Gate ee 2007 with solutions

GATE EE2007

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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 9.3270 100

mA=+

0. 250 mA=

In saturation, base current is given by, 10 (1) (1)I V IC CE E− − − 0=

210 IC (sat)= I I

V 0C E

CE

-

- IC (sat) 5= mA

IB(sat) IC (sat)

β=

.0501005 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= − = voltAt emitter (by applying KCL), IE I IB L= +

IE 1

.6 6 610

6kΩ Ω

= − + 0.6 amp-

10 6 4V V VCE C E= − = − = voltPower dissipated in transistor is given by. PT V ICE C#= 4 0.6#= 0.6I IC E` - = amp 2.4= WHence (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 transformerWhich 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= MWFor optimum generator ?PA = , ?PB =a From curve, maximum incremental cost for generator A 600= at 450 MWand maximum incremental cost for generator B 800= at 400 MWminimum 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= MWHence (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.Hence (C) is correct option.

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.Hence (B) is correct option.

MCQ 1.7 The circuit shown in the figure is

(A) a voltage source with voltage R R

rV1 2<

(B) a voltage source with voltage R

r R V1

2<

(C) a current source with current R Rr R

rV

1 2

2<+c m

(D) a current source with current R R

RrV

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 vR

v 01

1

2

1− + − 0=

vR R

R R1 2

1 21

+c m R

V1

=

v1 VR R

R1 2

2=+c m

Since the op-amp is ideal therefore

iL irv

11= =



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iL rV

R RR

1 2

2=+c m

Hence (D) is correct option.

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

( )( )

( )

( )( )

ss

s

ss

121

11

21

=+ +

−−

+−

( )( )ss

31= +

−

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 ss

s

11

121

11

=+ − +

−−

( )( )

( )s s

s1 3

2= − ++



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One pole of the system is lying in right half of s -plane, so the system is unstable.Hence (D) is correct option.

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=Hence (D) is correct option.

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 1secm

kHzT1

11

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

iy

jz

k22

22

22= + +t t t

So V4: ( ) ( )cos cosx

x xy yy

y xy22

22= − + + +6 6@ @

( )sinz

z x y2 2 2

22 + +6 @

( ) ( )sin sin cosx xy y y xy x z z2 2=− − + − + cosz z2 2=Hence (A) is correct option.

MCQ 1.12 x x xx n1 2T

g= 8 B is an n-tuple nonzero vector. The n n# matrixV 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 cosLsVm

ω α α μ= − +

20 [ ( )]cos cosLs2 50230 2 25 25 10#

c c cπ= − +

Ls 0.0045 H=

V0 cosV LsI2 m 0

πα

πω= −

. .. .cos

3 142 230 2 25

3 142 3 14 50 4 5 10 203

# # # # # #c= −−

.187 73 9= − .178 74c=

Displacement factor V IV 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 athree-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 42050 1000#= .119 05=

RMS value of thyristor current . .3

119 05 68 73= =Hence (C) is correct option.

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.Hence (C) is correct option.

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.Hence (B) is correct option.

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.Hence (B) is correct option.

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.Hence (B) is correct option.

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 stableHence (C) is correct option.



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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.Hence (D) is correct option.

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 VV

4 CC

Pπ=

where VP " peak value of input signal VCC " supply voltagehere V 7P = volt, V 10CC = volt

so, η 1004 107

# #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.Hence (D) is correct option.

MCQ 1.23 The differential equation dtdx 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 v10 100

0out− + −+ + 0=

v+ v1110

out=

Due to zener diodes, 5 5vout# #− +

So, v+ (5)1110= 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− - 225=

2217 e /t RC= -

lnt RC1722= ` j . .1 10 01 10 0 2573 6# # # #= -

2.57 secμ=Voltage waveforms in the circuit is shown below



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Hence (B) is correct option.

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) RQr

41

03πε (B)

RQr

43

03πε

(C) rQ

41

02πε (D)

rQR

41

03πε

SOL 1.25 Assume a Gaussian surface inside the sphere ( )x R<

From gauss law ψ Qenclosed= ψ D ds Qenclosed:= =#

Qenclosed RQ r

RQr

34

34 3

33

3

#ππ= =

So, D ds:# RQr

3

3

=

or D r4 2# π

RQr

3

3

= RQr

41

03πε=

D E0a ε=

Hence (A) is correct option.

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.Hence (D) is correct option.

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 50045

&#π=

Suspectance of Line .1 2= pu .C 2 501

1 21

&#

#π=

Velocity of wave propagation 3 105#= Km/sec

Length of line l ?=We know velocity of wave propagation

VX LCl=

l V LCX=

l ..3 10 2 50

452 50

11 215

##

##

#π π=

l 172= KmHence (B) is correct option.

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 ofcircuit-breaker ‘4’ the sequence of circuit-breaker operation will be5, 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 Viii

0

31

31

31

0

31

31

31

0

an bn cn

a

b

c

−

−

−R

T

SSSSSSS

R

T

SSSS

8

V

X

WWWWWWW

V

X

WWWW

B , then the magnitude of

of R is(A) VI3 (B) VI

(C) . VI0 7 (D) 0

SOL 1.29

R [ ]V V Viii

0

31

31

31

0

31

31

31

0

an bn cn

a

b

c

= −

−

−R

T

SSSSSSS

R

T

SSSS

V

X

WWWWWWW

V

X

WWWW

By solving we get

R ( ) ( ) ( )V i i V i i V i i3 3 3

anb c

bnc a

ca b= − + − + −; E



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R 3( )VI= , where ( )i i

I3

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

Here P1 power before the tripping of one ckt"

P2 Power after tripping of one ckt"

P sinXEV δ=



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Since Pmax XEV=

P max2 XEX

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)

sinXEV

22δ Pm=

as 1 1.0 1.0P E Vpu, pu, pum = = =

. . sinX1 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 :fff

kfff

1 1 111

a

b

c

p

n

o

2

2αα

αα

=

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

where ej32

α =π and k and is a constant

Now, if it is given that : .

..

VVV

iii

0 500

00 50

00

2 0

p

n

o

p

n

0

=

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

and VVV

Ziii

a

b

c

a

b

c

=

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

then,

(A) ...

.

..

...

Z1 00 750 5

0 51 00 75

0 750 51 0

=

R

T

SSSS

V

X

WWWW

(B) ...

.

.

.

.

.

.Z

1 00 50 5

0 51 00 5

0 50 51 0

=

R

T

SSSS

V

X

WWWW

(C) 3...

...

...

Z k1 00 50 75

0 751 00 5

0 50 751 0

2=

R

T

SSSS

V

X

WWWW

(D) ...

...

.

..

Z k3

1 00 50 5

0 51 00 5

0 50 5

1 0

2= −

−

−

−

−−

R

T

SSSS

V

X

WWWW

SOL 1.31 Given that FP KAFS= ...(1)

where, Phase component FP

fff

a

b

c

=

R

T

SSSS

V

X

WWWW

, sequence component FS

fff

p

n

o

=

R

T

SSSS

V

X

WWWW

and A 1 1 1

11

2

2

αα

αα

=

R

T

SSSS

V

X

WWWW



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V

IP

P

KAV

KAIS

S

==

3 ...(2)

and VS [ ]Z IS= l ...(3)

where Zl .

0..

0 500

05

0

00

2 0=

R

T

SSSS

V

X

WWWW

We have to find out Z if VP ZIP= ...(4)From equation (2) and (3) VP [ ]KAZ IS= l

VP KAZ KA Ip

1

=−

lb l

VP AZ A Ip1= −l ...(5)

A 1 1 1

11

2

2

αα

αα

=

R

T

SSSS

V

X

WWWW

A 1− AAAdj=

AAdj 111 1 1

2

2αα

αα=

R

T

SSSS

V

X

WWWW

A 31=

A 1− 31

111 1 1

2

2αα

αα=

R

T

SSSS

V

X

WWWW

From equation (5)

Vp .

. I31

1 1 111

0 500

00 50

002

111 1 1

p2

2

2

2

αα

αα

αα

αα=

R

T

SSSS

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

V

X

WWWW

Vp ..

.

.

.

. I1

0 50 5

0 51

0 5

0 50 51

p=

R

T

SSSS

V

X

WWWW

...(6)

By comparison of equation (5) and (6)

Z ..

.

.

.

.1

0 50 5

0 51

0 5

0 50 51

=

R

T

SSSS

V

X

WWWW




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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=Hence (A) is correct option.

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. CDBinary equivalent

CA B D1 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 3 201 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

1ω ω ω= + +

By simplifying

( )G jω j jj

j jj

j jj1

11

11

21

22

# # #ω ωω

ω ωω

ω ωω= −

−+ −

−+ −

−c c cm m m

11

42j j j

2 2 2ωω

ωω

ωω= −

+−

+−

c c cm m m

( )( )

( )j j1 4

2 32 2 2

2

ω ω ωω ω ω=

+ +− − −

( )( ) ( )( )

( )j1 4

31 4

22 2 2

2

2 2 2

2

ω ω ωω

ω ω ωω ω=

+ +− +

+ +−

( )G jω x iy= +

x [ ( )]Re G j 1 43

43

0 #ω= = − =−

"ω +



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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 of 45c at the frequency of 3 rad/s

(D) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45c at the frequency of 3 rad/s

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 s1 9900

C= + +

Where ( )G sC " Response of compensatorGiven that gain-crossover frequency of compensated system is same as phase crossover frequency of un-compensated systemSo, ( )g compensatedω ( )p uncompensatedω= 180c− ( )H j pUC+ ω=

180c− 90 ( )tan tan 9pp1 1c ω ω=− − −− −

a k

90c tan1 9

9p

pp

12ω

ω ω=

−

+−

J

L

KKK

N

P

OOO

1 9p2ω− 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 1c c +ω ω ω= − − − +− −

45c 180 90 ( ) ( / ) ( )tan tan G j3 1 3 C g1 1c c + ω= − − − +− −

45c 90 ( )tan G j1 3 3

13 3

1

g1

Cc + ω= −−

++−

b l

R

T

SSSS

V

X

WWWW



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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 9003 9 1 9 81= + +

9003 30#= 10

1=

in dB ( )G gC ω log20 101= b l

20=− dB (attenuation)Hence (D) is correct option.

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]21− (D) [ , 2]2

1−

SOL 1.36 System response is given as

( )H z ( )

( )KG z

G z1

=−

[ ]g n [ ] [ ]n n1 2δ δ= − + −

( )G z z z1 2= +- -

So ( )H z ( )

( )K z zz z

1 1 2

1 2

=− +

+- -

- -

z Kz K

z 12=

− −+

For system to be stable poles should lie inside unit circle.



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z 1#

z 1K K K2

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) cosTt4

4ππ π−` j (B) cos

Tt

4 2 4π π π+` j

(C) sinTt4

4ππ π−` j (D) sin

Tt

4 2 4π 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.Hence (C) is correct option.

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



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(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 s8

32+

++

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/secHence (B) is correct option.

MCQ 1.39 The system shown in figure below

can be reduced to the form

with



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(A) , 1/( ),X c s c Y s a s a Z b s b0 12

0 1 0 1= + = + + = +

(B) 1, ( )/( ),X Y c s c s a s a Z b s b0 12

0 1 0 1= = + + + = +

(C) , ( )/( ), 1X c s c Y b s b s a s a Z1 0 1 02

1 0= + = + + + =

(D) , 1/( ),X c s c Y s a s a Z b s b1 02

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

sa

sa

sPb

sPb

sc P

sc P

1 0 0 1

1

12 2

20

=+ + − −

+

( ) ( )

( )s a s a P b sb

c c s P2

1 0 0 1

0 1=+ + − +

+

( )

( )

s a s aP b sb

s a s ac c s P

1

21 0

21 0

0 1

0 1

=−

+ ++

+ ++

^ h

from the given reduced form transfer function is given by

T.F YPZXYP

1= −

by comparing above two we have X ( )c c s0 1= +

Y s a s a

12

1 0=

+ + Z ( )b sb0 1= +Hence (D) is correct option.

MCQ 1.40 The value of ( )z

dz1

C2+

# where C is the contour /z i 2 1− = is

(A) i2π (B) π

(C) tan z1- (D) tani z1π -

SOL 1.40 Hence (A) is correct option.

Given z

dz1C

2+#

( )( )z i z idz

C

= + −#

Contour z i2− 1=

P(0, 1) lies inside the circle 1z i2− = and ( , )P 0 1 does not lie.

So by Cauchy’s integral formula

z

dz1C

2+# 2 ( )

( )( )limi z i

z i z i1

z iπ= − + −"

limi z i2 1z i

π= +"



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i i2 21

#π=

π=

MCQ 1.41 A single-phase voltages source inverter is controlled in a singlepulse-width modulated mode with a pulse width of 150c in each half cycle. Total harmonic distortion is defined as

THD 100VV Vrms

1

212

#= −

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 100VV Vrms

1

12

2

#= −

Pulse width is 150c

Here Vrms 180150 0.91V Vs s= =b l

V1 Vrms(fundamental)= . sinV2

0 4 75s

#c

π= . V0 8696 s=

THD ( . )

( . ) ( . )V

V V0 87

0 91 0 87s

s s2

2 2

= − . %31 9=

Hence (C) is correct option

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



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SOL 1.43 When losses are neglected,

cos3 2 440# #π α K 60

750 2m #

# π=

Here back emf ε with φ is constant ε V Km m0 ω= =

440 K 601500 2

m ## π=

Km .2 8= cosα .0 37=at this firing angle

Vt (0.37)3 2 440##π= 219.85= V

Ia .4401500 34 090= =

Isr / .I 2 3 27 83a= =

p.f. V I

V I3 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 .4230 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) . [ ( )]cosV2 19 041 2 2m 1 1

#π θ ε π θ= − −

. [ ( )]cosV2 19 041 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



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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.Hence (B) is correct option.

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 of10 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 currentopen circuit line voltage#

=

3

3 415100 1000415

##

#=

c m

.1 722=Hence (A) is correct option.

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-stackVariable reluctance step motor has 20-poles

Step angle 3 20360 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%



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(C) 98.276% (D) 99.241%

SOL 1.48 Given 1-φ transformer50P kVA= , 250V V/500 V=

Two winding transformer efficiency 95% at full load unity power factor.

Efficiency 95% W W5050 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 631150= +

. %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 factorI 0N = through the use of pure inductor and capacitorThan L ?= , ?C = a IN 0 I I IA B C= = + + ...(1)Network and its Phasor is being as



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Here the inductor is in phase B and capacitor is in Phase C.We know P VI=

So Ia .VP

2304 10 17 39

3#= = = Amp.

From equation (1) IA ( )I IB C=− + I Ib ca -

IA I I23

23

B C# #=− +c m

IA I I3 3B C= =

IB . 10I3

17 39C- -= Amp

Now XC IV

10230 23

C- Ω= =

and XC fC21

π=

C& fX21

2 50 231

C # #π π= =

139.02 Fμ=

XL IV

10230 23

L- Ω= =

2 fLπ=

L& 72.95fX2 2 100

23L

#π π= = = mH

So L = 72.95 mH in phase B C = 139.02 Fμ in phase CHence (B) is correct option.

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



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(A) 1.4 3.75dtdI V I V

45

X1

1=− − + (B) 1.4 3.75dtdI I V

45VX

11= − −

(C) 1.4 3.75dtdI V I V

45

X1

1=− + + (D) 1.4 3.75dtdI I V

45VX

11=− + −

SOL 1.50 By writing KVL for both the loops

3( ) 0.5V I I V dtdI

x1 21− + − − 0=

3 3 0.5V I I V dtdI

x1 21− − − − 0=

...(1)in second loop

. .I V dtdI5 0 2 0 5x2

1− + + 0=

I2 0.04 0.1V dtdI

x1= + ...(2)

Put I2 from eq(2) into eq(2)

3 3 . . 0.5V I V dtdI V dt

dI0 04 0 1x x11 1− − + − −: D 0=

0.8 dtdI1 1.12 3V I Vx 1=− − +

dtdI1 1.4 3.75V I V4

5x 1=− − +

Hence (A) is correct option

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 t21 − −

(C) ( 1) ( 1)t u t21 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 + 1es2

s= c m (Time-shifting property)and [ ( )]r tL 1/s2=

( )r t 2L − 1es2

s2= -c m



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(Time-shifting property)

So ( )H s 1 1es

es2

s s2= -` cj m; ;E E

( )H s 1es3

s= -c m

Taking inverse Laplace transform

( )h t ( ) ( )t u t21 1 12= − −

Hence (C) is correct option.

MCQ 1.52 The integral ( )sin cost d21

0

2

πτ τ τ−

π# equals

(A) sin cost t (B) 0

(C) ( / )cos t1 2 (D) (1/2)sin t

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 z1Za − -

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= − + −- - - -



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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= − − − + − − −Hence (C) is correct option.

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=

41

81

41

# #=

1281=

In option (A) ( , , )P 3 4 5 ( ) ( ) ( )P P P3 4 5=

81

81

81

# #=

5121=

In option (B) ( , , )P 1 2 5 ( ) ( ) ( )P P P1 2 5=

41

81

81

# #=

2561=


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

detx,xy,x

x,yy,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



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(D) is zero only when either x or y is zero

SOL 1.55 Hence (D) is correct option.

det x xy x

x yy 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

( ) Mxxx

L x1

2

3

=

R

T

SSSS

V

X

WWWW

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



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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=Hence (B) is correct option.

MCQ 1.59 In the circuit of adjacent figure the diode connects the ac source to a pure inductance L .



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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



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SOL 1.61 At 0t = +, when switch positions are changed inductor current and capacitor voltage does not change simultaneouslySo at 0t = +

(0 )vc+ (0 ) 10vc= =− V

(0 )iL + (0 ) 10iL= =− AThe equivalent circuit is

by applying KCL

(0 ) (0 ) (0 )v v v10 10

L L c+ −+ + +

(0 ) 10iL= =+

2 (0 ) 10vL −+ 100=Voltage across inductor at 0t = +

(0 )vL+ 552

100 10= + = V

So, current in capacitor at 0t = +

(0 )iC + ( ) ( )v v

100 0L c= −+ +

4.51055 10= − = A

Hence (D) is correct option.

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



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SOL 1.62 Impedance of the given network is

Z R j L C1ω ω= + −b l

Admittance

Y Z1=

R j L C1

1ω ω

=+ −b l

R j L C R j L C

R j L C1

11

1

#ω ω ω ω

ω ω=

+ − − −

− −

b b

b

l l

l

R L C

R j L C1

1

22

ω ω

ω ω=

+ −

− −

b

b

l

l

R L C

R

R L C

j L C1 1

1

22

22

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.



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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



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VX V 0c+=

( )RV V

V j C2 0y

yc+ ω− + 0=

( )V j CR1y ω+ V2 0c+=

Vy j CRV

12 0c+

ω= +

VYX V VX Y= −

V j CRV

12

ω= − +

R 0" , VYX V V V2= − =−R " 3, VYX 0V V= − =Hence (B) is correct option.

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 2INL2− INL

2= 3INL

2 3= & INL 1= A VNL (1) 12= = VSo power dissipated in the non-linear resistance P V INL NL= 1 1 1#= = WHence (A) is correct option.

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



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(A) 1 (B) 0.25

(C) 0.1 (D) 0

SOL 1.65 For the given system Z is given by

Z ( )E s sKi=

Where ( )E s " steady state error of the systemHere

( )E s ( ) ( )( )

limG s H ssR s

1s 0= +"

Input ( )R s s1= (Unit step)

( )G s s s s2K Ki

p 2 2

2

ξω ωω= +

+ +b el o

( )H s 1= (Unity feed back)So,

Z

( )

lim

sK K

s s

s ssK

12

1

s ip

i

02 2

2

ξω ωω

=+ +

+ +"

b

bb

l

ll

R

T

SSSS

V

X

WWWW

( )

( )

lims K K s

s s

K

2s

i p

i

02 2

2

ξω ωω

=+ +

+ +" > H

1KK

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



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TStart 1.5TFL= Tmax 3TFL=Then

TT

max

Start 21= ...(1)

TT

max

Start S

S1

2max

max2 2=

+ ...(2)

from equation (1) and (2)

S

S1

2max

max2 +

21=

4 1S Smax max2 − + 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 T1 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.

101

0

11

00

1001

01

01

011

0

001

1−

−

−

−− −

R

T

SSSSS

V

X

WWWWW

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

nnnn

101

0

11

00

1001

01

01

011

0

001

1

1 2 3 4 5 6

1

2

3

4

−−

−

−− −

R

T

SSSSSS

V

X

WWWWWW

In option (C) E AV=



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e e e e T1 2 3 48 B V V V

101

0

11

00

1001

01

01

011

0

001

1

T1 2 6= −

−

−

−− − −−

R

T

SSSSSS

8

V

X

WWWWWW

B

eeee

1

2

3

4

R

T

SSSSSS

V

X

WWWWWW

V V VV V VV V VV V V

1 2 3

2 4 5

1 5 6

3 4 6

=

+ +− − +− − −− + +

R

T

SSSSSS

V

X

WWWWWW

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= HzIf additional load of 3.5 MW is connected than ?f =a Change in Frequency w.r.t to power is given as

fΔ rated powerdrop out frequency Change in power#=

. . %155 3 5 1 16#= =

1.16 0.5810050 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.



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(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 XX

RR

4

1

4

1&= = ...(2)

from eq (1) and (2) it is clear that for balancing the bridge first balance R4 and then R1.Hence (A) is correct option

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



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(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=

TT

Fl

St TI x S

22

Fl

StFl# #= b l

.1 5 ( ) .x7 0 052 2# #=

x . %78 252=Hence (C) is correct option.

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 motorSo

TT

Fl

St II S3

1 2

Fl

StFl# #= b l

.31 7 0 052# #=

TT

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,

TT

Fl

St II Ssc

2

FlFl#= b l

0.5 0.05IIsc

2

Fl#= b l

Per unit starting current

IIsc

Fl .

.0 050 5= 3.16= A




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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 appliedso 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 IVg

m

max= 150

7mA= .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 iLso,

Vs ( )Ldtdi R V 0i T .= +



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ia [ ]e1200 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-sHence (A) is correct option.

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 lN A0

2μ=

( )( ) ( )1 10

4 10 400 16 103

7 2 4

#

# # # #π= −

− −

321.6 mH= V IXL=

fL2230π= 2X fLL` π=

. .2 3 14 50 321 6 10

2303

# # # #= −

.2 28= AHence (D) is correct option.

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



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SOL 1.77 Energy stored is inductor

E LI21 2=

E 321.6 10 (2.28)21 3 2# # #= −

Force required to reduce the air gap of length 1 mm is

F 0.835lE

1 10 3#

= = −

835= NHence (A) is correct option.

Statement for Linked Answer Questions 78 and 79Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix

A32

20=

−−= 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

12

0λ

λ− −

− −> H 0=

or ( )( )3 2λ λ− − − + 0= ( )( )1 2λ λ+ + 0=According to Cayley-Hamiliton theorem ( )( )A I A I2+ + 0=Hence (C) is correct option.

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= − − + +



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255 254A I=− − A9 255 254A A2=− − 255( 3 2 ) 254A I A=− − − − A I511 510= +Hence (A) is correct option.

Statement for Linked Answer Questions 80 and 81: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 se se s

i

0= R Ls Cs

Cs1

1

=+ +b

b

l

l

( )H s 1LCs RCs

12=+ +

( )H s s L

R s LC

LC1

1

2=

+ +

b l

Characteristic equation is given by,

s LR s LC

12 + + 0=

Here natural frequency LC1

nω =

2 nξω LR=

Damping ratio ξ LR LC2=

ξ RLC

2=

Here



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ξ 0.5210

10 101 10

6

3

#

#= =−

− (under damped)

So peak overshoot is given by

% peak overshoot e 1001 2 #= ξπξ−

−

100e ( . ).

1 0 50 5

2 #=#π

−−

%16=Hence (C) is correct option

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.

Statement for Linked Answer Questions 82 and 83: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


Statement for Linked Answer Questions 84 and 85:

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 tdω= 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.Hence (C) is correct option.

MCQ 1.85 ( )G z z z1 3α 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 3α β= (D) ( / )1 3α β= -

SOL 1.85 Hence (A) is correct option. ( )G z

z ej= ω e ej j3α 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)

**********

Engineering

Gate ee 2007 with solutions