Gate ee 2005 with solutions

GATE EE
2005
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Q.1 - 30 Carry One Mark Each
MCQ 1.1 In the figure given below the value of R is
(A) 2.5 Ω (B) 5.0 Ω
(C) 7.5 Ω (D) 10.0 Ω
SOL 1.1 The Correct option is (C).
Current in the circuit
I
( || )
8
R 10 10
100=
+
= A (given)
&
R 5
100
+
8=
Or R 7.5
8
60 Ω= =
MCQ 1.2 The RMS value of the voltage ( )u t ( )cos t3 4 3= + is
(A) 17 V (B) 5 V
(C) 7 V (D) (3 2 )2 V+
SOL 1.2 Rms value is given as
rmsμ
( )
3
2
42
2
= +
9 8 17 V= + =
Hence (A) is correct option.
MCQ 1.3 For the two port network shown in the figure the Z -matrix is given by

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(A)
Z
Z Z
Z Z
Z
1
1 2
1 2
2+
+
= G (B)
Z
Z Z
Z
Z
1
1 2
1
2+= G
(C)
Z
Z
Z
Z Z
1
2
2
1 2+= G (D)
Z
Z
Z
Z Z
1
1
1
1 2+= G
SOL 1.3 By writing KVL in input and output loops
( )V i i Z1 1 2 1− + 0=
V1 Z i Z i1 1 1 2= + ...(1)
Similarly
( )V i Z i i Z2 2 2 1 2 1− − + 0=
V2 ( )Z i Z Z i1 1 1 2 2= + + ...(2)
From equation (1) and (2) Z -matrix is given as
Z
Z
Z
Z
Z Z
1
1
1
1 2
=
+> H
Hence (D) is correct option.
MCQ 1.4 In the figure given, for the initial capacitor voltage is zero. The switch is closed at
t 0= . The final steady-state voltage across the capacitor is
(A) 20 V (B) 10 V
(C) 5 V (D) 0 V
SOL 1.4 In final steady state the capacitor will be completely charged and behaves as an
open circuit
Steady state voltage across capacitor
( )vc 3 (10)
10 10
20=
+

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10= V
Hence (B) is correct option.
MCQ 1.5 If Ev is the electric intensity, ( )E4 4# v is equal to
(A) Ev (B) Ev
(C) null vector (D) Zero
SOL 1.5 We know that divergence of the curl of any vector field is zero
( )E4 4# 0=
MCQ 1.6 A system with zero initial conditions has the closed loop transfer function.
( )T s
( )( )s s
s
1 4
42
=
+ +
+
The system output is zero at the frequency
(A) 0.5 rad/sec (B) 1 rad/sec
(C) 2 rad/sec (D) 4 rad/sec
SOL 1.6 Closed loop transfer function of the given system is,
( )T s
( )( )s s
s
1 4
42
=
+ +
+
( )T jω
( )( )
( )
j j
j
1 4
42
ω ω
ω
=
+ +
+
If system output is zero
( )T jω
1 ( )j j 4
4 2
ω ω
ω
=
+ +
−
^ h
0=
4 2
ω− 0=
2
ω 4=
& ω 2= rad/sec
Hence (C) is correct option.
MCQ 1.7 Figure shows the root locus plot (location of poles not given) of a third order
system whose open loop transfer function is

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(A)
s
K
3
(B)
( )s s
K
12
+
(C)
( )s s
K
12
+
(D)
( )s s
K
12
−
SOL 1.7 From the given plot we can see that centroid C (point of intersection) where
asymptotes intersect on real axis) is 0
So for option (a)
( )G s
s
K
3=
Centroid 0
n m 3 0
0 0Poles Zeros
=
−
−
=
−
− =
//
MCQ 1.8 The gain margin of a unity feed back control system with the open loop transfer
function
( )G s
( 1)
s
s
2=
+
is
(A) 0 (B)
2
1
(C) 2 (D) 3
SOL 1.8 Open loop transfer function is.
( )G s
( )
s
s 1
2=
+
( )G jω
j 1
2
ω
ω
=
−
+
Phase crossover frequency can be calculated as.
( )G j p+ ω 180c=−
( )tan p
1
ω−
180c=−
pω 0=
Gain margin of the system is.
G.M
( )G j
1
1
1
p
p
p
2
2ω
ω
ω
= =
+
G.M
1
0
p
p
2
2
ω
ω
=
+
=
MCQ 1.9 In the matrix equation x qP = , which of the following is a necessary condition for
the existence of at least on solution for the unknown vector x
(A) Augmented matrix [ ]qP must have the same rank as matrix P
(B) Vector q must have only non-zero elements

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(C) Matrix P must be singular
(D) Matrix P must be square
SOL 1.9 The Correct option is (D).
For two random events conditional probability is given by
( )probability P Q+ ( ) ( )probability probabilityP Q=
( )Qprobability
( )
( )
P
P Q
1
probability
probability +
#=
so ( )P Qprobability + ( )Pprobability#
MCQ 1.10 If P and Q are two random events, then the following is TRUE
(A) Independence of P and Q implies that probability ( )P Q 0+ =
(B) Probability ( )P Q, $ Probability (P) + Probability (Q)
(C) If P and Q are mutually exclusive, then they must be independent
(D) Probability ( )P Q+ # Probability (P)
SOL 1.10 Option (D) is correct.
for two random events conditional probability is given by
( )probability P Q+ ( ) ( )probability probabilityP Q=
( )Qprobability
( )
( )
P
P Q
1
probability
probability +
#=
so ( )P Qprobability + ( )Pprobability#
MCQ 1.11 If S x dx3
1
=
3 -
# , then S has the value
(A)
3
1− (B)
4
1
(C)
2
1 (D) 1
SOL 1.11 Hence (C) is correct option
S x dx3
1
=
3 −
#
x
2
2
1
=
−
3−
: D
2
1=
MCQ 1.12 The solution of the first order differential equation '( ) ( )x t x t3=− , (0)x x0= is
(A) ( )x t x e t
0
3
= -
(B) ( )x t x e0
3
= -
(C) ( )x t x e /
0
1 3
= -
(D) ( )x t x e0
1
= -

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SOL 1.12 Hence (A) is correct option.
We have ( )x to ( )x t3=−
or ( ) 3 ( )x t x t+o 0=
A.E. D 3+ 0=
Thus solution is ( )x t C e t
1
3
= −
From ( )x x0 0= we get C1 x0=
Thus ( )x t x e t
0
3
= −
MCQ 1.13 The equivalent circuit of a transformer has leakage reactances , 'X X1 2 and magnetizing
reactance XM . Their magnitudes satisfy
(A) 'X X X>> >> M1 2 (B) 'X X X<< << M1 2
(C) 'X X X>> M1 2. (D) 'X X X<< M1 2.
The leakage reactances X1, and X2l are equal and magnetizing reactance Xm is
higher than X1, and X2l
X1 X X<< m2. l
MCQ 1.14 Which three-phase connection can be used in a transformer to introduce a phase
difference of 30c between its output and corresponding input line voltages
(A) Star-Star (B) Star-Delta
(C) Delta-Delta (D) Delta-Zigzag
SOL 1.14 The Correct option is (B).
Three phase star delta connection of transformer induces a phase difference of 30c
between output and input line voltage.
MCQ 1.15 On the torque/speed curve of the induction motor shown in the figure four points
of operation are marked as W, X, Y and Z. Which one of them represents the
operation at a slip greater than 1 ?
(A) W (B) X
(C) Y (D) Z
SOL 1.15 The Correct option is (A).
Given torque/speed curve of the induction motor

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When the speed of the motor is in forward direction then slip varies from 0 to 1 but
when speed of motor is in reverse direction or negative then slip is greater then 1.
So at point W slip is greater than 1.
MCQ 1.16 For an induction motor, operation at a slip s, the ration of gross power output to
air gap power is equal to
(A) ( )s1 2
− (B) ( )s1 −
(C) ( )s1 − (D) ( )s1 −
For an induction motor the ratio of gross power output to air-gap is equal to ( )s1 −
So
airgap power
gross power
( )s1= −
MCQ 1.17 The p.u. parameter for a 500 MVA machine on its own base are:
inertia, M 20= p.u. ; reactance, X 2= p.u.
The p.u. values of inertia and reactance on 100 MVA common base, respectively,
are
(A) 4, 0.4 (B) 100, 10
(C) 4, 10 (D) 100, 0.4
Given that pu parameters of 500 MVA machine are as following
M 20= pu, X 2= pu
Now value of M and X at 100 MVA base are
for inertia ( )M
( )pu new (pu)
new MVA
old MVA
old #=
( )Mpu new ( )M
100
500
Pu old #=
20 100
1
5
#= = pu
and for reactance (X )
( )pu new ( )pu
old MVA
new MVA
old #=
( )Xpu new ( )X
500
100
pu old #=

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( )XPu new 2 0.4
5
1
#= = pu
MCQ 1.18 An 800 kV transmission line has a maximum power transfer capacity of P. If it is
operated at 400 kV with the series reactance unchanged, the new maximum power
transfer capacity is approximately
(A) P (B) P2
(C) /P 2 (D) /P 4
800 kV has Power transfer capacity P=
At 400 kV Power transfer capacity = ?
We know Power transfer capacity
P sin
X
EV δ=
P V2

So if V is half than Power transfer capacity is
4
1 of previous value.
MCQ 1.19 The insulation strength of an EHV transmission line is mainly governed by
(A) load power factor (B) switching over-voltages
(C) harmonics (D) corona
In EHV lines the insulation strength of line is governed by the switching over
voltages.
MCQ 1.20 High Voltage DC (HVDC) transmission is mainly used for
(A) bulk power transmission over very long distances
(C) inter-connecting two systems with same nominal frequency
(C) eliminating reactive power requirement in the operation
(D) minimizing harmonics at the converter stations
For bulk power transmission over very long distance HVDC transmission preferably
used.
MCQ 1.21 The Q-meter works on the principle of
(A) mutual inductance (B) self inductance
(C) series resonance (D) parallel resonance
Q-meter works on the principle of series resonance.

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At resonance VC VL=
and I
R
V=
Quality factor Q
R
L
CR
1ω
ω
= =
Q
R I
L I
E
V
E
VL C
#
#ω= = =
Thus, we can obtain Q.
MCQ 1.22 A PMMC voltmeter is connected across a series combination of DC voltage source
2V1 = V and AC voltage source ( ) 3 (4 )sinV t t2 = V. The meter reads
(A) 2 V (B) 5 V
(C) ( / )2 3 2+ V (D) ( / )V17 2
PMMC instruments reads DC value only so it reads 2 V.
MCQ 1.23 Assume that D1 and D2 in figure are ideal diodes. The value of current is
(A) 0 mA (B) 0.5 mA
(C) 1 mA (D) 2 mA
From the circuit we can observe that Diode D1 must be in forward bias (since
current is flowing through diode).
Let assume that D2 is in reverse bias, so equivalent circuit is.

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Voltage Vn is given by
Vn 1 2 2 Volt#= =
Vp 0=
V V>n p (so diode is in reverse bias, assumption is true)
Current through D2 is
ID2 0=
MCQ 1.24 The 8085 assembly language instruction that stores the content of H and L register
into the memory locations 2050H and 2051H, respectively is
(A) SPHL 2050H (B) SPHL 2051H
(C) SHLD 2050H (D) STAX 2050H
SHLD transfers contain of HL pair to memory location.
SHLD 2050 L M[2050H]
H M[2051H]
& "
"
MCQ 1.25 Assume that the N-channel MOSFET shown in the figure is ideal, and that its
threshold voltage is 1.0+ V the voltage Vab between nodes a and b is
(A) 5 V (B) 2 V
(C) 1 V (D) 0 V
This is a N-channel MOSFET with
VGS 2 V=
VTH 1 V=+
VDS(sat) V VGS TH= −
VDS(sat) 2 1 1 V= − =
Due to 10 V source V V>DS DS(sat) so the NMOS goes in saturation, channel

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conductivity is high and a high current flows through drain to source and it acts
as a short circuit.
So, V 0ab =
MCQ 1.26 The digital circuit shown in the figure works as
(A) JK flip-flop (B) Clocked RS flip-flop
(C) T flip-flop (D) Ring counter
Let the present state is Q(t), so input to D-flip flop is given by,
D ( )Q t X5=
Next state can be obtained as,
( )Q t 1+ D=
( )Q t 1+ ( )Q t X5=
( )Q t 1+ ( ) ( )Q t X Q t X= +
( )Q t 1+ ( )Q t= , if 1X =
and ( )Q t 1+ ( )Q t= , if 0X =
So the circuit behaves as a T flip flop.
MCQ 1.27 A digital-to-analog converter with a full-scale output voltage of 3.5 V has a
resolution close to 14 mV. Its bit size is
(A) 4 (B) 8
(C) 16 (D) 32
Resolution of n-bit DAC
V
2 1n
fs
=
−
So 14 mv 3.5
2 1
V
n=
−
2 1n
− .
14 10
3 5
3
#
= −
2 1n
− 250=
2n
251=
n 8= bit
MCQ 1.28 The conduction loss versus device current characteristic of a power MOSFET is
best approximated by
(A) a parabola

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(B) a straight line
(C) a rectangular hyperbola
(D) an exponentially decaying function
The conduction loss v/s MOSFET current characteristics of a power MOSFET is
best approximated by a parabola.
MCQ 1.29 A three-phase diode bridge rectifier is fed from a 400 V RMS, 50 Hz, three-phase
AC source. If the load is purely resistive, then peak instantaneous output voltage
is equal to
(A) 400 V (B) 400 2 V
(C) 400
3
2 V (D)
3
400 V
In a 3-φ bridge rectifier
Vrms 400 V= , 50f Hz=
This is purely resistive then
instantaneous voltage V0 400V2 2rms= = V
MCQ 1.30 The output voltage waveform of a three-phase square-wave inverter contains
(A) only even harmonics (B) both odd and even harmonic
(C) only odd harmonics (D) only triple harmonics
A 3-φ square wave (symmetrical) inverter contains only odd harmonics.
Q.31 - 80 Carry Two Marks Each
MCQ 1.31 The RL circuit of the figure is fed from a constant magnitude, variable frequency
sinusoidal voltage source Vin . At 100 Hz, the Rand L elements each have a voltage
drop RMSμ .If the frequency of the source is changed to 50 Hz, then new voltage
drop across R is
(A)
8
5 uRMS (B)
3
2 uRMS
(C)
5
8 uRMS (D)
2
3 uRMS

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At 100f1 = Hz, voltage drop across R and L is RMSμ
RMSμ .
R j L
V Rin
1ω
=
+
( )
R j L
V j Lin
1
1
ω
ω
=
+
So, R L1ω=
at 50f2 = Hz, voltage drop across R
RMSμl .
R j L
V Rin
2ω
=
+
RMS
RMS
μ
μ
l R j L
R j L
1
2
ω
ω
=
+
+
R L
R L
2
1
2 2
2
2
2 2
ω
ω=
+
+
L L
L L
1
2 2
1
2 2
1
2 2
2
2 2
ω ω
ω ω=
+
+ , R L1ω=
2 1
2
1
2
2
2
ω
ω ω= +
f
f f
2 1
2
1
2
2
2
=
+
( )
( ) ( )
2 100
100 50
8
5
2
2 2
=
+
=
RMSμl
5
8
RMSμ=
MCQ 1.32 For the three-phase circuit shown in the figure the ratio of the currents : :I I IR Y B is
given by
(A) : :1 1 3 (B) : :1 1 2
(C) : :1 1 0 (D) : : /1 1 3 2
In the circuit
I B 0 120I IR yc c+ += +
IB
2
2 cosI I I I
2
120
R y R y
2 2 c= + + b l
IB
2
I I I IR y R y
2 2
= + +
a IR Iy=
so, IB
2
I I IR R R
2 2 2
= + + 3IR
2
=

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IB I3 R= I3 y=
: :I I IR y B 1:1: 3=
MCQ 1.33 For the triangular wave from shown in the figure, the RMS value of the voltage is
equal to
(A)
6
1 (B)
3
1
(C)
3
1 (D)
3
2
RMS value is given by
Vrms ( )
T
V t dt1 T
2
0
= #
Where
( )V t
,
,
T
t t T
T t T
2 0
2
0
2
<
# #
#
=
` j
*
So ( )
T
V t dt1 T
2
0
# ( )
T T
t dt dt1 2 0
/
/
T
TT 2
20
2
= +` j= G##
T T
t dt1 4 /T
2
2
0
2
$= #
T
t4
3
/T
3
3
0
2
= ; E
T
T4
243
3
#=
6
1=
Vrms
6
1 V=
MCQ 1.34 The circuit shown in the figure is in steady state, when the switch is closed at t 0=
.Assuming that the inductance is ideal, the current through the inductor at t 0= +
equals

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(A) 0 A (B) 0.5 A
(C) 1 A (D) 2 A
Switch was opened before 0t = , so current in inductor for 0t <
(0 )iL
−
1
10
10= = A
Inductor current does not change simultaneously so at t 0= when switch is closed
current remains same
(0 )iL
+
(0 ) 1 AiL= =−
MCQ 1.35 The charge distribution in a metal-dielectric-semiconductor specimen is shown in
the figure. The negative charge density decreases linearly in the semiconductor as
shown. The electric field distribution is as shown in

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Electric field inside a conductor (metal) is zero. In dielectric charge distribution os
constant so electric field remains constant from x1 to x2. In semiconductor electric
field varies linearly with charge density.
MCQ 1.36 In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at
the terminals P-Q, is given by
(A) (2 5 )V, Ω (B) (2 , 7.5 )V Ω
(C) (4 , 5 )V Ω (D) (4 , 7.5 )V Ω
Thevenin voltage:

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nodal analysis at P
V V
10
4
10
th th− + 0=
2 4Vth − 0=
& Vth 2= V
Thevenin resistance:
Rth 10 || 10 5Ω Ω Ω= =
MCQ 1.37 A unity feedback system, having an open loop gain
( ) ( )G s H s
( )
( )
s
K s
1
1
=
+
−
,
becomes stable when
(A) K 1> (B) K 1>
(C) K 1< (D) K 1< −
SOL 1.37 Characteristic equation for the given system
( ) ( )G s H s1 + 0=
( )
( )
K
s
s
1
1
1
+
+
−
0=
(1 ) (1 )s K s+ + − 0=
( ) ( )s K K1 1− + + 0=
For the system to be stable, coefficient of characteristic equation should be of same
sign.
K1 0>− , K 1 0>+
K 1< , K 1> −
1− K 1< <
K 1<
Hence (C) is correct option

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MCQ 1.38 When subject to a unit step input, the closed loop control system shown in the
figure will have a steady state error of
(A) .1 0− (B) .0 5−
(C) 0 (D) 0.5
SOL 1.38 In the given block diagram
Steady state error is given as
ess ( )limsE s
s 0
=
"
( )E s ( ) ( )R s Y s= −
( )Y s can be written as
( )Y s ( ) ( ) ( )R s Y s
s
R s
s
3
2
2= − −
+: D" ,
( )Y s ( )
( )
( )
( )
R s
s s s
Y s
s s2
6
2
2
2
6=
+
−
+
−
+; ;E E
( )
( )
Y s
s s
1
2
6+
+; E ( )
( )
R s
s s
s
2
6 2=
+
−
; E
( )Y s ( )
( )
( )
R s
s s
s
2 6
6 2
2=
+ +
−
So, ( )E s ( )
( )
( )
( )R s
s s
s
R s
2 6
6 2
2= −
+ +
−
( )E s ( )R s
s s
s s
2 6
4
2
2
=
+ +
+
; E
For unit step input ( )R s
s
1=
Steady state error ess ( )limsE s
s 0
=
"
ess
1
( 2 6)
( 4 )
lim s
s s s
s s
2
2
s 0
=
+ +
+
"
= G
0=
MCQ 1.39 Inthe ( ) ( )G s H s -plane,theNyquistplotofthelooptransferfunction ( ) ( )G s H s s
e . s0 25
= π -

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passes through the negative real axis at the point
(A) ( 0.25, 0)j− (B) ( 0.5, 0)j−
(C) 0 (D) 0.5
SOL 1.39 When it passes through negative real axis at that point phase angle is 180c− .
So ( ) ( )G j H j+ ω ω 180c=−
0.25j
2
ω π− − π=−
0.25jω−
2
π=−
0.25j ω
2
π=
j ω
.2 0 25#
π=
s 2jω π= =
Put s 2π= in given open loop transfer function we get
( ) ( )G s H s s 2π=
.e
2
0 5
.0 25 2
π
π= =−
# π−
So it passes through ( 0.5, 0)j−
MCQ 1.40 If the compensated system shown in the figure has a phase margin of 60c at the
crossover frequency of 1 rad/sec, then value of the gain K is
(A) 0.366 (B) 0.732
(C) 1.366 (D) 2.738
SOL 1.40 Open loop transfer function of the system is given by.
( ) ( )G s H s ( 0.366 )
( 1)
1K s
s s
= +
+; E
( ) ( )G j H jω ω
( )
.
j j
K j
1
0 366
ω ω
ω
=
+
+
Phase margin of the system is given as
60PM cφ = 180 ( ) ( )G j H jg gc + ω ω= +
Where gaing "ω cross over frequency 1= rad/sec
So,
60c
.
( )tan tan
K
180
0 366
90g
g
1 1
c c
ω
ω= + − −− −
b l

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. ( )tan tan
K
90 0 366 11 1
c= + −− −
b l
.tan
K
90 45 0 3661
c c= − + −
b l
15c .tan
K
0 3661
= −
b l
.
K
0 366 tan15c=
K
.
.
0 267
0 366= 1.366=
MCQ 1.41 For the matrix p
3
0
0
2
2
0
2
1
1
=
−
−
R
T
S
S
SS
V
X
W
W
WW
, one of the eigen values is equal to 2−
Which of the following is an eigen vector ?
(A)
3
2
1
−
R
T
S
S
SS
V
X
W
W
WW
(B)
3
2
1
−
−
R
T
S
S
SS
V
X
W
W
WW
(C)
1
2
3
−
R
T
S
S
SS
V
X
W
W
WW
(D)
2
5
0
R
T
S
S
SS
V
X
W
W
WW
SOL 1.41 Hence (D) is correct option.
For eigen value λ 2=−
( )
( )
( )
x
x
x
3 2
0
0
2
2 2
0
2
1
1 2
1
2
3
− − −
− − −
− −
R
T
S
S
SS
R
T
S
S
SS
V
X
W
W
WW
V
X
W
W
WW
0
0
0
=
R
T
S
S
SS
V
X
W
W
WW
x
x
x
5
0
0
2
0
0
2
1
1
1
2
3
−R
T
S
S
SS
R
T
S
S
SS
V
X
W
W
WW
V
X
W
W
WW
0
0
0
=
R
T
S
S
SS
V
X
W
W
WW
x x x5 21 2 3− + 0=
Only option (D) satisfies this equation
MCQ 1.42 If R
1
2
2
0
1
3
1
1
2
=
−
−
R
T
S
S
SS
V
X
W
W
WW
, then top row of R 1-
is
(A) 5 6 48 B (B) 5 3 1−8 B
(C) 2 0 1−8 B (D) /2 1 1 2−8 B
SOL 1.42 Hence (B) is correct option.

PUBLISHING FOR GATE
C11 ( )2 3 5= − − =
C21 ( ( ))0 3 3=− − − =−
C31 ( ( ))1 1= − − =
R ( )C C C1 2 211 21 31= + +
5 6 2= − + 1=
MCQ 1.43 A fair coin is tossed three times in succession. If the first toss produces a head, then
the probability of getting exactly two heads in three tosses is
(A)
8
1 (B)
2
1
(C)
8
3 (D)
4
3
SOL 1.43 If the toss produces head, then for exactly two head in three tosses three tosses
there must produce one head in next two tosses. The probability of one head in two
tosses will be 1/2.
MCQ 1.44 For the function ( )f x x e x2
= -
, the maximum occurs when x is equal to
(A) 2 (B) 1
(C) 0 (D) 1−
SOL 1.44 Hence (A) is correct option.
We have ( )f x x e x2
= −
or '( )f x xe x e2 x x2
= −− −
( )xe x2x
= −−
''( )f x ( )x x e4 2 x2
= − + −
Now for maxima and minima, '( )f x 0=
( )xe x2x
−−
0=
or x ,0 2=
at x 0= ''( )f 0 1( )ve= +
at x 2= ''( )f 2 2 ( )e ve2
=− −−
Now ''( )f 0 1= and ''( )f e2 2 0<2
=− −
. Thus x 2= is point of maxima
MCQ 1.45 For the scalar field u x y
2 3
2 2
= + , magnitude of the gradient at the point (1, 3) is
(A)
9
13 (B)
2
9
(C) 5 (D)
2
9
SOL 1.45 Hence (C) is correct option.
u4
x y
ui j
2
2
2
2= +t t
c m
x
u
y
ui j
2
2
2
2= +t t

PUBLISHING FOR GATE
x yi j
3
2= +t t
At (1, 3) magnitude is u4 x y
3
22
2
= +b l
1 4= +
5=
MCQ 1.46 For the equation ''( ) '( ) ( )x t x t x t3 2 5+ + = ,the solution ( )x t approaches which of
the following values as t " 3 ?
(A) 0 (B)
2
5
(C) 5 (D) 10
SOL 1.46 Hence (B) is correct option.
( )
dt
d x
dt
dx x t3 22
2
+ + 5=
Taking laplace transform on both sides of above equation.
( ) ( ) ( )s X s sX s X s3 22
+ +
s
5=
( )X s
( )s s s3 2
5
2=
+ +
From final value theorem
( )limx t
t " 3
( )limX s
s 0
=
"
( )
lims
s s s3 2
5
s 0 2=
+ +"
2
5=
MCQ 1.47 The Laplace transform of a function ( )f t is ( )
( 2 2)
F s
s s s
s s5 23 6
2
2
=
+ +
+ + as , ( )t f t" 3
approaches
(A) 3 (B) 5
(C)
2
17 (D) 3
By final value theorem
( )limf t
t " 3
( )lims F s
s 0
=
"
( )
( )
lims
s s s
s s
2 2
5 23 6
s 0 2
2
=
+ +
+ +
"
2
6= 3=
MCQ 1.48 The Fourier series for the function ( ) sinf x x2
= is
(A) sin sinx x2+ (B) cos x1 2−

PUBLISHING FOR GATE
(C) sin cosx x2 2+ (D) . . cos x0 5 0 5 2−
( )f x sin x2
=
cos x
2
1 2= −
. . cos x0 5 0 5 2= −
( )f x cos sinA a n x b n xn
n
n0 0
1
0ω ω= + +
3
=
/
( ) sinf x x2
= is an even function so b 0n =
A0 .0 5=
an
. ,
, otherwise
n0 5 1
0
=
− =
)
0ω
T T
2 2 2
0
π π= = =
MCQ 1.49 If ( )u t is the unit step and ( )tδ is the unit impulse function, the inverse z -transform
of ( )F z 1z
1
= + for k 0> is
(A) ( ) ( )k1 k
δ− (B) ( ) ( )k 1 k
δ − −
(C) ( ) ( )u k1 k
− (D) ( ) ( )u k 1 k
− −
Z-transform ( )F z
z 1
1=
+
1
z
z
1
= −
+
1
z1
1
1= −
+ −
so, ( )f k ( ) ( 1)k k
δ= − −
Thus ( 1)k
−
z1
1
1
Z
+ −
MCQ 1.50 Two magnetic poles revolve around a stationary armature carrying two coil
( , )c c c c1 1 2 2− −l l as shown in the figure. Consider the instant when the poles are in
a position as shown. Identify the correct statement regarding the polarity of the
induced emf at this instant in coil sides c1 and c2.

PUBLISHING FOR GATE
(A) cin 19 , no emf in c2 (B) cin 17 , no emf in c2
(C) cin 29 , no emf in c1 (D) cin 27 , no emf in c1
Given that two magnetic pole revolve around a stationary armature.
At c1 the emf induced upward and no emf induced at c2 and c2l
MCQ 1.51 A 50 kW dc shunt is loaded to draw rated armature current at any given speed.
When driven
(i) at half the rated speed by armature voltage control and
(ii) at 1.5 times the rated speed by field control, the respective output powers
delivered by the motor are approximately.
(A) 25 kW in (i) and 75 kW in (ii)
(B) 25 kW in (i) and 50 kW in (ii)
(C) 50 kW in (i) and 75 kW in (ii)
(D) 50 kW in (i) and 50 kW in (ii)
Given A 50 kW DC shunt motor is loaded, then
at half the rated speed by armature voltage control
So
P N
Pnew 25
2
50 kW= =
At 1.5 time the rated speed by field control
P constant=
So
P 50 kW=
MCQ 1.52 In relation to DC machines, match the following and choose the correct combination
List-I List-II
Performance Variables Proportional to
P. Armature emf (E ) 1. Flux(φ), speed ( )ω and
armature current ( )Ia

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Q. Developed torque (T ) 2. φ and ω only
R. Developed power (P) 3. φ and Ia only
4. Ia and ω only
5. Ia only
Codes:
P Q R
(A) 3 3 1
(B) 2 5 4
(C) 3 5 4
(D) 2 3 1
In DC motor
E PN
A
Zφ= b l
or
E K nωφ=
So
Armature emf E depends upon φ and ω only.
and torque developed depends upon
T
A
PZ I
2
a
π
φ
=
So, torque(T ) is depends of φ and Ia and developed power(P) is depend of flux φ
, speed ω and armature current Ia .
MCQ 1.53 In relation to the synchronous machines, which on of the following statements is
false ?
(A) In salient pole machines, the direct-axis synchronous reactance is greater than
the quadrature-axis synchronous reactance.
(B) The damper bars help the synchronous motor self start.
(C) Short circuit ratio is the ratio of the field current required to produces the
rated voltage on open circuit to the rated armature current.
(D) The V-cure of a synchronous motor represents the variation in the armature
current with field excitation, at a given output power.
In synchronous machine, when the armature terminal are shorted the field current
should first be decreased to zero before started the alternator.
In open circuit the synchronous machine runs at rated synchronous speed. The field
current is gradually increased in steps.
The short circuit ratio is the ratio of field current required to produced the rated
voltage on open to the rated armature current.

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MCQ 1.54 Under no load condition, if the applied voltage to an induction motor is reduced
from the rated voltage to half the rated value,
(A) the speed decreases and the stator current increases
(B) both the speed and the stator current decreases
(C) the speed and the stator current remain practically constant
(D) there is negligible change in the speed but the stator current decreases
SOL 1.54 The Correct option is ( )
MCQ 1.55 A three-phase cage induction motor is started by direct-on-line (DOL) switching at
the rated voltage. If the starting current drawn is 6 times the full load current, and
the full load slip is 4%, then ratio of the starting developed torque to the full load
torque is approximately equal to
(A) 0.24 (B) 1.44
(C) 2.40 (D) 6.00
Given a three-phase cage induction motor is started by direct on line switching at
rated voltage. The starting current drawn is 6 time the full load current.
Full load slip %4=
So
T
T
Fl
St
b l I
I S
2
Fl
St
Fl#= b l
( ) . .6 0 04 1 442
#= =
MCQ 1.56 In a single phase induction motor driving a fan load, the reason for having a high
resistance rotor is to achieve
(A) low starting torque (B) quick acceleration
(C) high efficiency (D) reduced size
SOL 1.56 Given single-phase induction motor driving a fan load, the resistance rotor is high
So
Eb V I Ra a= − ...(1)
a Pmech E Ia a=
τ P
m
mech
ω
= ...(2)
From equation (1) and (2) the high resistance of rotor then the motor achieves
quick acceleration and torque of starting is increase.
MCQ 1.57 Determine the correctness or otherwise of the following assertion[A] and the
reason[R]
Assertion [A] : Under /V f control of induction motor, the maximum value of
the developed torque remains constant over a wide range of speed in the sub-

PUBLISHING FOR GATE
synchronous region.
Reason [R] : The magnetic flux is maintained almost constant at the rated value by
keeping the ration /V f constant over the considered speed range.
(A) Both [A] and [R] are true and [R] is the correct reason for [A]
(B) Both [A] and [R] are true and but [R] is not the correct reason for [A]
(C) Both [A] and [R] are false
(D) [A] is true but [R] is false
Given /V f control of induction motor, the maximum developed torque remains
same
we have,
E 4.44K f Tw 11
φ=
If the stator voltage drop is neglected the terminal voltage E1. To avoid saturation
and to minimize losses motor is operated at rated airgap flux by varying terminal
voltage with frequency. So as to maintain ( / )V f ratio constant at the rated value,
the magnetic flux is maintained almost constant at the rated value which keeps
maximum torque constant.
MCQ 1.58 The parameters of a transposed overhead transmission line are given as :
Self reactance 0.4 /X kmS Ω= and Mutual reactance 0.1 /X kmm Ω= The positive
sequence reactance X1 and zero sequence reactance X0, respectively in /kmΩ are
(A) 0.3, 0.2 (B) 0.5, 0.2
(C) 0.5, 0.6 (D) 0.3, 0.6
Parameters of transposed overhead transmission line
XS 0.4 /kmΩ= , .1X 0 /kmm Ω=
ve+ sequence reactance X1 ?=
Zero sequence reactance X0 ?=
We know for transposed overhead transmission line.
ve+ sequence component X1 X XS m= −
0.4 0.1 .30 /kmΩ= − =
Zero sequence component X0 2X XS m= +
.4 2( .1) .60 0 0 /kmΩ= + =
MCQ 1.59 At an industrial sub-station with a 4 MW load, a capacitor of 2 MVAR is installed
to maintain the load power factor at 0.97 lagging. If the capacitor goes out of
service, the load power factor becomes
(A) 0.85 (B) 1.00
(C) 0.80 lag (D) 0.90 lag

PUBLISHING FOR GATE
Industrial substation of 4 MW load = PL
QC 2 MVAR= for load .970p.f. = lagging
If capacitor goes out of service than load . ?p.f =
cosφ .970=
tanφ ( .97) .25tan cos 0 01
= =−
P
Q Q
L
L C−
.0 25=
Q
4
2L −
0.25 3Q MVARL&= =
φ
4
3 36tan tan
P
Q
L
L1 1
c= = =− −
c bm l
cosφ .cos36 0 8c= = lagging
MCQ 1.60 The network shown in the given figure has impedances in p.u. as indicated. The
diagonal element Y22 of the bus admittance matrix YBUS of the network is
(A) 19.8j− (B) 20.0j+
(C) 0.2j+ (D) 19.95j−
Y22 ?=
I1 ( )V Y V V Y1 11 1 2 12= + −
0.05 10( ) 9.95 10V j V V j V j V1 1 2 1 2= − − =− +
I2 ( ) ( )V V Y V V Y2 1 21 2 3 23= − + −
. .j V j V j V10 9 9 0 11 2 3= − −
Y22 Y Y Y11 23 2= + +
9.95 9.9 0.1j j j=− − −
19.95j=−
MCQ 1.61 A load centre is at an equidistant from the two thermal generating stations G1 and
G2 as shown in the figure. The fuel cost characteristic of the generating stations
are given by
F a bP cP1 1 1
2
= + + Rs/hour
F a bP cP22 2 2
2
= + + Rs/ hour

PUBLISHING FOR GATE
Where P1 and P2 are the generation in MW of G1 and G2, respectively. For most
economic generation to meet 300 MW of load P1 and P2 respectively, are
(A) 150, 150 (B) 100, 200
(C) 200, 100 (D) 175, 125
F1 a bP cP1 1
2
= + + Rs/hour
F2 2a bP cP2 2
2
= + + Rs/hour
For most economical operation
P P1 2+ 300= MW then , ?P P1 2 =
We know for most economical operation
P
F
1
1
2
2
P
F
2
2
2
2=
2cP b1 + 4cP b2= +
P1 P2 2= ...(1)
P P1 2+ 300= ...(2)
from eq (1) and (2)
P1 200 MW= , 100P MW2 =
MCQ 1.62 Two networks are connected in cascade as shown in the figure. With usual notations
the equivalent , ,A B C and D constants are obtained. Given that, 0.025 45C c+= ,
the value of Z2 is
(A) 10 30c+ Ω (B) 40 45c+ Ω−
(C) 1 Ω (D) 0 Ω
We know that ABCD parameters
V
I
1
1
> H
A
C
B
D
V
I
2
1
= > >H H
B
I
V
V2
1
02
=
=
, C
V
I
I2
1
02
=
=
In figure C
Z Z
V Z
Z Z
V
Z
1
1 2
1
2
1 2
1
2
#
=
+
+ =
or Z2
C
1=

PUBLISHING FOR GATE
.0 025 45
1 40 45
c
c
+
+= = −
MCQ 1.63 A generator with constant 1.0 p.u. terminal voltage supplies power through a step-
up transformer of 0.12 p.u. reactance and a double-circuit line to an infinite bus bar
as shown in the figure. The infinite bus voltage is maintained at 1.0 p.u. Neglecting
the resistances and suspectances of the system, the steady state stability power
limit of the system is 6.25 p.u. If one of the double-circuit is tripped, then resulting
steady state stability power limit in p.u. will be
(A) 12.5 p.u. (B) 3.125 p.u.
(C) 10.0 p.u. (D) 5.0 p.u.
Given
Steady state stability Power Limit .6 25= pu
If one of double circuit is tripped than
Steady state stability power limit ?=
Pm1
.
6.25
X
EV
X0 12
2
1 1#= =
+
=
. . X0 12 0 5
1
+
.6 25=
X& .0 008= pu
If one of double circuit tripped than
Pm2
. . .X
EV
X0 12
1 1
0 12 0 08
1#= =
+
=
+
Pm2
.
5
0 2
1= = pu
MCQ 1.64 The simultaneous application of signals ( )x t and ( )y t to the horizontal and vertical
plates, respectively, of an oscilloscope, produces a vertical figure-of-8 display. If P
and Q are constants and ( ) (4 30 )sinx t tP c= + , then ( )y t is equal to
(A) (4 30 )sin tQ c− (B) (2 15 )sin tQ c+
(C) (8 60 )sin tQ c+ (D) (4 30 )sin tQ c+

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We can obtain the frequency ratio as following
f
f
X
Y
meeting points of vertical tangents
meeting points of horizontal tangents
=
f
f
X
Y
4
2=
fY f
2
1
X=
There should exist a phase difference(15c) also to produce exact figure of-8.
MCQ 1.65 A DC ammeter has a resistance of 0.1 Ω and its current range is 0-100 A. If
the range is to be extended to 0-500 A, then meter required the following shunt
resistance
(A) 0.010 Ω (B) 0.011 Ω
(C) 0.025 Ω (D) 1.0 Ω
The configuration is shown below
It is given that I 100m = A
Range is to be extended to 0 500− A,
I 500= A
So,
I Rm m ( )I I Rm sh= −
.100 0 1# (500 100)Rsh= −
Rsh
.
400
100 0 1#=
.0 025= Ω
MCQ 1.66 The set-up in the figure is used to measure resistance R .The ammeter and voltmeter
resistances are 0.01Ω and 2000 Ω, respectively. Their readings are 2 A and 180

PUBLISHING FOR GATE
V, respectively, giving a measured resistances of 90 Ω The percentage error in the
measurement is
(A) 2.25% (B) 2.35%
(C) 4.5% (D) 4.71%
The configuration is shown below
Current in voltmeter is given by
IV .E
2000 2000
180 09= = = A
I IV+ 2= amp
So I . .2 09 1 91= − = V
R
.
.
I
E
1 91
180 94 24= = = Ω
Ideally R0
2
180 90= = Ω
% error 100
R
R R
0
0
#= −
.
90
94 24 90 100#= −
.4 71= %
MCQ 1.67 A 1000 V DC supply has two 1-core cables as its positive and negative leads : their
insulation resistances to earth are 4 MΩ and 6 MΩ, respectively, as shown in the
figure. A voltmeter with resistance 50 kΩ is used to measure the insulation of the
cable. When connected between the positive core and earth, then voltmeter reads

PUBLISHING FOR GATE
(A) 8 V (B) 16 V
(C) 24 V (D) 40 V
The measurement system is shown below
Voltmeter reading
V
6 50 4
1000
M k M
(50 k 4 M )
z
z
Ω Ω Ω
Ω Ω=
+b l
.
.
6 049
1000 049#=
+
.8 10= V
MCQ 1.68 Two wattmeters, which are connected to measure the total power on a three-phase
system supplying a balanced load, read 10.5 kW and .2 5− kW, respectively. The
total power and the power factor, respectively, are
(A) 13.0 kW, 0.334 (B) 13.0 kW, 0.684
(C) 8.0 kW, 0.52 (D) 8.0 kW, 0.334
Total power P P P1 2= +
10.5 2.5= −
8= kW
Power factor cosθ=
Where
θ tan
P P
P P31
2 1
2 1
=
+
−−
b l; E
tan 3
8
131
#= −−
: D
.70 43c=−
Power factor .cos 0 334θ= =

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MCQ 1.69 The common emitter amplifier shown in the figure is biased using a 1 mA ideal
current source. The approximate base current value is
(A) 0 Aμ (B) 10 Aμ
(C) 100 Aμ (D) 1000 Aμ
Since the transistor is operating in active region.
IE IB. β
IB
IE
β
=
100
1 mA= 10 Aμ=
MCQ 1.70 Consider the inverting amplifier, using an ideal operational amplifier shown in the
figure. The designer wishes to realize the input resistance seen by the small-signal
source to be as large as possible, while keeping the voltage gain between 10− and
25− . The upper limit on RF is 1 MΩ. The value of R1 should be
(A) Infinity (B) 1 MΩ
(C) 100 kΩ (D) 40 kΩ
Gain of the inverting amplifier is given by,
Av
R
RF
1
=−
Av
R
1 10
1
6
#=− , 1R MF Ω=
R1
A
1 10
v
6
#=− 10Av =− to 25− so value of R1

PUBLISHING FOR GATE
R1
10
106
= 100 kΩ= for A 10v =−
R'
1
25
106
= 40 kΩ= for A 25v =−
R1 should be as large as possible so 100R k1 Ω=
MCQ 1.71 The typical frequency response of a two-stage direct coupled voltage amplifier is as
shown in figure
Direct coupled amplifiers or DC-coupled amplifiers provides gain at dc or very low
frequency also.
MCQ 1.72 In the given figure, if the input is a sinusoidal signal, the output will appear as
shown

PUBLISHING FOR GATE
Since there is no feedback in the circuit and ideally op-amp has a very high value
of open loop gain, so it goes into saturation (ouput is either V+ or V− ) for small
values of input.
The input is applied to negative terminal of op-amp, so in positive half cycle it
saturates to V− and in negative half cycle it goes to V+ .
MCQ 1.73 Select the circuit which will produce the given output Q for the input signals X1
and X2 given in the figure
SOL 1.73 (check)
From the given input output waveforms truth table for the circuit is drawn as
X1
X2
Q
1 0 1
0 0 1
0 1 0
In option (A), for 1, 0X Q1 = = so it is eliminated.
In option (C), for 0, 0X Q1 = = (always), so it is also eliminated.

PUBLISHING FOR GATE
In option (D), for 0, 1X Q1 = = , which does not match the truth table.
Only option (B) satisfies the truth table.
MCQ 1.74 If X1 and X2 are the inputs to the circuit shown in the figure, the output Q is
(A) X X1 2+ (B) X X1 2:
(C) X X1 2: (D) X X1 2:
In the given circuit NMOS Q1
and Q3
makes an inverter circuit. Q4
and Q5
are in
parallel works as an OR circuit and Q2
is an output inverter.
So output is
Q X X1 2= + .X X1 2=
MCQ 1.75 In the figure, as long as X 11 = and X 12 = , the output Q remains
(A) at 1 (B) at 0
(C) at its initial value (D) unstable
Let ( )Q t is the present state then from the circuit,
So, the next state is given by
( )Q t 1+ ( )Q t= (unstable)
MCQ 1.76 The figure shows the voltage across a power semiconductor device and the current

PUBLISHING FOR GATE
through the device during a switching transitions. If the transition a turn ON
transition or a turn OFF transition ? What is the energy lost during the transition?
(A) Turn ON, ( )VI t t
2 1 2+ (B) Turn OFF, ( )VI t t1 2+
(C) Turn ON, ( )VI t t1 2+ (D) Turn OFF, ( )VI t t
2 1 2+
In Ideal condition we take voltage across the device is zero.
average power loss during switching ( )VI t t
2 1 2= + (turn ON)
MCQ 1.77 An electronics switch S is required to block voltage of either polarity during its
OFF state as shown in the figure (a). This switch is required to conduct in only one
direction its ON state as shown in the figure (b)
Which of the following are valid realizations of the switch S?
(A) Only P (B) P and Q
(C) P and R (D) R and S

PUBLISHING FOR GATE
So in P thyristor blocks voltage in both polarities until gate is triggered and also
in R transistor along with diode can do same process.
MCQ 1.78 The given figure shows a step-down chopper switched at 1 kHz with a duty ratio
.D 0 5= . The peak-peak ripple in the load current is close to
(A) 10 A (B) 0.5 A
(C) 0.125 A (D) 0.25 A
Duty ratio α .0 5=
here T 10
1 10
1
3
3
#
= =−
−
sec
Ta 40
R
L
5
200 mH= = = msec
Ripple
( )( )
R
V
e
e e
1
1 1
/
/ ( ) /
s
T T
T T T T1
s
s a
=
−
− −α α
−
− − −
= G
( )I maxT
fL
V
4 4 10 200 10
100s
3 3
# # #
= = −
0.125 A=
MCQ 1.79 An electric motor, developing a starting torque of 15 Nm, starts with a load torque
of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec2
, the moment of inertia
of the system must be (neglecting viscous and coulomb friction)
(A) 0.25 kg-m2
(B) 0.25 Nm2
(C) 4 kg-m2
(D) 4 Nm2
Tst 15 Nm=
TL 7 Nm=
α 2 rad/sec2
=
T Iα=
so T 8T T NmLst= − =
I 4
2
8 kgm2
= =
MCQ 1.80 Consider a phase-controlled converter shown in the figure. The thyristor is fired at
an angle α in every positive half cycle of the input voltage. If the peak value of the
instantaneous output voltage equals 230 V, the firing angle α is close to

PUBLISHING FOR GATE
(A) 45c (B) 135c
(C) 90c (D) 83.6c
We know that Vrms 230 V=
so, Vm 230 2 V#=
If whether α 90c1
Then Vpeak sinV 230m α= =
sin230 2 α 230=
sinα
2
1=
angle α 135c=
Linked Answer Questions : Q.81 to Q.90 Carry Two Marks Each
Statement for Linked Answer Questions 81 and 82
A coil of inductance 10 H and resistance 40 Ω is connected as shown in the figure.
After the switch S has been in contact with point 1 for a very long time, it is moved
to point 2 at, t 0= .
MCQ 1.81 If, at t = 0+
, the voltage across the coil is 120 V, the value of resistance R is
(A) 0 Ω (B) 20 Ω
(C) 40 Ω (D) 60 Ω
SOL 1.81 When the switch is at position 1, current in inductor is given as

PUBLISHING FOR GATE
(0 )iL
−
20 40
120=
+
2= A
At 0t = , when switch is moved to position 1,inductor current does not change
simultaneously so
(0 )iL
+
(0 )iL= −
=2 A
Voltage across inductor at 0t = +
(0 )vL
+
120= V
By applying KVL in loop
120 2(40 20)R= + +
120 120 R= +
R 0 Ω=
MCQ 1.82 For the value as obtained in (a), the time taken for 95% of the stored energy to be
dissipated is close to
(A) 0.10 sec (B) 0.15 sec
(C) 0.50 sec (D) 1.0 sec
SOL 1.82 Let stored energy and dissipated energy are E1 and E2 respectively. Then
Current
i
i
1
2
2
2
E
E
1
2
= 0.95=
i2 . .i i0 95 0 971 1= =
Current at any time t, when the switch is in position (2) is given by
( )i t i e1 L
R t
= −
2 2e et t
10
60 6
= =− −
After 95% of energy dissipated current remaining in the circuit is
i 2 2 0.97 0.05#= − = A
So, 0.05 2e t6
= −
t 0.50. sec

PUBLISHING FOR GATE
A state variable system ( ) ( ) ( )t t tX X u
0
0
1
3
1
0
=
−
+o = =G G with the initial condition
(0) [ , ]X 1 3 T
= − and the unit step input ( )u t has
MCQ 1.83 The state transition matrix
(A)
( )e
e
1
0
1 t
t
3
1 3
3
− −
−= G (B)
( )e e
e
1
0
t t
t
3
1 3
−− −
−> H
(C)
( )e e
e
1
0
t t
t
3
1 3 3
3
−− −
−> H (D)
( )e
e
1
0
1 t
t
− −
−> H
SOL 1.83 Given state equation.
( )tXo ( ) ( )t tX u
0
0
1
3
1
0
=
−
+> >H H
Here
A ,B
0
0
1
3
1
0
=
−
=> >H H
State transition matrix is given by,
( )tφ [( ) ]sI AL 1 1
= −− −
[ ]sI A−
s
s0
0 0
0
1
3
= −
−> >H H
s
s0
1
3
=
−
+> H
[ ]sI A 1
− −
( )s s
s
s3
1 3
0
1
=
+
+
> H
( )
( )
s s s
s
1
0
3
1
3
1=
+
+
R
T
S
S
S
SS
V
X
W
W
W
WW
( )tφ [( ) ]sI AL 1 1
= −− −
e
1
0
t
t
3
1 3
3=
−
−
( )e1 −
> H
MCQ 1.84 The state transition equation
(A) ( )t
t e
e
X
t
t=
− -
-= G (B) ( )t
e
e
X
1
3
t
t3=
− -
-= G
(C) ( )t
t e
e
X
3
t
t
3
3=
−
-= G (D) ( )t
t e
e
X
t
t
3
=
− -
-= G
SOL 1.84 State transition equation is given by
( )sX ( ) (0) ( ) ( )s s B sX UΦ Φ= +

PUBLISHING FOR GATE
Here ( )s "Φ state transition matrix
( )sΦ
( )
( )
s s s
s
1
0
3
1
3
1=
+
+
R
T
S
S
S
SS
V
X
W
W
W
WW
(0)X " initial condition
(0)X
1
3
=
−
> H
B
1
0
= > H
So ( )sX
( )
( )
( )s s s
s
s s s
s
s
1
0
3
1
3
1
1
3
1
0
3
1
3
1
1
0
1=
+
+
−
+
+
+
R
T
S
S
S
SS
R
T
S
S
S
S
> >
V
X
W
W
W
WW
V
X
W
W
W
W
H H
( )s s s
s
s
s
1
3
3
0
3
3
1
0
1=
− +
+
+
+
+
R
T
S
S
S
S
>
V
X
W
W
W
W
H
s
s
s
3
1
3
3
1
0
2
=
−
+
+
+
R
T
S
S
S
S
>
V
X
W
W
W
W
H
( )sX s s
s
1
3
1
3
3
2
=
−
+
+
R
T
S
S
S
S
V
X
W
W
W
W
Taking inverse Laplace transform, we get state transition equation as,
( )tX
t e
e3
t
t
3
3=
− −
−> H
A 1000 kVA, 6.6 kV, 3-phase star connected cylindrical pole synchronous generator
has a synchronous reactance of 20 Ω. Neglect the armature resistance and consider
operation at full load and unity power factor.
MCQ 1.85 The induced emf(line-to-line) is close to
(A) 5.5 kV (B) 7.26 kV
(C) 9.6 kV (D) 12.5 kV
SOL 1.85 Given
P 1000 kVA= , 6.6 kV
Reactance 20 Ω= and neglecting the armature resistance at full load and unity

PUBLISHING FOR GATE
power factor
So
P V I3 L L=
I
.
.
3 6 6
1000 87 47
#
= = A
So,
IX . .87 47 20 1 75#= = kV
Eph
2 . ( . )
3
6 5 1 75
2
2
= +c m
Eph
. ( . )
3
6 5 1 75
2
2
= +c m
Eph 4.2 kV=
EL E3 ph= a Star connection
EL . .1 732 4 2#=
EL 7.26 kV=
MCQ 1.86 The power(or torque) angle is close to
(A) 13.9c (B) 18.3c
(C) 24.6c (D) 33.0c
SOL 1.86 Hence (C) is correct option.
Torque angle zα tan
R
X
a
s1
= −
b l
zα
.
.tan
6 6
3 1 751 #= −
c m
zα .24 6c=

PUBLISHING FOR GATE
At a 220 kV substation of a power system, it is given that the three-phase fault
level is 4000 MVA and single-line to ground fault level is 5000 MVA Neglecting the
resistance and the shunt suspectances of the system.
MCQ 1.87 The positive sequence driving point reactance at the bus is
(A) 2.5 Ω (B) 4.033 Ω
(C) 5.5 Ω (D) 12.1 Ω
SOL 1.87 Given data
Substation Level 220= kV
3-φ fault level 4000= MVA
LG fault level 5000= MVA
Positive sequence reactance:
Fault current If
3 220
4000
#
=
X1 /V Iph f=
3 220
4000
3
220
4000
220 220
#
#= =
12.1 Ω=
MCQ 1.88 The zero sequence driving point reactance at the bus is
(A) 2.2 Ω (B) 4.84 Ω
(C) 18.18 Ω (D) 22.72 Ω
SOL 1.88 Zero sequence Reactance ?X0 =
If
3 220
5000
#
=
Ia1 I I
I
3 3 3 220
5000
a a
f
2 0
#
= = = =
X X X1 2 0+ +
I
V
220 3 3
5000
3
220
a
ph
1
#
= =
X X X1 2 0+ + .
3 5000
220 220 29 04
#
# Ω= =
X1 .X 12 12 Ω= =
X0 29.04 12.1 12.1= − −
4.84 Ω=

PUBLISHING FOR GATE
Assume that the threshold voltage of the N-channel MOSFET shown in figure is +
0.75 V. The output characteristics of the MOSFET are also shown
MCQ 1.89 The transconductance of the MOSFET is
(A) 0.75 ms (B) 1 ms
(C) 2 ms (D) 10 ms
SOL 1.89 Trans-conductance of MOSFET is given by
gm
V
i
GS
D
2
2=
( )
( )
1
2 1
2 1
V
mA
=
−
−
= mS
MCQ 1.90 The voltage gain of the amplifier is
(A) 5+ (B) .7 5−
(C) 10+ (D) 10−
SOL 1.90 Voltage gain can be obtain by small signal equivalent circuit of given amplifier.
vo g v Rm gs D=−
vgs vin=
So, vo g R vm D in=−

PUBLISHING FOR GATE
Voltage gain Av
v
v
i
o
= g Rm D=−
(1 mS)(10 k )Ω=−
10=−
Answer Sheet
1. (C) 19. (B) 37. (C) 55. (B) 73. (B)
2. (A) 20. (A) 38. (C) 56. (B) 74. (D)
3. (D) 21. (C) 39. (B) 57. (A) 75. (D)
4. (B) 22. (A) 40. (C) 58. (D) 76. (A)
5. (D) 23. (A) 41. (D) 59. (C) 77. (C)
6. (C) 24. (C) 42. (B) 60. (D) 78. (C)
7. (A) 25. (D) 43. (B) 61. (C) 79. (C)
8. (A) 26. (C) 44. (A) 62. (B) 80. (B)
9. (D) 27. (B) 45. (C) 63. (D) 81. (A)
10. (D) 28. (A) 46. (B) 64. (B) 82. (C)
11. (C) 29. (B) 47. (A) 65. (C) 83. (A)
12. (A) 30. (C) 48. (D) 66. (D) 84. (C)
13. (D) 31. (C) 49. (B) 67. (A) 85. (B)
14. (B) 32. (A) 50. (A) 68. (D) 86. (C)
15. (A) 33. (A) 51. (B) 69. (B) 87. (D)
16. (B) 34. (C) 52. (D) 70. (C) 88. (B)
17. (D) 35. (A) 53. (C) 71. (B) 89. (B)
18. (D) 36. (A) 54. (*) 72. (C) 90. (D)

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