GATE EE 2003 Questions and Solutions

GATE EE
2003
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Q.1 - 30 Carry One Mark Each
MCQ 1.1 Figure Shows the waveform of the current passing through an inductor of resistance
1 Ω and inductance 2 H. The energy absorbed by the inductor in the first four
seconds is
(A) 144 J (B) 98 J
(C) 132 J (D) 168 J
SOL 1.1 The Correct option is (C).
Energy absorbed by the inductor coil is given as
EL Pdt
t
0
= #
Where power P VI= I L
dt
dI= b l
So, EL LI
dt
dI dt
t
0
= b l#
For0 4t# # sec
EL 2 I
dt
dI dt
0
4
= b l#
EL 2 ( ) 2 ( )I dt I dt3 0
0
2
2
4
= +# # ,
,
,dt
dI t
t
3 0 2
0 2 4< <
a # #=
=
*
EL 6 .I dt
0
2
= # =6(area under the curve ( )i t t− )

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6 2 6
2
1
# # #=
36= J
Energy absorbed by 1 Ω resistor is
ER I Rdt
t
2
0
= #
( ) ( )t dt dt3 1 62
0
2
2
2
4
#= +# # ,
A
I t t
t
3 0 2
6 2 4
# #
# #
=
=
)
9 36t
3
3
0
2
2
4
[ ]t#= +: D
24 72= + =96 J
Total energy absorbed in 4 sec
E E EL R= +
36 96= +
132= J
MCQ 1.2 A segment of a circuit is shown in figure 5 , 4 2sinv V v tR c= = .The voltage vL is
given by
(A) 3 8 cos t2− (B) 2sin t32
(C) sin t16 2 (D) 16 2cos t
SOL 1.2 The Correct option is (B).
Applying KCL at center node

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iL 1 2iC= + +
iL 3iC= +
iC C
dt
dvc
=−
1 [4 2 ]sin
dt
d t=−
8 2cos t=−
so iL 8 2 3cos t=− + (current through inductor)
Voltage across inductor
vL L
dt
diL
=
2 [3 8 2 ]cos
dt
d t#= −
32 2sin t=
MCQ 1.3 In the figure, 10 60 , 10 60 , 50 53.13 .Z Z Z1 2 3+ + += − = =% % %
Thevenin impedance
seen form X-Y is
(A) .56 66 45+ %
(B) 60 30+ %
(C) 70 30+ %
(D) .34 4 65+ %
SOL 1.3 Thevenin impedance can be obtain as following

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Zth ( || )Z Z Z3 1 2= +
given that Z1 10 60c+= − 10
j
2
1 3
=
−
c m
5(1 )j3= −
Z2 10 60c+=
10
j
2
1 3
=
+
c m
5(1 )j3= +
Z3 50 53.13c+=
50
j
5
3 4
=
+
b l
10(3 4 )j= +
So, Zth 10(3 4 )
( ) ( )
( ) ( )
j
j j
j j
5 1 3 5 1 3
5 1 3 5 1 3
= + +
− + +
− +
10(3 4 )
( )
j
10
25 1 3
= + +
+
30 40 10j= + +
40 40j= +
Zth 40 452 c+ Ω=
Hence (A) is correct option.
MCQ 1.4 Two conductors are carrying forward and return current of +I and I− as shown in
figure. The magnetic field intensity H at point P is
(A)
d
I Y
π
(B)
d
I X
π
(C)
d
I Y
2π
(D)
d
I X
2π
SOL 1.4 Due to the first conductor carrying I+ current, magnetic field intensity at point
P is
H1
d
I Y
2π
= (Direction is determined using right hand rule)
Similarly due to second conductor carrying I− current, magnetic field intensity is
H2 ( )
d
I Y
2π
= − −

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d
I Y
2π
=
Total magnetic field intensity at point P.
H H H1 2= +
d
I
d
IY Y
2 2π π
= +
d
I Y
π
=
MCQ 1.5 Two infinite strips of width w m in x -direction as shown in figure, are carrying
forward and return currents of +I and I− in the z - direction. The strips are
separated by distance of x m. The inductance per unit length of the configuration
is measured to be L H/m. If the distance of separation between the strips in snow
reduced to x/2 m, the inductance per unit length of the configuration is
(A) L2 H/m (B) /L 4 H/m
(C) /L 2 H/m (D) L4 H/m
SOL 1.5 The Correct option is ( ).
MCQ 1.6 A simple phase transformer has a maximum efficiency of 90% at full load and unity
power factor. Efficiency at half load at the same power factor is
(A) 86.7% (B) 88.26%
(C) 88.9% (D) 87.8%
SOL 1.6 Given that 1-φ transformer, maximum efficiency 90% at full load and unity power
factor
So
η
cos
cos
V I P P
V I
i c2 2 2
2 2 2
φ
φ
=
+ +
( . )
( . )
cos
cos
L F P P
L F
c2
2
i(Pu)φ
φ
=
+ +
where . .L F is the load fator.
At full load, load factor is

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. .L F 1
P
P
c
i
= =
cos 2φ 1= at unity power factor
so, 90%
P1 2
1 1
i
#=
+
Pi .0 0555= MVA
At half load, load factor is
L.F .
2
1 5= =
So,
η
. . ( . ) .
.
0 5 0 0555 0 5 0 0555
0 5 1 1002
# #
# #=
+
η . %87 8=
Hence (D) is correct option.
MCQ 1.7 Group-I lists different applications and Group-II lists the motors for these
applications. Match the application with the most suitable motor and choose the
right combination among the choices given thereafter
Group-I Group-II
P. Food mixer 1. Permanent magnet dc motor
Q. Cassette tape recorder 2. Single-phase induction motor
R. Domestic water pump 3. Universal motor
S. Escalator 4. Three-phase induction motor
5. DC series motor
6. Stepper motor
Codes:
P Q R S
(A) 3 6 4 5
(B) 1 3 2 4
(C) 3 1 2 4
(D) 3 2 1 4
SOL 1.7 In food mixer the universal motor is used and in cassette tap recorder permanent
magnet DC motor is used. The Domestic water pump used the single and three
phase induction motor and escalator used the three phase induction motor.
Hence (C) is correct option.
MCQ 1.8 A stand alone engine driven synchronous generator is feeding a partly inductive load.
A capacitor is now connected across the load to completely nullify the inductive
current. For this operating condition.
(A) the field current and fuel input have to be reduced

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(B) the field current and fuel input have to be increased
(C) the field current has to be increased and fuel input left unaltered
(D) the field current has to be reduced and fuel input left unaltered
SOL 1.8 Given a engine drive synchronous generator is feeding a partly inductive load. A
capacitor is connected across the load to completely nullify the inductive current.
Then the motor field current has to be reduced and fuel input left unaltered.
MCQ 1.9 Curves X and Y in figure denote open circuit and full-load zero power factor(zpf)
characteristics of a synchronous generator. Q is a point on the zpf characteristics
at 1.0 p.u. voltage. The vertical distance PQ in figure gives the voltage drop across
(A) Synchronous reactance (B) Magnetizing reactance
(C) Potier reactance (D) Leakage reactance
SOL 1.9 The Correct option is (A).
Given open circuit and full-load zero power factor of a synchronous generator. At
point Q the zero power factor at 1.0 pu voltage. The voltage drop at point PQ is
across synchronous reactance.
MCQ 1.10 No-load test on a 3-phase induction motor was conducted at different supply voltage
and a plot of input power versus voltage was drawn. This curve was extrapolated
to intersect the y-axis. The intersection point yields
(A) Core loss (B) Stator copper loss
(C) Stray load loss (D) Friction and windage loss
SOL 1.10 Given no load test on 3-φ induction motor, the graph between the input power
and voltage drop is shown in figure, the intersection point yield the friction and
windage loss.

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MCQ 1.11 Bundled conductors are mainly used in high voltage overhead transmission lines to
(A) reduces transmission line losses
(B) increase mechanical strength of the line
(C) reduce corona
(D) reduce sag
In case of bundled conductors, We know that self GMD of conductor is increased and
in a conductor critical disruptive voltage of line depends upon GMD of conductor.
Since GMD of conductor is increased this causes critical disruptive voltage is being
reduced and if critical disruptive voltage is reduced, the corona loss will also be
reduced.
MCQ 1.12 A power system consist of 300 buses out of which 20 buses are generator bus, 25
buses are the ones with reactive power support and 15 buses are the ones with fixed
shunt capacitors. All the other buses are load buses. It is proposed to perform a
load flow analysis in the system using Newton-Raphson method. The size of the
Newton Raphson Jacobian matrix is
(A) 553 553# (B) 540 540#
(C) 555 555# (D) 554 554#
Given that no. of buses n 300=
Generator bus 20=
Reactive power support buses 25=
Fixed buses with Shunt Capacitor 15=
Slack buses (ns ) 20 25 15 30= + − =
a Size of Jacobian Matrix is given as
2( ) 2( )n n n ns s#= − −
( ) ( )2 300 30 2 300 30#= − −
540 540#=
MCQ 1.13 Choose two appropriate auxiliary components of a HVDC transmission system
from the following
P. D.C line inductor

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Q. A.C line inductor
R. Reactive power sources
S. Distance relays on D.C line
T. Series capacitance on A.C. line
(A) P and Q (B) P and R
(C) Q and S (D) S and T
SOL 1.13 Auxiliary component in HVDC transmission system are DC line inductor and
reactive power sources.
Hence (B) is correct option.
MCQ 1.14 A round rotor generator with internal voltage .E 2 01 = pu and .X 1 1= pu is
connected to a round rotor synchronous motor with internal voltage .E 1 32 = pu
and 1.X 2= pu. The reactance of the line connecting the generator to the motor
is 0.5 pu. When the generator supplies 0.5 pu power, the rotor angle difference
between the machines will be
(A) 57.42c (B) 1c
(C) 32.58c (D) 122.58c
a Exchanged electrical power is being given as follows
P ( )sin
X
EV
d
1 2δ δ= −6 @ ...(1)
Given that P sPower upply by generator" 0.5= pu
E Voltage for rotar generator" 2.0= pu
V Voltage of motor rotor" 1.3= pu
Xd X Reactance of generator Reactance of motoreq= = +
Recatance of connecting line+
Xd 1.1 1.2 .5 2.80= + + =
1 2δ δ− Rotor angle difference ?= =
from eq(1), .0 5
.
. ( )sin
2 8
2 1 3
1 2
# δ δ= −
1 2& δ δ−
.
. .sin
2 6
2 8 0 51 #= −
b l
1 2& δ δ− .32 58=
MCQ 1.15 The interrupting time of a circuit breaker is the period between the instant of
(A) initiation of short circuit and the arc extinction on an opening operation
(B) energizing of the trip circuit and the arc extinction on an opening operation
(C) initiation of short circuit and the parting of primary arc contacts
(D) energizing of the trip circuit and the parting of primary arc contacts
SOL 1.15 Time period between energization of trip circuit and the arc extinction on an

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opening operation is known as the interrupting time of Circuit breaker.
MCQ 1.16 The variation of drain current with gate-to-source voltage (I VD GS− characteristic)
of a MOSFET is shown in figure. The MOSFET is
(A) an n-channel depletion mode device
(B) an n-channel enhancement mode device
(C) an p-channel depletion mode device
(D) an p-channel enhancement mode device
SOL 1.16 Since there exists a drain current for zero gate voltage ( )V 0GS = , so it is a depletion
mode device.
ID increases for negative values of gate voltages so it is a p-type depletion mode
device.
MCQ 1.17 In the circuit of figure, assume that the transistor has 99hfe = and .V 0 7BE = V.
The value of collector current IC of the transistor is approximately
(A) [3.3/3.3] mA (B) [3.3/(3.3+3.3)] mA
(C) [3.3/.33] mA (D) [3.3(33+3.3)] mA
Applying KVL in input loop,
( ) ( . )
4 (33 10 ) 0.7 (3.3 10 )( 1) 0
I V I
I h I
4 33 10 3 3 10 0B BE E
B fe B
3 3
3 3
# #
# #
− − − =
− − − + =
( 1)I h IE fe Ba = +
3.3 ( ) ( . )( ) I33 10 3 3 10 99 1 B
3 3
# #= + +6 @

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IB
.
.
33 10 3 3 10 100
3 3
3 3
# # #
=
+
IC h Ife B=
IC
[ . . ]
.
0 33 3 3 100
99 3 3 mA
#
#=
+
IC
. .
.
0 33 3 3
3 3 mA=
+
MCQ 1.18 For the circuit of figure with an ideal operational amplifier, the maximum phase
shift of the output vout with reference to the input vin is
(A) 0c (B) 90c−
(C) 90c+ (D) 180! c
SOL 1.18 Let the voltages at positive and negative terminals of op-amp are v vand+ -
respectively. Then by applying nodal equations.
R
v v
R
v vin out
1 1
− + −- -
0=
2 v-- v vin out= + ..(1)
Similarly,
R
v v
j C
v
1
0in
ω
−
+
−+ +
c m
0=
( )v v v j CRin ω− ++ + 0=
(1 )v j CRω++ Vin= ..(2)
By equation (1) & (2)
j CR
v
1
2 in
ω+
v vin out= + v va =+ -" (ideal op-amp)
v
j CR1
2 1in
ω+
−; E vout=
vout
( )
v
j CR
j CR
1
1
in
ω
ω
=
+
−
Phase shift in output is given by
θ ( ) ( )tan tanCR CR1 1
ω ω= − −- -

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θ ( ) ( )tan tanCR CR1 1
π ω ω= − −- -
θ 2 ( )tan CR1
π ω= − -
Maximum phase shift
θ π=
MCQ 1.19 Figure shows a 4 to 1 MUX to be used to implement the sum S of a 1-bit full
adder with input bits P and Q and the carry input Cin . Which of the following
combinations of inputs to , ,I I I0 1 2 and I3 of the MUX will realize the sum S ?
(A) ;I I C I I Cin in0 1 2 3= = = = (B) ;I I C I I Cin in0 1 2 3= = = =
(C) ;I I C I I Cin in0 3 1 2= = = = (D) ;I I C I I Cin in0 3 1 2= = = =
SOL 1.19 In given circuit MUX implements a 1-bit full adder, so output of MUX is given by.
F Sum A Q Cin5 5==
Truth table can be obtain as.
P Q Cin Sum
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
Sum is.
Sum P Q C PQ C P Q C P Q Cin in in in= + + +
Output of MUX can be written as
F P Q I PQ I PQ I PQ I0 1 2 3$ $ $ $= + + +
Inputs are,
I0 , , ,C I C I C I Cin in in in1 2 3= = = =
MCQ 1.20 When a program is being executed in an 8085 microprocessor, its Program Counter
contains

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(A) the number of instructions in the current program that have already been
executed
(B) the total number of instructions in the program being executed.
(C) the memory address of the instruction that is being currently executed
(D) the memory address of the instruction that is to be executed next
SOL 1.20 Program counter contains address of the instruction that is to be executed next.
MCQ 1.21 A control system is defined by the following mathematical relationship
( )
dt
d x
dt
dx x e6 5 12 1 t
2
2
2
+ + = − -
The response of the system as t " 3 is
(A) x 6= (B) x 2=
(C) .x 2 4= (D) x 2=−
SOL 1.21 Given system equation is.
6 5
dt
d x
dt
dx x2
2
+ + ( )e12 1 t2
= − −
Taking Laplace transform on both side.
( ) ( ) ( )s X s sX s X s6 52
+ + 12
s s
1
2
1= −
+: D
( ) ( )s s X s6 52
+ + 12
( 2)
2
s s
=
+; E
System transfer function is
( )X s
( )( )( )s s s s2 5 1
24=
+ + +
Response of the system as t " 3 is given by
( )lim f t
t " 3
lim ( )sF s
s 0
=
"
(final value theorem)
( 2)( 5)( 1)
24lims
s s s ss 0
=
+ + +" ; E
2 5
24
#
=
.2 4=
MCQ 1.22 A lead compensator used for a closed loop controller has the following transfer
function
(1 )
(1 )K
b
s
a
s
+
+
For such a lead compensator
(A) a b< (B) b a<

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(C) a Kb> (D) a Kb<
SOL 1.22 Transfer function of lead compensator is given by.
( )H s
b
s
K
a
s
1
1
=
+
+
a
a
k
k
( )H jω K
j
b
j
a
1
1
ω
ω
=
+
+
a
a
k
k
R
T
S
S
SS
V
X
W
W
WW
So, phase response of the compensator is.
( )hθ ω tan tan
a b
1 1ω ω= −− −
a ak k
1
tan
ab
a b
2
1
ω
ω ω
=
+
−−
J
L
K
KK
N
P
O
OO
( )
tan
ab
b a1
2
ω
ω
=
+
−−
; E
hθ should be positive for phase lead compensation
So, ( )hθ ω
( )
0tan
ab
b a
>1
2
ω
ω
=
+
−−
; E
a b<
Hence (A) is correct option
MCQ 1.23 A second order system starts with an initial condition of
2
3= G without any external
input. The state transition matrix for the system is given by
e
e0
0t
t
2-
-= G. The state of
the system at the end of 1 second is given by
(A)
.
.
0 271
1 100= G (B)
.
.
0 135
0 368= G
(C)
.
.
0 271
0 736= G (D)
.
.
0 135
1 100= G
SOL 1.23 Since there is no external input, so state is given by
( )tX ( ) (0)t Xφ=
( )t "φ state transition matrix
[0]X "initial condition
So ( )x t
e
e0
0 2
3
t
t
2
=
−
−> >H H
( )x t
e
e
2
3
t
t
2
=
−
−> H
At t 1= , state of the system
( )x t t 1=
e
e
2
2
2
1=
−
−> H
.
.
0 271
1 100
= > H

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Hence (A) is correct option
MCQ 1.24 A Manganin swap resistance is connected in series with a moving coil ammeter
consisting of a milli-ammeter and a suitable shunt in order to
(A) minimise the effect of temperature variation
(B) obtain large deflecting torque
(C) reduce the size of the meter
(D) minimise the effect of stray magnetic fields
SOL 1.24 In PMCC instruments, as temperature increases the coil resistance increases.
Swamp resistors are connected in series with the moving coil to provide temperature
compensation. Swamping resistors is made of Manganin, which has a zero-
temperature coefficient.
MCQ 1.25 The effect of stray magnetic field on the actuating torque of a portable instrument
is maximum when the operating field of the instrument and the stray fields are
(A) perpendicular (B) parallel
(C) inclined at 60%
(D) inclined at 30%
Effect of stray magnetic field is maximum when the operating field and stray fields
are parallel.
MCQ 1.26 A reading of 120 is obtained when standard inductor was connected in the circuit
of a Q-meter and the variable capacitor is adjusted to value of 300 pF. A lossless
capacitor of unknown value Cx is then connected in parallel with the variable
capacitor and the same reading was obtained when the variable capacitor is
readjusted to a value of 200 pF. The value of Cx in pF is
(A) 100 (B) 200
(C) 300 (D) 500
Let C1 300= pF
Q
C R
120 1
1ω
= =
Now when Cx is connected in parallel with variable resistor 'C 2001 = pF

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Q
( ' )C C R
120 1
x1ω
= =
+
So C1 'C Cx1= +
300 C200 x= +
Cx 100= pF
MCQ 1.27 Figure shows a thyristor with the standard terminations of anode (A), cathode (K),
gate (G) and the different junctions named J1, J2 and J3. When the thyristor is
turned on and conducting
(A) J1 and J2 are forward biased and J3 is reverse biased
(B) J1 and J3 are forward biased and J2 is reverse biased
(C) J1 is forward biased and J2 and J3 are reverse biased
(D) J1, J2 and J3 are all forward biased
SOL 1.27 When thyristor turned on at that time J2 junction will break. So J1, J2, J3 all are
in forward bias.
MCQ 1.28 Figure shows a MOSFET with an integral body diode. It is employed as a power
switching device in the ON and OFF states through appropriate control. The ON
and OFF states of the switch are given on the V IDS S− plane by

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SOL 1.28 The ON-OFF state of switch is given on V IDS S− plane as following
When V veDS =+ , diode conducts and I 0S =
V veDS =− , diode opens, but I 0S = , D ve"− potential.
MCQ 1.29 The speed/torque regimes in a dc motor and the control methods suitable for the
same are given respectively in List-II and List-I
List-I List-II
P. Field Control 1. Below base speed
Q. Armature Control 2. Above base speed
3. Above base torque
4. Below base torque
Codes:
(A) P-1, Q-3 (B) P-2, Q-1
(C) P-2, Q-3 (D) P-1, Q-4
P. Field control-Above base speed
Q. Armature control-below base torque
MCQ 1.30 A fully controlled natural commutated 3-phase bridge rectifier is operating with
a firing angle 30cα = , The peak to peak voltage ripple expressed as a ratio of the
peak output dc voltage at the output of the converter bridge is
(A) 0.5 (B) /3 2
(C) 1
2
3−c m (D) 3 1−
As we know in fully controlled rectifier.
VPP ( / )cosV V 6m m π α= − + 30a cα =
or VPP [1 ( /6 30 )]cosVm cπ= − +

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or
V
V
m
PP
.0 5=
Q. 31 - 90 carry two marks each
MCQ 1.31 In the circuit of figure, the magnitudes of VL and VC are twice that of VR. Given
that 50f Hz= , the inductance of the coil is
(A) 2.14 mH (B) 5.30 H
(C) 31.8 mH (D) 1.32 H
SOL 1.31 Given that magnitudes of VL and VC are twice of VR
VL VC= 2VR= (Circuit is at resonance)
Voltage across inductor
VL i j LR # ω=
Current iR at resonance
iR
R
5 0+=
%
5
5= 1 A=
so, VL Lω= 2VR=
Lω 2 5#= 5VR = V, at resonance
2 50 L# # #π 10=
L
314
10= 31.8= mH
MCQ 1.32 In figure, the potential difference between points P and Q is
(A) 12 V (B) 10 V

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(C) 6 V− (D) 8 V
Applying nodal analysis in the circuit
At node P
2 V V
2
10
8
P P
+ − + 0=
16 4 40V VP P+ − + 0=
5 24VP − 0=
& VP
5
24= Volt
At node Q
2
V V
4
10
6
0Q Q
=
−
+
−
24 3 30 2V VQ Q= − +
5 54VQ − 0=
& VQ
5
54= V
Potential difference between P-Q
VPQ V VP Q= −
5
24
5
54= − 6=− V
MCQ 1.33 Two ac sources feed a common variable resistive load as shown in figure. Under
the maximum power transfer condition, the power absorbed by the load resistance
RL is
(A) 2200 W (B) 1250 W
(C) 1000 W (D) 625 W
SOL 1.33 First obtain equivalent Thevenin circuit across load RL
Thevenin voltage

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j
V
j
V
6 8
110 0
6 8
90 0th thc c+ +
+
− +
+
− 0=
2 200 0Vth c+− 0=
& Vth 100 0c+= V
Thevenin impedance
Zth (6 8 ) || (6 8 )j jΩ Ω= + +
(3 4 )j Ω= +
For maximum power transfer
RL Zth= 3 42 2
= + 5 Ω=
Power in load
P i Reff L
2
=
P 5
j3 4 5
100 2
#=
+ +
( )
5
80
100 2
#= 625= Watt
MCQ 1.34 In figure, the value of R is
(A) 10 Ω (B) 18 Ω
(C) 24 Ω (D) 12 Ω
SOL 1.34 By applying mesh analysis in the circuit

PUBLISHING FOR GATE
10I1 = A, 5I2 =− A
Current in 2 Ω resistor
I2Ω ( )I I1 2= − −
10 ( 5)= − − 15= A
So, voltage VA 15 2 30#= = Volt
Now we can easily find out current in all branches as following
Current in resistor R is 5 A
5
R
100 40= −
R
5
60= 12 Ω=
MCQ 1.35 A balanced delta connected load of (8 6)j Ω+ per phase is connected to a 400 V,
50 Hz, 3-phase supply lines. If the input power factor is to be improved to 0.9 by
connecting a bank of star connected capacitor the required kVAR of the of the
bank is
(A) 42.7 (B) 10.2
(C) 28.8 (D) 38.4
Given data
A balanced delta connected load j8 6 2= + =
V2 400= volt
Improved Power Factor cos 2φ .90=
1φ 6/8 36.85tan 1
c= =−
^ h

PUBLISHING FOR GATE
2φ ( .9) 25.84cos 01
c= =−
I
.
40 36.86
Z
V
j8 6
400
10 36 86
400
c
c
+
+= =
+
= = −
I j32 24= −
Since Power factor is Improved by connecting a Y-connected capacitor bank like as
Phasor diagram is being given by as follows
In figure oa cos cosI I2 1φ φ= =l
.cosI 25 84cl 32=
.I 0 9#l 32=
Il .35 55=
ac 24= Amp. ( )sinIac 1φ=
ab 35.55 25.84sin sinI 2 cφ= =l
ab .15 49= Amp
Ic bc ac ab 24 15.49 8.51= = − = − = Amp
KVAR of Capacitor bank V I
1000
3 C# #=
.
1000
3 400 8 51# #=
.10 2= KVAR
MCQ 1.36 In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage
across the inductance at t 0= +
, is

PUBLISHING FOR GATE
(A) 2 V (B) 4 V
(C) 6− V (D) 8 V
SOL 1.36 Before 0t = , the switch was opened so current in inductor and voltage across
capacitor for 0t < is zero
(0 ) 0vc =−
, ( )i 0 0L =-
at 0t = , when the switch is closed, inductor current and capacitor voltage does not
change simultaneously so
(0 ) (0 ) 0v vc c= =+ −
, (0 ) (0 ) 0i iL L= =+ −
At 0t = +
the circuit is
Simplified circuit
Voltage across inductor (at 0t = +
)
(0 )vL
+
2
3 2
10
#=
+
4= Volt
MCQ 1.37 The h-parameters for a two-port network are defined by
E
I
h
h
h
h
I
E
1
2
11
21
12
22
1
2
== = =G G G
For the two-port network shown in figure, the value of h12 is given by
(A) 0.125 (B) 0.167
(C) 0.625 (D) 0.25

PUBLISHING FOR GATE
SOL 1.37 The Correct option is (D).
Given that E1 h I h E11 1 12 2= +
and I2 h I h E21 1 22 2= +
Parameter h12 is given as
h12
E
E
( )open circuitI2
1
01
=
=
At node A
E E E E E
2 2 4
A A A1 2− + − + 0=
E5 A E E2 21 2= + ...(1)
Similarly
E E E
2 2
A1 1− + 0=
2E1 EA= ...(2)
From (1) and (2)
5(2 )E1 2 2E E1 2= +
8E1 2E2=
h12
E
E
2
1
=
4
1=
MCQ 1.38 A point charge of +I nC is placed in a space with permittivity of .8 85 10 12
#
-
F/m
as shown in figure. The potential difference VPQ
between two points P and Q at
distance of 40 mm and 20 mm respectively fr0m the point charge is
(A) 0.22 kV (B) 225− V
(C) .2 24− kV (D) 15 V
VPQ V VP Q= −
KQ KQ
OP OQ
= −
40 10
9 10 1 10
20 10
9 10 1 10
3
9 9
3
9 9
#
# # #
#
# # #= −−
−
−
−

PUBLISHING FOR GATE
9 10
40
1
20
13
#= −: D 225=− Volt
MCQ 1.39 A parallel plate capacitor has an electrode area of 100 mm2
, with spacing of 0.1 mm
between the electrodes. The dielectric between the plates is air with a permittivity
of .8 85 10 12
#
-
F/m. The charge on the capacitor is 100 V. The stored energy in the
capacitor is
(A) 8.85 pJ (B) 440 pJ
(C) 22.1 nJ (D) 44.3 nJ
Energy stored in Capacitor is
E CV
2
1 2
=
C
d
A0ε=
.
.
0 1 10
8 85 10 100 10
3
12 6
#
# # #= −
− −
.8 85 10 12
#= −
F
E . ( )
2
1 8 85 10 10012 2
# # #= −
.44 3= nJ
MCQ 1.40 A composite parallel plate capacitor is made up of two different dielectric material
with different thickness (t1 and t2) as shown in figure. The two different dielectric
materials are separated by a conducting foil F. The voltage of the conducting foil is
(A) 52 V (B) 60 V
(C) 67 V (D) 33 V
SOL 1.40 The figure is as shown below
The Capacitor shown in Figure is made up of two capacitor C1 and C2 connected
in series.

PUBLISHING FOR GATE
C1 ,
t
A C
t
Ar r
1
0 1
2
2
0 2ε ε ε ε= =
Since C1 and C2 are in series charge on both capacitor is same.
Q1 Q2=
( )C V1001 − C V2= (Let V is the voltage of foil)
( )
t
A V100r
1
0 1ε ε −
t
A Vr
2
0 2ε ε=
.
( )V
0 5
3 100 − V
1
4=
V300 3− V2=
300 5 60V V&= = Volt
MCQ 1.41 Figure shows an ideal single-phase transformer. The primary and secondary coils
are wound on the core as shown. Turns ratio /N N 21 2 = .The correct phasors of
voltages ,E E1 2, currents ,I I1 2 and core flux Φ are as shown in
SOL 1.41 Correct Option is ( )
MCQ 1.42 To conduct load test on a dc shunt motor, it is coupled to a generator which is
identical to the motor. The field of the generator is also connected to the same
supply source as the motor. The armature of generator is connected to a load
resistance. The armature resistance is 0.02 p.u. Armature reaction and mechanical
losses can be neglected. With rated voltage across the motor, the load resistance

PUBLISHING FOR GATE
across the generator is adjusted to obtain rated armature current in both motor
and generator. The p.u value of this load resistance is
(A) 1.0 (B) 0.98
(C) 0.96 (D) 0.94
SOL 1.42 Hence (C) is correct option.
Given that:
The armature resistance in per unit is 0.2
so,
Ra .0 2=
back emf equation of motor is
Eb V I Ra a= −
given that no mechanical losses and armature reaction is neglected, so per unit
value of emf induced by motor is
Eb .0 98=
The DC shunt motor is mechanically coupled by the generator so the emf induced
by motor and generator is equal
Eg Eb=
so voltage generated by the generator is
V . . .0 98 1 0 2 0 96#= − =
per unit value of load resistance is equal to 0.96
MCQ 1.43 Figure shows a Y3− connected, 3-phase distribution transformer used to step
down the voltage from 11000 V to 415 V line-to-line. It has two switches S1 and
S2. Under normal conditions S1 is closed and S2 is open. Under certain special
conditions S1 is open and S2 is closed. In such a case the magnitude of the voltage
across the LV terminals a and c is
(A) 240 V (B) 480 V
(C) 415 V (D) 0 V

PUBLISHING FOR GATE
SOL 1.43 Given that when the switch S1 is closed and S2 is open then the 11000 V is step
down at 415 V output
Second time when the switch S1 is open and switch S2 is closed then 2-phase supply
is connected to the transformer then the ratio of voltage is
V
V
2
1
26.50
N
N
415
11000
2
1
= = =
The output terminal a and c are in opposite phase so cancelled with each other and
terminal is equal to zero volt.
MCQ 1.44 Figure shows an ideal three-winding transformer. The three windings 1, 2, 3 of the
transformer are wound on the same core as shown. The turns ratio : :N N N1 2 3 is
4:2:1. A resistor of 10 Ω is connected across winding-2. A capacitor of reactance
2.5 Ω is connected across winding-3. Winding-1 is connected across a 400 V, ac
supply. If the supply voltage phasor 400 0V1 += %
, the supply current phasor I1 is
given by
(A) ( 10 10)j− + A (B) ( 10 10)j− − A
(C) (10 10)j+ A (D) (10 10)j− A
SOL 1.44 Given that
N1: :N N2 3 is : :4 2 1
Resistance R 10 Ω=
V1 400 V=
so,
V
V
2
1
N
N
2
4
2
1
= =
V2
V
4
2 1
=
200 V=
and
V
V
3
1
N
N
1
4
3
1
= =
V3 100 V=
so current in secondary winding
I2
R
V
10
2002
= =

PUBLISHING FOR GATE
I2 20 A=
The current in third winding when the capacitor is connected
so
I3
.jX
V
j
j
2 5
100 40
c
3
=
−
=
−
=
When the secondary winding current I2 is referred to primary side i.e I'
1
So
I
I'
2
1
N
N
4
2
1
2
= =
I'
1
20 10
2
A= =
and winding third current I3 is referred to Primary side i.e I''
1. I3 flows to opposite
to I1
So
I
I''
3
1
− N
N
4
1
1
3
= =
I''
1 j10=−
So total current in primary winding is
I1 I I'' ''
1 2= +
10 10j A= −
MCQ 1.45 Following are some of the properties of rotating electrical machines
P. Stator winding current is dc, rotor winding current is ac.
Q. Stator winding current is ac, rotor winding current is dc.
R. Stator winding current is ac, rotor winding current is ac.
S. Stator has salient poles and rotor has commutator.
T. Rotor has salient poles and sliprings and stator is cylindrical.
U. Both stator and rotor have poly-phase windings.
DC machines, Synchronous machines and Induction machines exhibit some of the
above properties as given in the following table.
Indicate the correct combination from this table
DC
Machine
Synchronous
Machines
Induction
Machines
(A) P,S
(B) Q,U
(C) P,S
(D) R,S
Q,T
P,T
R,U
Q,U
R,U
R,S
Q,T
P,T
Given that:
P Stator winding current is dc, rotor winding current is ac

PUBLISHING FOR GATE
Q Stator winding current is ac, rotor winding current is dc
R Stator winding current is ac, rotor winding current is ac
S Stator has salient pole and rotor has commutator
T Rotor has salient pole and slip rings and stator is cylindrical
U Both stator and rotor have poly-phase windings
So
DC motor/machines:
The stator winding is connected to dc supply and rotor winding flows ac current.
Stator is made of salient pole and Commutator is connected to the rotor so rotor
winding is supply ac power.
Induction machines:
In induction motor the ac supply is connected to stator winding and rotor and
stator are made of poly-phase windings.
Synchronous machines:
In this type machines the stator is connected to ac supply but rotor winding is
excited by dc supply. The rotor is made of both salient pole and slip rings and
stator is made of cylindrical.
MCQ 1.46 When stator and rotor windings of a 2-pole rotating electrical machine are excited,
each would produce a sinusoidal mmf distribution in the airgap with peal values
Fs and Fr respectively. The rotor mmf lags stator mmf by a space angle δ at any
instant as shown in figure. Thus, half of stator and rotor surfaces will form one pole
with the other half forming the second pole. Further, the direction of torque acting
on the rotor can be clockwise or counter-clockwise.
The following table gives four set of statement as regards poles and torque. Select
the correct set corresponding to the mmf axes as shown in figure.
Stator
Surface
ABC forms
Stator
Surface
CDA forms
Rotor
Surface
abc forms
Rotor
surface
cda forms
Torque
is

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(A) North Pole
(B) South Pole
(C) North Pole
(D) South Pole
South Pole
North Pole
South Pole
North Pole
North Pole
North Pole
South Pole
South Pole
South Pole
South Pole
North Pole
North Pole
Clockwise
Counter
Clockwise
Counter
Clockwise
Clockwise
SOL 1.46 Given that
Fs is the peak value of stator mmf axis. Fr is the peak value of
rotor mmf axis. The rotor mmf lags stator mmf by space angle
δ. The direction of torque acting on the rotor is clockwise or counter clockwise.
When the opposite pole is produced in same half portion of stator and rotor
then the rotor moves. So portion of stator is north-pole in ABC and rotor abc is
produced south pole as well as portion surface CDA is produced south pole and the
rotor cda is produced North pole.
The torque direction of the rotor is clock wise and torque at surface is in counter
clockwise direction.
MCQ 1.47 A 4-pole, 3-phase, double-layer winding is housed in a 36-slot stator for an ac
machine with 60c phase spread. Coil span is 7 short pitches. Number of slots in
which top and bottom layers belong to different phases is
(A) 24 (B) 18
(C) 12 (D) 0
SOL 1.47 Given that:
A 4-pole, 3-φ, double layer winding has 36 slots stator with 60c phase spread, coil
span is 7 short pitched
so,
Pole pitch
pole
slot=
4
36 9= =
Slot/pole/phase 3=
so,
3-slots in one phase, if it is chorded by 2 slots then
Out of 3 2" have different phase
Out of 36 24" have different phase.
MCQ 1.48 A 3-phase induction motor is driving a constant torque load at rated voltage and
frequency. If both voltage and frequency are halved, following statements relate to
the new condition if stator resistance, leakage reactance and core loss are ignored
1. The difference between synchronous speed and actual speed remains same

PUBLISHING FOR GATE
2. The airgap flux remains same
3. The stator current remains same
4. The p.u. slip remains same
Among the above, current statements are
(A) All (B) 1, 2 and 3
(C) 2, 3 and 4 (D) 1 and 4
3-φ induction motor is driving a constant load torque at rated voltage and frequency.
Voltage and frequency are halved and stator resistance, leakage reactance and core
losses are ignored.
Then the motor synchronous speed and actual speed difference are same.
Ns
P
f120
=
The leakage reactance are ignored then the air gap flux remains same and the
stator resistance are ignored then the stator current remain same.
MCQ 1.49 A single-phase induction motor with only the main winding excited would exhibit
the following response at synchronous speed
(A) Rotor current is zero
(B) Rotor current is non-zero and is at slip frequency
(C) Forward and backward rotaling fields are equal
(D) Forward rotating field is more than the backward rotating field
1-φ induction motor main winding excited then the rotating field of motor changes,
the forward rotating field of motor is greater then the back ward rotating field.
MCQ 1.50 A dc series motor driving and electric train faces a constant power load. It is
running at rated speed and rated voltage. If the speed has to be brought down to
0.25 p.u. the supply voltage has to be approximately brought down to
(A) 0.75 p.u (B) 0.5 p.u
(C) 0.25 p.u (D) 0.125 p.u
A dc series motor driving a constant power load running at rated speed and rated
voltage. It’s speed brought down 0.25 pu. Then
Emf equation of dc series motor
E ( )V R Ra se= − +
R Ra se+ R=

PUBLISHING FOR GATE
so, E V IR= −
K Nφ=
then N
K
E
φ
=
In series motor Iφα
so, N
KI
V IR= −
At constant power load
E I# T W Const#= = ...(1)
T K I KI 2
φ= = ...(2)
If W is decreased then torque increases to maintain power constant.
T I 2

W
4
1= then T 4=
So current is increased 2 time and voltage brought down to 0.5 pu.
MCQ 1.51 The ABCD parameters of a 3-phase overhead transmission line are
0.9 0 , 200 90A D Bc c+ + Ω= = = and 0.95 10 90C S3
# += %-
. At no-load condition
a shunt inductive, reactor is connected at the receiving end of the line to limit the
receiving-end voltage to be equal to the sending-end voltage. The ohmic value of
the reactor is
(A) 3 Ω (B) 2000 Ω
(C) 105.26 Ω (D) 1052.6 Ω
Given that ABCD parameters of line as
0.9 0A D c+= = , 200 90B + Ω= %
, 0.95 10 90C S3
# += %-
.
at no-load condition,
Receiving end voltage( )VR sending end voltage(V )S=
ohmic value of reactor ?=
We know VS AV BIR R= +
aVS VR=
VR AV BIR R= +
(1 )V AR − BIR=
I
V
R
R
A
B
1
=
−
.1 0 9 0
200 90
c
c
+
+=
−
I
V
R
R
200 900 c+=
The ohmic value of reactor 2000 Ω=

PUBLISHING FOR GATE
MCQ 1.52 A surge of 20 kV magnitude travels along a lossless cable towards its junction
with two identical lossless overhead transmission lines. The inductance and the
capacitance of the cable are 0.4 mH and 0.5 Fμ per km. The inductance and
capacitance of the overhead transmission lines are 1.5 mH and 0.015 Fμ per km.
The magnitude of the voltage at the junction due to surge is
(A) 36.72 kV (B) 18.36 kV
(C) 6.07 kV (D) 33.93 kV
Surge impedance of cable
Z1
C
L= ; .L 0 4= mH/km, 0.5C Fμ= /km
.
. 28.284
0 5 10
0 4 10
6
3
#
#= =−
−
surge impedance of overhead transmission line
Z2 Z
C
L
3= = ; .L 1 5= mm/km, 0.015C F/kmμ=
Z Z2 3=
.
. 316.23
0 015 10
1 5 10
6
5
#
#= =−
−
Now the magnitude of voltage at junction due to surge is being given by as
Vl
Z Z
V Z2
2 1
2# #=
+
V 20= kV
.
.
316 28 284
2 20 10 316 233
# # #=
+
.36 72= kV
MCQ 1.53 A dc distribution system is shown in figure with load current as marked. The two
ends of the feeder are fed by voltage sources such that 3V VP Q− = V. The value
of the voltage VP for a minimum voltage of 220 V at any point along the feeder is
(A) 225.89 V (B) 222.89 V
(C) 220.0 V (D) 228.58 V
SOL 1.53 Let that current in line is I amp than
from figure current in line section PR is ( 10)I − amp
current in line section RS is ( 10 20) ( 30)I I− − = − amp
current in SQ Section is ( ) ( )I I30 30 60− − = − amp

PUBLISHING FOR GATE
Given that VP and VQ are such that
V VP Q− 3 V=
by applying KVL through whole line
V VP Q− ( 10)0.1 ( 30)0.15 ( 60) .2I I I 0#= − + − + −
3& .45 17.5I0= −
I
.
. 45.55
0 45
20 5= = amp
Now the line drop is being given as
( 10) .1 ( 30) .15 ( 60) .2I I I0 0 0= − + − + −
(33.55)0.1 (15.55)0.15 (14.45)0.2= + +
.8 58= V
The value of VP for minimum voltage of 220 V at any feeder is
220 Line voltage= +
220 8.58= +
228.58= V
MCQ 1.54 A 3-phase 11 kV generator feeds power to a constant power unity power factor
load of 100 MW through a 3-phase transmission line. The line-to line voltage at the
terminals of the machine is maintained constant at 11 kV. The per unit positive
sequence impedance of the line based on 100 MVA and 11 kV is 0.2j . The line to
line voltage at the load terminals is measured to be less than 11 kV. The total
reactive power to be injected at the terminals of the load to increase the line-to-line
voltage at the load terminals to 11 kV is
(A) 100 MVAR (B) 10.1 MVAR
(C) 100− MVAR (D) .10 1− MVAR
Given Load Power 100= MW
V VS R= 11= kV
Impedance of line ZL
MV
p.u. (kV)2
#=
. ( )j
100
0 2 11 2
#= .242j0 Ω=
We know PL
sin
X
V VS R δ
=
100 106
# .
sin
0 242
11 10 11 103 3
# # # δ=
.
121
100 0 242# sinδ=

PUBLISHING FOR GATE
δ (0.2) 11.537sin 1
c= =−
Reactive Power is being given by
QL cos
X
V V
X
VS R R
2
δ= −
.
(11.537 )
.
( )
cos
0 242
11 10 11 10
0 242
11 103 3 3 2
# # # #c= −
.
[ (11.537 ) 1]cos
0 242
121 106
# c= −
.10 1=− MVAR
MCQ 1.55 The bus impedance matrix of a 4-bus power system is given by
Zbus
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
0 3435
0 2860
0 2723
0 2277
0 2860
0 3408
0 2586
0 2414
0 2723
0 2586
0 2791
0 2209
0 2277
0 2414
0 2209
0 2791
=
R
T
S
S
S
S
S
V
X
W
W
W
W
W
A branch having an impedance of 0.2j Ω is connected between
bus 2 and the reference. Then the values of Z22,new and Z23,new of the bus impedance
matrix of the modified network are respectively
(A) .j0 5408 Ω and .j0 4586 Ω
(B) 0.j 1260 Ω and .j0 0956 Ω
(C) .j0 5408 Ω and .j0 0956 Ω
(D) 0.j 1260 Ω and 0.1j 630 Ω
Given the bus Impedance Matrix of a 4-bus Power System
Zbus
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
0 3435
0 2860
0 2723
0 2277
0 2860
0 3408
0 2586
0 2414
0 2723
0 2586
0 2791
0 2209
0 2277
0 2414
0 2209
0 2791
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
Now a branch os .j0 2 Ω is connected between bus 2 and reference
ZB(New) Z
Z Z
Z
Z
Z Z1
(B
ij b
ij
nj
ji jnOld) h g= −
+
R
T
S
S
SS
8
V
X
W
W
WW
B
New element 0.2Z jb Ω= is connected in jth
and reference bus
2j = , n 4= so
Z Z
Z
Z
Z
Z
Z Z Z Z1
ij b
12
22
23
24
21 22 23 24
+
R
T
S
S
S
S
SS
8
V
X
W
W
W
W
WW
B

PUBLISHING FOR GATE
( . ) .
.
.
.
.
. . . .
j j
j
j
j
j
j j j j
0 3408 0 2
1
0 2860
0 3408
0 2586
0 2414
0 2860 0 3408 0 2586 0 2414=
+
R
T
S
S
S
S
SS
6
8
V
X
W
W
W
W
WW
@
B ...(1)
Given that we are required to change only ,Z Z22 23
So in equation (1) Z 22l
( . )
( . )
.2147
j
j
j
0 5408
0 3408
0
2 2
= =
Z 23l
.
( . )( . )
.16296
j
j
0 5408
0 3408 0 2586
0
2
= =
Z22( )New Z Z22( ) 22Old= − l .3408 .2147 .1260j j j0 0 0= − =
Z23( )New Z Z23 23(Old)= − l 0.2586 0.16296 0.0956j j j= − =
MCQ 1.56 A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line.
The per unit positive-sequence, negative-sequence and zero-sequence impedances
of the alternator are .1j0 , .1j0 and 0.j 04 respectively. The neutral of the alternator
is connected to ground through an inductive reactor of 0.j 05 p.u. The per
unit positive-, negative- and zero-sequence impedances of transmission line are
.1j0 , .1j0 and 0.j 3, respectively. All per unit values are based on the machine
ratings. A solid ground fault occurs at one phase of the far end of the transmission
line. The voltage of the alternator neutral with respect to ground during the fault is
(A) 513.8 V (B) 889.9 V
(C) 1112.0 V (D) 642.2 V
SOL 1.56 Total zero sequence impedance, ve+ sequence impedance and ve− sequence
impedances
Z0 ( ) ( ) 0.04 0.3 0.34Z Z j j j pu0 0Line Generator= + = + =
Z1 ( ) ( ) 0.1 0.1 .2Z Z j j j0 pu1 1Line Generator= + = + =
Z2 ( ) ( ) 0.1 0.1 .2Z Z j j j0 pu2 2Line Generator= + = + =
Zn .05j0 pu=
for L-G fault
Ia1
Z Z Z Z
E
3 n
a
0 1 2
=
+ + +
. . . .
.
j j j j0 2 0 2 0 34 0 15
0 1=
+ + +
1.12j pu=−
IB
3 generator kV
generator MVA
=
.
175
3 6 6 10
20 10 0 Amp3
6
# #
#= =
Fault current
If ( )I I3 a B=
3( 1.12)(1750) 5897.6j j Amp= − =−
Neutral Voltage
Vn I Zf n=

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and Zn Z ZB pu#=
( . )
. .
20
6 6
0 05 0 1089
2
# Ω= =
Vn 5897.6 0.1089#=
Vn .642 2= V
MCQ 1.57 Incremental fuel costs (in some appropriate unit) for a power plant consisting of
three generating units are
. , . ,IC P IC P IC20 0 3 30 0 4 301 1 2 2 3= + = + =
Where P1 is the power in MW generated by unit i for ,i 1 2= and 3. Assume
that all the three units are operating all the time. Minimum and maximum loads
on each unit are 50 MW and 300 MW respectively. If the plant is operating on
economic load dispatch to supply the total power demand of 700 MW, the power
generated by each unit is
(A) 242.86 ; 157.14 ; 300P P PMW MW and MW1 2 3= = =
(B) 157.14 ; 242.86 ; 300P P PMW MW and MW1 2 3= = =
(C) 300 ; 300 ; 100P P PMW MW and MW1 2 3= = =
(D) 233.3 ; 233.3 ; 233.4P P PMW MW and MW1 2 3= = =
We know that Optimal Generation
IC1 IC2= , and P 3003 = MW (maximum load)
IC3 30= (Independent of load)
. P20 0 3 1+ . P30 0 4 2= +
. .P P0 3 0 41 2− 10= ...(1)
P P P1 2 3+ + 700=
P P 3001 2+ + 700=
P P1 2+ 400= ...(2)
From equation (1) and (2)
P1 242.8= MW
P2 .157 14= MW
MCQ 1.58 A list of relays and the power system components protected by the relays are given
in List-I and List-II respectively. Choose the correct match from the four choices
given below:
Group I Group II
P Distance relay 1. Transformers
Q Under frequency relay 2. Turbines
R Differential relay 3. Busbars
S Buchholz relay 4. Shunt capacitors

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5. Alternators
6. Transmission lines
Codes:
P Q R S
(A) 6 5 3 1
(B) 4 3 2 1
(C) 5 2 1 6
(D) 6 4 5 3
For transmission line protection-distance relay
For alternator protection-under frequency relay
For bus bar protection-differential relay
For transformer protection-Buchholz relay
MCQ 1.59 A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive
network. The maximum power that could be delivered by the generator is 2.0 p.u.
A three-phase fault occurs at the terminals of the generator which reduces the
generator output to zero. The fault is cleared after tc second. The original network
is then restored. The maximum swing of the rotor angle is found to be 110maxδ =
electrical degree. Then the rotor angle in electrical degrees at t tc= is
(A) 55 (B) 70
(C) 69.14 (D) 72.4
We know by equal area criteria
( )PS m 0δ δ− sinP dmax
C
m
δ δ=
δ
δ
#
( )sinPmax m0 0δ δ δ− [ ]cos cosPmax m0δ δ= − ...(1)
Pmax 2=
P0 sinP 1max 0δ= =
0δ 30c=
maxδ 110c= (given)
Now from equation (1)
( )sin2 30 110 30
180
c π− [ ]cos cos2 110c cδ= −

PUBLISHING FOR GATE
.50
180
80
#
π .342cos 0cδ= +
cos cδ .698 .3420 0= −
cδ .69 138c=
MCQ 1.60 A three-phase alternator generating unbalanced voltages is connected to an
unbalanced load through a 3-phase transmission line as shown in figure. The neutral
of the alternator and the star point of the load are solidly grounded. The phase
voltages of the alternator are 10 0 , 10 90 , 10 120E E EV V Va b cc c c+ + += = − = .
The positive-sequence component of the load current is
(A) 1.310 107c+ − A (B) 0.332 120c+ − A
(C) 0.996 120c+ − A (D) 3.510 81c+ − A
a Both sides are granted
So, Ia 5 90
Z
E
j2
10 0
a
a c c+ += = = −
Ib .
Z
E
j3
10 90 3 33 180
b
b c c+ += = − = −
Ic .
Z
E
j4
10 120 2 5 30
c
c c c+ += = =
We know Ia1
[ ]I I I
3
1
a b c
2
α α= + +
where 1 120 1 2402
&c c+ +α α= =
Ia1 [5 90 3.33 180 120 2.5 240 30 ]
3
1 c c c c c+ + += − + − + + +^ ^h h
Ia1 [ . . ]
3
1 5 90 3 33 60 2 5 270c c c+ + += − + − +
[ . . . ]j j j
3
1 5 1 665 2 883 2 5= − + − −
[ . . ]j
3
1 1 665 10 383= −
. .3 5 80 89c+= −
MCQ 1.61 For the n-channel enhancement MOSFET shown in figure, the threshold voltage

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2Vth = V. The drain current ID of the MOSFET is 4 mA when the drain resistance
RD is 1 kΩ.If the value of RD is increased to 4 kΩ, drain current ID will become
(A) 2.8 mA (B) 2.0 mA
(C) 1.4 mA (D) 1.0 mA
SOL 1.61 For a n-channel enhancement mode MOSFET transition point is given by,
V ( )DS sat V VGS TH= − 2V voltTHa =
V ( )DS sat V 2GS= −
From the circuit,
VDS VGS=
So V ( )DS sat 2V V V 2( )DS DS DS sat&= − = +
VDS V> ( )DS sat
Therefore transistor is in saturation region and current equation is given by.
ID ( )K V VGS TH
2
= −
4 ( 2)K VGS
2
= −
VGS is given by
V VGS DS= I R10 D D= −
10 4 1#= − 6 Volt=
So, 4 (6 2)K 2
= −
K
4
1=
Now RD is increased to 4 kΩ, Let current is I'
D and voltages are V V' '
DS GS=
Applying current equation.
I'
D ( )K V V'
GS TH
2
= −
I'
D ( 2)V
4
1 '
GS
2
= −
V V' '
GS DS= 10 I R' '
D D#= −
10 4I'
D= −
So,
4I'
D (10 4 2)I'
D
2
= − −
4I'
D (8 4 )I'
D
2
= −
4I'
D 16(2 )I'
D
2
= −
I'
D 4(4 ' 4 )I I'
D D
2
= + −
4 'I 17 16D
2
− + 0=

PUBLISHING FOR GATE
'I D
2
2.84 mA=
MCQ 1.62 Assuming the operational amplifier to be ideal, the gain /v vout in for the circuit
shown in figure is
(A) 1− (B) 20−
(C) 100− (D) 120−
SOL 1.62 Let the voltages at input terminals of op-amp are v- and v+ respectively.
So, 0v v= =+ - (ideal op-amp)
Applying node equation at negative terminal of op-amp,
v v
1
0
10
0in x− + − 0= ...(1)
At node x
v v v v
10
0
10 1
0x x out x− + − + − 0=
10v v v vx x out x+ − + 0=
12 vx vout=
vx
v
12
out
=
From equation (1),
v v
1 10
in x
+ 0=
vin
v
120
out
=−
v
v
in
out
120=−

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MCQ 1.63 A voltage signal sin t10 ω is applied to the circuit with ideal diodes, as shown in
figure, The maximum, and minimum values of the output waveform Vout of the
circuit are respectively
(A) 10 V+ and 10 V− (B) 4 V+ and 4 V−
(C) 7 V+ and 4 V− (D) 4 V+ and 7 V−
SOL 1.63 In the positive half cycle (when 4V V>in ) diode D2 conducts and D1 will be off so
the equivalent circuit is,
Vout 4 Volt= +
In the negative half cycle diode D1 conducts and D2 will be off so the circuit is,
Applying KVL
10I IV 4 10in − + − 0=
V
20
4in + I=
10V Vin =− (Maximum value in negative half cycle)
So, I
20
10 4
10
3 mA= − + =−

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V V
10
in out− I=
V
10
10 out− −
10
3=−
Vout ( )10 3=− −
Vout 7=− volt
MCQ 1.64 The circuit of figure shows a 555 Timer IC connected as an astable multi-vibrator.
The value of the capacitor C is 10 nF. The values of the resistors RA and RB for a
frequency of 10 kHz and a duty cycle of 0.75 for the output voltage waveform are
(A) 3.62 , 3.62R Rk kA BΩ Ω= =
(B) 3.62 , 7.25R Rk kA BΩ Ω= =
(C) 7.25 , 3.62R Rk kA BΩ Ω= =
(D) 7.25 , 7.25R Rk kA BΩ Ω= =
SOL 1.64 In the circuit, the capacitor charges through resistor ( )R RA B+ and discharges
through RB . Charging and discharging time is given as.
TC 0.693( )R R CA B= +
TD 0.693 R CB=
Frequency
. ( )
f
T T T R R C
1 1
0 693 2
1
D C A B
= =
+
=
+
. ( )R R0 693 2 10 10
1
A B
9
# #+ -
10 103
#=
.14 4 103
# R R2A B= + ...(1)
Duty cycle
T
TC
= .0 75=
. ( )
. ( )
R R C
R R C
0 693 2
0 693
A B
A B
+
+
4
3=
R R4 4A B+ R R3 6A B= +

PUBLISHING FOR GATE
RA R2 B= ...(2)
From (1) and (2)
R2 A .14 4 103
#=
RA 7.21 kΩ=
and RB 3.60 kΩ=
MCQ 1.65 The simplified block diagram of a 10-bit A/D converter of dual slope integrator
type is shown in figure. The 10-bit counter at the output is clocked by a 1 MHz
clock. Assuming negligible timing overhead for the control logic, the maximum
frequency of the analog signal that can be converted using this A/D converter is
approximately
(A) 2 kHz (B) 1 kHz
(C) 500 Hz (D) 250 Hz
SOL 1.65 Maximum frquency of input in dual slop A/D converter is given as
Tm T2n
C=
where fm
T
1 maximum frquency of input
m
"=
fC
T
1 clock frequency
C
"=
so fm
f
2n
C
= , n 10=
1
1024
10 kHz
6
= = (approax)
MCQ 1.66 The boolean expression X Y Z XY Z XYZ XY Z XYZ+ + + + can be simplified to
(A) XZ XZ YZ+ + (B) XY Y Z YZ+ +
(C) XY YZ XZ+ + (D) XY YZ XZ+ +
Given boolean expression can be written as,
F XYZ X Y Z XY Z XYZ XYZ= + + + +
( ) ( )X YZ Y Z X X XY Z Z= + + + +
XYZ Y Z XY= + +
( ) ( )( )Y Z Y X X Z A BC A B A Ca= + + + = + +

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( )( )Y Z Y X X X Z= + + +
( )Y Z Y X Z= + +
Y Z YX YZ= + +
MCQ 1.67 The shift register shown in figure is initially loaded with the bit pattern 1010.
Subsequently the shift register is clocked, and with each clock pulse the pattern
gets shifted by one bit position to the right. With each shift, the bit at the serial
input is pushed to the left most position (msb). After how many clock pulses will
the content of the shift register become 1010 again ?
(A) 3 (B) 7
(C) 11 (D) 15
X = X X1 05 , Y X2=
Serial Input [ ]Z X Y X X X1 0 25 5 5= =
Truth table for the circuit can be obtain as.
Clock pulse Serial Input Shif register
Initially 1 1010
1 0 1101
2 0 0110
3 0 0011
4 1 0001
5 0 1000
6 1 0100
7 1 1010
So after 7 clock pulses contents of the shift register is 1010 again.
MCQ 1.68 An X-Y flip-flop, whose Characteristic Table is given below is to be implemented

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using a J-K flip flop
(A) ,J X K Y= = (B) ,J X K Y= =
(C) ,J Y K X= = (D) ,J Y K X= =
Characteristic table of the X-Y flip flop is obtained as.
X Y Qn
Qn+1
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 0
1 1 0 0
Solving from k-map
Characteristic equation of X-Y flip flop is
Qn 1+ Y Q XQn n= +
Characteristic equation of a J-K flip-flop is given by
Qn 1+ KQ J Qn n= +
by comparing above two characteristic equations
J Y= , K X=
MCQ 1.69 A memory system has a total of 8 memory chips each with 12 address lines and 4
data lines, The total size of the memory system is

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(A) 16 kbytes (B) 32 kbytes
(C) 48 kbytes (D) 64 kbytes
Total size of the memory system is given by.
( )2 4 812
# #= bits
2 814
#= bits
214
= Bytes
16 K= bytes
MCQ 1.70 The following program is written for an 8085 microprocessor to add two bytes
located at memory addresses 1FFE and 1FFF
LXI H, 1FFE
MOV B, M
INR L
MOV A, M
ADD B
INR L
MOV M, A
XOR A
On completion of the execution of the program, the result of addition is found
(A) in the register A
(B) at the memory address 1000
(C) at the memory address 1F00
(D) at the memory address 2000
SOL 1.70 Executing all the instructions one by one.
LXI H,1FFE ( (H 1F) , L FE)H H& = =
MOV B,M B& Memory[HL] Memory[1FFE]= =
INR L L& ( (L 1) FF)H H= + =
MOV A,M A& Memory[HL] Memory[1FFF]= =
ADD B A& A B= +
INR L L& ( ( (L 1) FF) 1) 00H H H= + = + =
MOV M,A Memory[HL]& A=
Memory[1F00] A=
XOR A A& A XOR A=
0=
So the result of addition is stored at memory address 1F00.
MCQ 1.71 A control system with certain excitation is governed by the following mathematical
equation

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dt
d x
dt
dx x
2
1
18
1
2
2
+ + 10 5 2e et t4 5
= + +− −
The natural time constant of the response of the system are
(A) 2 sec and 5 sec (B) 3 sec and 6 sec
(C) 4 sec and 5 sec (D) 1/3 sec and 1/6 sec
SOL 1.71 Given equation
dt
d x
dt
dx x
2
1
18
1
2
2
+ + e e10 5 2t t4 5
= + +− −
Taking laplace on both sides we have
( ) ( ) ( )s X s sX s X s
2
1
18
12
+ +
s s s
10
4
5
5
2= +
+
+
+
( ) ( )s s X s
2
1
18
12
+ +
( )( )
( )( ) ( ) ( )
s s s
s s s s s s
4 5
10 4 5 5 5 2 4
=
+ +
+ + + + + +
System response is,
( )X s
( )( )
( )( ) ( ) ( )
s s s s s
s s s s s s
4 5
2
1
18
1
10 4 5 5 5 2 4
2
=
+ + + +
+ + + + + +
b l
( )( )
( )( ) ( ) ( )
s s s s s
s s s s s s
4 5
3
1
6
1
10 4 5 5 5 2 4
=
+ + + +
+ + + + + +
b bl l
We know that for a system having many poles, nearness of the poles towards
imaginary axis in s-plane dominates the nature of time response. So here time
constant given by two poles which are nearest to imaginary axis.
Poles nearest to imaginary axis
s1
3
1=− , s
6
1
2 =−
So, time constants
sec
sec
3
6
1
2
τ
τ
=
=
)
Hence (B) is correct option
MCQ 1.72 The block diagram shown in figure gives a unity feedback closed loop control
system. The steady state error in the response of the above system to unit step
input is
(A) 25% (B) 0.75 %
(C) 6% (D) 33%
SOL 1.72 Steady state error for a system is given by

PUBLISHING FOR GATE
ess
( ) ( )
( )
lim
G s H s
sR s
1s 0
=
+"
Where input ( )R s
s
1= (unit step)
( )G s
s s15
3
1
15=
+ +b bl l
( )H s 1=
(unity feedback)
So ess
( )( )
1
lim
s s
s
s
1
15 1
45s 0
=
+
+ +
"
b l
15 45
15=
+ 60
15=
%ess
60
15 100#= 25%=
MCQ 1.73 The roots of the closed loop characteristic equation of the system shown above
(Q-5.55)
(A) 1− and 15− (B) 6 and 10
(C) 4− and 15− (D) 6− and 10−
SOL 1.73 Characteristic equation is given by
( ) ( )G s H s1 + 0=
Here ( )H s 1= (unity feedback)
( )G s
s s15
3
1
15=
+ +b bl l
So, 1
s s15
3
1
15+
+ +b bl l 0=
( )( )s s15 1 45+ + + 0=
16 60s s2
+ + 0=
( 6)( )s s 10+ + 0=
s 6,=− 10−
MCQ 1.74 The following equation defines a separately excited dc motor in the form of a
differential equation
dt
d
J
B
dt
d
LJ
K
LJ
K Va
2 2
ω ω ω+ + =

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The above equation may be organized in the state-space form as follows
dt
d
dt
d P dt
d
QVa
2
2
ω
ω
ω
ω
= +
R
T
S
S
S
SS
>
V
X
W
W
W
WW
H
Where the P matrix is given by
(A)
1 0
J
B
LJ
K2
− −
= G (B)
0 1
LJ
K
J
B2
− −
= G
(C)
0 1
LJ
K
J
B2
− −= G (D)
1 0
J
B
LJ
K2
− −= G
SOL 1.74 Given equation can be written as,
dt
d
2
2
ω
J dt
d
LJ
K
LJ
K Va
2
β ω ω=− − +
Here state variables are defined as,
dt
dω x1=
ω x2=
So state equation is
x1o
J
B x
LJ
K x
LJ
K Va1
2
2=− − +
x2o
dt
dω= x1=
In matrix form
x
x
1
2
o
o> H
/ / /B J K LJ x
x
K LJ
V
1 0 0 a
2
1
2
=
− −
+> > >H H H
dt
d
dt
d
2
2
ω
ω
R
T
S
S
S
S
V
X
W
W
W
W
P
d
dt
QVa
ω
= +> H
So matrix P is
/ /B J K LJ
1 0
2
− −
> H
MCQ 1.75 The loop gain GH of a closed loop system is given by the following expression
( )( )s s s
K
2 4+ +
The value of K for which the system just becomes unstable is
(A) K 6= (B) K 8=
(C) K 48= (D) K 96=

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SOL 1.75 Characteristic equation of the system is given by
GH1 + 0=
( )( )s s s
K1
2 4
+
+ +
0=
( )( )s s s K2 4+ + + 0=
s s s K6 83 2
+ + + 0=
Applying routh’s criteria for stability
s3
1 8
s 2
6 K
s 1
K
6
48−
s 0
K
System becomes unstable if K
6
48 0− = 48K& =
MCQ 1.76 The asymptotic Bode plot of the transfer function /[1 ( / )]K s a+ is given in figure.
The error in phase angle and dB gain at a frequency of 0.5aω = are respectively
(A) 4.9c, 0.97 dB (B) 5.7c, 3 dB
(C) 4.9c, 3 dB (D) 5.7c, 0.97 dB
SOL 1.76 The maximum error between the exact and asymptotic plot occurs at corner
frequency.
Here exact gain(dB) at 0.5aω = is given by
( )gain dB . a0 5ω =
log logK
a
20 20 1 2
2
ω= − +
( . )
log logK
a
a
20 20 1
0 5 /
2
2 1 2
= − +; E

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.logK20 0 96= −
Gain(dB) calculated from asymptotic plot at . a0 5ω = is
logK20=
Error in gain (dB) 20 (20 0.96)log logK K dB= − − .0 96= dB
Similarly exact phase angle at 0.5aω = is.
( ) .
h
a0 5
θ ω ω =
tan
a
1 ω=− −
a k
.tan
a
a0 51
=− −
b l
2 .566 c=−
Phase angle calculated from asymptotic plot at ( 0.5 )aω = is 22.5c−
Error in phase angle
. ( . )22 5 26 56c=− − − .4 9c=
MCQ 1.77 The block diagram of a control system is shown in figure. The transfer function
( ) ( )/ ( )G s Y s U s= of the system is
(A)
s s18 1
12
1
3
1
+ +` `j j
(B)
s s27 1
6
1
9
1
+ +` `j j
(C)
s s27 1
12
1
9
1
+ +` `j j
(D)
s s27 1
9
1
3
1
+ +` `j j
SOL 1.77 Given block diagram
Given block diagram can be reduced as

PUBLISHING FOR GATE
Where G1
s
s
1 1
1
3=
+b
b
l
l
s 3
1=
+
G2
s
s
1 1 12
1
=
+b
b
l
l
s 12
1=
+
Further reducing the block diagram.
( )Y s
( )G G
G G
1 2 9
2
1 2
1 2
=
+
( ) ( )
( )
s s
s s
1 2
3
1
12
1 9
2
3
1
12
1
=
+
+ +
+ +
b b
b b
l l
l l
( )( )s s3 12 18
2=
+ + +
s s15 54
2
2=
+ +
( 9)( 6)s s
2=
+ +
s s27 1
9
1
6
1=
+ +a ak k
MCQ 1.78 The items in Group-I represent the various types of measurements to be made with
a reasonable accuracy using a suitable bridge. The items in Group-II represent the
various bridges available for this purpose. Select the correct choice of the item in
Group-II for the corresponding item in Group-I from the following
List-I List-II
P. Resistance in the milli- 1. Wheatstone Bridge
ohm range
Q. Low values of Capacitance 2. Kelvin Double Bridge
R. Comparison of resistance 3. Schering Bridge

PUBLISHING FOR GATE
which are nearly equal
S. Inductance of a coil with 4. Wien’s Bridge
a large time-constant
5. Hay’s Bridge
6. Carey-Foster Bridge
Codes :
(A) P=2, Q=3, R=6, S=5 (B) P=2, Q=6, R=4, S=5
(C) P=2, Q=3, R=5, S=4 (D) P=1, Q=3, R=2, S=6
Kelvin Double bridge is used for measuring low values of resistances. ( 2)P "
Low values of capacitances is precisely measured by schering bridge ( 3)Q "
Inductance of a coil with large time constant or high quality factor is measured by
hay’s bridge ( 5)R "
MCQ 1.79 A rectifier type ac voltmeter of a series resistance Rs , an ideal full-wave rectifier
bridge and a PMMC instrument as shown in figure. The internal. resistance of the
instrument is 100 Ω and a full scale deflection is produced by a dc current of 1 mA.
The value of Rs required to obtain full scale deflection with an ac voltage of 100 V
(rms) applied to the input terminals is
(A) 63.56 Ω (B) 69.93 Ω
(C) 89.93 Ω (D) 141.3 kΩ
SOL 1.79 Full scale deflection is produced by a dc current of 1 mA
( )Idc fs 1= mA
For full wave reactifier
( )Idc fs
I2 m
π
= , Im "peak value of ac current
1 mA
.
I
3 14
2 m
=
Im .1 57= mA
Full scale ac current
( )Irms fs
. .
2
1 57 1 11= = mA

PUBLISHING FOR GATE
V ( )( )R R Is m rms fs= +
100 ( 100)(1.11 )R mAs= +
(1.11 )
100
mA
R 100s= +
100 900# R 100s= +
Rs .89 9= kΩ
MCQ 1.80 A wattmeter reads 400 W when its current coil is connected in the R-phase and
its pressure coil is connected between this phase and the neutral of a symmetrical
3-phase system supplying a balanced star connected 0.8 p.f. inductive load. This
phase sequence is RYB. What will be the reading of this wattmeter if its pressure coil
alone is reconnected between the B and Y phases, all other connections remaining
as before ?
(A) 400.0 (B) 519.6
(C) 300.0 (D) 692.8
First the current coil is connected in R-phase and pressure coil is connected between
this phase and the neutral as shown below
reading of wattmeter
W1 cosI VP P 1θ= , . .cos 0 8 36 861 1& cθ θ= =
400 cosI V
3
L
L
1θ=
400 .I V
3
0 8L L
#= ...(1)
Now when pressure coil is connected between B and Y-phases, the circuit is

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phasor diagram
angle 2θ 23.14 30 .54 14c c c= + =
now wattmeter reading W2 cosV IYB L 2θ=
from equation (1) V IL L
.0 8
400 3#=
so W2
.
.cos
0 8
400 3 53 14# # c=
.519 5= W
MCQ 1.81 The inductance of a certain moving-iron ammeter is expressed as
10 3 ( /4)L H2
θ θ μ= + − , where θ is the deflection in radians from the zero position.
The control spring torque is 25 10 6
#
-
Nm/radian. The deflection of the pointer in
radian when the meter carries a current of 5 A, is
(A) 2.4 (B) 2.0

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(C) 1.2 (D) 1.0
SOL 1.81 In a moving-iron ammeter control torque is given as
cτ K I
d
dL
2
1 2
θ
θ
= =
Where
K " control spring constant
"θ deflection
Given that L 10 3
4
2
θ θ= + −
d
dL
θ
3
2
H/radθ μ= −b l
So,
(25 10 )c
6
#τ θ= −
( )
2
1 5 3
2
102 6
#
θ= − −
b l
2θ 3
2
θ= −
2
5 3 &θ θ= .
5
6 1 2= = rad.
MCQ 1.82 A 500A/5A, 50 Hz transformer has a bar primary. The secondary burden is a pure
resistance of 1 Ω and it draws a current of 5 A. If the magnetic core requires 250
AT for magnetization, the percentage ratio error is
(A) 10.56 (B) .10 56−
(C) 11.80 (D) .11 80−
Magnetizing current Im
1
250 250= = amp
Primary current Ip 500= amp
Secondary current Is 5= amp
Turn ratio n
I
I
5
500 100
s
p
= = =
Total primary current ( )IT [primary current(I )]p
2
= +
[magnetising current( )]I 2
m
IT I Ip m
2 2
= +
( ) ( )500 2502 2
= + .559 01= amp
Turn ratio n' . .
I
I
5
559 01 111 80
s
T
= = =
Percentage ratio error n3 100
n
n n
#= −
l
l

PUBLISHING FOR GATE
.
.
111 80
100 111 80 100#= − . %10 55=−
MCQ 1.83 The voltage-flux adjustment of a certain 1-phase 220 V induction watt-hour meter
is altered so that the phase angle between the applied voltage and the flux due to
it is 85c(instead of 90c). The errors introduced in the reading of this meter when
the current is 5 A at power factor of unity and 0.5 lagging are respectively
(A) 3.8 mW, 77.4 mW (B) .3 8− mW, .77 4− mW
(C) .4 2− W, .85 1− W (D) 4.2 W, 85.1 W
Power read by meter ( )sinP VIm 3 φ= −
Where
"3 Phase angle between supply voltage and pressure coil flux.
"φ Phase angle of load
Here
3 85 , 60c cφ= = .cos 0 5a φ ="
So measured power
Pm ( )sin200 5 85 60# c c= −
sin1100 25c=
.464 88= W
Actual power PO cosVI φ=
.220 5 0 5# #=
550= W
Error in measurement P Pm O= −
.464 88 550= −
.85 12=− W
For unity power factor cosφ 1=
φ 0c=
So Pm ( )sin220 5 85 0# c c= −
.1095 81= W
PO cos220 5 0# c=
1100=
Error in Measurement
.1095 81 1100= −
.4 19=− W
MCQ 1.84 Group-II represents the figures obtained on a CRO screen when the voltage signals
sinV V tx xm ω= and ( )sinV V ty ym ω Φ= + are given to its X and Y plates respectively
and Φ is changed. Choose the correct value of Φ from Group-I to match with the
corresponding figure of Group-II.

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Group-I
P. 0Φ =
Q. /2πΦ =
R. /3 2< <π πΦ
S. /3 2πΦ =
Group-II
Codes :
(A) P=1, Q=3, R=6, S=5 (B) P=2, Q=6, R=4, S=5
(C) P=2, Q=3, R=5, S=4 (D) P=1, Q=5, R=6, S=4
SOL 1.84 We can obtain the Lissaju pattern (in X-Y mode) by following method.
For 0cφ = , Vx sinV txm ω=
Vy ( 0 )sin sinV t tym cω ω= + =
Draw Vx and Vy as shown below

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Divide both Vy and Vx equal parts and match the corresponding points on the
screen.
Similarly for φ 90c=
Vx sinV txm ω=
Vy ( 90 )sinV tym cω= +
Similarly for
2
3φ π=

PUBLISHING FOR GATE
we can also obtain for 0
2
3< <φ π
MCQ 1.85 In the circuit shown in Fig. Q.85, the current gain ( )β of the ideal transistor is 10.
The operating point of the transistor ( , )V ICC C is
(A) (40 , 4 )V A (B) (40 , 5 )V A
(C) ( , 4 )V A0 (D) (15 ,4 )V A
SOL 1.85 The Correct option is ().
MCQ 1.86 A phase-controlled half-controlled single-phase converter is shown in figure. The
control angle 30cα =
The output dc voltage wave shape will be as shown in

PUBLISHING FOR GATE
SOL 1.86 The Correct option is ( ).
MCQ 1.87 A chopper is employed to charge a battery as shown in figure. The charging current
is 5 A. The duty ratio is 0.2. The chopper output voltage is also shown in the figure.
The peak to peak ripple current in the charging current is
(A) 0.48 A (B) 1.2 A
(C) 2.4 A (D) 1 A
In the chopper during turn on of chopper V -t area across L is,
V dtL
T
0
on
# L
dt
di dt Ldi
T
i
i
0 min
maxon
= =b l# #
( )L i i L Imax min Δ= − = ^ h
V -t are applied to ‘L’ is (60 12)Ton= −
T48 on=

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So now volt area
IΔ
L
T48 on
= . 0.48
20 10
48 0 2 10
3
3
#
# #= =−
−
A
MCQ 1.88 An inverter has a periodic output voltage with the output wave form as shown in
figure
When the conduction angle 120cα = , the rms fundamental component of the
output voltage is
(A) 0.78 V (B) 1.10 V
(C) 0.90 V (D) 1.27 V
Output voltage V0
4 /sin sin sin
n
V nd n t n 2
, ,
S
n 1 3 5 π
ω π=
3
=
b ^ ^ ^l h h h/
` RMS value of fundamental component
Vrms(fundamental) sinV d
2
4 1S
#
π
=
120cα = , 120 60d d2 &c c= =
Vrms(fundamental) sinV
2
4 60S
# c
π
=
0.78 0.78V VS= =
MCQ 1.89 With reference to the output wave form given in above figure , the output of the
converter will be free from 5th
harmonic when
(A) 72cα = (B) 36cα =
(C) 150cα = (D) 120cα =
After removing 5th
harmonic
d5 , ,0 2π π=

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` Pulse width , ,d2 0
5
2
5
4α π π= = =
, ,0 72 144c c c=
MCQ 1.90 An ac induction motor is used for a speed control application. It is driven from an
inverter with a constant /V f control. The motor name-plate details are as follows
(no. of poles = 2)
:415 :3 :50 :2850V V f NV V Hz rpmPh
The motor runs with the inverter output frequency set at 40 Hz, and with half the
rated slip. The running speed of the motor is
(A) 2400 rpm (B) 2280 rpm
(C) 2340 rpm (D) 2790 rpm
NSa 3000 rpm=
Na 2850 rpm=
SFL .
3000
3000 2850 0 05= − =
where by ( / )V f control
Nsb 3000
50
40= b l
2400 rpm=
` N2 new running speed of motor=
1 .2400
2
0 05= −b l
2340 rpm=

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

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