set 3 ee -2014 gate solved

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GATE SOLVED PAPER

Electrical Engineering

2014 (Set-3)

Copyright By NODIA & COMPANY

Information contained in this book has been obtained by authors, from sources believes to be reliable. However,neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor itsauthors shall be responsible for any error, omissions, or damages arising out of use of this information. This bookis published with the understanding that Nodia and its authors are supplying information but are not attemptingto render engineering or other professional services.

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GATE SOLVED PAPER - EE

2014 (SET-3)

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General Aptitude

Q. 1-5 carry one mark each

Q. 1 WhiletryingtocollectI

an envelope fromunder thetableII

, .Mr X fell downIII

and

sinwaslo gconsciousnessIV

Which one of the above underlined parts of the sentence is NOT appropriate ?(A) I (B) II

(C) II I (D) IV

Sol. 1 Correct option is (D).

IV underlined part is not correct. T he correct statement for the IV underline part

will be lost consciousness.

Q. 2 If she _ _ _ _ how to calibrate the instrument, she _ _ _ _ _ done the experiment.(A) knows, will have (B) knew, had

(C) had known, could have (D) should have known, would have

Sol. 2 Correct option is (C).For a conditional sentence, the rule is as given below.

If + past perfect tense

Would/ could have + past participles.

Thus, the complete sentence is

If she had known how to calibrate the instrument, she could have done the

experiment.

Q. 3 Choose the word that is opposite in meaning to the word coherent.(A) sticky (B) well-connected

(C) rambling (D) friendly

Sol. 3 Correct option is (C)

Coherent capable of thinking and expressing yourself in a clear and consistent

manner. rambling spreading out in different directions or distributed irregularly.

Sticky covered with an adhesive material.

So, coherent and rambling are opposite to each other.

Q. 4 Which number does not belong in the series below ?2, 5, 10, 17, 26, 37, 50, 64(A) 17 (B) 37

(C) 64 (D) 26

Sol. 4 Correct option is (C).2, 5, 10, 17, 26, 37, 50, 64


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GATE SOLVED PAPER - EE 2014 (SET-3)

The difference between successive terms is in AP with initial term 3 and the

common difference 2. With this logic 64 is the wrong term and the correct number

in place of 64 is 65.

Q. 5 The table below has question-wise data on the performance of students in anexamination. The marks for each question are also listed. There is no negative orpartial marking in the examination.

Q. No. Marks Answered Correctly Answered Wrongly Not Attempted

1 2 21 17 6

2 3 15 27 2

3 2 23 18 3

What is the average of the marks obtained by the class in the examination ?(A) 1.34 (B) 1.74

(C) 3.02 (D) 3.91

Sol. 5 Correct option is (C).

Total students in class 44=

Total marks scored by class 21 2 15 3 23 2# # #= + +

42 45 46= + +

133=

Average marksdeTotal stu nts

Total marks=

44133=

.302=

Q. 6 - Q. 10 carry one mark each

Q. 6 A dance programme is scheduled for 10.00 a.m. Some students are participating

in the programme and they need to come an hour earlier than the start of theevent. These students should be accompanied by a parent. Other students andparents should come in time for the programme. The instruction you think thatis appropriate for this is

(A) Students should come at 9.00 a.m. and parents should come at 10.00 a.m.

(B) Participating students should come at 9.00 a.m. accompanied by a parent,and other parents and students should come by 10.00 a.m.

(C) Students who are not participating should come by 10.00 a.m. and theyshould not bring their parents. Participating students should come at 9.00

a.m.

(D) Participating students should come before 9.00 a.m. Parents who accompany

them should come at 9.00 a.m. All others should come at 10.00 a.m.

Sol. 6 Correct option is (B).(i) Dance program is scheduled for 10 a.m.

(ii) Participating student should come at 9 a.m. accompanied by a parent.

(iii) non-participating student should come at 10 a.m. with their parents.

Combining these three instructions (B) option is most appropriate.


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Q. 7 By the beginning of the 20thcentury, several hypotheses were being proposed,suggesting a paradigm shift in our understanding of the universe. However, theclinching evidence was provided by experimental measurements of the position of

a star which was directly behind our sun. Which of the following inference(s) maybe drawn from the above passage ?(i) Our understanding of the universe changes based on the positions of stars

(ii) Paradigm shifts usually occur at the beginning of centuries

(iii) Stars are important objects in the universe

(iv) Experimental evidence was important in confirming this paradigm shift

(A) (i), (ii) and (iv) (B) (iii) only

(C) (i) and (iv) (D) (iv) only

Sol. 7 Correct option is (D).(i) Clinching evidence was provided by experimental measurement of position of

star.(ii) Our understanding of universe changes, so experimental evidence are

important.

So, by these (iv) is correct option.

Q. 8 The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013.For international comparison, the GDP is compared in US Dollars (USD) afterconversion based on the market exchange rate. During the period 2012-2013 the

exchange rate for the USD increased from Rs. 50/ USD to Rs. 60/ USD. IndiaaGDP in USD during the period 2012-2013.(A) increased by 5% (B) decreased by 13%

(C) decreased by 20% (D) decreased by 11%


Let GDP in 2011-12 be x

GDP in 2012-13 is . x107 (7% grew)

During 2012-13 exchange rate increase from Rs. 50/USD to Rs.60/USD.

So, GDIP in comparison to USD is obtained as

at initial of 2012-13 GDP/ USD x50

=

and at final 2012-13 GDP/ USD . x60

107=

So, increase and decrease %100.

x

x x

50

60107

50#= -

. %1083=

%11=

Q. 9 The ratio of male to female students in a college for five years is plotted in thefollowing line graph. I f the number of female students in 2011 and 2012 is equal,what is the ratio of male students in 2012 to male students in 2011 ?


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(A) 1 : 1 (B) 2 : 1

(C) 1.5 : 1 (D) 2.5 : 1


Let number of female students in 2011 be x.

So, Number of male students in 2011 x= (ratio is 1)

Number of female students in 2012 x= (Given)

Number of male students in 2012 . x15= (ratio is 1.5)

Ratio of male students in 2012 to male students in 2011 is

. . :x

x15 15 1= =

Q. 10 Consider the equation Y7526 43648 8 8- =^ ^ ^h h h , where X N^ h stands for X to thebase N . Find Y .(A) 1634 (B) 1737

(C) 3142 (D) 3162


Given equation is

Y7526 8 8-^ ^h h 4364 8= ^ hor Y

8^ h 7526 43648 8= -^ ^h hOctal subtraction is done in same way as decimal subtraction. The only difference

is that while obtaining carry we get 8 instead of 10.

7526

4364

3142

8

8

8

-

^

^

h

hh


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Electrical Engineering


Q. 1 Two matrices A and Bare given below :

A p

r

q

s= > H; B p q

pr qs

pr qs

r s

2 2

2 2= +

+

+

+> H

If the rank of matrix A is N , then the rank of matrix Bis(A) /N 2 (B) N 1-

(C) N (D) N2


Given the two matrices,

A pr

qs

= > Hand B

p q

pr qs

pr qs

r s

2 2

2 2= +

+

+

+> H

To determine the rank of matrix A , we obtain its equivalent matrix using the

operation, a i2 a aa

ai i211

211! - as

A p q

spr

q0= -> HIf s

pr

q- 0=

or ps r q - 0=

then rank of matrix A is 1, otherwise the rank is 2.

Now, we have the matrix

Bp q

pr qs

pr qs

r s

2 2

2 1= +

+

+

+> H

To determine the rank of matrix B, we obtain its equivalent matrix using the

operation, a i2 a aa

ai i211

211! - as

B

p q pr qs

r sp q

pr qs 0

2 2

2 22 2

2=

+ +

+ -+

+^ ^h hR

T

SSSS

V

X

WWWW

If r sp q

pr qs 2 22 2

2

+ -+

+^ ^h h ps r q 2= -^ h 0=or ps r q - 0=

then rank of matrix Bis 1, otherwise the rank is 2.

Thus, from the above results, we conclude that

If ps r q 0- = , then rank of matrix A and Bis 1.

If ps r q 0!- , then rank of A and Bis 2.

i.e. the rank of two matrices is always same. If rank of A is N then rank of B

also N .


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Q. 2 A particle, starting from origin at st 0= , is traveling along x-axis with velocity

v /cos m st2 2p p= a k

At st 3= , the difference between the distance covered by the particle and the

magnitude of displacement from the origin is _ _ _ _ _ _ .

Sol. 2 Correct answer is (C).

Given the velocity of particle,

v cos t2 2p p=

Since, the velocity of the particle is along x-axis. So, we have

vdtdx=

Therefore, the displacement of the particle from t 0= (origin) to t 3= is

x vdtt

t

0

3

= =

=

# cos tdt2 2tt

0

3p p

= =

=

# sin t2 0

3= p p=6 @ 1=-

or | |x 1=

i.e. the magnitude of displacement is 1.

Now, we have to determine the distance covered by the particle. For calculating

distance, we have to consider speed and speed cant be negative, so the distance

is given by

x0 v dtt

t

0

3

==

=# cos t dt

2 2t

t

0

3 p p==

=# cos cost tdt

2 2 2 2t

t

t

t

0

1

1

3p p p p= -=

=

=

=# # sin sint t

2 2t tt

0

1

1

3p p= -= =

=9 9C C 1 1 1= - - -^ h 3=

Thus, the difference between the distance covered and magnitude of displacement

is

xD | |x x 3 1 20= - = - =

Q. 3 Let f x y y z z x v 2 2 2:d = + +^ h , where f and v are scalar and vector fieldsrespectively. If y z xv i j k = + + , then fv : d is(A) x y y z z x 2 2 2+ + (B) xy yz zx 2 2 2+ +

(C) x y z+ + (D) 0

Sol. 3 Correct option is (A).

Given the relation

fv:d ^ h x y y z z x 2 2 2= + + ...(i)where f, vare scalar and vector field respectively. From the properties of vector

field, we have

f v:d ^ h v f f v : :d d= -^ ^h h


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or v f: d h f v f v : :d d= +^ ^h h ...(ii)Again, we have

v y z xi j k= + +

So, v:d xy

yz

zx

22

22

22= + +

0=

Therefore, we get

f v:d h 0=Substituting it in equation (ii), we get

fv : d h fv:d= ^ hThus, fv : d h x y y z z x 2 2 2= + + [from equation (i)]

Q. 4 Lifetime of an electric bulb is a random variable with density f x kx 2=

^ h, where x

is measured in years. I f the minimum and maximum lifetimes of bulb are 1 and 2years respectively, then the value of kis _ _ _ _ _ .

Sol. 4 Correct answer is 0.43.

Given that the life time of an electric bulb is a random variable with density,

f x^ h kx2= .where xis the lifetime measured in years.

From the property of random variable, we have

f x d x 3

3

-^ h# 1=

Since, the minimum and maximum lifetimes of bulbs are 1 and 2 years respectively.

So, we have 1 x 2# #

Therefore, equation (i) becomes

f x d x 1

2 ^ h# 1=or kx dx2

1

2# 1=

or k x3

3

1

2: D 1=or k

38 1-

b l 1=

Thus, k .73 043= =

Q. 5 A function f t^ his shown in the figure.

The Fourier transform F w^ hof f t^ his


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(A) real and even function of w

(B) real and odd function of w

(C) imaginary and odd function of w

(D) imaginary and even function of w


We have the waveform of function ( )f t as

From the waveform, we observe that ( )f t is an odd function, i.e.

( )f t- ( )f t=-

Hence, the Fourier transform of the function is imaginary.

Q. 6 The lineA to neutral voltage is V10 15c for a balanced three phase star connectedload with phase sequence A B C. The voltage of line Bwith respect to line C is

given by

(A) V10 3 105c (B) V10 105c

(C) V10 3 75c- (D) V10 3 90c-

Sol. 6 Correct option is (C).We sketch the phasor diagram for three phase star connected load as

Since, line A to neutral voltage is

A V10 15c=

So, for balance 3-phase, we have

B 10 105c= -

and C 10 225c= -

Therefore, the voltage of line Bwith respect to line C is

vBC v vB C= -

10 105 10 225c c= - - -

10 3 75c= -


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Solving the above circuit, we get the equivalent driving point impedance as

Z s^ h ||ss

ss

1 1= + +b l ||s s s

s1 12

= + +

b l s

s ss

s ss

1 1

1 1

2

2

#= +

+ +

+

ss s

s

21

3

2

= +++

or Z s^ hs s

s s

23 1

3

4 2

=+

+ +

Q. 9 A signal is represented by

x t^ ht

t

1 1

0 1

= *The Fourier transform of the convolved signal /y t x t x t 2 2= *^ ^ ^h h his(A) sin sin4

222w

wwa ^k h (B) sin4 22w wa k

(C) sin4 22ww^ h (D) sin42 2w w


Given the signal,

( )x t ,

,

t

t

1 1

0 1

= *

The waveform for the given signal is drawn as

Since, the waveform of rectangular function ( )rect t is

So, comparing the two waveforms, we get

( )x t rect t2

= a kNow, the Fourier transform pair for rectangular function is

sinrect t C2

*t

tpwtc cm m

so, sinrect t

C2 2*

pw

b cl msin2ww

=


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Similarly, sinrect t C4

4 2*pwc bm l sin sin24 2 2 2w w w w= =^ ^h h

and ( ) sinrect t C2

*pw

c m

/

/sin

2

2

w

w=

^ hTherefore, we have the fourier transforms

( )x t2F" , ( )rect tF= " ,

/

/sin

2

2

w

w=

^ hand { ( / )}x t 2F { ( / )}

( )sinrect t 4

2 2F

ww

= =

Thus, we obtain the fourier transform of signal ( )y t as

{ ( )}y tF { ( ) ( / )}x t x t 2 2F= * { ( )}. { ( / )}x t x t 2 2F F=

/

/ ( )sin sin2

2 2 2ww

ww= ^ h

( )sin sin42

22ww w= b l

Q. 10 For the signal sin sin sinf t t t t 3 8 6 12 14p p p= + +^ h , the minimum samplingfrequency (in Hz) satisfying the Nyquist criterion is _ _ _ _ _ .

Sol. 10 Correct answer is 14 Hz.

Given the signal,

f t^ h sin sin sint t t3 8 6 12 14p p p= + +According to Nyquist criterion the sampling frequency is

fs f2> maxwhere fmaxis the maximum frequency of the message signal. For the given signal,

we have maximum frequency as

maxw 14p=

or f2 maxp 14p=

or fmax 7=

Thus, the sampling frequency is

fs 2 7> #

or fs 14>

i.e. minimum sampling frequency is 14Hz.

Q. 11 In a synchronous machine, hunting is predominantly damped by

(A) mechanical losses in the rotor (B) iron losses in the rotor

(C) copper losses in the stator (D) copper losses in the rotor


After sudden change of the load in three phase synchronous machine, the rotor

has to search or hunt for its new equilibrium position. It causes the hunting,

which can be damped by rotor copper losses.

Q. 12 A single phase induction motor is provided with capacitor and centrifugal switchin series with auxiliary winding. The switch is expected to operate at a speed of

0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speedcharacteristic of the motor is represented by


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Sol. 12 Correct answer is (C).

If the switch operates correctly, then the torque-speed characteristic is

But, the switch does not operate due to malfunctioning. So, there will be no

discontinuity in the characteristic curve. T hus, we have the modified characteristic

curve of the motor as

Q. 13 The no-load speed of a 230 V separately excited dc motor is 1400 rpm. Thearmature resistance drop and the brush drop are neglected. The field current is

kept constant at rated value. T he torque of the motor in Nm for an armaturecurrent of 8 A is _ _ _ _ _ _ _ _ .


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Sol. 13 Correct answer is 12.55 N-m.

A dc motor circuit is shown below.

Given

No load speed, N rpm1400=

Armature current, Ia A8=

Excitation voltage, V V230=

So, we obtain the torque of the motor as

t V IN

V Ia a

602

#w= = p

1400230 8

602

#

#= p

.9551400

230 8#

#=

. N m1255 -=

Q. 14 In a long transmission line with r , l, gand c are the resistance, inductance,shunt conductance and capacitance per unit length, respectively, the condition fordistortionless transmission is

(A) lgr c = (B) /r l c=(C) r g lc = (D) /g c l=


For a transmission line to be distortionless, the required condition is

LR

CG=

where G=shunt conductance (in siemens/meter)

C=capacitance (in Farad/meter)

R=Resistance (in ohm/meter)

L =inductance (in henries per meter)

So, for the given parameters of long transmission line, the required condition for

distortionless transmission becomes

lr

c

g=

or r c lg=

Q. 15 For a fully transposed transmission line(A) positive, negative and zero sequence impedances are equal

(B) positive and negative sequence impedances are equal

(C) zero and positive sequence impedances are equal

(D) negative and zero sequence impedances are equal


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Sol. 15 Correct option is (B).

A fully transposed transmission line is completely symmetrically and therefore

the phase impedance offered by it is independent of phase sequence of a balanced

set of current. In other word impedance offered by it to positive and negative

sequence currents are equal.

Q. 16 A 183 bus power system has 150 PQ buses and 32 PV buses. In the general case, toobtain the load flow solution using Newton-Raphson method in polar coordinates,the minimum number of simultaneous equations to be solved is _ _ _ _ _ .

Sol. 16 Correct answer is 332.

Given, number of PQ buses 150=

number of PV buses 32=

For PQ buses, 2 equations are formed (one for P and other for Q)

For PV buses only 1 equation is formed (for Q only)

Thus, the minimum number of simultaneous equations, required to obtain theload, is

Total number of equations =Number of equation for PQ buses +

Number of

equation for PV buses.

2#= (number of PQ buses)

1#+ (number of PV buses)

( ) ( )2 150 1 32# #= +

332=

Q. 17 The signal flow graph of a system is shown below. U s^ his the input and C s^ histhe output.

Assuming, h b1 1= and h b b a 0 0 1 1= - , the input-output transfer function,

G sU s

C s=^ ^h hh of the system is given by

(A) G ss a s a

b s b2

0 1

0 1=+ +

+^ h (B) G ss b s b

a s a2

1 0

1 0=+ +

+^ h(C) G s

s a s a

b s b2

1 0

1 0=+ +

+^ h (D) G ss b s b

a s a2

0 1

0 1=+ +

+^ hSol. 17 Correct option is (C).

We have the signal flow graph of the system as


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Q. 18 A single-input single output feedback system has forward transfer function G s^ hand feedback transfer function H s^ h. It is given that G s H s 1


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Q. 20 The two signals S1and S2, shown in figure, are applied to Y and X deflectionplates of an oscilloscope.

The waveform displayed on the screen is


We consider the two signals S1and S2applied to Y and X deflection plates for

the period T as


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observing the waveforms, we condude that

For /t T0 2< < , S0 1< >1 - and S 12 =-

Since, S1is applied to Y -deflection plate and S2is applied to X -deflection plate.

So, we haveFor /t T0 2< < ; ,y x0 1 1< < =

At /t T 2= ; x y 0= =

For /T t T2< < ; y1 0<


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For the state diagram, we form the truth table (irrespective of the current state)

as

A B X

0 0 1

0 1 1

1 0 1

1 1 0

Thus, from the truth table, it is clear that the logic gate is NAND gate.

Q. 22 An operational amplifier circuit is shown in the figure.

The output of the circuit for a given input viis

(A)

R

Rvi

1

2-

b l (B)

R

Rv1 i

1

2- +

b l(C)RR

v1 i1

2+b l (D) Vsat+ or Vsat-Sol. 22 Correct option is (D).

We have the op-amp circuit as

For an operation amplifier, we know that1. If v V> sati ; then output voltage will be Vsat

2. If v V< sati - ; then the output voltage will be Vsat-

3. If V v V<


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So, we obtain the output for the corresponding input vi. Applying KCL at non-

inverting terminal of first op-amp, we have

R

vR

v x0i i1 2

-+

- 0=

or vR R1 1

i1 2

+b l Rx2=or x

RR R

vi1

1 2= +b l ...(i)where xis the voltage at non-inverting terminal of second op-amp. Again, applying

KCL at non-inverting terminal of second op-amp, we get

R

xR

x v0 0- + - 0=

or v0 x2=

or v0R

vR R

2 i1

1 2= +

^ h [substituting equation (i)]

vRR2 1i

1

2= +b lThere is no single option so our assumption is wrong and v V> sati + and v V< sati -

. So, the output voltage is Vsat+ and Vsat- .

Q. 23 In 8085A microprocessor, the operation performed by the instruction LHLD

2100His(A) H 21H!^ h , L 00H!^ h(B) H M 2100 H!^ ^h h , L M 2101H!^ ^h h(C) H M 2101H!^ ^h h, L M 2100H!^ ^h h(D) H 00H!^ h , L 21H!^ h


For the 8085A microprocessor, given instruction is

LHLD 2100HThis operation loads registers register L and H with the content in memory at

locations 2100Hand 2101Hrespectively, i.e.

H^ h M 2101H! ^ h and L M 2100 H!^ ^h hQ. 24 A non-ideal voltage sourceVshas an internal impedance of Zs. If a purely resistive

load is to be chosen that maximizes the power transferred to the load, its value

must be(A) 0 (B) real part of Zs

(C) magnitude of Zs (D) complex conjugate of Zs


Consider the circuit shown below with non-ideal voltage source Vs, an internal

impedance Z R j X s s s= + , and purely resistive load RL .


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Current through the circuit is given by

I R R j X

V

s L s

s=+ +

or I R R X

V

s L s

s

2= + +^ ^h hTherefore, the power transferred to load is

P I RL2=

R R X

V R

s L s

s L2 2

2

=+ +^^ h h

For maximum power, we have

dRdP

L

0=

or[ ]

R R X

R R X R R R 1 2

s L s

s L s L s L

2 2 2

2 2

+ +

+ + - +

^^^ ^h hh h 0=

or R R R R X R R R 2 2 2s L s L s s L L2 2 2 2+ + + - - 0=

or R Xs s2 2+ RL

2=

or RL R Xs s2 2= +

i.e. to maximise the power transferred to pure resistive load (RL ), its value must

be equal to the magnitude of Zsof internal circuit.

Q. 25 The torque speed characteristics of motor TM^ hand load TL^ hfor two cases areshown in the figures (a) and (b). The load torque is equal to motor torque atpoints P, Q, Rand S.

The stable operating points are(A) Pand R (B) Pand S

(C) Qand R (D) Qand S


Since, increase in torque accelerates the motor, so the slope of torque speed

characteristic should be negative to restore the speed to original value, which in

turn makes the system stable.

Thus, for stable operating point, the slope of torque-speed characteristic should

be negative.

Hence, the points P and S are stable operating points.


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Q. 26 Integration of the complex function f zz

z

122

=-

^ h , in the counterclockwisedirection, around z 1 1- = , is

(A) ip- (B) 0

(C) ip (D) i2p


Given the complex function,

f z^ hz

z

122

=-

and the contour, along which integration to be performed, is

C: | |z 1- 1=

Here, z 1 1- = is a circle with radius 1 and centre at point (1,0). So, we sketch

the contour Cas

Now, we have the complex function

f z

^ h

z

z

122

=-

So, the poles and zeros of the function are

poles 1= , 1-

zeros 0= , 0

Therefore, only pole (z=1) lies within the curve C . So, the residue at z 1= pole

is

Residue limz

z

1z 1

2

=+" ^ h

limz

z1 1 1

121

z 1

2

=+

=+

="

Thus, the integral of complex function along the contour is obtained as

f z d z C ^ h# i2p= (sum of residues)

i221p= b l

ip=

Q. 27 The mean thickness and variance of silicon steel laminations are 0.2 mm and0.02 respectively. T he varnish insulation is applied on both the sides of the

laminations. The mean thickness of one side insulation and its variance are 0.1mm and 0.01 respectively. If the transformer core is made using 100 such varnishcoated laminations, the mean thickness and variance of the core respectively are

(A) 30 mm and 0.22 (B) 30 mm and 2.44(C) 40 mm and 2.44 (D) 40 mm and 0.24


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Q. 28 The function f x e 1x= -^ h is to be solved using Newton-Raphson method. If theinitial value of x0is taken as 1.0, then the absolute error observed at 2

nditeration

is _ _ _ _ _ .Sol. 28 Correct answer is 0.06005.

We have the function,

f x^ h e 1x= -So, its first derivative is obtained as

f xl h ex=By Newton-Raphson method, we have

xi 1+ xf x

f xi

i

i= -

l hhSince, we have the initial value,

x0 1=So, we obtain the 1st iteration as

x1 xf x

f x0

0

0= -

l hh x

e

e 1x

x

0 0

0

= - -

e

e1 111

= - - .e 36781= =-

Therefore, the 2nd iteration is

x2 x

f x

f x1

1

1= -

l

^^

hh

xe

e 1x

x

1 1

1

= - -

x e1 x1 1= - + -

. e03678 1 .03678= - + -

.06005=

Q. 29 The Nortons equivalent source in amperes as seen into terminals X and Y is_ _ _ _ .

Sol. 29 Correct answer is 2.

To obtain the Norton equivalent along X -Y terminal, we redraw the given circuit

as


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In the network, 5W resistor is short-circuit. So, the equivalent circuit is

Now, writing the KVL equation in two loops, we have

I I I5 5 5 sc1 1- - -^ h 0= ...(i) . .I I I5 25 25sc sc 1- - - +^ h 0= ...(ii)Solving equation (i),

I I10 5 sc1- 5=

I2 1 I 1sc= + ...(iii)

Solving equation (ii),

. I I75 5sc 1- .25=

or I I3 2sc 1- 1=

or I I3 1sc sc - - 1= [using equation (iii)]

or I2 sc 2=

Thus, Isc A1=

i.e. the Norton equivalent source as seen into terminals X and Y is 1 A.

Q. 30 The power delivered by the current source, in the figure, is _ _ _ _ .

Sol. 30 Correct answer is 3 W.

To obtain the required unknown, we redraw the given circuit as


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Applying KCL at nodeVx, we have

V V1

11

x x- + 2=

or V2 1x - 2=

or Vx2

3=

Therefore, the power delivered by current source is obtained as

=Voltage across current source #current through

current source

W23 2 3#= =

Q. 31 A perfectly conducting metal plate is placed in x-y plane in a right handed

coordinate system. A charge of 32 20pe+ coulombs is placed at coordinate, ,0 0 2^ h. 0e is the permittivity of free space. Assume i , j , kto be unit vectors

along x, yand z axes respectively. At the coordinate , ,2 2 0^ h, the electricfield vector E (Newtons/ Coulomb) will be

(A) k2 2 (B) k2-

(C) k2 (D) k2 2-


We have the given system of a charge and a perfectly conducting metal plate as

shown below.


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Due to presence of conducting plane, we can assume an image charge for the

system as shown below

So, we can observe from the figure that electric field vector along x-y planeis cancelled, i.e. zero. Therefore, the electric field vector acts along negative z

-direction and given as

E"

E E k2 21 2= + - td ^n h ...(i)

where E1and E2are the field intensity due to the charge Qand Q- at point

( , , )2 2 0 . So, we obtain

E1( ( ) ( ) ( ) )r

Q

41

41

2 2 2

32 20

20 2 2 2 2

0

pe pepe

= =+ +

2=

Similarly, we get, E 22 =

Substituting these values in equation (i), we obtain

E"

k22

22= + - te ^o h

k2=- t

Q. 32 A series RLC circuit is observed at two frequencies. At /krad s11w = , we notethat source voltage VV 100 01 c= results in a current . AI 003 311 c= . At

/krad s22w = , the source voltage VV 100 02 c= results in a current AI 2 02 c=. The closest values for R, L , C out of the following options are

(A) R 50W= ; mHL 25= ; FC 10m= ; (B) R 50W= ; mHL 10= ; FC 25m= ;

(C) R 50W= ; mHL 50= ; FC 5m= ; (D) R 50W= ; mHL 5= ; FC 50m= ;


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Given the current and voltage at different frequencies as

At krad11w = / s, V1 100 0= .I 003 311 c=

At krad22w = / s, V2 100 0= I 2 02 c=From the given data, it is clear that voltage and current are in phase at 22w = .

So, it satisfies the resonance condition. Therefore, we have

RI

V2

100 50W= = =

Also, for the resonance circuit, we have

LwC1w

=

or L10 23 # # C2 101

3# #

=

or L4 106#

C

1= ...(i)

Again, at 1w = krad/s, we have

cos31cZR=

or cos31cR L

C

R

12 2ww

=+ -b l

or cos31cL

C2500 10

101

503

3

=+ -b l

...(ii)

Solving equations (i) and (ii), we get

L mH10= and FC 25. m

Q. 33 A continuous-time LT I system with system function H w^ hhas the following pole-zero plot. For this system, which of the alternatives is TRUE ?

(A) H 0 > w^ ^h h ; 0>w(B) H w^ h has multiple maxima, at 1w and 2w(C) H H0 < w^ ^h h ; 0>w(D) H w =^ h constant; <


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From the pole-zero plot, we obtain the transfer function as

H w^ h( )( )( )( )

( )( )( )( )K

S P S P S P S P

S Z S Z S Z S Z * *

* *

1 1 2 2

1 1 2 2=

- - - -

- - - -

So, | ( ) |H w( )( )( )( )

( )( )( )( )K

S P S P S P S P

S Z S Z S Z S Z * *

* *

1 1 2 2

1 1 2 2=- - - -

- - - - ...(i)

Since, we can observe from the pole-zero plot that

| |P1 | |Z2=

and | |P2 | |Z1=

Thus, the numerator and denominator terms of equation are cancelled. Hence,

we get

| ( ) |H w K=

Q. 34 A sinusoid x t^ hof unknown frequency is sampled by an impulse train of period20 ms. The resulting sample train is next applied to an ideal lowpass filter withcutoff at Hz25 . The filter output is seen to be a sinusoid of frequency 20 Hz. This

means that x t^ h(A) 10 Hz (B) 60 Hz(C) 30 Hz (D) 90 Hz


Given the period of sampling train, msT 20S =

So, the sampling frequency is Hzf20 10

1 50S 3#

= =-

Let the frequency of signal ( )x t be fxAfter sampling the signal, the sampled signal has the frequency

f f

S x- f

50 x= -and f fS x+ f50 x= +

Now, the sampled signal is applied to an ideal low-pass filter with cut off frequency,

fc Hz25=

Since, the output of the filter carried a single frequency component of 20 Hz. So,

it is clear that only lower component ( )f fs x- passes through the filter, i.e.

f f 25


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Q. 35 A differentiable non constant even function x t^ h has a derivative y t^ h, andtheir respective Fourier Transforms are X w^ hand Y w^ h. Which of the followingstatments is TRUE ?

(A) X w

^ hand Y w

^ hare both real (B) X w

^ his real and Y w

^ his imaginary

(C) X w^ hand Y w^ hare both imaginary (D) X w^ his imaginary and Y w^ his realSol. 35 Correct option is (B).

Given the relation,

( )y t ( )dtd

x t=

From the properties of Fourier transform we know that

if ( )y t ( )dtd

x t=

Then ( )Y w ( )j Xw w= ...(i)

where ( )X w and ( )Y w are the Fourier transform of ( )x t and ( )y t respectively.

Now, we have the function ( )x t , which is differentiable, non-constant, even

function. So, its frequency response will be real, i.e. ( )X w is real

Hence, equation (i) gives the result that ( )Y w is imaginary.

Q. 36 An open circuit test is performed on 50 Hz transformer, using variable frequency

source and keeping V/ f ratio constant, to separate its eddy current and hysteresislosses. The variation of core loss/ frequency as function of frequency is shown inthe figure

The hysteresis and eddy current losses of the transformer at 25 Hz respectively

are(A) 250 W and 2.5 W (B) 250 W and 62.5 W

(C) 312.5 W and 62.5 W (D) 312.5 W and 250 W


Since, we know that

PC P f P f H e2= + ...(i)

where PC=total core losses of the transformer

f=frequency

PH =Hysteresis losses constant

Pe=eddy current losses constantNow, we have the plot of core loss/ frequency versus frequency as


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So, we rewrite equation (i) as

f

PC P P fH e= + ...(ii)

i.e. the plot between /P fC and fis a straight line. From the graph, we have thevalues of PH and Peas

PH 10= ; P 101

e =

From equation (i), we have the hysteresis and eddy current losses as

Pe P f P f Hysteresisloss Eddycurrentloss

H e2= +

S S

Thus, we get

P fH W10 25 250#= =

P fe2 . W

10

1 25 6252#= =

^ hQ. 37 A non-salient pole synchronous generator having synchronous reactance of 0.8pu is supplying 1 pu power to a unity power factor load at a terminal voltage of1.1 pu. Neglecting the armature resistance, the angle of the voltage behind the

synchronous reactance with respect to the angle of the terminal voltage in degreesis _ _ _ _ _ _ _ .

Sol. 37 Correct answer is 33.47.

For the non-salient synchronous generator, we have

Synchronous reactance, XS . . .p u08=

Supplied power, P . .p u1=

Power, factor, cosf 1=Terminal voltage, Vt . . .p u11=

Since, power factor is unity, so phase angle of armature current is also zero. So,

we get

Ia ( . ) ( ) .cosVP

11 11

111

t f= = =

or Ia .111 0< c=

Again, we define the armature current as

Ia j XE V 0

S

f t+ +d= -

So, Ef+d ( )I j X V 0


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.

. .11 0

1 08 90 11 0c+

+ += +

...11 0

1108 90c+ += +

or cos sinE j Ef fd d+ . ..

j111108= + ...(i)

Now, we have the power equation

P sinX

E V

S

f t d=

1.( . )

sinE

0811f

d=

sinEf d ..

1108= ...(ii)

So, from equation (i), we get

cosEf d .11= ...(iii)

Dividing equation (ii) by equation (iii), we get

cossin

E

E

f

f

dd

. .

.11 11

08#

=

tand..

12108=

d ..

.tan 12108

33471 c= =- b lQ. 38 A separately excited 300 V DC shunt motor under no load runs at 900 rpm

drawing an armature current of 2 A. T he armature resistance is .05W and leakageinductance is 0.01 H. When loaded, the armature current is 15 A. Then, the speed

in rpm is _ _ _ _ _ .

Sol. 38 Correct answer is 880 rpm.At no load, we have

Armature resistance, ra .05W=


Leakage inductance, L . H001=

So, we draw the circuit as

Since, we have 300 V dc shunt motor, so we obtain

E ( )noload 1 V I Ra a= -

or E1 .300 2 0 5= -^ h V300 1 299= - =

Again, when the load is applied, we have

Armature resistance, ra .05W=


So, we draw the circuit as


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Therefore, we obtain

E ( )load 2 V I Ra a= -

or E2 .300 15 05#= -

. .300 75 2925= - =

Since, we know that

E N\

So, EE

2

1

NN

2

1

=

or.2925

299 N900

2= (given rpmN 9001 = )

or N2 rpm880=

Q. 39 The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.

Assuming the transformer to be ideal, the value of the reactance X to improvethe input power factor to unity is _ _ _ _ _ _ _ _ _ .

Sol. 39 Correct answer is 23.618 ohm.

We have the transformer circuit,

Given the power absorbed by load; kW WP 4 40002 = = .

and the power factor is .cos lagging0892f =

From the given circuit, voltage across load is VV 1102 =

So, the current through the load is

I2 cosVP

2 2

22+f

f= -

.

.cos110 0 89

4000 0891#

+= - -

. .40858 27 13c+= -

This is the current through secondary winding of transformer. So, the current

through primary winding is given by

II

2

1

NN

1

2=


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where N1= number of turns in primary winding

N2= Number of turns in secondary winding.

From the given circuit, we have

NN

2

1

12

=

So,II

2

121=

Therefore, I1. .I

2 240 858 27 132 c+= = -

. .2043 2713c+=

Since, it is required to improve the input power factor to unity. So, the input

current should be red. Therefore, the imaginary part of I1should pass through

reactance X . T hus, current through X is

IX { }I Im 1=

. ( . )sin20 43 27 13c= .931=

Hence, the reactance is

X .I

220931220

X

= =

.23618W=

Q. 40 The parameters measured for a 220 V/ 110 V, Hz50 , single phase transformer

are :Self inductance of primary winding mH45=Self inductance of secondary winding mH30=

Mutual inductance between primary and secondary windings mH20=Using the above parameters, the leakage ,L Ll l1 2^ hand magnetizing L m^ hinductanceas referred to primary side in the equivalent circuit respectively, are(A) 5 mH, 20 mH and 40 mH (B) 5 mH, 80 mH and 40 mH

(C) 25 mH, 10 mH and 20 mH (D) 45 mH, 30 mH and 20 mH


For long transmission lines (over 200 km), parameters of a line are not lumped

but distributed uniformally throughout its length. The A B C D parameter of

equivalent p circuit are given as

A

C

B

D> HY Z

Y Y Z

Z Y Z

Y Z2

1

16

1

61

2

=

+

+

+

+b bl l

R

T

S

SSSS

V

X

W

WWWW ...(i)

The p-equivalent circuit

A B C D parameter of above circuit

AC

BD> H Y ZY Y Y Y Z ZY Z1 121 2 1 2 1= ++ + +

l

> H ...(ii)


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comparing equation (i) and (ii), we get

Zl B Z Y Z1

6= = +b l

where Z zl= = total series impedance/ phase

Y yl= = total shunt admittance/ phase

So, .Z j j05 400 200# W= =

Y j j5 10 400 2 106 3# # # W= =- -

So, Zl

( . )j j j

j200 16

2 20 200200 0066

3# #= + = -

-c m . .j j200 0934 1868# W= =

| |Zl .1868W=

Primary leakage reactance L l1 ( )L M2 45 2 201= - = -

mH5=Self inductance of secondary winding refers to primary side

2Ll NN

L VV

L2

12

22

12

2= =b bl l 110

22030

2

#= b lSo, 2Ll mH120=

Leakage inductance of secondary referred to primary side

L l2 2L M2= -l

( ) mH120 2 20 80= - =

magnetizing inductance, ( ) mmL M

2 2 20 40m= = =

Q. 41 For a 400 km long transmission line, the series impedance is . . / kmj00 05W+^ h and the shunt admittance is . . /mho kmj00 50 m+^ h . The magnitude of the seriesimpedance (in W) of the equivalent p circuit of the transmission line is _ _ _ _ .

Sol. 41 Correct answer is 187

Q. 42 The complex power consumed by a constant-voltage load is given by P j Q1 1+^ h,where, .kW kWP1 151# # and . kVAR kVARQ05 11# #

A compensating shunt capacitor is chosen such that . kVARQ 025# , where

Q is the net reactive power consumed by the capacitor load combination. Thereactive power (in kVAR) supplied by the capacitor is _ _ _ _ .

Sol. 42 Correct answer is 0.75kVAR.Given the constant voltage load,

P j Q1 1+

where kW1 . kWP 15<


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Q. 43 The figure shows the single line diagram of a single machine infinite bus system.

The inertia constant of the synchronous generator H 5= MW-s/MVA. Frequencyis 50 Hz. Mechanical power is 1 pu. The system is operating at the stable equilibrium

point with rotor angle d equal to 30c. A three phase short circuit fault occurs ata certain location on one of the circuits of the double circuit transmission line.During fault, electrical power in pu is sinPmax d . If the values of d and /d dtd atthe instant of fault clearing are 45cand 3.762 radian/s respectively, then Pmax(in

pu) is _ _ _ _ _ _ .

Sol. 43 Correct answer is 0.23 .Q. 44 The block diagram of a system is shown in the figure

If the desired transfer function of the system is

R s

C s

^^hh

s s

s

12=

+ +then G s^ his(A) 1 (B) s(C) / s1 (D)

s s s

s

23 2+ - --


We have the block diagram of given system as

By using minimization technique, we get

The closed loop transfer function for the dotted box is

G s

G s1+ ^^ hh


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Again, we obtain the closed loop transfer function for the dotted box as

s G s

sG s

1 1+ +^ ^^h hhSo, the block diagram reduces to

Thus, we have the overall transfer function as

T.F.

s s G s

s s G s s G s

s G s

G s

1 1

1 12

+ +

+ + +

+ +

^ ^

^ ^ ^^ ^

^

h h

h h hh h

h

6 @s s G s s

sG s

12=

+ + +^ ^h hhALTERNATIVE METHOD :

For the given block diagram, we draw the signal flow graph as

From the signal flow graph, we have forward paths

p1 sG s s G s

1= =^ ^h hand the closed loops

L 11 sG s

s

G s1 1= - =-^ ^ ^h h h ; L 12 G s G s 1= - =-^ ^ ^h h h;and L 13 G s s sG s 1= - =-^ ^ ^h h hSo, the transfer function is obtained as

T.F.p1 1

T

T=


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s

G sG s sG s

G s

1

1=

- - - -^ ^ ^

^ ^h h hh h

; E T.F.

s G s s s sG s

12=

+ + +^ ^^h hhFor any value of G s^ hgiven in the options, the desired system transfer functioncan not be achieved.

Q. 45 Consider the system described by following state space equations

x

x

1

2

o

o> H x

x u

0

1

1

1

0

11

2=

- - +> > >H H H ; y x

x1 0

1

2= 8 >B H

If uis unit step input, then the steady state error of the system is(A) 0 (B) 1/ 2

(C) 2/ 3 (D) 1


We have the state space equation,

x

x

1

2

o

o> H x

x u

0

1

1

1

0

11

2=

- - +> > >H H H

yx

x1 0

1

2= 8 >B H

From the state-space equation, we obtain the matrices

A 0

1

1

1=

- -> H

B 01= > H C 1 0= 8 BSo, we obtain

SI A-6 @ SS

S

S0

0 0

1

1

1 1

1

1= -

- - =

-

+> > >H H H

Therefore,

SI A1- -6 @

S S

S

S11 1

1

12= + +

+

-> H

Hence, we obtain the transfer function of the system as

T.F. [ ]C SI A B 1= - -

S S

S

S1 0

11 1

1

1 0

12=

+ +

+

-8 > >B H H

S S S1

1 1 01

2= + +8 >B H

S S 1

12= + +

This is the closed loop transfer function,

So we get

( )

( )

G S

G S

1+

S S 1

12=

+ + ( )G S

S S

12= +


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Thus, the steady state error for unit step input is

eSS ( )limG S11

S 0

=+"

lim1 1

1

0 2d d

=+

+"d

11 0

3=

+ =

Q. 46 The magnitude Bode plot of a network is shown in the figure

The maximum phase angle mf and the corresponding gain Gmrespectively, are

(A) 30c- and 1.73 dB (B) 30c- and 4.77 dB

(C) 30c+ and 4.77 dB (D) 30c+ and 1.73 dB


Given the magnitude Bode plot,

Since, slope at 1/ 3 is increasing and slope at 1 is decreasing. So, we can write

frequency response function as

G jw^ hj

j

1

31

w

w

=+

+

^b

hl

or G jw^ h /jj

11 1 33

1

w

w

=+

+c h mor G jw^ h

/

/

j b

j a

1

1ba

w

w=

+

+^ hhTherefore, the maximum phase angle is

mf /tan a b90 21

c= - - _ i /tan90 2 1 31c= - - ^ h 90 60 30c c c= - =

and the maximum phase occur at

w /ab 1 3= =Thus, we obtain the gain at the frequency as


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Gm

log log203

131 20 1

31 /2 2 2 2 1 2= + - +c bf cem lp mo

.476=Q. 47 A periodic waveform observed across a load is represented by

V t^ h sinsin

t t

t t

1 0 6

1 6 12

iFor this case, the diode conducts. Since, the voltage at negative and positive

terminals are same for ideal op-amp. So, for the 1st op-amp, we have

V V1 1=- + 0=

Therefore, the output of first op-amp is zero, which is applied to negative terminal

of second op-amp.

Thus we get the output v 0o = .

Case II : when v 0

set 3 ee -2014 gate solved

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