EE-GATE 2018d1zttcfb64t0un.cloudfront.net/.../Gateforum_EE_GATE-2018_solutions.… · EE-GATE 2018...

EE-GATE 2018

© All rights reserved by Thinkcell Learning Solutions Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

1

EE-GATE 2018


2

Section-I: General Ability

1. Functions F(a,b) and G(a,b) are defined as follows:

F(a,b) = (a-b)2 and G(a,b) =|a-b|, where |x| represents the absolute value of x.

What would be the value of G(F(1,3), G(1,3))?

(A) 2 (B) 4 (C) 6 (D) 36

Key: (A)

Exp: Given,

2

2

F a,b a b and G a,b a b

F 1,3 1 3 ; G 1,3 1 3

F 1,3 4 ; G 1,3 2

G F 1,3 , G 1,3 G 4,2 F 1,3 4&G 1,3 2

4 2 G a,b a b

G F 1,3 , G 1,3 2

2. “Since you have gone off the ____________, the _________ sand is likely to damage the car.”

The words that best fill the blanks in the above sentence are

(A) course, coarse (B) course, course (C) coarse, course (D) coarse, coarse

Key: (A)

3. “A common misconception among writers is that sentence structure mirrors thought; the more

_______ the structure, the more complicated the ideas.”

The word that best fills the blank in the above sentence is

(A) detailed (B) simple (C) clear (D) convoluted

Key: (D)

4. For what values of k given below is

2k 2

k 3

an integer?

(A) 4,8,18 (B) 4, 10, 16 (C) 4,8,28 (D) 8, 26, 28

Key: (C)

Exp: Actually we can find an infinite no. of values of k, for which

2k 2

k 3

to be an integer

From the given choices;

2 2k 2 4 2

If k 4, then 36 integer.k 3 4 3

EE-GATE 2018


3

2 2

2 2

k 2 8 2 100If k 8, then 20 integer

k 3 8 3 5

28 2 30 30 30If k=28, then 36 integer.

28 3 25 25

5. The three roots of the equation f(x) = 0 are x = {-2, 0, 3}. What are the three values of x for which

f(x-3) = 0?

(A) -5, -3, 0 (B) -2, 0, 3 (C) 0, 6, 8 (D) 1, 3, 6

Key: (D)

Exp: Given that;

X= -2, 0, 3 are the 3 roots of f x 0.

i.e., f 2 0, f 0 0 and f 3 0.

Now we should find the values of x for which

f x 3 0 x 3 2; x 3 0; x 3 3

f 2 0; f 0 0; f 3 0

x 2 3; x 3; x 6

x 1, 3, 6

6. A class of twelve children has two more boys than girls. A group of three children are randomly

picked from this class to accompany the teacher on a field trip. What is the probability that the

group accompanying the teacher contains more girls than boys?

(A) 0 (B) 325

864 (C)

525

864 (D)

5

12

Key: (B)

Exp: Given, a class of 12 children has no two more boys than girls.

Let, the no.of girls x

Then no.of boys x 2

x x 2 12

2x 10

x 5

No.of Grils 5

No.of Boys 7

Total no. of ways of selection of 3 children = 312C n s

Let A be the event that no. of girls are more than boys

EE-GATE 2018


4

i.e.

Favorable no. of ways of A 3 0 2 17 75 5C C C C

2 35 510 70 80 C C 10

312

n A 80 80 3!Prob A

n S C 12 11 10

80 6 8 4 3250.3636

12 11 10 22 11 864

325Req Prob

864

7. An e-mail password must contain three characters. The password has to contain one numeral from 0

to 9, one upper case and one lower case character from the English alphabet. How many distinct

passwords are possible?

(A) 6,760 (B) 13,520 (C) 40,560 (D) 1,05,456

Key: (C)

Exp: We know that;

Total no. of digits 0,1,2,....9

Total no. of upper case English alphabets 26 A,B,C,....,Z

Total no. of lower case English alphabets 26 a,b,c,....,z

No. of ways of selecting a digit out of 11010 C 10

No. of ways of selecting an upper case alphabet 126C 26

No. of ways of selecting an lower case alphabet 126C 26

Permutation = selection + arrangement

Case I

3G 0B&

or Case II

2G 1B&

Two cases possible

Selection

6760 6 3! 6

40,560

a1 AbB2..3 ... ...ZZ9

arrangement of3things

arrangement

Total no.of distinct passwords 10 26 3!26

EE-GATE 2018


5

8. In a certain code, AMCF is written as EQGJ and NKUF is written as ROYJ. How will DHLP be

written in that code?

(A) RSTN (B) TLPH (C) HLPT (D) XSVR

Key: (C)

Exp: Given

9. A designer uses marbles of four different colours for his designs. The cost of each marble is the

same, irrespective of the colour. The table below shows the percentage of marbles of each colour

used in the current design. The cost of each marble increased by 25%. Therefore, the designer

decided to reduce equal numbers of marbles of each colour to keep the total cost unchanged. What

is the percentage of blue marbles in the new design?

Blue Black Red Yellow

40% 25% 20% 15%

(A) 35.75 (B) 40.25 (C) 43.75 (D) 46.25

Key: (C)

Exp: Assume that,

Total no. of marbles = 100

C.P of 1 marble =Rs 100

Total C.P of 100 marbles 100 100

New cost price of 1 marble = Rs 125 [Given]

No. of marbles with new price 100 100

80125

Blue Black Red Yellow

No.of reduced marbles 100 80 20 5 5 5 5 20

Now; the no. of blue marbles = 40-5=35

Total no. of marbles =10-20=80

DH L P H LP T

4 4

4 4

A M C F E Q G J & N K U F RO Y J

4

4 4

4 4

4 4

4

EE-GATE 2018


6

% of blue marbles in new design 35

100 %80

35 5%

4

43.75%

10. P,Q,R and S crossed a lake in a boat that can hold a maximum of two persons, with only one set of

oars. The following additional facts are available.

(i) The boat held two persons on each of the three forward trips across the lake and one person on

each of the two return trips.

(ii) P is unable to row when someone else is in the boat.

(iii) Q is unable to row with anyone else except R.

(iv) Each person rowed for at least one trip.

(v) Only one person can row during a trip.

Who rowed twice?

(A) P (B) Q (C) R (D) S

Key: (C)

Exp: Section of 2 persons out of 4 (P, Q, R, S), i.e., 24C = 6ways

From the fact (i), there are

3 forward trips Two person can travel

2 returned trips only 1 person travel

From the facts (ii) and (iii); P and Q should not travel.

To satisfy (iv) fact; only Q and R should travel in 1st trip.

P

P,RR

ndSecond trip : In 2 trip only R & P should travel.

R Rowed in forward trip

P Rowedin return trip

R,QQ

R

First trip : Q Rowed in forward trip

R Rowed in return trip

PQ PR PS QR QS RS

EE-GATE 2018


7

Section-II: Electrical Engineering

1. Four power semiconductor devices are shown in the figure along with their relevant terminals. The

devices (s) that can carry dc current continuously in the direction shown when gated appropriately

is (are)

(A) Triac only (B) Triac and MOSFET

(C) Triac and GTO (D) Thyristor and Triac

Key: (B)

Exp: SCR: conducts only from anode to cathode

Triac:

(1) Triac is combination of two anti parallel SCRs

P,S S

rdThird trip : In3 triponlyP &Swill travel

S must rowed in forward trip From ii fact

R rowed twice.

A

K

G

I

A

GK

I

Thyristor

1MT

G 2MT

Triac

I

A

GK

I

GTO

D

S

G

MOSFET

I

Triac

G

2MT

1MT

EE-GATE 2018


8

(2) 1 2When MTis ve &MT is ve

(3) 1 2When MT is ve & MT is ve

GTO: GTO conducts only from anode to cathode

MOSFET

(1) When D is + ve & S is –ve, diode is OFF

(2) When D is –ve & S is +ve, diode is ON

I

G

G

D

S

G

D

S

D ve

I S ve

D

S

I

I

A

K

G

EE-GATE 2018


9

2. The op-amp shown in the figure is ideal. The input impedance in

in

v

t is given by

(A) 1

2

RZ

R (B) 2

1

RZ

R (C) Z (D) 1

1 2

RZ

R R

Key: (B)

Exp: Given op- Amp is ideal

By virtual ground concept

in

in 0in

0 2in

1 2

10 in

2

1in in in

2 1in in

2

in 2

in 1

V V

V Vi

Z

By voltage divider principle

V RV V

R R

RV V 1

R

RV V V

R Ri V

Z ZR

V RZ

i R

Z

ini

inVV

1R

2R

0V

Zini

inV

1R

2R

0V

EE-GATE 2018


10

3. Two wattmeter method is used for measurement of power in a balanced three phase load supplied

from a balanced three phase system. If one of the wattmeters reads half of the other (both positive),

then the power factor of the load is

(A) 0.532 (B) 0.632 (C) 0.707 (D) 0.866

Key: (D)

Exp: In two Wattmeter of power measurement we know

1 21

1 2

3tan well known standard formula

It is given that one meter reads half of the other

1 22 1

11

1

11

1

1

1

1

or wecan take2 2

32

tan

2

32

tan

32

1tan 30

3

power factor cos

3cos 30 0.866

2

4. The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph

is

(A) 11 (B) 12 (C) 13 (D) 14

Key: (B)

Exp: we know in a network

b n 1

Where : number of loops = 5 (given)

b :number of branches (to be found)

h : number of nodes = 8 (given)

b n 1

5 b 8 1

b 12

EE-GATE 2018


11

5. A continuous time input signal x(t) is an eigenfunction of an LTI system, if the output is

(A) kx(t), where k is an eigenvalue

(B) j tke x t, where k is an eigenvalue j te is a complex exponential signal

(C) x(t) j te ,

where j te is a complex exponential signal

(D) kH , where k is an eigenvalue and H is a frequency response of the system

Key: (A)

Exp: Consider for example eigen function, stx t e

Where H(s) is Laplace transform of h(t).

At specific 0S S , 0H S becomes a constant quantity.

Thus sty t K.e or K.x t

Where „K’ is eigen value.

Thus option (A) is correct.

In option (B) extra frequency forms one getting generated at output which is not possible for LTI

system.

Option (C) and Option (D) one also not the correct answer due to extra frequency term getting

generated in output.

6. In the logic circuit shown in the figure, Y is given by

(A) Y=ABCD (B) Y = (A+B) (C+D) (C) Y=A+B+C+D (D) Y=AB+CD

Key: (D)

Exp: y MN

y AB. CD

y AB CD

ste h t sty t H s e

Y

A

B

C

D

Y

A

B

C

D

M AB

N CD

EE-GATE 2018


12

7. The value of the integral 2

C

z 1

z 4

dz in counter clockwise direction around a circle C of radius 1

with center at the point z = - 2 is

(A) i

2

(B) 2 i (C)

i

2

(D) 2 i

Key: (A)

Exp: 2

C

z 1dz

z 4

2z 4 0

z 2

z 2 lies inside & z 2 lies outside 'C'

By Cauchy's integralformula,

2

C C C

z 1

z 1 z 1 z 2dz dz dz

z 4 z 2 z 2 z 2

2 if ( 2)

2 1 i2 i

2 2 2

8. Let f be a real valued function of a real variable defined as f(x) = x-[x], where [x] denotes the

largest integer less than or equal to x. The value of 1.25

0.25

f x dx is ______(up to 2 decimal places).

Key: (0.5)

Exp: We know that

x x x where x integral partof x,

1.25 1.25

0.25 0.25

1 1.25

0.25 1

1.2512 2

0.25 1

x x x where x fractionalpart of 'x '

f x dx x dx

x dx x 1 dx

x xx

2 2

1

2

0.5

2x

4 1 0 1 2

y

C

3

y

1

1 0 1 2 3 x

EE-GATE 2018


13

9. The value of the directional derivative of the function 2 2 2x,y,z xy yz zx at the point (2,-1,

1) in the direction of the vector p = i + 2j + 2k is

(A) 1 (B) 0.95 (C) 0.93 (D) 0.9

Key: (A)

Exp: i j Kx y z

2 2 2

2, 1,1

i y 2zx j 2xy z k 2yz x

5i 3 j 2k

i 2 j 2kRequired directional deivative 5i 3 j 2k .

1 4 4

5 6 41

3

10. Match the transfer functions of the second order systems with the nature of the systems given

below.

Transfer functions Nature of system

P. 2

15

s 5s 15 I. Overdamped

Q. 2

25

s 10s 25 II. Critically damped

R. 2

35

s 18s 35 III. Underdamped

(A) P-I, Q-II, R-III (B) P-II, Q-I, R-III (C) P-III, Q-II, R-I (D) P-III, Q-I, R-II

Key: (C)

Exp:

2

2 2 2

n n

n

n

2

n

2 2 2

n n

15 nP

s 5s 15 s s

15

2 5

51 underdamped

2 15

25Q

s 10s 25 s 2 s

n 25

2 5 10

1 critically damped

EE-GATE 2018


14

2

n

2 2 2

n n

n

n

35R

s 18s 35 s 2 s

35

18

181 overdamped

2 35

P III Q II R I

11. A 1000×1000 bus admittance matrix for an electric power system has 8000 non –zero elements. The

minimum number of branches (transmission lines and transformers) in this system are ______ (up to 2

decimal places).

Key: (3500)

Exp: Given 1000 1000 bus matrix

3 3 6

6

6

N 1000

Total root Buses 10 10 10

given no of non zero elements 8000

No of zero elementsSparsity s

Total noof elements

10 80000.992

10

No of transmission lines and transformers

2 2N 1 s N 1000 1 0.992 1000

35002 2

12. A single phase 100kVA, 1000 V/100V. 50Hz transformer has a voltage drop of 5% across its series

impedance at full load. Of this, 3% is due to resistance. The percentage regulation of the transformer at

full load with 0.8 lagging power factor is

(A) 4.8 (B) 6.8 (C) 8.8 (D) 10.8

Key: A

Exp: puImpedance 5% 0.05pu Z

pu

2 2

pu pu pu

2 2

pu pu

Resistance 3% 0.03pu R

Reactance, X z R

0.05 0.03 0.04pu

VR 0.8 p.f lag R cos X sin

0.03 0.8 0.04 0.6

0.048pu 4.8%

EE-GATE 2018


15

13. Let f be a real valued function of a real variable defined as f(x) = x2 for x 0 , and 2f x x for x 0 .

Which one of the following statements is true?

(A) f(x) is discontinuous at x=0.

(B) f(x) is continuous but not differentiable at x=0

(C) f(x) is differentiable but its first derivative is not continuous at x=0.

(D) f(x) is differentiable but its first derivative is not differentiable at x=0.

Key: (D)

Exp: Given

f x x x

f x in continuals&differentiable

x xf ' x x

x

2 x

f x is differentiable but its first derivative is not differentiable at x=0

14. A positive charge of 1nC is placed at (0, 0, 0.2) where all dimensions are in metres. Consider the x-y

plane to be a conducting ground plane. Take 12

0 8.85 10 F / m. The Z component of the E field at

(0, 0, 0.1) is closest to

(A) 899.18V/m (B) -899.18V/m (C) 999.09V/m (D) -999.09 V/m

Key: (D)

Exp:

1212 3

0 12

99 Zz

12 3 312 12

5 5

z

z z

QRE

4 R

1 10 0.3 aˆ1 10 0.1aE

4 8.854 10 0.1 4 8.854 10 0.3

10 10a

4 8.854 4 8.854 9

898.774 99.863 a 999.09a V m

15. A separately excited dc motor has an armature resistance Ra=0.05 . The field excitation is kept

constant. At an armature voltage of 100V, the motor produces a torque of 500Nm at zero speed.

Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of

150 V is _____ (up to 2 decimal places).

Key: (600)

Exp: Torque

Produced by motor = 500N-m

EE-GATE 2018


16

At zero speed, such that

L

b

a

a

a

nL

nL

b

b

b

E K 0

V E 100 0I 2000A

R 0.05

Also,T k I

500 5 1k T I

2000 20 4

For armature voltage of 150V, no load speed in

At no load I 0

V E 150V

E k Field excitation or flux is constant, k constant

E 150600 rad s

K 1 4

ec

16. In a salient pole synchronous motor, the developed reluctance torque attains the maximum value when

the load angle in electrical degrees is

(A) 0 (B) 45 (C) 60 (D) 90

Key: (B)

Exp: Reluctance torque sin 2

Torque ix maximum when

2 90 45

17. Consider a lossy transmission line with V1 and V2 as the sending and receiving end voltages,

respectively. Z and X are the series impedance and reactance of the line, respectively. Z and X are the

series impedance and reactance of the line, respectively. The steady-state stability limit for the

transmission line will be

(A) greater than 1 2V V

X (B) less than 1 2V V

X

(C) equal to 1 2V V

X (D) equal to 1 2V V

Z

Key: (B)

Exp: Power with impudence Z,

1

2

1 2 2

2

1 2 2max

V V VP cos cos

Z Z

V V VP cos for

Z Z

Power with reactence X,

0.05100V

EE-GATE 2018


17

2

2 1

2

1 2

1 2max

max max

1 2max

V VP sin

X

V VP for 90

X

i.e P P

V VSteady state stability limit for transmission line P

X

18. A single phase fully controlled rectifier is supplying a load with an anti-parallel diode as shown in the

figure. All switches and diodes are ideal. Which one of the following is true for instantaneous load

voltage and current?

(A) 0 00 &i 0 (B) 0 00 &i 0

(C) 0 00 &i 0 (D) 0 00 &i 0

Key: (C)

Exp: During forward bias

Let in m 0 nV t sin t ,V t sin t positive

During reverse bias

off

ON

ON

off

oV

0i

Load

Free wheeling diode is off

ON

Off

off

ON

Load

o m

Free wheeling diode isoff

V t V sin t positive

0i

L

O

A

D

0i

0V

EE-GATE 2018


18

Key concept:

SCR allows current only in one direction, Anode to cathode 0i 0

19. Consider a unity feedback system with forward transfer function given by

1G s

s 1 s 2

The steady state error in the output of the system for a unit step input is _________ (upto 2 decimal

places).

Key: (0.67)

Exp: ps 0

ss

p

1unit stepinput, K limG s

2

A 1for unit step input, e 2 3 0.67

1 K 1 1 2

20. In the two port network shown, the h11 parameter 111 2

1

VWhere, h ,when V 0

I

in ohms is

___________ (up to 2 decimal places).

Key: (0.5)

Exp: when 2V 0 the output port is short circuited

1I

1

12I

11V

2V

1

inV t

2

2

mV

mV

mV

0

0

EE-GATE 2018


19

Writing KVL on the input loop we have

1 1 1 2

1 1 2

V 1 I 1 I I 0

V 2I I ... 1

Writing KVL on the output loop we have

2 1 1 2

2 1

2 1

1 I 2I 1 I I 0

2I 3I 0

3I I ....(2)

2

Using equation 2 in equation 1, we have

2

1 1 1

1 1

111

1 v 0

1

3V 2I I

2

V 0.5I

Vh 0.5

I

h 0.5

21. The waveform of the current drawn by a semi-converter from a sinusoidal AC voltage source is shown

in the figure. If I0=20 A, the rms value of fundamental component of the current is ___________ A (up 2

decimal places).

1I 1

12I12I

2I

2V 01 2I I 1

2 1I 2I

1V

1

mV sin t

voltage and

current

0

30 180

210

0I t

0I

EE-GATE 2018


20

Key: (17.39)

Exp:

Fourier series representation of io(t) is

00

n 1,3,5

4I n ni t cos sin n t

n 2 2

RMS value of nth harmonic is 00

4I n 2 2 nn cos I cos

2 n 22n

RMS value of fundamental is S1 0

2 2I I cos

2

2 2 30

20cos 17.39A2

22. In the figure, the voltages are 1 2t 100cos t , t 100cos t /18 and

3 t 100cos t / 36 . The circuit is in sinusoidal steady state, and R << L. P1, P2 and P3 are

the average power outputs. Which one of the following statements is true?

(A) P1 = P2 = P3 = 0 (B) P1 < 0, P2 >0, P3 > 0

(C) P1 < 0, P2 > 0, P3 < 0 (D) P1 >0, P2 < 0, P3 >0

Key: (C)

Exp: 1V t 100 cos t

2V t 100 cos t18

0I

0

0I

2

t

0i t

1v t

1P

2v t

2P 3P

3v t

R L L R

EE-GATE 2018


21

3

1 2 3

1

2

3

V t 100 cos t36

V t V t V t 100

V t 0

180V t 10

18 18

V t 536

2V t is leading 3 1V t andV t by10 and 5 .

2V t is source & delivering power 2P 0

1 3V t andV t are sinks & absorbing power 3 1P 0 , P 0

23. Consider a non singular 2×2 square matrix A. If trace (A) = 4 and trace (A2) = 5, the determinant of the

matrix A is _____ (up to 1 decimal place).

Key: (5.5)

Exp: a b

Let A Trace of A a dc d

2

2

2

2 2 2

2 2

a b a b a bc ab bd&A

c d c d ac cd bc d

Trace of A a bc bc d

a 2bc d

Given a d 4 ... 1 &

2 2

2 2 2 2

a 2bc d 5 ... 2

1 a d 16 a d 2ad 16

5 2bc 2ad 16 from 2

2 ad bc 11

11ad bc 5.5

2

24. The series impedance matrix of a short three phase transmission line in phase coordinates is

s m m

m s m

m m s

Z Z Z

Z Z Z

Z Z Z

. If the positive sequence impedance is (1+j10) , and the zero sequence is (4+j31) .

Then the imaginary part of Zm (in ) is _______ (up to 2 decimal places).

2

3

1

V t 10

V t 5

V t 0

EE-GATE 2018


22

Key: (7)

Exp: 1 s mZ Z Z 1 j10 .... 1

0 s m

0 1 m

m

m

Z Z 2Z 4 j31 .... 2

2 1

Z Z 3Z 3 j21

Z 1 j7

Im Z 7

25. The positive, negative and zero sequence impedances of a 125 MVA, three phase, 15.5kV, star-

grounded, 50Hz generator are j0.1 pu, j0.05 pu and j0.01 pu respectively on the machine rating base.

The machine is unloaded and working at the rated terminal voltage. If the grounding impedance of the

generator is j0.01 pu, then the magnitude of fault current for a b-phase to ground fault (in kA) is

____________ (up to 2 decimal places).

Key: (73.51)

Exp: f

1 2 0 n

3I pu

Z Z Z 3Z

3 3

j0.1 j0.05 j0.01 3 0.01 0.19

f f bpu base

MVA

b

bKV

I actual I I

S3kA

0.19 3V

3 12573.51kA

0.19 3 15.5

26. The signal energy of the continuous time signal

x t t 1 u t 1 t 2 u t 2 t 3 u t 3 t 4 u t 4 is

(A) 11/3 (B) 7/3 (C) 1/3 (D) 5/3

Key: (D)

Exp: x t t 1 u t 1 t 2 u t 2 t 3 u t 3 t 4 u t 4

1 2 3 4

1

a b

c

0

EE-GATE 2018


23

2 3 42 2 2

1 2 3

2 42 2

1 3

2 32 2

1 2

2 32 2

1 2

23

1

Energyof x t a dt b dt c .dt

a dt c dt

Energy of x t 2 a dt b dt

2 t 1 dt 1 dt

t 12 1 5 3.

3

27. A 0-1 Ampere moving iron ammeter has an internal resistance of 50 m and inductance of 0.1mH. A

shunt coil is connected to extend its range to 0-10 Ampere for all operating frequencies. The time

constant in milliseconds and resistance in m of the shunt coil respectively are

(A) 2, 5.55 (C) 2,1 (C) 2.18,0.55 (D) 11.1,2

Key: (A)

Exp:

It is given that

m

m

R 50m

L 0.1mH

Existing range 0 1ampere

Reauired range 0 10 ampere

Required 10Scalingfactor m 10

Existing 1

The required value shunt coil resistance is given by

3

msn

R 50 10R 5.55m .

m 1 10 1

To make the meter independent of frequency, the time constant of both parallel branch

should be same i.e.

mR mL

snR snL

MI ammeter

Shunt coil

EE-GATE 2018


24

sn m

sn m

3

sn 3

sn

sn

L L

R R

0.1 100.002 2m.sec

50 10

2 m.sec

R 5.55 mA.

28. The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a

frequency =100 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the

magnitude of the impedance Z (in ) seen between the terminals a and b is _________ (up to 2 decimal

places).

Key: (50)

Exp: when the Source voltage and current are in phase, then the circuit is under resonance and the

impedance is purely real.

Transforming the given network into its equivalent phasor domain we have

L

C

R

at 100

Z j L j100L

jZ j100

C

Z 100

eq

4

4

Z j100L 100 || j100

j10j100L

100 j100

j 10 100 j100j100L

100 j100 100 j100

j100L

j100 100

eqZ 100 rad sec, given

100 F

100Z

a

b

v t

i t

L

EE-GATE 2018


25

6 6

2 2 2 2

10 10j 100L

100 100 100 100

Since the circuit is under resonance, the imaginary part of eqZ 0, hence

6

eq 2 2

6 6 2

2 4

10Z

100 100

10 10 1050

2 100 2 10 2

29. Which one of the following statements is true about the digital circuit shown in the figure?

(A) It can be used for dividing the input frequency by 3.

(B) It can be used for dividing the input frequency by 5.

(C) It can be used for dividing the input frequency by 7.

(D) It cannot be reliably used as a frequency divider due to disjoint internal cycles.

Key: (B)

Exp:

0 1 2D Q Q

0 21

1 2 0 1 0 1 2D DD

Clk Q Q Q Q Q Q Q

0 0 0

1 1 0 0 1 0 0

2 1 1 0 1 1 0

3 1 1 1 1 1 1

4 0 1 1 0 1 1

5 0 0 1 0 0 1

6 1 0 0 1 0 0 Repeated

io

i.e., MOD 5

fHence, f

5

D Q

C

D Q D Q OUTf

INf

C C

0D 0Q

C

1D 1Q 2D 2Q OUTf

INf

C C

EE-GATE 2018


26

30. The voltage across the circuit in the figure. And the current through it, are given by the following

expressions:

v t 5 10cos t 60 V

i t 5 Xcos t A

Where 100 radian/s. If the average power delivered to the circuit is zero, then the value of X (in

Ampere) is _____ (up to 2 decimal places).

Key: (10)

Exp: If the voltage and currents of electrical circuit is in the form of

1 2

1 2

0 1 1 v 2 2 v

0 1 1 i 2 2 i

V t V V cos t V cos t ....

i t i i cos t i cos t ....

Then the average power is given by

1 1 2 2

T

avg

0

1 1 2 20 0 v i v i

1P V t i t dt,

T

V i V IV i cos cos ...

2 2 2 2

its a very well known standard result

avg

avg

It is given that

V t 5 10cos t 60

5 10cos t 120

i t 5 x cos t 0

10 xthen P 5 5 cos 120 0 ,

2 2

10x0 25 cos 120 P 0, given

2

0 25 5x 1 2

5x 25

2

x 10

i t

v tElectrical

Circuit

EE-GATE 2018


27

31. The equivalent impedance Zeq for the infinite ladder circuit shown in the figure is

(A) j12 (B) -j12 (C) j13 (D) 13

Key: (A)

Exp:

eq eq

eq

eq

Z j9 j4 || Z

j4 Zj9

j4 Z

eq eq eq eq

2

eq eq eq eq

2

eq eq

j4 Z Z j9 j4 Z j4 Z

j4 Z Z 36 j9Z j4 Z

Z j9 Z 36 0

2

eq

j9 j9 4 36 1Z

2 1

j9 81 144

2

j9 225

2

j9 j15

2

j9 j15 jq j15or

2 2

j12 or j3

j5

j1 j1

j5

eqZ

j9 j9

eqZ

j9 j9

eqZ

j4 j4

eqZ

j9

eqZ

j4

j5

j1 j1

j5eqZ

j9 j9

. . .

EE-GATE 2018


28

Since the nature of the network is overally inductive the reactance must should be positive (for

capacitive case the reactance could have –ve).

Hence we are discarding –j3.

eqZ j12

32. A transformer with toroidal core of permeability is shown in the figure. Assuming uniform flux

density across the circular core cross-section of radius r < < R, and neglecting any leakage flux, the best

estimate for the mean radius R is

(A)

2 2

pVr N

I

(B)

2

p sIr N V

V

(C)

2 2

pVr N

2I

(D)

2 2

pIr N

2V

Key: (D)

Exp: The inductance of primary is 2Np

LS

Where S is reluctance of magnetic path

22

2 2 2

2

2 2

L

2 2

L

2 2

2 R 2RS

A rr

Np Np rL

2R 2R

r

Np rX L

2R

I. .Np . rV IX

2R

.I.r .Np .R

2V

33. The number of roots of the polynomial, 7 6 5 4 3 2s s 7s 14s 31s 73s 25s 200 in the open left

half of the complex plane is

(A) 3 (B) 4 (C) 5 (D) 6

pi I sin ax si 0

SN Sv

Pv V cos t

PNR

r

EE-GATE 2018


29

Key: (A)

Exp:

4 2

3

A s 8s 48 s 200

dA s32s 96 s

ds

No. of sign changes 4

Right hand roots 4

No. of imaginary roots order of auxillary eqn 2 No.of sign changes below zero th row

4 2 2 0

left hand roots

7 4 3

34. The equivalent circuit of a single phase induction motor is shown in the figure, where the parameters are

1 2 l 2 MR R X X 12 , X 240 and s is the slip. At no load, the motor speed can be

approximated to be the synchronous speed. The no-load lagging power factor of the motor is ____ (up to

3 decimal places).

Key: (0.106)

Exp: At no load, 2m s

'R. S 0.

2S

The modified equivalent Circuit is

7

6

5

4

3

2

1

0

s 1 7 31 25

s 1 14 73 200

s 7 42 175

s 8 48 200

s 1 3 0

s 24 200

s 128 0

s 200

2

sign

change

2

sign

change

1R 1jX

2'R

2s

2'X

j2

2'R

2 2 s

2'X

j2

MXj

2

MXj

2

V 0

EE-GATE 2018


30

2 2

eq

' 'R R

2 2 s 4

j120 3 j6Z 12 j12 j120

3 j6 j120

j120 3 j612 j132

3 j126

14.72 j137.78

138.56183.9

No load p.f cos cos83.9 0.106 lag.

35. A 3-phase 900kVA, 3kV/ 3 kV / Y , 50 Hz transformer has primary (high voltage side) resistance

per phase of 0.3 and secondary (low voltage side) resistance per phase of 0.02 , Iron loss of the

transformer is 10kW. The full load % efficiency of the transformer operated at unity power factor is


Key: (97.36)

Exp:

ph

Hv 1

2

1eq 1 2

3a 3 ;

3 3

900I I 100A

3 3

R R a R

0.3 9 0.02 0.48

240j

2

240j

2

12

4

j12

2

12 j12

eqZ

EE-GATE 2018


31

2

1 1eq

2

Cu loss 3 3I R

3 100 0.48

14400W 14.4kW

Core loss 10kW

900% 97.36%

900 10 14.4

36. A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t=0 as shown

in the figure. Assume i(0)=0, v(0)=0. Which one of the following circular loci represents the plot of i(t)

versus v(t)?

(A) (B)

(C) (D)

Key: (B)

Exp: Since the network is of 2nd order, to obtain the expression of v(t)and i(t) we should use

Laplace transform approach.

Once we get the expression for v(t) and i(t) we can plot the loci by observing them at

different time instant.

Transforming the given time domain network into its equivalent s-domain form for t>0, we

have

S

i t

C 1F

5V v t

t 0 L 1H

i t

0, 5

v t i t

5,0 v t

i t

0,5

v t

5,0

i t

v t

EE-GATE 2018


32

2

5 s 5I s

s 1 s s 1

i t 5sin t

1 s

V s 5 ss 1 5

2

2

5

s s 1

5 2.5 2.5

s s j1 s j1

5 5s

s s 1

v t 5 5 cos

Finaly we have

i t 5sint

v t 5 5cos t

t i t v t Coordinate

0 0 0 0,0

2

5 5 5,5

0 10 0,10

3

2

-5 -5 5,5

2 0 0 0,0

37. A three phase load is connected to a three phase balanced supply

as shown in the figure.

If

an bn cnV 100 0 V, V 100 120 V and V 100 240 V

(angles are considered positive in the anti clock wise direction).

The value of R for zero current in the neutral wire is


5

s

I s

V s1

s

s

t 2

t

3t

2

5, 5 t 0

0,0 0,10

5,5

a

n

c

b

j10

j10

R

EE-GATE 2018


33

Key: (5.77)

Exp:

By KCL at node x, we have

a b c N

an bn cnN

I I I I

V V VI

R j10 j10

N

100 0 100 -120 100 -2400 I 0 given

R 10 90 10 -90

10010 -210 10 -150 0

R

10010 -210 10 -150

R

100R

10 -210 10 -150

R 5.77

38. The unit step response y(t) of a unity feedback system with open loop transfer function G(s) H(s)

2

K

s 1 s 2

is shown in the figure. The value of K is ___________ (up to 2 decimal places).

a

n

c

b

NI

aI

CI

j10 j10

R

bI

X

y t 1.4

1.2

1

0.8

0.6

0.2

00 2 4 6 8 10 12 14 16 18 20

t sec

0.4

EE-GATE 2018


34

Key: (8)

Exp: Steady state output =0.8

Input is unit step input =1

Stead state error ess = 0.2

For unit step input

ps 0

2s 0

ss

p

K limG s H s

Klim K 2

s 1 s 2

Ae

1 K

10.2

1 K 2

1 K 2 5 K 2 4; K 8

39. The Fourier transform of a continuous- time signal x(t) is given by

1X , .

10 j

where j 1 and denotes frequency. Then the value of nx t at t 1 _______(up to 1 decimal

place.) (ln denotes the logarithm to base e).

Key: (10)

Exp:

2

1Given X ,

10 j

Converting to s domain,

2

1X s

s 10

10t

10t

2

10t

10t

1we know e u t

s 10

1t.e

s 10

x t t.e u t

n x t n t.e u t

n t 10t n c

n 1 10 1 1 10 10.0

40. Consider a system governed by the following equations

12 1

21 2

dx tx t x t

dt

dx tx t x t

dt

EE-GATE 2018


35

The initial conditions are such that 1 2x 0 x 0 . Let

1f 1t

x limx t

and 2f 2

tx limx t

.

Which one of the following is true?

(A) 1f 2fx x (B) 2f 1fx x

(C) 1f 2fx x (D) 1f 2fx x

Key: (C)

Exp:

1

2 1

dx tx t x t .... 1

dt

2

1 2

2

1 2 1

2

1

1 2

1 1

1 1

dx tx t x t .... 2

dt

Differentiating 1 w.r.t ' t '

d x t dx t dx t

dt dt dt

dx tx t x t from 2

dt

dx t dx tx t x t from 1

dt dt

2

1 1

2

2

d x t dx t2 0

dt dt

D 2D 0

D D 2 0

2t

1 1 2

1f 1 1t

D 0, 2

Solution, x t C C e

x limx t C

Similarly, Differently (2) w.r.t „t‟ using (1) & (2)

we get 2

2 2

2

d x t 2dx t0

dt dt

2t

2 1 2

2f 2 1t

1f 2f

solution,x t C C e

x lim x t C

x x

41. Consider the two bus power system network with given loads as shown in the figure.

2G1G

1.0 1.0 0j0.1

lossQG120 jQ G215 jQ

15 j5 20 j10

EE-GATE 2018


36

All the values shown in the figure are in per unit. The reactive power supplied by generator G1 and G2

are QG1 and QG2 respectively. The per unit values of G1 G2Q ,Q , and line reactive power loss (Qloss)

respectively are

(A) 5.00, 12.68, 2.68 (B) 6.34, 10.00, 1.34

(C) 6.34, 11.34, 2.68 (D) 5.00, 11.34, 1.34

Key: (C)

Exp: S RV VP sin

x

S

s S R

1 15 sin

0.1

sin 0.5

30

V 1Q V V cos 1 cos 30 1.34pu

X 0.1

R

R S R

loss S R

V 1Q V cos V cos30 1 1.34pu

x 0.1

Q Q Q

G1 load S

1.34 1.34 2.68 pu

Q Q Q

G2 R load

5 1.34 6.34pu

Q Q Q

G2 R loadQ Q Q 1.34 10 11.34pu

42. A phase controlled single phase rectifier, supplied by an AC source, feeds power to an R-L-E load as

shown in the figure. The rectifier output voltage has an average value given by m0

VV 3 cos ,

2

where Vm = 80 volts and is the firing angle. If the power delivered to the lossless battery is 1600W,

in degree is ________ (up to 2 decimal places).

mV sin t 10mH

80VBattery

2

0V

EE-GATE 2018


37

Key: (90)

Exp: Power delivered to lossless battery 0EI =1600W

0

0

80 I 1600

I 20A

From given 0 01 rectifier, V E I R

m0

V3 cos E I R

2

803 cos 80 20.2

2

40 3 cos 120

3 cos 3

cos 0

90

43. The positive, negative and zero sequence impedances of a three phase generator are Z1, Z2 and Z0

respectively. For a line to line fault with fault impedance Zf, the fault current is f 1 fI kI , where If is the

fault current with zero fault impedance. The relation between Zf and k is

(A)

1 2

f

Z Z 1 kZ

k

(B)

1 2

f

Z Z 1 kZ

k

(C) 1 2

f

Z Z kZ

1 k

(D)

1 2

f

Z Z kZ

1 k

Key: (A)

Exp: f

1 2

3I L L fault

Z Z

1f

f

1 2 f

f1 f

1 2 f 1 2

1 2 f 1 2

f 1 2

1 2

f

I L L fault

3with fault impedance, Z

Z Z Z

I kI given

3 3 3

Z Z Z Z Z

Z Z k Z k Z Z

Z k Z Z 1 k

Z Z 1 kZ

k

EE-GATE 2018


38

44. As shown in the figure, C is the arc from the point (3, 0) to the point (0, 3) and the circle x2+ y2=9. The

value of the integral 2 2

Cy 2yx dx 2xy x dy is _________ (up to 2 decimal places).

Key: (0)

Exp: 2 2Given x y 9

2 2

2

2 2

0

2 2 2 2 2 2

2C x 3

y 9 x

y 9 x

1 xdy 2xdx dx

2 9 x 9 x

xdxy 2yx dx 2xy x dy 9 x 2x 9 x dx 2x 9 x x

9 x

0 a32

23 t 0

x 9 t dtConsider, dx, put 9 x t 18

2t9 x

from 1 , 18 18 0

45. Digital input signals A,B,C with A as the MSB and C as the LSB are used to realize the Boolean

function F = m0 + m2 + m3 + m5 + m7, where mi denotes the ith minterm. In addition, F has a don‟t care

for m1. The simplified expression for F is given by

(A) AC BC AC (B) A C

(C) C A (D) AC BC AC

Key: (B)

Exp: F m 0,2,3,5,7 d 1

A AC

A A A C Distributive law

A C

01

12 023 3 3

23

3

9 xx x x9x 2 dx ... 1

13 3 9 x12

1 x 1 1

0 1 3 2

4 1 15 7 6

A

A

BC BC BC BC

y

x

0,3 C

3,0

EE-GATE 2018


39

46. Let 3 2f x 3x 7x 5x 6. The maximum value of f(x) over the interval [0,2] is ________ (up to 1

decimal place).

Key: (12)

Exp: 3 2f x 3x 7x 5x 6

2f ' x 9x 14x 5 0

x 1, 5 9

f '' x 18x 14

f '' 1 0 &

f '' 5 9 0

At x 5 9, f x has local maximum &

5 1733f 7.13

9 243

f 0 6

f 2 12

The maximum value of f x in 0,2 is"12"

47. The per unit power output of salient-pole generator which is connected to an infinite bus, is given by the

expression, P=1.4sin +0.15sin 2 , where is the load angle. Newton Raphson method is used to

calculate the value of for P=0.8pu. If the initial guess is 30°, then its value (in degree) at the end of the

first iteration is

(A) 15° (B) 28.48° (C) 28.74° (D) 31.20°

Key: (C)

Exp: P 0.8 1.4 sin 0.15sin2

0

1 0

0

1

f 1.4sin 0.15sin 2 0.8

By, Newton Raphson method

f

'f

1.4sin30 0.15sin 60 0.86

1.4 cos30 2 0.15 cos60

0.50164 rad

28.74

48. Consider the two continuous time signals defined below:

1 2

| t |, 1 t 1 1 | t |, 1 t 1x t , x t

0, otherwise 0, otherwise

EE-GATE 2018


40

These signals are sampled with a sampling period of T=0.25 seconds to obtain discrete time signals x1[n]

and x2[n], respectively. Which one of the following statements is true?

(A) The energy of x1[n] is greater than the energy of x2[n].

(B) The energy of x2[n] is greater than the energy of x1[n].

(C) x1[n] and x2[n] have equal energies.

(D) Neither x1[n] nor x2[n] is a finite energy signal.

Key: (A)

Exp: Given

1

2

x n 1, 0.75, 0.5,0.25,0,0.25,0.5,0.75,1

x n 0,0.25, 0.5,0.75,1,0.75,0.5,0.25,0

We can see clearly, 1x n has greater energy than 2x n

Energy in 2 2 22 2

1x n 2 1 2 0.75 2 0.5 2 0.25 0

Energy in 2 2 22 2

2x n 2 0 2 0.75 2 0.25 2 0.5 0

49. The figure shows two buck converters connected in parallel. The common input dc voltage for the

converters has a value of 100V. The converters have inductors of identical value.

1x t

1 1

t

2x t

1 1t

1

| t |, 1 t 1x t

0, otherwise

2

1 | t |, 1 t 1x t

0, otherwise

100V

S2 L

S1 LC 1

S1i

S1

S2

Switch control signals

t

0 0.5ms 1ms t

EE-GATE 2018


41

The load resistance is 1 . The capacitor voltage has negligible ripple. Both converters operate in the

continuous conduction mode. The switching frequency is 1kHz, and the switch control signals are as

shown. The circuit operates in the steady state. Assuming that the converters share the load equally, the

average value of iS1, the current of switch S1 (in Ampere), is ______ (up to 2 decimal places).

Key: (12.5)

Exp: 0 sV DV 0.5 100 50V

00

in out

S S out out

S

S

sS1 S2

V 50I 50A.

R 1

Sence losser negligible, P P

V F V I

100 I 50 50

I 25A

II I 12.5A

2

50. Let

1 0 1

A 1 2 0

0 0 2

and B = A3-A2-4A+5I, where I is the 3 3 identity matrix. The determinant of B

is _______ (up to 1 decimal place).

Key: (1)

Exp:

1 0 1

Given A 1 2 0

0 0 2

Characteristic equation A is A I 0

1 0 1

1 2 0 0

0 0 2

By Cayley Hamilton theorem,

3 2

3 2

A A 4A 4I 0

A A 4A 5I I

B I According to the given question

B I 1

EE-GATE 2018


42

51. A 200V DC series motor, when operating from rated voltage while driving a certain load, draws 10A

current and runs at 1000 r.p.m. The total series resistance is 1 . The magnetic circuit is assumed to be

linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.pm

(rounded to the nearest integer) is _________.

Key: (825)

Exp: 200V, 10A

1 1

a f

b a a f

R R 1

E V I R R

200 10 190V

When torque becomes 1.44 times

2 1T 1.44T

Magnetic circuit linear, aI

2

1

2 1

2 2

2

1

2

1

2 1

1 2

2

a2 2a 2

1 a

2a a

1

b a a f

b 2 2

b 1 1

a2

1 a

b a

2 1

b a

ITT I

T I

TI I

T

1.44 10 12A,

E V I R R

200 12 188V

E N

E N

IN

N I

E I 188 10N N 1000

E I 190 12

824.56rpm

825 rpm nearest integer

52. The capacitance of an air filled parallel – plate capacitor is 60pF. When a dielectric slab whose thickness

is half the distance between the plates, is placed on one of the plates converting it entirely, the

capacitance becomes 86pF. Neglecting the fringing effects, the relative permittivity of the dielectric is

_______ (up to 2 decimal places).

Key: (2.53)

Exp: 1

0 0A 2 ACapacitance, C 2 60pF 120pF

d / 2 d

1

200V

EE-GATE 2018


43

0 r2 r r

1 2 req

1 2 r

r

r

r

2 Aand C 2 60 pF 120 pF

d

C C 120 120Now, C

C C 120 1

86

120 1

86or, 2.53

34

53.

2

22

C

zIf C is circle | z | 4 and f z . then f z dz is

z 3z 2

(A) 1 (B) 0 (C) -1 (D) -2

Key: (B)

Exp:

2 2

2 2 22

z zf z

z 1 z 2z 3z 2

z 1& 2 are poles of order 2, both lie inside z 4

22

2 2z 1

2

2z 1

z 1

Residue of f z at z 1

1 d zRes f z lim z 1

z 1 dz z 1 z 2

d zlim

dz z 2

2 2

4z 1

22

2 2z 2

2

2z 2

C

z 2

z 2 2z z .2 z 2lim 4

z 2

Residue of f z at z 2

1 d zRes f z lim z 2

z 1 ! dz z 1 z 2

d zlim 4

dz z 1

By Cauchy Residue theorem, f z dz 2 i 4 4 0

54. A dc to dc converter shown in the figure is charging a battery bank. B2 whose voltage is constant at 150

V. B1 is another battery bank whose voltage is constant at 50V. The value of the inductor, L is 5mH and

the ideal switch, S is operated with a switching frequency of 5kHz with a duty ratio of 0.4. Once the

EE-GATE 2018


44

circuit has attained steady state and assuming the diode D to be ideal, the power transferred from B1 to

B2 (in Watt) is ___________ (up to 2 decimal places)

Key: (12)

Exp: Given in outV 50V, V 150V

0

in

V 1

V 1 D

150 1

50 1 D

2D 0.666

3

But given duty cycle = 0.4

The boost converter is operating in discontinuous mode.

on

1T 200 sec

f

t DT 80 sec.

0 s

S

on

3

6

max min min

max

V VD

150 50 0.60.4

di IV L L

dt t

I50 5 10 I 0.8A

80 10

I I I I 0

I I

6

max

L avg 6

1 2 in L avg

1 1I T 0.8 0.6 200 10

2 2I 0.24AT 200 10

Power transfered from B to B V I 50 0.24 12W

LI

maxI

t0 DT BT T

50 V 150 V

LiL 5mH

B2B1

S

D

EE-GATE 2018


45

55. In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain 100 . The

base-emitter voltage drop is a constant, VBE=0.7V. The value of the Thevenin equivalent resistance Rth

(in ) as shown in the figure is _____ (up to 2 decimal places).

Key: (90.9)

Exp: Form BJT amplifier, ThR can be written as

OCThTh

SC S

B B

B

OC E

B

SC B

Th

VVR it havedependent sources

I I

10.7 0.7 10kI 1k 1 I 0

10 10mI

10k 101k 111

1V I 1k 10m 1k

111

10.7 0.7 10I 1m

10k 10k

I 1 I 101 1m

10 1kR 90.09

111

15V

10

10 k

10.7V

1k ThR

b

a

EE-GATE 2018


46

EE-GATE 2018d1zttcfb64t0un.cloudfront.net/.../Gateforum_EE_GATE-2018_solutions.… · EE-GATE 2018...

Documents

Transcript of EE-GATE 2018d1zttcfb64t0un.cloudfront.net/.../Gateforum_EE_GATE-2018_solutions.… · EE-GATE 2018...