GATE- EE
(Electrical Engg)
Previous Year: Fully
Solved Papers
(2011-14)
(Total- 6 Tests)
GATE- EE 2014
Paper-01
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Q. No. 1 – 5 Carry One Mark Each
1.
Which of the following options is the closest in meaning to the phrase underlined in the
sentence below?
It is fascinating to see life forms cope with varied environmental conditions.
(A) Adopt to
(B) Adapt to
(C) Adept in
(D) Accept with
Answer: (B)
Choose the most appropriate word from the options given below to complete the following
sentence.
He could not understand the judges awarding her the first prize, because he thought that her
performance was quite ___________________.
(A) Superb
(B) Medium
(C) Mediocre
(D) Exhilarating
Answer: (C)
2.
3.
In a press meet on the recent scam, the minister said, “The buck stops here”. What did the
minister convey by the statement?
(A) He wants all the money
(B) He will return the money
(C) He will assume final responsibility
(D) He will resist all enquiries
Answer: (C)
4.
Exp:
If ( z + 1 / z ) = 98, compute ( z 2 + 1 / z 2 )
2
96
Expanding
1
1
1
z 2 + 2 + 2.z. = 98 ⇒ z 2 + 2 = 96
z
z
z
The roots of ax 2 + bx + c = 0 are real and positive a, b and c are real. Then ax 2 + b x + c = 0
has
(A) No roots
(B) 2 real roots
(C) 3 real roots
(D) 4 real roots
Answer: (D)
Exp: ax2+bx+c=0
for roots to be real & +ve
b2-4ac>0
This will have 2 real positive roots.
ax 2 + b x + c = 0
5.
This can be written as;
ax 2 + bx + c
Discri min ant = b 2 − 4ac > 0
ax 2 − bx + c
(− b) 2 − 4ac
⇒ b 2 − 4ac
Is also >0 This will have real roots
⇒ This will have 4 real roots.
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Q. No. 6 – 10 Carry One Mark Each
6.
The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the
Western Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons
for the formation of this gap are not clear. It results in the neighbouring regions of Tamil
Nadu getting more rainfall from the South West monsoon and the neighbouring regions of
Kerala having higher summer temperatures.
What can be inferred from this passage?
(A) The Palghat gap is caused by high rainfall and high temperatures in southern Tamil Nadu
and Kerala
(B) The regions in Tamil Nadu and Kerala that are near the Palghat Gap are low–lying
(C) The low terrain of the Palghat Gap has a significant impact on weather patterns in
neighbouring parts of Tamil Nadu and Kerala
(D) Higher summer temperatures result in higher rainfall near the Palghat Gap area
Answer: (B)
7.
Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses
such as depression and schizophrenia, and consequently, that doctors will be able to eradicate
these diseases through early identification and gene therapy.
On which of the following assumptions does the statement above rely?
(A) Strategies are now available for eliminating psychiatric illnesses
(B) Certain psychiatric illnesses have a genetic basis
(C) All human diseases can be traced back to genes and how they are expressed
(D) In the future, genetics will become the only relevant field for identifying psychiatric
illnesses
Answer: (B)
8.
Round–trip tickets to a tourist destination are eligible for a discount of 10% on the total fare.
In addition, groups of 4 or more get a discount of 5% on the total fare. If the one way single
person fare is Rs 100, a group of 5 tourists purchasing round–trip tickets will be charged Rs
___________
Answer:
Exp:
850
One way force =100
Two way fare per person=200
5 persons=1000/Total discount applicable=10+5=15%
Discount amount =
15
× 1000 = 150
100
Amount to be paid=1000-150=850
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9.
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In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were
further asked to mention whether they own a car or scooter or both. Their responses are
tabulated below. What percent of respondents do not own a scooter?
Own vehicle
Men
Women
Car
40
34
Scooter
30
20
Both
60
46
20
50
Do not own vehicle
Answer:
Exp:
48%
Total respondents=300
Those who don’t have scooter
⇒ Men= 40+20=60
women = 34 + 50 =
84
144
144
× 100
300
= 48%
%=
When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by
straight lines to its corners, how many (new) internal planes are created with these lines?
_______________________
Answer:
6
10.
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Q. No. 1 – 25 Carry One Mark Each
1.
Given a system of equations:
x + 2y + 2z = b1
5x + y + 3z = b 2
Which of the following is true regarding its solutions?
(A) The system has a unique solution for any given b1 and b 2
(B) The system will have infinitely many solutions for any given b1 and b 2
(C) Whether or not a solution exists depends on the given b1 and b 2
(D) The system would have no solution for any values of b1 and b 2
Answer: (B)
Exp:
1
2

1
[ A / B] =  5
2 b1 

3 b2 
1
R 2 → R 2 − 5R 1 
0
2
−9
b1 

− 7 b 2 − 5b1 
2
∴ rank ( A ) = rank ( A / B) < number of unknowns, for all values of b1 and b 2
∴ The equations have infinitely many solutions, for any given b1 and b 2
2.
Let f ( x ) = x e − x . The maximum value of the function in the interval ( 0, ∞ ) is
(A) e −1
Answer:
(C) 1 − e −1
(B) e
(D) 1 + e −1
(A)
Exp:
f ' ( x ) = 0 ⇒ e − x (1 − x ) = 0 ⇒ x = 1 and f '' ( x ) < 0 at x = 1
∴ M ax imum value is f (1) = e −1
3.
The
solution
x ( 0 ) = 1 and
dx
dt
for
differential
equation
d2x
= −9 x
dt 2
with
initial
conditions
= 1, is
t =0
(A) t 2 + t + 1
(C)
the
1
sin 3t + cos3t
3
1
2
(B) sin 3t + cos3t +
3
3
(D) cos3t + t
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Exp:
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(C)
A.E : m 2 + 9 = 0 ⇒ m = ±3i
∴ Solution is x = a cos3t + bsin 3t
..... (1)
dx
= −3a sin 3t + 3b cos 3t ...... ( 2 )
dt
dx
= 1, (1) and ( 2 ) gives
U sin g x ( 0 ) = 1 and
dt t = 0
and
1 = a and 1 = 3b ⇒ b =
1
3
1
∴ x = cos3t + sin 3t
3
4.
Let X ( s ) =
3s + 5
be the Laplace Transform of a signal x(t). Then, x ( 0+ ) is
s + 10s + 21
2
(A) 0
Answer:
(B) 3
(C) 5
(D) 21
(B)
Exp:
x  0+  = lim
s →∞
s. 3s + 5
[ u sin g initial value therorem ]
( s + 10s + 21)
2
 5
s 2 3 + 
s

= lim
=3
s →∞
10
21 
2 
s 1 + + 2 
s s 

5.
Let S be the set of points in the complex plane corresponding to the unit circle. (That is,
S = {z : z = 1} . Consider the function f (z) = z z* where z* denotes the complex conjugate of
z. The f (z) maps S to which one of the following in the complex plane
(A) Unit circle
(B) Horizontal axis line segment from origin to (1, 0)
(C) The point (1, 0)
(D) The entire horizontal axis
Answer:
Exp:
(C)
f ( Z ) = Z.Z* where Z* is conjugate of Z
= Z = 1 = 1 + i.0
2
∴ f ( Z ) maps S to the point (1,0 ) in the complex plane
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6.
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The three circuit elements shown in the figure are part of an electric circuit. The total power
absorbed by the three circuit elements in watts is ________________
10 A
8A
100 V
80 V
15 V
Answer: 330 watts.
Exp:
10 A
8A
80 V
100 V
2A
15V
By KCL, current through 15V source is 2A.
→ When current entering in to +ve terminal of a battery means, it is absorbing the power.
→ When current entering in to –ve terminal, means, delivering the power.
→ 100V source is absorbing the power
i.e. (10) (100) = 1000 watts.
→ 80V source is delivering the power
i.e. (8) (80) = 640 watts
→ 15V source is delivering the power
i.e. (2) (15) = 30 watts
∴ The total power absorbed by the circuit elements ie = 1000 - (640+30) = 330 watts.
7.
C0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If,
half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity ∈r , the
expression for the modified capacitance is
( a)
(A)
C0
(1+ ∈r )
2
(B)
( C0 + ∈r )
(b )
(C)
C0
∈r
2
(D) C0 (1+ ∈r )
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Answer:
Exp:
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(A)
Co.
C0 =
Aε 0
d
A 1d
A 2 ε2
ε1
d
C = C1 + C2
A1ε1 A 2 ε 2
+
d
d
Aε
Aε ε
= 0+ r 0
2d
2d
Aε 0
C=
(1 + ε r )
2d
C
C = 0 (1 + ε r )
2
C=
8.
A combination of 1µF capacitor with an initial voltage vc ( 0 ) = −2V in series with a 100Ω
resistor is connected to a 20 mA ideal dc current source by operating both switches at t = 0s
as shown. Which of the following graphs shown in the options approximates the voltage vs
across the current source over the next few seconds?
+
t =0
VC
−
+
−
VS
t =0
VS
VS
(A)
(B)
t
t
−2
−2
VS
(C)
VS
t
−2
(D)
t
−2
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Answer: (C)
Exp:
Under steady state,
+
20
mA
3
•
+
•
−
ϑc = 2v
−
1µF
100 Ω
ϑs
•
•
+
−
When switch is opened:
I
•
+
•
−
−
+
1µF
−2
s
ϑs
+
100 Ω
−
•
•
By using Laplace transform approach,
ϑs
( )
1 1   1 
−2
VS ( S ) =   
 +  100 +
S
S
SC
S
 
  
VS ( S ) =
20 × 103
S2
⇒ VS ( t ) =
2 × 10
S2
t
−2
VS ( t ) = 2 × 10 t
4
9.
x(t) is nonzero only for Tx < t < T 'x , and similarly, y(t) is nonzero only for Ty < t < T 'y . Let
z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE?
(A) z(t) can be nonzero over an unbounded interval
(B) z(t) is nonzero for t < Tx + Ty
(C) z(t) is zero outside of Tx + Ty < t < T 'x + T 'y
(D) z(t) is nonzero for t > T 'x + T 'y
Answer:
Exp:
(C)
Given that z ( t ) is x ( t ) ∗ y ( t )
Range of z(t) is [sum of lower limits of x ( t ) and y ( t ) to sum of upper limit of x(t) and y(t)].
Tx + Ty < t < Tx + Ty
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10.
For a periodic square wave, which one of the following statements is TRUE?
(A) The Fourier series coefficients do not exist
(B) The Fourier series coefficients exist but the reconstruction converges at no point
(C) The Fourier series coefficients exist and the reconstruction converges at most points.
(D) The Fourier series coefficients exist and the reconstruction converges at every point
Answer:
(C)
Exp: For a periodic square wave, fourier series coefficients value decreases as the ‘k’ increases. At
some value of ‘k’ coefficient becomes zero, thus no conveyance otherwise it converges at
most points.
An 8–pole, 3–phase, 50 Hz induction motor is operating at a speed of 700 rpm. The frequency
of the rotor current of the motor in Hz is _______________
Answer:
3.33Hz
Exp: Given, P = 8, F = 50Hz, N=700 rpm
4 Frequency of Rotor current = s.f
N −N
750 − 700
=
= 0.067 ∴ f r = ( 0.067 ) × 50 = 3.33Hz
S = s
Ns
750
11.
12.
For a specified input voltage and frequency, if the equivalent radius of the core of a
transformer is reduced by half, the factor by which the number of turns in the primary should
change to maintain the same no load current is
(A) 1/4
(B) 1/2
(C) 2
(D) 4
Answer:
(C)
Exp: If the equivalent Radius of the core of a transformer is reduced by half then the
reluctance of the core becomes double.
4 to maintain same no-load current the primary turns should be double. Then flux remains
same.
13.
A star connected 400V, 50Hz, 4 pole synchronous machine gave the following open circuit
and short circuit test results:
Open circuit test: Voc = 400V (rms, line–to–line) at field current, If = 2.3A
Short circuit test: Isc = 10 A (rms, phase) at field current, If = 1.5A
The value of per phase synchronous impedance in Ω at rated voltage is _________________
Answer:
15.06 Ω / ph
Exp:
Given, O.C. test: VOC = 400V ( L − L )1 If = 2.3A
S.C test: Isc = 10A ( phase ) , If = 1.5A
4 Per phase synchronous impedance, Zs = ?
`
We know, Zs =
Voc
Isc If issame
∴ Isc at If = 2.3A ⇒
2.3
×10 = 15.33A
1.5
400
∴ Zs =
3 = 15.06 = 15.06 Ω / ph
15.33
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14.
The undesirable property of an electrical insulating material is
(A) High dielectric strength
(B) High relative permittivity
(C) High thermal conductivity
(D) High insulation resistivity
Answer:
(B)
15.
Three–phase to ground fault takes place at locations F1 and F2 in the system shown in the
figure
I F1
F1
E A ∠δ
I F2
A
B
F2
VF1
E B ∠0
VF2
If the fault takes place at location F1 , then the voltage and the current at bus A are VF1 and IF1
respectively. If the fault takes place at location F2 , then the voltage and the current at bus A are
VF2 and I F2 respectively. The correct statement about voltages and currents during faults at
F1 and F2 is
(A) VF1 leads IF1 and VF2 leads I F2
(B) VF1 leads IF1 and VF2 lags IF2
(C) VF1 lags IF1 and VF2 leads IF2
(D) VF1 lags IF1 and VF2 lags I F2
Answer:
Exp:
(C)
When fault takes place at F1
A
F1
B
F2
Current is feeding into the BUS A. It is like a generator delivering power to Bus (A)
When fault takes place at F2 , F2 point is like load, taking power from generator.
16.
A 2–bus system and corresponding zero sequence network are shown in the figure.
Bus 1
T1
T2
Bus 2
X Gn
X Mn
(a )
X 0G
X 0T1
X 0L
3X Gn
X 0T 2
X 0M
3X Mn
(b)
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The transformers T1 and T2 are connected as
and
(A)
and
(C)
Answer: (B)
Exp:
and
(B)
and
(D)
XOTI
Transformer
zerosequence network
Transformer
zerosequence network
In the formation of Routh–Hurwitz array for a polynomial, all the elements of a row have
zero values. This premature termination of the array indicates the presence of
(A) Only one root at the origin
(B) Imaginary roots
(C) Only positive real roots
(D) Only negative real roots
Answer: (B)
Exp: If all elements of a row have zero values. Which leads to auxiliary equation formation and
roots of auxiliary equations gives imaginary roots.
17.
18.
The root locus of a unity feedback system is shown in the figure
jω
K=0
K=0
−2
−1
σ
The closed loop transfer function of the system is
(A)
(C)
C (s )
R (s)
C (s )
R (s)
=
K
( s + 1)( s + 2 )
(B)
=
K
s
+
1
s
( )( + 2 ) − K
(D)
C (s )
R (s)
C (s )
R (s)
=
−K
( s + 1)( s + 2 ) + K
=
K
s
+
1
s
( )( + 2 ) + K
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Answer: (C)
Exp:
C(s)
R (s)
=
k
will give the root locus given the diagram.
( s + 1)( s + 2 ) − k
19.
Power consumed by a balanced 3–phase, 3–wire load is measured by the two wattmeter
method. The first wattmeter reads twice that of the second. Then the load impedance angle in
radians is
π
π
π
π
(B)
(C)
(D)
(A)
12
8
6
3
Answer: C
Exp: When load impedance is π radians. The first wattmeter reads twice that if the second
6
wattmeter.
20.
In an oscilloscope screen, linear sweep is applied at the
(A) Vertical axis
(B) Horizontal axis
(C) Origin
(D) Both horizontal and vertical axis
Answer: (B)
21.
A cascade of three identical modulo–5 counters has an overall modulus of
(A) 5
(B) 25
(C) 125
(D) 625
Answer:
(C)
Exp: When more than one modulus counter is cascaded then their overall modulus will be product
of modulus of each individual .So, in this question overall modular of the counter
= 5 × 5 × 5 = 125
22.
In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when
C1
R1
+ Vcc
+
−
R3
− Vcc
C2
R2
R4
(A)
R 3 R1
1
=
, ω=
R4 R2
R1C1R 2 C 2
(B)
R 2 C2
1
=
, ω=
R1 C1
R 1C1R 2 C 2
(C)
R 3 R1 C 2
1
=
+
, ω=
R 4 R 2 C1
R1C1R 2 C 2
(D)
R 3 R1 C2
1
+
=
, ω=
R 4 R 2 C1
R 1C1R 2 C 2
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Answer: (C)
Exp: When bridge is balanced, z1z 4 = z 2 z3
⇒
23.
R 3 R1 C2
1
=
+
,f =
R 4 R 2 C1
2π R1R 2 C1C 2
The magnitude of the mid–band voltage gain of the circuit shown in figure is (assuming h fe
of the transistor to be 100)
+ Vcc
10 k
C
C
10 k
Vi
(A) 1
Answer: (D)
Exp:
A Vs =
=
=
A Vs =
h fe = 100
1k
(B) 10
V0
C
(C) 20
(D) 100
Vo
Vs
− h fe I b (10 k )
10k
Ib (10 k ) + I b h ie
− h fe (10 k )
I3
hfc
VS
10 k + h ie
IC
hfcIb
V0
10k
−100 × 10k
( neglecting hie )
10k
A Vs = 100
24.
The figure shows the circuit of a rectifier fed from a 230–V (rms), 50–Hz sinusoidal voltage
source. If we want to replace the current source with a resistor so that the rms value of the
current supplied by the voltage source remains unchanged, the value of the resistance (in
ohms) is ______________ (Assume diodes to be ideal.)
10 A
230V, 50Hz
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23Ω
Vs = 230V, 50 Hz
Exp:
The rms volt of given
Re ctifier Vor = Vs
∴ Vor = 230V
Ior =
10 A
Vs
Vor
V
230
⇒ R = or =
R
Ior
10
∴ R = 23Ω
25.
Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches can block
voltages of either polarity (applied between terminals ‘a’ and ‘b’) when the active device is in
the OFF state?
a
b
b
(i )
( ii )
a
a
b
b
( iii )
( iv )
(A) (i), (ii) and (iii)
(B) (ii), (iii) and (iv)
(C) (ii) and (iii)
(D) (i) and (iv)
Answer:
Exp:
a
(C)
Switch (i)
blocks the voltage in only forward direction
Switch (ii)
blocks the voltage in both forward and reverse directions
Switch (iii)
blocks voltage in both direction
Switch (iv)
blocks voltage in only forward direction
Hence Ans. is (C) because switch (i) & (ii) only satisfy the given requirement
Q. No. 26 – 55 Carry Two Marks Each
26.
Let g : [ 0, ∞ ) → [ 0, ∞ ) be a function defined by g ( x ) = x − [ x ] , where [ x ] represents the
integer part of x. (That is, it is the largest integer which is less than or equal to x).
The value of the constant term in the Fourier series expansion of g(x) is ______________
1/2
Answer:
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Clearly, g(x) is a periodic function with period ‘1’
consider, g ( x ) = x − [ x ] for 0 < x < 1
The constant term in the fourier series expansion of
g ( x ) is A 0 =
1
a0
= ∫ g ( x ) dx
0
2
1
1
 x2 
1
= ∫ x dx − ∫ [ x ] dx =   − ∫ ( 0 ) dx =
0
0
2
 2 0 0
1
27.
1
A fair coin is tossed n times. The probability that the difference between the number of heads
and tails is ( n − 3) is
(A) 2− n
(B) 0
(C)
n
C n −3 2− n
(D) 2− n + 3
Answer:
(B)
Exp: Let X = difference between the number of heads and tails.
Take n = 2 ⇒ S = {HH,HT,TH,TT} and X = −2,0,2; Here, n − 3 = −1 is not possible
Take n = 3 ⇒ S = {HHH, HHT, HTH, HTT,THH,THT,TTH,TTT} and X = −3, −1,1,3
Here n − 3 = 0 is not possible
Similarly, if a coin is tossed n times then the difference between heads and tails is n − 3 is not
possible
∴ required probability is 0
28.
The line integral of function F = yzi, in the counterclockwise direction, along the circle
x 2 + y 2 = 1 at z = 1 is
(A) −2π
Answer:
Exp:
(B) −π
(C) π
(D) 2π
(B)
Line int egral = ∫ F.dr
C
= ∫ yzdx
C
=∫
2π
0
 C is circle x 2 + y 2 = 1 at Z = 1



⇒ x = cos θ, y = sin θ and θ = 0 to 2π 
( sin θ )(1)( − sin θdθ )
2π
2 π  cos 2θ − 1 
1  sin 2θ

=∫ 
− θ  = −π
 dθ = 
0
2
2 2


0
29.
An incandescent lamp is marked 40W, 240V. If resistance at room temperature
( 26o C ) is 120Ω and temperature coefficient of resistance is 4.5 × 10−3 / o C , then its ‘ON’ state
filament temperature in o C is approximately ________________
Answer:
2471
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P = 40W
V = 240V
V 2 2402
=
= 1440Ω
P
40
Αt t = 26o , R = 120Ω, α = 4.5 × 10−3 / o C
R lamp =
R lamp = R 1 + α ( θ2 − θ1 ) 
1440 = 120 1 + 4.5 × 10−3 [ θ2 − 26]
θ2 = 2470.44o C
30.
In the figure, the value of resistor R is ( 25 + I / 2 ) ohms, where I is the current in amperes.
The current I is _______________
I
300V
R
Answer: 10A
Exp:
I

Given R =  25 + 
2

V = IR
I

300 = I  25 + 
2

⇒ I 2 + 50 I − 600 = 0
I = 10A 80 − 60A.
The current 10A is correct based on the given direction.
31.
In an unbalanced three phase system, phase current Ia = 1∠ ( −90o ) pu, negative sequence
current I b2 = 4∠ ( −150o ) pu, zero sequence current IC0 = 3∠90o pu. The magnitude of phase
current I b in pu is
(A) 1.00
Answer:
(C)
Exp:
(B) 7.81
(C) 11.53
(D) 13.00
Ia =1 −90 p.u
Ib2 = 4 − 150 p.u
Ic0 = 3 90 p.u
Ia = Ia1 + Ia 2 + Ia0
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sin ce Ib2 = αIa 2
4 −150 = (1120 ) Ia 2
Ia 2 = 4 −270
Ia 0 = Ibo = Ico = 3 90°
Ia1 =1 −90 = Ia1 + 4 −270 + 3 90
Ia1 = 8 −90
I b1 = α 2 Ia1 = (1 240 )( 8 −90 ) = 8 +150
I b = Ib0 + Ib1 + Ib2 = 8 +150 + 4 −150 + 3 90
= 11.53 + 154.3 p.u
32.
The following four vector fields are given in Cartesian co–ordinate system. The vector field
which does not satisfy the property of magnetic flux density is
(A) y 2 a x + z 2 a y + x 2 a z
(B) z 2 a x + x 2 a y + y 2 a z
(C) x 2 a x + y 2 a y + z 2 a z
(D) y 2 z 2 a x + x 2 z 2 a y + x 2 y 2 a z
Answer:
Exp:
(C)
For magnetic fields ∇ .B = 0
By verification
(a) ∇.B = ( 0 ) + 0
(b) ∇.B = 0
(c) ∇ .B = 2x + 2y + 2z ≠ 0
So C is correct
33.
The function shown in the figure can be represented as
(A) u ( t ) − u ( t − T ) +
(B) u ( t ) +
Answer:
Exp:
T
(t − T) −
( t − 2T ) u
T
( t − 2T )
1
t
t
u ( t − T ) − u ( t − 2T )
T
T
(C) u ( t ) − u ( t − T ) +
(D) u ( t ) +
(t − T) u
(t − T) u
T
(t − T) u
T
(t − T) − 2
(t) −
( t − 2T ) u
T
( t − 2T ) u
T
(t)
t
0
T
2T
( t − 2T )
(A)
t−T
 t − 2T 
x (t) = u (t) − u (t − T) + 
u (t − T) − 
 u ( t − 2T )
 T 
 T 
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34.
1
be the Z–transform of a causal signal x[n]. Then, the values of
1 − z −3
x [ 2] and x [3] are
Let X ( z ) =
(A) 0 and 0
Answer:
(B)
Exp:
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(B) 0 and 1
Given x ( z ) =
(C) 1 and 0
(D) 1 and 1
1
1 − z −3
x ( z ) can be written as
= 1 + z −3 + z −6 + 2−9
x [ 2] correspond to coefficient z −2 = 0
x [3] correspond to coefficient of z −3 = 1
35.
Let f ( t ) be a continuous time signal and let F ( ω) be its Fourier Transform defined by
∞
F ( ω) = ∫ f ( t ) e − jωt dt
Define g ( t ) by
−∞
∞
g ( t ) = ∫ F ( u ) e − jut du
−∞
What is the relationship between f ( t ) and g ( t ) ?
(A) g(t) would always be proportional to f(t)
(B) g(t) would be proportional to f(t) if f(t) is an even function
(C) g(t) would be proportional to f(t) only if f(t) is a sinusoidal function
(D) g(t) would never be proportional to f(t)
Answer:
(B)
Exp: We know the fourier transform relationship
∞
F ( ω) =
∫ f ( t ).e
− jω t
dt
−∞
∞
1
jwt
∫ F ( ω) e dω
2π −∞
ω can be replaced by u
∞
1
jut
f (t) =
∫ F ( u ) e du
2π −∞
and f ( t ) =
Now g ( t ) =
∞
∫ F ( u ).e
jut
du
....... (1)
....... ( 2 )
−∞
replace t by − t in (1)
∞
1
− jut
∫ F ( u ) e du .....( 3)
2π −∞
1
f ( −t ) =
g(t)
2π
if f ( t ) = f ( − t ) f ( t ) is even function 
f ( −t ) =
⇒ g ( t ) = 2π f ( t )
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36.
The core loss of a single phase, 230/115V, 50Hz power transformer is measured from 230 V
side by feeding the primary (230V side) from a variable voltage variable frequency source
while keeping the secondary open circuited. The core loss is measured to be 1050 W for
230V, 50Hz input. The core loss is again measured to be 500W for 138V, 30Hz input. The
hysteresis and eddy current losses of the transformer for 230V, 50Hz input are respectively,
(A) 508 W and 542 W
(B) 468 W and 582 W
(C) 498 W and 552 W
(D) 488 W and 562 W
Answer:
(A)
Exp:
Given data, 1 − φ
230
, 50 Hz
115 V
Care loss = 1050W at 230V, 50 Hz
Care loss = 500W at 138V, 30 Hz
4 In both cases V
Hence,
Wi
↓
CORE LOSS
f
ratio is constant
= A.f + B.f 2 = Wn + We
at 50 Hz ⇒ 1050 = A ( 50 ) + B ( 50 )
at 30 Hz ⇒ 500 = A ( 30 ) + B ( 30 )
2
2
A =10.167
B = 0.217
∴Wn at 50 Hz = (10.167 ) × 50 = 508 W
We at 50 Hz = ( 0.217 ) × ( 50 ) = 542 W
2
A 15kW, 230V dc shunt motor has armature circuit resistance of 0.4Ω and field circuit
resistance of 230Ω . At no load and rated voltage, the motor runs at 1400 rpm and the line
current drawn by the motor is 5 A. At full load, the motor draws a line current of 70A.
Neglect armature reaction. The full load speed of the motor in rpm is ________________
Answer: 1240 rpm.
Exp: Given 15kW, 230V . dc shunt motor
37.
Armature resistance R a = 0.4Ω
Field Resistance, R sh = 230Ω
5A
1A
1A
4A
70A
69A
230V
230Ω
∴ E b1 = 230 − 4 × 0.4
= 228.4V
230V
∴ E b2 = 230 − 69 × 0.4
= 202.4V
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∴
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N2
E
φ
= b2 × 1
N1
E b1 φ2
=
E b2
( ∵ Flux is constant )
E b1
N 2 = 1400 ×
202.4
= 1240 rpm
228.4
A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1Ω and reactance of
0.92Ω . Neglect the voltage drop in stator and assume that the rotor resistance is constant.
Given that the full load slip is 3%, the ratio of maximum torque to full load torque is
(A) 1.567
(B) 1.712
(C) 1.948
(D) 2.134
Answer:
(C)
Exp: Given, P = b, f = 50Hz
38.
R 2 = 0.1Ω, X 2 = 0.092Ω
Sfl = 3% = 0.03
Tmax = ?, we know that ⇒
∴
Tmax S2m + Sfl2
=
Tfl
2Sm Sfl
Tmax
S2 + Sfl2
= m
Tfl
2Sm Sfl
where s m =
R2
0.1
=
= 0.108 = 0.03
x2
0.92
( 0.108) + ( 0.03)
T
∴ max =
Tfl
2 ( 0.108 )( 0.03)
2
39.
2
= 1.938 1.94
A three phase synchronous generator is to be connected to the infinite bus. The lamps are
connected as shown in the figure for the synchronization. The phase sequence of bus voltage
is R–Y–B and that of incoming generator voltage is R '− Y '− B' .
R Y
B
R'
Y'
B'
La
Infinite Bus
Lb
Inco min g Generator
Lc
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It was found that the lamps are becoming dark in the sequence L a − L b − L c . It means that the
phase sequence of incoming generator is
(A) Opposite to infinite bus and its frequency is more than infinite bus
(B) Opposite to infinite bus but its frequency is less than infinite bus
(C) Same as infinite bus and its frequency is more than infinite bus
(D) Same as infinite bus and its frequency is less than infinite bus
Answer:
Exp:
(A)
According to given connection of Lamp’s. Tthey are becoming dark in the sequence La-
Lb-Lc. Hence the phase sequences are different, and also the frequency is more than infinite
bus.
40.
A distribution feeder of 1km length having resistance, but negligible reactance, is fed from
both the ends by 400V, 50Hz balanced sources. Both voltage sources S1 and S2 are in phase.
The feeder supplies concentrated loads of unity power factor as shown in the figure.
S1
400m
400V
50Hz
200m
200A
200m
100A
200m
200A
S2
400V
50Hz
The contributions of S1 and S2 in 100A current supplied at location P respectively, are
(A) 75A and 25A
(B) 50A and 50A
(C) 25A and 75A
(D) 0A and 100A
Answer: (D)
Exp:
source−1
400 m
200 m
I
B
200 A
200 m
100 A
200 m
source−2
200 A
Let I be current supplied by source – 1
‘r’ be resistance/ length
400 = ( 400r ) I + ( 200r )( I − 200 ) + ( 200 r )( I − 300 ) + ( 200r ) ( I − 500 ) + 400
0 = 400rI + 200rI − 40000r + 200rI − 60000 + 200 rI −100000r
100 Ir = 20000r
I = 200A
Current in branch- B is I − 200 = 200 − 200 = 0
4 point p, source-1 supplies 0A current
Source-2 supplies 100A current
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41.
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A two bus power system shown in the figure supplies load of 1.0 + j0.5p.u.
G1
Bus 1
Bus 2
V1∠0o
1 ∠δ 2
j0.1
1.0 + j0.5
j2
The values of V1 in p.u. and δ 2 respectively are
(A) 0.95 and 6.00 ο
(B) 1.05 and -5.44 ο
(C) 1.1 and -6.00 ο
(D) 1.1 and -27.12 ο
Answer: (B)
Exp:
A B 
1 2 
1
 C D  = 0 1  =  0





 Vs 
A B 
  = 

 C D
 Is 
j0.1
 1 0.1 90 
= 

1 
1 
0
Vs = V1 0, B = 0.1 90
 Vr 
 
Ir 
Is
+
Z
Ir
+
Vs
Vr
−
−
Vr =1 δ 2 , A =1 0
Vs = AVr + BIr
Ir =
Vs A
V 0
− Vr = 1
( 1( δ 2 ) )
B B
0.1 90
= 10 V1 −90 −10 δ 2 − 90
I*r =10V1 90 −10 90 − δ2
( ) [ 10V 90 − δ ]
Sr = Pr + jQ r = Vr I*r = 1 δ 2
1
2
= 10V1 90 + δ 2 −10 90
= 10  V1 cos ( 90 + δ 2 ) + jV1 sin ( 90 + δ 2 )  − j10
=10 [ − V1 sin δ 2 + jV1 cos δ 2 ] − j10
= [ − 10V1 sin δ 2 ] + j[10V1 cos δ 2 −10]
Given Sr =1+ j0.5
−10V1 sin δ 2 = + 1 
→ 10 V1 sin δ2 = − 1
−1

⇒ δ 2 = − 5.44
 tan δ 2 =
10.5
10V1 cos δ 2 − 10 = 0.5 
→ 10V1 cos δ 2 = 10.5
From 10 V1 sin δ 2 = − 1
10V1 sin [5.44] = − 1
V1 = 1.054
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42.
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The fuel cost functions of two power plants are
Plant P1 : C1 = 0.05Pg12 + APg1 + B
Plant P2 : C 2 = 0.10 Pg 22 + 3APg 2 + 2B
Where, Pg1 and Pg 2 are the generated powers of two plants, and A and B are the constants. If the
two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio
of load shared by plants P1 and P2 is
(B) 2 : 3
(A) 1: 4
(C) 3 : 2
(D) 4 :1
Answer: (D)
Exp:
C1 = Pg12 ( 0.05 ) + APg1 + B
Pg1 + Pg 2 =1000
C 2 = 0.1Pg 22 + 3A Pg 2 + B
dc1 dc 2
=
=1000
dpg1 dp 2
dc1
= 2 × 0.05Pg1 + A = 100
dpg1
dc 2
= 2 × 0.1Pg 2 + 3A = 100
dpg 2
0.1Pg1 + A = 100  0.3Pg1 − 0.2 Pg 2 = 200
0.2 Pg 2 + 3A =100  Pg1 + Pg 2 = 1000

Solving, Pg1 = 800 MW
Pg 2 = 200 MW
Pg1
4
=
Pg 2
1
43.
The over current relays for the line protection and loads connected at the buses are shown in
the figure
A
C
B
RA
300 A
RB
200 A
100 A
The relays are IDMT in nature having the characteristic
t op =
0.14 × Time Multiplier Setting
( Plug Setting Multiplier )
0.02
−1
The maximum and minimum fault currents at bus B are 2000 A and 500 A respectively.
Assuming the time multiplier setting and plug setting for relay R B to be 0.1 and 5A
respectively, the operating time of R B (in seconds) is _______________
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Answer: 0.23
Exp:
A
C
B
RA
RB
300A
t op =
For Re lay B
T.M.S = 0.1
0.14 × T.M.S
( P.S.M )
0.02
100A
200A
−1
P.S.M = 5A
Ifault ( max ) = 2000A
Ifault ( min ) = 500A
Maximum load current at C = 100A
For a setting current of 1A, plug setting is 100%
Max fault current
2000
=
= 20
CT ratio × current setting
100 ×1
0.14 × T.M.S
0.14 × 0.1
=
=
= 0.227 ≈ 0.23sec
0.02
0.02
( P.S.M ) − 1 ( 20 ) − 1
PSM =
t op
44.
For the given system, it is desired that the system be stable. The minimum value of α for this
condition is ___________________
R ( s)
+
−
( s + α)
s3 + (1 + α) s2 + ( α − 1) s + (1 − α)
C(s)
Answer: 0.618
Exp: The characteristic equation is 1 + G(s) = 0
1+
s
3
(s + α )
=0
+ (1 + α ) s + ( α − 1) s + (1 − α )
2
⇒ s3 + (1 + α ) s 2 + αs + 1 = 0
For stable system α should be 0.618
By R- H criteria, (1 + α ) α > 1.
(α
2
+ α − 1) > 0
α = 0.618 & − 0.618
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The Bode magnitude plot of the transfer function G ( s ) =
K (1 + 0.5s )(1 + as )
I shown
s 
 s

s  1 +  (1 + bs )  1 + 
 8
 36 
below:
Note that − 6dB / octave = −20dB / decade.The value of
a
is ____________________
bK
0dB / Octave
−6dB / Octave
6dB / Octave
0dB / Octave
−6dB / Octave
dB
−12dB / Octave
0
0.01
2
4
8
24
36
→ ω ( rad / s )
Answer: 0.75
Exp: By observing the magnitude plot,
 S  S
K 1 +  1 + 
 2  4
G (s ) =
S 
S
 S 
S 1 +  1 +  1 + 
 8   24   36 
By comparing with given transfer function,
a= 1 ; b= 1 .
4
24
For finding K:
(M)ind B
(−20 dB/ dec)
ω1
bg ω
K = ( ω1 ) : where n is no.of poles from the given plot; K = ( 8 )
n
1
i.e.K = 8
So,
1
a
4 ⇒ 0.75
=
bk 1 .8
24
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A system matrix is given as follows
1 − 1
 0

A =  −6 − 11 6 
 −6 − 11 5 
The absolute value of the ratio of the maximum eigen value to the minimum eigen value is
___________________
Answer:
1/3
Exp:
Characteristic equation is A − λI = 0
−λ
i.e., −6
1
− 11 − λ
−1
6
− 11
5−λ
−6
=0
⇒ λ 3 + 6λ 2 + 11λ + 6 = 0
⇒ λ = −1, −2, −3 are the eigen values of A
λ max = −1 and λ min = −3
∴
47.
λ max
−1 1
=
=
λ min
−3 3
The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is
____________________
R = 0.5 Ω
1 / jΩ
1jΩ
100sin ( ωt )
V
1jΩ
1 / jΩ
Answer:
Exp:
142
Net z = j1 − j1 = 0, acts as short circuit
i(t) =
100sin ( ωt )
0.5
i(t)
= 200sin ( ωt )
= 100sin ( ωt ) = i V2
2
V1 = ( − j1 )100sin ( ωt )
i V1 =
0.5 Ω
V2 = ( j1 )100sin ( ωt )
100 sin ( ωt )
j1
− j1
V1
− j1
V2
j1
V = V1 − V2 = − j200sin ωt
VRMS =
Vm
2
=
200
= 141.42 Volts
2
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The dc current flowing in a circuit is measured by two ammeters, one PMMC and another
electrodynamometer type, connected in series. The PMMC meter contains 100 turns in the
coil, the flux density in the air gap is 0.2 Wb / m 2 , and the area of the coil is 80 mm 2 . The
electrodynamometer ammeter has a change in mutual inductance with respect to deflection of
0.5 mH/deg. The spring constants of both the meters are equal. The value of current, at which
the deflections of the two meters are same, is ________________
Answer: 3.2
Exp:
→ Given pmmc and electro dynamometer type meters are connectsed in series.
→ Both meters are carrying same current. And both are having same spring constants.
→ Both are reflecting same readings. i.e. we should equate the reflecting torques.
For pmmc, T def = BAN.I.
Electrodynometer, T def = I 2 .
BAN.I = I2 .
dM
dθ
dm
dθ
( 0.2 ) × (80 × 10−6 ) × 100 × I = I2 × 0.5 × 10−3
⇒ I = 3.2
49.
Given that the op–amps in the figure are ideal, the output voltage V0 is
(A) ( V1 − V2 )
V2
+
R
−
R
(B) 2 ( V1 − V2 )
R
−
V0
2R
(C)
+
( V1 − V2 ) / 2
R
R
R
−
(D) ( V1 + V2 )
V1
+
Answer: (B)
Exp:
V2
If R
+
−
I1
R
V2
V02
R V01 / 2
−
R
V0
V1
−
V1
+
R V0 / 2
1
+
V01
R
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V2 − V1 V2 − V02
V1 − V2 V1 − V01
+
=0
+
=0
2R
R
2R
R
3V − V1
3V − V2
⇒ V02 = 2
........ (1) ⇒ V01 = 1
........ ( 2 )
2
2
∵ I1 = If
V01   V01


 V02 − 2   2 − V0 

=

R
R
 3V − V2   3V2 − V1 
⇒ V0 = V01 − V02 =  1
 −

2
2

 

from (1) & ( 2 )
V0 = 2 ( V1 − V2 )
50.
Which of the following logic circuits is a realization of the function F whose Karnaugh map is
shown in figure.
AB
0
00
01
1
1
11
10
C
1
1
1
A
A
(A)
(B)
B
B
C
C
A
A
C
(C)
(D)
B
B
Answer:
Exp:
C
(C)
AC
AB
C
0
1
00
01
1
1
1
11
10
1
BC
F = A C + BC
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A
A
A. (B.C )
(A)
B
C
BC
A
A
A + BC
(B)
B
C
BC
A
A
(C)
A C
C
C
A C + BC
BC
B
A
A
A C
(D)
A C + BC
C
BC
B
Among all the options, option (C) is matching with function F = A C + BC
51.
In the figure shown, assume the op–amp to be ideal. Which of the alternatives gives the correct
V ( ω)
Bode plots for the transfer function 0
?
V1ω
Vi
+ Vcc
1kΩ
+
1µF
V0
−
− Vcc
Rf
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(A)
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φ
 V (ω) 
20 log  0

 Vi ( ω ) 


0
3
10 10 2 10
0
1
−10
10 10 2 10 3
ω
ω
1
−π / 4
−20
−π / 2
−30
φ
(B)
 V (ω) 
20 log  0

 Vi ( ω ) 


0
π/2
π/4
0
10 3
1
−10
10 10 2
ω
10 3
1
−π / 4
−20
10 10 2
ω
−π / 2
−30
φ
(C)
 V (ω) 
20 log  0

 Vi ( ω ) 


0
10
1
−10
10
π/2
π/4
0
2
10 3 ω
−20
3
10 10 2 10
1
ω
1 10 10 2 10 3
ω
−π / 4
−π / 2
−30
φ
 V (ω) 
20 log  0

 Vi ( ω ) 


0
π/2
π/4
0
1
(D)
−10
−20
10 10 2 10 3 ω
−π / 4
−π / 2
−30
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Answer: (A)
Exp:
+
Vin
−
Rf
−
1K
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V0
+
1 µF
Act as Buffer
The given circuit is LPF (Low Pass Filter)
V0
1
1
1
1
1000
=
=
=
=
=
Vin 1 + SCR 1 + jω (103 )(10−6 ) 1 + jω (10 −3 ) 1000 + jω 1 + S
1000
V0
( dB)
Vin
Actual Curve
1000
ω
o
−3
Slope = −20 dB decade
Low frequency asymptote curve
Corner frequency is at ω =1000
Low frequency gain=1{ ω =0}
52.
An output device is interfaced with 8–bit microprocessor 8085A. The interfacing circuit is
shown in figure
AB
8
BDB
A15
A14
A13
A12
I2
I1
I0
E1
A11
IO / M
3L × 8L Decoder
8
0
1
2
8
3
4
E2
5
6
E3
7
Output Device
WR
BCB
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The interfacing circuit makes use of 3 Line to 8 Line decoder having 3 enable lines E1 , E 2 , E 3 .
The address of the device is
(A) 50H
Answer:
Exp:
(B) 500H
(C) A0H
(D) A000H s
(B)
To enable 3L × 8L decoder, three enable lines E1 (which is connected as an output of ANDI
gate) should be HIGH and E 2 and E 3 should be active low, it means 0 should be active low
m
which is indicating that it is memory mapped I/O interfacing. So, address of the device will be
in 16-bits. To select output port through decoder 2nd line the status of
A15 ( I 2 ) A14 ( I1 ) A13 ( I0 ) = 010 and to enable decoder through E1 enable line
A12 = 1 and A11 = 0 and by default as a starting address other address lines
should be zero .So, overall port address is
A15 A14 A13 A12 A11 A10 A 9 A8 A 7 A 6 A 5 A 4 A 3 A 2 A1 A 0
0
53.
1
0
1
0
0
0
0
0
0
0
0
0
0
0
( A10 ......A 0 )
0 = 5000H
The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10Ω and
an inductance 0.05H connected in series. Assuming ideal thyristor and ideal diode, the
thyristor firing angle (in degree) needed to obtain an average load voltage of 70V is
_______________
+
325sin ( 314t ) V
Answer:
Exp:
Load
−
α = 69.3°
Vo = 70v;
Load, R = 10Ω;
L = 0.05H
Firing angled = ?
The o/p volt of given converter is
Vm
( 1+ cos α )
2π
525
70 =
( 1+ cos α )
2π
α = 69.3°
T
+
+
D
V0 =
−
V0
load
−
Vs
π
0
2π
3π
V0
α
π
2π+2
3π
u(t)
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Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit in figure (i) is
switched such that the voltage v D across the diode has the wave shape as shown in figure (ii).
The capacitance C is large so that the voltage across it is constant. If switch S and the diode
are ideal,
the peak to peak ripple (in A) in the inductor current is _____________
S
1mH
+
100V
C
VD
Load
−
Figure ( i )
VD
100V
0
0.05
0.1
0.15
0.2
t ( ms )
Figure ( ii )
Answer:
Exp:
2.5A
V0
Peak to peak ripple in inductor current in
V − Vo
Tm
∆I = s
L
0.05 

100 −  100 ×

0.1 

=
× 0.05 ×10−3
1×10−3
= 50 × 0.05 ⇒ ∆I = 2.5 A
100V
∆I
I1
Ton
TON
55.
0.1
0.05
I2
T
TOFF
The figure shows one period of the output voltage of an inverter. α should be chosen such
that 60o < α < 90o . If rms value of the fundamental component is 50V, then α in degree is
________________
100V
0
100V
α
180 − α
100V
180
180 + α
360 − α
360
ωt
( deg ree )
−100V
−100V
−100V
Answer: 76 to 77
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π

4  α
Vs  ∫ sin θdθ − ∫ 2 sin θ d θ 
0
α
π 

4Vs
[1 − cos α − cos α + 0]
π
4V
= s [1 − 2cos α ]
π
=
RMS Value of V01 =
⇒ 50 =
400
π 2
α = 77.15o
4Vs
π 2
(1 − 2cos α )
(1 − 2cos α)
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GATE- EE 2014
Paper-02
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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)
Who ___________ was coming to see us this evening?
(A) you said
(B) did you say
(C) did you say that
Answer: (B)
2.
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
= 2.5 m=2.5f
=3
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
…(1)
L1 + L2 + 2M = 380 H
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
Given,
Exp:
1000V
100V
3
50 ×10
= 500
I2 =
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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Exp:
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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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20.
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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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23.
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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
⇒ 1− D =
400 =
1− D
400
3
= 0.375
D=
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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25.
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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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27.
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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
(B) 2.0 and 4.20
(C) 2.5 and 3.50
(A) 2.0 and 1.98
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
(
π
2πρ
2π (1)
Hy =
−1
I
2
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
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
35.
[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
of 125 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.
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=
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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48.
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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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51.
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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- EE 2014
Paper-03
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Q.No. 1 – 5 Carry One Mark Each
1.
While trying to collect an envelope from under the table, Mr. X fell down and
I
II
III
was losing consciousness.
IV
Which one of the above underlined parts of the sentence is NOT appropriate?
(A) I
(B) II
(C) III
(D) IV
Answer: (D)
2.
If she _______________ how to calibrate the instrument, she _______________ done the
experiment.
(A) knows, will have
(B) knew, had
(C) had known, could have
(D) should have known, would have
Answer: (C)
3.
Choose the word that is opposite in meaning to the word “coherent”.
(A) sticky
(B) well-connected
(C) rambling
(D) friendly
Answer: (C)
4.
Which number does not belong in the series below?
2, 5, 10, 17, 26, 37, 50, 64
(A) 17
(B) 37
(C) 64
(D) 26
Answer: (C)
5.
The table below has question-wise data on the performance of students in an examination.
The marks for each question are also listed. There is no negative or partial marking in the
examination.
No
Marks
Answered
Correctly
Answered
Wrongly
Not
Attempted
1
2
21
17
6
2
3
15
27
2
3
2
23
18
3
What is the average of the marks obtained by the class in the examination?
(A) 1.34
(B) 1.74
(C) 3.02
(D) 3.91
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Answer: (C)
Exp:
Total question
44×2=88
44×3=132
144 = 88
132 308
Total marks obtained= (21×2) + (15×3) + (23×2) =133
Total Number of students=44
133
= 3.02
Average =
44
Q.No. 6 – 10 Carry One Mark Each
6.
A dance programme is scheduled for 10.00 a.m. Some students are participating in the
programme and they need to come an hour earlier than the start of the event. These students
should be accompanied by a parent. Other students and parents should come in time for the
programme. The instruction you think that is appropriate for this is
(A) Students should come at 9.00 a.m. and parents should come at 10.00 a.m.
(B) Participating students should come at 9.00 a.m. accompanied by a parent, and other
parents and students should come by 10.00 a.m.
(C) Students who are not participating should come by 10.00 a.m. and they should not bring
their parents. Participating students should come at 9.00 a.m.
(D) Participating students should come before 9.00 a.m. Parents who accompany them should
come at 9.00 a.m. All others should come at 10.00 a.m.
Answer: (B)
7.
By the beginning of the 20th century, several hypotheses were being proposed, suggesting a
paradigm shift in our understanding of the universe. However, the clinching evidence was
provided by experimental measurements of the position of a star which was directly behind
our sun.
Which of the following inference(s) may be drawn from the above passage?
(i) Our understanding of the universe changes based on the positions of stars
(ii) Paradigm shifts usually occur at the beginning of centuries
(iii) Stars are important objects in the universe
(iv) Experimental evidence was important in confirming this paradigm shift
(A) (i), (ii) and (iv)
Answer: (D)
8.
(B) (iii) only
(C) (i) and (iv)
(D) (iv) only
The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For
international comparison, the GDP is compared in US Dollars (USD) after conversion based
on the market exchange rate. During the period 2012-2013 the exchange rate for the USD
increased from Rs. 50/ USD to Rs. 60/ USD. India’s GDP in USD during the period 20122013
(A) increased by 5 %
(B) decreased by 13%
(C) decreased by 20%
(D) decreased by 11%
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Answer: (D)
Exp: Per 100 Rs final value 107 Rs
100
107
⇒ Per
Dollars final value
50
60
for 100 dollars____?
100 × 50 107
=
×
= 89.16
100
60
Discrased by 11%.
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 in 2011 and 2012 is equal, what is the ratio of male
students in 2012 to male students in 2011?
Ra
tio
of
m
al
e
to
fe
m
al
3.5
3
2.5
2
1.5
1
0.5
0
2008
2009
2010
2011
(A) 1:1
(B) 2:1
(C) 1.5:1
Answer: (C)
Exp: Take number of female students in 2011=100
∴ Number of male in 2011=100
No. of female in 2012=100
No. of male in 2012=150
150
ratio =
100
2012
(D) 2.5:1
Consider the equation: (7526)8 - (Y)8 = (4364)8 , where (X) N stands for X to the base N. Find
Y.
(A) 1634
(B) 1737
(C) 3142
(D) 3162
Answer: (C)
10.
Exp:
( 7526 )8 − ( y )8 = ( 4364 )8
⇒ y8 = ( 7526 )8 − ( 4364 )8
7
4
3
4
( 8+ 2 =10 )
5
3
2
6
6
4
1
4
2
When we have base 8, we borrow 8 instead of 10 as done in normal subtraction
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Q.No. 1 – 25 Carry One Mark Each
1.
Two matrices A and B are given below:
 p 2 + q 2 pr + qs
B= 
2
2
 pr + qs r + s 
p q 
A=
;
r s
If the rank of matrix A is N, then the rank of matrix B is
(A) N /2
(B) N-1
(C) N
(D) 2 N
Answer: (C)
Exp:
Rank of a matrix is unaltered by the elementary transformations i.e., Row/column operations
(Here B is obtained from A by applying row/column operations on A)
Since rank of A is N
∴ rank of B is also N
2.
A particle, starting from origin at t = 0 s, is traveling along x-axis with velocity
v=
π
π 
cos  t  m / s
2 
2
At t = 3 s, the difference between the distance covered by the particle and the magnitude of
displacement from the origin is _________.
Answer:
Exp:
2
At t = 3s , the distance covered by the particle is 1 + 1 + 1 = 3m and displacement from the
origin is − 1
∴ difference between the distance covered by the particle and the magnitude of displacement
from the origin is 3 − −1 = 2
3.
Let, ∇ ( f v ) = x 2 y + y 2 z + z 2 x; where f and v are scalar and vector fields respectively. If
v = yi + zj + xk, then v = ∇ f is
(A) x 2 y + y 2 z + z 2 x
(B) 2xy + 2yz + 2zx (C) x + y + z
(D) 0
Answer: (A)
Exp:
∇.( f v ) = f ( ∇ .v ) + ∇f.v
Now ∇.v = 0 + 0 + 0 = 0
.....(1)
∴ (1) becomes x 2 y + y 2 z + z 2 x = f ( 0 ) + v.∇f
∴ v.∇f = x 2 y + y 2 z + z 2 x
4.
Lifetime of an electric bulb is a random variable with density f(x) = kx2, where x is measured
in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then
the value of k is ________.
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Answer: 0.43
f ( x ) = kx 2 , 1 < x < 2 and f ( x ) = 0, otherwise sin ce f ( x ) is a p.d.f
Exp:
2
 x3 
∴ ∫ f ( x ) dx = 1 ⇒ k   = 1
1
 3 1
⇒ k = 3 = 0.428 0.43
7
2
5.
A function f(t) is shown in the figure.
f (t)
1/ 2
T/ 2
−T / 2
t
0
−1/ 2
The Fourier transform F ( ω ) of f(t) is
(A) real and even function of ω
(B) real and odd function of ω
(C) imaginary and odd function of ω
(D) imaginary and even function of ω
Answer:
Exp;
(C)
Since f(t) is odd and real
f ( t ) = −f ( − t )
∫ F ( ω) is imaginary and odd [symmetry property of fourier Transform]
6.
The line A to neutral voltage is 10 ∠15O V for a balanced three phase star-connected load
with phase sequence ABC. The voltage of line B with respect to line C is given by
( A)
10 3 ∠1050 V
( B)
10 ∠1050 V
( C)
10 3∠ − 75O V
( D)
− 10 3∠ − 90O V
Answer: (C)
VL = 3 Vph
Exp:
If
VA
− VC
= 3 × 10 = 10 3
VA = 10 0
60
30
then VBC = 10 3 −90
120
VBC
given VA = 10 15
o
∴
VBC = 10 3 −90 + 15 = 10 3 −75
VC
VB
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A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux
coming out of the concentric spherical surface of radius R ( > r) is
( A ) 4πε 0 r
( B) 4πε 0 r 2
( C) 4πε 0 R
( D ) 4πε 0 R 2
Answer: (A)
8.
The driving point impedance Z(s) for the circuit shown below is
1H
Z ( s)
( A)
s4 + 3s2 + 1
s3 + 2s
( B)
1H
1F
s 4 + 2s 2 + 4
s2 + 2
1F
(C)
s2 + 1
s4 + s 2 + 1
( D)
s3 + 1
s4 + s2 + 1
Answer: (A)
Exp: S-Domain representation of the given circuit.
1  1
s+
s  s 
Z (s) = s +
1  1
+ s + 
s  s
Z (s) =
9.
S
S
1
s
1
s
s 4 + 3s 2 + 1
s3 + 2s
Z(s)
A signal is represented by
1
x ( t) = 
1
t <1
t <1
The Fourier transform of the convolved signal y(t)=x(2t)*x(t/2) is
( A)
4
 ω
sin   sin ( 2ω )
2
 2
ω
4
sin ( 2ω )
ω2
Answer: (A)
Exp:
( C)
( B)
4
 ω
sin  
2
 2
ω
( D)
4
sin 2 ω
2
ω
x(t)
t
t
x ( t ) = rect   = rect  
τ
2
t F
 ωτ 
∵ rect   
→ τ sa 

τ
 2 
2sin ω
t F
x ( f ) = rect   
→ 2 sa ( ω) =
ω
2
1
−1
t
+1
τ=2
By using the scaling property of Fourier transforms,
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 ω
sin  
1  ω
2
F{x ( 2t )} = F  rect ( t )  = 2 × sa   =
2 2
 ω
 
2
  t 

4sin 2ω
 t 
Similarly, F  x    = F  rect    = 2 × 2sa ( 2ω) =
2ω
 4 
  2 

t
The convolved signal y ( f ) = x ( 2t ) × x  
2
  ω
 sin  2    4sin 2 ω  4
  t 
 ω
∴ F  y ( t )  = F x ( 2t ) F  x    =     
= 2 sin   sin ( 2ω)

  ω    2ω  ω
2
  2 
  2  
For the signal f ( t ) = 3 sin 8πt + 6sin12πt + sin14 πt, the minimum sampling frequency (in
Hz) satisfying the Nyquist criterion is _________.
Answer: 14 sample / sec
Exp: Maximum frequency content in 7Hz
⇒ Nyquist rate = 2 × 7 = 14 sample / sec
10.
11.
In a synchronous machine, hunting is predominantly damped by
(A) mechanical losses in the rotor
(B) iron losses in the rotor
(C) copper losses in the stator
(D) copper losses in the rotor
Answer: (D)
Exp: In a synchronous machine the hunting will be damped by the copper losses in the rotor
12.
A single phase induction motor is provided with capacitor and centrifugal switch in series
with auxiliary winding. The switch is expected to operate at a speed of 0.7 Ns, but due to
malfunctioning the switch fails to operate. The torque-speed characteristic of the motor is
represented by
Torque
Torque
( A)
( B)
NS
0.7Ns
Speed
Torque
NS
0.7Ns
Speed
Torque
( C)
(D)
NS
NS
0.7Ns
0.7Ns
Speed
Speed
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Answer: (C)
Exp:
In 1 − φ induction motor the centrifugal switch is expected to operate at 0.7 N S but it fails to
operate.
Then there is no increase in the torque after 0.7 N S Hence (C) is correct
13.
The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The armature
resistance drop and the brush drop are neglected. The field current is kept constant at rated
value. The torque of the motor in Nm for an armature current of 8 A is ____________.
Answer: 12.5Nm
Exp:
Given, V = 230 V, N = 1400 rpm
Ra = 0
∴ Torque, T =
14.
Ia = 8A
V.Ia
230 × 8
=
= 12.5 Nm
2π × 1400
W
60
In a long transmission line with r,l,g and c are the resistance, inductance, shunt conductance
and capacitance per unit length, respectively, the condition for distortionless transmission is
( A)
rc = lg
( B)
( C)
rc = l/ c
( D)
rg = lc
g = c/l
Answer: (A)
R G
= ⇒ RC = GL
L C
Exp
For distortionless transmission line,
15.
For a fully transposed transmission line
(A) positive, negative and zero sequence impedances are equal
(B) positive and negative sequence impedances are equal
(C) zero and positive sequence impedances are equal
(D) negative and zero sequence impedances are equal
Answer: (B)
Exp:
Ia
b
Ib
2s
2m
Where Zn → neutral impedance
2m
2s
ZS → Self impedance
Z0 → zero sequence impedance
Ic
c
Vb
Z1 → + ve sequence impedance
Va '
2m
Va
Zm → mutual impedance
Let
a
2s
VC
Vb '
VC '
Z2→ − ve sequence impedance
 Z0 00   Zs + 2Zm + 3Zn

 
Z012 =  0Z1 0  = 
0
 00 Z2  
0
∴Ζ1 = Z2 = Zs − Z m
0
Zs − Z m
0


0

Zs − Z m 
0
Zn
In = Ia + Ib + Ic
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16.
A 183-bus power system has 150 PQ buses and 32 PV buses. In the general case, to obtain
the load flow solution using Newton-Raphson method in polar coordinates, the minimum
number of simultaneous equations to be solved is ___________.s
Answer: 332
Exp: No of load | PQ buses = 150
No of generator |PV buses =32
Minimum no of simultaneous equations = 2 × 150 + 32 = 332
17.
The signal flow graph of a system is shown below. U(s) is the input and C(s) is the output
h1
h0
U ( s)
1
s
1
1
s
1
1
C ( s)
− a1
−a 0
C ( s)
Assuming, h1 = b1 and h0 = b0 – b1 a1, the input-output transfer function, G(s) =
of the
U ( s)
system is given by
( A)
G ( s) =
b0s + b1
s 2 + a 0s + a 1
( B)
G ( s) =
a1s + a 0
a 2 + b1s + b0
( C)
G ( s) =
b1s + b0
a + a1s + a 0
( D)
G ( s) =
a 0s + a 1
a + b0s + b1
2
2
Answer: (C)
Exp:
From the signal flow graph, G ( s ) =
C (s )
U (s)
By mason’s gain relation,
Transfer function =
P1 =
P1∆1 + P2 ∆ 2 + ...
∆
h
h1
; P2 = o
S
s2
 a 
∆1 = 1 + 1  ; ∆ 2 = 1 ;
s

∆ =1+
a1 a 0
+
s s2
h1  a 1  b 0
1+  + 2
b s + b0
s 
s s
Transfer function = 
= 2 1
a1 a 0
s + a1s + a 0
1+ + 2
s s
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A single-input single-output feedback system has forward transfer function G(s) and feedback
transfer function H(s) It is given that | G(s).H(s)|<1. Which of the following is true about the
stability of the system?
(A) The system is always stable
(B) The system is stable if all zeros of G(s).H(s) are in left half of the s-plane
(C) The system is stable if all poles of G(s).H(s) are in left half of the s-plane
(D) It is not possible to say whether or not the system is stable from the information given
Answer: (A)
19.
An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150 V and 75
V, and two current settings 5 A and 10 A. The full scale reading is 150. If the wattmeter is
used with 150 V voltage setting and 10 A current setting, the multiplying factor of the
wattmeter is _________.
Answer: 2
Exp: In LPF wattmeter, Td on the moving system is small owing to low power factor even when
the current and potential coils are fully excited. Also the errors introduced due to inductance
of pressure coil tend to be large at low power factors. So for calculating multiplying factor for
a low p.f. wattmeter, p.f. mentioned on the wattmeter should be taken into account.
Therefore,
Multiplying Factor = (Current range used*Voltage range used*p.f)/Power at FSD
Given, Power at Full scale reading = 150
Current Range used
= 10 A
Voltage Range used
= 150 V
Power Factor
= 0.2
10 × 150 × 0.2
=2
Therefore, m =
150
20.
The two signals S1 and S2, shown in figure, are applied to Y and X deflection plates of an
oscilloscope.
1
S1
1
v
V
2T
T
S2
t
Y
The waveform displayed on the screen is
( A)
2T
T
t
Y
1
( B)
1
X
X
−1
−1
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Y
1
( C)
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Y
1
( D)
X
X
−1
−1
Answer: (A)
Exp:
+1
S2
S1
+1
•
t
T
t
0
−1
−1
X Y mode
Vector sum (in X-Y mode)
y
1
−1
•
•
•
•
•
x
1
−1
φ = 100 −1
Points
Y( s1 )
X (s2 )
A
0
1
B
1
1
C
0
0
0
0
−1
2
225o
−1
D
x 2 + y2
0
1
2
45ο
E
21.
A state diagram of a logic gate which exhibits a delay in the output is shown in the figure,
where X is the don’t care condition, and Q is the output representing the state.
0x /1. 10 /1
11/ 0
Q=0
Q =1
0x /1.10 /1
11/ 0
The logic gate represented by the state diagram is
(A) XOR
(B) OR
(C) AND
(D) NAND
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Answer: (D)
Exp: True Table
A
B
Y
Q
1
1
0
0
0
0
0
1
1
1
1
1
1
1
0
1
1
0
1
0
If you will observe this true table corresponding to state diagram, then if any input is 0 output
is 1 and if all the inputs are one output is zero it means it corresponds to NAND gate.
22.
An operational-amplifier circuit is shown in the figure.
−
Vi
R
+ Vsat
R
+ Vsat
−
+
+
− Vsat
R1
Vo
− Vsat
R2
The output of the circuit for a given input vi is
( A)
R 
−  2  vt
 R1 
( B)
 R 
− 1 + 2  v t
R1 

( C)
 R2 
 1 + R  v t
1
( D)
+ Vsat or − Vsat
Answer: (D)
Exp:
Output of first Op-Amp is + Vsat
2nd Op-Amp circuit is non-inverting amplifier
 R
∴ V0 =  1 +  Vsat > Vsat
 R
∴ V0 = ± Vsat
23.
In 8085A microprocessor, the operation performed by the instruction LHLD 2100H is
( A ) ( H ) ← 2IH , ( L ) ← 00H
( B) ( H ) ← M ( 2I00H ) , ( L) ← M ( 2101H )
( C) ( H ) ← M ( 2I01H ) , ( L) ← M ( 2100H )
( D ) ( H ) ← 00H , ( L ) ← 21H
Answer: (C)
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Exp:
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Instruction given is:
LHLD
2100H
The operation performed by this instruction is load HL register pair from the specified
address in the instruction, directly. HL register pair is required 2-Byte data, but in 8085 at one
address it contains only one-byte data, so this instruction will access two memory locations.
So, first byte address (i.e., 2100H) is mentioned in instruction itself and by default second
byte data is accessed from the next location (i.e., 2101H). Lower address data will be copied
to lower byte ( i.e., ( L ) ← M ( 2100H ) ) and higher address data will be copied to higher byte
(i.e, ( H ) ← M ( 2101H ) )
24.
A non-ideal voltage source VS has an internal impedance of ZS If a purely resistive load is to
be chosen that maximizes the power transferred to the load, its value must be
(A) 0
(B) real part of Zs
(C) magnitude of Zs
(D) complex conjugate of Zs
Answer: (C)
•
Exp: For minimum power transferred to the load,
jx
Its value must be R L = Zs
s
Rs
Vs
∼
RL
R L = R S2 + XS2
25.
•
The torque-speed characteristics of motor ( TM ) and load ( TL ) for two cases are shown in the
figures (a) and (b). The load torque is equal to motor torque at points P, Q, R and S
TM
Speed
Speed
S
Torque
P
R
Q
Torque
TL
(a)
The stable operating points are
(A) P and R
(B) P and S
Answer: (B)
Exp:
TL
( b)
(C) Q and R
(D) Q and S
From the given torque speed characteristics of motor ( Tm ) and load ( TL ) at the points ‘P’ and
‘S’ the motor stable. Since at low value of slip the motor is stable.
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Q.No. 26 – 55 Carry Two Marks Each
26.
Integration of the complex function f ( z ) =
z2
, , in the counterclockwise direction, around
z2 − 1
z − 1 = 1, is
(A) −πi
Answer: (C)
Exp:
(B) 0
(C) πi
(D) 2 πi
Z = −1,1 are the simple poles of f ( z ) and z = 1 lies inside C : z − 1 = 1
∴ ∫ f ( z ) dz = 2π i ×  Res f ( z ) 
C
 z =1

= 2π i ×  Lt ( z − 1) .f ( z ) 
 z →1

2

z 
= 2π i ×  Lt
 = πi
z →1 z + 1


27.
The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02
respectively. The varnish insulation is applied on both the sides of the laminations. The mean
thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the
transformer core is made using 100 such varnish coated laminations, the mean thickness and
variance of the core respectively are
(A) 30 mm and 0.22
(B) 30 mm and 2.44
(C) 40 mm and 2.44
(D) 40 mm and 0.24
Answer: (D) [Key form IIT website]
Exp: mean thickness of silicon steel laminations= 0.2 mm
variance of silicon steel lamination = 0.02
Vanish insulation applied both sides of laminations
Mean thickness of one side insulation=0.1 mm
variance of one side insulation=0.01
Transformer core made with 100 vanish coated laminations
Mean thickness of two side insulation applied to core
=2×0.1=0.2mm
∴ mean thickness of one lamination
= mean thickness of silicon steel + mean thickness of two side insulation
= 0.2+0.2=0.4 mm
100 laminations are using so
Mean thickness of core=0.4×100=40 mm
To find thickness of each lamination
From variance
( d1 − 0.2 )
2
+ ( d 2 − 0.2 ) + ...(d100 − 0.2) 2
2
100
= 0.02
2

(x − x)
∑
∵ Varience =

n


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All have same thickness
100 × (d − 0.2) 2
= 0.02
100
(d-0.2)=0.1414
Each lamination d=0.3414mm
Thickness
Similarly
∴
(x1 − 0.1)2 + ... + (x100 0.1)2
= 0.01
100
Each side lamination is equal (Assume)
100 × (x − 0.1) 2
= 0.01
100
x − 0.1 = 0.1
Each side x=0.2 mm
Insulation thickness
Two side insulation thickness =0.4 mm
Each lamination (insulation  = 0.3414 + 0.4

+ si steel) thickness
 = 0.7414 mm
Variance of overall core
100 × (0.7414 − 0.4) 2
= 0.1166
100
∴ mean thickness & variance of the core=40mm and 0.1166
[No option is matching]
=
The function f ( x ) = e x − 1 is to be solved using Newton-Raphson method. If the initial value
of x0 is taken as 1.0, then the absolute error observed at 2nd iteration is __________.
Answer: 0.06
28.
Exp:
Clearly, x = 0 is a root of the equation f ( x ) = ex − 1 = 0
f ' ( x ) = ex and x 0 = 1.0
Using Newton raphson method, x1 = x 0 −
f ( x1 )
1
= −
and x 2 = x1 − 1
f ( x1 ) e
f ( x0 )
f ( x0 )
'
=1−
( e − 1) = 1
e
e
(e − 1) = 1 + 1 − 1 = 0.37 + 0.69 − 1 = 0.06
1
e
1
e e 1e
e e
∴ absolute error at 2nd iteration is 0 − 0.06 = 0.06
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29.
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The Norton’s equivalent source in amperes as seen into the terminals X and Y is _______.
2.5Ω
X
2.5Ω
5Ω
5Ω
5Ω
5V
Y
2.5 V
Answer: 2
[Key form IIT website]
X
2.5Ω
Exp:
5Ω
5Ω
5Ω
5V
Y
2.5 V
X
−+
X
2.5 Ω
5Ω
2.5 Ω
2.5 V
ISC ( I N ) =
+
−
5Ω
5V
5Ω
+
−
5
5
Isc
Y
Y
ISC = 1A
[Answer is not matching]
30.
The power delivered by the current source, in the figure, is ________.
1V
+−
1Ω
1V +
−
1Ω
↑ 2A
1Ω
+−
Answer: 3 Watts
Exp: KCL at node Vx:
1 − Vx
V
+2= x
1
1
1V
Vx
1Ω
+
−
1V
•
1Ω
↑
1Ω
2A
Vx = 1.5V
Power delivered by current source is = 2× 1.5 = 3 watts
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31.
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A perfectly conducting metal plate is placed in x-y plane in a right handed coordinate system.
A charge of +32πε 0 2 columbs is placed at coordinate (0, 0, 2). ∈0 is the permittivity of free
space. Assume ˆl, ˆj, kˆ to be unit vectors along x, y and z axes respectively. At the coordinate .
2, 2, 0 , the electric field vector E (Newtons/Columb) will be
(
)
( A) 2
z
2 kˆ
( B) − 2 kˆ
32nε 0 2
columbs
y
(
( 0,0,0)
2, 2, 0
Perfectly conducting
metal plate
)
( C) 2 kˆ
( D) − 2
Answer:
Exp:
2 kˆ
Z
(B)
(0,0,2 )
Q
(
2, 2,0
X
−X
(0,0, −2 )
)
−Q
−Z
E = E1 + E 2 =
1  Q1R 1 Q 2 R 2 
+


4πε 0  R 13
R 23 
R1 =
Q 2 = −Q1
(
)
2, 2,0 − ( 0,0,2 )
R 1 = 2 a x + 2 a y − 2a z
R2 =
(
)
2, 2,0 − ( 0,0, −2 )
= 2 a x + 2 a y + 2a z
E=
=
1  Q1
4πε 0 16 2
Q
16 2 × 4πε 0
(
)
2 a x + 2 a y − 2a z −
[ −4a z ] =
32 2πε 0
16 2πε 0
Q1
16
(
2
)

2 a x + 2 a y + 2a z 

( −a z )
E = −2a z
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32.
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A series RLC circuit is observed at two frequencies. At ω1 = 1k rad / s, we note that source
voltage V1 = 100∠00 V results in current I1 = 0.03∠31O A. At ω 2 = 2 krad/s, the source
voltage V2 = 100∠0o V results in a current I 2 = 2∠0o V A. The closest values for R,L,C out
of the following options are
( A ) R = 50Ω; L = 25mH;C = 10µF;
( C ) R = 50Ω;L = 50 mH;C = 5µF;a
( B) R = 50Ω;L = 10 mH;C = 25µF;
( D ) R = 50Ω; L = 5 mH;C = 50µF;
Answer: B
Exp:
Given V1 = 100 0o V; I1 = 0.03 31o at ω1 = 1000 r sec
V2 = 100 0o V; I 2 = 2 0o at ω2 = 2000 r sec
R=
i.e.
Z=
ϑ2 100
=
⇒ 50Ω
I2
2
v1
100 0o
⇒
⇒ R + j( X L − X C )
I1
0.03 31o
 X − XC 
φ = 31o ⇒ tan −1  L

R


 X − XC 
⇒ tan 31o =  L

R



1 
 ω1L −

ω1c 
tan 31o = 
R


1 
1 
⇒  ω1L −
 = 0.600 × 50 ⇒  ω1L −
 = 30.04 q
ω1C 
ω1C 


ω2 L −
1
=0
ω2 C
….(1)
….(2)
ω1 = 1000 r / sec ; ω2 = 2000 r / sec
from 1 and 2,C = 25µF
L = 10mH
33.
A continuous-time LTI system with system function H ( ω ) has the following pole-zero plot.
For this system, which of the alternatives is TRUE?
ω
( A)
( B)
H ( ω ) has multiple max ima,at ω1 and ω 2
( C)
H (ω ) < H (ω ) ; (ω ) > 0
( D)
H ( ω) = cons tan t; −∞ < ω < ∞
Answer: (D)
H ( 0) > H ( ω ) ; ( ω ) > 0
X
O
O
O
ω2
ω1
( 0,0 )
X
X
σ
X
O
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Exp:
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The transfer function can be written as
H (s ) = K
jω
( s − z1 )( s − z1 *)( s − z 2 )( s − z 2 *)
p2
( s − p1 )( s − p1 *)( s − p2 )( s − p 2 *)
H ( jω )
=K
z1
z2
ω2 + z1
s + p1
2
2
2
ω2 + z1
2
ω + p1
2
2
ω2 + z 2
2
ω + p2
2
2
ω2 + z 2
2
ω + p2
2
2
p1
σ
p1 *
z2 *
from figure z1 = p 2
z1 *
p2 *
z 2 = p1
⇒ ( H ( jω) ) = K [ cons tan t ]
A sinusoid x(t) of unknown frequency is sampled by an impulse train of period 20 ms. The
resulting sample train is next applied to an ideal lowpass filter with a cutoff at 25 Hz. The
filter output is seen to be a sinusoid of frequency 20 Hz. This means that x(t) has a frequency
of
(A) 10 Hz
(B) 60 Hz
(C) 30 Hz
(D) 90 Hz
Answer: (C)
Exp: Sampling rate 50Hz
34.
Let x(t) has a frequency of fx Hz,
−fx
fx
f
After sampling with 50Hz
Spectrurm will be
LPF of cut off = 25Hz
output = 20Hz
⇒ 50 − f x = 20
− (50 + fx )
− fx2 5
2J
−50 + fx
fx
50 − fx
50 + fx
⇒ f x = 30Hz
35.
A differentiable non constant even function x(t) has a derivative y(t), and their respective
Fourier Transforms are X ( ω ) and Y ( ω ) . Which of the following statements is TRUE?
(A) X ( ω ) and Y ( ω ) are both real.
(B) X ( ω) is real and Y ( ω) is imaginary.
(C) X ( ω ) and Y ( ω ) are both imaginary.
(D) X ( ω ) is imaginary and Y ( ω ) is real.
Answer: (B)
Exp:
y(t) =
d
x (t)
dt
↓
Y ( ω) = jω× ( ω)
⇒ if X ( ω) is real, Y ( ω) is imaginary
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36.
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An open circuit test is performed on 50 Hz transformer, using variable frequency source and
keeping V/f ratio constant, to separate its eddy current and hysteresis losses. The variation of
core loss/frequency as function of frequency is shown in the figure
15
Pc / f
10
( W / Hz )
5
25
f ( Hz )
50
The hysteresis and eddy current losses of the transformer at 25 Hz respectively are
(A) 250 W and 2.5 W
(B) 250 W and 62.5W
(C) 312.5 W and 62.5 W
(D) 312.5 W and 250 W
Answer: (B)
15
Exp: Given 50Hz transformer
Here
PC
−v
ratio maintains constant
f
f
θ
∴ Wn = Af ; We = B − f 2
10
5
= 0.1
50
∴ Wn at 25Hz = 10 × 25 = 250 W
where, A = 10 & B = tan θ =
5
25
We at 25Hz = 0.1 × ( 25 ) = 62.5 W
2
37.
f
A non-salient pole synchronous generator having synchronous reactance of 0.8 pu is
supplying 1 pu power to a unity power factor load at a terminal voltage of 1.1 pu. Neglecting
the armature resistance, the angle of the voltage behind the synchronous reactance with
respect to the angle of the terminal voltage in degrees is ________.
Answer:
Exp:
50
33.61o
Given, P = 1 P.u;
Pf = 1.
∴ Ia =
Vt = 1.1 P.u;
Xs = 0.8 P.u
P
1
=
= 0.91
V cos φ 1.1 × 1
∴ E=
E=
( V cos φ + Ia R a )
(1.1× 1 + 0 )
2
2
+ ( V sin φ + Ia X s )
+ ( 0 + 0.91 × 0.8 ) = 1.314 V
∴ From power equation, P =
⇒1=
2
2
EV
sin δ
Xs
1.314 × 1.1
sin δ ⇒ δ = 33.61o
0.8
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38.
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A separately excited 300 V DC shunt motor under no load runs at 900 rpm drawing an
armature current of 2 A. The armature resistance is 0.5Ω and leakage inductance is 0.01 H.
When loaded, the armature current is 15 A. Then the speed in rpm is _____
Answer: 880 rpm
Exp:
Given 300V1
N1 = 900rpm
R a = 0.5Ω
Ia1 = 2A; Ia 2 = 15A
∴ Eb1 = 300 − 2 × 0.5 = 299V
Eb 2 = 300 − 15 × 0.5 = 292.5V
∴
39.
N 2 Eb 2
2925
=
⇒ N 2 = 900 ×
= 880 rpm
N1 Eb1
29
The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.
1Ω
2 :1
50 Hz
~
ac source
x
110 V
ZL
Assuming the transformer to be ideal, the value of the reactance X to improve the input
power factor to unity is ___________.
Answer: 24
Exp:
Given, V2 = 110
Load power, P2 = 4 kW
pf = cos θ2 = 0.89
I2 =
load power
4000
=
= 40.858 A
V2 × cos θ2 110 × 0.89
primary current, I1 = K I 2
N 
1
=  2  I2 = × 40.858
2
 N1 
= 20.429 A
to improve input power factor units,
Reactive power = φ=
∴ Reactance, X=
V12
X
V12
2202
=
= 23.6Ω.
φ 220 × 20.429 × sin ( 27 )
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40.
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The parameters measured for a 220V/110V, 50 Hz, single-phase transformer are:
Self inductance of primary winding = 45 mH
Self inductance of secondary winding = 30 mH
Mutual inductance between primary and secondary windings = 20 mH
Using the above parameters, the leakage (Ll1 , Ll2) and magnetizing (Lm) inductances as
referred to primary side in the equivalent circuit respectively, are
(A) 5mH, 20mH and 40mH
(B) 5mH, 80mH and 40mH
(C) 25mH, 10mH and 20mH
(D) 45mH, 30mH and 20mH
Answer: (B)
Exp:
Given, 220/110v, 50Hz.
L1 = 45mH, L 2 = 30mH
M=20 mH.
Leakage inductancy are,
Ll1 = L1 − 2(m) = 45 − 2(20)
= 5 mH
Ll2 = L 2 − 2(m)
= 30 − 2(20) = 10mH
But refered to primary,
Ll'2 =
30
− 2(20)
2
1
 
2
= 120 − 40 = 80mH
41.
For a 400 km long transmission line, the series impedance is ( 0.0 + j0.5) Ω / km and the shunt
admittance is (0.0 + j5.0) µmho / km . The magnitude of the series impedance (in O) of the
equivalent p circuit of the transmission line is ________.
Answer: 186.66
Exp:
For a long transmission line,
 yz
 yz  
1+
z 1 +  

2
6 
 AB  

CD  = 
yz 
  y 1 + yz 
1+
 


6
2 
 
Where z= total series impedance = j0.5Ω / km × 400 km = j200Ω
y = j5s 10−6 / km = j5 × 10−6 × 400s = j2 × 10−3 simen
ABCD parameters of a π circuit is
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Z
y2
y1
 AB  1 + y 2 z
CD  =  y + y + y y z
   1
2
1 2
 1 + yz 
∴Z = B = Z

 6 
 ( j200 ) j2 × 10−3
= j200 1 +

6

1 − 0.4 
= j200 
 = j186.66
 6 
(
42.


1 + y1z 
z
) 


The complex power consumed by a constant-voltage load is given by (P1 + jQ1) where,
1kW ≤ P1 ≤ 1.5kW and 0.5kVAR ≤ Q1 ≤ 1kVAR
A compensating shunt capacitor is chosen such that Q ≤ 0.25kVAR where Q is the net
reactive power consumed by the capacitor-load combination. The reactive power (in kVAR)
supplied by the capacitor is _________.
Answer: 0.75
Exp: Net reactive power consumed by capacitor-load combination is 0.25 KVAR.
Reactive power required for load is 1KVAR
Reactive power supplied by capacitor is 0.75 KVAR.
43.
The figure shows the single line diagram of a single machine infinite bus system.
Infinite
bus
The inertia constant of the synchronous generator H =5 MW-s/MVA. Frequency is 50 Hz.
Mechanical power is 1 pu. The system is operating at the stable equilibrium point with rotor
angle δ equal to 30O. A three phase short circuit fault occurs at a certain location on one of
the circuits of the double circuit transmission line. During fault, electrical power in pu is Pmax
sin δ If the values of δ and d δ /dt at the instant of fault clearing are 45O and 3.762 radian/s
respectively, then Pmax (in pu) is _______.
Answer: 0.24
Exp:
Given,
H = 5 MW – S
MVA
f = 50 Hz
Pm =1pu
δ0 = 30°
At the instant of fault clearing,
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dδ
= 3.762 rad / s
dt
δ = 45° ;
∴
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d 2δ 1
= [ p m − p3 ]
dt 2 M
2
dδ
= 
dt
 M
 2
=
 H πf

∫δ ( pm − pe ) ds 
0

δ
1
2
d

 dt

∫30 (1 − p max sin δ ) ds 
45
1
2
 ds 
 dδ   d δ  
=
2
 
   2 
 dt 
 dt   dt  
2
2
45
 2 ×π× 50
3.762 = 
( δ + p max cos δ )30 
5


1
2
⇒ 14.152 = 20 π { ( 45° − 30° ) + p max ( cos 45° − cos30° )}
⇒ p max = 0.24 p.u
44.
The block diagram of a system is shown in the figure
Rs
()
+−
+−
1
s
+−
G ( s)
C ( s)
s
If the desired transfer function of the system is
C (s )
R (s )
=
s
s + s +1
2
then G(s) is
(A) 1
(B) s
(C) 1/s
( D)
−s
s +s −s−2
3
2
Answer: (B)
Exp:
+
R(s)
−
+
1
s
−
G(s)
S
C(s)
−
If G ( s ) = S.
C(s)
R (s )
=
S
s +s+2
2
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45.
Consider the system described by following state space equations
 x 1   0 1   x1  0
 x1 
 x  =  −1 −1  x  + 1 u; y = [1 0]  x 
  2  
 2 
 2
If u is unit step input, then the steady state error of the system is
(A) 0
(B) 1/2
(C) 2/3
Answer: (A)
Exp:
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(D) 1
Transfer function ⇒ C [SI − A ] .B.
−1
S
= [1 0] 
1
−1
−1   0 
( s + 1) 1 
Transfer function =
G (s)
1 + G (s)
=
1
s + s +1
2
1
s + s +1
2
1
s2 + s
Steady state error for unit step
A
ess =
1 + Kp
⇒ G (s) =
ess =
1
1 + lim G ( s )
s→0
ess =
1
1 + lim
s→0
ess =
1
s +s
2
1
1+ ∞
ess = 0
46.
The magnitude Bode plot of a network is shown in the figure
G ( jω )
dB
0
Slope 20 dB / decade
1
3
log10 ω
1
The maximum phase angle .m and the corresponding gain Gm respectively, are
( A)
( C)
− 30O and 1.73dB
+ 30O and 4.77dB
( B)
( D)
− 30O and 4.77dB
+ 30O and 1.73dB
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Answer: C
Exp:
G ( s ) = k.
(1 + 3s )
(1 + s )
 1
3k. s + 
3

G (s ) =
+
s
1
( )
Here k = 1
1 1
1
= ⇒
=1
T 3 αT
ωm = 1
3
G ( s ) ω= 1
Gm
in dB
;α = 1
=
3
3
4
⇒ 3
4
3
= 20log 3 = 4.77 dB
1 − α 
φm = sin −1 

1 + α 
1 −
α = 1 = sin −1 
3
1 +

1 
3  = sin −1  1 
 
1 
2
3
φm = 30o
47.
A periodic waveform observed across a load is represented by
 1 + sin ωt 0 ≤ωt < 6π
V ( t) = 
 −1 + sin ωt 6π ≤ ωt < 12 π
The measured value, using moving iron voltmeter connected across the load, is
( A)
3
2
( B)
2
3
( C)
3
2
( D)
2
3
Answer: (A)
Exp:
M.I instrument reads RMS value
=
(1)
2
+  1 
2

2
= 1+ 1 ⇒ 3
2
2
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48.
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In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance
C1 in microfarad is _________.
C1
35k
Vsin ( t )
G
0.1F
105k
Answer: 0.3 µF
Exp:
Bridge is balanced
z1z 4 = z 2 z3
35k.
1
1
= 105k.
jω0.1µF
jωc1
C1 = 0.3 µF
49.
Two monoshot multivibrators, one positive edge triggered ( M1 ) and another negative edge
triggered ( M 2 ), are connected as shown in figure
+5V
M1
10k
Q1
Q1
10µF
M2
Q2
Vo
Q1
The monoshots M1 and M2 when triggered produce pulses of width T1 and T2 respectively,
where T1>T2. The steady state output voltage vo of the circuit is
( A)
Vo
T1
T1
T2
T1
T2
t
( B)
Vo
T1
T1
T1
T1
t
( C)
Vo
T2
T2
T1
T2
T1
t
( D)
Vo
T2
T2
T2
T2
T2
T2
t
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Answer: (C)
Exp:
10 k
V2 (t)
M1
•
•
+ (5 V)
10 µF
www.gateforum.com
M2
Q1 (t)
Q1
Q2
Q1
Q2
V0 (t)
V2 (t) =
5V
5τ = 50sec
t
→ given M1 mono-stable multivibrator generates pulse width T1.
T1
M2 mono-stable multivibrator generates public width T2.
(Because T1 > T2 )
T2
(1) Assume Initially if Q 2 = 1 (high state), then Q 2 =0 (low state)
Then output of AND gate is low, M1(multi vibrator) it does not generates pulse width
T1(Because it is positive edge triggered),
(2) Output (Q2), after T2 duration , it is low (comes to stable state then Q 2 is high, the output
of And gate is high now, then M1 multivibrator generates pulse width T1 (Because it
positive edge triggered), At this time Q2 does not generates pulse width T2 (Because it
negative edge Triggered) then,at the end of T1 pulse, M2 multi vibrator generates T2
pulse width (Because it is negative edge triggered)
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(1)
Q 2 (t) = V0 (c)
T2
Q1 (t)
T1
Then again Q 2 (t) is high at the end of T1 pulse
Q 2 (t)
T2
Overall output wave form
T1
T2
So ......on
T2
50.
The transfer characteristic of the Op-amp circuit shown in figure is
R
Vi
R
+ Vsat
+ V sat
R
−
+
R
R
−
+
− V sat
R
− Vsat
Vo
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( B)
( A)
−1
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1
V0
V0
Vi
Vi
( D)
( C)
V0
V0
Vi
Vi
1
Answer: (C)
Exp:
When Vi > 0; V0 = 0 ⇒ slope = 0
−1
V0
When Vi < 0; V0 = Vi ⇒ slope = 1
Slope = 0
Transfer characteristics are
Vi
Slope = 1
51.
A 3-bit gray counter is used to control the output of the multiplexer as shown in the figure.
The initial state of the counter is 0002. The output is pulled high. The output of the circuit
follows the sequence
A2
3 − bit gray A
1
counter A
0
+ 5V
S0
CLK
(A) I0, 1, 1, I1, I3, 1, 1, I2
(C) 1, I0, 1, I1, I2, 1, I3,1
Answer: (A)
I0
I1
I2
I3
E
0
1
2
3
S1
4 X1
MUX
R
Output
(B) I0, 1, I1, 1, I2, 1, I3, 1
(D) I0, I1, I2, I3, I0, I1, I2, I3
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Exp:
Decimal
Binary
Gray
A 2 A1 A 0
Output
0
1
2
3
4
5
6
7
000
0 01
0 10
011
1 00
101
11 0
111
000
0 01
0 11
010
1 10
1 11
101
10 0
I0
1
1
I1
I3
1
1
I2
A 0 is mapped to E of 4 :1 MUX it means when A 0 ( E ) will be low then MUX will be
enabled and as per S0 ( A1 ) and S2 ( A 2 ) will produce the output and when A 0 ( E ) will be
high then 4 :1 MUX will be disabled and disabled output will be 1.
52.
A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the
figure.
10k
+5V
VO
0.1µF
If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for
which output is LOW, is __________.
Answer: 0.66
Exp:
Given LTP = 0.9
UTP = 1.7
VC ( t ) = Vmax + ( Vinitial − Vmax ) e − t RC
LTP = 0 + (1.7 − 0 ) e − t /RC = 0.9
⇒ t = 0.635 ms ( Given R = 10k, C = 0.1 µF ) .
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53.
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A three-phase fully controlled bridge converter is fed through star-delta transformer as shown
in the figure.
IR
R
IO
1: K
Y
Y
The converter is operated at a firing angle of 300. Assuming the load current (I0) to be
virtually constant at 1 p.u. and transformer to be an ideal one, the input phase current
waveform is
( A)
IR
0
( C)
2 / 3K
2 / 3K
1/ 3K
( B)
π
IR K / 3
0
2π
2K / 3
2π
0
2π
2 / 3K
( D)
IR
π
IR
0
2π
π
Answer: (B)
54.
A diode circuit feeds an ideal inductor as shown in the figure. Given v s = 100 sin ( ωt ) V ,
where ω = 100πrad / s, , and L = 31.83 mH. The initial value of inductor current is zero.
Switch S is closed at t = 2.5 ms. The peak value of inductor current iL (in A) in the first cycle
is ________.
t = 2.5ms
iL
S
+
Vs ~
−
L
Answer: 17.07A
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Exp:
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When S is closed
VL = Vs
∴L
di
= Vm sin ωt
dt
t = 2.5ms
D
S
+
−
iL
+
∼
VL
L = 31.83mH
−
Vs = 100 sin ωt
di =
Vm
sin ωt.dt but ω = 100π rad / sec
L
f =50 Hg ⇒ T = 20 msa
Integrate on both sides
Vs
t(ms)
20
0
10
iL
I L max
10
2.5
20
t(ms)
10×10−3
Vm 10×10−3
sin ωt.dt
−3
∫2.5×10
L ∫2.5×10
Current Ic is changing from 0 to mgx density their period.
V
V
2.5×10−3
∴ Imax = − m [ cos ωt ]2.5×10−3 = − m  cos(100π × 10 × 10−3 ) − cos(100π × 2.5 × 10−3 
ωL
ωL
di =
−3
=−
I max =
Vm 
π
 cos π − cos 
ωL 
4
100
[1 + 0.707]
100π × 31.83 × 10−3
∴ Imax = 17.07A
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55.
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A single-phase voltage source inverter shown in figure is feeding power to a load. The
triggering pulses of the devices are also shown in the figure.
S1
S2
C
A
O
VDC
C
iL
S3
S1 ,S4
Load
θ
B
S4
π−θ
S2 ,S3
π
π+θ
2π
2π − θ
If the load current is sinusoidal and is zero at 0, x, 2x...., the node voltage VAO has the
waveform
( B)
( A)
VDC / 2
VDC / 2
VAO
θ
π−θ π
VAO
2π
θ
π−θ π
2π
− VDC / 2
( C)
( D)
VDC / 2
VAO
θ
π−θ π
VDC / 2
VAO
2π
θ π−θ
π
2π
| − VDC / 2
Answer: (D)
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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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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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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)
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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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42.
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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
(B) θ2 = 0, θ3 = −0.1
(A) θ2 = −0.1, θ3 = −0.2
(C) θ2 = 0.1, θ3 = 0.1
(D) θ2 = 0.1, θ3 = 0.2
Answer: (C)
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)
53.
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.
www.gateforum.com
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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65.
www.gateforum.com
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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Paper
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7
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7
%
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7
Ω
≡
7
+ 1
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7
7
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4
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4
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( ) = % − 0 < /= U4
( ) = %4 − 0 > /
+
+-%
(+)
∴
+
%1 − .× %% + %4 × % + 8 = %8
"
+
'>
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1>
=
D
8 + %&
−8 − 1
%
/
&=
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/
/
1
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*+ #$ D
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*
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10
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1.
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(
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/
/
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48
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8
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8.
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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
written permission.
GATE- EE 2011
Paper
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; X C = 1Ω
1Ω
+
1 Sin t
−
I ( t) =
Sin t
1
A
= Sin t; Irms =
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
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)
5.
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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2
EE-Paper Code-A GATE 2011
8.
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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 = 30MW + losses
(D) P2 = 45MW
P1 = 40MW
(C) P2 = 40MW
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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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
(B) 4µC
(C) 8µC
(D) 10µC
(A) 2µC
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; xq = 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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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


7.5
(B) j  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 
0 
5
 −10


7.5 12.5
(D) j  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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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
(B) 01
(A) 00
Answer: - (C)
EXP:-
(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
(A)
Vo
Vo
4.3
4.3
(B)
4.3
4.3
Vi
10
Vi
10
(C)
Vo
5.7
Vo
(D)
−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
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.
(B) ascertained
(C) Texacerbated
(D) Analysed
(A) Identified
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.
57.
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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.
(D) Available
(A) Similar
(B) Most
(C) Uncommon
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
(C) teacher : classroom
Answer: - (D)
(B) commuter: train
(D) lawyer : courtroom
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
(C) R
(B) Q
(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
(C) 48
(B) 31
(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
(C) 1
(B) -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
10n+1 − 9n − 10
=
− n =
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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