Gate ee 2007 with solutions

GATE EE
2007
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Q.1 - Q.20 carry one mark each
MCQ 1.1 The common emitter forward current gain of the transistor shown is 100Fβ = .
The transistor is operating in
(A) Saturation region (B) Cutoff region
(C) Reverse active region (D) Forward active region
SOL 1.1 If transistor is in normal active region, base current can be calculated as following,
By applying KVL for input loop,
10 (1 ) 0.7 270I I10 10C B
3 3
# #− − − 0=
270I IB Bβ + .9 3= mA, I IC B` β=
( 270)IB β + .9 3= mA
IB
9.3
270 100
mA=
+
0. 250 mA=
In saturation, base current is given by,
10 (1) (1)I V IC CE E− − − 0=
2
10 IC (sat)= I I
V 0
C E
CE
-
-
IC (sat) 5= mA
IB(sat)
IC (sat)
β
=
.050
100
5 mA= =

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I I )B B(sat1 , so transistor is in forward active region.
Hence (D) is correct option.
MCQ 1.2 The three-terminal linear voltage regulator is connected to a 10 Ω load resistor as
shown in the figure. If Vin is 10 V, what is the power dissipated in the transistor ?
(A) 0.6 W (B) 2.4 W
(C) 4.2 W (D) 5.4 W
SOL 1.2 In the circuit
We can analyze that the transistor is operating in active region.
VBE(ON) 0.6= volt
V VB E− .0 6=
. V6 6 E− .0 6=
VE 6.6 0.6 6= − = volt
At emitter (by applying KCL),
IE I IB L= +
IE
1
.6 6 6
10
6
kΩ Ω
= − + 0.6 amp-
10 6 4V V VCE C E= − = − = volt
Power dissipated in transistor is given by.
PT V ICE C#=
4 0.6#=
0.6I IC E` - = amp
2.4= W
Hence (B) is correct option.
MCQ 1.3 Consider the transformer connections in a part of a power system shown in the

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figure. The nature of transformer connections and phase shifts are indicated for all
but one transformer
Which of the following connections, and the corresponding phase shift θ, should be
used for the transformer between A and B ?
(A) Star-star ( 0 )θ = %
(B) Star-Delta ( )30θ =− %
(C) Delta-star ( )30θ = %
(D) Star-Zigzag ( )30θ = %
SOL 1.3 a Equal Phase shift of point A & B with respect to source from both bus paths.
So the type of transformer Y-Y with angle 0c.
Hence (A) is correct option.
MCQ 1.4 The incremental cost curves in Rs/MWhr for two generators supplying a common
load of 700 MW are shown in the figures. The maximum and minimum generation
limits are also indicated. The optimum generation schedule is :
(A) Generator A : 400 MW, Generator B : 300 MW
(B) Generator A : 350 MW, Generator B : 350 MW
(C) Generator A : 450 MW, Generator B : 250 MW
(D) Generator A : 425 MW, Generator B : 275 MW
SOL 1.4 Given incremental cost curve

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P PA B+ 700= MW
For optimum generator ?PA = , ?PB =
a From curve, maximum incremental cost for generator A
600= at 450 MW
and maximum incremental cost for generator B
800= at 400 MW
minimum incremental cost for generator B
650= at 150 MW
a Maximim incremental cost of generation A is less than the minimum incremental
constant of generator B. So generator A operate at its maximum load 450= MW
for optimum generation.
PA 450= MW
PB (700 450)= −
250= MW
Hence (C) is correct option.
MCQ 1.5 Two regional systems, each having several synchronous generators and loads are
inter connected by an ac line and a HVDC link as shown in the figure. Which of
the following statements is true in the steady state :
(A) Both regions need not have the same frequency
(B) The total power flow between the regions ( )P Pac dc+ can be changed by

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controlled the HDVC converters alone
(C) The power sharing between the ac line and the HVDC link can be changed by
controlling the HDVC converters alone.
(D) The directions of power flow in the HVDC link (Pdc) cannot be reversed
SOL 1.5 Here power sharing between the AC line and HVDC link can be changed by
controlling the HVDC converter alone because before changing only grid angle we
can change the power sharing between the AC line and HVDC link.
MCQ 1.6 Considered a bundled conductor of an overhead line consisting of three identical
sub-conductors placed at the corners of an equilateral triangle as shown in the
figure. If we neglect the charges on the other phase conductor and ground, and
assume that spacing between sub-conductors is much larger than their radius, the
maximum electric field intensity is experienced at
(A) Point X (B) Point Y
(C) Point Z (D) Point W
SOL 1.6 We have to find out maximum electric field intensity at various points. Electric
field intensity is being given by as follows

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From figures it is cleared that at point Y there is minimum chances of cancelation.
So maximum electric field intensity is at point Y.
MCQ 1.7 The circuit shown in the figure is
(A) a voltage source with voltage
R R
rV
1 2<
(B) a voltage source with voltage
R
r R
V
1
2<
(C) a current source with current
R R
r R
r
V
1 2
2<
+c m
(D) a current source with current
R R
R
r
V
1 2
2
+c m
SOL 1.7 This is a voltage-to-current converter circuit. Output current depends on input
voltage.
Since op-amp is ideal v v v1= =+ -
By writing node equation.
R
v v
R
v 0
1
1
2
1− + − 0=
v
R R
R R
1 2
1 2
1
+
c m R
V
1
=
v1 V
R R
R
1 2
2
=
+c m
Since the op-amp is ideal therefore
iL i
r
v
1
1
= =

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iL
r
V
R R
R
1 2
2
=
+c m
MCQ 1.8 The system shown in the figure is
(A) Stable
(B) Unstable
(C) Conditionally stable
(D) Stable for input u1, but unstable for input u2
SOL 1.8 For input u1, the system is ( )u 02 =
System response is
( )H s1
( )
( )
( )
( )
( )
s
s
s
s
s
1
2
1
1
1
2
1
=
+
+
−
−
+
−
( )
( )
s
s
3
1
=
+
−
Poles of the system is lying at s 3=− (negative s-plane) so this is stable.
For input u2 the system is ( )u 01 =
System response is
( )H s2
( )( )
( )
( )
s s
s
s
1
1
1
2
1
1
1
=
+
− +
−
−
( )( )
( )
s s
s
1 3
2
=
− +
+

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One pole of the system is lying in right half of s-plane, so the system is unstable.
MCQ 1.9 Let a signal ( )sina t1 1ω φ+ be applied to a stable linear time variant system. Let the
corresponding steady state output be represented as ( )a F t2 2 2ω φ+ . Then which of
the following statement is true?
(A) F is not necessarily a “Sine” or “Cosine” function but must be periodic with
1 2ω ω=
(B) F must be a “Sine” or “Cosine” function with a a1 2=
(C) F must be a “Sine” function with 1 2ω ω= and 1 2φ φ=
(D) F must be a “Sine” or “Cosine” function with 1 2ω ω=
SOL 1.9 For an LTI system input and output have identical wave shape (i.e. frequency of
input-output is same) within a multiplicative constant (i.e. Amplitude response is
constant)
So F must be a sine or cosine wave with 1 2ω ω=
MCQ 1.10 The frequency spectrum of a signal is shown in the figure. If this is ideally sampled
at intervals of 1 ms, then the frequency spectrum of the sampled signal will be

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SOL 1.10 The spectrum of sampled signal ( )s jω contains replicas of ( )U jω at frequencies nfs! .
Where n , , .......0 1 2=
fs 1
secm
kHz
T
1
1
1
s
= = =

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Hence (D) is correct Option
MCQ 1.11 Divergence of the vector field
( , , ) ( ) ( ) ( )cos cos sinV x y z x xy y i y xy j z x y k2 2 2
=− + + + + +t t t is
(A) cosz z2 2
(B) sin cosxy z z2 2
+
(C) sin cosx xy z− (D) None of these
SOL 1.11 Divergence of a vector field is given as
Divergence V4:=
In cartesian coordinates
4
x
i
y
j
z
k
2
2
2
2
2
2= + +t t t
So V4: ( ) ( )cos cos
x
x xy y
y
y xy
2
2
2
2= − + + +6 6@ @
( )sin
z
z x y2 2 2
2
2 + +6 @
( ) ( )sin sin cosx xy y y xy x z z2 2
=− − + − +
cosz z2 2
=
MCQ 1.12 x x xx n1 2
T
g= 8 B is an n-tuple nonzero vector. The n n# matrix
V xxT
=
(A) has rank zero (B) has rank 1
(C) is orthogonal (D) has rank n
SOL 1.12 Hence ( ) is correct option.
MCQ 1.13 A single-phase fully controlled thyristor bridge ac-dc converter is operating at a

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firing angle of 25c and an overlap angle of 10c with constant dc output current of
20 A. The fundamental power factor (displacement factor) at input ac mains is
(A) 0.78 (B) 0.827
(C) 0.866 (D) 0.9
SOL 1.13 Hence (A) is correct option.
Firing angle α 25c=
Overlap angle μ 10c=
so, I0 [ ( )]cos cos
Ls
Vm
ω
α α μ= − +
` 20 [ ( )]cos cos
Ls2 50
230 2 25 25 10
#
c c c
π
= − +
` Ls 0.0045 H=
V0
cosV LsI2 m 0
π
α
π
ω= −
. .
. .cos
3 14
2 230 2 25
3 14
2 3 14 50 4 5 10 203
# # # # # #c= −
−
.187 73 9= − .178 74c=
Displacement factor
V I
V I
s s
0 0
=
.
230 20
178 25 20
#
#= .0 78=
MCQ 1.14 A three-phase, fully controlled thyristor bridge converter is used
as line commutated inverter to feed 50 kW power 420 V dc to a
three-phase, 415 V(line), 50 Hz ac mains. Consider dc link current to be constant.
The rms current of the thyristor is
(A) 119.05 A (B) 79.37 A
(C) 68.73 A (D) 39.68 A
SOL 1.14 Given that
P 50 1000 W#=
Vd 420=
So P V Id d#=
Id
420
50 1000#= .119 05=
RMS value of thyristor current . .
3
119 05 68 73= =
MCQ 1.15 In a transformer, zero voltage regulation at full load is
(A) not possible

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(B) possible at unity power factor load
(C) possible at leading power factor load
(D) possible at lagging power factor load
SOL 1.15 In transformer zero voltage regulation at full load gives leading power factor.
MCQ 1.16 The dc motor, which can provide zero speed regulation at full load without any
controller is
(A) series (B) shunt
(C) cumulative compound (D) differential compound
SOL 1.16 Speed-armature current characteristic of a dc motor is shown as following
The shunt motor provides speed regulation at full load without any controller.
MCQ 1.17 The probes of a non-isolated, two channel oscillocope are clipped to points A,
B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable
low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the
“Signal” and “Ground” probes S G,1 1and ,S G2 2 of Ch1 and Ch2 respectively are
connected to points :
(A) A, B, C, A (B) A, B, C, B
(C) C, B, A, B (D) B, A, B, C

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SOL 1.17 Since both the waveform appeared across resistor and inductor are same so the
common point is B. Signal Probe S1 is connecte with A, S2 is connected with C and
both the grount probes G1 and G2 are connected with common point B.
MCQ 1.18 A single phase full-wave half-controlled bridge converter feeds an inductive load.
The two SCRs in the converter are connected to a common DC bus. The converter
has to have a freewheeling diode.
(A) because the converter inherently does not provide for free-wheeling
(B) because the converter does not provide for free-wheeling for high values of
triggering angles
(C) or else the free-wheeling action of the converter will cause shorting of the AC
supply
(D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause
a high load current in the other SCR.
SOL 1.18 Single phase full wave half controlled bridge converter feeds an Inductive load. The
two SCRs in the converter are connected to a common dc bus. The converter has
to have free wheeling diode because the converter does not provide for free wheeling
for high values of triggering angles.
MCQ 1.19 The electromagnetic torque Te of a drive and its connected load torque TL are as
shown below. Out of the operating points A, B, C and D, the stable ones are
(A) A, C, D (B) B, C
(C) A, D (D) B, C, D
SOL 1.19 From the given characteristics point A and D are stable

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MCQ 1.20 “Six MOSFETs connected in a bridge configuration (having no other power device)
must be operated as a Voltage Source Inverter (VSI)”. This statement is
(A) True, because being majority carrier devices MOSFETs are voltage driven.
(B) True, because MOSFETs hav inherently anti-parallel diodes
(C) False, because it can be operated both as Current Source Inverter (CSI) or a
VSI
(D) False, because MOSFETs can be operated as excellent constant current sources
in the saturation region.
SOL 1.20 If we connect the MOSFET with the VSI, but the six MOSFETs are connected in
bridge configuration, in that case they also operated as constant current sources in
the saturation region so this statement is false.
Q.21 to Q. 75 carry two marks each
MCQ 1.21 The input signal Vin shown in the figure is a 1 kHz square wave voltage that
alternates between 7+ V and 7− V with a 50% duty cycle. Both transistor have
the same current gain which is large. The circuit delivers power to the load resistor
RL
. What is the efficiency of this circuit for the given input ? choose the closest
answer.
(A) 46% (B) 55%
(C) 63% (D) 92%
SOL 1.21 This is a class-B amplifier whose efficiency is given as
η
V
V
4 CC
Pπ=
where VP " peak value of input signal
VCC " supply voltage
here V 7P = volt, V 10CC = volt
so, η 100
4 10
7
# #
π= 54.95%= %55-

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Hence (B) is correct option
MCQ 1.22 , ,A B C and D are input, and Y is the output bit in the XOR gate circuit of the
figure below. Which of the following statements about the sum S of , , ,A B C D and
Y is correct ?
(A) S is always with zero or odd
(B) S is always either zero or even
(C) 1S = only if the sum of , ,A B C and D is even
(D) 1S = only if the sum of , ,A B C and D is odd
SOL 1.22 In the circuit output Y is given as
Y [ ] [ ]A B C D5 5 5=
Output Y will be 1 if no. of 1’s in the input is odd.
MCQ 1.23 The differential equation dt
dx x1
= τ
-
is discretised using Euler’s numerical integration
method with a time step T 0>3 . What is the maximum permissible value of T3
to ensure stability of the solution of the corresponding discrete time equation ?
(A) 1 (B) /2τ
(C) τ (D) 2τ
SOL 1.23 Hence ( ) is correct option
MCQ 1.24 The switch S in the circuit of the figure is initially closed, it is opened at time t 0=
. You may neglect the zener diode forward voltage drops. What is the behavior of
vout for t 0> ?
(A) It makes a transition from 5 V− to 5+ V at 12.98t sμ=

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(B) It makes a transition from 5 V− to 5 V+ at 2.57t sμ=
(C) It makes a transition from 5 V+ to 5 V− at 12.98t sμ=
(D) It makes a transition from 5 V+ to 5 V− at 2.57t sμ=
SOL 1.24 In the circuit the capacitor starts charging from 0 V (as switch was initially closed)
towards a steady state value of 20 V.
for t 0< (initial) for t " 3 (steady state)
So at any time t, voltage across capacitor (i.e. at inverting terminal of op-amp) is
given by
( )v tc ( ) [ (0) ( )]v v v ec c c
RC
t
3 3= + −
-
( )v tc ( )e20 1 RC
t
= −
-
Voltage at positive terminal of op-amp
v v v
10 100
0out−
+
−+ +
0=
v+ v
11
10
out=
Due to zener diodes, 5 5vout# #− +
So, v+ (5)
11
10= V
Transistor form 5− V to 5+ V occurs when capacitor charges upto v+.
So ( )e20 1 /t RC
− -
11
10 5#=
e1 /t RC
− -
22
5=
22
17 e /t RC
= -
lnt RC
17
22= ` j . .1 10 01 10 0 2573 6
# # # #= -
2.57 secμ=
Voltage waveforms in the circuit is shown below

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MCQ 1.25 A solid sphere made of insulating material has a radius R and has a total charge
Q distributed uniformly in its volume. What is the magnitude of the electric field
intensity, E , at a distance ( )r r R0 < < inside the sphere ?
(A)
R
Qr
4
1
0
3πε
(B)
R
Qr
4
3
0
3πε
(C)
r
Q
4
1
0
2πε
(D)
r
QR
4
1
0
3πε
SOL 1.25 Assume a Gaussian surface inside the sphere ( )x R<
From gauss law
ψ Qenclosed=
ψ D ds Qenclosed:= =#
Qenclosed
R
Q
r
R
Qr
3
4
3
4 3
3
3
3
#
π
π= =
So, D ds:#
R
Qr
3
3
=
or D r4 2
# π
R
Qr
3
3
=
R
Qr
4
1
0
3πε
=
D E0a ε=
MCQ 1.26 The figure below shows a three phase self-commutated voltage source converter
connected to a power system. The converter’s dc bus capacitor is marked as C

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in the figure. The circuit in initially operating in steady state with 0δ = and the
capacitor dc voltage is equal to Vdc0. You may neglect all losses and harmonics.
What action should be taken to increase the capacitor dc voltage slowly to a new
steady state value.
(A) Make δ positive and maintain it at a positive value
(B) Make δ positive and return it to its original value
(C) Make δ negative and maintain it at a negative value
(D) Make δ negative and return it to its original value
SOL 1.26 To increase capacitive dc voltage slowly to a new steady state value first we have
to make veδ =− than we have to reach its original value.
MCQ 1.27 The total reactance and total suspectance of a lossless overhead EHV line, operating
at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity of wave
propagation is 3 105
# km/s, then the approximate length of the line is
(A) 122 km (B) 172 km
(C) 222 km (D) 272 km
SOL 1.27 Given that
Reactance of line .0 045= pu .L
2 50
045&
#π
=
Suspectance of Line .1 2= pu
.
C
2 50
1
1 2
1&
#
#π
=
Velocity of wave propagation 3 105
#= Km/sec
Length of line l ?=
We know velocity of wave propagation
VX
LC
l=
l V LCX=
l .
.
3 10
2 50
45
2 50
1
1 2
15
#
#
#
#
#π π
=
l 172= Km
MCQ 1.28 Consider the protection system shown in the figure below. The circuit breakers
numbered from 1 to 7 are of identical type. A single line to ground fault with zero
fault impedance occurs at the midpoint of the line (at point F), but circuit breaker
4 fails to operate (‘‘Stuck breaker’’). If the relays are coordinated correctly, a valid

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sequence of circuit breaker operation is
(A) 1, 2, 6, 7, 3, 5 (B) 1, 2, 5, 5, 7, 3
(C) 5, 6, 7, 3, 1, 2 (D) 5, 1, 2, 3, 6, 7
SOL 1.28 Due to the fault ‘F’ at the mid point and the failure of
circuit-breaker ‘4’ the sequence of circuit-breaker operation will be
5, 6, 7, 3, 1, 2 (as given in options)
(due to the fault in the particular zone, relay of that particular zone must operate
first to break the circuit, then the back-up protection applied if any failure occurs.)
Hence (C) is correct option
MCQ 1.29 A three phase balanced star connected voltage source with frequency ω rad/s
is connected to a star connected balanced load which is purely inductive. The
instantaneous line currents and phase to neutral voltages are denoted by ( , , )i i ia b c
and ( , , )V V Van bn cn respectively, and their rms values are denoted by V and I .
If R = V V V
i
i
i
0
3
1
3
1
3
1
0
3
1
3
1
3
1
0
an bn cn
a
b
c
−
−
−
R
T
S
S
S
S
S
SS
R
T
S
S
SS
8
V
X
W
W
W
W
W
WW
V
X
W
W
WW
B , then the magnitude of
of R is
(A) VI3 (B) VI
(C) . VI0 7 (D) 0
SOL 1.29
R [ ]V V V
i
i
i
0
3
1
3
1
3
1
0
3
1
3
1
3
1
0
an bn cn
a
b
c
= −
−
−
R
T
S
S
S
S
S
S
S
R
T
S
S
SS
V
X
W
W
W
W
W
W
W
V
X
W
W
WW
By solving we get
R ( ) ( ) ( )V i i V i i V i i
3 3 3
an
b c
bn
c a
c
a b= − + − + −; E

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R 3( )VI= , where
( )i i
I
3
b c−
= and V Van =
MCQ 1.30 Consider a synchronous generator connected to an infinite bus by two identical
parallel transmission line. The transient reactance ‘x’ of the generator is 0.1 pu
and the mechanical power input to it is constant at 1.0 pu. Due to some previous
disturbance, the rotor angle (δ) is undergoing an undamped oscillation, with the
maximum value of ( )tδ equal to 130%
.One of the parallel lines trip due to the
relay maloperation at an instant when ( )t 130δ = %
as shown in the figure. The
maximum value of the per unit line reactance, x such that the system does not lose
synchronism subsequent to this tripping is
(A) 0.87 (B) 0.74
(C) 0.67 (D) 0.54
SOL 1.30 Hence (C) is correct option
HereP1 power before the tripping of one ckt"
P2 Power after tripping of one ckt"
P sin
X
EV δ=

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Since Pmax
X
EV=
` P max2
X
EX
2
= , here, [ (0.1 )( )]X X pu2 = +
To find maximum value of X for which system does not loose synchronism
P2 Pm= (shown in above figure)
` sin
X
EV
2
2δ Pm=
as 1 1.0 1.0P E Vpu, pu, pum = = =
. . sin
X
1 0 1 0 130
2
# c 1=
X2& .0 77=
( . )X0 1& + .0 77=
X& 0.67=
MCQ 1.31 Suppose we define a sequence transformation between ‘‘a-b-c’’ and ‘‘p-n-o’’ variables
as follows :
f
f
f
k
f
f
f
1 1 1
1
1
a
b
c
p
n
o
2
2
α
α
α
α
=
R
T
S
S
SS
R
T
S
S
SS
R
T
S
S
SS
V
X
W
W
WW
V
X
W
W
WW
V
X
W
W
WW
where ej
3
2
α =
π
and k and is a constant
Now, if it is given that :
.
.
.
V
V
V
i
i
i
0 5
0
0
0
0 5
0
0
0
2 0
p
n
o
p
n
0
=
R
T
S
S
SS
R
T
S
S
SS
R
T
S
S
SS
V
X
W
W
WW
V
X
W
W
WW
V
X
W
W
WW
and
V
V
V
Z
i
i
i
a
b
c
a
b
c
=
R
T
S
S
SS
R
T
S
S
SS
V
X
W
W
WW
V
X
W
W
WW
then,
(A)
.
.
.
.
.
.
.
.
.
Z
1 0
0 75
0 5
0 5
1 0
0 75
0 75
0 5
1 0
=
R
T
S
S
SS
V
X
W
W
WW
(B)
.
.
.
.
.
.
.
.
.
Z
1 0
0 5
0 5
0 5
1 0
0 5
0 5
0 5
1 0
=
R
T
S
S
SS
V
X
W
W
WW
(C) 3
.
.
.
.
.
.
.
.
.
Z k
1 0
0 5
0 75
0 75
1 0
0 5
0 5
0 75
1 0
2
=
R
T
S
S
SS
V
X
W
W
WW
(D)
.
.
.
.
.
.
.
.
.
Z k
3
1 0
0 5
0 5
0 5
1 0
0 5
0 5
0 5
1 0
2
= −
−
−
−
−
−
R
T
S
S
SS
V
X
W
W
WW
SOL 1.31 Given that
FP KAFS= ...(1)
where, Phase component FP
f
f
f
a
b
c
=
R
T
S
S
SS
V
X
W
W
WW
, sequence component FS
f
f
f
p
n
o
=
R
T
S
S
SS
V
X
W
W
WW
and A
1 1 1
1
1
2
2
α
α
α
α
=
R
T
S
S
SS
V
X
W
W
WW

PUBLISHING FOR GATE
`
V
I
P
P
KAV
KAI
S
S
=
=
3 ...(2)
and VS [ ]Z IS= l ...(3)
where Zl
.
0.
.
0 5
0
0
0
5
0
0
0
2 0
=
R
T
S
S
SS
V
X
W
W
WW
We have to find out Z if VP ZIP= ...(4)
From equation (2) and (3)
VP [ ]KAZ IS= l
VP KAZ
K
A Ip
1
=
−
lb l
VP AZ A Ip
1
= −
l ...(5)
` A
1 1 1
1
1
2
2
α
α
α
α
=
R
T
S
S
SS
V
X
W
W
WW
` A 1−
A
AAdj
=
AAdj
1
1
1 1 1
2
2
α
α
α
α=
R
T
S
S
S
S
V
X
W
W
W
W
A
3
1=
A 1−
3
1
1
1
1 1 1
2
2
α
α
α
α=
R
T
S
S
S
S
V
X
W
W
W
W
From equation (5)
Vp
.
. I
3
1
1 1 1
1
1
0 5
0
0
0
0 5
0
0
0
2
1
1
1 1 1
p
2
2
2
2
α
α
α
α
α
α
α
α=
R
T
S
S
SS
R
T
S
S
SS
R
T
S
S
S
S
V
X
W
W
WW
V
X
W
W
WW
V
X
W
W
W
W
Vp .
.
.
.
.
. I
1
0 5
0 5
0 5
1
0 5
0 5
0 5
1
p=
R
T
S
S
SS
V
X
W
W
WW
...(6)
By comparison of equation (5) and (6)
Z .
.
.
.
.
.
1
0 5
0 5
0 5
1
0 5
0 5
0 5
1
=
R
T
S
S
SS
V
X
W
W
WW

PUBLISHING FOR GATE
MCQ 1.32 Consider the two power systems shown in figure A below, which are initially not
interconnected, and are operating in steady state at the same frequency. Separate
load flow solutions are computed individually of the two systems, corresponding to
this scenario. The bus voltage phasors so obtain are indicated on figure A.
These two isolated systems are now interconnected by a short transmission line as
shown in figure B, and it is found that P P Q Q 01 2 1 2= = = = .
The bus voltage phase angular difference between generator bus X and generator
bus Y after interconnection is
(A) 10c (B) 25c
(C) 30c− (D) 30c
SOL 1.32 Given that the first two power system are not connected and separately loaded.
Now these are connected by short transmission line.
as P P Q Q 01 2 1 2= = = =
So here no energy transfer. The bus bar voltage and phase angle of each system
should be same than angle difference is

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θ 30 20c c= −
10c=
MCQ 1.33 The Octal equivalent of HEX and number AB.CD is
(A) 253.314 (B) 253.632
(C) 526.314 (D) 526.632
SOL 1.33 First convert the given number from hexadecimal to its binary equivalent, then
binary to octal.
Hexadecimal no. AB. CD
Binary equivalent
CA B D
1 010 1 011 11 0 0 11 01$ A CBBBSS S
To convert in octal group three binary digits together as shown
52 3 6 3 2
01 0 1 01 011 11 0 011 01 0$SSSSSS
So,( . )AB CD H ( . )253 632 8=
MCQ 1.34 If [ ( )]Rex G jω= , and [ ( )]Imy G jω= then for 0"ω +
, the Nyquist plot for
( ) / ( )( )G s s s s1 1 2= + + is
(A) x 0= (B) /x 3 4=−
(C) /x y 1 6= − (D) /x y 3=
SOL 1.34 Given function is.
( )G s
( )( )s s s1 2
1=
+ +
( )G jω
( )( )j j j1 2
1
ω ω ω
=
+ +
By simplifying
( )G jω
j j
j
j j
j
j j
j1
1
1
1
1
2
1
2
2
# # #ω ω
ω
ω ω
ω
ω ω
ω
=
−
−
+ −
−
+ −
−
c c cm m m
1
1
4
2j j j
2 2 2
ω
ω
ω
ω
ω
ω
= −
+
−
+
−
c c cm m m
( )( )
( )j j
1 4
2 3
2 2 2
2
ω ω ω
ω ω ω
=
+ +
− − −
( )( ) ( )( )
( )j
1 4
3
1 4
2
2 2 2
2
2 2 2
2
ω ω ω
ω
ω ω ω
ω ω
=
+ +
− +
+ +
−
( )G jω x iy= +
x [ ( )]Re G j
1 4
3
4
3
0 #
ω= = − =−"ω +

PUBLISHING FOR GATE
MCQ 1.35 The system 900/ ( 1)( 9)s s s+ + is to be such that its gain-crossover frequency
becomes same as its uncompensated phase crossover frequency and provides a 45c
phase margin. To achieve this, one may use
(A) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45c
at the frequency of 3 3 rad/s
(B) a lead compensator that provides an amplification of 20 dB and a phase lead
of 45c at the frequency of 3 rad/s
(C) a lag-lead compensator that provides an amplification of 20 dB and a phase lag
(D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead
SOL 1.35 Let response of the un-compensated system is
( )H sUC
( )( )s s s1 9
900=
+ +
Response of compensated system.
( )H sC
( )( )
( )
s s s
G s
1 9
900
C=
+ +
Where ( )G sC " Response of compensator
Given that gain-crossover frequency of compensated system is same as phase
crossover frequency of un-compensated system
So,
( )g compensatedω ( )p uncompensatedω=
180c− ( )H j pUC+ ω=
180c− 90 ( )tan tan
9p
p1 1
c ω
ω
=− − −− −
a k
90c tan
1
9
9
p
p
p
1
2
ω
ω
ω
=
−
+−
J
L
K
K
K
N
P
O
O
O
1
9
p
2
ω
− 0=
pω 3= rad/sec.
So,
( )g compensatedω 3= rad/sec.
At this frequency phase margin of compensated system is
PMφ 180 ( )H j gCc + ω= +
45c 180 90 ( ) ( / ) ( )tan tan G j9g g C g
1 1
c c +ω ω ω= − − − +− −
45c 180 90 ( ) ( / ) ( )tan tan G j3 1 3 C g
1 1
c c + ω= − − − +− −
45c 90 ( )tan G j
1 3
3
1
3
3
1
g
1
Cc + ω= −
−
+
+−
b l
R
T
S
S
SS
V
X
W
W
WW

PUBLISHING FOR GATE
45c 90 90 ( )G j gCc c + ω= − +
( )G j gC+ ω 45c=
The gain cross over frequency of compensated system is lower than un-compensated
system, so we may use lag-lead compensator.
At gain cross over frequency gain of compensated system is unity so.
( )H j gC ω 1=
( )G j
1 81
900
g g g
g
2 2
C
ω ω ω
ω
+ +
1=
( )G j gC ω
900
3 9 1 9 81= + +
900
3 30#=
10
1=
in dB ( )G gC ω log20
10
1= b l
20=− dB (attenuation)
MCQ 1.36 Consider the discrete-time system shown in the figure where the impulse response
of ( )G z is (0) 0, (1) (2) 1, (3) (4) 0g g g g g g= = = = = =
This system is stable for range of values of K
(A) [ 1, ]2
1
− (B) [ 1, 1]−
(C) [ , 1]2
1
− (D) [ , 2]2
1
−
SOL 1.36 System response is given as
( )H z
( )
( )
KG z
G z
1
=
−
[ ]g n [ ] [ ]n n1 2δ δ= − + −
( )G z z z1 2
= +- -
So ( )H z
( )
( )
K z z
z z
1 1 2
1 2
=
− +
+
- -
- -
z Kz K
z 1
2
=
− −
+
For system to be stable poles should lie inside unit circle.

PUBLISHING FOR GATE
z 1#
z 1K K K
2
42
! #= +
2K K K42
! #+
4K K2
+ 2 K# −
4K K2
+ 4 4K K2
# − +
8K 4#
K 1/2#
MCQ 1.37 A signal ( )x t is given by
( )
, / /
, / /
( )
x t
T t T
T t T
x t T
1 4 3 4
1 3 4 7 4
<
<
#
#=
−
−
− +
*
Which among the following gives the fundamental fourier term of ( )x t ?
(A) cos
T
t4
4π
π π−` j (B) cos
T
t
4 2 4
π π π+` j
(C) sin
T
t4
4π
π π−` j (D) sin
T
t
4 2 4
π π π+` j
SOL 1.37 Given signal has the following wave-form
Function x(t) is periodic with period T2 and given that
( )x t ( )x t T=− + (Half-wave symmetric)
So we can obtain the fourier series representation of given function.
MCQ 1.38 If the loop gain K of a negative feed back system having a loop transfer function
( 3)/( 8)K s s 2
+ + is to be adjusted to induce a sustained oscillation then
(A) The frequency of this oscillation must be 4 3 rad/s

PUBLISHING FOR GATE
(B) The frequency of this oscillation must be 4 rad/s
(C) The frequency of this oscillation must be 4 or 4 3 rad/s
(D) Such a K does not exist
SOL 1.38 Characteristic equation for the given system,
1
( )
( )
s
K s
8
3
2+
+
+
0=
( 8) ( 3)s sK2
+ + + 0=
(16 ) (64 3 )s K s K2
+ + + + 0=
By applying Routh’s criteria.
s2
1 64 3K+
s1
16 K+ 0
s0
64 3K+
For system to be oscillatory
16 0K+ = 16K& =−
Auxiliary equation ( ) (64 3 ) 0A s s K2
= + + =
& ( )s 64 3 162
#+ + − 0=
s 64 482
+ − 0=
s 162
=− j j4& ω =
ω 4= rad/sec
MCQ 1.39 The system shown in figure below
can be reduced to the form
with

PUBLISHING FOR GATE
(A) , 1/( ),X c s c Y s a s a Z b s b0 1
2
0 1 0 1= + = + + = +
(B) 1, ( )/( ),X Y c s c s a s a Z b s b0 1
2
0 1 0 1= = + + + = +
(C) , ( )/( ), 1X c s c Y b s b s a s a Z1 0 1 0
2
1 0= + = + + + =
(D) , 1/( ),X c s c Y s a s a Z b s b1 0
2
1 1 0= + = + + = +
SOL 1.39 From the given block diagram we can obtain signal flow graph of the system.
Transfer function from the signal flow graph is written as
T.F
s
a
s
a
s
Pb
s
Pb
s
c P
s
c P
1 0 0 1
1
1
2 2
2
0
=
+ + − −
+
( ) ( )
( )
s a s a P b sb
c c s P
2
1 0 0 1
0 1
=
+ + − +
+
( )
( )
s a s a
P b sb
s a s a
c c s P
1
2
1 0
2
1 0
0 1
0 1
=
−
+ +
+
+ +
+
^ h
from the given reduced form transfer function is given by
T.F
YPZ
XYP
1
=
−
by comparing above two we have
X ( )c c s0 1= +
Y
s a s a
1
2
1 0
=
+ +
Z ( )b sb0 1= +
MCQ 1.40 The value of
( )z
dz
1C
2
+
# where C is the contour /z i 2 1− = is
(A) i2π (B) π
(C) tan z1-
(D) tani z1
π -
Given
z
dz
1C
2
+
# ( )( )z i z i
dz
C
=
+ −
#
Contour z i
2
− 1=
P(0, 1) lies inside the circle 1z i
2
− = and ( , )P 0 1 does not lie.
So by Cauchy’s integral formula
z
dz
1C
2
+
# 2 ( )
( )( )
limi z i
z i z i
1
z i
π= −
+ −"
limi
z i
2 1
z i
π=
+"

PUBLISHING FOR GATE
i
i
2
2
1
#π=
π=
MCQ 1.41 A single-phase voltages source inverter is controlled in a single
pulse-width modulated mode with a pulse width of 150c in each half cycle. Total
harmonic distortion is defined as
THD 100
V
V Vrms
1
2
1
2
#= −
where V1 is the rms value of the fundamental component of the output voltage. The
THD of output ac voltage waveform is
(A) 65.65% (B) 48.42%
(C) 31.83% (D) 30.49%
SOL 1.41 Given that, total harmonic distortion
THD 100
V
V Vrms
1
1
22
#= −
Pulse width is 150c
Here Vrms
180
150 0.91V Vs s= =b l
V1 Vrms(fundamental)= . sinV
2
0 4 75s
#
c
π
= . V0 8696 s=
THD
( . )
( . ) ( . )
V
V V
0 87
0 91 0 87
s
s s
2
2 2
=
−
. %31 9=
MCQ 1.42 A voltage source inverter is used to control the speed of three-phase, 50 Hz, squirrel
cage induction motor. Its slip for rated torque is 4%. The flux is maintained at
rated value. If the stator resistance and rotational losses are neglected, then the
frequency of the impressed voltage to obtain twice the rated torque at starting
should be
(A) 10 Hz (B) 5 Hz
(C) 4 Hz (D) 2 Hz
SOL 1.42 Hence ( ) is Correct Option
MCQ 1.43 A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15
kW, 1500 rpm separately excited dc motor with a ripple free continuos current in
the dc link under all operating conditions, Neglecting the losses, the power factor
of the ac mains at half the rated speed is
(A) 0.354 (B) 0.372
(C) 0.90 (D) 0.955

PUBLISHING FOR GATE
SOL 1.43 When losses are neglected,
cos3 2 440# #
π
α K
60
750 2
m # # π=
Here back emf ε with φ is constant
ε V Km m0 ω= =
440 K
60
1500 2
m # # π=
Km .2 8=
cosα .0 37=
at this firing angle
Vt (0.37)3 2 440# #π
= 219.85= V
Ia .
440
1500 34 090= =
Isr / .I 2 3 27 83a= =
p.f.
V I
V I
3 s sr
t s
= 0.354=
Hence (A) is correct option
MCQ 1.44 A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying
power to a half-wave uncontrolled ac-dc converter used for charging a battery(12 V
dc) with the series current limiting resistor being 19.04 Ω. The charging current is
(A) 2.43 A (B) 1.65 A
(C) 1.22 A (D) 1.0 A
SOL 1.44 Hence (D) is correct option.
Vs .
4
230 57 5= =
Here charging current I=
sinVm θ 12=
1θ 8.486 0.148 radian= =
Vm .81 317= V
ε 12 V=
There is no power consumption in battery due to ac current, so average value of
charging current.
Iav(charging)
.
[ ( )]cosV
2 19 04
1 2 2m 1 1
#π
θ ε π θ= − −
.
[ ( )]cosV
2 19 04
1 2 12 2m 1 1
#
# #π
θ π θ= − −
1.059 /AΩ=
MCQ 1.45 A three-phase synchronous motor connected to ac mains is running at full load
and unity power factor. If its shaft load is reduced by half, with field current held

PUBLISHING FOR GATE
constant, its new power factor will be
(A) unity (B) leading
(C) lagging (D) dependent on machine parameters
SOL 1.45 When the 3-φ synchronous motor running at full load and unity power factor and
shaft load is reduced half but field current is constant then it gives leading power
factor.
MCQ 1.46 A 100 kVA, 415 V(line), star-connected synchronous machine
generates rated open circuit voltage of 415 V at a field current
of 15 A. The short circuit armature current at a field current of
10 A is equal to the rated armature current. The per unit saturated synchronous
reactance is
(A) 1.731 (B) 1.5
(C) 0.666 (D) 0.577
SOL 1.46 Given star connected synchronous machine, 100P kVA=
Open circuit voltage 415V V= and field current is 15 A, short circuit armature
current at a field current of 10 A is equal to rated armature current.
So,
Line synchronous impedance
3 short ckt phase current
open circuit line voltage
#
=
3
3 415
100 1000
415
#
#
#
=
c m
.1 722=
MCQ 1.47 A three-phase, three-stack, variable reluctance step motor has 20 poles on each
rotor and stator stack. The step angle of this step motor is
(A) 3c (B) 6c
(C) 9c (D) 81 c
SOL 1.47 Given 3-φ, 3-stack
Variable reluctance step motor has 20-poles
Step angle
3 20
360 6
#
c= =
MCQ 1.48 A single-phase, 50 kVA, 250 V/500 V two winding transformer has an efficiency of
95% at full load, unity power factor. If it is re-configured as a 500 V/750 V auto-
transformer, its efficiency at its new rated load at unity power factor will be
(A) 95.752% (B) 97.851%

PUBLISHING FOR GATE
(C) 98.276% (D) 99.241%
SOL 1.48 Given 1-φ transformer
50P kVA= , 250V V/500 V=
Two winding transformer efficiency 95% at full load unity power factor.
Efficiency 95%
W W50
50 1 1
cu i#
# #=
+
So W Wcu i+ .2 631=
Reconfigured as a 500 V/750 V auto-transformer
auto-transformer efficiency
η
.150 2 631
150=
+
. %98 276=
Hence (C) is corret option.
MCQ 1.49 A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC.
A unity power-factor load of 4 kW is connected between phase A and neutral N.
It is desired to achieve zero neutral current through the use of a pure inductor and
a pure capacitor in the other two phases. The value of inductor and capacitor is
(A) 72.95 mH in phase C and 139.02 Fμ in Phase B
(B) 72.95 mH in Phase B and 139.02 Fμ in Phase C
(C) 42.12 mH in Phase C and 240.79 Fμ in Phase B
(D) 42.12 mH in Phase B and 240.79 Fμ in Phase C
SOL 1.49 Given that,
230 V, 50 Hz, 3-φ, 4-wire system
Load 4 kwP = = at unity Power factor
I 0N = through the use of pure inductor and capacitor
Than L ?= , ?C =
a IN 0 I I IA B C= = + + ...(1)
Network and its Phasor is being as

PUBLISHING FOR GATE
Here the inductor is in phase B and capacitor is in Phase C.
We know P VI=
So Ia .
V
P
230
4 10 17 39
3
#= = = Amp.
From equation (1)
IA ( )I IB C=− + I Ib ca -
` IA I I
2
3
2
3
B C# #=− +c m
` IA I I3 3B C= =
IB
. 10I
3
17 39
C- -= Amp
Now XC
I
V
10
230 23
C
- Ω= =
and XC
fC2
1
π
=
C&
fX2
1
2 50 23
1
C # #π π
= =
139.02 Fμ=
XL
I
V
10
230 23
L
- Ω= =
2 fLπ=
L& 72.95
f
X
2 2 100
23L
#π π
= = = mH
So L = 72.95 mH in phase B
C = 139.02 Fμ in phase C
MCQ 1.50 The state equation for the current I1 in the network shown below in terms of the
voltage VX and the independent source V , is given by

PUBLISHING FOR GATE
(A) 1.4 3.75
dt
dI V I V
4
5
X
1
1=− − + (B) 1.4 3.75
dt
dI I V
4
5VX
1
1= − −
(C) 1.4 3.75
dt
dI V I V
4
5
X
1
1=− + + (D) 1.4 3.75
dt
dI I V
4
5VX
1
1=− + −
SOL 1.50 By writing KVL for both the loops
3( ) 0.5V I I V
dt
dI
x1 2
1
− + − − 0=
3 3 0.5V I I V
dt
dI
x1 2
1
− − − − 0=
...(1)
in second loop
. .I V
dt
dI5 0 2 0 5x2
1
− + + 0=
I2 0.04 0.1V
dt
dI
x
1
= + ...(2)
Put I2 from eq(2) into eq(2)
3 3 . . 0.5V I V
dt
dI V
dt
dI0 04 0 1x x1
1 1
− − + − −: D 0=
0.8
dt
dI1
1.12 3V I Vx 1=− − +
dt
dI1
1.4 3.75V I V
4
5
x 1=− − +
MCQ 1.51 If ( ), ( )u t r t denote the unit step and unit ramp functions respectively and ( ) * ( )u t r t
their convolution, then the function ( ) * ( )u t r t1 2+ − is given by
(A) ( 1) ( 1)t u t2
1
− − (B) ( 1) ( 2)t u t2
1
− −
(C) ( 1) ( 1)t u t2
1 2
− − (D) None of the above
SOL 1.51 Given Convolution is,
( )h t ( ) ( )u t r t1 2)= + −
Taking Laplace transform on both sides,
( )H s [ ( )] [ ( 1)] [ ( 2)]h t u t r tL L L)= = + −
We know that, [ ( )] /u t s1L =
[ ( 1)]u tL + 1e
s2
s
= c m
(Time-shifting property)
and [ ( )]r tL 1/s2
=
( )r t 2L − 1e
s2
s2
= -
c m

PUBLISHING FOR GATE
(Time-shifting property)
So ( )H s 1 1e
s
e
s2
s s2
= -
` cj m; ;E E
( )H s 1e
s3
s
= -
c m
Taking inverse Laplace transform
( )h t ( ) ( )t u t
2
1 1 12
= − −
MCQ 1.52 The integral ( )sin cost d
2
1
0
2
π
τ τ τ−
π
# equals
(A) sin cost t (B) 0
(C) ( / )cost1 2 (D) (1/2)sint
SOL 1.52 Hence ( ) is correct Option
MCQ 1.53 ( ) 1 3 , ( ) 1 2X z z Y z z1 2
= − = +- -
are Z transforms of two signals [ ], [ ]x n y n
respectively. A linear time invariant system has the impulse response [ ]h n defined by
these two signals as [ ] [ ] * [ ]h n x n y n1= − where * denotes discrete time convolution.
Then the output of the system for the input [ ]n 1δ −
(A) has Z-transform ( ) ( )z X z Y z1-
(B) equals [ ] [ ] [ ] [ ]n n n n2 3 3 2 4 6 5δ δ δ δ− − − + − − −
(C) has Z-transform 1 3 2 6z z z1 2 3
− + −- - -
(D) does not satisfy any of the above three
SOL 1.53 Impulse response of given LTI system.
[ ]h n [ ] [ ]x n y n1 )= −
Taking z -transform on both sides.
( )H z ( ) ( )z X z Y z1
= -
[ 1] ( )x n z x z1Z
a − -
We have ( ) 1 3 ( ) 1 2andX z z Y z z1 2
= − = +- -
So
( )H z ( )( )z z z1 3 1 21 1 2
= − +- - -
Output of the system for input [ ] [ 1] isu n nδ= − ,
( )y z ( ) ( )H z U z=
[ ] ( )U n U z z 1Z
= -
So
( )Y z (1 3 )(1 2 )z z z z1 1 2 1
= − +- - - -
( )z z z z1 3 2 62 1 2 3
= − + −- - - -
z z z z3 2 62 3 4 5
= − + −- - - -

PUBLISHING FOR GATE
Taking inverse z-transform on both sides we have output.
[ ]y n [ ] [ ] [ ] [ ]n n n n2 3 3 2 4 6 5δ δ δ δ= − − − + − − −
MCQ 1.54 A loaded dice has following probability distribution of occurrences
Dice Value 1 2 3 4 5 6
Probability 1/4 1/8 1/8 1/8 1/8 1/4
If three identical dice as the above are thrown, the probability of occurrence of
values 1, 5 and 6 on the three dice is
(A) same as that of occurrence of 3, 4, 5
(B) same as that of occurrence of 1, 2, 5
(C) 1/128
(D) 5/8
SOL 1.54 Probability of occurrence of values 1,5 and 6 on the three dice is
( , , )P 1 5 6 ( ) ( ) ( )P P P1 5 6=
4
1
8
1
4
1
# #=
128
1=
In option (A)
( , , )P 3 4 5 ( ) ( ) ( )P P P3 4 5=
8
1
8
1
8
1
# #=
512
1=
In option (B)
( , , )P 1 2 5 ( ) ( ) ( )P P P1 2 5=
4
1
8
1
8
1
# #=
256
1=
MCQ 1.55 Let x and y be two vectors in a 3 dimensional space and x,y< > denote their dot
product. Then the determinant
det
x,x
y,x
x,y
y,y
< >
< >
< >
< >= G
(A) is zero when x and y are linearly independent
(B) is positive when x and y are linearly independent
(C) is non-zero for all non-zero x and y

PUBLISHING FOR GATE
(D) is zero only when either x or y is zero
SOL 1.55 Hence (D) is correct option.
det
x x
y x
x y
y y
$
$
$
$> H ( )( ) ( )( )x x y y x y y x: : : := −
0= only when x or y is zero
MCQ 1.56 The linear operation ( )L x is defined by the cross product L(x) b x#= , where
b 0 1 0
T
= 8 B and x x xx 1 2 3
T
= 8 B are three dimensional vectors. The 3 3# matrix
M of this operations satisfies
( ) M
x
x
x
L x
1
2
3
=
R
T
S
S
SS
V
X
W
W
WW
Then the eigenvalues of M are
(A) , ,0 1 1+ − (B) , ,1 1 1−
(C) , ,i i 1− (D) , ,i i 0−
MCQ 1.57 In the figure, transformer T1 has two secondaries, all three windings having the
same number of turns and with polarities as indicated. One secondary is shorted
by a 10 Ω resistor R, and the other by a 15 Fμ capacitor. The switch SW is opened
( )t 0= when the capacitor is charged to 5 V with the left plate as positive. At
t 0= + the voltage VP and current IR are
(A) 25 , 0.0V A− (B) very large voltage, very large current
(C) 5.0 V, 0.5 A (D) 5.0 , 0.5V A− −
MCQ 1.58 IC 555 in the adjacent figure is configured as an astable multi-vibrator. It is enabled
to to oscillate at t 0= by applying a high input to pin 4. The pin description is : 1
and 8-supply; 2-trigger; 4-reset; 6-threshold 7-discharge. The waveform appearing
across the capacitor starting from t 0= , as observed on a storage CRO is

PUBLISHING FOR GATE
SOL 1.58 In a 555 astable multi vibrator circuit, charging of capacitor occurs through
resistor ( )R RA B+ and discharging through resistor RB only. Time for charging and
discharging is given as.
TC . ( )R R C0 693 A B= +
TD . R C0 693 B=
But in the given circuit the diode will go in the forward bias during charging, so
the capacitor will charge through resistor RA only and discharge through RB only.
a RA RB=
So TC TD=
MCQ 1.59 In the circuit of adjacent figure the diode connects the ac source to a pure inductance
L.

PUBLISHING FOR GATE
The diode conducts for
(A) 90c (B) 180c
(C) 270c (D) 360c
SOL 1.59 Conduction angle for diode is 270c as shown in fig.
MCQ 1.60 The circuit in the figure is a current commutated dc-dc chopper where, ThM
is the
main SCR and ThAUX
is the auxiliary SCR. The load current is constant at 10 A.
ThM
is ON. ThAUX
is trigged at t 0= . ThM
is turned OFF between.
(A) 0 25ts s< #μ μ (B) 25 50ts s< #μ μ
(C) 50 75ts s< #μ μ (D) 75 100ts s< #μ μ
SOL 1.60 Hence ( ) is correct option.
MCQ 1.61 In the circuit shown in figure. Switch SW1 is initially closed and SW2 is open.
The inductor L carries a current of 10 A and the capacitor charged to 10 V with
polarities as indicated. SW2 is closed at t 0= and SW1 is opened at t 0= . The
current through C and the voltage across L at ( )t 0= +
is
(A) 55 A, 4.5 V (B) 5.5 A, 45 V
(C) 45 A, 5.5 A (D) 4.5 A, 55 V

PUBLISHING FOR GATE
SOL 1.61 At 0t = +
, when switch positions are changed inductor current and capacitor voltage
does not change simultaneously
So at 0t = +
(0 )vc
+
(0 ) 10vc= =−
V
(0 )iL
+
(0 ) 10iL= =−
A
The equivalent circuit is
by applying KCL
(0 ) (0 ) (0 )v v v
10 10
L L c
+
−+ + +
(0 ) 10iL= =+
2 (0 ) 10vL −+
100=
Voltage across inductor at 0t = +
(0 )vL
+
55
2
100 10= + = V
So, current in capacitor at 0t = +
(0 )iC
+ ( ) ( )v v
10
0 0L c
=
−+ +
4.5
10
55 10= − = A
MCQ 1.62 The R-L-C series circuit shown in figure is supplied from a variable frequency
voltage source. The admittance - locus of the R-L-C network at terminals AB for
increasing frequency ω is

PUBLISHING FOR GATE
SOL 1.62 Impedance of the given network is
Z R j L
C
1ω
ω
= + −b l
Admittance
Y
Z
1=
R j L
C
1
1
ω
ω
=
+ −b l
R j L
C
R j L
C
R j L
C
1
1
1
1
#
ω
ω
ω
ω
ω
ω
=
+ − − −
− −
b b
b
l l
l
R L
C
R j L
C
1
1
2
2
ω
ω
ω
ω
=
+ −
− −
b
b
l
l
R L
C
R
R L
C
j L
C
1 1
1
2
2
2
2
ω
ω
ω
ω
ω
ω
=
+ −
−
+ −
−
b b
b
l l
l
( ) ( )Re ImY Y= +
By varying frequency for ( )Re Y and ( )Im Y we can obtain the admittance-locus.

PUBLISHING FOR GATE
MCQ 1.63 In the figure given below all phasors are with reference to the potential at point '' ''O
. The locus of voltage phasor VYX as R is varied from zero to infinity is shown by
SOL 1.63 In the circuit

PUBLISHING FOR GATE
VX V 0c+=
( )
R
V V
V j C
2 0y
y
c+
ω
−
+ 0=
( )V j CR1y ω+ V2 0c+=
Vy
j CR
V
1
2 0c+
ω
=
+
VYX V VX Y= −
V
j CR
V
1
2
ω
= −
+
R 0" , VYX V V V2= − =−
R " 3, VYX 0V V= − =
MCQ 1.64 A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear
resistance characterized by the relation V INL NL
2
= . The power dissipated in the non
linear resistance is
(A) 1.0 W (B) 1.5 W
(C) 2.5 W (D) 3.0 W
SOL 1.64 The circuit is
Applying KVL
3 2 INL
2
#− VNL=
3 2INL
2
− INL
2
=
3INL
2
3= & INL 1= A
VNL (1) 12
= = V
So power dissipated in the non-linear resistance
P V INL NL=
1 1 1#= = W
MCQ 1.65 Consider the feedback system shown below which is subjected to a unit step input.
The system is stable and has following parameters 4, 10, 500K Kp i ω= = = and
.0 7ξ = .The steady state value of Z is

PUBLISHING FOR GATE
(A) 1 (B) 0.25
(C) 0.1 (D) 0
SOL 1.65 For the given system Z is given by
Z ( )E s
s
Ki
=
Where ( )E s " steady state error of the system
Here
( )E s
( ) ( )
( )
lim
G s H s
sR s
1s 0
=
+"
Input ( )R s
s
1= (Unit step)
( )G s
s s s2
K Ki
p 2 2
2
ξω ω
ω= +
+ +b el o
( )H s 1= (Unity feed back)
So,
Z
( )
lim
s
K K
s s
s
s
s
K
1
2
1
s i
p
i
0
2 2
2
ξω ω
ω
=
+ +
+ +
"
b
b
b
l
l
l
R
T
S
S
S
S
V
X
W
W
W
W
( )
( )
lim
s K K s
s s
K
2
s
i p
i
0
2 2
2
ξω ω
ω
=
+ +
+ +
"
> H
1
K
K
i
i
= =
MCQ 1.66 A three-phase squirrel cage induction motor has a starting torque of 150% and a
maximum torque of 300% with respect to rated torque at rated voltage and rated
frequency. Neglect the stator resistance and rotational losses. The value of slip for
maximum torque is
(A) 13.48% (B) 16.42%
(C) 18.92% (D) 26.79%
SOL 1.66 Given a 3-φ squirrel cage induction motor starting torque is 150% and maximum
torque 300%
So

PUBLISHING FOR GATE
TStart 1.5TFL=
Tmax 3TFL=
Then
T
T
max
Start
2
1= ...(1)
T
T
max
Start
S
S
1
2
max
max
2 2=
+
...(2)
from equation (1) and (2)
S
S
1
2
max
max
2
+ 2
1=
4 1S Smax max
2
− + 0=
So
Smax . %26 786=
Hence (D) is correct option
MCQ 1.67 The matrix A given below in the node incidence matrix of a network. The columns
correspond to branches of the network while the rows correspond to nodes. Let
[ ..... ]V V V V T
1 2 6= denote the vector of branch voltages while [ ..... ]I i i i T
1 2 6= that
of branch currents. The vector [ ]E e e e e T
1 2 3 4= denotes the vector of node voltages
relative to a common ground.
1
0
1
0
1
1
0
0
1
0
0
1
0
1
0
1
0
1
1
0
0
0
1
1
−
−
−
−
− −
R
T
S
S
S
S
S
V
X
W
W
W
W
W
Which of the following statement is true ?
(A) The equations ,V V V V V V0 01 2 3 3 4 5− + = + − = are KVL equations for the
network for some loops
(B) The equations ,V V V V V V0 01 3 6 4 5 6− − = + − = are KVL equations for the
network for some loops
(C) E AV=
(D) AV 0= are KVI equations for the network
SOL 1.67 In node incidence matrix
b b b b b b
n
n
n
n
1
0
1
0
1
1
0
0
1
0
0
1
0
1
0
1
0
1
1
0
0
0
1
1
1 2 3 4 5 6
1
2
3
4
−
−
−
−
− −
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
In option (C)
E AV=

PUBLISHING FOR GATE
e e e e T
1 2 3 48 B V V V
1
0
1
0
1
1
0
0
1
0
0
1
0
1
0
1
0
1
1
0
0
0
1
1
T
1 2 6=
−
−
−
−
− −
−−
R
T
S
S
S
S
SS
8
V
X
W
W
W
W
WW
B
e
e
e
e
1
2
3
4
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
V V V
V V V
V V V
V V V
1 2 3
2 4 5
1 5 6
3 4 6
=
+ +
− − +
− − −
− + +
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
which is true.
MCQ 1.68 An isolated 50 Hz synchronous generator is rated at 15 MW which is also the
maximum continuous power limit of its prime mover. It is equipped with a speed
governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each
at 50 Hz. One of these loads is programmed to trip permanently if the frequency
falls below 48 Hz .If an additional load of 3.5 MW is connected then the frequency
will settle down to
(A) 49.417 Hz (B) 49.917 Hz
(C) 50.083 Hz (D) 50.583 Hz
SOL 1.68 Maximum continuous power limit of its prime mover with speed governor of 5%
droop.
Generator feeded to three loads of 4 MW each at 50 Hz.
Now one load Permanently tripped
` f 48= Hz
If additional load of 3.5 MW is connected than ?f =
a Change in Frequency w.r.t to power is given as
fΔ
rated power
drop out frequency
Change in power#=
. . %
15
5 3 5 1 16#= =
1.16 0.58
100
50 Hz#= =
System frequency is 50 .580= −
49.42 Hz=
MCQ 1.69 Which one of the following statements regarding the INT (interrupt) and the BRQ
(but request) pins in a CPU is true?
(A) The BRQ pin is sampled after every instruction cycle, but the INT is sampled
after every machine cycle.
(B) Both INT and BRQ are samped after every machine cycle.

PUBLISHING FOR GATE
(C) The INT pin is sampled after every instruction cycle, but the BRQ is sampled
after every machine cycle.
(D) Both INT and BRQ are sampled after every instruction cycle.
SOL 1.69 Hence ( ) is correct Option
MCQ 1.70 A bridge circuit is shown in the figure below. Which one of the sequence given
below is most suitable for balancing the bridge ?
(A) First adjust R4, and then adjust R1
(B) First adjust R2, and then adjust R3
(C) First adjust R2, and then adjust R4
(D) First adjust R4, and then adjust R2
SOL 1.70 To balance the bridge
( )( )R jX R jX1 1 4 4+ − R R2 3=
( ) ( )R R X X j X R R X1 4 1 4 1 4 1 4+ + − R R2 3=
comparing real and imaginary parts on both sides of equations
R R X X1 4 1 4+ R R2 3= ...(1)
X R R X1 4 1 4− 0
X
X
R
R
4
1
4
1
&= = ...(2)
from eq (1) and (2) it is clear that for balancing the bridge first balance R4 and
then R1.
Common Data Questions
Common Data for Questions 71, 72, 73:
A three phase squirrel cage induction motor has a starting current of seven times
the full load current and full load slip of 5%
MCQ 1.71 If an auto transformer is used for reduced voltage starting to provide 1.5 per unit
starting torque, the auto transformer ratio(%) should be

PUBLISHING FOR GATE
(A) 57.77 % (B) 72.56 %
(C) 78.25 % (D) 81.33 %
SOL 1.71 Given 3-φ squirrel cage induction motor has a starting current of seven the full load
current and full load slip is 5%
ISt 7IFl=
SFl %5=
T
T
Fl
St
T
I x S
2
2
Fl
St
Fl# #= b l
.1 5 ( ) .x7 0 052 2
# #=
x . %78 252=
MCQ 1.72 If a star-delta starter is used to start this induction motor, the per unit starting
torque will be
(A) 0.607 (B) 0.816
(C) 1.225 (D) 1.616
SOL 1.72 Star delta starter is used to start this induction motor
So
T
T
Fl
St
I
I S
3
1 2
Fl
St
Fl# #= b l
.
3
1 7 0 052
# #=
T
T
Fl
St
.0 816=
MCQ 1.73 If a starting torque of 0.5 per unit is required then the per unit starting current
should be
(A) 4.65 (B) 3.75
(C) 3.16 (D) 2.13
SOL 1.73 Given starting torque is 0.5 p.u.
So,
T
T
Fl
St
I
I Ssc
2
Fl
Fl#= b l
0.5 0.05
I
Isc
2
Fl
#= b l
Per unit starting current
I
Isc
Fl .
.
0 05
0 5= 3.16= A

PUBLISHING FOR GATE
Common Data for Questions 74, 75:
A 1:1 Pulse Transformer (PT) is used to trigger the SCR in
the adjacent figure. The SCR is rated at 1.5 kV, 250 A with
50I 2L = mA, I 150H = mA, and I 150maxG = mA, 1 0I 0minG = mA. The SCR is
connected to an inductive load, where L 150= mH in series with a small resistance
and the supply voltage is 200 V dc. The forward drops of all transistors/diodes and
gate-cathode junction during ON state are 1.0 V
MCQ 1.74 The resistance R should be
(A) 4.7 kΩ (B) 470 kΩ
(C) 47 Ω (D) 4.7 Ω
SOL 1.74 Here, Vm = maximum pulse voltage that can be applied
so 10 1 1 1 7 V= − − − =
Here 1 V drop is in primary transistor side, so that we get 9V pulse on the secondary
side. Again there are 1 V drop in diode and in gate cathode junction each.
Ig max 150 mA=
So R
I
V
g
m
max
=
150
7
mA
= .46 67 Ω=
MCQ 1.75 The minimum approximate volt-second rating of pulse transformer suitable for
triggering the SCR should be : (volt-second rating is the maximum of product of
the voltage and the width of the pulse that may applied)
(A) 2000 μV-s (B) 200 μV-s
(C) 20 μV-s (D) 2 μV-s
SOL 1.75 We know that the pulse width required is equal to the time taken by ia to rise upto
iL
so,
Vs ( )L
dt
di R V 0i T .= +

PUBLISHING FOR GATE
ia [ ]e
1
200 1 / .t 0 15
= − −
Here also t T= , 0.25i ia L= =
0.25 [ ]e200 1 / .T 0 5
= − −
T 1.876 10 4
#= −
187.6 sμ=
Width of pulse 187.6 sμ=
Magnitude of voltage 10 V=
Vsec rating of P.T. 10 187.6 s# μ=
1867= μV-s is approx to 2000 μV-s
Linked Answer Questions : Q-76 to Q-85 carry two marks each
Statement for Linked Answer Questions 76 and 77:
An inductor designed with 400 turns coil wound on an iron core of 16 cm2
cross
sectional area and with a cut of an air gap length of 1 mm. The coil is connected
to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and
leakage inductance, ( 4 10 )H/M0
7
#μ π= -
MCQ 1.76 The current in the inductor is
(A) 18.08 A (B) 9.04 A
(C) 4.56 A (D) 2.28 A
SOL 1.76 Inductance is given as
L
l
N A0
2
μ
=
( )
( ) ( )
1 10
4 10 400 16 10
3
7 2 4
#
# # # #π
= −
− −
321.6 mH=
V IXL=
fL2
230
π
= 2X fLL` π=
. .2 3 14 50 321 6 10
230
3
# # # #
= −
.2 28= A
MCQ 1.77 The average force on the core to reduce the air gap will be
(A) 832.29 N (B) 1666.22 N
(C) 3332.47 N (D) 6664.84 N

PUBLISHING FOR GATE
SOL 1.77 Energy stored is inductor
E LI
2
1 2
=
E 321.6 10 (2.28)
2
1 3 2
# # #= −
Force required to reduce the air gap of length 1 mm is
F 0.835
l
E
1 10 3
#
= = −
835= N
Statement for Linked Answer Questions 78 and 79
Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic
equation. Consider a matrix
A
3
2
2
0
=
−
−= G
MCQ 1.78 A satisfies the relation
(A) A I A3 2 01
+ + =-
(B) 2 2 0A A I2
+ + =
(C) ( )( )A I A I2+ + (D) ( ) 0exp A =
SOL 1.78 For characteristic equation
3
1
2
0
λ
λ
− −
− −> H 0=
or ( )( )3 2λ λ− − − + 0=
( )( )1 2λ λ+ + 0=
According to Cayley-Hamiliton theorem
( )( )A I A I2+ + 0=
MCQ 1.79 A9
equals
(A) 511 510A I+ (B) 309 104A I+
(C) 154 155A I+ (D) ( )exp A9
SOL 1.79 According to Cayley-Hamiliton theorem
( )( )A I A I2+ + 0=
or A A I3 22
+ + 0=
or A2
( )A I3 2=− +
or A4
( ) ( )A I A A I3 2 9 12 42 2
= + = + +
9( 3 2 ) 12 4A I A I= − − + +
15 14A I=− −
A8
( )A I A A15 14 225 420 1962 2
= − − = + +
225( 3 2 ) 420 196A I A I= − − + +

PUBLISHING FOR GATE
255 254A I=− −
A9
255 254A A2
=− −
255( 3 2 ) 254A I A=− − − − A I511 510= +
Consider the R-L-C circuit shown in figure
MCQ 1.80 For a step-input ei , the overshoot in the output e0 will be
(A) 0, since the system is not under damped
(B) 5 %
(C) 16 %
(D) 48 %
SOL 1.80 System response of the given circuit can be obtained as.
( )
( )
( )
H s
e s
e s
i
0
=
R Ls
Cs
Cs
1
1
=
+ +b
b
l
l
( )H s
1LCs RCs
1
2=
+ +
( )H s
s
L
R s
LC
LC
1
1
2
=
+ +
b l
Characteristic equation is given by,
s
L
R s
LC
12
+ + 0=
Here natural frequency
LC
1
nω =
2 nξω
L
R=
Damping ratio ξ
L
R LC
2
=
ξ R
L
C
2
=
Here

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ξ 0.5
2
10
10 10
1 10
6
3
#
#= =−
−
(under damped)
So peak overshoot is given by
% peak overshoot e 1001 2
#= ξ
πξ
−
−
100e ( . )
.
1 0 5
0 5
2
#=
#π
−
−
%16=
MCQ 1.81 If the above step response is to be observed on a non-storage CRO, then it would
be best have the ei as a
(A) Step function (B) Square wave of 50 Hz
(C) Square wave of 300 Hz (D) Square wave of 2.0 KHz
SOL 1.81 Hence ( ) is Correct Option.
The associated figure shows the two types of rotate right instructions ,R R1 2
available in a microprocessor where Reg is a 8-bit register adn C is the carry bit.
The rotate left instructions L1 and L2 are similar except that C now links the most
significant bit of Reg instead of the least significant one.
MCQ 1.82 Suppose Reg contains the 2’s complement number 11010110. If this number is
divided by 2 the answer should be
(A) 01101011 (B) 10010101
(C) 11101001 (D) 11101011
MCQ 1.83 Such a division can be correctly performed by the following set of operatings
(A) , ,L R R2 2 1 (B) , ,L R R2 1 2
(C) , ,R L R2 1 1 (D) , ,R L R1 2 2
MCQ 1.84 A signal is processed by a causal filter with transfer function ( )G s

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For a distortion free output signal wave form, ( )G s must
(A) provides zero phase shift for all frequency
(B) provides constant phase shift for all frequency
(C) provides linear phase shift that is proportional to frequency
(D) provides a phase shift that is inversely proportional to frequency
SOL 1.84 Output is said to be distortion less if the input and output have identical wave
shapes within a multiplicative constant. A delayed output that retains input
waveform is also considered distortion less.
Thus for distortion less output, input-output relationship is given as
( )y t ( )Kg t td= −
Taking Fourier transform.
( )Y ω ( )KG e j td
ω= ω-
( )Y ω ( ) ( )G Hω ω=
( )H &ω transfer function of the system
So, ( )H ω Ke j td
= ω-
Amplitude response ( )H Kω =
Phase response
( )nθ ω tdω=−
For distortion less output, phase response should be proportional to frequency.
MCQ 1.85 ( )G z z z1 3
α β= +- -
is a low pass digital filter with a phase characteristics same as
that of the above question if
(A) α β= (B) α β=−
(C) ( / )1 3
α β= (D) ( / )1 3
α β= -
( )G z z ej
= ω
e ej j3
α β= +ω ω− −
for linear phase characteristic α β= .

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

Gate ee 2007 with solutions

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Gate ee 2007 with solutions