d
4A 0
C2 
d
4A 0
CC
Ceq  1 2 
C1  C2
5d
air
0
30kV cm
1
5mm
 2
5mm
C1
C2
glass
40
30kV cm
Q CV Ceqv


A A
A
4A 0
Dn 
V
5dA
 4 
Dn   0  V
 5d 
D
4
E1  n  V
0 5d
D n  s 
4V
30  5  5  103  105
V
4d
4
V  18.75kV
30  105 
41.
A sustained three-phase fault occurs in the power system shown in the figure. The current
and voltage phasors during the fault (on a common reference), after the natural transients
have died down, are also shown. Where is the fault located?
I3
I1
Transmission line
V1
P
V2
S
Q
Transmission line
R
I4
I2
V2
V1
I3
I2
I1
I4
(A) Location P
(B) Location Q
(C) Location R
(D) Location S
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Answer:
Exp:
www.gateforum.com
(D)
V2
I3
Since V2 leads I3
42.
 3 0 2 


The maximum value of “a” such that the matrix  1 1 0  has three linearly
 0 a 2 


independent real eigenvectors is
(A)
Answer:
Exp:
2
(B)
3 3
1
(C)
3 3
1 2 3
3 3
(D)
1 3
3 3
(B)
The characteristic equation of A is
|A-XI| = 0
f(x) = x3+6x2+11x+6+2a
= (x+1)(x+2)(x+3)+2a = 0
f(x) cannot have all 3 real roots (if any) equal
for if f(x) = (x-k)3, then comparing coefficients, we get
6 = -3k, 3k2 = 11
No such k exists
(a) Thus f(x) = 0 has repeated (2) roots (say) α,α,β
or
(b) f(x) = 0 has real roots (distance)(say) α,β,δ
Now f '  x   0  x1 
6  3
 2.577a;
3
x2 
6  3
 1.422
3
At x1, f(x) has relative max.
At x2, f(x) has relative min.
The graph of f(x) will be as below
y
y
Max.
x1
Max.
Min.
x
x2
x
x1
 x1 is repeated root 
x2
Min.
 x 2 is repeated root 
Case (a) repeated roots (α,α,β)
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y
Max.
x
x2
x1
Min.
Case (b)
distinct roots
Note that the graph of f(x) cannot be like the one given below
y
3realroots not possible
x
x1
x2
Thus in all possible cares we have

3
1
f(x2)≤0  2  a 
  0  a 
9 
3 3

43.
The open loop poles of a third order unity feedback system are at 0,-1,-2. Let the frequency
corresponding to the point where the root locus of the system transits to unstable region be K.
Now suppose we introduce a zero in the open loop transfer function at -3, while keeping all the
earlier open loop poles intact. Which one of the following is TRUE about the point where the
root locus of the modified system transits to unstable region?
(A) It corresponds to a frequency greater than K
(B) It corresponds to a frequency less than K
(C) It corresponds to a frequency K
(D) Root locus of modified system never transits to unstable region
Answer:
(D)
44.
A 200/400V, 50 Hz, two-winding transformer is rated at 20 kVA. Its windings are
connected as an auto-transformer of rating 200/600V. A resistive load of 12 is
connected to the high voltage (600V) side of the auto-transformer. The value of
equivalent load resistance (in Ohm) as seen from low voltage side is _________.
Answer: (4)8
V1
200
K
 K  0.5
Exp:
V2
400
2
 1 
R L1  R L2 
  R 4  12  4  48
 1  0.5 
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45.
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Consider the economic dispatch problem for a power plant having two generating units.
The fuel costs in Rs/MWh along with the generation limits for the two units are given
below:
C1  P1   0.01P12  30P1  10; 100MW  P1  150MW
C2  P2   0.05P22  10P2  10; 100MW  P2  180MW
The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is _____.
Answer: (30)
dC1
Exp:
 2  0.01P1  30  0.02 P1  30
dP1
dC2
 2  0.05P2  10  0.1P2  10
dP2
dC1 dC2

dP1 dP2
0.02P1  30  30  0.1P2  10
2P1  3000  10P2  1000  2P1  2000  10P2  P1  P2  200
P2  200; P1  0
dC1
 30Rs / Mwh
dP1
46.
An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the
magnitude of V0 be maximum?
1  x 
(A)
(B) (1+x)
(D)
Answer:

V0

1  x 
(C) 1
R 1  x 
pR
pR
1  x 
(A)
R
E
Exp:
R 1  x 
PR

V0

PR

PR
PR
E 

V0
R
R
R 1  x   Ry

E
Let1  x  y
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RE
E

R  PR 1  P
 y 
R.y
V0   
.E  
E
Ry  PR
py
 y
1 
V0  V0    V0    

E
p  y 1 p 
dV0   y
1 


E  0
dp   p  y 2 1  p 2 


1
y

1  P 2  p  y 2
V0   
y
1

 py y p y
1 p p  y


p 1  y  y  y  y 1  y 
p   y   1 x
47.
A separately excited DC motor runs at 1000 rpm on no load when its armature terminals
are connected to a 200V DC source and the rated voltage is applied to the field winding.
The armature resistance of this motor is 1. The no-load armature current is negligible.
With the motor developing its full load torque, the armature voltage is set so that the rotor
speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the
same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational
losses, the full load armature current (in Ampere) is _______.
Answer:
(100)
Exp:
N 0  1000 rpm  E  N

E 0  200V  200 1000

N full  500 rpm  E full 500
E full  100V  V  Ia ra  200  Ia
Ia  100A
48.
A solution of the ordinary differential equation
and y 1  
Answer:
Exp:
d2 y
dy
 5  6y  0 is such that y(0) = 2
2
dt
dt
1  3e
dy
. The value of
 0  is ___________.
3
e
dt
(-3)
Roots, 3, 2
y  t   C1e 3t  C2 e 2t
y  0   C1  C2  2
 1  3e 
y 1    3   e 3  3e 2  C1e 3  C2 e 2
 e 
So, C1  1, C2  3
So,
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y  t   e3t  3e 2t
dy  t 
dt
49.
 3e3t  6e2t ,
dy  0 
dt
 3  6  3
C
The op-amp shown in the figure has a
finite gain A = 1000 and an infinite input
R
resistance. A step-voltage Vi = 1 mV is
applied at the input at time t = 0 as
1k
vi
shown. Assuming that the operational 1mV


amplifier is not saturated, the time
constant (in millisecond) of the output t  0s
voltage V0 is
(A) 1001
(B) 101
(C) 11
(D) 1
Answer:
1F

A  1000


Vo

(D)
Exp:
Time constant = RC= 1103 1106  1ms
50.
A 3-phase 50 Hz square wave (6-step) VSI feeds a 3-phase, 4 pole induction motor. The
VSI line voltage has a dominant 5th harmonic component. If the operating slip of the
motor with respect to fundamental component voltage is 0.04, the slip of the motor with
respect to 5th harmonic component of voltage is ________.
Answer:
5.8
Exp: Slip of motor w.r.t. 5th harmonic = 6-5s = 6-5×0.04= 5.8
51.
An 8 bit unipolar Successive Approximation Register type ADC is used to convert 3.5V
to digital equal output. The reference voltage is +5V. The output of ADC at end of 3 rd
clock pulse after the start of conversion is ________.
(A) 1010 0000
(B) 1000 0000
(C) 0000 0001
(D) 0000 0011
Answer: (A)
Exp: The block diagram of SAR type ADC is as follows
Vin

VDAC

Control logic
Start of conversion
CLOCK
1st CP  2.56 V
2nd CP  3.84 V
Output Register
3rd CP  3.2 V
8bit DAC
Unipolar means all the voltages will be +ve i.e. nothing is –ve.
The functionality of SAR type DAC is, it will load a value to output register with MSB=1
and remaining bit=0, and it will cross check a logic as follows.
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if Vin  VDAC  ma int ain the loaded bit
Vin  VDAC  clear the loaded bit.
This process continues upto 8 number of clock pulses
The output of DAC=(Resolution)×(Decimal equivalent of applied binary).
From the given information
Resolution 
5
 20 mV.
2 1
8
when SOC is applied
on 1st clock the value located to output register is ' 10000000 2 '  (128)10
then VDAC  128  20mv  2.56V
So 3.5>2.56V  maintain the bit
So at the end of 1st clock pulse the output is 10000000.
On second clock pulse the value loaded to output register is (10100000)2  (192)10
then VDAC  195  20mv  3.84V
So 3.5  3.84V  clear the loaded bit
So at the end of 2nd clock pulse output is (10000000)2 .
On third clock pulse the value loaded to output register is (10100000)2  (160)10
then VDAC  160  20mv  3.2V
So 3.5  3.2V  ma int ain the loaded bit
So at the end of 3rd clock pulse output is (10100000)2 .
52.
The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output
frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of
50 kHz and modulation index of 0.7. For V m = 100 V DC, L = 9.55 mH, C = 63.66 μF,
and R = 5, the amplitude of the fundamental component in the output voltage V0 (in
volt) under steady-state is __________.



Vin
Answer:
Full  bridge V
n
VSI

L
C
R

Vo

(56.72V)
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Exp:
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
 pulse width

    0.7  180  126

2Vdc

sin

2
2  100 126

sin
 56.72V

2
The amplitude of fundamental component in V0 
53.
f(A,B,C,D) = Πm (0,1,3,4,5,7,9,11,12,13,14,15) is a maxterm representation of a Boolean
function f(A,B,C,D) where A is the MSB and D is the LSB. The equivalent minimized
representation of this function is


(A) A  C  D A  B  D

(B) ACD  ABD
(C) ACD  ABCD  ABCD



(D) B  C  D A  B  C  D A  B  C  D
Answer:
Exp:

(C)
f  A,B,C,D   ACD  ABD
In option (C)
f  A, B,C, D   ACD  ABCD  ABCD

 ACD  ABD C  C

 ACD  ABD.1
 ACD  ABD
CD
00
01
11
10
00
0
0
0
1
01
0
0
0
1
11
0
0
0
0
10
1
0
0
1
AB
ACD
ABD
54.
A 50Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating
in steady state at synchronous speed, and producing 1 pu of real power. The initial value
of the rotor angle δ is 5o , when a bolted three phase to ground short circuit fault occurs at
the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the
value of δ in degrees, 0.02 second after the fault is ________.
Answer: (0.9)
Exp:
M
Pavg

2 1
1

PU
180  50 4500
 0.5
0.5
 2250
1 4500 
  45deg sec  1   t  45o  0.2  45o  0.9o
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The circuit shown in the figure has two sources connected in
series. The instantaneous voltage of the AC source (in volt) is
given by (t) = 12 sin t. If the circuit is in steady-state, then
the rms value of the current (in Ampere) flowing in the circuit
is ______.
Answer: (10)
1
1
Exp: Y  S 

Z  S 1  j
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55.
Y  S 
1
1 
2
vt
8V


1
1H
  tan 1  u 
in  t   8  12sin t
1
  tan 1  0  
1 0
12 
1
it  8 
 cos t
 sin t.
2
2
i  t   8  6sin t  6cos t
i  t   8.
2
1.2
sin  t  45 
11
1 

2
2
 6   6 
I rms  82  
 
  10
 2  2
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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
A generic term that includes various items of clothing such as a skirt, a pair of trousers
and a shirt as
(A) fabric
(B) textile
(C) fibre
(D) apparel
Answer:
2.
(D)
Choose the statement where underlined word is used correctly.
(A) The industrialist had a personnel jet.
(B) I write my experience in my personnel diary.
(C) All personnel are being given the day off.
(D) Being religious is a personnel aspect.
Answer:
3.
(C)
Based on the given statements, select the most appropriate option to solve the given
question.
What will be the total weight of 10 poles each of same weight?
Statements:
(I) One fourth of the weight of a pole is 5 kg
(II) The total weight of these poles is 160 kg more than the total weight of two poles.
(A) Statement II alone is not sufficient
(B) Statement II alone is not sufficient
(C) Either I or II alone is sufficient
(D) Both statements I and II together are not sufficient.
Answer: (C)
4.
Consider a function f  x   1  x on 1  x  1. The value of x at which the function
attains a maximum, and the maximum value of function are:
(A) 0, 1
Answer:
Exp:
(B) 1,0
(C) 0, 1
(D) -1, 2
(C)
f  x   1  x on  1  x  1
f  1  1  1  1  1  0
f  0.5   1  0.5  1  0.5  0.5
f  0  1  0  1
f  0.5   1  0.5  0.5
f 1  1  1  1  1  0
 maximum value occurs at x = 0 and maximum value is 1.
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5.
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We ____________ our friend’s birthday and we ___________ how to make it up to him.
(A) completely forgot --- don’t just known
(B) forget completely --- don’t just know
(C) completely forget --- just don’t know
(D) forgot completely --- just don’t know
Answer:
(C)
Q. No. 6 – 10 Carry Two Marks Each
6.
In a triangle PQR, PS is the angle bisector of QPR and QPS  60o. What is the length
of PS?
P
r
q
s
Q
R
p
(A)
Answer:
7.
q  r 
(B)
qr
qr
q
  r
(C)
 q2  r 2 
(D)
q  r 
2
qr
(B)
Four branches of a company are located at M,N,O, and P. M is north of N at a distance of
4 km; P is south of O at a distance of 2 km; N is southeast of O by 1 km. What is the
distance between M and P in km?
(A) 5.34
Answer:
(B) 6.74
(C) 28.5
(D) 45.49
(A)
Exp:
N
M
0
45
2
4
1
E
W
N
S
P
8.
If p, q, r, s are distinct integers such that:
f (p, q, r, s) = max (p, q, r, s)
g (p, q, r, s) = min (p, q, r, s)
h (p, q, r, s) = remainder of (p × q)/(r × s) if (p × q) > (r × s) or remainder of (r × s)/(p × q)
if (r × s) > (p × q)
Also a function fgh (p, q, r, s) = f(p, q, r, s) × g(p, q, r, s) × h(p, q, r, s)
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Also the same operations are valid with two variable function of the form f(p, q).
What is the value of fg (h(2, 5, 7, 3), 4, 6, 8)?
Answer:
Exp:
8
f g (h(2,5,7,3),4,6,8)
=fg(1,4,6,8)
=f(1,4,6,8)xg(1,4,6,8)=8x1=8
9.
If the list of letters, P, R, S, T, U is an arithmetic sequence, which of the following are
also in arithmetic sequence?
I.
2P,2R,2S,2T,2U
II.
P  3,R  3,S 3,T 3, U 3
III. P2 ,R 2 ,S2 ,T2 , U2
(A) I only
(B) I and II
(C) II and III
(D) I and III
Answer:
10.
(B)
Out of the following four sentences, select the most suitable sentence with respect to
grammer and usage:
(A) Since the report lacked needed information, it was of no use to them.
(B) The report was useless to them because there were no needed information in it.
(C) Since the report did not contain the needed information, it was not real useful to them
(D) Since the report lacked needed information, it would not had been useful to them.
Answer: (A)
Electrical Engineering
Q. No. 1 – 25 Carry One Mark Each
1.
Find the transformer ratios a and b that the impedance (Zin) is resistive and equal 2.5
when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s.
C  10F
L  1mH
R  2.5
1: b
1: a
(A) a = 0.5, b = 2.0
(B) a = 2.0, b = 0.5
(C) a = 1.0, b = 1.0
(D) a = 4.0, b = 0.5
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Answer:
Exp:
(B)
 j20
X c   j20 
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j5
X L  j 5
2.5
a2
1: b
Zin 
 2.5

 a 2  j5  j20


1
b2
5
0
b2
b  0.5
 20 
2.5
 2.5  a  2
a 2 b2
2.
The synchronous generator shown in the figure is supplying active power to an infinite
bus via two short, lossless transmission lines, and is initially in steady state. The
mechanical power input to the generator and the voltage magnitude E are constant. If one
line is tripped at time t1 by opening the circuit breakers at the two ends (although there is
no fault), then it is seen that the generator undergoes a stable transient. Which one of the
following waveforms of the rotor angle  shows the transient correctly?
Line1
Synchronous generator
Xs
~
Infinte Bus
10
E
(A)
Line 2

(B)

0
0
t1
0
Answer:
Exp:
time
time

(C)
t1

(D)
t1
time
0
t1
time
(A)
For generator  is +Ve. After fault   increases
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3.
In the following circuit, the input voltage Vin is 100 sin 100t  . For 100RC  50, the
average voltages across R (in volts) under steady-state is nearest to
C
Vin



R
C
(A) 100
Answer:
Exp:
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(B) 31.8

(C) 200
(D) 63.6
(D)
Given circuit is voltage doubler
Vm = Voltage across each capacitor = 100V
Voltage across two capacitors (in steady state)
= 2Vm = 200V
Average voltageacross'R' 
4.
A 4-pole, separately excited, wave wound DC machine with negligible armature
resistance is rated for 230 V and 5 kW at a speed of 1200 rpm. If the same armature coils
are reconnected to forms a lap winding, what is the rated voltage (in volts) and power (in
kW) respectively at 1200 rpm of the reconnected machine if the field circuit is left
unchanged?
(A) 230 and 5
(B) 115 and 5
(C) 115 and 2.5
(D) 230 and 2.5
Answer:
Exp
200
 63.6V

E
(C)
1
no. of parallel paths
230 4

E
2
E  115V
PE
 P  2.5kW
5.
Given f  z   g  z   h  z  , where f, g, h are complex valued functions of a complex
variable z. Which one of the following statements is TRUE?
(A) If f(z) is differential at z0, then g(z) and h(z) are also differentiable at z0.
(B) If g(z) and h(z) are differentiable at z0, then f(z) is also differentiable at z0.
(C) If f(z) is continuous at z0, then it is differentiable at z0.
(D) If f(z) is differentiable at z0, then so are its real and imaginary parts.
Answer:
Exp:
(B)
Given f  z   g  z   h  z 
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f  z  ,g  z  ,h  z  are complex variable functions
(c) is not correct, since every continuous function need not be differentiable
(D) is also not correct
Let g(z) = x h(z) = iy
 g  z   x  i0
ux v0
x
x
1
0
x
x
  0   0
h  z   0  iy
u0 vy
y

0
0
x
x
x

0
1
y
y
Cauchy – rieman equation as of g(z), h(z) are failed.
 f(z) and g(z) are not differentiable
But f  z   x  iy
ux
u
1
x
u u

x y
vy
v
0
x
u v

y x
 f(z) in differentiable
 i.e, f(z) is differential need not imply g(z) and h(z) are differentiable
 Ans (B)
i.e, g(z) and h(z) are differentiable then f(z) = g(z) + h(z) is differentiable.
6.
A 3-bus power system network consists of 3 transmission lines. The bus admittance
matrix of the uncompensated system is
j4 
  j6 j3
 j3  j7 j5  pu.


 j4
j  j8
If the shunt capacitance of all transmission line is 50% compensated, the imaginary part
of the 3rd row 3rd column element (in pu) of the bus admittance matrix after compensation
is
(A)  j 7.0
Answer:
(B)
Exp:
 y10  y12  y13
   y12
  y13
y Bus
(B)  j 8.5
 y12
y 20  y 21  y 23
 y 23
(C)  j 7.5
 y13


 y 23

y30  y31  y32 
(d)  j 9.0
y31  y13   j4
y32  y 23   j5
y30  y31  y32   j8
y30  ( j4)  ( j5)   j8
y30  j1
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after compensating, y30 
www.gateforum.com
j1
2
 y30  y31  y32 (new)  j0.5  j4  j5   j8.5
7.
A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is
200 radians/second. At its rated torque of 500 Nm, its speed is 180 radian/second. The
motor is used to directly drive a load whose load torque T L depends on its rotational speed
(in radians/second), such that TL  2.78  T . Neglecting rotational losses, the steady-state
speed (in radian/second) of the motor, when it drives this load is _________.
Answer:
Exp:
179.86
Under steady state load torque = motor torque
500  2.78  T
 T 178.88 rad/sec
8.
A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-axis
as shown in the figure. The value of  0 is 4107 in SI unit. If a uniform magnetic field

intensity H  107 zˆ A/m is applied, then the peak value of the inducted voltage,
Vturn  in volts  , is __________.
Z
H
Vturn X
Answer:
Exp:
39.45

   V  B .dl
Vemf  E m .dl
Vemf
V  r  1  6.28a 
B  0  107 a z
Vemf 
  6.28a


107 0 a z .dl
 4  107  6.28  107.
Vemf 
1
2
78.91
 39.45V
2
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www.gateforum.com
0.1 F
9.
The operational amplifier shown in the
figure is ideal. The input voltage (in
Volt) is Vi  2sin  2 2000t . The
amplitude of the output voltage Vi
Vo  in Volt  is _________.
Answer:
1k
1 k

Vo

1.96
Exp:
1
103
0.1 106 s 
Z1  1k; Z2 
1
j0.1 103   1
103 
6
0.1 10 s
1 103 
  2000 rad / sec
Z2 
103
103

1  j0.1  2000  103 1  j0.2
V0 
 Z2 Vi
103  2sin(2  2000t) 2sin(2  2000t)


Z1
(1  j0.2)  103
1.0198
V0  1.96sin  22000t 
Amplitude of the output voltage =1.96V
10.
In the following circuit, the transistor is in active mode and VC  2V. To get VC  2V. To
get VC  4V, we replace RC with R C . Then the ratio R C R C is _________.
10V
RC
RB
Answer:
Exp:
VC
0.75
we have Vc  2V;
Ic R c  10  2  8
… (i)
We have Vc=4V
Ic R 'c  10  2  8
… (ii)
(2) Ic R c' 6 R c' 3

 ;
  0.75
(1) Ic R c 8 R c 4
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11.
The Laplace transform of f  t   2 t  is s 3 2 . The Laplace transform of g  t   1 t is
(B) s 1 2
(A) 3s5 2 2
Answer:
Exp:
www.gateforum.com
(C) s1 2
(D) s3 2
(B)
Given that laplace transform of f  t   2
Given as g  f  
 gt 
t
is s 3 2 .

1
t
2 t  f t

2t
2t
f t 1
L g  t   L 

 2t  2
 f  t ds
0
s

 3 1 
1  3 2
1 s 2 

s ds  

2 s
2  3  1 
 2
s
1
1
  2  0  s 1 2   s 1 2 
2
s

12.
A capacitive voltage divider is used to measure the bus voltage V bus in a high-voltage 50
Hz AC system as shown in the figure. The measurement capacitor C1 and C2 have
tolerances of 10% on their normal capacitance values. If the bus voltage Vbus is 100 kV
rms, the maximum rms output voltage Vout (in kV), considering the capacitor tolerance, is
__________.
C1
1F  10%
v bus
C2
Answer:
Exp:
9F  10%
vout
10
 Xc2

Vout  
 Vbus
 Xc1  Xc2 
1

 C
2

1
1

C C
2
 1



 C
1
 Vbus  
1
1



C C
2

 1


1.1 
 Vbus  
  10kV

 1.1  9.9 


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13.
Match the following
P. Stokes’s Theorem
1.

 D.ds  Q
Q. Gauss’s Theorem
2.
 f  z  dx  0
R. Divergence Theorem
3.
 .A  dv  
 A.ds
S. Cauchy’s Integral Theorem
4.
   A .ds   A.dl
(A) P-2, Q-1, R-4, S-3
(B) P-4, Q-1, R-3, S-2
(C) P-4, Q-3, R-1, S-2
(D) P-3, Q-4, R-2, S-1
Answer:
14.
www.gateforum.com
(B)
Consider the following Sum of Products expression, F.
F  ABC  ABC  ABC  ABC  ABC
The equivalent Product of Sums expression is
(A) F   A  B  C   A  B  C  A  B  C 
(B) F   A  B  C   A  B  C   A  B  C 
(C) F   A  B  C  A  B  C   A  B  C 
(D) F   A  B  C  A  B  C   A  B  C 
Answer: (A)
Exp: Given minterm is
F  m  0,1,3,5,7 
F  m  2, 4,6 
So product of sum expression is
F   A  B  C   A  B  C  A  B  C 
15.
A series RL circuit is excited at t = 0 by closing a switch as shown in the figure.
d 2i
Assuming zero initial conditions, the value of 2 at t = 0+ is
dt
R
V
(A)
Answer:
V
L
L
(B)
V
R
(C) 0
(D)
RV
L2
(C)
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Exp:
i L (0 )  0  i L (0 )
at t  0
www.gateforum.com
P


 L (0 )  V

di L (0 ) V

dt
L
2

d i L (0 )
0
dt 2
16.
 L (0 )
V

We have a set of 3 linear equations in 3 unknowns. 'X  Y' means X and Y are
equivalent statements and 'X  Y' means X and Y are not equivalent statements.
P: There is a unique solution.
Q: The equations are linearly independent.
R: All eigenvalues of the coefficient matrix are nonzero.
S: The determinant of the coefficient matrix is nonzero.
Which one of the following is TRUE?
(A) P  R  Q  S
(C) P  Q  R  S
Answer:
17.
(B) P  R  Q  S
(D) P  Q  R  S
(A)
Match the following:
Instrument Type
Used for
P. Permanent magnet moving coil
1. DC only
Q. Moving iron connected through current
R. Rectifier
transformer
2. AC only
3.AC and DC
S. Electrodynamometer
P 1
(A)
Answer:
18.
Q2
R 1
P 1
(B)
S3
(C)
Q3
R 1
P 1
(C)
S2
Q2
R 3
P3
(D)
S3
Q 1
R 2
S 1
Two semi-infinite dielectric regions are separated by a plane boundary at y=0. The
dielectric constant of region 1 (y<0) and region 2 (y>0) are 2 and 5, Region 1 has uniform

electric field E  3aˆ x  4aˆ y  2aˆ z , where aˆ x ,aˆ y , and aˆ z are unit vectors along the x, y and
z axes, respectively. The electric field region 2 is
(A) 3aˆ x  1.6aˆ y  2aˆ z
(B) 1.2 aˆ x  4 aˆ y  2aˆ z
(C) 1.2aˆ x  4aˆ y  0.8aˆ z
(D) 3aˆ x  10aˆ y  0.8aˆ z
Answer:
Exp:
(A)
(1)
(2)
y0
y>0
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E1  3ax  4ay
www.gateforum.com
Ez= y=0
+2az
2
E 2  3a x  (4a y )  2a z
5
E z  3a x  1.6a y  2a z
19.
The filters F1 and F2 having characteristics as shown in Figures (a) and (b) are connected
as shown in Figure (c).
F1
Vo / Vi
F1
Vo / Vi
V0
Vi
V0
Vi
f2
f1
f
f
(b)
(a)
R/2
R

F1
 Vsat

Vi
V0
Vsat
R
F2
(c)
The cut-off frequencies of F1 and F2 are f1 and f 2 respectively. If f1  f 2 , the resultant
circuit exhibits the characteristics of a
(A) Band-pass filter (B) Band-stop filter (C) All pass filter
(D) High-Q filter
Answer:
20.
The figure shows the per-phase equivalent circuit of a two-pole three-phase induction
motor operating at 50 Hz. The “air-gap” voltage, Vg across the magnetizing inductance, is
210 V rms, and the slip, is 0.005. The torque (in Nm) produced by the motor is ______.
Answer:
Exp:
(B)
333.27
Voc across j6.28  Vg  210V
0.0375
j0.2127
j0.22

j6.28  (0.4  j0.22)
(0.04  j0.22)  j6.28
 0.21680.04
R th 
 0.0375  j0.2127
Rotor circuit is
Vg
I2

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Vg
I2 
z

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210
(0.0375  1)  j(0.2127  j0.22)
I2  186.81A
Te 
3
r2
3
 I 22 . 
 186.812  1
s
s 314.15
 333.27 Nm
21.
Nyquist plot of two functions G1 (s) and G 2 (s) are shown in figure.
Im

Im



0
G 2 (s)
G1 (s)
Re
Re


0
Nyquist plot of the product of G1 (s) and G 2 (s) is
(A)
(B)
Im
Im
0
 
(C)
1
Re
(D)
Im
Im
Re




Re
Re

0
Answer:
Exp:
(B)
Im
1
G1 (s)  ; G 2 (s)  5
s
G1G 2 (s)  1
Re
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22.
A 3-phase balanced load which has a power factor of 0.707 is connected to balanced
supply. The power consumed by the load is 5kW. The power is measured by the twowattmeter method. The readings of the two wattmeters are
(A) 3.94 kW and 1.06 kW
(B) 2.50 kW and 2.50 kW
(C) 5.00 kW and 0.00 kW
(D) 2.96 kW and 2.04 kW
Answer:
Exp:
www.gateforum.com
(A)
P1  VL IL cos(30  )
P2 VL IL cos(30  )

3  p1  p 2  
cos   cos  tan 1

p1  p 2 


 3.94  1.06  
o
 cos   tan1 3 
   45
5



satisfiying only for(A)
An open loop control system results in a response of e2t  sin 5t  cos5t  for a unit
impulse input. The DC gain of the control system is _________.
Answer: 0.241
23.
Exp:
g(t)  e2 sin 5t  cos5t 
G(s) 
5
s2

2
2
(s  2)  5 s  2  52
DC gain means G(s) s  0
G(0) 
24.
5
2
7
 2

2
2
2 5
2 5
29
2
When a bipolar junction transistor is operating in the saturation mode, which one of the
following statement is TRUE about the state of its collector-base (CB) and the baseemitter (BE) junctions?
(A) The CB junction is forward biased and the BE junction is reverse biased.
(B) The CB junction is reversed and the BE junction is forward biased.
(C) Both the CB and BE junctions are forward biased.
(D) Both the CB and BE junctions are reverse biased.
Answer:
25.
(C)
The current i (in Ampere) in the 2 resistor of the given network is ____ .
1
i
5V


1
1
2
1
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Answer:
Exp:
www.gateforum.com
0
The Network is balanced Wheatstone bridge.
 i  0 Amp
Q. No. 26 – 55 Carry Two Marks Each
26.
A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated
voltage and frequency. The stator resistance, magnetizing reactance, and core loss are
negligible. The maximum torque produced by the rotor is 225% of full load torque and it
occurs at 15% slip. The actual rotor resistance is 0.03 / phase. The value of external
resistance (in Ohm) which must be inserted in a rotor phase if the maximum torque is to
occur at start is ________.
Answer:
Exp:
0.17
Smt 
r2
x2
0.15 
r2 0.03
 x 2  0.2

x2
x2
For Test=Temax,
Test
2

 1  SmT  1
1
Tem
 SmT
s mT
1
r2 '
 r2'  x 2  0.2 
x2
Extra resistance  0.2  0.03  0.17 / p4
27.
Two three-phase transformers are realized using single-phase transformers as shown in
the figure.
A2
B2
C2
A2
A1
a1
B1
b1
C1
c1
A1
a1
B2
B1
C2
C1
a2
b2
V1
c2
b1
c1
b2
V2
c2
The phase different (in degree) between voltage V1 and V2 is ________.
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Answer:
Exp:
www.gateforum.com
30
Upper transformer secondary is connected in 
Bottom transformer secondary is connected in Y
Phase angle between delta voltage & star voltage is 30o.
28.
A balanced (positive sequence) three-phase AC voltage source is connected to a balanced,
start connected through a star-delta transformer as shown in the figure. The line-to-line
voltage rating is 230 V on the star side, and 115 V on the delta side. If the magnetizing
current is neglected and Is  1000o A, then what is the value of I p in Ampere?
Ip
Is
a
A
R

B

R
R

C
b
C
(A) 5030o
Answer:
Exp:
(B) 50  30o
(C) 50 330o
(D) 200300
(B)
kVA is constant (per phase)
 230

 100 
30  I p  115 0  


 3

 3
Ip  50  30
29.
Two semi-infinite conducting sheets are placed at
right angles to each other as shown in the figure. A
point charge of +Q is placed at a distance of d from
both sheets. The net force on the charge is
Q2 K
, where K is given by
40 d 2
(A) 0
1
1
(B)  ˆi  ˆj
4
4
1 1
(C)  ˆi  ˆj
8 8
(D)
y
d
Q
d
1 2 2 ˆ 1 2 2 ˆ
i
j
8 2
8 2
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Answer:
Exp:
www.gateforum.com
(D)
F  F1  F2  F3
1


2da x  2da y  

 2da x  2da y 
2 2


F
1 Q2
40 (2d)3
F
1 Q 2 1  2 2
1 2 2 
ax 
ay 
2 
40 d  8 2
8 2

y
Q
Q(d,d,0)
X
X
So, Ans : (D)
Q
Q
y
30.
The volume enclosed by the surface f (x, y)  ex over the triangle bounded by the line
x=y; x=0; y=1 in the xy plane is _____.
Answer:
Exp:
Triangle is banded by x = y, x = 0, y = 1 is xy plane.
yx
Required volume 
  f  x, y  dxdy
 0,1
0AB
1
1
1,1
y 1
x0
  e dxdy

A
B
x
x 0 y  x
1


 0,0 
e x .  y  x dx
1
0
1,0 
x 0
1


e x 1  x  dx 
x 0
 e
x

 xe x dx
x 0
   e
 ex
1
1
0

x
 x  1 0
1

 e1  1  0   1   e  2
31.
For the system governed by the set of equations:
dx1 / dt  2x1  x 2  u
dx 2 / dt  2x1  u
y  3x1
the transfer function Y(s)/U(s) is given by
(A) 3(s  1) / (s2  2s  2)
(B) 3(2s  1)(s2  2s  1)
(C) (s  1) / (s2  2s  1)
(D) 3(2s  1)(s2  2s  2)
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Answer:
Exp:
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(A)
dx1
 2x1  x 2  4
dt
dx 2
 2x1  4
dt
y  3x1
Considering the standard equation
x i  Ax  BU
y  Cx  DU
 x 1   2 1   x1  1
 x    2 0   x   1 [4]
 2  
 2 
x 
y  [3 0]  1 
x2 
Transform function C SI  A  B
1
1
 s 0   2 1   1
G(s)  3 0  

  
 0 s   2 0   1
1
s  2 1 1
[3 0] 

s  1
2
s
3 0 2

1  1
s  2  1
s2  2s  2
32.

s  1

1
3
0




2
s s 2
 2  s  2 

s  1 
1
3
0




s 2  2s  2
s  4 

3(s  1)
s  2s  2
2
For linear time invariant systems, that are Bounded Input Bounded stable, which one of
the following statement is TRUE?
(A) The impulse response will be integral, but may not be absolutely integrable.
(B) The unit impulse response will have finite support.
(C) The unit step response will be absolutely integrable.
(D) The unit step response will be bounded.
Answer:
(B)
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33.
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In the following sequential circuit, the initial state (before the first clock pulse ) of the
circuit is Q1Q0  00. The state (Q1Q0 ), immediately after the 333rd clock pulse is
Q1
Q0
J0
Q0
J1
Q1
K0
Q0
K1
Q1
CLK
(A) 00
Answer:
(B) 01
(C) 10
(D) 11
(B)
Exp:
J1  Q0  K1  Q0  J 0  Q1  K 0  Q1  Q1 Q0

0

1

1

0
0
0
0
1
1
1
0
0
0
1
1
0
0
0
1
1
1
1
0
1
0
0
If is a Johnson (MOD-4) counter. Divide 333 by 4, so it will complete 83 cycle and
remainder clock is 1, at the completion of cycles output’s in at Q1Q0  00 so, next at
333rd clock pulse output is at Q1Q0  01.
34.
A three-phase, 11 kV, 50 Hz, 2 pole, star connected, cylindrical rotor synchronous motor
is connected to an 11 kV, 50 Hz source, Its synchronous reactance is 50 per phase, and
its stator resistance is negligible. The motor has a constant field excitation. At a particular
load torque, its stator current is 100A at unity power factor. If the load torque is increased
so that the stator current is 120 A, then the load angle (in degrees) at this node is ____.
Answer:
47.27
Exp:
Ef = Vt - Ia×S

11
kV  j100  50  6350  j5000
3
E f  8082.23
 Ia  S  E f 2  Vt2  2E f Vt cos 
120  50 2  8082.232  63502  2  8082.23  6350  cos 
2
  47.27
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35.
Two coins R and S are tossed. The 4 joint events HR HS ,TR TS ,HR TS ,TR HS have
probabilities 0.28, 0.18, 0.30, 0.24, respectively, where H represents head and T
represents tail. Which one of the following is TRUE?
(A) The coin tosses are independent
(B) R is fair, R it not.
(C) S is fair, R is not
Answer:
Exp:
www.gateforum.com
(D) The coin tosses are dependent
(D)
Given events HRHS, TRTS, HRTS, TRHS
If coins are independent
Corresponding probabilities will be
1 1 1 1 1 1 1 1
. , . , . , .
2 2 2 2 2 2 2 2
1 1 1 1
 , , , respectively
4 4 4 4
But given probabilities are 0.28, 0.18, 0.3, 0.24 respectively we can decide whether R is
fair or S is fair
 The coin tosses are dependent.
36.
A composite conductor consists of three conductors of radius R each. The conductors are
arranged as shown below. The geometric mean radius (GMR) (in cm) of the composite
conductor is kR. The value of k is ___________.
3R
R
60
60
Answer:
Exp:
1.193
GMR  0.7788R  3R  3R 
13
 1.9137R  kR
k  1.913
37.
The z-Transform of a sequence x[n] is given as X(z) = 2z+4-4/z+3/z2. If y[n] is the first
difference of x[n], then Y(z) is given by
(A) 2z+2-8/z+7/z2-3/z3
(B) -2z+2-6/z+1/z2-3/z3
(C) -2z-2+8/z-7/z2+3/z3
(D) 4z-2-8/z-1/z2+3/z3
Answer:
Exp:
(A)
y(n) is first difference of x(n) So
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g(n)  x(n)  x(n  1)
 Y(z)  x(Z)(1  z 1 )  X(z)  z 1X(z)
Y(z)   2z  4  4z 1  3z 2    2  4z 1  4z 2 
 2z  4  4z 1  3z 2  2  4z 1  4z 2  3z 3
 2z  2  8z 1  7z 2  3z 3
38.
An open loop transfer function G(s) of a system is
G s 
K
s  s  1 s  2 
For a unity feedback system, the breakaway point of the root loci on the real axis occurs
at,
(A) -0.42
(C) -0.42 and -1.58
Answer:
Exp:
(B) -1.58
(D) None of the above
(A)
1  G(s)  0
s  s 2  3s  2   12  0
 k  s3  3s 2  2s
dK
0
ds
3s 2  6s  s  0
S=-0.42 is the solution makes k>0
39.
In the given rectifier, the delay angle of the thyristor T 1 measured from the positive going
zero crossing of Vs is 30°. If the input voltage Vs is 100 sin(100πt)V, the average voltage
across R (in Volt) under steady-state is _______.
Vs D3
T1

D4
Answer:
Exp:

D2

R
V0

61.52
  30
Vin  100sin 100t 
Vm
3  cos  
2
100

 3  cos30   61.52V
2
V0 
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40.
Two identical coils each having inductance L are placed together on the same core. If an
overall inductance of αL is obtained by interconnecting these two coils, the minimum
value of α is ________.
Answer:
41.
www.gateforum.com
0
In the given network V1 = 100∠0°V, V2 = 100∠-120°V, V3 = 100∠+120°V. The phasor
current i (in Ampere) is
 j1
V1
j1
V2
V3
(A) 173.2∠-60°
Answer:
(A)
Exp:
i 
(B) 173.2∠-120°
i
(C) 100.0∠-60°
(D) 100.0∠-120°
 V1  V3    V2  V3 
j
j
1000o  100120o 100  120o  100120o

1  90o
190o
i  173.2  60o
i 
42.
A differential equation
di
 0.2i  0 is applicable over -10< t <10. If i(4) = 10, then i(-5)
dt
is ________.
Answer: 1.65
Exp:
di
 0.2i  0
dt
 D  0.2  i(t)  0
D  0.2
i(t)  k.e0.2t ,  10  t  10
t  u;
10  K.e0.8
K=4.493
 i(5)  4.493  e 1
i(5)  1.65
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43.
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A 3-phase transformer rated for 33 kV/11 kV is connected in delta/star as shown in figure.
The current transformers (CTs) on low and high voltage sides have a ratio of 500/5. Find
the currents i1 and i2, if the fault current is 300A as shown in figure.
a
b
300A
c
i2
i1
(A) i1  1
3 A,i 2  0A
(C) i1  0A,i 2  1
Answer:
Exp:
(B) i1  0A,i 2  0A
(D) i1  1
3A
3A,i 2  1
3A
(A)
i2 = 0 since entire current flows through fault
Primary kVA = Secondary kVA
5 

3  33000  I L  3  11000   300 

500 

I L  1A
I L  3 I Ph
I ph  i1 
44.
1
A
3
A Boolean function f(A,B,C,D) = ∏(1,5,12,15) is to be implemented using an 8×1
multiplexer (A is MSB). The inputs ABC are connected to the select inputs S 2S1S0 of the
multiplexer respectively.
0
1
2
3
4
5
6
7
f  A,B,C,D 
S2 S1 S0
A B C
Which one of the following options gives the correct inputs to pins 0,1,2,3,4,5,6,7 in
order?
(A) D,0,D,0,0,0,D,D
(B) D,1,D,1,1,1,D,D
(C) D,1,D,1,1,1,D,D
(D) D,0,D,0,0,0,D,D
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Answer:
Exp:
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(B)
Given maxterm f  A,B,C,D    1,5,12,15 so minterm
f  A,B,C,D   m  0,2,3,4,6,7,8,9,10,11,13,14 
I0 I1
0 2
1 3
D 1
D  0
D 1
45.
I 2 I3
4 6
5 7
D 1
I 4 I5 I 6 I 7
8 10 12 14
9 11 13 15
1 1 D D
The incremental costs (in Rupees/MWh) of operating two generating units are functions
of their respective powers P1 and P2 in MW, and are given by
dC1
 0.2 P1  50
dP1
dC2
 0.24 P2  40
dP2
Where
20MW≤P1≤150 MW
20MW≤P2≤150MW
For a certain load demand, P1 and P2 have been chosen such that dC1/dP1 = 76 Rs/MWh
and dC2/dP2 = 68.8 Rs/MWh. If the generations are rescheduled to minimize the total
cost, then P2 is _____________.
Answer:
Exp:
136.36

 P1  P2  250


dc2
 68.8  0.24 P2  40  P2  120 

dP2

dc1 dc2
For total cost minimization,

dp1 dp 2
dc1
 76  0.2P1  50  P1  130
dP1
0.2p1  50  0.24p 2  40
0.2  250  P2   50  0.24P2  40
P2  136.36
46.
The unit step response of a system with the transfer function G  s  
1  2s
is given by
1 s
which one of the following waveforms?
(A)
(B)
y
1
y
2
t
0
1
5
t
0
2
5
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y
2
(C)
(D)
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y
1
1
t
0
t
0
5
5
0.75
2
Answer:
(A)
y(t)  ?
Exp:
G(s)
1
1
Y(s)  G(s)  U(s)
(1  2s) 1
.
(1  s) s
A
B
Y(s) 

(s) (s  1)
A  1, B  3
Y(s) 
t
0
5
2
y(t)  u(t)  3e  t u(t)
y(t)  1  3e  t  u(t)
47.
Answer:
48.
C
L
A symmetrical square wave of 50% duty cycle
has amplitude of ±15V and time period of 0.4π
ms. This square wave is applied across a series
RLC circuit with R = 5, L = 10 mH, and C =
4μF. The amplitude of the 5000 rad/s component
of the capacitor voltage (in Volt) is __________.


R
15
For the switching converter shown in the following figure, assume steady-state operation.
Also assume that the components are ideal, the inductor current is always positive and
continuous and switching period is T5. If the voltage VL is as shown, the duty cycle of the
switch S is _______.

Vin


VL

V0

S
C

15V
R
VL
TS
t
45V
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Answer:
0.75
Exp: VS  15V
VS  V0  45  V0  VS  45  60V
VS
1 D
15
60 
 D  3 4  0.75
1 D
V0 
49.
With an armature voltage of 100V and rated field winding voltage, the speed of a
separately excited DC motor driving a fan is 1000 rpm, and its armature current is 10A.
The armature resistance is 1. The load torque of the fan load is proportional to the
square of the rotor speed. Neglecting rotational losses, the value of the armature voltage
(in Volt) which will reduce the rotor speed to 500 rpm is ___________.
Answer:
Exp:
97.5
TL  N 2  2
 N  Ia
T  Ia 
2
Ia
 500 

   Ia  2.5V
 1000  10
E b  100  2.5  1  97.5V
50.
The saturation voltage of the ideal op-amp shown below is ±10V. The output voltage 0 of
the following circuit in the steady-state is
1k
10V
0.25 F


0
2k
10V
2k
(A) Square wave of period 0.55 ms
(B) Triangular wave of period 0.55 ms
(C) Square wave of period 0.25 ms
(D) Triangular wave of period 0.25 ms
Answer: (A)
Exp: Astable multivibrator produces square wave.

R2
2
  0.5
R1  R 2 4
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T  2R c log
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(1  )
 1  0.5 
 2 1103  0.25  106  log 

(1  )
 1  0.5 
T= 0.55 ms
Square wave of period 0.55 ms.
51.
A three-winding transformer is connected to an AC voltage source as shown in the figure.
The number of turns are as follows: N1 = 100, N2 = 50. If the magnetizing current is
neglected, and the currents in two windings are I2  230 Aand I3  2150 A, then
what is the value of the current I1 in Ampere?
I1
N2
I2
I3
N1
(A) 190
Answer:
Exp:
(B) 1270
N3
(C) 490
(D) 4270
(A)
I1N1 = I2N2 + I3N3
I1.100  2 30  50  2 150  50
I1  1 30  1150
 1 90
52.
The coils of a wattmeter have resistances 0.01 and 1000; their inductances may be
neglected. The wattmeter is connected as shown in the figure, to measure the power
consumed by a load, which draws 25A at power factor 0.8. The voltage across the load
terminals is 30V. The percentage error on the wattmeter reading is _________.
Load
Answer:
Exp:
0.15
Pload = 30×25×08 = 600W
Wattmeter measures loss in pressure coil circuit
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loss in Pc 
error 
53.
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V 2 302

 0.9W
R P 1000
0.9
 100  0.15%
600
Consider a signal defined by
j10t

e for t  1
xt  

0 for t  1
Its Fourier Transform is
(A)
(C)
Answer:
Exp:
2sin    10 
(B)
  10
2sin 
  10
(D)
2e j10 sin    10 
  10
e
j10 
2sin 

(A)
1
1
1
1
X()   e j10t .e jt dt   e j(10)t dt
1
e j(10)t
2sin(  10)


j(10  ) 1
(  10)
54.
A buck converter feeding a variable resistive load is shown in the figure. The switching
frequency of the switch S is 100 kHz and the duty ratio is 0.6. The output voltage V 0 is
36V. Assume that all the components are ideal, and that the output voltage is ripple-free.
The value of R (in Ohm) that will make the inductor current (i L) just continuous is
_________.
S
iL
5mH
60V 

36V
Answer:
Exp:
V0
R
1250
f = 100kHz
D = 0.6
V0 = 36V
Vin = 60V
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I  I 2
I 
VS D 1  D 
fL
 I2
60  0.6  0.4
36

3
3
R
100  10  5  10
R  1250
55.
The following discrete-time equations result from the numerical integration of the
differential equations of an un-damped simple harmonic oscillator with state variables x
and y. The integration time step is h.
x k 1  x k
 yk
h
y k 1  y k
 x k
h
For this discrete-time system, which one of the following statements is TRUE?
(A) The system is not stable for h>0
(B) The system is stable for h 
1

(C) The system is stable for 0  h 
(D) The system is stable for
Answer:
1
2
1
1
h
2

(A)
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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
(D)
In a quadratic function, the value of the product of the roots  ,   is 4. Find the value of
Key:
5.
 n  n
  n   n
(A) n 4
(B) 4 n
Key:
(B)
Exp:
Given   4
(C) 2 2n 1
(D) 4n 1
 n  n
 n  n

1
1
  n   n
 n
n




n
 n   n n

n
 n 
 () n  4n
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Q. No. 6 – 10 Carry Two Mark Each
6.
Key:
Exp:
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
(A)
F  Facebook, W  WhatsApp, E  Total faculties
given
E, n(E)  150
n(E)  150, n F  W  30


n  F  W   n(E)  N  F  W   150  30
85
55
n  F  W   120
n  f  w   n  f    n(w)  n(F w 
120  n(F)  85
F
n(F)  35
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.
Key:
8.
Key:
Fw
n F  w
 20
W
n(w)  60
 30
F  W
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)
(D)
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)
(D)
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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)
A
B
C
D
E
2: ACMK, ADNK 2: ECMD,EBLO 2: ACHF,ADIF
F
G
H
I
J
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.
25
K
L
M
N
O
f x
2
1.5
1
0.5
0
4
3
2
1
0.5
0
1
2
3
X4
1
1.5
2
25
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
Key:
Exp:
(D) f  x   2  x  1
(C)
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
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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 ( x)  x( x 1)( x  2)
 x  1
 f ( x)  x 3  3 x 2  2 x  f 1 ( x)  0  3 x 2  6 x  2  0
1
3
But x  1 
1
only lies on the interval [1,2]
3
1 11
1 

;f ( x)  6 x  6  6 1 
6  0
3
3

1
 x  1
is a point of minimum
3
 f(x) = x(x-1)(x-2) = 0 at either ends x =1 & x =2
At x  1 
 Max value = 0
2.
Key:
Exp:
Consider a 3  3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
3
Consider A 33
1 1 1
 1 1 1
1 1 1
It’s only non-zero Eigen value λ = 1 order of the matrix = 1 3  3
The Laplace Transform of f  t   e2t sin  5t  u  t  is
3.
(A)
Key:
Exp:
5
s 2  4s  29
(A)
Consider
(B)
5
s 5
2
(C)
s2
s 2  4s  29
(D)
5
s5

5
 X(S)
s +25
By frequency shifting property, 1 X(S-S0 )  x(t)eS0t
x(t) = sin5t u(t)
thus at S0  2
2t
f(t) = e sin5tu(t)
5
 F(s)  2
s  4s  29
2

5
 F(s)
(s-2)2 +25
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A function y  t  , such that y  0   1 and y 1  3e1 , is a solution of the differential equation
4.
d2 y
dy
2
 y  0 . Then y  2  is
2
dt
dt
Key:
(A) 5e1
(B)
(B) 5e2
(C) 7e1
Exp:
2
The operator function of given D.E is (D +2D+1)y = 0
(D) 7e2
 A.E is D2 +2D+1 = 0  D = -1,-1
 y = e-x C1 + C2   (1)
 Given y(0)=1 & y(1) =3e-1
From ; y(0) = 1 i.e , y = 1 at x = 0 1 = C1
y(1) = 3e-1 i.e, y = 3/e at x=1
3 -1
 e 1  C2 (1)  3  1  C2  C2  2
e
 y = e-x [1+2x]  y(2) = e-2 [1+4] = 5e-2
5.
The value of the integral
2z  5
dz over the contour z  1. taken in the anti1 2
 z    z  4z  5 
2

 
C
clockwise direction, would be
(A)
Key:
Exp:
24i
13
(B)
48i
13
(C)
24
13
(D)
12
13
(B)
Singular points are obtained by
1 2
1

 z    z -4z+5   0  z = & z =2  i
2
2

1
Out of these z = only lies inside C. z =1
2
By Cauchy’s integrated formula,
 
C
2z+5
1 2
 z -   z -4z+5   0
 2
dz =

C


1
2   5


2z+5
2


  48πi
dz = 2πi
2
2
 1 
z -4z+5
 1   13

4



   5
1
2


2 

z2
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The transfer function of a system is
Y s 
R s 

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s
, The steady state output x(t) is A cos  2t   for
s2
the input cos  2t  . The value of A and , respectively are
(A)
1
2
Key:
(B)
Exp:
H(s) =
H( ) =
,  45O
(B)
1
2
,  45O
(C)
2,  45O
(D)
2,  45O
S
S+2

 
900  tan 1  
2
 2 +4
If input is COS 2t i.e.,  = 2
1
450
2
1
 y(t) =
cos(2t  450 )
2
1
by comparision A=
&   450
2
H( ) =
7.
The phase cross-over frequency of the transfer function G(s) =
(A)
3
(B)
1
3
100
 s  1
(C) 3
Key:
(A)
Exp:
Phase Crossover frequency PC : GH =PC  1800
3
in rad/s is
(D) 3 3
GH  3tan-1  1800  3tan -1w PC  WPC  tan 600  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
Key:
Exp:
(A) linear and time-invariant
(C) linear and time-varying
(C)
Linearity
y1 (t)= x1 (t)cost;
(B) Non-linear and time-invariant
(D) Non-linear and time-varying
y 2 (t)= x 2 (t)cost
 y1 (t) + y2 (t) = x1 (t)cost + x 2 (t)cost
y1 (t) + y2 (t) = [x1 (t) + x 2 (t)]cost
If x1 (t) + x 2 (t) = x(t) then y1 (t) + y 2 (t) = y(t)
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Thus the system is linear
Time-invariance
consider y1 (t)  x1 (t) cost ; If x1 (t)  x(t - τ)
 y1 (t)  x(t - τ) cost
Define y(t - τ)  x(t - τ)cos(t - τ)
 y1 (t)  y(t - τ) system is time-varient
9.
The value of
A



e t   2t  2  dt. where   t  is the Dirac delta function, is
1
2e
 B
2
e
C
1
e2
 D
1
2e 2
Key:
(A)
10.
Key:
A temperature in the range of -40OC to 55OC is to be measured with a resolution of 0.1OC. 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
(B)
Exp:
Usually when voltage information is given we use the formula R =
Vmax -Vmin
2n
Here based on the given information we can think the systems as following block diagram
Temp

Temperature
to
voltage
converter
analog
voltage
nbit
ADC
.
.
.
Here the Vmax and Vmin will be the equivalent of Tmax and Tmin respectively
So we can still use the above relation
Tmax -Tmin
5-(-40)
 2n =
 2n = 950
n
2
0.1
 n  log 2 (950)  9.89  10
 Resolution =
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
0
F
Y
1 S
A
B
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(D)
We can redraw the max circuit as follows
A
0
A
1
F
B
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 ________.
10V
4.7k
220
0.5mA
5.3V
Key:
19
Exp:
5.3 = 0.6 + 470 IE
470
I E  0.01A
10 - 5.3
 1mA
4.7 103
I B =1  0.5  0.5mA
IX =
IE  (1  β)IB
 0.01  (1  β)  0.5 103  (1  β) 
0.01
 20  β  19
0.5 103
13.
In cylindrical coordinate system, the potential produced by a uniform ring charge is given
Key:
(A) increases with r
(C) is 3
(B)
Exp:
Since the charge is not varying with time the field E is static So   E  0

  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
(B) is 0
(D) decreases with z
 
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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
Key:
Exp:
points located just inside and just outside the toroid, is ________.
100
The field inside and outside the toroid is due to long straight conductor only
Let the two points almost at same distance
μ 0μ r I
2πr
μI
The flux density outside toroid = 0 T
2πr
μ 0μ r I
2πr  μ  100
Ratio =
r
μ0I
2πr
The flux density inside toroid =
15.
R A and R B 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?
2
RA
2
1
2
RB
(A) R A  R B
Key:
Exp:
1
1
1
2
1
(B) R A  R B  0
1
2
1
(C) R A  R B
(D) R B  R A 1  R A 
(D)
By comparing 2 network on the input side
we can say that R B =1//R A  R B =
16.
2
RA
1 RA
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
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Key:
(C)
Exp:
SmT 
r2
r
r2
 2 
x2
j L 2
j2L 2 .f
SmT 
1
f
NS 
120f
P
 SMT 
17.
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1
NS
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 ______.
4
6
5
Key:
Exp:
2
Here the power information regarding the resistor is given because
E Calorie

=watt
t
sec
 P5 = 10
P=
V5Ω
= 10  V5Ω = 50
5
 P4Ω is asked

 V4Ω 
2
1 4

P4Ω =
 
50 
4
4 4  6

1 16
calorie
 
 50  2
4 100
sec
18.
2
In the given circuit, the current supplied by the battery, in ampere, is ________.
l1
`
1
1

1
l2
1V
Key:
Exp:
l2
0.5
If we write KCL at node × then
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I1 = 2 I2  I 2 
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I1
2
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1
X
1
Write KVL in the outer boundary of network
1  (1 I1 )  (2  I 2 )  0  1  I1  2 I 2
1
I 
1 =I1  2  1   1  2J1  I1   0.5A
2
2
19.
Key:
Exp:
20.
Key:
Exp:
1V
I1
I2
I2
1
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
(B)
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
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 threephase fault at bus B, in pu, is _____.
0.84
Voltage at ith bus when fault is at kth bus is

zik 
Vi  E 1 

z

kk  z f 
If 
Vproduct
X  p.u 
1
10  
 X A  0.1P.U
xn
1
8  
 B  0.125P.U
XB
 Z 
VB  E 1  AB 
 ZAA 
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 Z 
0.8  1 1  AB   ZAB  0.02
0.1 

 2AB 
0.02 

VA  1 
 1  1 
  0.84 P.U
2
0.125



BB 
21.
Key:
22.
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
(A)
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 VL
during a complete switching interval. Assuming all devices are ideal, the duty cycle of the buck
converter is ______
VL
30V
M

Vg

TOFF
TON

VL

0
D
C
V0
R
t
20V
a 
Key:
Exp:
Ta
b
0.4
When M is ON, VS  VL  V0
 VL  30  VS  V0
When M is OFF,
VL  V0  20  V0  V0  20
 30  VS  20  Vs  50V
D
V0
2
  0.4
VS 5
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23.
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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 _____
Is  A 
I0  A 
V0  0.7V
S
D
dV dl  0.02
V0  1V
VS  Volt 
VS  Volt 
100 A
V  l characteristic of diode
c
V  l characteristic of 1GBT
a 
dV dl  0.01
b
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
40A
I
(A) I
(B)
10A
t
t
(C)
(D)
I
10A
T  25ms
Key:
Exp:
10A
I
T  25ms
T  25ms
T  25ms
10A
10A
(C)
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, t 
0.1
 25m sec
4 poles
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25.
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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
n :1
`
L
Port 1
(A) nL
Port 2
(B) n2L
(C)
n
L
(D)
n2
L
Key:
(B)
Exp:
The property of an ideal transformer is of port2 is terminated by an impedance
Z L then the
ZL
impedance seen from port is
 NZ / NL 
In the given problem Zin 
L
1/n 
2
2
 n 2L
Q. No. 26 – 50 Carry Two Mark Each
26.
Key:
Exp:
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 ______.
0.06
Total possible options = 43
Favorable choices {BBB, BlBlBl, GGG, RRR} = 4
4
1
Probability = 3   0.06
4 16
27.
Let S   n n where   1. The value of  in the range 0    1 . Such that S  2 is ______.

n 0
Key:
0.293
Exp:
S =  n αn

n=0
 2α  α  2α 2 + 3α3    ] \
 2α  α[1  2α  3α 2    ]
 2α  α[1  α]2 if α  1
 (1  α) 2  2  α  1 
1
 α  0.293
2
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Let the eigenvalues of a 2 x 2 matrix A be 1, -2 with eigenvectors x1 and x 2 respectively. Then
the eigenvalues and eigenvectors of the matrix A2  3A  4l would, respectively, be
(A) 2,14; x1 , x 2
(B) 2,14; x1  x2 , x1  x2
(D) 2,0; x1  x 2 , x1  x 2
(C) 2,0; x1 , x 2
Key:
Exp:
(A)
Matrix A
Eigen values 1 , -2
Matrix A 2 -3A+4I
12  3(1)  4, (2) 2  3(2)  4
Eigen values 2 , 14 respectively
 A & P(A) = a 0 I + a1A + a 2 A 2 have same eigen vectors
Key:
Let A be a 4  3 real matrix with rank 2. Which one of the following statement is TRUE?
(A) Rank of ATA is less than 2
(B) Rank of ATA is equal 2
(C) Rank of ATA is greater than 2
(D) Rank of ATA can be any number between 1 and 3
(B)
Exp:
Given that ρ  A43   2
29.


From the properties of Rank; we have ρ AAT = ρ(A)
 Rank of AAT = Rank of AT A = Rank of A  2
30.
Consider the following asymptotic Bode magnitude plot   is in rad s  .
magnitude
 dB
12dB
20dB dec
0.dB
8
0.5
Which one of the following transfer function is best represented by the above Bode magnitude plot?
2s
(A)
1  0.5s 1  0.25s 
(C)
1  2s 1  4s 
2s
2
(B)
(D)
4 1  0.5s 
s 1  0.25s 
4s
1  2s 1  4s 
2
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Key:
Exp:
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(A)
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
Slope =
M 2 - M1
log2 - log1
magnitude
 dB
0  12
 40 
log 8 - log2
12
40
12
 log2 = log 840
12dB
20dB dec
0.dB
 log 8- log2 
0.5
1
2
8
 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 1+
31.
s
 having 2 poles
4
Consider the following state-space representation of a linear time-invariant system
1 0 
1
1
x  t   
x  t  , y  t   cT x  t  , c    and x  0    

0 2
1
1
The value of y(t) for t  loge 2 is _________.
Key:
6
32.
Loop transfer function of a feedback system is G  s  H  s  
s3
.Take the Nyquist contour in
s 2  s  3
the clockwise direction. Then the Nyquist plot of G(s) H(s) encircles – 1 + j0
Key:
(A) Once in clockwise direction
(C) Once in anticlockwise direction
(A)
Exp:
GH =
S+3
S(S-3)
GH 
( 2 +a)1/2
1
 2
2
2
1/2
 ( +a)

(B) Twice in clockwise direction
(D) Twice in anticlockwise direction
 



GH   tan 1   1800  1800  tan 1   2 tan 1
3 
3
3

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 GH =
1

2
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
3
at   0, GH =  0
0
at   , GH = 0 180
1 0
90
9
So the plot start at 00 and goes to 1800 through 900 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
at   3, GH =
actual plot started in clockwise direction. So the plot will be
-1+io
w=0
w=
So the Nyquist plot of G(s) H(s)
encircles – 1 + j0
once in clockwise direction
M= -1
33.
Given the following polynomial equation s3  5.5s2  8.5s  3  0 , the number of roots of the
Key:
polynomial, which have real parts strictly less than – 1, is _______
2
Exp:
The polynomial is S3 +5.5S2 +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
(Z-1)3 +5.5(Z-1)2 +8.5(Z-1)+3=0
 Z3 -3Z2 +3Z-1+5.5(Z2 +1-2Z)+8.5Z-8.5+3=0
 Z3 +Z2 (3  5.5)+Z(3+8.5-11)+(-1+5.5-8.5+3)=0
 Z3 +2.5Z2  0.52 1  0
Using RH table
Z3 1
2
Z 2.5
0.5
1
1
Z 0.9
Z0 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.
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Suppose x1  t  and x 2  t  have the Fourier transforms as shown below.
34.
X1  
X 2  
1
1
0.5
0.5
0.3
1
0.3
0
1
2

2
1
0
1

Which one of the following statements is TRUE?
(A) x1  t  and x 2  t  are complex and x1  t  x 2  t  is also complex with nonzero imaginary part
(B) x1  t  and x 2  t  are real and x1  t  x 2  t  is also real
(C) x1  t  and x 2  t  are complex but x1  t  x 2  t  is real
(D) x1  t  and x 2  t  are imaginary but x1  t  x 2  t  is real
Key:
(C)
Exp:
x1 (t) & x 2 (t) are complex functions
x 2 ( ) = x1 (- ) , x 2 (t) = x1 (-t)
 x1 (t) x 2 (t)will be real
The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where x  t  is
35.
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
Key:
Exp:
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
(D)
Consider the Eigen signal Z(t) = Cos t
For Cost, y(t) 




 cos(t-τ)h(τ)dτ 

 (cos t cos τ +sin t sin τ)h(τ)dτ

= cos t  cos τh(τ)dτ  sin t  sin τh(τ)dτ


 h(τ)is an even signal

 sinτh(τ)dτ = 0

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
 y(t) = cos t  cos τh(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  cos τh(τ)dτ  cos t  sin τh(τ)dτ
Thus y(t)= sin t
 cos τh(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
5V
J
K
QA
QO
J
QV
K
CLK
Key:
Exp:
(A) 00
(C)
(B) 01
(C) 11
(D) 10
Logic 1
JA
QA
KA
QA
JB
QB
KB
QB
J A = K A =1
J B = K B =QA
It is given initially QA QB = 0
Since it is a synchronous counter, when clock is applied both flipflop will change there state
simultaneously based on JK FF state table
  JA =1, KA =1 , QA =0  QA  = 1
JB =1, KB =1, QB =0  QB = 1


So next state Q A Q B is 11
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A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is 0  VIN  3 Volts.
37.
The expression for the LSB of the output B0 as a Boolean function of X2 , X1 , and X0 is
3V
100


X2
200


X1

X0
B1
Digital
circuit
B0
200

100
VIN
Key:
38.
(A) X0 X2  X1 
(B) X0 X2  X1 
(C) X0  X2  X1 
(D) X 0  X 2  X1 
(A)
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
(C) x 2  y2  2
(B) y  2
Key:
(D)
Exp:
The potential due to Q 4-2Q at (x, y) is 
q
2
k x +y
If potential at (x, y) = 0 where k=
q
2
k x +y
2

2q
2
k (x-6) + y
2
2

(D)  x  2   y 2  16
2
2q
k (x-6) 2 + y 2
1
4πε
 0  (x-6)2 + y2  2 x 2 + y2  0
 (x-6) 2 + y 2  4(x 2 + y 2 )  x 2 +36 -12x + y 2 = 4x 2  4y 2
 x 2  y 2 - 4x-12 = 0
2
2
2
2
2
2
Option:D (x+2) + y  16  x +4x +4+ y = 16  x  y + 4x-12 = 0
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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 _______.
S1
1
S2
2
1H
3V
3V
Key:
2
Exp:
At t = 0- Network is in steady state with S1 opens S 2 (closed) So we can say i L (0- )=
3
 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
VL (0 + )

Writing Nodes equation at VL (0 ) node
1 1  3 3
VL (0 )=       1.5
1 2  1 2
 VL (0  ) = 2
1
3V
2
1.5
3V
di L (0+ )
di(0+ )
L
2
 2A/sec
dt
dt
at t = 0+
40.
Key:
41.
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 _________.
47 to 49
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  Xn  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 X n  in p.u. that will
Key:
Exp:
limit the positive sequence generator current to 4270 A is ________.
1.108
X1  0.15 P.U
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Ibase 
30 106
3  13.8 103
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 1255.109
X2  0.15P.U
X0  0.05P.U
Xf  0.1P.U
I  P.U  
I actual
4270

3.4  P.U 
I base
1255.109
Ia1 
Ea
Z1   Z2 || ZO  3zf  3zn 
3.4 
1
0.15   Z2 || Z0  3zf  3z n 
Z2 ||  20  3z f  3z n   0.144
1
1
1


Z2
20  3z f  3zln 0.144
20  3zf  3zn  3.675
3zn  3.675  0.05  3  0.1
Zn 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
a
(A) VAB leads Vab by 60O
(B) VAB lags Vab by 60O
(C) VAB leads Vab by 30O
(D) VAB lags Vab by 30O
B
N
c
b
C
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
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44.
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The switches T1 and T2 are shown in Figure (a).

C
T1
Vdc 2  250V
iL
1

0.8

Vdc
R  12
XL  16 at
100Hz
2  250V
t
T2
Vm

b
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  i L  , in ampere, is ______.
Key:
10
Exp:
ma 
Vm
 0.8
Vcarrier
V0l  max   ma.
Vds
 0.8  250  200V
2
2L  162 122  20
I L  max  
45.
200
 10A
20
The voltage  s  across and the current  lg  through a semiconductor switch during a turn-ON
transition are shown in figure. The energy dissipated during the turn – ON transition, in mJ, is
_________.
vs
600V
0
t
is
50A
100A
0
T1  1s
T2  1s
t
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Key:
75
Exp:
1

E   Pdt   V  t  tdt  600  150  106  + 100
2


46.
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1
6 
 2  600  10   75mJ


Key:
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
1.4423
Exp:
Since
V  400 200


 8  is constant, Bm is constant

f  50
2.5

Pcore  Ph  Pe  k1f  k 2 f 2
Ph  k1f
5000  50k1  502. k 2  50k1  2500k 2
Pe  K 2 f 2
2000  25k1  25k1  252 k 2  25k1  625k 2
k1  50k 2  100 k 2  0.8

k1  25k 2  80  k1  60
∴ Pcore = 60f   0.8 f 2
Ph  60f
Pe  0.8f 2
When f  52Hz, Ph  60  52  3120
Pe  0.8  522  2163.2
Ph
3120

 1.4423
Pe
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
Key:
Exp:
efficiency of the DC generator is ______.
86.84
Pout  220  45  990W
IL  45A
If
220
If 
 5A
44
Ia  IL  If  45  5  50A
44
Armature losses = Ia 2 ra  502  0.01  25W

ra  0.01 220V

Field losses = If 2 R f  52  44  1100W
Stray losses = 375W
Total losses = 1500W
1 
losses    1 Pout
 
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1

1500    1 9900   86.84%
 
A three-phase, 50Hz salient-pole synchronous motor has a per-phase direct-axis reactance  Xd 
48.
of 0.8pu and a per-phase quadrature-axis reactance  X q  of 0.6pu . Resistance of the machine is
Key:
Exp:
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 ______.
1.608
Xd  0.8P.U P.f  0.8 leading Ra  0
Xq  0.6PU
Vt  1P.U
cos   0.8    36..86
Ia  136.86
tan  
I a  q cos   I a R a sin 
Ia  q cos 
1 0.6  0.8


 0.3529
Vt  I a  q sin   I a R a cos  Vt  Ia  q sin  1  1 0.6  0.6
  19.44O
     36.86 19.44  56.3
Id  Ia sin   1 sin56.4  0.832
Iq  Ia cos   1 0.5547  0.5547
Ef  V cos   Id  d  Iq R a 0  Vcos   Id  d  1.cos19.44  0.832  0.8  1.608pu
EF  1.608PU
49.
Key:
A single-phase, 22kVA, 2200V/220, 50Hz, distribution transformer is to be connected as an autotransformer 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
(C)
Exp:
Rated LV current =
KVA max 
2200 10
 242kVA
1000
Or
2420  100

 242kVA
1000
50.
Key:
100A
22kVA
 100A
220

110A
2200VLOV

220V

  10A
2200V

2420V

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
(A)
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Exp:
VO  t  
V01 
51.
Key:
Exp:
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n
 4Vs 
sin n  t 

 cos
n

6

n 1,3,s 


1  4VS 

 cos  6  233.9  234V
2   
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 _______.
1.439
For single phase
L1
1m
 1 
D 
 0.2n  m   0.2n  
 Ds 
 Ds 
For three phase
L3    0.2n
DM
 0.2n
DS
 D23

 Ds





D
D
L3   1.05L1  
D m  3 D.D.1  D
 D2/3 
 1 
0.2  
  1.05  0.2  n 

D
 DS 
 S 
2
3
1m
 D  1.439 mts
In the circuit shown below, the supply voltage is 10sin 1000t  volts. The peak value of the steady
52.
state current through the 1 resistor, in amperes, is _________.
2F
4
240F
500mH
1
5
4mH
~
10sin 1000t 
Key:
Exp:
1
W =1000, the various impedance at this frequency are
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-j


 Z250μf Z4mH  
( j 1000  4 103 )
6 
1000  250 10 
 ( j4) (j4)    open circuit
-j


 Z24f Z500mH  
( j 1000  500 103 )
6 
1000

250

10


 ( j500) (j500)    open circuit
Since both LC pair parallel combination becomes
open then the circuit can be redrawn as
10sin1000t
 sin1000t
4+1+5
 So peak value of I1Ω  1A
 I1Ω 
4
1
5
V
10sin1000t
A dc voltage with ripple is given by   t   100  sin  t   5sin  3t   volts. Measurements of
53.
this voltage   t  , made by moving-coil and moving-iron voltmeters, show readings of V1 and V2
respectively. The value of V2  V1 , in volts, is _______.
Key:
Exp:
0.312
V1  100V
2
2
 10   5 
V2  100  
 
  100.312V
 2  2
2
V2  V1  0.312V
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 ________.
54.
100F
R
R
0.02H
Key:
Exp:
~
14.14
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
C
L
R

R
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get the answer.
j
R+
1
1
R-j L
C  mg[Y ]  0
Y

 2

eq
2
2
1
R+j L RR +( L)
2  1 
R +

C
 C 
1
L
C
 2

2
R +( L) 2
 1 
R2 +

 C 
 Cross multiplying
2
 1   1  2
2 1
(L)R  L 
 
 R +(L)
C
 C   C 
1   1 
1 

 R 2  L
0
  L
C   C 
C 

2
L 
1 

  R 2    L0
C
C 

Now by looking into above equation we can say that if
L
then it will have no depending on frequency
C
L
for resonance  R 2 
C
R2 
So R 
55.
L
0.02

 10 2  14.14
C
100 106
In the circuit shown below, the node voltage VA is _________V.
A
I1
5
5
5A


10I1

5
5
10V 
Key:
11.42
Exp:
All the branch currents are expressed interval of VA now writing KCL at node A

VA
V  10I1 VA  10
5 A

0
5
5
5
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1 1 1 
 VA      2I1  5  1
 5 5 10 
VA
 2 1   V  10 
 VA     2  A
6
 5 10   10 
2 1 2 
 VA      6  2
 5 10 10 
80
7
 VA    8  VA 
 11.42V
7
 10 
5A
5
VA
5
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I1
5
5
5
VA  10I1
5
10I1


10
VA -10
10
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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
Key:
2.
Key:
3.
Key:
Exp:
4.
The chairman requested the aggrieved shareholders to ___________him.
(A) bare with
(B) bore with
(C) bear with
(D) bare
(C)
Identify the correct spelling out of the given options:
(A) Managable
(B) Manageable
(C) Manageble
(B)
Pick the odd one out in the following
13, 23, 33, 43, 53
(A) 23
(B) 33
(C) 43
(B)
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.
(D) Managible
(D) 53
Key:
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.
(B)
5.
If 9y  6  3, then y2  4y 3 is _________.
(A) 0
Key:
Exp:
(B) 1 3
(C) 1 3
(D) undefined
(C)
9Y  6  3
Possibility (A): 9y  9  y  1
Possibility (B): 9y  3  y 
1
3
When y=1
4y 2 4(1) 3  4
1
y2 
1 


3
3
3
3
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1
3
1
4 
2
1
 3   1  4  3   1
  
3
9 9 9
3
 3
When y 
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
300
250
Installed Capacity
Actual Pr oduction
250
230
220
190
180
200
160
Capacity
production 150
 tonnes 
100
200
200
190
190
150
160
160
150
140
120
100
5
6
120
50
0
1
2
3
4
7
8
9
Plant Number
Key:
Exp:
120
Largent plant
Installed Capacity
Actual production
Plant number
220
160
1
200
190
4
250
230
8
200
190
9
Total production of larger plants = 160+190+230+190=770 tonnes
Smaller Plants
Installed Capacity
Actual production
Plant number
180
150
190
160
160
120
150
100
140
120
Total production of smallest plants = 150+160+120+100+120= 650tonnes
Difference = 770-650 = 120 tonnes
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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.
Key:
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
(A)
10.
The numeral in the units position of 211870  146127  3424 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
BC
00
A
(B) A  C
0
(C) A  C
1
11
10
0
0
1
1
1
1
01
1
1
(D) A  C
Key:
Exp:
(B)
A
BC
BC
1
0
0
1
1
1
BC
BC
F  A, B, C   A  C
1
1
A
R2
2.
The circuit shown below is an example of a
C
(A) low pass filter
15V
(B) band pass filter
Vin
(C) high pass filter

Vout

(D) notch filter
Key:
R1
15V
(A)
1
R2
cs 
Z1  R i , Z2 
1 1  R 2 CS
R2 
cs
Z
R2
TF   2  
Z1
1  R 2CS R1
R2 
Exp:
When S  0 ; TF  
R2
 non zero value 
R1
When S  ; TF  0
Given circuit is an example of low pass filter
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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 _____.
XL  10
(10)
Exp:
Where ZL 
I
R  80k
XC  40k
I
100V ~
Key:
I :100
 80  j40 
3

 10    8  4i  
2
 100 1 
Zsec ondary
n2
100
100
100


 10 36.86
j10  ZL
J10  8  j4 8  j6
Since all the calculation done with respect to RMS, I also in RMS
4.
dy  t 
Consider a causal LTI system characterized by differential equation
dt

1
y  t   3x  t  . The
6
1

3
response of the system to input x  t   3e u  t  , where u(t) denotes the unit step function, is
(A) 9e
(C) 9e
Key:
Exp:

t
3
t

3
ut
(B) 9e
t

6
u  t   6e u  t 

t
6
ut
(D) 54e

t
6

t
3
u  t   54e u  t 
(D)
1

 S   y s   3  s 
6

3
 H s 
1

s  
6

Similarly X  s  
 y s 
3
 1
s  
 3
9
54
54


 1  1   1   1 
 s   s    s    s  
6 
6  3
 3 
Thus y  t   54 e
t
6
u  t   54e t 3 u  t   0
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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
Key:
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
(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
A 3  3 matrix P is such that, P3  P. Then the eigenvalues of P are
7.
(A) 1, 1, - 1
(B) 1,0.5  j0.866,0.5  j0.866
(C) 1, 0.5  j0.866,0.5  j0.866
Key:
Exp:
(D) 0, 1, - 1
(D)
If  is an Eigen value of p then 3 is an Eigen
value of p3.
 p3  p
 3  
 3    0    2  1  0    0;  2  1
   0;    1
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The solution of the differential equation, for t  0, y"  t   2y '  t    y  t   0 with initial
8.
conditions y  0  = 0 and y '  0   1, is u  t  denotes the unit step function),
(A) te  t u  t 
(B) e t  te t u  t 


(C) e t  te t u  t 
(D) e t u  t 
Key: (A)
Exp: The operator form of the of given D.E is
D2  2D  1 y  0
The A.E is D2  2D  1  0
  D  1  0  D  1,  1.
2
 y  t   e  t  C1  C2 t 
Given y  0   0 & y '  0   1 i.e, t  0; y  0
from (s); 0  C1  C1  0  y '  0   1
dy
 e  t  C1  C2 t   e  t C2 
dt
dy
At t  0;
 y' 1
dt
From (1),
1  0  C2  0  1C2   1  C2  C2  1
From (1) y(t)  e t (t)  te t u(t)
9.
  2xy dx  2x ydy  dz  along a path joining the origin (0,0,0) and
2
The value of the line integral
2
c
the point (1,1,1) is
(A) 0
Key: (B)
Exp:
(B) 2
(C) 4
(D) 6
Let f  2xy 2 i  2x 2 y j  k
i

 curl f    f 
x
2xy 2
j

y
2x 2 y
k

0
z
1
 f is irrotational
 Consider a straight line passing through  0,0,0  & 1,1,1
i.e,
x y z
   t  x  y  z  t  dx  dy  dz  dt
1 1 1
 2xy dx  2x
2
C
2
ydy  dz 

1
t 0
2t 3 dt  2t 3 dt  dt  
1
t 0
 4t
3
 1 dt   t 4  t   2
1
0
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Let f(x) be a real, periodic function satisfying f   x   f  x  . The general form of its Fourier
10.
series representation would be
(A) f  x   a 0   k 1 a k cos  kx 
(B) f  x    k 1 bk sin  kx 

(C) f  x   a 0  k 1 a 2k cos  kx 

Key:
Exp:

(D) f  x    k 1 a 2k 1 sin  2k  1x

(B)
We know that a periodic function f(x) defined in (-c, c) can be represented by the poisoins series


a
nx
nx
f  x   o   a n cos
  bn sin
2 n 1
c
c
n 1
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
 VR  and coil  VC 
respectively, in volts, are
VR
~
VC
VS
Key:
(A) 65, 35
(C)
(B) 50, 50
12.
The voltage (v) and current (A) across a load are as follows.

(C) 60, 90
(D) 60, 80

v  t  100 sin   t  , i  t   10sin t  60O  2sin 3t   sin 5t 
Key:
Exp:
The average power consumed by the load, in W, is ________.
250
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
 since Pavg 
 P  t  dt , in the above expression
0
st
Only 1 term will result non zero answer
Remaining 2 terms will be 0.
 so directly consider
P  t   100sin t  10 sin t  60
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Pavg  Vrms I rms cos  V  I  
13.
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100 10
1000 1
cos  60 
 250 watt
2 2
2 2
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 R1 to R 6 . Assuming a mix of
directional and nondirectional relays at appropriate locations, the remote backup relays for R 4 are
R1
R2
S1
~
(A) R1 , R 2
R5
R3
R6
~
R4
(B) R 2 , R 6
S2
(C) R 2 , R 5
(D) R1 , R 6
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:
Exp:
(A)
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.
Key:
Exp:
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
(B)
XL  jL  j314 1.6 103  j0.5024
XC 
j
j

  j31847.3376
C 314 10 109
Since XC  XL , 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
A
r  1
B
r  4
2cm
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(B) 2
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Key:
(A) 1
(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 E 
(C) 4
(D) 16
V
d
Electric field in region B = 4kV cm
17.
Key:
Exp:
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 _________.
5
Vprefault
ISC 
X  P.U 
ISC  initial 
IScfinal

1
5
0.2
Consider a liner time-invariant system transfer function H  s  
18.
Key:
Exp:
1
. If the input is cos(t) and
 s  1
the steady state output is Acos  t   , then the value of A is _________.
0.707
1
H   
 tan 1 
2
 1
1
H   
45O
2
So when input is cost then O/P
1
yt 
cos t  45O
2
1
 0.707
So A 
2


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: IO  100A
RMS, diode current =
100
 57.73A
3
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20.
GATE-2016-PAPER-02
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.

24V
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Load
36V
S

2mH

The peak voltage across the solid-state switch (S), in
volt, is _________.
Key: 60
For the network shown in the figure below, the frequency  in rad s  at which the maximum
21.
9
phase lag occurs is.______.
1
in
0
1F
Key:
Exp:
0.316
The given circuit is standard lag compensator
Whose Transfer function
1
1
s  s  1  1  s
G s
1 1  10s
1  s
a 1
s
So   1,   10 the frequency at which maximum phase lag happen
1
1
m 

 0.316 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
2
10V


5A

Vouts

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(A) 0
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(B) 5
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(C) 10
(D) 20
Key: (D)
Exp:
10V 
Writing KVL
VO  10  10  0
10V


5A
VO  20V

VO

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
D
C
B
(A) A
Key:
(A)
Exp:
At A
(B) B
(C) C
(D) D
Fields are additive F1  F2
At C
C

Fields are subtractive F1  F2
At D field is due to one electrode F2
At B field make an angle
F12  F22  2F1F2 cos 
So maximum electric field is at ‘A’
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Q. No. 26 – 50 Carry Two Mark Each


26.
The Boolean expression a  b  c  d   b  c  simplifies to
Key:
(A) 1
(D)
Exp:
F
27.
For the circuit shown below, taking the opamp is ideal, the output voltage Vout in terms of the
=
(C) a.b
(B) a.b
a  b  c  d  b  c

= a d  bb  cc
input voltages V1 , V2 and V3 is
1
V3
V1
1
 
(D) 0

= a  d 1  1 = 1 = 0
9
Vx
Vx


Vout
4
V2
Key:
Exp:
(A) 1.8V1  7.2V2  V3
(B) 2V1  8V2  9V3
(C) 7.2V1  1.8V2  V3
(D) 8V1  2V2  9V3
(D)
Vx  V1 Vx  V2
4V  V2

 0; Vx  1
1
4
5
 Vx  V3  
Vx  Vout
0
1
9
 4V1  V2   V
out
 4V1  V2   V 
5
0
3
5
9
4V  V  5V
 4V1  V2  5V3    1 92 out   0
36V1  9V2  45V3  4V1  V2  5Vout  0
40V1  10V2  45V3
5
 8V1  2V2  9V3
Vout 
Vout
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Let x1  t   X1   and x 2  t   X 2   be two signals whose Fourier Transforms are as shown
28.
in the figure below. In the figure h(t) = e
2 t
denotes the impulse response.
X1  
B1
 B1
2
X 2  
B1
2


 B2
B1
B2
x1  t 
h t  e
2 t
yt
x2  t 
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) 2B1
(B) 2  B1  B2 
Key:
(B)
Exp:
y  t    x1  t  x 2  t  * h  t 
(C) 4  B1  B2 
(D) 
In frequency duration
y    X1   *X2   H  
Max. Frequency of X1    B1
Max, frequency of X 2    B2
Max frequency of X1   *X 2     B1  B2 
Max frequency of H    
Thus max, freq of y     B1  B2 
Max frequency
 Nyquist frequency = 2  B1  B2 
  sin 2t 
The value of the integral 2 
dt is equal to

 t 
(A) 0
(B) 0.5
(C) 1
29.
Key:
(D) 2
(D)
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
Exp:

sin 2 t
 sin 2 t 
2 
dt
 dt  2  2 
t 
t
 
0

 4  e0.t
0
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 sin 2 t

 t is an even function 


sin 2 t
.dt
t
By the defining of L.T; we have
 sin 2t 
 4L 
 ; where S  0
 t 

4
4  sin 2t 
1  2 
L
 ; wheres  0  tan   ; where s  0

  t 
 s 
  sin at 
1  a  
L  t   tan  s  

 
 
Putting s = 0; than

4
4
tan 1        2
2



2
30.
 sin 2t 
dt  2
t 

 
Let y(x) be the solution of the differential equation
y  0   0 and
Key:
Exp:
dy
dx
d2 y
dy
4
 4y  0 with initial conditions
2
dx
dx
 1. Then the value of y (1) is _________.
x 0
7.398
The operate form of given D.E is
D2  4D  4 y  0
The A.E is D2  4D  4  0
  D  2  0  D  2,2
2
  D  2  0  D  2,2
2
 The solution is y  e 2x  C1  C2 x   1
Given that
 from 1 y  e
y  0  0
&
y '0  1
i.e x  0  y  0
i.e at x  0, y'  1
 from (1); 0  1  C1  0
from 1  y1  e2x C2    C1  C2 x  2c 2x
 C1  0
 1  C2  0  C2  1
2x
 0  1.x   y  xe
 y 1  1e 21  e 2  y 1  e 2
2x
 or  y 1  7.389
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
The line integral of the vector field F = 5xziˆ  3x 2  2y ˆj  x 2 zkˆ along a path from (0,0,0) to
31.


(1,1,1) parametrized by t, t 2 , t is _________.
Key:
Exp:
4.4167
F  5xzi  3x 2  2y  j  x 2 z  k
x  t; y  t 2 ; z  t
 dx  dt d  2t dt;  dz  dt
 The line integral of the vector field is
 F.dr   5xzdx  3x
2
 2y dy   x 2 z  dz
C
1

 5t
2
dt  10t 3dt  t 3dt
2
dt  11t 3dt
t 0
1

 5t
t 0
1
1
 t3 
 t4 
 5    11  
 3 0
 4 0
20  23 53
 5  11 
  4.4167
3
4
`12
12
32.
Key:
Exp:
a
x
3 1 
x
Let P = 
. Consider the set S of all vector   such than a 2  b 2  1 where    p   .

b
 y
1 3
 y
Then S is
1
(A) A circle of radius 10
(B) a circle of radius =
10
1
1
(C) an ellipse with major axis along  
(D) an ellipse with minor axis along  
1
1
(D)
3 1 
P 

1 3
a 
 x   a  3 1  x 
 b   P  y    b   1 3  y  3x  y  a  x  3y  b
 
    
 
a 2  b2  1
  3x  y    x  3y   1
2
2
 10x 2  10y2 12xy  1
a  10; b  10; h  6
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 It represents ellipse

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
The length of semi-axes is ab  h 2 r 4   a  b  r 2  1  0
1
1
 or  r 2 
4
16
2
Both r values are positive, so it represents ellipse
1
1
 r   or  r 
2
4
Length of major axis = 2r  1
 64r 4  20r 2  1  0  r 2 
1
Length of minor axis = 2r  2    1
2
4

1
 Equation of the major axis is  a  2  x  hy  0
r1 

 10  4  x  6y  0  x  y  0

1
Equation of the minor axis is  a  2  x  hy  0
r2 

  10  16  x  6y  0  y  x  0
Major axis exists along y = -x and minor axis exists along y = x.
1
 The vector   Lies on the line y = x
1
33.
Let the probability density function of random variable, X, be given as:
3
f x  x   e 3x u  x   ae 4x u   x 
2
where u(x) is the unit step function. Then the value of ‘a’ and Prob X  0 , respectively, are
(A) 2,
Key:
1
2
(B) 4,
1
2
(C) 2,
1
4
(D) 4,
1
4
(A)

Exp:
we have
 f  x dx  1
x


0

 f  x  dx   f  x  dx  1
x
x


0
0


0
4x
 a e dx 
3
2 e
3x
dx  1
 u  x   1 for x  0
 0, other wise
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u   x   1 for x  0
0,otherwise

 ae

4x

0

0
 e4x 
3
3  e3x 
dx   e 3x dx  1  a    
 1
2
 4   2  3  0
0
a 1
a 1
1
 1
 a   0  0  1  1   1    a  2
4 2
4 2
4
 2
Prob  x  0  
0
 f x  x  dx 

0
4x
 a.e dx  a

0
e
4x
dx

0
 e4x 
2
1
 2    1  0 
2
 y   4
The driving point input impedance seen from the source V S of the circuit shown below, in  , is
__________.
34.

IS

VS 
Key:
Exp:
V1
2

2
3
4V1
4
20
The Driving point impedance is nothing but the ratio of voltage to current from the defined port.
V
In this case it is S
Vx
2
V1 
IS

 Writing KCL at node x
V
V
 IS  x  4V1  x  0
3
6
Substituting these in Eq(1)
V  2Is
V  2Is
 IS  S
 8Is  S
0
3
6
2
2
1 1

 VS     IS 1   8  
3
6
3 6

IS

VS 
2
3
4
4V1
Vx
3
V1
6
I1
 VS  2  1  IS  6  4  48  2 

VS 60

 20
IS
3
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The z-parameters of the two port network shown in the figure are Z11  40, Z12  60 ,
35.
Z21  80, Z22  100 . The average power delivered to R L  20, in watts, is ____.
I2
10 I1

20V 

Key:
Exp:
V1

 Z
V2

RL

35.55
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 R L as load let first obtain the thevenin equivalent of 2 port
→ Thevenin equivalent means Vth & R th
Vth  V2 I
2 0
i.e., O.C voltage of port 2
ISC   I2 V2  0 i.e., s.c current of port 2,
R in 
Vth
Isc.
→ Evaluation of Vth .
The Z matrix equation is
V1  40I1  60I2
V2  80I1  100I2
In the above two equations if I2  0 then
V1  40I1
(1)
V2  80I1
(2)
From the input side we can say  V1  20 10I1 
 20  10I1  40I1
2
 I1  2 A Then equation 2 becomes
5
3
2
 V2  80 I1  80   32 V
5
so Vth  V2  32
Evaluation of ISC
In the Z matrix equation if we put V2  0 then
V1  40I1  60I2
…(5)
0  80I1  100I2
…(4)
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10
100
I 2 & V1  20  10I1  20 
I2
8
8
Using these V1 & I1 in equation 3
I1  
100
400
I2  
I 2  60I 2
8
8
 160  100I2   400I2  480I2
20 
 160  20I2  I2  8A
ISC   I2  8A
 R in 
R th  4
Vin 32

 4
Isc
8
 Now the ckt is from port 2is
R L  20
Vth  32
P20   I20  20
2
 32 

 20  35.55watt
 4  20 
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 _____.
C
L
C
L
C
L
Key:
Exp:
IL
3.04
IL  0  2ph  
 jL   j 



X L .XC
3  c 
Zph 

jL
j
X L  XC

3
c
L
1

b
c
3
314 
10  103  C
C  3.04 mF
L
3
C
L3
L3
C
C
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S1
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 _______.
4V
5
5F
Key:
6.99
Exp:
It is given VC 0  4
 
1
 40000
RC
Since it is a source free network we can say
VC  t   VC 0 e t  ; t  0  4e40000 t
R  5, C  54f so
→
 
→
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 
Q  C VC  25 sec   VC 100 sec 
→ VC  4e40000t
6
 VC
 4e40000 2510  1.47
 VC
 4e 40000 10010  0.073
t  25 sec
6
t 100  sec
 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 ______
Generator int ernal bus
inf inite bus
j0.1
j0.5
Line 1
j0.2
~
j0.5
Line 2
j0.1
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Key:
Exp:
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2.286
During fault
Before fault
0.6
0.5
0.1
0.2
0.2
~
~
0.5
0.1
Xeq  0.2 
0.1  0.25
0.1  0.5
 0.5 P.U
2
0.6
0.2
~
EV
Pe  max  
 2EV
Xeq
0.35
Converting Y  
c
0.175
xe
0.35
0.125
C
b
xc
C
0.25
b
0.6
b
xb
 0.25
a
0.375
0.125
a
C
Xac
C

0.0729
0.25
0.0729
0.375  0.125  0.125  0.0729  0.0729  0.375
 1.143
0.0729
Pe  Prefault 
EV
1.143
Pe  max  


 2.286
Pe
during
fault
1.143

 20.5
X ac 
39.
The open loop transfer function of a unity feedback control system is given by
K  s  1
G s 
, K  0, T  0.
s 1  Ts 1  2s 
The closed loop system will be stable if
(A) 0  T 
(C) 0  K 
Key:
Exp:
4  K  1
K 1
T2
T 1
(B) 0  K 
(D) 0  T 
4  T  2
T2
8  K  1
K 1
(C)
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
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 S  TS2  1  2S  KS  K  0
 S  2S2  TS2  2TS3  KS  K  0
 S3  2T   S2  2  T   S 1  k   k  0
k
 2  T  2  k 1
 s3  
0
s  
 s
2T
 2T 
 2T 
 for stability using R  t  criterion
 2  T   K 1 K

 

 2T   2T  2T
K 1
1

  T  2   K  1  K 
K
T2
1
1
1
 T 1 
T  2


`1    
 K  


K T2
K
T2
 T 1 
40.
Key:
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 __________.
294.84
 2  2  1010 
Exp: VS  Vr 1 

18



3142   2  1010 
400  420 1 
    294.84km
18


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
2 400
Exp: cos 2  1
100
2  0
1  36.86
36.86
P
S
P1  400; P2  100
cos 1 
Qmotor  P1 tan 1   P1  P2  tan 2  400 tan36.86  500 tan   300 kW
Smotor  100  j300
cos m  0.3162
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The flux linkage    and current (i) relation for an electromagnetic system is  
42.
 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.
Key:
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 _________.
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
20A
Exp:


`
max
30A
100V

10A

200V

45.
300V
KVA rating  300  200
 6kVA

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 IL a smooth dc current. Switches are
lossless. The real power fed back to the source. In kW is ________.
Load
IL
T1
~ Vin
L
T3
R  20
Bridge
`
T4
T2

E  800V

Key: 6
Exp: V0 
2Vm
200
cos   2 
cos120   200V


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E  V0 800  200

 30A
R
20
Pfedback  V0 I0   200  30  6kW .
I0 
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
46.
devices, the power consumed by the load, in kW, is _________.

30
30
600V
Key:
24
Exp: Vph 
R ph
47.
30
2
2
Vdc 
 600  200 2 V
3
3

200 2
R 30
3Vph 2


 10  PLoad 
 3
3
3
R ph
10

2
 24kW
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 ______.
10mH
Load
S
360V
1mF

400 V

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
Vs  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  VS 
and fundamental AC rms VSC voltage  VC1  , in degree, is________.
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5mH

IS
220VAC ~
1mH
XS
 400V
IS
VS

VC1
I1X1
Key:
9.1 to 9.3
49.
Consider a linear time invariant system x  Ax, with initial condition x(0) at t = 0. Suppose 
 2  2  matrix A corresponding to distinct eigenvalues 1 and 2
respectively. Then the response x  t  of the system due to initial condition x (0) =  is
and  are eigenvectors of
(A) e1t 
Key:
Exp:
(B) e1t 
(C) e2 t 
(D) e2 t   e2 t 
(A)
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
e1t   e
2t
but we are asking w.r.t
Initial condition x  0    only so the response should be e1t
50.
A second-order real system has the following properties:
(a) the damping ratio   0.5 and undamped natural frequency n  10 rad s,
Key:
Exp:
(b) the steady state value of the output, to a unit step input, is 1.02.
The transfer function of the system is
1.02
102
100
(A) 2
(B) 2
(C) 2
s  5s  100
s  10s  100
s  10s  100
(B)


n 2
The standard 2nd order T/F is  K 2
2 
 s  2n s  n 
(D)
102
s  5s  100
2
it is given that   0.5 & n 10
G s  K
100
s  10s  100
2
Now to satisfy the steady state O/P 1.02
y     t
s 0
G s 

1
100
 2
 K  1.02  K  1.02
s  s  10s  100 
1.02  100
102
 2
s  10s  100 s  10s  100
2
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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
Key:
(A)
Exp:
N1 : N2  110 :
(B) 10 3 :1
(C) 1:10
(D)
3 :10
11
 10 3 :1
3
I1N1  I2 N2


Ii 10 3  I 2 .1
I1
1

I2 10 3
52.
The gain at the breakaway point of the root locus of a unity feedback system with open loop
Ks
transfer function G  s  
is
 s  1  s  4 
Key:
(A) 1
(A)
Exp:
G s
(B) 2
(C) 5
(D) 9
Ks
 s  1  s  4 
To find Break away point
dk
 0 where
We need to find the root of
ds
 s  1  s  4     s 2  5s  4 
K


s
s



1

4
d
 d 2
s
s  5s  4    s 2  5s  4 
 s  
dk  ds 
ds


ds 
s2



 S  2S  5  S2  5S  4   0
 2S2  5S  S2  5S  4  0
 S2  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

KS
KS
1 
1 K 1
 s  1 s  4 s2
1 2 
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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
0.4
0.3
Z2 
 0.15P.U
 0.2 P.U ;
2
2
3Vprefault
3
If 

 6p.u
ZO  Z1  Z2 0.15  0.2  0.15
Exp: Z0  0.15 P.U ; Z1 
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
20revolutions
Exp: K  1200re v kwh 
 30 
P  kW   
 hr
 3600 
 P  2kW
54.
z
55.
Key:
Exp:
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 ______.
9.55
Voltage =  B 
Velocity  
B
1m
1m
Voltage
1

 1m s
B
11
i.e, 1m takes
 2  21  2 m Takes 2r
 2r for 1 rotation
 in 1 minute =
Velocity =
60
2
60
 9.55 rpm
2
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Electrical Engineering
Q. No. 1 – 25 Carry One Mark Each
1.

 t   t  , t  0
Consider g  t   
, where t  
 t   t  , otherwise

Here,  t  represents the largest integer less than or equal to t and  t  denotes the smallest
integer greater than or equal to t. The coefficient of the second harmonic component of the
Fourier series representing g(t) is _________.
Key: 0 to 0
t   t 

0 

Exp: Given g  t     


t   t  otherwise 

If we plot the above signal, we get
gt
1
2
3
1
Since this wave form contain hidden half wave symmetry, even harmonics does not exist.
Thus coefficient of second harmonic component of Fourier series will be zero.
2.
A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure
below. The instantaneous voltage and current in phase-a are Van=220sin 100t  V and
i a  10sin 100t  A, respectively. Similarly for phase-b the instantaneous voltage and current
are Vbn  220cos 100t  V and i b  10cos 100t  A, respectively.
ia
a'
Van i b
b'
a

b
Load

Source
n
 
Vbn
n'
The total instantaneous power flowing form the source to the load is
(A) 2200 W
(B) 2200sin2 100t  W
(C) 440 W
(D) 2200sin 100t  cos 100t  W
Key: (A)
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Exp: Pins tan eous  Van ia  Vbn ib
 220sin 100t 10sin 100t  t  220cos 100t 10cos 100 t 
 2200sin 2 100t   2200cos 2 100t 
 2200W
3.
A three-phase, 50Hz, star-connected cylindrical-rotor synchronous machine is running as a
motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor
(UPF). The synchronous reactance of the motor is 30  per phase. The load angle is 30o. The
power delivered to the motor in kW is _______.
Key:
835 to 842
Exp:
Vt  6.6kV
30
V I a Xs E
b
Ia
V
6600 3

cos 
cos30o
E b  4400 volts
Eb 
Ia

E b 30o
E b cos   V
E b cos
~
cos   1
P  3.
E b .V
4400  3810.51
sin   3.
 sin 30o
Xs
30
P  838.31 kW
4.
For a complex number z, lim
z i
(A) -2i
Key: (D)
Exp: lim
z i
z2  1
is
z3  2z  i  z 2  2 
(B) -i
(C) i
(D) 2i
z2  1
zi
2i
 lim 2

 2i
3
2
z

i
z  2 1  2
z  2z  i  z  2 
5.
Consider an electron, a neutron and a proton initially at rest and placed along a straight line such
that the neutron is exactly at the center of the line joining the electron and proton. At t=0, the
particles are released but are constrained to move along the same straight line. Which of these
will collide first?
(A) The particles will never collide
(B) All will collide together
(C) Proton and Neutron
(D) Electron and Neutron
Key: (D)
mp
mn
me
Exp:
e
n
p
q  e
q  e
q0
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The Gravitational force of alteration between any two particles shown above is made much
negligible when compared to coloumbic force of alteration between electron and proton.
Force of alteration
F
e 2
F
, accelleration q  ;q e  q p
2
4o r
M
Due to this force, the electron as well as the proton will move towards each other, since
me  m p , the speed and acceleration of the electron will be much greater than that of proton.
This causes electron to collide with the neutron faster when compared to proton.
6.
Let z  t   x  t   y  t  , where “  ” denotes convolution. Let C be a positive real-valued
constant. Choose the correct expression for z (ct).
(A) c.x  ct   y  ct 
(B) x  ct   y  ct 
(C) c.x  t   y  ct 
(D) c.x  ct   y  t 
Key: (A)
Exp: z  t   x  t  * y  t 
 z  s   x  s  .y  s 
Converting into Laplace transform and applying time sealing property.
1
z  ct   z  s / c 
c
1
  s / c y s / c
c
1
1
 c  s / c y s / c 
c
c
z  ct   c.x  ct  * y  ct 
7.
A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus
matrix are Y11   j12pu, Y22   j15pu and Y33   j7pu
Bus  1
Bus  2
jq
jr
jp
Bus  3
The per unit values of the line reactances p, q and r shown in the figure are
(A) p  0.2, q  0.1, r  0.5
(B) p  0.2, q  0.1, r  0.5
(C) p  5, q  10, r  2
(D) p  5, q  10, r  2
Key: (B)
Exp: Y11  y10  y12  y13   j12    jq1  jr1 
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 q1  r1  12
Y22   j15  y 20  y 21  y 23    jq1  jp1 
 p1  q1  15
Y33   j7  y30  y31  y32    jp1  jr1 
 p1  r1  7
solving P1  5,q1  10, r1  2  admit tan ces 
P  0.2,q  0.1, r  0.5  reac tan ces 
The equivalent resistance between the terminals A and B is ______  .
8.
2
1
1
A
6
3
1
6
0.8
3
B
Key: 2.9 to 3.1
Exp: The Ckt will become
2
1
1
1
A
3
1 
3
6
6
1.2
B
0.8
0.8
R AB  1  1.2  0.8  3
The Boolean expression AB  AC BC simplifies to
9.
(A) BC  AC
Key: (A)
(B) AB  AC  B
Exp: AB  AC  BC
A
0
1
 BC  AC
10.
BC 00
01
11
(C) AB  AC
(D) AB  BC
10
1
1
1
BC
1
AC
The following measurements are obtained on a single phase load:
V  220V  1%, I  5.0A  1% and W  555W  2%. If the power factor is calculated using
these measurements, the worst case error in the calculated power factor in percent is ________.
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Key:
4 to 4
Exp:
P  VIcos 
cos  
11.
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P
555  2%
555  2%


 0.504  4%
V.I  220  1%  5  1%  1100  2%
The transfer function of a system is given by,
Vo  s 
Vi  s 

1 s
Let the output of the system be
1 s
vo  t   Vm sin  t    for the input vi  t   Vm sin  t  . Then the minimum and maximum
values of  (in radians) are respectively
(A)


and
2
2
(B)

and 0
2
(C) 0 and

2
(D)  and 0
Key: (D)
Exp:
Vo  s 
Vi  s 

1 s
 H s
1 s
H    1  2Tan 1, If   0,   0
If   ,   
Vo  t   Vm sin  t  2Tan 1   
12.
3
2

The matrix A   0
1

2
1
2

1 0  has three distinct eigenvalues and one of its eigenvectors is
3
0

2
Which one of the following can be another eigenvector of A?
0
(A)  0 
 1
0
 1
(B)  0 
 0 
1
(C)  0 
 1
1 
0
 
1 
1
(D)  1
 1 
Key: (C)
1
Exp: By the properties of Eigen values and Eigen vectors, another eigen vector of A is  0 
 1
The eigen vectors corresponding to distinct eigen values of a real symmetric matrix are
orthogonal i.e., pair wise dot product is zero.
13.
For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the
following statements is TRUE?
(A) All of the four are majority carrier devices.
(B) All the four are minority carrier devices
(C) IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority
carrier devices.
(D) MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier
devices.
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Key: (D)
Exp: MOSFET → Majority carrier device (NMOS, PMOS)
Diode → both majority & minority carrier device
Transister → Npn, pnp
IGBT → input is MOSFET, Output is BJT
14.
Consider the unity feedback control system shown.
The value of K that results in a phase margin of the
system to be 30o is _______.
Key: 1.01 to 1.06
Exp: PM  180  G gc
G s 
U s  

Ke  s
s
Y s
Ke s
s
For gc | G  s  |
K
1

gc  K  G  s   
180
 90o

180o
 90o

180o

K
 60  K   1.047

3
30o  180o  K 
15.
A solid iron cylinder is placed in a region containing a uniform magnetic field such that the
cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the
cylinder will
(A) bend closer to the cylinder axis
(B) bend farther away from the axis
(C) remain uniform as before
(D) cease to exist inside the cylinder
Key: (A)
Exp: Flux always chooses less reluctance path. So flux tried to flow inside the conductor and closer
to the axis of the cylinder.
16.
Let I  c   R xy2dxdy, where R is the region shown in the figure and c  6 104. The value of I
equals________.
y
10
R
2
1
5
x
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Key: 0.99 to 1.01
Exp:
 xy dxdy    xy dxdy    xy dxdy
2
2
R
2
R1
R2
5

2
 
x 1 y  0
5
2
2x
 x 2   y3  5  y3 
2
xy
dx

      x   dx

 2 1  3 0 1  3  2
y 2
5
xy 2 dxdy 
2x

x 1
5
5
5
 x2  
1   x5 
8 1
3
 12      x  8x  8  dx  32  8    8   
3   5 1
3 3 1
 2 1 

1  24992 
24992
 32  
  32 
3 5 
15
2
 C xy 2 dxdy   24992   104  0.99968  1
5
R
OR
 2x 2 
R xy dxdy  x1  y0 xy dy dx


5
2
2x
5
5
 y3 
8
  x   dx   x  x 3  dx
3 
31
1 
5
8  x5 
8
24992
     3124  
3  5 1 15
15
24992
2
 C  xy 2 dxdy 
 104  6   2.4992  0.9968  1
15
5
R
17.


Consider the system with following input-output relation y  n   1   1 x  n 
n
where, x[n] is the input and y[n] is the output. The system is
(A) invertible and time invariant
(B) invertible and time varying
(C) non-invertible and time invariant
(D) non-invertible and time varying
Key: (D)


Exp: Given y  n   1   1 x  n 
n
For time invariance


y' n   1   1 x  n  n o   (1)
n

y  n  n o   1   1
n no
 x  n  n   (2)
o
Since (1) is not equal to (2)
System is time variant
For inverse system
For each unique x  n  , there should be unique y  n 
If x  n     n  1
n
y  n   1   1    n  1


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 y 1  0
if x  n   2  n  1
y  n   1   1  2  n  1


y(1)=0
For two different inputs we have same output. Thus one to one mapping is not possible. Hence
the systems is non invertible
n
18.
The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep
generator has a response time of 50 ns. In order to display correctly, a delay line of
(A) 150 ns has to be inserted into the y-channel
(B) 150 ns has to be inserted into the x-channel
(C) 150 ns has to be inserted into both x and y channels
(D) 100 ns has to be inserted into both x and y channels
Key: (A)
Exp: The delay line should be inverted in VDP (Y-channel) only.
19.
A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure
below. For a star connected resistive load of 20  per phase, the load power for 120o device
conduction, in kW is __________.
20
600V

20

20
Key: 8.5 to 9.5
Exp: Vdc  600V
R L  20 / Ph 120o mod e
RMS value of phase voltage  VP   0.4082Vdc  244.92V
Load power 
20.
3Vph 2 3  244.922

 8.99kW  9kW
R
20
3
2
A closed loop system has the characteristic equation given by s  Ks   K  2  s  3  0. For
this system to be stable, which one of the following conditions should be satisfied?
(A) 0 < K < 0.5
(B) 0.5 < K < 1
(C) 0 < K < 1
(D) K > 1
Key: (D)
3
2
Exp: Given CE  s  ks   k  2  s  3  0
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For stable
k  k  2  3
k 2  2k  3  0
 k  1 k  3  0
k  1  k  3
k 1
(OR)
By R-H criteria
s3
1
k2
2
k
3
s
k  k  2  3
s
0
k
s0
3
k  0   k  3 k  1  0
K  0  k  1  k  3  k  1
21.
A 4 pole induction machine is working as an induction generator. The generator supply
frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in
RPM is
(A) 1350
(B) 1650
(C) 1950
(D) 2250
Key: (C)
Exp:
Ns 
120  60
 1800 rpm
4
Rotor speed should be greater than syn.speeed, to ge inductance generator mode.
N  NS
S r
NS
fr
5
1

f r  sf  s  f  60  12 


 N r  NS 1  S
1

N r  1800 1    1950 rpm
 12 
22.
For the circuit shown in the figure below, assume that diodes D1, D2 and D3 are ideal.
D1
R
 v1 
v  t    sin 100t  V
D2
D3
R

v2

The DC components of voltages v1 and v2, respectively are
(A) 0 V and 1 V
(B) -0.5 V and 0.5 V
(C) 1 V and 0.5 V
(D) 1 V and 1 V
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Key: (B)
Exp: For half wave Rectifier Vdc 
Vm

V1  Vdc for  ve pulse  Vdc for  ve pulse 
V2 
    1
     1  0.5V
2    2

 0  0.5V
2
23.
A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load
buses indexed as L1, L2, L3, L4, L5, L6. The generator bus G1 is considered as slack bus, and
the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply
the required reactive power demand, and hence it is operating at its maximum reactive power
limit. The number of non-linear equations required for solving the load flow problem using
Newton-Raphson method in polar form is _______.
Key: 14 to 14
Exp: Total no of buses=10
Given G1=slack bus, G2=generator/PQ bus
 G 3 ,G 4 are PV buses
PQ buses  L1 , L 2 , L5 , L 6 (4)
Voltagecontrolled PV buses  L3 , L 4 (2)
Minimum no of nonlinear equations to be solved  2  10  2  4  14
R1
I
24.
The power supplied by the 25 V source in the
figure shown below is ________W.
Key: 248 to 252
Exp: KCL
25V
I  0.4I  14
 17V 



R2
14A
0.4I

 1.4I  14
 I  10A
The power supplied by 25 V = 25  10  250W
25.
In the converter circuit shown below, the switches are controlled such that the load voltage vo(t)
is a 400 Hz square wave.
S3
S1
220V

LOAD

S4
 v t 
0
S2
The RMS value of the fundamental component of vo(t) in volts is _______.
Key : 196 to 200
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Exp: Vo  t  
4Vs
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
4VS
1
1


sin

t

sin
3

t

sin
5

t


sin  nt 



3
5

 n 1,3,5 n
4 VS
2 n
4 VS
Vol 

 0.9VS
2 
 0.9  220  198.069V
Vo  t 
Von 
Vs  220V

2
t
Vs  220V
Q. No. 26 – 55 Carry Two Marks Each
26.
The output expression for the Karnaugh map shown below is
CD
AB
(A) BD  BCD
Key: (D)
CD
Exp:
00
AB
00
01
11
10
00
0
0
0
0
01
1
0
0
1
11
1
0
1
1
10
0
0
0
0
(B) BD  AB
01
11
10
00
0
0
0
0
01
1
0
0
1
11
1
0
1
1
10
0
0
0
0
(C) BD  ABC
(D) BD  ABC
BD
ABC
27.
A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field
circuit resistances are 0.4  and 0.1  respectively. The load torque varies as the square of the
speed. The flux in the motor may be taken as being proportional to the armature current. To
reduce the speed of the motor by 50% the resistance in ohms that should be added in series with
the armature is _________.
Key: 9.5 to 12
Exp:
E b1  220  30  0.5   205 volts
E b2  220  I a 2  0.5  R X 
Given T  N2 and   IR
We know that, in series motor  T  Ia2
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T  Ia2  N 2
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Ia2
30A
Ia 2
N2

 Ia 2  0.5. Ia1  15 Amp
N1 Ia1
Eb

N 2 E b2 1


N1 E b1 2
N
0.1
0.1
0.4
220V
Rx
220V
0.4
E b1
0.5N1 220  15  0.5  R x  30


N1
205
15
E b2
 R x  10.75 
28.
The transfer function of the system Y(s)/U(s) whose state-space equations are given below is:
 x 1  t   1 2   x1  t   1 


    ut

 x 2  t   2 0  x 2  t  2
x t
y  t   10  1 
 x 2  t 
(A)
s
s  2
2
 2s  2 
(B)
s
s  2
2
 s  4
(C)
s
s  4
2
 s  4
(D)
s
s  4
2
 s  4
Key: (D)
Exp: Given
 x 1  t   1 2   x1  t   1 


    u(t)

 x 2  t   2 0   x 2  t  2
x  t   Ax  Bu
Transfer function = CSI  A B  D
1
Here D = 0
C  1 0
1 2 
1 
A
B 

2 0
2
1
s  1 2  1 
T / F  1 0 
s   2 
 2
2  1 
s
 2 s  1  2 
 
 1 0  2
s s4
 s4 
 2  2s  2 

 1 0  2
s s4
s4
T/F 2
s s4
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29.
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The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB
sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in
position N, the three watt-meters W1, W2 and W3 read 577.35 W each. If the switch is moved to
position Y, the readings of the watt-meters in watts will be:
(A) W1  1732and W2  W3  0
(B) W1  0, W2  1732and W3  0
(C) W1  866, W2  0, W3  866
(D) W1  W2  0and W3  1732
Key: (D)
Exp:
R
3
Y
Supply
B
W1
3
load
W2
W3
Y
S
N
N
If the switch is connected to Neutral, then each wattmeter will read 1  power.
W1  W2  W3  3.Vph .I ph cos   1732.05
VR
 cos   0.51agg.    60
V4
Given that, load drawing Apparent power of 3464 VA.
3VL I L  3464
60
IR
30
VB4
3464
IL 
 5A
3  400
If the switch connected to “Y”, then W2=0
V4
VB
I4
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W1  Vpc Icc cos  Vpc & Icc   VRy I R cos  VRy & I R 
 400  5cos  90   0W
W3  VBy IB cos  VBy & IB   400  5  cos30  1732watts
30.
Two passive two-port networks are connected in cascade as shown in figure. A voltage source is
connected at port 1.
Given V1  A1V2  B1I 2
I1  C1V2  D1I 2
V2  A 2 V3  B2 I3
I 2  C 2 V3  D 2 I3
A1 ,B1 ,C1 ,D1 ,A2 ,B2 ,C2 and D2 are the generalized circuit constants. If the Thevenin equivalent
circuit at port 3 consists of a voltage source VT and impedance ZT connected in series, then
V
A B  B1D2
V1
A B  B1D2
(A) VT  1 , ZT  1 2
(B) VT 
, ZT  1 2
A1A2
A1A2  B1C2
A1A2  B1C2
A1A2
(C) VT 
V1
A B  B1D2
, ZT  1 2
A1  A2
A1  A2
(D) VT 
V1
A B  B1D2
, ZT  1 2
A1A2  B1C2
A1A2  B1C2
Key: (D)
Exp: We can write V1 ,I1 in terms of V3  I3
 V1   A1 B1   A 2 B2   V3 
 I   C D  C D   I 
1  2
2 3 
 1  1
 V1   A1A 2  B1C2  V3   A1B2  B1D 2  I3
I1   C1A 2  D1C 2  V3   C1B2  D1D 2  I3
To find Vth  or  Voc I3  0
V1   A1A 2  B1C2  Vth
 Vth  Voc 
V1
A1A 2  B1C 2
To find ISC V3  0
V1   A1B2  B1D2  ISC  ISC 
To find R th R th 
V1
A1B2  B1D2
VOC A1B2  B1D2

ISC A1A2  B1C2
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31.
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The circuit shown in the figure uses matched transistors with a thermal voltage VT  25mV.
The base currents of the transistors are negligible. The value of the resistance R in k  that is
required to provide 1 A bias current for the differential amplifier block shown is ______.
Key: 170 to 174
Exp: R 
VT  IC1 
ln 
;
IC2  IC2 
IC1  1mA;IC2  1A
R
32.
25  103 1 103 
ln 
 172.7k
6 
1 106
1 10 
The figure below shows an uncontrolled
diode bridge rectifier supplied form a 220
V, 50 Hz 1-phase ac source. The load
draws a constant current Io  14A. The
conduction angle of the diode D1 in
degrees is___________.
Key: 220 to 230
Exp: Average reduction in output voltage due
to Ls
Vo  4f s Ls Io  4  50  10  103   14  28V
Vm
cos   cos     
 
for a diode,   0
V
Vo  m 1  cos  

220 2
28 
1  cos     44.17o

 conduction angle of diode  180    224.17 o
Vo 
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t
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 81
dy
 5ty  sin  t  with y 1  2 . There exists a
dt
unique solution for this differential equation when t belongs to the interval
(A) (-2,2)
(B) (-10,10)
(C) (-10,2)
(D) (0,10)
Key: (A)
33.
Consider the differential equation
sin  t 
dy
5t
 2
y 2
is a first order linear eq.
dt t  81
t  81
Exp: D.E is
5t
I.F = e
2
 t 2 81dt
5
 e2


ln t 2 81
  t 2  81
5/2
 Solution is y  t 2  81
t
y
2

5/ 2
 81 .sin tdt
5/ 2
3/ 2
sin t 2
t  81    t 2  81 sin tdt  c

t  81
2
3/2
t
2
 81
5/2

C
t
2
 81
5/2
If t  9,9 then the solution exists.
Options (b), (c), (d) contain either -9 or 9 or both. So answer is option A
34.
A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running
at 800 RPM. The armature resistance of the generator is 0.1  . If the speed of the generator is
increased to 1000 RPM, the current in amperes supplied by the generator to the DC grid is
_______.
Key:
548 to 552

Exp:
145V


150A
Grid
Ra
0.1
E g1
0.1
E g2
800 rpm
1000 rpm
E g2  Ia 2  0.1  145
E g1  150  0.1  145
E g1  160V
N  E g    constant 
N 2 E g2

N1 E g1
E g2 

150A
Grid
Ra
145V
200  145  Ia 2  0.1 
55
 Ia 2
0.1
Ia 2  550 amps
1000
 160  200 volts
800
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GATE-2017-PAPER-I
In the circuit shown below, the maximum power transferred to the resistor R is _______ W.
3
Key: 3 to 3.1
Exp: To find Vth
5
5  6  10 21

 2.1A
10
10
Vth  5  5  2.1  5.5V
5
6V
5V

Vth

To find R th

I

10V

55
 2.5
55
The maximum power transferred to
 R th 
R


I
36.
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Vth2
5.52

 3.025W
4R th 4  2.5
5

5
R th

Let a causal LTI system be characterized by the following differential equation, with initial rest
condition
dx  t 
d2 y
dy
 7  10y  t   4x  t   5
2
dt
dt
dt
Where x(t) and y(t) are the input and output respectively. The impulse response of the system is
(u(t) is the unit step function)
(A) 2e 2t u(t)  7e 5t u  t 
(B) 2e2t u  t   7e 5t u  t 
(C) 7e2t u  t   2e 5t u  t 
(D) 7e2t u  t   2e5t u  t 
Key: (B)
Exp: Given causal LTI system
d2 y  t 
7dy  t 
5dx(t)
dt
dt
dt
2
 s y  s   7sy  s   10y  s   u x  s   5sx(s)
2


Y s
X s

 10y  t   ux  t  
4  5s
5s  4
 H s 
s  7s  10
 s  2  s  5 
2
Inverse Laplace transform will give h  t  (impulse response).
2
7

s2 s5
h  t   2e2t u  t   7e 5t u  t 
H s 
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37.
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The approximate transfer characteristic for the circuit shown below with an ideal operational
amplifier and diode will be
Key: (A)
38.
The switch in the figure below was closed for a long time. It is opened at t = 0. The current in
the inductor of 2 H for t  0, is
(A) 2.5e4t
Key: (A)
(B) 5e4t
Exp: at t  0


(D) 5e0.25t
8
I 6
50V
(C) 2.5e0.25t
8
IL  O 
50
 5A
64
5
IL  0    2.5A
2
I
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8
For t  0
8
32
T
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32
2H
2.5A
L
2 1
 
Req 8 4
IL     0  bc3 it is a sourcefreeckt 
i L  t   I L      I L  0   I L    e  t /T  2.5e 4t
39.
j20
j20 
  j39.9

 j39.9
j20  pu
The bus admittance matrix for a power system network is  j20
 j20
j20
 j39.9 
There is a transmission line, connected between buses 1 and 3, which is represented by the
circuit shown in figure.
If this transmission line is removed from service, What is the modified bus admittance matrix?
j20
0 
  j19.9

 j39.9
j20  pu
(A)  j20
 0
j20
 j19.9 
j20
0 
  j39.95

 j39.9
j20  pu
(B)  j20
 0
j20
 j39.95
j20
0 
  j19.95

 j39.9
j20  pu
(C)  j20
 0
j20
 j19.95
j20
j20 
  j19.95

 j39.9
j20  pu
(D)  j20
 j20
j20
 j19.95 
Key: (C)
Exp: When the line 1-3 is removed
z13  0.05  z31
y13 
1
  j20,
0.05
y13  y31  0
'
y13
 Half line shunt susceptance = j0.05
2
y'
y11  new  = y11  old   y13  13   j39.9    j20   j0.05   j19.95
2
y'
y33  new  = y33  old   y13  13   j39.9    j20   j0.05   j19.95
2
j20
0 
  j19.95

Modified Bus admittance matrix: yBus new   j20
 j39.9
j20 

 0
j20
 j19.95 pu
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In the system whose signal flow graph is shown in the figure, U1(s) and U2 (s) are inputs. The
Y(s)
transfer function
is
U1 (s)
(A)
(C)
k1
JLs  JRs  k1k 2
(B)
2
k1  U 2  R  sL 
JLs   JR  U 2 L  s  k1k 2  U 2 R
2
(D)
k1
JLs  JRs  k1k 2
2
k1  U 2  sL  R 
JLs   JR  U 2 L  s  k1k 2  U 2 R
2
Key: (A)
Exp:
Y s
U1  s  U
2
s 0
By Masons gain formula
Y s
U1  s 

P11
1   L1  L 2 
Here P1 
1 k1
.
k 2 LJ
1  1
R1
Ls
1 1
L 2  . 2  k 2  k1
LJ s
1 k1
.
2
Y s
s
LJ

R
1
1
U1  s  1  .  . 1 k k
2 1
L s LJ s 2
k1
T/F 2
s LJ  sRJ  k1k 2
L1  
s  1
, a unit step input is applied at time t=0.
s 1
The value of the response of the system at t=1.5 sec is __________.
Key: 0.550 to 0.556
For a system having transfer function G  s  
41.
Exp:
Y s
R s 

s  1
s 1
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1 s 1
.
s 1 s
1
2
Y s  
s s 1
Apply Inverse L.T
Y s 
y  t   u  t   2e  t u  t 
y 1.5   1  2e1.5  1  0.44626
y 1.5   0.5537
42.
The magnitude of magnetic flux density (B) in micro Teslas  T  at the center of a loop of wire
wound as a regular hexagon of side length 1m carrying a current (I=1A), and placed in vacuum
as shown in the figure is __________.
Key: 0.65 to 0.75
Exp: For a finite length conductor B at a point P
I
B  o  cos 1  cos  2 
1
4r
2
P
2
For a given hexagon 4 for a side
3
2
1   2  60o
r
60o
o
60
3
2 60o
I
o I
 cos 1  cos  2 
4r
4  107  1
 6
cos 60o  cos 60o 

4  3 / 2
 6.9  107
Total flux density B  6 
 0.69  107 Tesla
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The figure shows the single line diagram of a power system with a double circuit transmission
line. The expression for electrical power is 1.5 sin  , where  is the rotor angle. The system is
operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the
transmission line circuits is removed, the maximum value of  as the rotor swings, is 1.221
radian. If the expression for electrical power with one transmission line circuit removed is
Pmax sin , the valueof Pmax, in pu is _________.
Key: 1.21 to 1.23
Exp: Given m  1.22rad  69.958
PC
P 
1  sin 1  m 
 1.5 
 1 
 sin 1 
  41.81  0.729 rad
 1.5 
Using equal area criterion
A1=A2
2
m
  Pm0  Pm1 sin  d  
1
P
max1
2
sin   Pm0  d
1.5sin 
1.5
Pm sin 
Pm =1
Pm0 =1
1 2  m

By solving above integration
Pmax1 
44.
Pm0  m  1 
cos 1  cos m

11.221  0.7297 
1.22pu
cos  41.81  cos  69.95 
A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants
for the main and auxiliary windings (at starting): Zm  12.50  j15.75   (main winding),
Za   24.50  j12.75  (auxiliary winding). Neglecting the magnetizing branch the value of the
capacitance (in F ) to be added in series with the auxiliary winding to obtain maximum torque
at starting is _______.
Key: 95 to 100
Exp:
 x  xe 
X 
tan 1  m   tan 1  a
  90
 Rm 
 Ra 
 15.75 
1  12.75  X c 
tan 1 
  tan 
  90
 12.5 
 24.5 
 12.75  X c 
 12.75  X c 
51.562  tan 1 
 90   tan 1 

  38.43
 24.5 
 24.5 
12.75  X c
 0.793  X c  32.194
24.5
1
Xc 
 98.87F
2  50  32.194
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x

x 1
e ,
A function f(x) is defined as f (x)  
where x   . Which one of the
2
1nx

ax

bx,
x

1


following statements is TRUE?
(A) f(x) is NOT differentiable at x=1 for any values of a and b.
(B) f(x) is differentiable at x = 1 for the unique values of a and b
(C) f(x) is differentiable at x = 1 for all values of a and b such that a + b = e
(D) f(x) is differentiable at x = 1 for all values of a and b.
Key: (A) Not matching with IIT key
45.
Exp: Lf 1 1  Lt
f  x   f 1
x 1
x 1
x 1
 Lt
x 1
ex   a  b 
x 1
does not exists, for any values of a and b
 f  x  is not differentiable at x  1 , for any values of a and b.
46.
Consider a causal and stable LTI system with rational transfer function H(z). Whose
5
corresponding impulse response begins at n = 0. Furthermore, H 1  . The poles of H(z) are
4
Pk 
  2k  1  
1
n
exp  j
 for k = 1,2,3,4. The zeros of H(z) are all at z = 0. Let g[n] = j h[n].
4
2


The value of g[8] equals ___________.
Key: 0.06 to 0.065
Only one of the real roots of f  x   x 6  x  1 lies in the interval 1  x  2 and bisection method
47.
is used to find its value. For achieving an accuracy of 0.001, the required minimum number of
iterations is ________.
Key: 10 to 10
ba
Exp: a  1, b  2and n  0.001 using bisection method
2
 2n  1000  n  10 is the minimum number of iterations
48.
Two parallel connected, three-phase, 50Hz, 11kV, star-connected synchronous machines A and
B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid.
Current supplied by both the machines are equal. Synchronous reactances of machine A and
machine B are 1 and 3 respectively. Assuming the magnetic circuit to be linear, the ratio of
excitation current of machine A to that of machine B is ________.
11kV
Key: 2.05 to 2.13
Exp:
 syn. Condencors
 current‟s supplied both the machines
50 MVAR
1
are same
2624.31
 1312.159 Amps
2
As the two motors, supplying reactive power
only, the phasor diagaram will be
3
 I1  I 2 
I1
I2
~
~
A
B
IL 
50 106
 2624.31
3 11103
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E  jIa Xs  Vt
I a Xs
E  V  jIa Xs
Consider magnitudes  E2   V  Ia Xs 
E
 V  Ia XS 
E
Vt
2

2
90o
EA 
 6350.85  1312.159 1
EB 
 6350.85  1312.159  3
2
2
 5038.7 volts
Ia
 2414.14Volts
IfA E A 5038.7


 2.086
IfB E B 2414.14
49.
The positive, negative and zero sequence reactances of a wye-connected synchronous generator
are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal
voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected
between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to
ground fault occurs at the terminals is _______ (assume fault impedance to be zero).
Key: 0.1 to 0.1
3Ef
Exp: If 
Z0  Z1  Z2  3Zn
3 1
0.1  0.2  0.2  3Zn
Zn  0.1P.U
3.75 
50.
Let the signal x  t  

  1
k 
k
k 

 t 
 be passed through an LTI system with frequency
 2000 
response H   , as given in the figure below
The Fourier series representation of the output is given as
(A) 4000+4000cos  2000t   4000cos  4000t 
(B) 2000  2000cos  2000t   2000cos  4000t 
(C) 4000cos  2000t 
(D) 2000cos  2000t 
Key: (C)
Exp: Given x  t  is a periodic signal for which Fourier transform x   is to be calculated
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
x    2  Dn     n
n 
D n is exponential Fourier series coefficient for x(t)
xt
3
2000
1
2000
3
2000
1
2000
2
2000
t
2
2000
1
1 

x t   t    t 

 2000 
Define x  t  over one period
1
sec; o  2000 rad/sec
1000
1
1
 D n  1  e  jno t o ; t o 
To
2000
n
D n  1000 1  e  jn   D n  1   1 1000
Where as To 
At n  0,2,4,...........Dn =0
for even values of n 
0


i.e., Dn  

2000 for odd values of n 
 x    2  D0  D1      2000   D 1    2000   D 2    4000   D 2    4000 
 ..................]
x  
D3
D1
D1
t
6000 4000 2000
Given x   is
D3
2000 4000 6000
x  
500
500

Thus the filtered output is
y    2  D1    2000    D1    2000 
 D1  D1  2000
y    4000        2000       2000  
 y  t   4000cos  2000t 
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The logical gate implemented using the circuit shown below where. V 1 and V2 are inputs (with
0 V as digital 0 and 5 V as digital 1) and VOUT is the output is
(A) NOT
Key: (B)
Exp:
V1
V2
0
0
0
1
1
0
1
1
(B) NOR
Q1
OFF
OFF
ON
ON
Q2
OFF
ON
OFF
ON
Vout
5V
0V
0V
0V
(C) NAND
(D) XOR
Logic Level
1
0
0
0
So, this logic level o/p is showing the functionality of NOR-gate.
52.
A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q
consumed by the load are such that 1 kW  P  2kW and 1kVAR  Q  kVAR . A capacitor
connected across the load for power factor correction generates 1 kVAR reactive power. The
worst case power factor after power factor correction is
(A) 0.447 lag
(B) 0.707 lag
(C) 0.894 lag
(D) 1
Key: (B)
P
Exp: Under worst case,
2
Pmax  2kW
1
Q max  2kVAR
Q
1  tan 1  45o
P
cos 45  0.707lag
53.
The input voltage VDC of the buck-boost converter
shown below varies from 32 V to 72 V. Assume that
all components are ideal, inductor current is
continuous, and output voltage is ripple free. The
range of duty ratio D of the converter for which the
magnitude of the steady state output voltage remains
constant at 48 V is
2
3
2
3
(A)  D 
(B)  D 
(C) 0  D  1
5
5
3
4
Key: (A)
(D)
1
2
D
3
3
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Exp:
Vdc  32V
Vdc  72V
Vo  48V
Vo  48V
Vo
D

Vdc 1  D
Vo
D

Vdc 1  D
48
D

32 1  D
3
D

2 1 D
3  3D  2D
2
D

3 1 D
3D  2  2D
3  5D  D 
5D  2
2
D
5
3
5
2
3
D
5
5
A three-phase, three winding  /  / Y (1.1kV/6.6kV/400 V) transformer is energized from AC
mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV
winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line
current in ampere drawn by the 1.1 kV winding from the mains is _______.
54.
Key:
623 to 627
Exp:
3  , 3  winding T F
//Y
per phase representation

 6.6 kV
I1
1.1 kV
3
~
I2
3
900  103
 78.73 Amps
3  6.6  103
I
Iph  2  45.45 36.87o
3
I3
I2 
I2  KI 2 
900 kVA
0.8

300 kVA
0.6
400
3
6.6  103
 45.45  I2  272.7 36  .87 o
1.1  103
300  103
 433.01 Amp
3  400
I ph  I L  I3  433.01 53.13
I3 
I3 
400 3
 433.01  I3  90.91 53.13o
3
1.1  10
I1  I2  I3  I1  360.87 40.91  I1  3 I1  625.05 40.91
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Consider the line integral I   C  x 2  iy2  dz where z = x + iy. The line C is shown in the figure
below.
The value of I is
1
2
(A) i
(B) i
2
3
Key: (B)
Exp: curve „C‟ is y  x  dy  dx

I   x2  i  x 
1
0
2

 dx  idx   1  i 
(C)
3
i
4
(D)
4
i
5
1
2
 x3  2
2
x
dx

2i
    i
0
 3 0 3
1
General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
Research in the workplace reveals that people work for many reasons_________.
(A) money beside
(B) beside money
(C) money besides
(D) besides money
Key: (D)
2.
The probability that a k-digit number does NOT contain the digits 0.5, or 9 is
(A) 0.3k
(B) 0.6k
(C) 0.7k
(D) 0.9k
Key: (C)
k digits
Each digit can be filled in 7 ways as 0, 5 and 9 is not allowed, so each of these places can be
filled by 1, 2, 3, 4, 6, 7, 8.
k
 7
So required probability is   or 0.7 k.
 10 
Find the smallest number y such that y  162 is a perfect cube.
(A) 24
(B) 27
(C) 32
Key: (D)
3.
(D) 36
Exp: Factorization of 162 is 2  3 3 3 3
y 162 is a perfect cube
y  2  3  3  3  3  Perfect cube
For perfect cube 2's & 3's are two more required each.
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4.
After Rajendra Chola returned from his voyage to Indoneisa, he _______ to visit the temple in
Thanjavur.
(A) was wishing
(B) is wishing
(C) wished
(D) had wished
Key: (C)
5.
Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of
Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite
each other?
(A) Rahul and Murali
(B) Srinivas and Anil
(C) Srinivas and Murali
(D) Srinivas and Rahul
Key: (C)
Exp:
Srinivas
Rahul
Arul
Murali
Q. No. 6 – 10 Carry Two Marks Each
6.
Six people are seated around a circular table. There are at least two men and two women. There
are at least three right-handed persons. Every woman has a left-handed person to her immediate
right. None of the women are right-handed. The number of women at the table is
(A) 2
(B) 3
(C) 4
(D) Cannot be determined
Key: (A)
Exp: Out of six people, 3 place definitely occupied by right handed people as atleast 2 women are
there so these two will sit adjacently. Now as only one seat is left it will be occupied by a left
handed man because on right side of this seat is sitting an right handed man.
R  m
R  m
Lw
Lw
R  m
?
Therefore, answer should be 2 women.
7.
The expression
 x  y  | x  y |
2
(A) the maximum of x and y
(C) 1
Key: (B)
is equal to
(B) the minimum of x and y
(D) none of the above
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Exp: If x  y
Exp 
x  y   x  y
2
 y min
If x  y
Exp 
x  y  y  x
2
 The expression
8.
 x min
 x  y  x  y
2
is equal to min imum of x & y
A contour line joins locations having the same
height above the mean sea level. The following is
a contour plot of a geographical region. Contour
lines are shown at 25m intervals in this plot. If in
a flood, the water level rises to 525m. Which of
the villages P,Q,R,S,T get submerged?
(A) P, Q
(C) R,S,T
Key: (C)
(B) P,Q,T
(D) Q,R,S
Exp: The given contour is a hill station, the peak point of this hill station is P, it is under a contour of
550. At floods, the water level is 525m. So the village of R, S and T are under a contour of
500. Therefore these villages are submerged.
9.
Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts coloured
red, pink, blue and white respectively. Arun dislikes the colour red and Shweta dislikes the
colour white, Gulab and Neel like all the colours. In how many different ways can they choose
the shirts so that no one has a shirt with a colour he or she dislikes?
(A) 21
(B) 18
(C) 16
(D) 14
Key: (D)
Exp: As there are 4 people A,G,N,S and 4 colours so without any restriction total ways have to be
4  4  16
Now, Arun  dislikes Red and
Shweta  dislikes white
So 16-2=14 ways
“The hold of the nationalist imagination on our colonial past is such that anything inadequately
or improperly nationalist is just not history.”
Which of the following statements best reflects the author‟s opinion?
(A) Nationalists are highly imaginative.
(B) History is viewed through the filter of nationalism.
(C) Our colonial past never happened
(D) Nationalism has to be both adequately and properly imagined.
Key: (B)
10.
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Electrical Engineering
Q. No. 1 – 25 Carry One Mark Each
In the circuit shown, the diodes are ideal, the inductance is small, and Io  0. Which one of the
1.
following statements is true?
(A) D1 conducts for greater than 180o and D 2 conducts for greater than 180o
(B) D 2 conducts for more than 180o and D1 conducts for 180o
(C) D1 conducts for 180o and D 2 conducts for 180o .
(D) D1 conducts for more than 180o and D 2 conducts for 180o
Key: (A)
2.
For a 3-input logic circuit shown below, the output Z can be expressed as
P
Z
Q
R
(A) Q  R
(B) PQ  R
(C) Q  R
(D) P  Q  R
Key: (C)
Exp:
 PQ.Q.QR
 PQ  Q  QR
 Q  QR
QR
3.
P
PQ
Z
 Q  QP  Q 
 A  AB  A  B
Q
PQ.Q.QR
R
QR
An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and
discarded without noticing its colour. The probability to get a red ball in the second draw is
1
4
5
6
(A)
(B)
(C)
(D)
2
9
9
9
49
Key: (A)
Exp:
1
2
4R, 5B
59
R
B
5R
5B
1
2
R
5R, 5B
B
59
R
49
B
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Here R is red ball, B is black ball
 The probability to get a red ball in the second draw is
4.
1 4 1 5 1
   
2 9 2 9 2
When a unit ramp input is applied to the unity feedback system having closed loop transfer
function
C s
R s

Ks  b
 a  0, b  0, K  0  , the steady state error will be
s  as  b
2
(A) 0
(B)
a
b
(C)
aK
b
(D)
aK
b
Key: (D)
Exp: Given T  s  
C s 
R s
Ct   r t   t 
Apply L.T to above equations
E  s   R  s  1  T  s 
ess  C     lt S.E  s   lt .s.
s 0
ess 
5.
s 0
 Ks  b   lt 1 s2  s a  K   lt s   a  K 
1
1



s 0 s 2  as  b
s 2  s 2  as  b  s 0 s s 2  as  b
aK
b
A three-phase voltage source inverter with ideal devices operating in 180o conduction mode is
feeding a balanced star-connected resistive load. The DC voltage input is Vdc . The peak of the
fundamental component of the phase voltage is
V
2Vdc
3Vdc
(A) dc
(B)
(C)



Key: (B)
Exp: Fourier series expansion of line to neutral voltage Vao is given by
(D)
4Vdc


 2Vs 

 sin  nt 
n  6k 1  n 
2V
for n  1, Vao  s  max value 

Vao 
6.

 

The figures show diagrammatic representations of vector fields X, Y and Z respectively.
Which one of the following choices is true?



(A) .X  0,   Y  0,   Z  0



(C) .X  0,   Y  0,   Z  0



(B) .X  0,   Y  0,   Z  0



(D) .X  0,   Y  0,   Z  0
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Key: (C)
Exp: for x Divergence not equal to zero    x  0
for y
Divergence  0 
  t  0
Curl  0

for z
Divergence  0 
  z  0
Curl  0

7.
Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green
(vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of
vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time
(in minutes) for the vehicle at the junction is ________.
Key: 0.9 to 0.9
8.
Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million
electrons are added to this sphere, these electrons will be distributed.
(A) uniformly over the entire volume of the sphere
(B) uniformly over the outer surface of the sphere
(C) concentrated around the centre of the sphere
(D) along a straight line passing through the centre of the sphere
Key:
Exp:
(B)
For a perfect conductor the charge is present only on the surface.
i.e,
Pu  0
 inside the conductor
E0 
The transfer function C  s  of a compensator is given below.
9.
s 
s 

1 
 1 

0.1  100 
C s  
s
1  s  1  
 10 
The frequency range in which the phase (lead) introduced by the compensator reaches the
maximum is
(A) 0.1    1
(B) 1    10
(C) 10    100
(D)   100
Key: (A)
10.
The figure show the per-phase representation of a phase-shifting transformer connected between
buses 1 and 2, where  is a complex number with non-zero real and imaginary parts.
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For the given circuit, Ybus and Zbus are bus admittance matrix and bus impedance matrix,
respectively, each of size 2  2. Which one of the following statements is true?
(A) Both Ybus and Zbus are symmetric
(B) Ybus is symmetric and Zbus is unsymmetric
(C) Ybus is unsymmetric and Zbus is symmetric
(D) Both Ybus and Zbus are unsymmetric
Key: (D)
 yt
 2
a
Exp: YBUS  
  yt

 a
z BUS  y bus 1
 yt 
a* 


yt 

11.
A phase-controlled, single-phase, full-bridge converter is supplying a highly inductive DC load.
The converter is fed from a 230 V, 50 Hz, AC source. The fundamental frequency in Hz of the
voltage ripple on the DC side is
(A) 25
(B) 50
(C) 100
(D) 300
Key: (C)
Exp:
Vo
Vm

 
o
2
2  
Q
For one input pulse, Vo has 2 pulses
 frequency of Vo ripple = 2f supply  2  50  100Hz
12.
Let x and y be integers satisfying the following equations
2x 2  y 2  34
x  2y  11
The value of  x  y  is ________.
Key: 7 to 7
Exp:
Clearly x = 3 and y = 4 satisfies the given two equation
x  y  7
13.
Consider a function f  x, y, z  given by
f  x, y,z    x 2  y2  2z 2  y2  z 2 
The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is
________
Key: 40 to 40
f
Exp:
  y 2  z 2   2x  at x  2, y  1, z  3  1  9  4   40
x


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14.
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For the given 2-port network, the value of transfer impedance Z21 in ohms is_______
Key: 3 to 3
Exp:
V
Z21  2
I1
I1
I1
I2  0

I1
 2I1  3I1
2
V
 Z21  2  3
I1
V2  2 
15.
I1
2
2
 2x
2
I1
4
V1

2


I1
2
2


V2
2I1

The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules
transferred from the DC source until steady state condition is reached equals ______. (Give the
answer up to one decimal place.)
Key: 99 to 101
Exp:
Before initial charge on the capacitor  0  Vc  0   0V
Final voltage Vc     10V
To find time constant 
10  10
 5
10  10
  R eq ; C  5
R eq 
5
5
5
Vc  t   VC      VC  0   VC     e  t   10  10e  t 
d
 i C  t   C C  2e t 5
dt
We know that i r  i C     2e

5

5
t 5
Instantaneous power p   i r  10  2e t 5  20e t 5


0
0
Energy transferred   pdt   20e

t 5
e t 5
dt  20
 100 0 1  100J
1 5 0
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The figure below shows the circuit diagram of a controlled rectifier supplied from a 230 V, 50
Hz, 1-phase voltage source and a 10:1 ideal transformer. Assume that all devices are ideal. The
firing angles of the thyristors T1 and T2 are 90o and 270o , respectively.
The RMS value of the current through diode D 3 in amperes is ________
Key: 0 to 0
Exp: 0A since D2 is OFF and it will not turn ON for R load.
17.
In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of
the Jacobian is 100  100. If there are 20 PV buses in addition to PQ buses and a slack bus, the
total number of buses in the system is ________.
Key: 61 to 61
Exp: Given the size of bus is 100*100.
so [J]= 100
we have formula for [J] = [2n-m-2]
100= [2n-20-2]
total no.of buses ,n = 61
18.
A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The
speed of the rotor flux in mechanical rad/sec, sensed by a stationary observer, is closest to
(A) 1500
(B) 1470
(C) 157
(D) 154
Key:
(C)
Exp:
3
4P
400V
S.C.I.M
s  0.02
r  N r
Rotor flux speed is same as stator flux speed.
120  50
 1500
4
2N 2  1500
Ws 

 157.08 rad sec
60
60
Ns 
19.
Two resistors with nominal resistance values R1 and R 2 have additive uncertainties
R1 and R 2 , respectively. When these resistances are connected in parallel, the standard
deviation of the error in the equivalent resistance R is
2
 R
  R

R1   
R 2 
(A)  
 R1
  R 2

2
2
 R
  R

R1   
R 2 
(B)  
 R 2
  R1

2
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2
2
 R 
 R 
(C)  
 R 2  
 R 1
 R1 
 R 2 
Key:
(A)
Exp:
R eq 
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2
2
 R 
 R 
(D)  
 R1  
 R 2
 R1 
 R 2 
R1R 2
R1  R 2
2
2
 R  2  R  2

  R1  
 R 2
 R1 
 R 2 
OR
2
 R
  R

R1   
R 2 

 R1
  R 2

2
The nominal-  circuit of a transmission line is shown in the figure.
20.
Impedance Z  100 80o  and reactance X  3300 . The magnitude of the characteristic
impedance of the transmission line, in  , is _______________. (Give the answer up to one
decimal place.)
Key: 404 to 408
y 1
Exp: 
2 x
2
2
y 
 6.06 10 4
x 3300
z
100
z0 

 406.2
y
6.06  104
21.
The pole-zero plots of three discrete-time systems P, Q and R on the z-plane are shown below.
Which one of the following is TRUE about the frequency selectivity of these systems?
(A) All three are high-pass filters.
(B) All three are band-pass filters.
(C) All three are low-pass filters.
(D) P is a low-pass filter, Q is a band-pass filter and R is a high-pass filter.
Key:
(B)
Exp:
  0 rad / samples reprsent lowfrequencies
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   rad / samples reprsent highfrequencies
Since zeros are located at   0 rad / samples and   rad / samples they cannot be high pass
and low pass filters. Thus they all replresent band pass filters.
The mean square value of the given periodic waveform f  t  is_________
22.
Key:
6 to 6
Exp: Mean square value 
f 2 t
Area under the squarred function
Period of the function
16
Area  16   0.7  0.3  4  2.7  0.7 
 16  8  24 volt  second
Period  2.7   1.3  4
Mean square value 
23.
4
24
6
4
1.3
0.3
0.7
2.7
A stationary closed Lissajous pattern on an oscilloscope has 3 horizontal tangencies and 2
vertical tangencies for a horizontal input with frequency 3 kHZ. The frequency of the vertical
input is
(A) 1.5 kHz
(B) 2 kHz
(C) 3 kHz
(D) 4.5 kHz
Key:
(D)
Exp:
3
9
f r   3   4.5 kHz
2
2
n 3
fy
fx

nx
ny
ny  2
24.
Let y 2  2y  1  x and x  y  5. The value of x  y equals _________. (Give the answer up
to three decimal places)
Key: 5.7 to 5.8
Exp: y 2  2y  1  x  x  y  1
 x  y  5 gives 2y  1  5  y  3
x  4
 x  y  4  1.732  5.732
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If a synchronous motor is running at a leading power factor, its excitation induced voltage  E f  is
25.
Key:
Exp:
(A) equal to terminal voltage Vt
(B) higher than the terminal voltage Vt
(C) less than terminal voltage Vt
(D) dependent upon supply voltage Vt
(B)
Higher than the terminal voltage.
Ef
iIa X s
V
Ei
Ia
Q 
Q. No. 26 – 55 Carry Two Marks Each
26.
Which of the following systems has maximum peak overshoot due to a unit step input?
100
100
(A) 2
(B) 2
s  10s  100
s  15s  100
100
100
(C) 2
(D) 2
s  20s  100
s  5s  100
Key: (C)

Exp: Peak over shoot  e
12
If   0,peak over shoot 100%  Maximum 
In General 
If   1 peak over shoot  0%  Minimum 
Here which of the following has '  ' value less, the system will have maximum over shoot.
Option „A‟, n  10, 2n  10    0.5
Option „B‟ n  10, 2n  15    0.75
Option „C‟ n  10, 2n  5    0.25
Option „D‟ n  10, 2n  20    1
So, option „C‟ is correct
(OR)
By Inspection, see all options n  cons tan t, 2n varies, so, 2n less means, that system
have maximum over shoot.
27.
Consider an overhead transmission line with 3-phase, 50 Hz balanced
system with conductors located at the vertices of an equilateral triangle
of length Dab  Dbc  Dca  1m as shown in figure below. The resistance
of the conductors are neglected. The geometric mean radius (GMR) of
each conductor is 0.01m. Neglecting the effect of ground, the magnitude
of positive sequence reactance in  / km (rounded off to three decimal
places) is ________
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Key: 0.271 to 0.301
Exp: Deq  3 Dab Dbc Dca  1m  GMD
DS  GMR  0.01m
Inductance/phase/m  2  107 ln
Dm
 1 
7
 2  107 ln 
  9.21 10 H
DS
 0.01 
Inductance/phase/km  9.21104 H
Reactance  L  2 50  9.21104  0.2892 / km
28.
Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and
4% respectively from no load to full load. Both the generating units are operating in parallel to
share a load of 600 MW. Assuming free governor action, the load shared by the larger unit is
_______ MW.
Key:
Exp:
395 to 405
Assume No – load speed regulations are equal
% Speed Reg
A
x 4%
B
G
6%
F
E
300MW
C
D

400MW
H
Power
Power
From similar triangles method
A

F
6%

E
x


B
D
300
x

G


C
400
B
4%

H
BG AB

CH AC
FB AB

ED AD
P1  300 
A

x
6
P2  400 
x
4
P2  100x
P1  50x
Given that P1  P2  600MW
150x  600
x4
 The load supplied by largest machine is P2=100×4=400MW
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For the network given in figure below, the Thevenin‟s voltage Vab is
29.
(A) -1.5 V
Key:
Exp:
(B) -0.5 V
(C) 0.5 V
(A)
The equivalent CKT is
10
5
(D) 1.5 V
Vth
Apply nodal
Vth  30 Vth Vth  16


0
15
10
10
2Vth  60  3Vth  3Vth  48  0
30V
a


 16V

10
b
 8Vth  12
 Vth  1.5V
30.
The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples
uniquely. The required minimum sampling rate is
(A) 1000 samples/s
Key:
Exp:
(B) 1500 sample/s
(C) 2000 samples/s
(D) 3000samples/s
(B)
500
f cos 1000t 
 Convolution 
x f 
500
f
500
500
f
Consider
1
F
cos 1000t  
   f  500     f  500 
2
Input signal to the LTI system is
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1
x  f     f  500     f  500 
2
1
1
W  f     f  500     f  500 
2
2
If the input signal is defined as w(t) then its Fourier transform can be drawn as follows:
 f 
1000
0
1000
f
H f 
sin 150t 
1500 sinc 1500t 
t
 f 
 H  f   rect 

 1500 
Given h  f  
 Y  f   W  f  H  f  has a max frequency 750 HZ
∴ Minimum sampling rate = 1500 HZ
750
f
750
A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance R a  0.02.
31.
When the motor operates at rated speed and with rated terminal voltage, the electromagnetic
torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the
current drawn by the motor from the 220 V supply is
(A) 34.2 A
(B) 30 A
(C) 22 A
(D) 4.84 A
Key:
Exp:
(B)
Separately excited d.c. motor
2NT
60
2  900  70
E b Ia 
 6597
60
6597
Ia 
...(1)
Eb
Ia
P
V  Ia R a  E b
220 
k
0.02
V  220V
Eb
6597
 0.02  E b
Eb
220E b   6597  0.02   E 2b
By solving above equation
We get E b1  219.39, E b2  0.61
Ia 
V  E b 220  E b
220  219.39

 Ia 
 Ia  30.5 Amps
Ra
0.02
0.02
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 
 
A cascade system having the impulse responses h1  n   1, 1 and h 2  n   1,1 is shown in the
32.


figure below, where symbol ↑ denotes the time origin.
The input sequence x  n  for which the cascade system produces an output sequence


y  n   1,2,1, 1, 2, 1 is

 
x  n   1,1,1,1
 
(D) x  n   1,2,2,1
(A) x  n   1,2,1,1
(B) x  n   1,1,2,2

(C)



Key: (D)
Exp:
Y1  
h(n)  h1[n]*h2[n]  {1, 0, 1}

Y[n]=h(n)*x(n)
Given
y[n]  {1, 2, 1,  1,  2,  1}
By observation x[n] should be{1,2,2,1}
2000
1000
For the circuit shown in the figure below, it is given that VCE 
33.

1000 2000
VCC
. The transistor has
2
  29and VBE  0.7V when the B-E junction is forward biased.
RB
is
R
(B) 92
For this circuit, the value of
(A) 43
Key:
(C) 121
(D) 129
(D)
Exp: Given VCE 
Vec 10
  5V
2
2
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10  1    I B  4R  I B R B  0.7  1    I B .R
10  30I B  4R  I B R B  0.7  30I B  R
9.3  I B 120R  30R  R B 
9.3  I B 150R  R B 
...(1)
10  4 R 1    I B  VCE  1    I B  R
10  120RI B  5  30I B .R
5
1

...(2)
150R 30R
Substituting equation (2) in equation (1)
1
9.3 
150  R B 
30R
R
RB
279  150  B ;
 279  150  129
R
R
IB 
34.
A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The
kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running
steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical
degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles,
the value of the power angle after 5 cycles is ________ electrical degrees.
Key: 12.5 to 12.9
Pa  Pm  Pe
Exp:
 60  0  60mw
GH
1000
1


180f 180  50 9
10
t  10cycles   0.25sec
50
5
t  5cycles   0.1sec
50
m
2
Pa t 2
60  0.1
. 

 2.7
1
m 2
2
9
New ratio,   10  2.7 12.7

35.
For the circuit shown below, assume that the OPAMP is ideal.
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Which one of the following is TRUE?
(A) vO  vS
(B) vO  1.5vS
Key:
Exp:
(C) vO  2.5vS
(D) vO  5vS
(C)
At node (1)
Vx 
Vs  2R Vs

4R
2
At node (2)
R
Vx Vx  Vy

0
R
R
2Vx  Vy ;
R
Vy 
2Vs
 Vs
2
R
Vy
R
V

y
 Vx 
V

y
 Vo 
R
R
V
Vs  Vs  s  Vs  Vo  0;
2
Vs
3Vs 
 Vo
2
5V
Vo  s ;
Vo  2.5Vs
2
R
3
Vy
2
At node (3)
36.
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
Vx
Vo

0
2R 1
Vx
Vs
2R
The root locus of the feedback control system having the characteristic equation
s  6Ks  2s  5  0 where K  0, enters into the real axis at
(A) s  1
(C) s  5
(B) s   5
(D) s  5
Key: (B)
Exp: C.E  s2  6ks  2s  5  0
6ks
1 2
0
s  2s  5
6ks
G  s   1  2
s  2s  5
  s 2  2s  5 
1
5
K
  s  2  
6s
6
s
dk
 5
 0  1  2   0
ds
 s 
j
2j
 5
1
2j
s2  5  0  s   5
S=  5 it enters
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For the synchronous sequential circuit shown below, the output Z is zero for the initial
conditions QA QBQC  Q'A Q'BQ'C  100.
The minimum number of clock cycles after which the output Z would again become zero is
________
Key: 6 to 6
Exp: Upper part of the circuit is ring counter and lower part of the circuit is Johnson counter as per
the connection established. Outputs of the Ring counter and Johnson counter is given to Ex-OR.
Gates, whose output is given to the three inputs OR-gate.
Ring counter output
Ring counter output
Jonson counter
output
QA
QB
QC
Q1A
Q1B
Q1C
Z
1
0
0
1
0
0
0  Inital valume
0
1
0
1
1
0
1
1 CP
0
0
1
1
1
1
1
2 CP
1
0
0
0
1
1
1
3CP
0
1
0
0
0
1
1
4CP
0
0
1
0
0
0
1
5CP
1
0
0
1
0
0
0
6CP
So, output Z will become again 1 after 6 clock pulses.
38.
In the circuit shown below, the value of capacitor C required for maximum power to be
transferred to the load is
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(B) 1 F
(A) 1 nF
Key:
Exp:
(C) 1 mF
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(D) 10 mF
(D)
To get the maximum power the load Ckt must be at resonance i.e. imaginary part of load
impedance is zero.
1
R 1  jRC 
R
jC
ZL  jL 
 jL 
 jL 
1
1  jRC
1  2  R 2  C2
R
jC
j term  0
R
R 2 C
R 2C
C

L

 5  103 
2
2 2
2 2 2
1   R C
1  R C
1  104 C 2
From options C  10mF will satisfy the about equation
 L 
39.
In the circuit shown all elements are ideal and the switch S is operated at 10 kHz and 60% duty
ratio. The capacitor is large enough so that the ripple across it is negligible and at steady state
acquires a voltage as shown. The peak current in amperes drawn from the 50 V DC source is
________. (Give the answer up to one decimal place.)
Key: 39 to 41
Exp: Given is Buckboost converter
DVS
V0 
1 D
Given Vs  50V, D  0.6, Vo  75V
Vo Is
D
0.6
 

1.5
Vs Io 1  D 1  0.6
Vo 75
 15A
R
5
D
3
Is 
. Io  15  22.5A
1 D
2
Since capacitor is very large, ic  0
Io 
i L avg  is avg  i o avg
I L  Is  Io  22.5 15  37.5A
IL 
DVS
0.6  50

 5A
fL
10,000  0.6 103 
IL
5
 37.5   40A
2
2
Peak current drawn from source is 40A
 i L peak  IL 
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In the circuit shown in the figure, the diode used is ideal. The input power factor is _______.
(Give the answer up to two decimal places.)
Key: 0.70 to 0.71
V
Exp: Vor  m
2
Vo
Vm  2VS
Vm
Vor Vm
Ior 

R 2R

0
2
VS
Vm
2

VS 
2
2
2
Vor Ior Vor
PLoad
VS
1
IPF 




 0.707
InputVA VS Ior
VS
2 VS
2
41.
3

Consider the system described by the following state space representation


 x1  t     0 1   x1  t     0  u  t 
 
 0 2   x  t   1 
 2 
 x 2  t  
 x1  t  
y  t   1 0 

 x 2  t  
 x1  0    1 
If u  t  is a unit step input and 
    , the value of output y  t  at t = 1 sec (rounded
 x 2  0   0 
off to three decimal places) is_________
Key: 1.280 to 1.287
0 1 
Exp: Given A  

0 2 
0 
B 
1 
C  1 0
 s  2  1
s  2 1


 0

  0
s   1   0 
s 1 
1
X  s    SI  A   x  0   Bu  s     2

1
/
S



s  2s  0  1 
s  s  2   s 

1

 s  2   s 


1

;
x s 
s s  2
y s  
s s  2  1
s2  s  2 
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1
1
1 1
1
1
y s   2
   2
s s  s  2  s 4s 2s
4 s  2
3
1
1
3 1 1
y  l   4  t   t4  t   e 2t 4  t     e 2
4
2
4
4 2 4
y 1  1.2838
42.
A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has
following equivalent circuit parameters per phase referred to the stator:
R1  0.3,R 2  0.3,X1  0.41,X2  0.41. Neglect shunt branch in the equivalent circuit.
The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3phase AC source is __________.
Key:
Exp:
69 to 71
Reactance offered by stator and Rotor will be changes, because of change in frequency.
R1
20
 0.41  0.13666 
60
20
x 2 
 0.41  0.1366
60
Vph
Ist  I ph  Isc 
Zeq
x1 

0.3
R2
0.3
80
v, 20Hz
3
X1
X2
0.4
0.4
Ist  Isc
80 3
46.18

 0.3  0.3  j  0.1366  0.1366   0.65
 71.4 24.46o Amps
A 25 kVA, 400 V,  - connected, 3-phase, cylindrical rotor synchronous generator requires a
field current of 5 A to maintain the rated armature current under short-circuit condition. For the
same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and
magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the
generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is _________.
43.
Key:
-15 to -14
Exp:
25kVA,
400V,   connection
Voc  360V
25  103
 36.084 Amps
3  400
I ph  20.833 Amps
IL 
ISC  I rated , If  5A
Xs  Zs 
E
Voc phase
360

 17.28
Isc phase 20.833
 V cos   Ia R a 
2
  Vsin   Ia Xs  
2
 400  0.8  0 
2
  400  0.6  20.833  17.28 
2
E ph  341.758 volts
341.758  400
 14.6%
400
Hint: Obtained regulation should be negative.
% Reg 
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If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA. Y   transformer (the per
phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is_____.
44.
Key:
37 to 39
Exp:
25KVA, Y  D, 3.3kV  
N1 : N 2  5 :1
25  103
3  3.3  103
IL  Iph  4.374 Amps
Ist 
Transformer is a constant-Power device
E 2 I2  E1I1
N 2 I2  N1I1  I2 
N1
5
.I1   4.374
N2
1
Iph  I2  21.869 Amps    side
IL  3  I2  3  21.869  37.879 Amps
45.
For the balanced Y-Y connected 3-Phase circuit shown in the figure below, the line-line voltage
is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading.
The approximate value of the impedance Z is
(A) 33  53.1o 
(B) 6053.1o 
(C) 60  53.1o 
(D) 180  53.1o 
Key:
(C)
Exp:
VL  208V, P  432W
cos   0.6 leading
P  3 Vph  I ph  cos 
2
 208 
3 
2
  0.6
Vph
3

P  3.
.cos z ph 
 60.08
z ph
432
cos 1 0.6  53.1o
z ph  60 53.1o
z ph 
Vph
o
I ph 53.1

Vph
Iph
53.1o
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A thin soap bubble of radius R = 1 cm, and thickness a  3.3m  a  R  , is at a potential of 1
46.
V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical
drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap
4
in the thin bubble is 4R 2a and that of the drop is r 3 . The potential in volts, of the resulting
3
single spherical drop with respect to the same reference point at infinity is __________. (Give
the answer up to two decimal places.)
Key:
Exp:
9.50 to 10.50
Charge must be same
 4R 2a  P   43 r3  P
r
3
3R 2 a
0.996  103
The potential of thin bubble is 1 V
Q
1
4E 0  1  102
Q  40  102 C
Potential of Soap drop
Q
V
40 r

40  102
40  0.9966  103
 10.03V
47.
The value of the contour integral in the complex-plane
z 3  2z  3
 z  2 dz
Along the contour |Z| = 3, taken counter-clockwise is
(A) 18i
(B) 0
(C) 14i
Key: (C)
Exp:
(D) 48i
z = 2 is the singularity lies inside C : z  3


C
z3  2z  3
dz  2i  z3  2z  3  14i
z2
z2
(Using Cauchy‟s Integral formula)
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1  x x  0
 x x  1
andf  x    2
Let g  x   
x  1, x  1
x , x  0
48.
Consider the composition of f and g, i.e.,  f  g  x   f  g  x . The number of discontinuities
in  f  g  x  present in the interval  ,0  is:
(A) 0
(B) 1
(C) 2
(D) 4
Key:
(A)
Exp:
1  x, x  1
Clearly  fog  x    2
is discontinuous at x  1  ,0 
 x , x 1
 The number of discontinuities in (fog) (x) present in the interval  , 0  is 0
Alternative Method:
f  x   1  x for x  0 and g  x    x for x < 0
 Both f(x) and g(x) are continuous when x < 0
  fog  x  is also continuous for x < 0
(Since the composite function of two continuous functions is continuous)
 The number of discontinuities in the interval  ,0  i.e., x < 0 is „0‟
49.
A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200
rpm. The armature resistance is 0.8  , and the shunt field resistance is 240 . The no load
speed, in rpm, is _______________.
Key: 1235 to 1250
Exp:
2A
0.5
0.8
240
0.8
120
240
120
1.5A
6.5
E b1
N  Eb
2A
0.5
No load, N1  ?
Eb2
Full load, N2  1200rpm
 120  1.5  0.8 
N1 E b1

 N1  
  1200  1241.82 rpm
N 2 E b2
120   6.5  0.8 
50.
A 10 ½ digit timer counter possesses a base clock of frequency 100 MHz. When measuring a
particular input, the reading obtained is the same in: (i) Frequency mode of operation with a
gating time of one second and (ii) Period mode of operation (in the x 10 ns scale). The
frequency of the unknown input (reading obtained) in Hz is _______.
Key: 10000 to 10000
51.
A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3
tosses. Let the random variable Y denote the number of heads. The value of var Y , where
var . denotes the variance, equals
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(A)
7
8
(B)
49
64
(C)
7
64
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(D)
105
64
Key: (C)
52.
The figure below shows a half-bridge voltage source inverter supplying an RL-load with
 0.3 
R  40 and L  
 H. The desired fundamental frequency of the load voltage is 50 Hz. The
  
switch control signals of the converter are generated using sinusoidal pulse width modulation
with modulation index. M = 0.6. At 50 Hz, the RL-load
draws an active power of 1.44 kW. The value of DC source
voltage VDC in volts is
(B) 500
(A) 300 2
Key: (C)
(C) 500 2
(D) 1000 2
0.3
 30

z  R L  jx L  40  j30  50 36.86
Exp: x L  L  100 
M  0.6
I Load 
PL
1440

 6A
RL
40
VA o1
VA o1
 6
z1
50
VA o1  300V  rms 
I Load 
VAo1  300 2 V  max 
VAo1  m.vdc
300 2  0.6  vdc
 Vdc  500 2
The range of K for which all the roots of the equation s3  3s2  2s  K  0 are in the left half of
the complex s-plane is
(A) 0 < K < 6
(B) 0 < K < 16
(C) 6< K < 36
(D) 6< K < 16
Key: (A)
53.
Exp: C.E  s3  3s2  2s  k  0
If system to be stable
K  06  K
0K6
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The eigen values of the matrix given below are
0
0

0
(A)
1 0
0 1 
3 4 
(0,-1,-3)
(B) (0,-2,-3)
(C) (0,2,3)
(D) (0,1,3)
Key: (A)
Exp:
 1
0
1 0
Characteristic equation is 0 
0 3 4  
   4  2  3  0      1   3  0    0, 1, 3 are the eigen values
55.
A 3-phase 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive
load through an overhead line. The per phase line resistance and reactance are
0.25  and 3.925  respectively. If the voltage at the generator terminal is 17.87 kV, the power
factor of the load is ________.
Key: 0.75 to 0.85
Exp: PS  3MW
Z  3.9329 86.35  ph
17.87
E f  17.32 KV
Vt 
 10317.249 V ph
3
Z   0.25  j3.925   ph
E f  10.000 V ph
V  17.87 KV
t
2  90  86.35  3.65
2
V 
EV
Pog  f t sin     2    t  ra
Zs
 Zs 
10000  10317.249
 10317.249 
sin    3.65   
  0.25
3.9329
 3.9329 
  2.3024
2
106 
 I a Zs 
2
 E f 2  Vt 2  2E f Vt cos 
Ia  131.43 A / ph
PS  3MW  3  17.87  103  131.43  cos 
cos   0.737
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General Aptitude
Q. No. 1 – 5 Carry One Mark Each
1.
There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to
the West of W, Z is to the East of X and the West of V, W is to the West of Y. Which is the
building in the middle?
(A) V
(B) W
(C) X
(D) Y
Key: (A)
Exp:
From the given data, the following is formed
X
Z
V
W
Y
N
W
East
West
E
S
 The building „V‟ is in the middle
2.
Saturn is _________ to be seen on a clear night with the naked eye.
(A) enough bright
(B) bright enough
(C) as enough bright
(D) bright as enough
Key: (B)
3.
Choose the option with words that are not synonyms.
(A) aversion, dislike
(B) luminous, radiant
(C) plunder, loot
(D) yielding, resistant
Key: (D)
4.
There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that
they are of the same colour is
(A) 1/5
(B) 7/30
(C) 1/4
(D) 4/15
Key: (D)
3
Exp:
C2  4 C2 3 C2 12 4


10
C2
45 15
5.
A test has twenty questions worth 100 marks in total. There are two types of questions.
Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each.
How many multiple choice questions does the exam have?
(A) 12
(B) 15
(C) 18
(D) 19
Key: (B)
Exp: x  y  20
x  MCQ
3x  11y  100
 x  15, y  5
y  essay type
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Q. No. 6 – 10 Carry Two Marks Each
6.
An air pressure contour line joins locations in a region having the same atmospheric pressure.
The following is an air pressure contour plot of a geographical region. Contour lines are shown
at 0.05 bar intervals in this plot.
If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region,
which of the following regions is most likely to have a thunderstorm?
(A) P
(B) Q
(C) R
(D) S
Key: (C)
Exp:
Region
Air pressure
difference
P
0.95 – 0.90 = 0.05
Q
0.80 – 0.75 = 0.05
R
0.85 – 0.65 = 0.20
S
0.95 – 0.90 = 0.05
In general thunder storms are occurred in a region where suddenly air pressure changes (i.e.,)
should rise (or) sudden fall of air pressure. From the given contour map in „R‟ region only
more changes in air pressure. So, the possibility of a thunder storms in this region.
So option (C) is correct.
7.
There are three boxes. One contains apples, another contains oranges and the last one contains
both apples and oranges. All three are known to be incorrectly labelled. If you are permitted to
open just one box and then pull out and inspect only one fruit, which box would you open to
determine the contents of all three boxes?
(A) The box labelled „Apples‟
(B) The box labelled „Apples and Oranges‟
(C) The box labelled „Oranges‟
(D) Cannot be determined
Key: (B)
Exp:
The person who is opening the boxes, he knew that all 3 are marked wrong.
Suppose if 3 boxes are labelled as below.
(1) Apples
(2) Oranges
(3) Apples & Oranges
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If he inspected from Box(1), picked one fruit, found orange, then he don‟t know whether box
contains oranges (or) both apples and oranges.
Similarly, if he picked one fruit from box(2), found apple then he don‟t know whether box
contain apples (or) both apples and oranges.
But if he picked one fruit from box(3), i.e., labelled is “apples and oranges‟, if he found apple
then he can decide compulsorily that box(3) contains apples and as he knew all boxes are
labelled as incorrect, he can tell box(2) contains both apples and oranges, box(1) contain
remaining oranges. So, he should open box labelled „Apples and Oranges‟ to determine
contents of all the three boxes.
“We lived in a culture that denied any merit to literary works, considering them important only
when they were handmaidens to something seemingly more urgent – namely ideology. This was
a country where all gestures, even the most private, were interpreted in political terms.”
8.
The author‟s belief that ideology is not as important as literature is revealed by the word:
(A) „culture‟
(B) „seemingly‟
(C) „urgent‟
(D) „political‟
Key: (B)
X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X3
will have
(A) 90 digits
(B) 91 digits
(C) 92 digits
(D) 93 digits
Key: (A)
9.
Exp:
X   47........... 30 digits
Suppose  47   2  2  2 digits in (47)3
3
Similarly  47   contains 30 + 30 + 30 digits = 90 digits.
3
The number of roots of ex  0.5x 2  2  0 in the range  5,5 is
10.
(A) 0
Key: (C)
Exp:
(B) 1
(C) 2
(D) 3
f  x   e x  0.5x 2  2
f  5  10.50; f  4   6.01, f  2   0.135; f  1  1.13;
f  0   1, f 1  1.21, f  2   7.38, f  3 , f  4  , f  5  also  ve.
 As there are 2 sign changes from +ve to –ve and –ve to +ve, two roots will be there in
the range [-5, 5].
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EE-GATE 2018
Section-I: General Ability
1.
Functions F(a,b) and G(a,b) are defined as follows:
F(a,b) = (a-b)2 and G(a,b) =|a-b|, where |x| represents the absolute value of x.
What would be the value of G(F(1,3), G(1,3))?
(A) 2
(B) 4
(C) 6
(D) 36
Key: (A)
Exp: Given,
F  a, b    a  b  and G  a, b   a  b
2
 F 1,3  1  3 ;  G 1,3  1  3
2
 F 1,3  4
;  G 1,3  2
 G  F 1,3 , G 1,3   G  4, 2    F 1,3  4&G 1,3  2 
 4  2   G  a, b   a  b 
 G  F 1,3 , G 1,3   2
2.
“Since you have gone off the ____________, the _________ sand is likely to damage the car.”
The words that best fill the blanks in the above sentence are
(A) course, coarse
(B) course, course
(C) coarse, course
(D) coarse, coarse
Key: (A)
3.
“A common misconception among writers is that sentence structure mirrors thought; the more
_______ the structure, the more complicated the ideas.”
The word that best fills the blank in the above sentence is
(A) detailed
(B) simple
(C) clear
(D) convoluted
Key: (D)
4.
 k  2 2
an integer?
k 3
(B) 4, 10, 16
(C) 4,8,28
For what values of k given below is
(A) 4,8,18
(D) 8, 26, 28
Key: (C)
Exp: Actually we can find an infinite no. of values of k, for which
 k  2
k 3
2
to be an integer
From the given choices;
If k  4, then
 k  2
k 3
2

 4  2
43
2
 36  integer.
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EE-GATE 2018
 k  2
If k  8, then
2
k 3
If k=28, then
5.
 28  2
8  2 

2

100
 20  integer
5

30  30
 36  integer.
25
83
2
28  3

 30
25
2
The three roots of the equation f(x) = 0 are x = {-2, 0, 3}. What are the three values of x for which
f(x-3) = 0?
(A) -5, -3, 0
(B) -2, 0, 3
(C) 0, 6, 8
(D) 1, 3, 6
Key: (D)
Exp: Given that;
X= -2, 0, 3 are the 3 roots of f  x   0.
i.e., f  2  0, f  0  0 and f 3  0.
Now we should find the values of x for which
f  x  3  0  x  3  2; x  3  0; x  3  3
 f  2   0; f  0   0; f  3  0 
 x  2  3; x  3; x  6
 x  1, 3, 6
6.
A class of twelve children has two more boys than girls. A group of three children are randomly
picked from this class to accompany the teacher on a field trip. What is the probability that the
group accompanying the teacher contains more girls than boys?
(A) 0
(B)
325
864
(C)
525
864
(D)
5
12
Key: (B)
Exp: Given, a class of 12 children has no two more boys than girls.
Let, the no.of girls  x
Then no.of boys  x  2
 x   x  2   12
 2x  10
x 5
No.of Grils  5
No.of Boys  7
 Total no. of ways of selection of 3 children = 12C3  n  s 
Let A be the event that no. of girls are more than boys
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3
EE-GATE 2018
Case   I 
i.e.
3G &
Case   II 
 or 
0B
2G &
Two cases possible
1B
Favorable no. of ways of A 5C3  7C0  5C2  7C1
 10  70  80 5C2  5C3  10
 Prob  A  
n  A  80
80  3!
 12 
n  S
C3 12  11  10
80  6
8
4


0.3636
12  11  10 22 11
325
 Req Pr ob 
864

7.
325
864
An e-mail password must contain three characters. The password has to contain one numeral from 0
to 9, one upper case and one lower case character from the English alphabet. How many distinct
passwords are possible?
(A) 6,760
(B) 13,520
(C) 40,560
(D) 1,05,456
Key: (C)
Exp: We know that;
Total no. of digits  0,1,2,....9
Total no. of upper case English alphabets  26 A,B,C,...., Z
Total no. of lower case English alphabets  26a,b,c,....,z
 No. of ways of selecting a digit out of 10  10C1  10
No. of ways of selecting an upper case alphabet  26C1  26
No. of ways of selecting an lower case alphabet  26C1  26
 Permutation = selection + arrangement
 Total no.of distinct passwords  10  26  26 
1 
 
2
3 
 
. 
. 
 
9 
A 
 
B 
. 
. 
 
. 

Z 

a 
 
b

. 

. 
. 
 

Z

3!
arrangement of
3things

arrangement
Selection
 6760  6  3!  6
 40,560
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4
EE-GATE 2018
8.
In a certain code, AMCF is written as EQGJ and NKUF is written as ROYJ. How will DHLP be
written in that code?
(A) RSTN
(B) TLPH
(C) HLPT
(D) XSVR
Key: (C)
Exp: Given
4
4
A MCFE QG J
4
4
&
N K U F R O Y J
4
4
4
4
4
4
 DH L P  H LP T
4
9.
4
A designer uses marbles of four different colours for his designs. The cost of each marble is the
same, irrespective of the colour. The table below shows the percentage of marbles of each colour
used in the current design. The cost of each marble increased by 25%. Therefore, the designer
decided to reduce equal numbers of marbles of each colour to keep the total cost unchanged. What
is the percentage of blue marbles in the new design?
(A)
35.75
Blue
Black
Red
Yellow
40%
25%
20%
15%
(B) 40.25
(C) 43.75
(D) 46.25
Key: (C)
Exp: Assume that,
Total no. of marbles = 100
C.P of 1 marble =Rs 100
Total C.P of 100 marbles  100  100
New cost price of 1 marble = Rs 125 [Given]
No. of marbles with new price 
100  100
 80
125
 No.of reduced marbles  100  80  20  5  5  5  5  20
 Blue Black Red Yellow

Now; the no. of blue marbles = 40-5=35
Total no. of marbles =10-20=80
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5
EE-GATE 2018
 35

 % of blue marbles in new design    100  %
80


 35  5 

%
 4 
 43.75%
10.
P,Q,R and S crossed a lake in a boat that can hold a maximum of two persons, with only one set of
oars. The following additional facts are available.
(i) The boat held two persons on each of the three forward trips across the lake and one person on
each of the two return trips.
(ii) P is unable to row when someone else is in the boat.
(iii) Q is unable to row with anyone else except R.
(iv) Each person rowed for at least one trip.
(v) Only one person can row during a trip.
Who rowed twice?
(A) P
(B) Q
(C) R
(D) S
Key: (C)
Exp: Section of 2 persons out of 4 (P, Q, R, S), i.e., 4C 2 = 6ways
PQ PR PS QR QS RS
3 forward trips Two person can travel
From the fact (i), there are  
2 returned trips only 1 person travel
From the facts (ii) and (iii); P and Q should not travel.
To satisfy (iv) fact; only Q and R should travel in 1st trip.
First trip : Q Rowed in forward trip
R Rowed in return trip
R,Q
Q
R
Second trip : In 2nd trip only R & P should travel.
R Rowed in forward trip
P, R
P Rowedin return trip
R
P
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6
EE-GATE 2018
P,S
Third trip : In 3rd triponly P & Swill travel
S
S must rowed in forward trip  From  ii  fact 
 R rowed twice.
Section-II: Electrical Engineering
1.
Four power semiconductor devices are shown in the figure along with their relevant terminals. The
devices (s) that can carry dc current continuously in the direction shown when gated appropriately
is (are)
A
A
MT1
D
G
K
I
G MT2
I
Thyristor
G
I
Triac
(A) Triac only
(C) Triac and GTO
G
K
S
I
MOSFET
GTO
(B) Triac and MOSFET
(D) Thyristor and Triac
Key: (B)
Exp: SCR: conducts only from anode to cathode
Triac:
A
I
G
K
(1) Triac is combination of two anti parallel SCRs
Triac
MT1
G
MT2
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7
EE-GATE 2018
(2) When MTis
1  ve & MT2 is  ve

I

(3) When MT1 is  ve & MT2 is  ve

G
I

GTO: GTO conducts only from anode to cathode
A
G
K
MOSFET
D
D

G
G
S
S
(1) When D is + ve & S is –ve, diode is OFF
D   ve 
S ve 
I
(2) When D is –ve & S is +ve, diode is ON
D
I
S
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8
EE-GATE 2018
2.
The op-amp shown in the figure is ideal. The input impedance
i in

Z

Vin
R1
R2
V0


R1
R2
(A) Z
vin
is given by
t in
(B)  Z
R2
R1
(C) Z
(D)  Z
R1
R1  R 2
Key: (B)
Exp: Given op- Amp is ideal
Z
i in




Vin
V0
V
R1
R2
By virtual ground concept
Vin  V
Vin  V0
Z
By voltage divider principle
V  R2
Vin  V  0
R1  R 2
iin 
 R 
V0  Vin 1  1 
 R2 
Vin  Vin  Vin
iin 
Z
R1
R2
 Vin
R1
ZR 2
Vin
R
 Z 2
iin
R1
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9
EE-GATE 2018
3.
Two wattmeter method is used for measurement of power in a balanced three phase load supplied
from a balanced three phase system. If one of the wattmeters reads half of the other (both positive),
then the power factor of the load is
(A) 0.532
(B) 0.632
(C) 0.707
(D) 0.866
Key: (D)
Exp:  In two Wattmeter of power measurement we know
 3  1  2  
  tan 1 
  well known standard formula 
 1  2 
 It is given that one meter reads half of the other
 2 
1 
 
or wecan take 1  2 

2
2 


   tan 1 




   tan 1 


   tan 1
 

3  1  1  
2


1  

 1  2  

 
  
3 1 
 2 
 
3  1  
 2 
1
   30
3
 power factor  cos 
 cos 30 
4.
3
 0.866
2
The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph
is
(A) 11
(B) 12
(C) 13
(D) 14
Key: (B)
Exp:  we know in a network
 b   n  1
Where
: number of loops = 5 (given)
b : number of branches (to be found)
h : number of nodes = 8 (given)
  b  n 1
 5  b  8 1
 b  12
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10
EE-GATE 2018
5.
A continuous time input signal x(t) is an eigenfunction of an LTI system, if the output is
(A) kx(t), where k is an eigenvalue
(B) ke jt x  t  , where k is an eigenvalue e jt is a complex exponential signal
(C) x(t) e jt , where e jt is a complex exponential signal
(D) kH   , where k is an eigenvalue and H   is a frequency response of the system
Key: (A)
Exp: Consider for example eigen function, x  t   est
est
ht
y  t   H  s  est
Where H(s) is Laplace transform of h(t).
At specific S  S0 , H S0  becomes a constant quantity.
Thus y  t   K.est or K.x  t 
Where „K’ is eigen value.
Thus option (A) is correct.
In option (B) extra frequency forms one getting generated at output which is not possible for LTI
system.
Option (C) and Option (D) one also not the correct answer due to extra frequency term getting
generated in output.
6.
In the logic circuit shown in the figure, Y is given by
A
B
Y
C
D
(A) Y=ABCD
(B) Y = (A+B) (C+D) (C) Y=A+B+C+D
Key: (D)
Exp: y  MN
A
B
M  AB
Y
y  AB. CD
y  AB  CD
(D) Y=AB+CD
C
D
N  CD
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11
EE-GATE 2018
7.
z 1
dz in counter clockwise direction around a circle C of radius 1
2
4
z
The value of the integral
C
with center at the point z = - 2 is
(A)
i
2
(C) 
(B) 2i
i
2
(D) 2i
Key: (A)
Exp:
z 1
dz
2
4
z
C
y
z2  4  0
 z  2
 z  2 lies inside & z  2 lies outside 'C'
 By Cauchy's integralformula,

C
C
43 2 1 0
1 2
x
 z 1 


z 1
z 1
 z  2 dz
dz

dz

C  z  2 z  2  C z  2
z2  4
 2if (2)
 2  1  i
 2i 

 2  2  2
8.
Let f be a real valued function of a real variable defined as f(x) = x-[x], where [x] denotes the
1.25
largest integer less than or equal to x. The value of
 f  x  dx is ______(up to 2 decimal places).
0.25
Key: (0.5)
Exp: We know that
where x  integral part of x,
x   x   x
 x   x   x where x  fractional part of 'x '
1.25
1.25
y
  f  x  dx   x dx
0.25
0.25
1


1.25
x dx 
0.25
1
1
2 1
x

2
  x  1 dx
1.25
 x2

  x 
2
1
0.25 
1 0
1
2
3
x
1
2
 0.5

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12
EE-GATE 2018
9.
The value of the directional derivative of the function   x, y,z   xy2  yz2  zx 2 at the point (2,-1,
1) in the direction of the vector p = i + 2j + 2k is
(A) 1
(B) 0.95
(C) 0.93
(D) 0.9
Key: (A)
Exp:   i



 j K
x
y
z
 i  y 2  2zx   j  2xy  z 2   k  2yz  x 2 
  2,1,1  5 i  3 j  2k
Required directional deivative   5 i  3 j  2k  .

10.
 i  2 j  2k 
1 4  4
56 4
1
3
Match the transfer functions of the second order systems with the nature of the systems given
below.
Transfer functions
Nature of system
P.
15
s  5s  15
I. Overdamped
Q.
25
s  10s  25
II. Critically damped
2
2
35
s  18s  35
(A) P-I, Q-II, R-III
Key: (C)
R.
Exp: P 
III. Underdamped
2
(B) P-II, Q-I, R-III
(C) P-III, Q-II, R-I
(D) P-III, Q-I, R-II
15
n 2

s 2  5s  15 s 2  n s  2n
n  15
2n  5
5
 1 underdamped 
2 15
2n
25
Q 2
 2
s  10s  25 s  2n s  2n

n  25
25  10
  1 critically damped 
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13
EE-GATE 2018
R
2n
35

s 2  18s  35 s 2  2n s  2n
n  35
n  18
18
 1 over damped 
2 35
P  III Q  II R  I

11.
A 1000×1000 bus admittance matrix for an electric power system has 8000 non –zero elements. The
minimum number of branches (transmission lines and transformers) in this system are ______ (up to 2
decimal places).
Key: (3500)
Exp: Given 1000  1000 bus matrix
N  1000
Total root Buses  103 103  106
given no of non  zero elements  8000
No of zero elements
Sparsity  s  
Total noof elements
106  8000
 0.992
106
No of transmission lines and transformers

N2 1  s   N 10002  1  0.992   1000


 3500
2
2
12.
A single phase 100kVA, 1000 V/100V. 50Hz transformer has a voltage drop of 5% across its series
impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at
full load with 0.8 lagging power factor is
(A) 4.8
(B) 6.8
(C) 8.8
(D) 10.8
Key: A
Exp: Impedance  5%  0.05pu  Zpu
Resistance  3%  0.03pu  R pu
2
2
Reactance, X pu  z pu
 R pu
 0.052  0.032  0.04pu
VR  0.8 p.f lag   R pu cos   X pu sin 
 0.03  0.8  0.04  0.6
 0.048pu  4.8%
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14
EE-GATE 2018
13.
Let f be a real valued function of a real variable defined as f(x) = x2 for x  0 , and f  x    x 2 for x  0 .
Which one of the following statements is true?
(A) f(x) is discontinuous at x=0.
(B) f(x) is continuous but not differentiable at x=0
(C) f(x) is differentiable but its first derivative is not continuous at x=0.
(D) f(x) is differentiable but its first derivative is not differentiable at x=0.
Key: (D)
Exp: Given
f x  x x
 f  x  in continuals &differentiable
f ' x   x 
xx
x
2 x
 f  x  is differentiable but its first derivative is not differentiable at x=0
14.
A positive charge of 1nC is placed at (0, 0, 0.2) where all dimensions are in metres. Consider the x-y
plane to be a conducting ground plane. Take 0  8.85  1012 F / m. The Z component of the E field at
(0, 0, 0.1) is closest to
(A) 899.18V/m
(B) -899.18V/m
(C) 999.09V/m
(D) -999.09 V/m
Key: (D)
Exp: E12 
E12 
QR12
40 R12
3
 110   0.3 a 

  0.1 4 8.854 10   0.3
9
1 109  0.1aˆ z 
4 8.854  1012
3
Z
12
3
 105

105


 az
 4  8.854  4 8.854  9 
  898.774  99.863 a z  999.09a z V m
15.
A separately excited dc motor has an armature resistance Ra=0.05  . The field excitation is kept
constant. At an armature voltage of 100V, the motor produces a torque of 500Nm at zero speed.
Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of
150 V is _____ (up to 2 decimal places).
Key: (600)
Exp: Torque
Produced by motor = 500N-m
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2
Preparation…………………………..
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EE-GATE 2018
At zero speed, such that
E L  K  0
I 
V  E b 100  0

 2000A
Ra
0.05
100V
0.05
Also,T  kIa
500
5 1


2000 20 4
For armature voltage of 150V, no load speed in nL
k  T Ia 
At no load I nL  0
 V  E b  150V
 E b  k  Field excitation or flux is constant, k  constant 

16.
E b 150

 600 rad sec
K 1 4
In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when
the load angle in electrical degrees is
(A) 0
(B) 45
(C) 60
(D) 90
Key: (B)
Exp: Reluctance torque  sin 2
Torque ix maximum when
2  90    45
17.
Consider a lossy transmission line with V1 and V2 as the sending and receiving end voltages,
respectively. Z and X are the series impedance and reactance of the line, respectively. Z and X are the
series impedance and reactance of the line, respectively. The steady-state stability limit for the
transmission line will be
(A) greater than
(C) equal to
V1V2
X
V1V2
X
(B) less than
V1V2
X
(D) equal to
V1V2
Z
Key: (B)
Exp: Power with impudence Z,
P
V1V2
V2
cos       2 cos 
Z
Z
Pmax1 
V1V2 V22

cos  for    
Z
Z
Power with reactence X,
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16
EE-GATE 2018
V1V2
sin 
X
VV
Pmax 2  1 2  for   90 
X
i.e Pmax 2  Pmax1
P
Steady state stability limit for transmission line  Pmax 2 
18.
V1V2
X
A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the
figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load
voltage and current?
i0

L
O
V
A 0
D

(A) 0  0 & i0  0
(B) 0  0 & i0  0
(C) 0  0 & i0  0
(D) 0  0 & i0  0
Key: (C)
Exp: During forward bias
ON
i0
off
L
o
a
d


off
 
ON
Vo
Free wheeling diode is off 
 
Let Vin  t   m sin  t  ,V0  t   n sin  t positive
During reverse bias
i0

off
ON

ON

Off
L
o
a
d
 Free wheeling diode isoff 
Vo  t   Vm sin  t   positive

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17
EE-GATE 2018
Vin  t 
Vm
0

0 

Vm
2

  2

Vm
Key concept:
SCR allows current only in one direction, Anode to cathode i0  0
19.
Consider a unity feedback system with forward transfer function given by
G s 
1
 s  1 s  2
The steady state error in the output of the system for a unit step input is _________ (upto 2 decimal
places).
Key: (0.67)
Exp: unit stepinput, K p  limG  s  
s 0
for unit step input, ess 
20.
1
2
A
1

 2 3  0.67
1  Kp 1  1 2
V


In the two port network shown, the h11 parameter  Where, h11  1 , when V2  0  in ohms is
I1


___________ (up to 2 decimal places).
2I1

1
1

I1
V1
1

V2

Key: (0.5)
Exp:  when V2  0 the output port is short circuited
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18
EE-GATE 2018
2I1
I1
1
2I1
1

I 2  2I1
V1
I1  I2
1
I2
V2  0

 Writing KVL on the input loop we have
V1  1 I1   1 I1  I2   0
...1
 V1  2I1  I2
 Writing KVL on the output loop we have
1 I2  2I1   1 I1  I2   0
 2I2  3I1  0
3
I1 ....(2)
2
 Using equation 2 in equation 1, we have
 I2 
 3 
V1  2I1    I1
 2 
V1  0.5I1
 h11 
V1
I1
 0.5
v2  0
h1  0.5
21.
The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown
in the figure. If I0=20 A, the rms value of fundamental component of the current is ___________ A (up 2
decimal places).
voltage and
current
Vm sin  t 
I0
0
30
t
I0
180
210
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19
EE-GATE 2018
Key: (17.39)
Exp:
i0  t 
I0
0


 
2
  t
I 0
Fourier series representation of io(t) is
i0  t  

 4I0 
n 
 n  
 sin nt  
2  
2 
  n  cos 
n 1,3,5
4I0
n 2 2
 n 
cos

I0 cos  
2
n
2n
 2 
RMS value of nth harmonic is n 
RMS value of fundamental is IS1 

22.
In
the
figure,
the
2 2

I0 cos

2
2 2
 30 
20cos    17.39A

 2
voltages
are
1  t   100cos  t  ,  2  t   100cos  t   / 18  and
3  t   100cos  t   / 36  . The circuit is in sinusoidal steady state, and R << L. P1, P2 and P3 are
the average power outputs. Which one of the following statements is true?
R

v1  t 
L
P1

(A) P1 = P2 = P3 = 0
(C) P1 < 0, P2 > 0, P3 < 0
L

v2  t 

R
P2

v3  t 
P3

(B) P1 < 0, P2 >0, P3 > 0
(D) P1 >0, P2 < 0, P3 >0
Key: (C)
Exp: V1  t   100 cos  t 


V2  t   100 cos  t  
18 

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20
EE-GATE 2018


V3  t   100 cos  t  
36 

V1  t   V2  t   V3  t   100
V1  t   0
 180

 10
18 18

V3  t  
 5
36
V2  t  
V2  t   10
V2  t  is leading V3  t  andV1  t  by10 and 5.
V3  t   5
 V2  t  is source & delivering power  P2  0 
V1  t  andV3  t  are sinks & absorbing power  P3  0 ,  P1  0
23.
V1  t    0 
Consider a non singular 2×2 square matrix A. If trace (A) = 4 and trace (A2) = 5, the determinant of the
matrix A is _____ (up to 1 decimal place).
Key: (5.5)
a b 
Exp: Let A  
  Trace of A  a  d
c d
a b  a b  a 2  bc ab  bd 
&A 2  


2
 c d   c d   ac  cd bc  d 
 Trace of A 2  a 2  bc  bc  d 2
 a 2  2bc  d 2
Given a  d  4
...1
a 2  2bc  d 2  5
... 2
1
2
&
  a  d   16  a 2  d 2  2ad  16
2
 5  2bc  2ad  16  from  2  
 2  ad  bc   11
 ad  bc 
24.
11
 5.5
2
The series impedance matrix of a short three phase transmission line in phase coordinates is
 Zs
Z
 m
 Zm
Zm
Zs
Zm
Zm 
Zm  . If the positive sequence impedance is (1+j10)  , and the zero sequence is (4+j31)  .
Zs 
Then the imaginary part of Zm (in  ) is _______ (up to 2 decimal places).
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21
EE-GATE 2018
Key: (7)
....1
Exp: Z1  Zs  Zm  1  j10
Z0  Zs  2Zm  4  j31 ....  2 
 2  1
Z0  Z1  3Zm  3  j21
Zm  1  j7
Im  Zm   7
25.
The positive, negative and zero sequence impedances of a 125 MVA, three phase, 15.5kV, stargrounded, 50Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base.
The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the
generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is
____________ (up to 2 decimal places).
Key: (73.51)
Exp: If  pu  

3
Z1  Z2  Z0  3Zn
3
3

j0.1  j0.05  j0.01   3  0.01 0.19
If  actual    If pu   I b base
MVA
 3

Sb


 kA
 0.19

3Vb


KV
3
125


 73.51kA
0.19
3  15.5
26.
The signal energy of the continuous time signal
x  t    t  1 u  t  1   t  2 u  t  2   t  3 u  t  3   t  4  u  t  4  is
(A)
11/3 (B)
7/3 (C)
1/3 (D)
5/3
Key: (D)
Exp: x  t    t  1 u  t  1    t  2  u  t  2    t  3 u  t  3   t  4  u  t  4 
b
a 
c
1
0
1
2
3
4
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22
EE-GATE 2018
2
3
4
Energyof x  t     a  dt    b  dt    c  .dt
2
2
1
2
 a 
1
2
2
3
4
2
dt    c  dt
2
3
2
3
 Energy of x  t   2   a  dt    b  dt
2
2
1
2
2
3
 2  t  1 dt  12 dt
2
1
 2
27.
2
 t  1
3
3 2
 1  5 3.
1
A 0-1 Ampere moving iron ammeter has an internal resistance of 50 m and inductance of 0.1mH. A
shunt coil is connected to extend its range to 0-10 Ampere for all operating frequencies. The time
constant in milliseconds and resistance in m  of the shunt coil respectively are
(A) 2, 5.55
(C) 2,1
(C) 2.18,0.55
(D) 11.1,2
Key: (A)
Exp:
MI ammeter
Rm
Lm
R sn
Lsn
 It is given that
Shunt coil
R m  50m
Lm  0.1 mH
Existing range  0  1ampere
Reauired range  0  10 ampere
Required 10
Scaling factor  m  
  10
Existing
1
 The required value shunt coil resistance is given by
R sn 
Rm
50 103

 5.55m.
m 1
10  1
 To make the meter independent of frequency, the time constant of both parallel branch
should be same i.e.
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23
EE-GATE 2018
Lsn Lm

R sn R m
0.1 103
 0.002  2m.sec
50  103
 sn  2 m.sec
 sn 
R sn  5.55 mA.
28.
The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a
frequency  =100 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the
magnitude of the impedance Z (in  ) seen between the terminals a and b is _________ (up to 2 decimal
places).
it
vt 
a
L
100
Z
100 F

b
Key: (50)
Exp:  when the Source voltage and current are in phase, then the circuit is under resonance and the
impedance is purely real.
Transforming the given network into its equivalent phasor domain we have
j100L
 j100
Zeq
100
  100 rad sec,  given 
at   100
ZL  jL  j100L
j
  j100
C
ZR  100
ZC 
 Zeq   j100L   100 ||   j100 
  j104 
  j100L  

100  j100 
  j100L 
 j 104  100  j100 
100  j100 100  j100 
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24
EE-GATE 2018

 

106
106

 j 100L 
2
2
2
2
100  100 
100  100  
Since the circuit is under resonance, the imaginary part of Zeq  0, hence
106
1002  1002
106
106
102



 50
2  1002
2  104
2
Zeq 
29.
Which one of the following statements is true about the digital circuit shown in the figure?
Q
D
D
C
Q
C
D
Q
fOUT
C
f IN
(A)
(B)
(C)
(D)
It can be used for dividing the input frequency by 3.
It can be used for dividing the input frequency by 5.
It can be used for dividing the input frequency by 7.
It cannot be reliably used as a frequency divider due to disjoint internal cycles.
Key: (B)
Exp:
D0
Q0
D1
C
Q1
C
D2
Q2
fOUT
C
f IN
D0  Q1 Q2
Clk Q1Q2
D0
1
2
3
4
5
6
1
1
1
0
0
1
Q0
Q1
D1
D2
0
1
1
1
0
0
0
0
1
1
1
0
Q0
0
1
1
1
0
0
1
Q1 Q2
0
0
1
1
1
0
0
0
0
i.e., MOD 5
0
f
1
Hence, fo  i
5
1
1
0 Repeated
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25
EE-GATE 2018
30.
The voltage across the circuit in the figure. And the current through it, are given by the following
expressions:
v  t   5  10cos  t  60 V
i  t   5  Xcos  t  A
Where   100 radian/s. If the average power delivered to the circuit is zero, then the value of X (in
Ampere) is _____ (up to 2 decimal places).
it

Electrical
Circuit
vt

Key: (10)
Exp: If the voltage and currents of electrical circuit is in the form of




V  t   V0  V1 cos 1t  v1  V2 cos 2 t  v2  ....




i  t   i0  i1 cos 1t  i1  i 2 cos 2 t  i2  ....
Then the average power is given by
T
Pavg 
1
V  t  i  t  dt,
T 0


V1 i1
V I
cos v1  i1   2 2 cos v2  i2  ...
2 2
2 2
 its a very well known standard result 
 V0i0 
 It is given that
V  t   5  10cos  t  60 
 5  10cos  t  120 
i  t   5  x cos  t  0 
 10  x 

then Pavg  5  5  

 cos  120  0,  
 2  2 

 10x 

 0  25  
 cos  120    Pavg  0, given 
 2 

 0  25   5x  1 2 
5
 x  25
2
 x  10
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26
EE-GATE 2018
31.
The equivalent impedance Zeq for the infinite ladder circuit shown in the figure is
j 9
j 9
j5
Zeq
j5 
 j1
(A) j12 
Key: (A)
Exp:
..
 j1 
(B) -j12 
j9
.
(C) j13 
(D) 13 
j9
j9
j5
j5
 j1
j9

j4
j4
 j1
Zeq
Zeq
Zeq
Zeq  j9   j4 || Zeq 
  j4   Zeq  

 j9  
 j4  Zeq 
j9
Zeq
  j4  Zeq  Zeq   j9   j4  Zeq     j4   Zeq 
j4
  j4   Zeq   Zeq 2  36  j9Zeq   j4   Zeq 
 Zeq 2   j9  Zeq  36  0
 Zeq 





   j9  
  j9   4  36 1
 2  1
Zeq
2
j9  81  144
2
j9  225
2
j9  j15
2
j9  j15 jq  j15
or
2
2
j12 or  j3
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27
EE-GATE 2018
 Since the nature of the network is overally inductive the reactance must should be positive (for
capacitive case the reactance could have –ve).
Hence we are discarding –j3.
 Zeq  j12
32.
A transformer with toroidal core of permeability  is shown in the figure. Assuming uniform flux
density across the circular core cross-section of radius r < < R, and neglecting any leakage flux, the best
estimate for the mean radius R is
r
ip  I sin ax

is  0 
R
NP
NS
vS
vP  V cos t

(A)

Vr 2 N p2 
I
(B)
Ir 2 N p Vs 
V
(C)
Vr 2 N p2 
2I
(D)
Ir 2 N 2p 
2V
Key: (D)
Np2
S
Where S is reluctance of magnetic path
Exp: The inductance of primary is L 
S
A
L 

2R
2R
 2
2
  r  r
Np 2 Np 2r 2

2R
2R
r 2
Np 2r 2
2R
I..Np 2 .r 2
V  IX L 
2R
2
2
.I.r .Np .
R
2V
X L  L 
33.
The number of roots of the polynomial, s7  s6  7s5  14s4  31s3  73s2  25s  200 in the open left
half of the complex plane is
(A) 3
(B) 4
(C) 5
(D) 6
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28
EE-GATE 2018
Key: (A)
Exp: s7
1
7
31
25
s 2
1
14
73 200
5
s sign 7 42 175
change
s4
8 48 200
3
s
1
3
0
2
s
24 200
2
1
s sign 128 0
s0 change 200
A  s   8s 4  48 s 2  200
6
dA  s 
 32s3  96 s
ds
No. of sign changes  4
Right hand roots  4
No. of imaginary roots  order of auxillary eqn  2  No.of sign changes below zero th row 
 4  2  2  0
 left hand roots
743
34.
The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are
R1  R 2  X l  X 2  12, XM  240 and s is the slip. At no load, the motor speed can be
approximated to be the synchronous speed. The no-load lagging power factor of the motor is ____ (up to
3 decimal places).
jX
R1
1
X
j M
2
R'2
2s
j
V0
j
XM
2
X' 2
2
R '2
22  s
j
X' 2
2
Key: (0.106)
R '2

2S
The modified equivalent Circuit is
Exp: At no load, m  s . S  0.
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29
EE-GATE 2018
j12
12
j
Zeq
j
240
2
240
2
12
4
j12
2
R '2
R'
 2
22  s 4
Zeq  12  j12  j120 
j120   3  j6 
 3  j6  j120 
 j120   3  j6  
 12  j132  

 3  j126 
 14.72  j137.78
 138.56183.9
No load p.f cos   cos83.9  0.106 lag.
35.
A 3-phase 900kVA, 3kV/ 3 kV   / Y  , 50 Hz transformer has primary (high voltage side) resistance
per phase of 0.3  and secondary (low voltage side) resistance per phase of 0.02, Iron loss of the
transformer is 10kW. The full load % efficiency of the transformer operated at unity power factor is
___________ (up to 2 decimal places).
Key: (97.36)
Exp: a ph 
3
 3;
3
900
I Hv  I1 
 100A
3 3
R1eq  R1  a 2 R 2
3
 0.3  9   0.02   0.48
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30
EE-GATE 2018
Cu loss  3     3I12 R1eq
 3  1002  0.48
 14400W  14.4kW
Core loss  10kW
900
% 
 97.36%
900  10  14.4
36.
A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t=0 as shown
in the figure. Assume i(0)=0, v(0)=0. Which one of the following circular loci represents the plot of i(t)
versus v(t)?
it
S
t 0


it
(A)
5V
vt
L  1H
C  1F


(B)
 0, 5
(C)
vt
it
 5,0
it
(D)
vt
it
 0,5
 5,0
vt
vt
Key: (B)
Exp:  Since the network is of 2nd order, to obtain the expression of v(t)and i(t) we should use
Laplace transform approach.
 Once we get the expression for v(t) and i(t) we can plot the loci by observing them at
different time instant.
 Transforming the given time domain network into its equivalent s-domain form for t>0,
have
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31
we
EE-GATE 2018
 I s  
s
5s
5
 2
s 1 s s 1
 i  t   5sin t
5
s
 1s 
 V s  
 5 s 
s 1 5



I s

V s

1
s
5
s  s  1
2
 5   2.5 2.5 
 


 s   s  j1 s  j1 
 5   5s 
  2 
 s   s  1
 v  t   5  5 cos
 Finaly we have
i  t   5sint
v  t   5  5cos t
t
it
vt
Coordinate
0
0
0
 0,0

2
5
5
 5,5

0
10
 0,10
-5
-5
 5,5
0
0
 0,0
3
2
2
37.
 t   2
 5,5
 0,0
 t  
 0,10
 t  0
 3 
 t  2 
A three phase load is connected to a three phase balanced supply
as shown in the figure.
If
Van  1000V, Vbn 100  120 V and V cn 100  240 V
(angles are considered positive in the anti clock wise direction).
The value of R for zero current in the neutral wire is
___________  (up to 2 decimal places).
 5, 5
a
R
n
j10
c
 j10
b
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32
EE-GATE 2018
Key: (5.77)
Exp: a
Ia
R
IN
n
X
j10
 j10
IC
c
b
Ib
By KCL at node x, we have
Ia  Ib  Ic  IN

Van Vbn Vcn


 IN
R
j10  j10
1000 100-120 100-240


 0  IN  0 given 
R
1090
10-90
100

 10-210  10-150  0
R
100

  10-210  10-150
R
100
R 
10-210  10-150

 R  5.77
38.
The unit step response y(t) of a unity feedback system with open loop transfer function G(s) H(s)

K
is shown in the figure. The value of K is ___________ (up to 2 decimal places).
 s  12  s  2 
y  t  1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
16 18
20
t  sec 
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33
EE-GATE 2018
Key: (8)
Exp: Steady state output =0.8
Input is unit step input =1
Stead state error ess = 0.2
For unit step input K p  limG  s  H  s 
s 0
 lim
s 0
ess 
K
 s  1  s  2 
2
K 2
A
1  Kp
1
1 K 2
1  K 2  5  K 2  4; K  8
0.2 
39.
The Fourier transform of a continuous- time signal x(t) is given by X   
where j  1 and  denotes frequency. Then the value of
1
,     .
10  j
nx  t  at t  1 _______(up to 1 decimal
place.) (ln denotes the logarithm to base e).
Key: (10)
Exp: Given X   
1
10  j
2
,
Converting to s domain, X  s  
we know e10t u  t  

t.e10t 

1
 s  10
2
1
s  10
1
 s  10 
 x  t   t.e10t u  t 
 n x  t   n t.e10t u  t 
 n t   10t n c 
2
 n 1  10  1  1  10  10.0
40.
Consider a system governed by the following equations
dx1  t 
 x 2  t   x1  t 
dt
dx 2  t 
 x1  t   x 2  t 
dt
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34
EE-GATE 2018
The initial conditions are such that x1  0   x 2  0    . Let x1f  limx1  t  and x 2f  limx 2  t  .
t 
Which one of the following is true?
(A) x1f  x 2f  
(B) x 2f  x1f  
(C) x1f  x 2f  
(D) x1f  x 2f  
t 
Key: (C)
Exp:
dx1  t 
 x 2  t   x1  t  ....1
dt
dx 2  t 
 x1  t   x 2  t  .... 2 
dt
Differentiating 1 w.r.t 't '

d 2 x1  t  dx 2  t  dx1  t 


dt 2
dt
dt
dx  t 
 x1  t   x 2  t   1  from  2  
dt
 dx  t 
 dx  t 
 x1  t    1
 x1  t    1  from 1 
dt
 dt

d 2 x1  t 
dx  t 

2 1
0
2
dt
dt
D2  2D  0
D  D  2  0
 D  0, 2
Solution, x1  t   C1  C2e2t
x1f  lim x1  t   C1
t 
Similarly, Differently (2) w.r.t „t‟ using (1) & (2)
we get
d 2 x 2  t  2dx 2  t 

0
dt 2
dt
 solution, x 2  t   C1  C2 e 2t
x 2f  lim x 2  t   C1
t 
 x1f  x 2f  
41.
Consider the two bus power system network with given loads as shown in the figure.
1.0
G1
j0.1
1.00
G2
Q loss
20  jQG1
15  j5
15  jQG 2
20  j10
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35
EE-GATE 2018
All the values shown in the figure are in per unit. The reactive power supplied by generator G1 and G2
are QG1 and QG2 respectively. The per unit values of QG1 ,QG 2 , and line reactive power loss (Qloss)
respectively are
(A) 5.00, 12.68, 2.68
(C) 6.34, 11.34, 2.68
(B) 6.34, 10.00, 1.34
(D) 5.00, 11.34, 1.34
Key: (C)
Exp: P 
VS VR
sin 
x
11
sin 
0.1
sin   0.5
  30
5
VS
1
VS  VR cos   
1  cos 30   1.34 pu

X
0.1
Qs 
VR
1
VS cos   VR  
 cos30  1  1.34pu

x
0.1
 QS  Q R
QR 
Qloss
 1.34   1.34   2.68 pu
QG1  Qload  QS
 5  1.34  6.34pu
QG 2  Q R  Qload
QG2  QR  Qload  1.34  10  11.34pu
42.
A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as
shown in the figure. The rectifier output voltage has an average value given by V0 
Vm
 3  cos   ,
2
where Vm = 80  volts and  is the firing angle. If the power delivered to the lossless battery is 1600W,
 in degree is ________ (up to 2 decimal places).

2
Vm sin  t 
10mH
V0


80V
 Battery
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36
EE-GATE 2018
Key: (90)
Exp: Power delivered to lossless battery  EI0 =1600W
80 I0  1600
I0  20A
From given 1   rectifier, V0  E  I0 R
Vm
3  cos   E  I0 R
2
80
3  cos   80  20.2
2
403  cos   120
3  cos   3
cos   0
  90
43.
The positive, negative and zero sequence impedances of a three phase generator are Z1, Z2 and Z0
respectively. For a line to line fault with fault impedance Zf, the fault current is If 1  kIf , where If is the
fault current with zero fault impedance. The relation between Zf and k is
(A) Zf 
(C) Zf 
 Z1  Z2  1  k 
(B) Zf 
k
 Z1  Z2  k
(D) Zf 
1 k
 Z1  Z2  1  k 
k
 Z1  Z2  k
1 k
Key: (A)
Exp: If  L  L fault  
3
Z1  Z2
If1  L  L fault 
with fault impedance, Zf 
3
Z1  Z2  Zf
If 1  kIf  given 

3
3 3

Z1  Z2  Zf Z1  Z2
 Z1  Z2  k  Zf k  Z1  Z2
Zf k   Z1  Z2 1  k 
 Z  Z2 1  k 
Z  1
f
k
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37
EE-GATE 2018
44.
As shown in the figure, C is the arc from the point (3, 0) to the point (0, 3) and the circle x2+ y2=9. The
value of the integral
 y
2
C
 2yx  dx   2xy  x 2  dy is _________ (up to 2 decimal places).
y
 0,3
C
x
 3,0
Key: (0)
Exp: Given x 2  y 2  9
 y2  9  x 2
 y  9  x2
1
x
 dy 
2xdx  
dx

2 9  x2
9  x2
   y2  2yx  dx   2xy  x 2  dy 
C
 9  x
 
0
2
 2x 9  x 2 dx  2x 9  x 2  x 2
x 3

 xdx
9  x2
0
1


1
2
0

x3 9  x  2
x3 
x3
 9x  
2  
dx ...1
1
3
3  3 9  x2

1
2

3
0
Consider,

x3
dx, put 9  x 2  t 
9x
from 1 ,  18   18   0
2
3
45.
9  t  dt 
    18
t  2
t 0
a

Digital input signals A,B,C with A as the MSB and C as the LSB are used to realize the Boolean
function F = m0 + m2 + m3 + m5 + m7, where mi denotes the ith minterm. In addition, F has a don‟t care
for m1. The simplified expression for F is given by
(A) AC  BC  AC
(B) A  C
(C) C  A
(D) AC  BC  AC
Key: (B)
Exp: F  m  0,2,3,5,7   d 1
A  AC
  A  A  A  C   Distributive law 
AC
BC
BC
0
A
BC
1
x
1
A
4
1
BC
3
1
5
1
2
1
7
6
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38
EE-GATE 2018
46.
Let f  x   3x3  7x 2  5x  6. The maximum value of f(x) over the interval [0,2] is ________ (up to 1
decimal place).
Key: (12)
Exp: f  x   3x3  7x 2  5x  6
 f '  x   9x 2  14x  5  0
 x  1, 5 9
f ''  x   18x  14
f '' 1  0
&
f ''  5 9  0
 At x  5 9, f  x  has local maximum &
 5  1733
f 
 7.13
 9  243
f  0  6
f  2   12
 The maximum value of f  x  in 0, 2is"12"
47.
The per unit power output of salient-pole generator which is connected to an infinite bus, is given by the
expression, P=1.4sin  +0.15sin 2  , where  is the load angle. Newton Raphson method is used to
calculate the value of  for P=0.8pu. If the initial guess is 30°, then its value (in degree) at the end of the
first iteration is
(A) 15°
(B) 28.48°
(C) 28.74°
(D) 31.20°
Key: (C)
Exp: P  0.8  1.4 sin   0.15 sin 2
f     1.4sin   0.15sin 2  0.8
By, Newton Raphson method
1  0 
 6
f  0 
f '  0 
1.4sin 30  0.15sin 60  0.8
1.4 cos30   2  0.15 cos 60
 0.50164 rad
1  28.74
48.
Consider the two continuous time signals defined below:
| t |,
x1  t   
0,
1  t  1
,
otherwise
1 | t |, 1  t  1
x2  t   
otherwise
0,
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39
EE-GATE 2018
These signals are sampled with a sampling period of T=0.25 seconds to obtain discrete time signals x1[n]
and x2[n], respectively. Which one of the following statements is true?
(A) The energy of x1[n] is greater than the energy of x2[n].
(B) The energy of x2[n] is greater than the energy of x1[n].
(C) x1[n] and x2[n] have equal energies.
(D) Neither x1[n] nor x2[n] is a finite energy signal.
Key: (A)
Exp: Given
| t |,
x1  t   
0,
1 | t |, 1  t  1
x2  t   
otherwise
0,
1  t  1
otherwise
x1  t 
x2  t 
t
1
t
1
1
1


x1  n   1, 0.75, 0.5,0.25,0,0.25,0.5,0.75,1





x 2  n   0,0.25, 0.5,0.75,1,0.75,0.5,0.25,0

We can see clearly, x1  n  has greater energy than x 2  n 
Energy in x1  n   2  12  2   0.75   2   0.5   2   0.25   0 2
2
2
2
Energy in x 2  n   2  02  2   0.75   2   0.25   2   0.5   02
2
49.
2
2
The figure shows two buck converters connected in parallel. The common input dc voltage for the
converters has a value of 100V. The converters have inductors of identical value.
iS1

S1
L
C
100V

1
Switch control signals
S1
t
S2
S2
L
0
0.5ms
1ms
t
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40
EE-GATE 2018
The load resistance is 1  . The capacitor voltage has negligible ripple. Both converters operate in the
continuous conduction mode. The switching frequency is 1kHz, and the switch control signals are as
shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the
average value of iS1, the current of switch S1 (in Ampere), is ______ (up to 2 decimal places).
Key: (12.5)
Exp: V0  DVs  0.5 100  50V
I0 
V0 50

 50A.
R
1
Sence losser negligible, Pin  Pout
VS FS  Vout Iout
100  IS  50  50
IS  25A
IS1  IS2 
50.
Is
 12.5A
2
 1 0 1
Let A   1 2 0  and B = A3-A2-4A+5I, where I is the 3  3 identity matrix. The determinant of B
 0 0 2 
is _______ (up to 1 decimal place).
Key: (1)
 1 0 1
Exp: Given A   1 2 0 
 0 0 2 
Characteristic equation A is A  I  0
1 
0
1
 1 2  
0 0
0
0
2  
By Cayley Hamilton theorem,
A3  A 2  4A  4I  0
 A3  A 2  4A  5I  I
 B  I  According to the given question 
 B  I 1
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41
EE-GATE 2018
51.
A 200V DC series motor, when operating from rated voltage while driving a certain load, draws 10A
current and runs at 1000 r.p.m. The total series resistance is 1  . The magnetic circuit is assumed to be
linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.pm
(rounded to the nearest integer) is _________.
Key: (825)
Exp: 200V, 10A
 R a  R f   1
E b  V  Ia  R a  R f 
1
1
1
 200  10  190V
When torque becomes 1.44 times
200V
T2  1.44T1
Magnetic circuit linear,   Ia
2
T2 Ia 2
TI   2
T1 Ia1
2
a
T2
 Ia1
T1
Ia 2 
 1.44  10  12A,
E b2  V  I a 2  R a  R f 
 200  12  188V
E b2
E b1


N 2 2

N1 1
N 2 Ia 2

N1 Ia1
N2 
E b2
E b1

Ia1
Ia 2
 N1 
188 10
  1000
190 12
 824.56rpm
825 rpm  nearest integer 
52.
The capacitance of an air filled parallel – plate capacitor is 60pF. When a dielectric slab whose thickness
is half the distance between the plates, is placed on one of the plates converting it entirely, the
capacitance becomes 86pF. Neglecting the fringing effects, the relative permittivity of the dielectric is
_______ (up to 2 decimal places).
Key: (2.53)
Exp: Capacitance, C1 
0 A 20 A

 2   60pF   120pF
d/2
d
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42
EE-GATE 2018
20  r A
  2  60   r pF  120 r pF
d
CC
120  120 r
Ceq  1 2 
C1  C2 120 1   r 
C2 
and
Now,

86
 r
120 1   r
r 
or,
53.
86
 2.53
34
If C is circle | z | 4 and f  z  
(A) 1
z2
 z2  3z  22
(B) 0
. then  f  z  dz is
C
(C) -1
(D) -2
Key: (B)
Exp: f  z  
z
z2
2
 3z  2 
2

z2
 z  1  z  2 
2
2
 z  1& 2 are poles of order 2, both lie inside z  4
Residue of f  z  at z  1
 d 

1
z2
2

lim   z  1
2
2
 z  1 z1  dz 
 z  1  z  2  
d  z2 

 lim 
2
z 1 dz 

  z  2 
Res f  z  
z 1
 z  2   2z   z 2 .2  z  2   4
 lim
4
z 1
 z  2
Residue of f  z  at z  2
2
 d 

1
z2
2

lim   z  2 
2
2 
 z  1! z2  dz 
z 2
 z  1  z  2  
d  z2 
  4
 lim 
2
z  2 dz 

z

1




Res f  z  
 By Cauchy Residue theorem,
54.
 f  z  dz  2i  4  4   0
C
A dc to dc converter shown in the figure is charging a battery bank. B2 whose voltage is constant at 150
V. B1 is another battery bank whose voltage is constant at 50V. The value of the inductor, L is 5mH and
the ideal switch, S is operated with a switching frequency of 5kHz with a duty ratio of 0.4. Once the
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43
EE-GATE 2018
circuit has attained steady state and assuming the diode D to be ideal, the power transferred from B1 to
B2 (in Watt) is ___________ (up to 2 decimal places)
iL
L  5mH
D

50 V


B1
S
B2
150 V

Key: (12)
Exp: Given Vin  50V, Vout  150V
V0
1

Vin 1  D
150
1

50 1  D
2
D   0.666
3
But given duty cycle = 0.4
 The boost converter is operating in discontinuous mode.
1
 200 sec
f
 DT  80 sec.
T
t on
  
V0  
 Vs
D
  
150  
 50    0.6
   0.4 
VS  L
IL
I max
0
DT BT
T
t
di
I
L
dt
t on
I
 I  0.8A
80  106
I  I max  I min  I min  0 
50  5  103 
I  I max
1
1
Imax T 
 0.8  0.6  200  106 
IL avg   2
2
 0.24A
T
200  106
Power transfered from B1to B2  Vin IL avg   50  0.24  12W
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44
EE-GATE 2018
55.
In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain   100 . The
base-emitter voltage drop is a constant, VBE=0.7V. The value of the Thevenin equivalent resistance Rth
(in  ) as shown in the figure is _____ (up to 2 decimal places).
a
10

10 k
15V


10.7V
R Th
1k

b
Key: (90.9)
Exp: Form BJT amplifier, R Th can be written as
R Th 
VTh VOC

ISC
IS

it havedependent sources
10.7  0.7  10kIB  1k    1 I B  0
10
10m

10k  101k 111
1     10m  1k
VOC  IE  1k 
111
10.7  0.7 10
IB 

 1m
10k
10k
ISC  1    IB  1011m
IB 
R Th 
10 1k
 90.09
111
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45
EE-2019
GENERAL APTITUDE
Q. No. 1 - 5 Carry One Mark Each
1.
The passengers were angry _______ the airline staff about the delay.
(A)
towards
(B)
on
(C)
with
(D)
about
(D)
2197
Key: (C)
2.
The missing number in the given sequence 343, 1331, ________, 4913 is
(A)
4096
(B)
3375
(C)
2744
Key: (D)
343 1331
2197



3
3
7
11
must be133
4913

173
[Since 7, 11, 13, 17  Prime number series].
3.
Newspapers are a constant source of delight and recreation for me. The _______ trouble is that I
read ______ many of them.
(A)
only, too
(B)
only, quite
(C)
even, too
(D)
even, quite
Key: (A)
4.
It takes two hours for a person X to mow the lawn. Y can mow the same lawn in four hours. How
long (in minutes) will it take X and Y, if they work together to mow the lawn?
(A)
60
(B)
80
(C)
120
(D)
90
Key: (B)
X can mow the lawn  2 hours  X’s 1 hour work 
1
2
Y can mow the lawn  4 hours  Y’s 1 hour work 
1
4
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EE-2019
1 1  2 1 3
 

4
4
2 4
1
4
4
Total time required 
 hours   60min  80min
3
3 3
 
4
 
 X  Y  's  1 hour work  
5.
I am not sure if the bus that has been booked will be able to ________ all the students.
(A)
deteriorate
(B)
sit
(C)
accommodate
(D)
fill
Key: (C)
Q. No. 6 - 10 Carry Two Marks Each
6.
Given two sets X = {1, 2, 3} and Y = {2, 3, 4}, we construct a set Z of all possible fractions where
the numerators belong to set X and the denominators belong to set Y. The product of elements
having minimum and maximum values in the set Z is _________.
(A)
1/12
(B)
3/8
(C)
1/8
(D)
1/6
Key: (B)
Given two sets
X  1,2,3 & Y  2,3,4
1 1 1 2 2 2 3 3 3 
Z   , , , , , , , , 
2 3 4 2 3 4 2 3 4
Set of all possible fractions
Where the numerators belongs to set x and the denominators belong to set y.




 1 1 1 2 2 3 2 3 3 
z
, , , , , , , ,

 4 3 2 4 3 4 2 3 2 
0.2
1.5
minimum

Maximum 


1 3 3
The required product   
4 2 8
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EE-2019
7.
Consider five people-Mita, Ganga, Rekha, Lakshmi and Sana. Ganga is taller than both Rekha and
Lakshmi. Lakshmi is taller than Sana. Mita is taller than Ganga.
Which of the following conclusions are TRUE?
1.
Lakshmi is taller than Rekha
2.
Rekha is shorter than Mita
3.
Rekha is taller than Sana
4.
Sana is shorter than Ganga
(A)
3 only
(B)
1 only
(C)
2 and 4
(D)
1 and 3
Key: (C)
From the given information, we can draw as follows
Mita
Ganga
Rekha & Lakshmi
Sana
Here we don’t know who is taller between Rekha & Lakshmi.
So A is false:B, C & D are true.
8.
How many integers are there between 100 and 1000 all of whose digits are even?
(A)
60
(B)
100
(C)
90
(D)
80
Key: (B)
Integers between 100 and 1000 all of the whole digits are even = 100; since
100  199  No integers
200  300  25 integers
[200, 202, 204, 206, 208, 220, 222, 224, 226, 228, 240, 242, 246, 248, 260, 262, 264, 266, 268,
280, 282, 284, 286, 288].
Similarly; 400  500  25 integers
600  700  25 integers
800  900  25 integers
Total  100 integers
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EE-2019
9.
An award-winning study by a group researchers suggests that men are as prone to buying on
impulse as women but women feel more guilty about shopping.
Which one of the following statements can be inferred from the given text?
(A)
Many men and women indulge in buying on impulse
(B)
All men and women indulge in buying on impulse
(C)
Few men and women indulge in buying on impulse
(D)
Some men and women indulge in buying on impulse
Key: (A)
10.
The ratio of the number of boys and girls who participated in an examination is 4:3. The total
percentage of candidates who passed the examination is 80 and the percentage of girls who passed
is 90. The percentage of boys who passed is _________.
(A)
90.00
(B)
80.50
(C)
55.50
(D)
72.50
Key: (D)
Given, ratio of number of boys to girls who participated in the exam=4:3
Let, total students participated in the examination = 7x;
Given, total pass percentage = 80%
Total number of students passed in the examination  7x 
80
 5.6x
100
And
The percentage of girls who passed = 90%
i.e., number of girls who passed  3x 
90
 2.7x [Since the number of girls participated =3x]
100

Number of boys who passed in the exam = 5.6x – 2.7x = 2.9x

Required % 
2.9x
 100  72.50
4x
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EE-2019
ELECTRICAL ENGINEERING
Q. No. 1 to 25 Carry One Mark Each
1.
Given, Vgs is the gate-source voltage, Vds is the drain source voltage, and Vth is the threshold
voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in
saturation are
(A)
Vgs  Vth ; Vds  Vgs  Vth
(B)
Vgs  Vth ; Vds  Vgs  Vth
(C)
Vgs  Vth ; Vds  Vgs  Vth
(D)
Vgs  Vth ;Vds  Vgs  Vth
Key: (D)
For creating inversion layer in an n-channel MOSFET we need Vgs > Vth
For operating the n-channel MOSFET in the saturation region, we need VdS   Vgs  Vth 
2.
kT
, where k is Boltzmann’s constant, T is the
C
absolute temperature, and C is a capacitance. The standard deviation of the random process is
The mean-square of a zero-mean random process is
kT
C
(A)
(B)
kT
C
(C)
C
kT
(D)
kT
C
Key: (D)
Mean square value 
kT
C
Standard deviation  mean square value 
3.
kT
C
The parameter of an equivalent circuit of a three-phase induction motor affected by reducing the
rms value of the supply voltage at the rate frequency is
(A)
magnetizing reactance
(B)
rotor leakage reactance
(C)
rotor resistance
(D)
stator resistance
Key: (B)
I
V
z
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When v/g alone reduces, the current drawn by I.M reduces, Torque reduces  Tem  v2 
As the torque reduces, slip increases to get steady state operation.
 Tem 
 sv2
R2
The change in slip causes change in reactance of rotor
x 2r  sx 2
4.
A co-axial cylindrical capacitor show in Figure (i) has dielectric with relative permittivity r1  2.
When one-fourth portion of the dielectric is replaced with another dielectric of relative permittivity
 r 2 , as shown in Figure (ii), the capacitance is doubled. The value of  r 2 is __________.
r2
R
R
r
r
r1  2
r1  2
Figure(i)
Figure(ii)
Key: (10)
The capacitance of a coaxial cable
Co 
2
n b
 a

2  2o 
2o r

n b
n b
a
a
 
 
c1
The two capacitors C1 & C2 are connected in parallel

Ceq  C1  C2
3  2o  o r2


2 n b
2 n b
a
a
 
C2
b
a
 
Given Ceq  2Co
r2
r1  2


o  r2
2  2 o  
3o

 2
 n b 
n b
2 n b

a
a
a 
r
3  2  8   r2  10
2
 
 
 
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5.
The output voltage of a single-phase full bridge voltage source inverter is controlled by unipolar
PWM with one pulse per half cycle. For the fundamental rms component of output voltage to be
75% of DC voltage, the required pulse width in degree (round off up to one decimal place) is
___________.
Key: (112.8)
4Vdc
sin  d 
 2
 d  56.41
0.75 Vdc 
Hence the required pulse width in degrees = 2d = 2 × 56.41 = 112.82°
6.
The current I flowing in the circuit shown below in amperes (round off to one decimal place) is
___________.
3
2
I
2A
20V


5I
Key: (1.4)
The given circuit is
I
2
3
x

20V
2A


5I
I2
By KCL at node x, the current through the dependent source is I + 2.
Writing KVL at outer loop
20  2I  3  I  2   5I  0
20  2I  3I  6  5I  0
 14  10I
 I  1.4A
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7.
In the circuit shown below, the switch is closed at t = 0. The value of  in degrees which will give
the maximum value of DC offset of the current at the time of switching is
R  3.77
L  10mH
v  t   150sin 377t  
~
(A)
–45
t 0
(B)
90
(C)
60
(D)
–30
Key: (A)
8.
The total impedance of the secondary winding, leads, and burden of a 5ACT is 0.01 . If the fault
current is 20 times the rated primary current of the CT, the VA output of the CT is ___________.
Key: (100)
The VA output of the CT is  5  20   0.01  100 VA
2
9.
A 5kVA, 50 V/100V, single-phase transformer has a secondary terminal voltage of 95V when
loaded. The regulation of the transformer is
(A)
5%
(B)
9%
(C)
4.5%
(D)
1%
Key: (A)
Percentage regulation of T/F 
10.
E 2  V2
100  95
5
 100 
 100 
 5%
E2
100
100
Five alternators each rated 5MVA, 13.2 kV with 25% of reactance on its own base are connected in
parallel to a busbar. The short-circuit level in MVA at the busbar is ___________.
Key: (100)
S.C.MVA 
Base MVA
x d4
5

 100 mVA
0.25 5
0.25
0.25
0.25
0.25
0.25
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11.
A six pulse thyristor bridge rectifier is connected to a balanced three-phase, 50 Hz AC source.
Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in
the AC input current is
(A)
100 Hz
(B)
150 Hz
(C)
250 Hz
(D)
300 Hz
Key: (C)
The lowest Harmonic content in the AC input current is 5th harmonics.
f  5  fs  50  5  250 Hz
12.
The characteristic equation of a linear time-invariant (LTI) system is given by
  s   s4  3s3  3s2  s  k  0
The system BIBO stable if
(A)
k>3
(B)
0k
8
9
(C)
0k
12
9
(D)
k>6
Key: (B)
The given characteristic equation of LTI system is
  s   s4  3s3  3s2  s  k, for
BIBO stability, we prefer R-H criterion.
s4
s3
s2
s1
s0
1
3
8
3
8 3  3k
83
k
3 k
1 0
k
For stability all elements of 1st column should be positive
8
 3k
3
0
83
8
  3k
and k  0
3
8
 3k 
3
8
k
9
8
8
 k  0    k    0  k 
9
9

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13.
A system transfer function is H  s  
a1s2  b1s  c1
. If a1  b1  0, and all other coefficients are
a 2s 2  b2s  c2
positive, the transfer function represents a
(A)
high pass filter
(B)
notch filter
(C)
low pass filter
(D)
band pass filter
Key: (C)
It is given that
H s 
a1s2  b1s  c1
a 2 s 2  b 2s  c 2
If a1  b1  then H  s  becomes
H s 
H  0 
c1
a 2s 2  b 2s  c 2
c1
 i.e., as low frequency s  0    0 
c2
H     0  i.e., as high frequency s       
So the system passes low frequency and blocks high frequency. So it represents a low pass filter.
14.
The symbols, a and T, represent positive quantities, and u(t) is the unit step function. Which one of
the following impulse response is NOT the output of a causal linear time-invariant system?
(A)
e a  t  T  u  t 
(B)
1  eat u  t 
(C)
e a  t T  u  t 
(D)
e at u  t 
Key: (B)
If a L.T.I system is causal, we must should have the condition.
h  t   0; for t  0
If we see the options a, c, d in these u(t) is multiplied that means their impulse response are zero
for –ve value of time, hence they are causal.
If we check option B
h  t   1  eat u  t 
We can see h(t) = 1; t < 0, hence it represents a non causal system.
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15.
 grad f.dr
If f  2x 3  3y 2  4z, the value of line integral
C
evaluated over contour C formed by the
segments  3, 3,2   2, 3,2   2,6,2   2,6, 1 is ___________.
Key: (139)
Given,
f  2x 3  3y2  4z
 f  ˆi 6x 2   ˆj64  kˆ  4
Curl f   0ˆ

 f is irrotational
 The value of line integral does not dependent on the path of integration, only depends on the
end points of integral.
 gradf .dr   6x dx  6xydy  4dz   d  2x

2
C


 3, 3,2 
C
 2,6, 1

 3, 3,2 
  2x 3  3y 2  4z 
16.
 3y 2  4z 
C
 2, 3,2 
  grad f.dr 
3
d  2x 3  3y 2  4z  
 2,6,2 

 2, 3,2 
d  2x 3  3y 2  4z  
 2,6, 1
  d  2x
 2,6,2
3
 3y 2  4z 
d  2x 3  3y 2  4z 
 2,6, 1
 3, 3,2 
 120   19   139
A three-phase synchronous motor draws 200 A from the line at unity power factor at rated load.
Considering the same line voltage and load, the line current at a power factor of 0.5 leading is
(A)
100A
(B)
300A
(C)
Key: (C)
Power drawn by load  3   .P  3VL IL cos 
400A
(D)
200A
p.fVs If
1
Ia VS If
VL  Vrated ; IL  200A]
cos   unity
0.5
400A
If the power factor reduces to 0.5 lead
whereas
200A
If
P, VL are same as earlier
I
1
cos  
 I  400A
Unity
lagg P.f
under excitation
lead p.f
over excitation
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17.
The inverse Laplace transform of H  s  
(A)
3te t  e t
(B)
s3
for t  0 is
s  2s  1
2
3e t
(C)
4te t  e t
(D)
2te t  e t
Key: (D)
Given, H  s  
s3
t0
s  2s  1
2
 s3 
s3 
1
1 
 L1 

 L H s  L  2
2

 s  2s  1 
  s  1 
s 1 2
 1
2 
 L1 

  L1 

2
2
  s  1 
 s  1  s  1 
 1  t
 1 
1
 L1 

2L

  e 1  2e  t t
2

 s  1
  s  1 
 1  1 

 L  s2   1 
 


 L1  H  s    2te  t  e t
18.
The open loop transfer function of a unity feedback system is given by
G s 
e0.25s
,
s
In G(s) plane, the Nyquist plot of G(s) passes through the negative real axis at the point.
(A)
 1.5, j0
(B)
 0.5, j0
(C)
 0.75, j0
(D)
 1.25, j0
Key: (B)
It is given that G  s  
e0.25s
, whe the Nyquist
s
Plot cuts the negative real axis, its phase becomes –180°
 180  90  0.25
180

180
 90

 0.5
 
   2
 0.25
 0.25
Magnitude at this frequency
G  2  
e0.25 j 2  1
  0.5
j2
2
At –ve real axis the co-ordinate becomes (–0.5, j0)
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19.
The partial differential equation
2u 2  2u 2u 
 c  2  2   0; where c  0 is known as
t 2
y 
 x
(A)
Wave equation
(B)
Poisson’s equation
(C)
Laplace equation
(D)
Heat equation
Key: (A)
Given partial D.E
2
2u
2u 
2 u

C

 2
  0, where c  0
t 2
y2 
 x

20.
2
2u
2u 
2 u

C

 2
 ; which is clearly two dimensional have equation.
t 2
y2 
 x
M is a 2 × 2 matrix with eigen values 4 and 9. The eigen values of M2 are
(A)
16 and 81
(B)
2 and 3
(C)
–2 and –3
(D)
4 and 9
Key: (A)
Given,
M is a 2 × 2 matrix with Eigen values 4 and 9.

i.e.,   4,9 [Where ‘  ’ represents Eigen values of M]
From the properties of Eigen values; we have if  is an Eigen value of the matrix M then 2 is
an Eigen value of the matrix M2.
 2  42 ,92  16, 81 are Eigen values of matrix M2.
21.
The Ybus matrix of a two-bus power system having two identical parallel lines connected between
  j8 j20
them in pu is given as Ybus  
.
 j20  j8
The magnitude of the series reactance of each line in pu (round off up to one decimal) place) is
____________.
Key: (0.1)
  j8 j20 
Given Ybus  

 j20  j8
j20
Y  each line  
 j10
2
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The magnitude of the series reactance of each line in
Pu 
22.
1
1
  0.1pu
j10 10
0 1 1 
The rank of the matrix, M  1 0 1  , is __________.
1 1 0 
Key: (3)
Linear algebra –Rank
Method-I
Given,
0 1 1 
M 1 0 1 
1 1 0 
Applying R 3  R 3  R 2 ; then
0 1 1 
M 1 0 1 
0 1 1
Applying
R 3  R 3  R1 there
0 1 1 
M ~ 1 0 1 
0 0 2 
Applying R 2  R 3 ;
1 0 1 
M ~ 0 1 1   Echelon form of the matrix
0 0 2 
  M   Number of non  zeros in echelonform
  M  3
Method-2
 
0 1
M
1 0
1 1

1
 1 0  1  11  0
1
0
 1  1  2  0.
 M33  0    M   3
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23.
The output response of a system is denoted as y(t), and its Laplace transform is given by
10
Y  S 
. The steady state value of y(t) is
s s2  s  100 2

(A)

1
100 2
(B)
(C)
10 2
1
10 2
(D)
100 2
Key: (C)
 It is given that
Y s 

10
s s  s  100 2
2

We need to find steady state value of y  t  i.e, y   
 By final value theorem
y     limsY  s   lim s
s 0

24.
s 0

10
s s  s  100 2
2

10
1

100 2 10 2
A current controlled current source (CCCS) has an input impedance of 10 and output impedance
of 100 K. When this CCCS is used in a negative feedback closed loop with a loop gain of 9, the
closed loop output impedance is
(A)
100 k
(B)
100 
(C)
(D)
10
1000 k
Key: (D)
For a current controlled current source, both input as well output signals are in current form. so,
the correct feedback technique will be current shunt feedback (or in another word it is called as
shunt series feedback)
R inf
R in  10
I1

V1
R 0  100K
CCCS


Vf

R of  Zof
Load
Feed back
Network
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Output impedance given loop gain AB  9, Z0  100k
Zof  Z0 1  AB  100k1  9  1000k
25.
Which one of the following functions is analytic in the region z  1?
(A)
z2  1
z  j0.5
(B)
z2  1
z2
(C)
z2  1
z  0.5
(D)
z2  1
z
Key: (B)
Given region is Z  1; which represents the region inside and on the unit circle z  1
The function given in the options 1,3,4 are not analytic functions in the region Z  1; Since the
singular points –j(0.5), 0.5, 0 lies inside
Z  1.
Let f  2  
Z2  1
Z2
A Z  2 is the singular point but this is lies on side
 Z 1

Z2  1
is analytic in the region Z  1.
Z2
Q. No. 26 - 55 Carry Two Marks Each
26.
In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent
circuit is a
X
Y
Z
(A)
XOR gate
(B)
NOR gate
(C)
XNOR gate
(D)
NAND gate
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Key: (A)
X
XY
Y
Z
XY
Z  XY  XY  X  Y
Given circuit is equivalent to XOR gate.
27.
The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The
mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible.
10cm
I
10cm
0.2cm
When the core relative permeability is assumed to be infinite, the magnetic flux density computed
in the air gap is 1tesla. With same Ampere-turns, if the core relative permeability is assumed to be
1000 (linear), the flux density in tesla (round off to three decimal places) calculated in the air gap
is________.
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Key: (0.834)
Given,
Air gap  lg   0.2 cm.
Mean length of core = 40. 0.2=39. 8 cm. (lc)
 r2  core   1000,  r1  core   .
B1  01 T
from
NI  constant  B.A Reluetance
 g

 g

c
c
 B1 A 



  B2 A 
 0 A 0 r1 A 
 0 A 0 r2 A 
 r1  , Hence
c
0 r1
 0.
 0.2 
 0.2 39.8 
B1    B2 


  100  
 0 
B1  0.2 
1  0.2
B2 

 0.83402 T.
39.8  0.2398

 0.2 

1000 

Hence, if the core relative permeability is assumed to be 1000, then the flux density in the air gap
is 0.834 T.
28.
A delta-connected, 3.7 kW, 400 V(line), three-phase, 4-pole, 50-Hz squirrel-cage induction motor
has the following equivalent circuit parameter per phase referred to the stator:
R1  5.39, R 2  5.72, X1  X2  8.22. Neglect shunt branch in the equivalent circuit. The
starting line current in amperes (round off to two decimal places) when it is connected to a 100V
(line), 10 Hz, three-phase AC source is__________.
Key: (14.94)
As frequency has changed to 10 Hz. Reactance will change but resistance will be same.
8.22
 10  1.644
50
R1  5.39 R 2  5.72
X1  X 2  at 10Hz  
Istarting  line  at 10Hz 
100 3
 5.39  5.72 
2
 1.644  2 
2
 14.94A
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29.
If A  2xi  3yj  4zk and u  x 2  y 2  z 2 , then div(uA) at (1, 1,1) is_______.
Key: (45)
Given ,
A  2x i  3y j  4z k and
u  x 2  y2  z2
div  uA   .  uA 
 div  uA   u  A   uA  Usin g vector for identities 
.... 1



 2x    3y    4z 
x
y
z
 23 4  9
A 
 A  9   2 
&
u 

u
u
u  j  K ; where u  x 2  y 2  z 2
x
y
z
 i  2x   j 2y   k  2z   2  xi  yi  2K 
u. A   2x  i   2y  j   2z  K.  2x  i   3y  i   4Z 
...  3
 u.A  4x 2  6y 2  8z 2
From 1 ,  2  &  3 , we have
div  uA    x 2  y 2  z 2  9   4x 2  6y 2  8z 2 
div  uA 
30.
1,1,1
 1  1  1 9   4  6  8   27  18  45.
The asymptotic Bode magnitude plot of a minimum phase transfer function G(s) is shown below.
G j dB
60
20dB/decade
40
40dB/decade
20
0
1
10
20
log scale
60dB/decade
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Consider the following two statements.
Statement I: Transfer function G(s) has three poles and one zero.
Statement II: At very high frequency      , the phase angle G  j  
3
.
2
Which one of the following option is correct?
(A)
Statement I is false and statement II is true.
(B)
Both the statements are true.
(C)
Both the statements are false.
(D)
Statement I is true and statement II is false.
Key: (A)
 From the given bode-plot, we can say
 At origin, their is a pole at origin, since the initial slope is -20db/dec.
 At w=1, the change in slope is 40   20  20db/sec, so it imply one pole at w=1.
 At w=20, the change in slope is 60   40  20db/dec, so it imply one pole at w=20.
 So in total the transfer function has 3 poles, hence at w   , the net phase contributed by
3 poles is 270 or 
3x
2
 Hence statement I is false and II is right
31.
1 

3 s 

aT 
The transfer function of a phase lead compensator is given by D  S  
.
1

s  
 T
The frequency (in rad/sec), at which D  j is maximum, is
(A)
3T 2
(B)
3
T2
(C)
3T
(D)
1
3T 2
Key: (D)
 It is given that transfer function of a lead compensator is
1 

 s  3T 
D s  3 

 s 1 
T 

1
T
1
3T
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 The frequency at which phase is maximum is given
by geometric mean of pole, zero location,
1
 1  1 
Wm     
3T 2
 3T  T 
32.
The voltage across and the current through a load are expressed as follows


v  t   170sin  377t   V
6



i  t   8 cos  377t   A
6

The average power in watts (round off to one decimal place) consumed by the load is ______.
Key: (588.89)
It is given that


 V  t   170sin  377t  
6

  170 sin  377t  30 
 170 cos  377t  60 


 i  t   8cos  377t    8 cos  377t  30 
6

To calculate the phase difference between V(t) and i(t) both of them should in either +ve cos or
+ve sin.
 Pavg  Vrms I rms cos  v  I 
 170   8 


 cos  60 30 
 2 2
170  8

cos  30   680 cos30  588.89 watts
2
33.
A DC-DC buck converter operates in continuous conduction mode. It has 48 V input voltage, and it
feed a resistive load of 24 . The switching frequency of the converter is 250 Hz. If switch-on
duration is 1 ms, the load power is
(A)
12W
(B)
6W
(C)
48W
(D)
24W
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Key: (D)
Given
Resistive load  24
Vin  48v, f s  250Hz, TON  1ms.
1
 4ms.
250
T
1
D  ON   0.25.
T
4
As load is resistive and they did not mention current is ripple free. Hence the load power can be
writer as.
T
Pout  load 

34.
V

 0.5  48
24
2

 
2
0(rms)
R
0.25  48

2
24
24  24
 24
24
A single-phase fully-controlled thyristor converter is used to obtain an average voltage of 180V
with 10 A constant current to feed a DC load. It is fed form single-phase AC supply of 230V, 50
Hz. Neglect the source impedance. The power factor (round off to two decimal places) of AC mains
is _____.
Key: (0.78)
V0 dC  180V Idc  10A
Vs  230V, 50 Hz
For, single- phase fully controlled thyristor converter if load is constant, then
2vm
cos   180v

180  
cos  
 0.8692
2  230 2
Vdc 
Ipf 
35.
2 2
2 2
cos  
 0.8692  0.782.


The closed loop line integral
z3  z 2  8
 z  2 dz
z 5
evaluated counter-clockwise, is
(A)
4j
(B)
4j
(C)
8j
(D)
8j
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Key: (C)
Let F  z  
z3  z 2  8
z2
singular point of F  2 is Z  2; which
Lies inside C: Z  5
Using Cauchy’s integral formula, we have
 F  z  dz 
C


C

C
z3  z 2  8
dz
z2
z3  z 2  8
dz
z   2 
 2j  z3  z 2  8
Z  2


f  2
dz  2j  f (z 0 ) 
 by cauchy 's formula; 
C z  z0


3
2
z  z 8
C z  2 dz  2j 8  4  8  8j
36.
A fully-controlled three-phase bridge converter is working form a 415V, 50 Hz, AC supply, It is
supplying constant current of 100 A at 400 V to a DC load. Assume large inductive smoothing and
neglect overlap. The rms value of the AC line current in amperes (round off tow two decimal
places) is _____.
Key: (81.64)
The rms value of the AC line (for three phase bridge converter)
 2 3 Ide  2 3 100  81.64A
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37.
The enhancement type MOSFET in the circuit below operates according to the square law.
 n Cox  100 A V 2 , the threshold voltage  VT  is 500 mV. Ignore channel length modulation. The
output voltage Vout is
VDD  2V
5A
Vout
W 10m

L
1m
(A)
2V
(B)
100 mV
(C)
500 mV
(D)
600 mV
Key: (D)
Given  n cox  100 A V 2 , Vt  500mv,   0
The MOSFET is following square law, Hence it is operating in the saturation region
1
2
W
ID   n cox    Vas  Vt 
2
L
VDD  2V
1
2
 10 
5  106   100  106    VGS  0.5
2
1
 
 VGS  0.5
2

VGS  0.6 volt
Vout  600 mV
38.
2  5  106
1000  106
ID
5A
Vout  VGS

VGs
W 10
  VDS
L
1

In a 132 kV system, the series inductance up to the point of circuit breaker location is 50 mH. The
shunt capacitance at the circuit breaker terminal is 0.05 F. The critical value of resistance in ohms
required to be connected across the circuit breaker contacts which will give no transient oscillation
is_______.
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Key: (500)
The critical value of resistance required to be connected across the circuit breaker contacts which
will give no transient oscillation
R cr 
39.
1 R 1 50 103

 500
2 C 2 0.05 106
The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit.
The probability of two or more defective resistors in the circuit (round off to two decimal places) is
______.
Key: (0.26)
Given,
The probability of a resistor being defective
i.e. P  0.02.
 Number of resistors n  50.
  np  50  0.02  50 
2
 1.
100
Let ‘x’ denote the number of defective resistors using Poison distribution we have
P  x  2  1  P  x  2  1  P  x  0   P  x  1
 e  0 e   
e  x 
1 

 P  n  

1!  
n! 
 0!
 1  e 1   
 1  e1 1  1  1  2 e  0.26
40.
The output expression for the Karnaugh map shown below is
PQ
RS 00
00 0
(A)
QR  S
01
1
11
1
10
0
01
1
1
1
1
11
1
1
1
1
10
0
0
0
0
(B)
QR  S
(C)
QR  S
(D)
QR  S
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Key: (C)
41.
PQ
RS 00
00 0
01 1
11 1
10 0
QR
01 11 10
1 1 0
S
1 1 1
1 1 1
0 0 0
A periodic function f  t  , with a period of 2, is represented as its Fourier series,
If f  t   a 01 a n cos nt  n 1bn sin nt. .
A sin t, 0  t  
f t
  t  2 '
0,
The Fourier series coefficients a1 and b1 of f (t) are
(A)
a1  0; b1  A 
(C)
a1 
A
; b1  0
2
A
; b1  0

(B)
a1 
(D)
a1  0; b1 
A
2
Key: (D)
As per the given description of f(t), if we draw its waveform, if looks like
f t
..........
A
0

2
3
4
 One way to obtain its C.T.T.S is by obtaining its odd and even part and then by obtaining their
individual C.T.F.S and finally we can add them to get complete C.T.F.S of f(t). However in this
case we can pick the correct option by eliminating others.
 f  t   f  t    f (t)  f  t  
 

2
2

 

 f  t   fo  t  1  fe  t   
A/2
A/2
..........

0
f t


..........
..........

2

0
f et

2
f t 
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A
 N
f o  t    sin w o t    a n cos w o t
2
 n 1
From this b1 
42.
A
, So only option D satisfy this.
2
A 0.1 F capacitor charged to 100 V is discharged through a 1 k resistor. The time in ms (round
off to two decimal places) required for the voltage across the capacitor to drop to 1V is________.
Key: (0.46)
It is given that
 since Vc     0 in this case
VC  t   Vc (0 )e  t /e
VC  t x   100e
 1  100e10000tx  e10000t x 
 10000t x  n  0.01
 tx 
43.
0.1f
 tx /0.11103
1k
1
100
VC0100V
n 0.01
 0.46  103 sec  0.46m.sec
10,000
A moving coil instrument having a resistance of 10, gives a full-scale deflection when the current
is 10 mA. What should be the value of the series resistance, so that it can be used as a voltmeter for
measuring potential difference up to 100 V?
(A) 9990 
(B)
990 
(C)
99
(D)
9
Key: (A)
Vm  10  10  103  0.1V
m
V 100

 1000
Vm 0.1
R Series   m  1 R m  999 10  9990 
44.
A three-phase 50 Hz, 400 kV transmission line is 300 km long. The line inductance is 1 mH/km per
phase, and the capacitance is 0.01 F km per phase. The line is under open circuit condition at the
receiving end and energized with 400 kV at he sending end, the receiving end line voltage in kV
(round off to two decimal places) will be______.
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Key: (418.59)
Given, line length = 300 km long (means long line).
VS  400kv  line to line  . L  1m H Km and C  0.01F km.
V  speed  
1
1

 316227.766km s.
LC
0.01 106  1 103
 2f 
 2  50  300 




B2
V
  1   316227.766   0.955.
A 1
1 
2
2
2
Vs
400
Vr  Noload  

 418.59 kV.
A 0.955
2
45.
2
In the circuit below, the operational amplifier is ideal. If V1  10 mV and V2  50 mV, the output
voltage  Vout  is
100k
10k
V1

V2

Vout
10k
100k
(A)
100 mV
(B)
600 mV
(C)
400 mV
(D)
Key: (C)
100
Va 
 0.05
100  10
1
Va  volt.
22
Va  Vout Va  0.01

0
100
10
Va  Vout  10Va  0.1  0
Vout  11Va  0.1
500 mV
100k
10k
V1  10mV
 0.01V
V2  50mV
Va
10k
Va

Vout

 0.05V
100k
 1 
 11   0.1  0.4 volt  400 mv
 22 
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46.
The current I flowing in the circuit shown below in amperes is ___________
50
40
25
I
20
20
200V
160V
100V
80V
Key: (0)
X
50
40
25
I
20
20
200V
→
160V
100V
80V
By Miliman's theorem, the network to the left of xy can be replaced by
Rm
X
I
Vm
I
Vm
R m  20
20
…(1)
Y
1  
1  
1  
1 

 200    160    100     80  
50  
40  
25  
20 
where Vm  
1
1 1 1
  
50 40 25 20
4444

0
1
1 1 1
  
50 40 25 20
0
 0A
→ Putting Vm in equation (1) becomes I 
R m  20
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47.
A 220V DC shunt motor takes 3A at no-load. It draws 25A when running at full-load at 1500 rpm.
The armature and shunt resistances are 0.5Ω and 220 Ω, respectively. The no-load speed in rpm
(round off to two decimal places) is _______ .
Key: (1579.32)
3A
25
2A
IA
0.5
220


IA
220V
24
0.5
220
Eb
No  load speed N1
E b1  V  Ia R a
220V
Eb
N 2  1500 r.p.m Full load
 220  2  0.5
 219Volt
E b2  220  24  0.5
 208Volt
N .E
N2 Eb2 1
1500  219


 N1  2 b1 
1579.32 rpm
N1 Eb1 2
Eb2
208
1  2
48.
In a DC-DC boost converter, the duty ratio is controlled to regulate the output voltage at 48V. The
input DC voltage is 24V. The output power is 120W. The switching frequency is 50kHz. Assume
ideal components and a very large output filter capacitor. The converter operates at the boundary
between continuous and discontinuous conduction modes. The value of the boost inductor (in μH)
is _________.
Key: (24)
Given, Vo  48V, Vin  24V, Pout  12000
fswitching  50kHz,
48 
V 
24 
Vout  in 

1 D 
1 D 
2  2D  1, 2D  1  D  0.5
120
Po  Vo .Io  Io 
48
120 48  48
R Load 

 19.2
48
120
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D 1  D  R 0.5  0.52 19.2

 24H
2f
2  50 103
2
LC 
Hence, the converter operates at the boundary between continuous and discontinuous conduction
modes, if the value of the boost inductor is 24μH.
49.
The line currents of a three-phase four wire system are square waves with amplitude of 100A.
These three currents are phase shifted by 120° with respect to each other. The rms value of neutral
current is
100
(A) 100A
(B) 0A
(C) 300A
(D)
A
3
Key: (A)
50.
A single-phase transformer of rating 25kVA, supplies a 12kW load at power factor of 0.6 lagging.
The additional load at unity power factor in kW (round off to two decimal places) that may be
added before this transformer exceeds its rated kVA is __________.
Key: (7.2)
Given, rating of transformer = 25kVA,
Existing load, S  12  j16
Let P is extra load with exceeding rated KVA.
 P  12 
51.
2
 16   252  P  7.209kW
2
Consider a state-variable model of a system
1   x1   0 
 x1   0
 x     2  x     r
 2  
 2 
x 
y  1 0  1 
x2 
Where y is the output, and r is the input. The damping ratio  and the undamped natural frequency
n  rad/sec of the system are given by
(A)
   ; n 


(B)


; n  

(C)
   ; n  
(D)


; n  

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Key: (D)
From the given state space model, we can say that
1 
0
0
A
, B    , C  1 0 , D  ??

  2
 
In order to calculate
, n . we need the transfer function of the system, which is given by
T  s   C SI  A  B
1
 S 0   0
1 
 1 0 
 

 0 S    2  
1
0
 
 
1
1   0 
S
 1 0 
  
 S  2   

S  2 1    0 
1
 1 0 


S   
 S  S  2     

 
1
 1 0  2
  
 S  2S    S

52.
1 0 S

 
S2  2 S  
=

, by comparing with standard
S  2S  

  
n
We can say  n
2
S  2 nS  n
2   2
2
2
In the single machine infinite bus system shown below, the generator is delivering the real power of
0.8 pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees
(round off to one decimal place) is _________.
Xt 0.2pu
G
Xd' 0.25pu
XL1 0.4pu
XL2 0.4pu
V 10

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Key: (20.5)
The generator is delivering the real power = 0.8pu at 0.8pf
0.8
 1pu. X net  0.25  0.2  0.2  0.65pu
0.8
E g  V  Ig z  10 1  0.6590  36.86
I
E g  10  0.6553.14  1   cos53.14  jsin 53.14  0.65
 1  0.389  j0.520
 1.389  j0.520
Hence load angle is 20.52°.
53.
A 30kV, 50Hz, 50MVA generator has the positive, negative, and zero sequence reactances of
0.25pu, 0.15pu, and 0.05pu, respectively. The neutral of the generator is grounded with a
reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the
generator terminal are equal. The value of grounding reactance in ohms (round off to one decimal
place) is _______.
Key: (1.8)
Given X1  0.25pu, X2  0.15pu, X0  0.05pu
According to question, LG  LLLG  current wise  
3
1

0.25  0.15  0.05  3X  0.25
3X   0.3  X   0.1pu
X   0.1 
54.
302
 1.8
50
A 220V (line) three-phase,Y-connected, synchronous motor has a synchronous impedance of
 0.25  j2.5  / phase.
The motor draws the rated current of 10A at 0.8 pf leading. The rms value
of line-to line internal voltage in volts (round off to two decimal places)is ________.
Key: (245.35)
Given, V=220V, Vph 
E

220
127.01V
3
 V cos   Ia R a    Vsin  Ia Xa 
2
2
127.01  0.8  10  0.25  127.01  0.6 10  2.5
2
2
 141.65V
Eline  141.65  3  245.34V
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55.
Consider a 2  2 matrix
M   v1 v2 , where v1 and v2 are the column vectors.
Suppose
u1T 
M   T  where u1T andu T2 are the row vectors. Consider the following statements:
u 2 
Statement 1: u1T v1  1 and u T2 v 2  1
1
Statement 2: u1T v 2  0 and u T2 v1  0
Which of the following options is CORRECT ?
(A) Statement 2 is true and statement 1 is false
(B) Statement 1 is true and statement 2 is false
(C) Both the statements are false
(D) Both the statements are true
Key: (C)
Given
x x2 
Let M 22   V1 V2    1

 y1 y 2 
 u1T   X1 Y1   u1T
1
M 2 2   T   

T
 u 2   X 2 Y2   u 2
We have, MM 1  I
 x x 2   X1 Y1  1 0
  1
 


 y1 y 2   X 2 Y2  0 1 
 x X  x 2 X 2 x1Y1  x 2 Y2  1 0 
 1 1
 

 y1X1  y 2 X 2 y1Y1  y 2 Y2  0 1 
x X  x 2 X 2  1; x1Y1  x 2 Y2  0 
 1 1
  1
y1X1  y 2 X 2  0; y1Y1  y 2 Y2  1
x 
x 
 u1T v1   X1 Y1   1  & u T2 v 2   X 2 Y2   2 
 y1 
 y2 
 u1T v1  x1 X1  y1Y1
 x 2 X 2  y 2 Y2  1
So statement 1 is false (From (1))
x 
u1T v 2   X1 Y1   2   x 2 X1  y 2 Y1  0
 y2 
x 
u T2 v1   X 2 Y2   1   x1 X 2  y1 Y2  0  From (1) 
 y1 
So statement 2 is also false
∴ Both the statements are false.

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GATE ESE PSU’s 2019-20
EEE ENGINEERING
GATE EEE OBJ. PAPER SOL.(2011-2019)
GATE (2011-2019) EEE 19 SET PAPER SOLUTION
CONTENT COVERED:
1.GATE EEE PAPER 2011 SOLUTION
2.GATE EEE PAPER 2012 SOLUTION
3.GATE EEE PAPER 2013 SOLUTION
4.GATE EEE PAPER 2014 (1 SET) SOLUTION
5.GATE EEE PAPER 2015 (2 SET) SOLUTION
6. GATE EEE PAPER 2016
(2 SET)
SOLUTION
7. GATE EEE PAPER 2017
(2 SET)
SOLUTION
8. GATE EEE PAPER 2018
(1 SET)
(1 SET)
SOLUTION
Page
1
9. GATE EEE PAPER 2019
SOLUTION
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