1372014 (paper - iii)

8/12/2019 1372014 (Paper - III)

1/51

No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical,photocopying, or otherwise without the prior permission of the author.

GATE SOLVED PAPERElectrical Engineering2014 (Paper - I II)

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.

NODIA AND COMPANYB-8, Dhanshree Tower I st, Central Spine, Vidyadhar Nagar, J aipur 302039Ph : +91 - 141 - 2101150www.nodia.co.inemail : [email protected]

8/12/2019 1372014 (Paper - III)

2/51

GATE SOLVED PAPER - EE

2014 (PAPER - III)

w w

w . n o d

i a . c

o . i n

General Aptitude set 8

Q. 1-5 carry one mark each

Q. 1 While tryingto collectI

an envelope fromunder thetableII

,

.Mr X fell downII I

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 partwill 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 tenseWould/ could have + past participles.

Thus, the complete sentence isIf she had known how to calibrate the instrument, she could have done theexperiment.

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 consistentmanner.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

8/12/2019 1372014 (Paper - III)

3/51

w w

w . n o

d i a .

c o . i n

GATE SOLVED PAPER - EE 2014 (PAPER - III)

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

The difference between successive terms is in AP with initial term 3 and thecommon difference 2. With this logic 64 is the wrong term and the correct numberin 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 Attempted1 2 21 17 62 3 15 27 23 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 marks de Total stu nts Total marks=

44133=

.3 02=

Q. 6 - Q. 10 carry one mark each

Q. 6 A dance programme is scheduled for 10.00 a.m. Some students are participatingin 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. T he 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 they

should not bring their parents. Participating students should come at 9.00a.m.

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

8/12/2019 1372014 (Paper - III)

4/51

w w

w . n o d

i a . c

o . i n


(iii) non-participating student should come at 10 a.m. with their parents.Combining these three instructions (B) option is most appropriate.

Q. 7 By the beginning of the 20 th century, 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 ofa 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%

Sol. 8 Correct option is (D).Let GDP in 2011-12 be x GDP in 2012-13 is . x 1 07 (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 x

50=

and at final 2012-13 GDP/ USD . x 601 07=

So, increase and decrease %100.

x

x x

50

60107

50 #= -

. %10 83= %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 ?

8/12/2019 1372014 (Paper - III)

5/51

w w

w . n o

d i a .

c o . i n


(A) 1 : 1 (B) 2 : 1(C) 1.5 : 1 (D) 2.5 : 1

Sol. 9 Correct option is (C).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 . x 1 5= (ratio is 1.5) Ratio of male students in 2012 to male students in 2011 is

. . :x x 1 5 1 5 1= =

Q. 10 Consider the equation Y 7526 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) 3162Sol. 10 Correct option is (C).

Given equation is Y 7526 8 8-^ ^h h 4364 8= ^ hor Y 8^ h 7526 43648 8= -^ ^h hOctal subtraction is done in same way as decimal subtraction. T he only differenceis that while obtaining carry we get 8 instead of 10.

75264364

3142

8

8

8

-^^

hhh

8/12/2019 1372014 (Paper - III)

6/51

w w

w . n o d

i a . c

o . i n



Q. 1 Two matrices A and B are given below :

A p r

q s = > H; B

p q pr qs

pr qs r s

2 2

2 2= +

+++> HIf the rank of matrix A is N , then the rank of matrix B is

(A) /N 2 (B) N 1-(C) N (D) N 2

Sol. 1 Correct option is (C).Given the two matrices,

A p r

q s = > H

and 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 theoperation, a i 2 a a

a a i i 211

211

! - as

A p q

s p r q 0= -> H

If s p r q - 0=

or ps r q - 0=

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

B p 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 i 2 a a a a i i 2

1121

1! - as

B p q pr qs

r s p q

pr qs 0

2 2

2 22 2

2=+ +

+ -++^ ^h h

R

T

SSSS

V

X

WWWW

If r s p q

pr qs 2 22 2

2

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

or ps r q - 0=then rank of matrix B is 1, otherwise the rank is 2.

Thus, from the above results, we conclude thatIf ps r q 0- = , then rank of matrix A and B is 1.If ps r q 0!- , then rank of A and B is 2.i.e. the rank of two matrices is always same. If rank of A is N then rank of B also N .

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

v /cos m st 2 2p p= a k

8/12/2019 1372014 (Paper - III)

7/51

w w

w . n o

d i a .

c o . i n


At st 3= , the difference between the distance covered by the particle and themagnitude of displacement from the origin is _ _ _ _ _ _ .

Sol. 2 Correct answer is (C).

Given the velocity of particle, v cos t 2 2

p p=

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

v dt dx =

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

x vdt t

t

0

3=

=

= # cos tdt 2 2t t

0

3 p p==

= # sin t 2 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 calculatingdistance, we have to consider speed and speed cant be negative, so the distanceis given by

x 0 v dt t

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 t

t

0

1

1

3p p= -= =

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

Thus, the difference between the distance covered and magnitude of displacementis x D | |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. I f y z x v i j k = + + , then f v : 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 f v :d ^ h x y y z z x 2 2 2= + + ...(i)where f , v are scalar and vector field respectively. From the properties of vectorfield, we have f v :d ^ h v f f v : :d d= -^ ^h hor v f : d

^ h f v f v : :d d= +

^ ^h h ...(ii)

Again, we have v y z x i j k = + +

8/12/2019 1372014 (Paper - III)

8/51

w w

w . n o d

i a . c

o . i n


So, v :d x y

y z

z x

22

22

22

= + +

0=

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

Thus, f v : 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 2

years respectively, then the value of k is _ _ _ _ _ .

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 kx 2= .where x is 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 dx 2

1

2 # 1=or k x 3

3

1

2: D 1=or k 3

8 1-b l 1= Thus, k .7

3 0 43= =

Q. 5 A function f t

^ h is shown in the figure.

The Fourier transform F w^ h of f t ^ h is(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

8/12/2019 1372014 (Paper - III)

9/51

w w

w . n o

d i a .

c o . i n


Sol. 5 Correct option is (A).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 line A to neutral voltage is V10 15 c for a balanced three phase star connectedload with phase sequence A B C . T he voltage of line B with respect to line C isgiven by(A) V10 3 105 c (B) V10 105 c

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

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 15 c=So, for balance 3-phase, we have

B 10 105 c= -and C 10 225 c= -

Therefore, the voltage of line B with respect to line C is v BC v v B C = - 10 105 10 225c c= - - -

10 3 75 c= -

Q. 7 A hollow metallic sphere of radius r is kept at potential of 1 Volt. T he totalelectric flux coming out of the concentric spherical surface of radius R r >^ h is(A) r 4 0pe (B) r 4 0 2pe(C) R 4 0pe (D) R 4 0

2pe

Sol. 7 Correct option is (A).

8/12/2019 1372014 (Paper - III)

10/51

w w

w . n o d

i a . c

o . i n


According to Gausss law, the total electric flux coming out of the concentricspherical surface is equal to the charge enclosed by the surface, i.e.

Q D ds := # Since, the electric potential is defined as v r

Q 4 0pe=

So, Q v r 4 0# pe= r 4 0pe= (since, v 1= volt)

Hence, D ds : # r 4 0pe=Q. 8 The driving point impedance Z s ^ h for the circuit shown below is

(A)s s

s s 2

3 13

4 2

++ + (B)

s s s

22 4

2

4 2

++ +

(C)s s

s 1

14 2

2

+ ++ (D)

s s s

11

4 2

3

+ ++

Sol. 8 Correct option is (A).We have the impedance circuit,

The capacitor and inductor can be replaced with its equivalent reactance (infrequency domain) as shown below.

So, we have the equivalent circuit as

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

Z s ^ h ||s s s s 1 1= + +b l ||s s s

s 1 12= + +b l

8/12/2019 1372014 (Paper - III)

11/51

w w

w . n o

d i a .

c o . i n


s

s s s

s s s

1 1

1 12

2#

= ++ +

+

s s s s 2132

= + ++

or Z s ^ h s s s s 23 134 2= ++ +Q. 9 A signal is represented by

x t ^ h t t 1 10 1= * The Fourier transform of the convolved signal /y t x t x t 2 2= *^ ^ ^h h h is(A) sin sin4 2 22w

w wa ^k h (B) sin4 22w wa k(C) sin

422w w^ h (D) sin

42

2

w wSol. 9 Correct option is (A).

Given the signal,

( )x t ,,

t

t

1 10 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 t 2= a kNow, the Fourier transform pair for rectangular function is sinrect t C 2

*

t t pwt c cm m

so, sinrect t C 2 2*

pwb cl m sin2 w w=

Similarly, sinrect t C 4 42*pwc bm l sin sin24 2 2 2w w w w= =^ ^h h

and ( ) sinrect t C 2*

pwc m / /sin 2 2ww= ^ h

Therefore, we have the fourier transforms ( )x t 2F " , ( )rect t F = " ,

8/12/2019 1372014 (Paper - III)

12/51

w w

w . n o d

i a . c

o . i n


//sin2

2w

w= ^ h

and { ( / )}x t 2F { ( / )} ( )sinrect t 4 2 2F ww= =

Thus, we obtain the fourier transform of signal ( )y t as { ( )}y t F { ( ) ( / )}x t x t 2 2F = * { ( )}. { ( / )}x t x t 2 2F F =

// ( )sin sin2

2 2 2w

ww

w= ^ h ( )sin sin4 2 22w

w w= b lQ. 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 t 3 8 6 12 14p p p= + +According to Nyquist criterion the sampling frequency is f s f 2> maxwhere f max is the maximum frequency of the message signal. For the given signal,we have maximum frequency as maxw 14 p=or f 2 maxp 14 p=or f max 7=

Thus, the sampling frequency is f s 2 7> #

or f s 14>i.e. minimum sampling frequency is 14 Hz.

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

Sol. 11 Correct option is (D).After sudden change of the load in three phase synchronous machine, the rotorhas 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. T he switch is expected to operate at a speed of0.7 Ns, but due to malfunctioning the switch fails to operate. T he torque-speedcharacteristic of the motor is represented by

8/12/2019 1372014 (Paper - III)

13/51

w w

w . n o

d i a .

c o . i n


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 nodiscontinuity in the characteristic curve. T hus, we have the modified characteristiccurve 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. T he field current iskept constant at rated value. T he torque of the motor in Nm for an armaturecurrent of 8 A is _ _ _ _ _ _ _ _ .

Sol. 13 Correct answer is 12.55 N-m.A dc motor circuit is shown below.

8/12/2019 1372014 (Paper - III)

14/51

w w

w . n o d

i a . c

o . i n


GivenNo load speed, N rpm1400=Armature current, I a A8=Excitation voltage, V V230=So, we obtain the torque of the motor as

t V I N

V I a a 602 #w= = p

1400230 8602 ##= p

.9 55 1400230 8# #=

. N m12 55 -=

Q. 14 In a long transmission line with r , l , g and c are the resistance, inductance,shunt conductance and capacitance per unit length, respectively, the condition fordistortionless transmission is(A) lg r c = (B) /r l c =(C) r g lc = (D) /g c l =

Sol. 14 Correct option is (A).For a transmission line to be distortionless, the required condition is

LR C

G =

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 fordistortionless transmission becomes

l r 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

Sol. 15 Correct option is (B).

A fully transposed transmission line is completely symmetrically and thereforethe phase impedance offered by it is independent of phase sequence of a balanced

8/12/2019 1372014 (Paper - III)

15/51

w w

w . n o

d i a .

c o . i n


set of current. I n other word impedance offered by it to positive and negativesequence currents are equal.

Q. 16 A 183 bus power system has 150 PQ buses and 32 PV buses. In the general case, to

obtain 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 ofequation 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 ^ h is the input and C s ^ h isthe output.

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

G s U s C s

=^ ^

^h h

h of the system is given by(A) G s

s a s a b s b 2

0 1

0 1=+ +

+^ h (B) G s s b s b a s a 2 1 01 0=

+ ++^ h

(C) G s s a s a

b s b 2

1 0

1 0=+ +

+^ h (D) G s s b s b a s a 2 0 10 1= + ++^ hSol. 17 Correct option is (C).

We have the signal flow graph of the system as

8/12/2019 1372014 (Paper - III)

16/51

w w

w . n o d

i a . c

o . i n


Now, we have to obtain the transfer function,

G s ^ h U s C s = ^hhFrom the SFG, we have two forward paths given as

p 1 h s s s h 1 1 1 1 10 20= =

p 2 h s s h 1 11 1= =

The graph determinant is given by T = 1 - (sum of all loop gains) + (sum of all gain products of 2 non-touching loops) - (sum of all gain products of 3 non-touching loops) ...+

So, T s a

s a 1 1 2 0= -

- + -a k s

a s a 1 1 20= + +

Also, we have the determinants for the forward paths p 1 and p 2 as 1T 1=

2T s a

s a 1 11 1= - - = +a k

Thus, using Masons gain formula, we obtain the transfer function as

T.F.p p p k k 1 1 2 2D

DD

D D= = +/

s a

s

a s h

s h

s a

1

1 1

120

20 1 1

=+ +

+ +^ ah k

s a s a h h s a

21 0

0 1 1=+ +

+ +^ hSince, we have h 1 b 1=and h 0 b b a 0 1 1= -Substituting it in equation (i), we get

T.F.s a s a

b b s 2

1 0

0 1=+ +

+

or G s V s C s

=^ ^

^h h

h s a s a

b b s 2

1 0

0 1=+ +

+

8/12/2019 1372014 (Paper - III)

17/51

w w

w . n o

d i a .

c o . i n


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

8/12/2019 1372014 (Paper - III)

18/51

w w

w . n o d

i a . c

o . i n


Q. 20 The two signals S 1 and S 2, shown in figure, are applied to Y and X deflectionplates of an oscilloscope.

The waveform displayed on the screen is

Sol. 20 Correct option is (A).We consider the two signals S 1 and S 2 applied to Y and X deflection plates forthe period T as

8/12/2019 1372014 (Paper - III)

19/51

w w

w . n o

d i a .

c o . i n


observing the waveforms, we condude thatFor /t T 0 2< < , S 0 1< >1 - and S 12 =-Since, S 1 is applied to Y -deflection plate and S 2 is applied to X -deflection plate.So, we haveFor /t T 0 2< < ; ,y x 0 1 1< < =At /t T 2= ; x y 0= =For /T t T 2 < < ; y 1 0<

8/12/2019 1372014 (Paper - III)

20/51

w w

w . n o d

i a . c

o . i n


as

A B X0 0 10 1 11 0 11 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 v i is(A) R

R v i 1

2- b l (B) R R v 1 i 12- +b l(C) R R v 1 i 1

2+b l (D) V sat+ or V sat-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 V sat2. If v V < sati - ; then the output voltage will be V sat-3. If V v V <

8/12/2019 1372014 (Paper - III)

21/51

w w

w . n o

d i a .

c o . i n


inverting terminal of first op-amp, we have

R v

R v x 0i i

1 2

- + - 0=

or v R R 1 1

i 1 2+b l R x

2=

or x R R R v i

1

1 2= +b l ...(i)where x is the voltage at non-inverting terminal of second op-amp. Again, applyingKCL at non-inverting terminal of second op-amp, we get

R x

R x v 0 0- + - 0=

or v 0 x 2=

or v 0 R v R R 2 i

11 2= +^ h [substituting equation (i)]

v R R 2 1i 12= +b l There is no single option so our assumption is wrong and v V > sati + and v V < sati -

. So, the output voltage is V sat+ and V sat- .

Q. 23 In 8085A microprocessor, the operation performed by the instruction LHLD2100 H is(A) H 21 H!^ h , L 00 H!^ h(B) H M 2100 H!^ ^h h, L M 2101 H!^ ^h h(C) H M 2101 H!^ ^h h, L M 2100 H!^ ^h h(D) H 00 H!

^ h, L 21 H!

^ hSol. 23 Correct option is (C).For the 8085A microprocessor, given instruction is LHLD 2100 H

This operation loads registers register L and H with the content in memory atlocations 2100 H and 2101 H respectively, i.e.

H h M 2101 H! ^ h and L M 2100 H!^ ^h hQ. 24 A non-ideal voltage source V s has an internal impedance of Z s . If a purely resistive

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

(A) 0 (B) real part of Z s (C) magnitude of Z s (D) complex conjugate of Z s

Sol. 24 Correct option is (C).Consider the circuit shown below with non-ideal voltage source V s , an internalimpedance Z R j X s s s = + , and purely resistive load R L .

Current through the circuit is given by

8/12/2019 1372014 (Paper - III)

22/51

w w

w . n o d

i a . c

o . i n


I R R j X V

s L s

s = + +

or I R R X

V

s L s

s 2= + +

^ ^h h Therefore, the power transferred to load is P I R L2=

R R X

V R s L s

s L2 2

2=

+ +^ h hFor maximum power, we have dR

dP L

0=

or [ ]R R X

R R X R R R 1 2s 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 X s s 2 2+ R L2=

or R L R X s s 2 2= +i.e. to maximise the power transferred to pure resistive load ( R L ), its value mustbe equal to the magnitude of Z s of internal circuit.

Q. 25 The torque speed characteristics of motor T M ^ h and load T L^ h for two cases areshown in the figures (a) and (b). T he load torque is equal to motor torque atpoints P , Q , R and S .

The stable operating points are(A) P and R (B) P and S (C) Q and R (D) Q and S

Sol. 25 Correct option is (B).Since, increase in torque accelerates the motor, so the slope of torque speedcharacteristic should be negative to restore the speed to original value, which inturn makes the system stable.

Thus, for stable operating point, the slope of torque-speed characteristic shouldbe negative.Hence, the points P and S are stable operating points.

8/12/2019 1372014 (Paper - III)

23/51

w w

w . n o

d i a .

c o . i n



Q. 26 Integration of the complex function f z z

z 12

2=

-^ h , in the counterclockwisedirection, around z 1 1- = , is(A) i p- (B) 0(C) i p (D) i 2p

Sol. 26 Correct option is (C).Given the complex function,

f z ^ h z z 12 2= -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 sketchthe contour C as

Now, we have the complex function

f z

^ h

z

z

12

2=

-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= poleis

Residue limz

z 1z 1

2=

+" ^ h lim z

z 1 1 1

121

z 1

2= + = + ="

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

f z d z C ^ h # i 2p= (sum of residues)

i 2 21p= b l

i p=

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 thelaminations. T he 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

8/12/2019 1372014 (Paper - III)

24/51

w w

w . n o d

i a . c

o . i n


Sol. 27 Correct option is (D).

Q. 28 The function f x e 1x = -^ h is to be solved using Newton-Raphson method. If theinitial value of x 0 is taken as 1.0, then the absolute error observed at 2nd iterationis _ _ _ _ _ .

Sol. 28 Correct answer is 0.06005.We have the function, f x ^ h e 1x = -So, its first derivative is obtained as f x l^ h e x =By Newton-Raphson method, we have

x i 1+ x f x f x

i i

i = - l^hhSince, we have the initial value,

x 0 1=So, we obtain the 1st iteration as

x 1 x f x f x

00

0= - l^hh x

e e 1

x

x

0 0

0

= - -

e

e 1 111

= - - .e 36781= =-

Therefore, the 2nd iteration is

x 2 x f x f x

11

1

= - l^hh x e

e 1x

x

1 1

1

= - -

x e 1 x 1 1= - + -

. e 0 3678 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 circuitas

8/12/2019 1372014 (Paper - III)

25/51

w w

w . n o

d i a .

c o . i n


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

Now, writing the KVL equation in two loops, we have I I I 5 5 5 sc 1 1- - -^ h 0= ...(i) . .I I I 5 2 5 2 5sc sc 1- - - +^ h 0= ...(ii)Solving equation (i), I I 10 5 sc 1 - 5= I 2 1 I 1sc = + ...(iii)Solving equation (ii), . I I 7 5 5sc 1- .2 5=or I I 3 2sc 1- 1=or I I 3 1sc sc - - 1= [using equation (iii)]or I 2 sc 2=

Thus, I sc 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

8/12/2019 1372014 (Paper - III)

26/51

w w

w . n o d

i a . c

o . i n


Applying KCL at node V x , we have

V V 11

1x x - + 2=

or V 2 1x - 2=

or V x 23=

Therefore, the power delivered by current source is obtained as = Voltage across current source # current throughcurrent source

W23 2 3#= =

Q. 31 A perfectly conducting metal plate is placed in x -y plane in a right handedcoordinate 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 , k to be unit vectorsalong x , y and z axes respectively. At the coordinate , ,2 2 0^ h, the electricfield vector E (Newtons/ Coulomb) will be

(A) k 2 2 (B) k 2-(C) k 2 (D) k 2 2-

Sol. 31 Correct option is (D).We have the given system of a charge and a perfectly conducting metal plate asshown below.

8/12/2019 1372014 (Paper - III)

27/51

w w

w . n o

d i a .

c o . i n


Due to presence of conducting plane, we can assume an image charge for thesystem as shown below

So, we can observe from the figure that electric field vector along x -y planeis cancelled, i.e. zero. T herefore, the electric field vector acts along negative z -direction and given as

E "

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

where E 1 and E 2 are the field intensity due to the charge Q and Q - at point( , , )2 2 0 . So, we obtain

E 1 ( ( ) ( ) ( ) )r

Q 4

14

12 2 2

32 20

20 2 2 2 2

0pe pe

pe= =+ +

2=Similarly, we get, E 22 =Substituting these values in equation (i), we obtain

E "

k 22

22= + - te ^ o h

k 2=- 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 50 W= ; mHL 25= ; FC 10 m= ; (B) R 50 W= ; mHL 10= ; FC 25 m= ;(C) R 50 W= ; mHL 50= ; FC 5 m= ; (D) R 50 W= ; mHL 5= ; FC 50 m= ;

8/12/2019 1372014 (Paper - III)

28/51

w w

w . n o d

i a . c

o . i n


Sol. 32 Correct option is (B).Given the current and voltage at different frequencies asAt krad11w = / s, V 1 100 0= .I 003 311 c=

At krad22w = / s, V 2 100 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

R I V

2100 50 W= = =

Also, for the resonance circuit, we have

Lw C 1

w=

or L10 23 # # C 2 10

13# #

=

or L4 10 6# C

1= ...(i)

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

cos31 c Z R =

or cos31 c R L

C

R 12 2w w

=+ -b l

or cos31 c L

C 2500 10

101

503

3

=+ -b l

...(ii)

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

L mH10= and FC 25. mQ. 33 A continuous-time LT I system with system function H w^ h has 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 H 0 < w^ ^h h; 0>w(D) H w =^ h constant; <

8/12/2019 1372014 (Paper - III)

29/51

w w

w . n o

d i a .

c o . i n


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

| |P 1 | |Z 2=and | |P 2 | |Z 1=

Thus, the numerator and denominator terms of equation are cancelled. Hence,we get | ( ) |H w K =

Q. 34 A sinusoid x t ^ h of 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 . T he filter output is seen to be a sinusoid of frequency 20 Hz. T hismeans that x t ^ h(A) 10 Hz (B) 60 Hz(C) 30 Hz (D) 90 Hz

Sol. 34 Correct option is (C).Given the period of sampling train, msT 20S =So, the sampling frequency is Hzf

20 101 50S 3#= =-

Let the frequency of signal ( )x t be f x After sampling the signal, the sampled signal has the frequency f f S x - f 50 x = - and f f S x + f 50 x = +Now, the sampled signal is applied to an ideal low-pass filter with cut off frequency, f c Hz25=Since, the output of the filter carried a single frequency component of 20 Hz. So,it is clear that only lower component ( )f f s x - passes through the filter, i.e. f f 25

8/12/2019 1372014 (Paper - III)

30/51

w w

w . n o d

i a . c

o . i n


Sol. 35 Correct option is (B).Given the relation,

( )y t ( )dt d x t =

From the properties of Fourier transform we know that

if ( )y t ( )dt d x t =

Then ( )Y w ( ) j X w 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, evenfunction. So, its frequency response will be real, i.e. ( )X w is realHence, 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 respectivelyare(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

Sol. 36 Correct option is (B).Since, we know that

P C P f P f H e 2= + ...(i)where P C = total core losses of the transformer f = frequency P H = Hysteresis losses constant P e = eddy current losses constantNow, we have the plot of core loss/ frequency versus frequency as

8/12/2019 1372014 (Paper - III)

31/51

w w

w . n o

d i a .

c o . i n


So, we rewrite equation (i) as

f P C P P f H e = + ...(ii)

i.e. the plot between /P f C and f is a straight line. From the graph, we have thevalues of P H and P e as

P H 10= ; P 101

e =

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

Hysteresis loss Eddycurrent loss

H e 2= +

S S

Thus, we get P f H W10 25 250#= =

P f e 2 . W10

1 25 62 52#= =

^ 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 thesynchronous 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 haveSynchronous reactance, X S . . .p u0 8=Supplied power, P . .p u1=

Power, factor, cos f 1= Terminal voltage, V t . . .p u1 1=Since, power factor is unity, so phase angle of armature current is also zero. So,we get

I a ( . ) ( ) .cosV P

1 1 11

1 11

t f= = =

or I a .1 11 0< c=

Again, we define the armature current as

I a j X E V 0

S

f t + +d= -

So, E f + d ( )I j X V 0

8/12/2019 1372014 (Paper - III)

32/51

w w

w . n o d

i a . c

o . i n


. . .1 1 01 0 8 90 1 1 0c+ + += +

. ..1 1 0 1 1

0 8 90 c+ += +

or cos sinE j E f f d d + . .. j 1 1 1 1

0 8= + ...(i)Now, we have the power equation

P sinX E V

S

f t d=

1 .( . ) sinE 0 81 1f d=

sinE f d ..

1 10 8= ...(ii)

So, from equation (i), we get cosE f d .1 1= ...(iii)

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

sinE E

f

f

d d . .

.1 1 1 1

0 8#

=

tan d ..

1 210 8=

d .. .tan 1 21

0 8 33 471 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 .0 5 W and leakageinductance is 0.01 H. When loaded, the armature current is 15 A. T hen, the speedin rpm is _ _ _ _ _ .

Sol. 38 Correct answer is 880 rpm.At no load, we haveArmature resistance, r a .0 5 W=Armature current, I a A2=Leakage inductance, L . H0 01=So, we draw the circuit as

Since, we have 300 V dc shunt motor, so we obtain E ( )noload 1 V I R a a = -or E 1 .300 2 0 5= - ^ h V300 1 299= - =Again, when the load is applied, we haveArmature resistance, r a .0 5 W=Armature current, I a A15=

So, we draw the circuit as

8/12/2019 1372014 (Paper - III)

33/51

w w

w . n o

d i a .

c o . i n


Therefore, we obtain E ( )load 2 V I R a a = -or E 2 .300 15 0 5#= - . .300 7 5 292 5= - =Since, we know that E N \

So, E E

2

1 N

N 2

1=

or .292 5299 N

9002

= (given rpmN 9001 = )

or N 2 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 lagging0 892f =From the given circuit, voltage across load is VV 1102 =So, the current through the load is

I 2 cosV P

2 2

22+f f= -

. .cos110 0 894000 0 891

# += - -

. .40 858 27 13 c+= - This is the current through secondary winding of transformer. So, the current

through primary winding is given by I

I 2

1N N

1

2=

8/12/2019 1372014 (Paper - III)

34/51

w w

w . n o d

i a . c

o . i n


where N 1 = number of turns in primary winding N 2 = Number of turns in secondary winding.From the given circuit, we have

N N

2

1 1

2=

So, I I

2

1 21=

Therefore, I 1 . .I 2 240 858 27 132 c+= = -

. .2043 2713 c+=Since, it is required to improve the input power factor to unity. So, the inputcurrent should be red. T herefore, the imaginary part of I 1 should pass throughreactance X . T hus, current through X is I X { }I I m 1=

. ( . )sin20 43 27 13 c= .9 31=Hence, the reactance is

X .I 220

9 31220

X = =

.23 618 W=

Q. 40 The parameters measured for a 220 V/ 110 V, Hz50 , single phase transformerare :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 l 1 2^ h and magnetizing L m ^ h inductanceas 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

Sol. 40 Correct option is (B).For long transmission lines (over 200 km), parameters of a line are not lumpedbut distributed uniformally throughout its length. The A B C D parameter ofequivalent p circuit are given as

AC

B D > H

Y Z

Y Y Z

Z Y Z

Y Z 2

1

1 6

16

1 2=

+

+

+

+b bl lR

T

SSSSS

V

X

WWWWW ...(i)

The p -equivalent circuit

ABCD parameter of above circuit

AC B D > H Y Z Y Y Y Y Z Z Y Z 1 121 2 1 2 1= ++ + +l

> H ...(ii)

8/12/2019 1372014 (Paper - III)

35/51

w w

w . n o

d i a .

c o . i n


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

Z l B Z Y Z 1 6= = +b lwhere Z zl = = total series impedance/ phase

Y yl = = total shunt admittance/ phaseSo, .Z j j 0 5 400 200# W= = Y j j 5 10 400 2 106 3# # # W= =- -

So, Z l

( . ) j j j j 200 1 62 20 200 200 0 066

3# #= + = -

-c m . . j j 200 0 934 186 8# W= = | |Z l .186 8 W=Primary leakage reactance L l 1 ( )L M 2 45 2 201= - = - mH5=Self inductance of secondary winding refers to primary side

2Ll N N

L V V

L2

12

2 2

12

2= =b bl l 110

220 302

#= b lSo, 2Ll mH120=Leakage inductance of secondary referred to primary side L l 2 2L M 2= -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 . . / km j 0 0 0 5 W+^ h and the shunt admittance is . . /mho km j 0 0 5 0 m+^ h . T he 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 Q 1 1+^ h,where, .kW kWP 1 1 51# # and . kVAR kVARQ 0 5 11# #A compensating shunt capacitor is chosen such that . kVARQ 0 25# , whereQ is the net reactive power consumed by the capacitor load combination. T hereactive power (in kVAR) supplied by the capacitor is _ _ _ _ .

Sol. 42Correct answer is 0.75kVAR.Given the constant voltage load, P j Q 1 1+where kW1 . kWP 1 5<

8/12/2019 1372014 (Paper - III)

36/51

w w

w . n o d

i a . c

o . i n


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. T he system is operating at the stable equilibriumpoint with rotor angle d equal to 30 c . 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 sinP max d . I f the values of d and /d dt d atthe instant of fault clearing are 45 c and 3.762 radian/ s respectively, then P max (inpu) 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 ^ h is(A) 1 (B) s (C) / s 1 (D)

s s s s

23 2+ - --

Sol. 44 Correct option is (B).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 s 1 + ^hh

8/12/2019 1372014 (Paper - III)

37/51

w w

w . n o

d i a .

c o . i n


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

s G s

sG s 1 1+ +^ ^h hh

So, 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 hh 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

p 1 s G s s G s 1= =^ ^h h

and the closed loops

L 11 s G s s G s 1 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. p 1 1TT

=

8/12/2019 1372014 (Paper - III)

38/51

8/12/2019 1372014 (Paper - III)

39/51

w w

w . n o

d i a .

c o . i n


( )G S S S

12= +

Thus, the steady state error for unit step input is

e SS ( )lim G S 11

S 0= + "

lim1 1

1

0 2d d

=+

+"d

11 03= + =

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

The maximum phase angle m f and the corresponding gain G m respectively, are(A) 30 c- and 1.73 dB (B) 30 c- and 4.77 dB(C) 30 c+ and 4.77 dB (D) 30 c+ and 1.73 dB

Sol. 46 Correct option is (C).Given the magnitude Bode plot,

Since, slope at 1/ 3 is increasing and slope at 1 is decreasing. So, we can writefrequency response function as

G j w^ h j j

131

w

w

= +

+

^b

hl

or G j w^ h / j j

1

1 1 331

w

w

=+

+

^c

hm

or G j w^ h // j b j a 11b a ww= ++^ hh Therefore, the maximum phase angle is m f /tan a b 90 2 1c= - - _ i /tan90 2 1 31c= - - ^ h 90 60 30c c c= - =and the maximum phase occur at

8/12/2019 1372014 (Paper - III)

40/51

w w

w . n o d

i a . c

o . i n


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

log log203

1 31 20 1 31 /2 2 2 2 1 2= + - +c bf ce m l p m o

.4 76=

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

V t ^ h sinsin t t t t 1 0 61 6 12

8/12/2019 1372014 (Paper - III)

41/51

w w

w . n o

d i a .

c o . i n


For balancing, we have

. j

35 100 1 10

136# # # #w - j C 105 10

13# ##w= or C . F0 3 m=

Q. 49 Two monoshot multivibrators, one positive edge triggered M 1^ h and anothernegative edge triggered M 2^ h

, are connected as shown in figure

The monoshots M 1 and M 2 when triggered produce pulses of width T 1 and T 2 respectively, where T T >1 2 . T he steady state output voltage v 0 of the circuit is

8/12/2019 1372014 (Paper - III)

42/51

w w

w . n o d

i a . c

o . i n


Sol. 49 Correct option is (C).We have the multivibrators circuit as shown below

When monoshot multivibrator M 1 is triggered, it produces a pulse of T 1 duration.

Similarly, when monoshot multivibrator M 2 is triggered, it produces a pulse of T 2 duration

Let, the output Q 2 is initially high.So, the output v o will be high for time period T 2.

For this period, Q 02 =So, the output of AND gate is low. Since, M 1 is positive triggered. So, M 1 will beOFF for duration T 2.

After T 2 time, Q 2 0=and Q 2 1=So, the output of AND gate is 1; i.e. M 1 is triggered. Hence, output Q 1 is highfor duration T 1. Since, M 2 is negative edge triggered, So M 2 will be off for theduration T 1 and hence, output v o will be as

After T 1 time, the output Q 1 becomes low, and the M 2 gets ON (due to negative

8/12/2019 1372014 (Paper - III)

43/51

w w

w . n o

d i a .

c o . i n


edge triggering). Hence, the output v o again goes high for duration T 2. Thus, thecomplete output waveform is

Q. 50 The transfer characteristic of the Op-amp circuit shown in figure is

Sol. 50 Correct option is (C).Given the op-amp circuit

8/12/2019 1372014 (Paper - III)

44/51

w w

w . n o d

i a . c

o . i n


For the given circuit, we consider the following two cased.Case I : when v 0>i For this case, the diode conducts. Since, the voltage at negative and positiveterminals are same for ideal op-amp. So, for the 1st op-amp, we have V V 1 1=- + 0=

Therefore, the output of first op-amp is zero, which is applied to negative terminalof second op-amp.

Thus we get the output v 0o = .Case II : when v 0

1372014 (paper - iii)

Documents