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1

General Aptitude

Q. No. 1 – 5 Carry One Mark Each

1. The man who is now Municipal Commissioner worked as _____________.

(A) the security guard at a university

(B) a security guard at the university

(C) a security guard at university

(D) the security guard at the university

Key: (B)

2. Nobody knows how the Indian cricket team is going to cope with the difficult and seamer-friendly

wickets in Australia.

Choose the option which is closest in meaning to the underlined phase in the above sentence.

(A) put up with (B) put in with (C) put down to (D) put up against

Key: (D)

3. Find the odd one in the following group of words.

Mock, deride, praise, jeer

(A) mock (B) deride (C) praise (D) jeer

Key: (C)

4. Pick the odd one from the following options.

(A) CADBE (B) JHKIL (C) XVYWZ (D) ONPMQ

Key: (D)

5. In a quadratic function, the value of the product of the roots , is 4. Find the value of

n n

n n

(A) 4n (B) n4 (C) 2n 12 (D) n 14

Key: (B)

Exp: Given 4

n n n n

n n

n n

n n n n

n n

n n

1 1

( ) 4




2

Q. No. 6 – 10 Carry Two Mark Each

6. Among 150 faculty members in an institute, 55 are connected with each other through Facebook

and 85 are connected through WhatsApp. 30 faculty members do not have Facebook or WhatsApp

accounts. The number of faculty members connected only through Facebook accounts is ______.

(A) 35 (B) 45 (C) 65 (D) 90

Key: (A)

Exp: F Facebook, WWhatsApp, ETotal faculties

given

n(E) 150, n F W 30

n F W n(E) N F W 150 30

n F W 120

n f w n f n(w) n(F w

120 n(F) 85

n(F) 120 85 35

55 n(F) n F W

n F w 55 n(F) 55 35 20

n(w) 85 20 65

7. Computers were invented for performing only high-end useful computations. However, it is no

understatement that they have taken over our world today. The internet, for example, is

ubiquitous. Many believe that the internet itself is an unintended consequence of the original

invention with the advent of mobile computing on our phones, a whole new dimension is now

enabled. One is left wondering if all these developments are good or more importantly, required.

Which of the statement(s) below is/are logically valid and can be inferred from the above paragraph?

(i) The author believes that computers are not good for us

(ii) Mobile computers and the internet are both intended inventions

(A) (i) (B) (ii) only

(C) both (i) and (ii) (D) neither (i) nor (ii)

Key: (D)

8. All hill-stations have a lake. Ooty has two lakes.

Which of the statement(s) below is/are logically valid and can be inferred from the above

sentences?

(i) Ooty is not a hill-station

(ii) No hill-station can have more than one lake.

(A) (i) Only (B) (ii) Only (C) both (i) and (ii) (D) neither (i) nor (ii)

Key: (D)

E, n(E) 150

55 85

F

n(F) 35

W

n(w) 60

F w

n F w

20

30

F W




3

9. In a 2 4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be

observed in the grid?

(A) 21 (B) 27

(C) 30 (D) 36

Key: (C)

Exp: 1: (AEOK)

2: (AEJF), (FJOK)

4: (ABLK), (BCML), (CDNM), (DEON)

2: ACMK, ADNK 2:ECMD,EBLO 2:ACHF,ADIF

2: ECHJ, EBGJ 2:FHMK,FINK 2: JHMD,JGLO

1: BDNL 2: BDIG,GINL

8: ABGF, BCHJ, CDIH, EDI, FGLK, GHML, HINM

Total = 1+2+4+2+2+2+2+2+2+1+2+8=30

10.

Chose the correct expression for f(x) given in the graph.

(A) f x 1 x 1 (B) f x 1 x 1

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

Key: (C)

Exp: Substituting the coordinates of the straight lines and checking all the four options given, we get

the correct option as C which is f(x)= 2 x 1

25

2

1.5

1

0.5

0

0.5

1

1.5

2

25

4 3 2 10 1 2 3 4X

f x

A B C D E

F G H I J

K L M N O




4

Electrical Engineering

Q. No. 1 –25 Carry One Mark Each

1. The maximum value attained by the function f x x x 1 x 2 in the interval 1, 2 is ____.

Key: 0

Exp: f ( ) ( 1)( 2) x x x x 3 2f ( ) 3 2 x x x x 1 2f ( ) 0 3 6 2 0 x x x

1

13

x

1But 1 only lies on the interval [1,2]

3 x

111At 1 ;f ( ) 6 6

3 x x x

16 1 6 0

3

1

1 is a point of minimum3

x

f(x) = x(x-1)(x-2) = 0 at either ends x =1 & x =2

Max value = 0

2. Consider a 3 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.

Key: 3

Exp: Consider 3 3

1 1 1

A 1 1 1

1 1 1

It’s only non-zero Eigen value λ = 1 order of the matrix = 1 3 3

3. The Laplace Transform of 2tf t e sin 5t u t is

(A) 2

5

s 4s 29 (B)

2

5

s 5 (C)

2

s 2

s 4s 29

(D)

5

s 5

Key: (A)

Exp: Consider

By frequency shifting property, 0S t1

0X(S-S ) (t)e x

0thus at S 2

2

5F(s)

s 4s 29

x(t) = sin5t u(t) 2

5X(S)

s +25

2tf(t) = e sin5tu(t)

2

5F(s)

(s-2) +25




5

4. A function y t , such that y 0 1 and 1y 1 3e , is a solution of the differential equation

2

2

d y dy2 y 0

dtdt . Then y 2 is

(A) 15e (B) 25e (C) 17e (D) 27e

Key: (B)

Exp: The operator function of given D.E is 2(D +2D+1)y = 0

2A.E is D +2D+1 = 0 D = -1,-1

-x

1 2y = e C + C (1)

-1Given y(0)=1 & y(1) =3e

From ; y(0) = 1 i.e , y = 1 at x = 0

11 = C

-1y(1) = 3e i.e, y = 3/e at x=1

-1

2

3e 1 C (1)

e 2 23 1 C C 2

-x -2 -2y = e [1+2x] y(2) = e [1+4] = 5e

5. The value of the integral

2C

2z 5dz

1z z 4z 5

2

over the contour z 1. taken in the anti-

clockwise direction, would be

(A) 24 i

13

(B)

48 i

13

(C)

24

13 (D)

12

13

Key: (B)

Exp: Singular points are obtained by

21z z -4z+5 0

2

1z = & z =2 i

2

Out of these

1z =

2only lies inside C. z =1

By Cauchy’s integrated formula,

2

2C C

2z+5 2z+5 dz = dz

1 z -4z+5z - z -4z+5 0

12z -

2

2

12 5

48πi2= 2πi

131 14 5

2 2




6

6. The transfer function of a system is

Y s s,

R s s 2

The steady state output x(t) is A cos 2t for

the input cos 2t . The value of A and , respectively are

(A) O1, 45

2 (B) O1

, 452

(C) O2, 45 (D) O2, 45

Key: (B)

Exp: S

H(s) = S+2

0 1

2H( ) = 90 tan

2+4

If input is COS 2t i.e., = 2

01

H( ) = 452

01

y(t) = cos(2t 45 )2

0

1by comparision A= & 45

2

7. The phase cross-over frequency of the transfer function G(s) =

3

100

s 1 in rad/s is

(A) 3 (B) 1

3 (C) 3 (D) 3 3

Key: (A)

Exp: Phase Crossover frequency PC

0

PC =: GH 180

-1 0 -1 0

PC PCGH 3tan 180 3tan w W tan60 3

8. Consider a continuous-time system with input x(t) and output y(t) given by

y t x t cos t . This system is

(A) linear and time-invariant (B) Non-linear and time-invariant

(C) linear and time-varying (D) Non-linear and time-varying

Key: (C)

Exp: Linearity

1 1 2 2y (t)= x (t)cost; y (t)= x (t)cost

1 2 1 2y (t) + y (t) = x (t)cost + x (t)cost

1 2 1 2y (t) + y (t) = [x (t) + x (t)]cost

1 2 1 2If x (t) + x (t) = x(t) then y (t) + y (t) = y(t)




7

Thus the system is linear

Time-invariance

consider 1 1 1y (t) x (t) cost ; If x (t) x(t - τ)

1y (t) x(t - τ)cost

Define y(t - τ) x(t - τ)cos(t - τ)

1y (t) y(t - τ) system is time-varient

9. The value of te 2t 2

dt. where t is the Dirac delta function, is

1

A2e

2

Be

2

1C

e

2

1D

2e

Key: (A)

10. A temperature in the range of -40OC to 55

OC is to be measured with a resolution of 0.1

OC. The

minimum number of ADC bits required to get a matching dynamic range of the temperature

sensor is

(A) 8 (B) 10 (C) 12 (D) 14

Key: (B)

Exp: Usually when voltage information is given we use the formula max min

n

V -VR =

2

Here based on the given information we can think the systems as following block diagram

Here the maxV and minV will be the equivalent of maxT and minT respectively

So we can still use the above relation

max min

n

T -TResolution =

2 n n5-(-40)

2 = 2 = 9500.1

2n log (950) 9.89 10

So minimum requirement is 10 bits

11. Consider the following circuit which uses a 2-to-1 multiplexer

as shown in the figure below. The Boolean expression for

output F in terms of A and B is

(A) A B (B) A B

(C) A B (D) A B

A B

S1

Y0

F

Temp

Temperatureto

voltageconverter

analogvoltage

nbit

ADC

.

.

.




8

Key: (D)

Exp: We can redraw the max circuit as follows

So the Boolean expression of F(A, B)=BA BA=A B=A B

12. A transistor circuit is given below. The Zener diode breakdown voltage is 53 V as shown The

base to emitter voltage droop to be 0.6V. The value of the current gain is ________.

Key: 19

Exp: E5.3 = 0.6 + 470 I

EI 0.01A

X 3

10 - 5.3I = 1mA

4.7 10

BI =1 0.5 0.5mA

E BI (1 β)I

30.01 (1 β) 0.5 10 3

0.01(1 β) 20

0.5 10

β 19

13. In cylindrical coordinate system, the potential produced by a uniform ring charge is given

f r, z , where f is a continuous function of r and z. Let E

be the resulting electric field. Then

the magnitude of E

(A) increases with r (B) is 0

(C) is 3 (D) decreases with z

Key: (B)

Exp: Since the charge is not varying with time the field E is static So E 0

4.7k

10V

220

470

0.5mA

5.3V

0 F

A

1

A

B




9

14. A soft-iron toroid is concentric with a long straight conductor carrying a direct current I. If the

relative permeability r of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent

points located just inside and just outside the toroid, is ________.

Key: 100

Exp: The field inside and outside the toroid is due to long straight conductor only

Let the two points almost at same distance

The flux density inside toroid = 0 rμ μ I

2πr

The flux density outside toroid = 0μ IT

2πr

Ratio =

0 r

r0

μ μ I

2πr μ 100μ I

2πr

15. A BR and R are the input resistances of circuits as shown below. The circuits extend infinitely in

the direction shown. Which one of the following statements is TRUE?

(A) A BR R (B)

A BR R 0 (C) A BR R (D) B A AR R 1 R

Key: (D)

Exp: By comparing 2 network on the input side

we can say that AB A B

A

RR =1//R R =

1 R

16. In a constant V/f induction motor drive, the slip at the maximum torque

(A) is directly proportional to the synchronous speed

(B) remains constant with respect to the synchronous speed

(C) has an inverse relation with the synchronous speed

(D) has no relation with the synchronous speed

2 2 2

1

1 1

BR

AR

2 2 2

1

1 1 1




10

Key: (C)

Exp: 2 2 2

2 2 2

r r rSmT

x j L j2 L .f

1

SmTf

S

120fN

P

1

SMTNS

17. In the portion of a circuit shown, if the heat generated in 5 resistance is 10 calories per second

then heat generated by the 4 resistance, the calories per second, is ______.

Key: 2

Exp: Here the power information regarding the resistor is given because

E Calorie

P = =wattt sec

5P = 10

5ΩV = 10

5

5ΩV = 50

4ΩP is asked

2 2

4Ω

4Ω

V 1 4P = 50

4 4 4 6

1 16 calorie50 2

4 100 sec

18. In the given circuit, the current supplied by the battery, in ampere, is ________.

`

Key: 0.5

Exp: If we write KCL at node × then

4 6

5

1V

2l1

1

11l

2l




11

11 2 2

II = 2 I

2 I

Write KVL in the outer boundary of network

1 21 (1 I ) (2 ) 0 I 1 21 I 2 I

11 1 1

I 1 1 =I 2 1 2J I 0.5A

2 2

19. In a 100bus power system, there are 10 generators. In a particular iteration of Newton Raphson

load flow technique (in polar coordinates). Two of the PV buses are converted to PQ type. In this

iteration.

(A) The number of unknown voltage angles increases by two and the number of unknown voltage

magnitudes increases by two.

(B) The number of unknown voltage angles remain unchanged and the number of unknown

voltage magnitudes increases by two

(C) The number of unknown voltage angles increases by two and the number of unknown voltage

magnitudes decreases by two

(D) The number of unknown voltage angles remains unchanged and the number of unknown

voltage magnitudes decreases by two

Key: (B)

Exp: Total no of Buses = 100

Generator Buses = 10

Load Buses = 100-10=90

Slack Bus = 1

If 2 of the PV buses are converted to PQ type the no of on voltage magnitudes increases by 2 with

constant unknown voltage angles

20. The magnitude of three-phase fault current at buses A and B of a power system are 10 pu and 8

pu, respectively. Neglect all resistance in the system and consider the pre-fault system to be

unloaded. The pre-fault voltage at all buses in the system is 1.0 pu. The voltage magnitude at bus

B during a three-phase fault as but. A is 0.8pu. The voltage magnitude at bus A during a three-

phase fault at bus B, in pu, is _____.

Key: 0.84

Exp: Voltage at ith bus when fault is at k

th bus is

productik

i f

kk f p.u

VzV E 1 I

z z X

n

1

Ax10 X 0.1P.U

B

1

X8 B 0.125P.U

AB

B

AA

ZV E 1

Z

1V 1I

2I 2I

1 1

1

X




12

AB

AB

Z0.8 1 1 Z 0.02

0.1

A

BB

2AB 0.02V 1 1 1

2 0.125

0.84 P.U

21. Consider a system consisting of a synchronous generator working at a lagging power factor, a

synchronous motor working at an overexcited condition and a directly grid-connected induction

generator. Consider capacitive VAr to be a source and inductive VAr to be a sink of reactive

power. Which one of the following statements is TRUE?

(A) Synchronous motor and synchronous generator are sources and induction generator is a sink

of reactive power.

(B) Synchronous motor and induction generator are sources and synchronous generator is a sink

of reactive power.

(C) Synchronous motor is a source and induction generator and synchronous generator are sinks

of reactive power

(D) All are sources of reactive power

Key: (A)

22. A buck converter, as shown in figure (a) below, is working in steady state. The output voltage and

the inductor current can be assumed to be ripple free. Figure (b) shows the inductor voltage LV

during a complete switching interval. Assuming all devices are ideal, the duty cycle of the buck

converter is ______

Key: 0.4

Exp: When M is ON, S L 0V V V

L S 0V 30 V V

When M is OFF, L 0 0 0V V 20 V V 20

S s30 V 20 V 50V

0

S

V 2D 0.4

V 5

M

LV

D C0V

R

a

gV

LV

0

20V

aT

ONT OFFT

t

b

30V




13

23. A steady dc current of 100A is flowing through a power module (S.D) as shown in Figure (a). The

V-I characteristics of the IGBT (S) and the diode (D) are shown in Figures (b) and (c),

respectively. The conduction power loss in the power module (S, D), in watts, is _____

Key: 169 to 171

24. A 4pole, lap-connected, separately excited dc motor is drawing a steady current of 40 A while

running at 600 rpm. A good approximation for the waveshape of the current in armature

conductor of the motor is given by

(A) (B)

(C) (D)

Key: (C)

Exp: no of parallel paths = 4

Armature current/conductor = 40

10A4

For linear commutation, the change from 10A to 10A is straight line

N = 600rpm

Time for 1 revolution =- 0.1 sec.

For 1 pole-pitch, 0.1

t 25msec4 poles

sI A

0V 1V

SV Volt

dV dl 0.02

V l characteristic of 1GBT

b

100 A

DS

a

0I A

0V 0.7V

SV Volt

dV dl 0.01

V l characteristic of diode

c

10A

t

II

40A

t

I10A

T 25ms

10AT 25ms

I

10A

T 25ms

T 25ms

10A




14

25. If an ideal transformer has an inductive load element at port 2 as shown in the figure below, the

equivalent inductance at port 1 is

`

(A) nL (B) n2L (C)

n

L (D)

2n

L

Key: (B)

Exp: The property of an ideal transformer is of port2 is terminated by an impedance LZ then the

impedance seen from port is

L

2

Z L

Z

N / N

In the given problem

2

in 2

LZ n L

1/n

Q. No. 26 – 50 Carry Two Mark Each

26. Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were

the permitted pen colours that the candidate could bring. The probability that a candidate comes

with all 3 pens having the same colour is ______.

Key: 0.06

Exp: Total possible options = 43

Favorable choices {BBB, BlBlBl, GGG, RRR} = 4

Probability =3

4 10.06

4 16

27. Let n

n 0

S n where 1.

The value of in the range 0 1 . Such that S 2 is ______.

Key: 0.293

Exp: n

n = 0

S = n α

2 32α α 2α + 3α ] \

22α α[1 2α 3α ]

22α α[1 α] if α 1

2 1

(1 α) 2 α 1 α 0.2932

Port 1 Port 2

n :1

L




15

28. Let the eigenvalues of a 2 x 2 matrix A be 1, -2 with eigenvectors 1 2x and x respectively. Then

the eigenvalues and eigenvectors of the matrix 2A 3A 4l would, respectively, be

(A) 1 22,14; x ,x (B)

1 2 1 22,14; x x ,x x

(C) 1 22,0; x ,x (D)

1 2 1 22,0; x x , x x

Key: (A)

Exp: Matrix A

Eigen values 1 , -2

Matrix 2A -3A+4I

2 21 3(1) 4, ( 2) 3( 2) 4

Eigen values 2 , 14 respectively

0 1 2 2 A & P(A) = a I + a A + a A have same eigen vectors

29. Let A be a 4 3 real matrix with rank 2. Which one of the following statement is TRUE?

(A) Rank of ATA is less than 2

(B) Rank of ATA is equal 2

(C) Rank of ATA is greater than 2

(D) Rank of ATA can be any number between 1 and 3

Key: (B)

Exp: Given that 4 3ρ A 2

From the properties of Rank; we have Tρ AA = ρ(A)

T TRank of AA = Rank of A A = Rank of A 2

30. Consider the following asymptotic Bode magnitude plot is in rad s .

Which one of the following transfer function is best represented by the above Bode magnitude plot?

(A)

2

2s

1 0.5s 1 0.25s (B)

4 1 0.5s

s 1 0.25s

(C)

2s

1 2s 1 4s (D)

2

4s

1 2s 1 4s

magnitude

dB

0.dB

20dB dec12dB

0.5 8




16

Key: (A)

Exp: By looking to the plot we can say that since the initial slope is +20 there must be a zero on the

origin

If we find 2 we can get the

answer by eliminating options

2 1

2 1

M - MSlope =

log - log

2

0 1240

log 8 - log

2

12log 8- log

40

2

12 log = log 8-

40

2 =4

So one of the corner frequency is 2 = 4s at this frequency 2 poles should exist because the

change in slope is -40db

From this we can say option A satisfies the condition

(i) A zero at origin

(ii) one of corner frequency 4H term will be s

1+ 4

having 2 poles

31. Consider the following state-space representation of a linear time-invariant system

T1 0 1 1

x t x t , y t c x t , c and x 00 2 1 1

The value of y(t) for et log 2 is _________.

Key: 6

32. Loop transfer function of a feedback system is 2

s 3G s H s

s s 3

.Take the Nyquist contour in

the clockwise direction. Then the Nyquist plot of G(s) H(s) encircles – 1 + j0

(A) Once in clockwise direction (B) Twice in clockwise direction

(C) Once in anticlockwise direction (D) Twice in anticlockwise direction

Key: (A)

Exp: S+3

GH = S(S-3)

2 1/2

2 2 1/2 2

( +a) 1GH

( +a)

1 0 0 1 1GH tan 180 180 tan 2 tan3 3 3

1 2

magnitude

dB

0.dB

20dB dec12dB

0.5 8




17

1

2

1GH = 2 tan

3

at 0, GH = 0

at0, GH = 0180

at 01

3, GH = 909

So the plot start at 00 and goes to

0180 through 090 Since there are 2 poles on origin we will get

2 radius semicircle those will start where the mirror image ends and will terminate where the

actual plot started in clockwise direction. So the plot will be

So the Nyquist plot of G(s) H(s)

encircles – 1 + j0

once in clockwise direction

33. Given the following polynomial equation 3 2s 5.5s 8.5s 3 0 , the number of roots of the

polynomial, which have real parts strictly less than – 1, is _______

Key: 2

Exp: The polynomial is 3 2S +5.5S +8.5S+3=0 , since we are interested to see the roots wrt S = -1 so in

the above equation replace S by z-1 then the equation is

3 2(Z-1) +5.5(Z-1) +8.5(Z-1)+3=0

3 2 2Z -3Z +3Z-1+5.5(Z +1-2Z)+8.5Z-8.5+3=0

3 2Z +Z ( 3 5.5)+Z(3+8.5-11)+(-1+5.5-8.5+3)=0

3 2Z +2.5Z 0.52 1 0

Using RH table

3

2

1

0

Z 1 0.5

Z 2.5 1

Z 0.9

Z 1

The single sign change in 1st column indicate that out of 3 roots 1 root lie on the right half of S = -

1 plane if memory remaining 2 lies on left half of S = -1 plane.

w =

-1+io

M= -1

w = 0




18

34. Suppose 1 2x t and x t have the Fourier transforms as shown below.

Which one of the following statements is TRUE?

(A) 1x t and 2x t are complex and 1 2x t x t is also complex with nonzero imaginary part

(B) 1 2 1 2x t and x t are real and x t x t is also real

(C) 1 2 1 2x t and x t are complex but x t x t is real

(D) 1 2 1 2x t and x t are imaginary but x t x t is real

Key: (C)

Exp: 1 2x (t) & x (t) are complex functions

2 1 2 1x ( ) = x (- ) , x (t) = x (-t)

1 2x (t) x (t)will be real

35. The output of a continuous-time, linear time-invariant system is denoted by T{x(t)} where x t is

the input signal. A signal z(t) is called eigen-signal of the system T, when T z t z t , ,

where is a complex number, in general, and is called an eigen value of T. suppose the impulse

response of the system T is real and even. Which of the following statements is TRUE?

(A) cos(t) is and eigen-signal but sin(t) is not

(B) cos(t) is and sin(t) are both eigen-signal but with different eigenvalues

(C) sin(t) is an eigen-signal but cos(t) is not

(D) cos(t) and sin(t) are both eigen-signal with identical eigenvalues

Key: (D)

Exp: Consider the Eigen signal Z(t) = Cos t

For Cost,

y(t) cos(t-τ)h(τ)dτ

(cos t cos τ +sin t sin τ)h(τ)dτ

=cos t cos τh(τ)dτ sin t sin τh(τ)dτ

h(τ)is an even signal

sinτh(τ)dτ =0

1X

1

0.3

0.5

1

0 1 2

2X

0.5

1

2 1 0 1

0.3




19

y(t) =cos t cos τh(τ)dτ

Thus the integration value will be an Eigen value γ .

Similarly consider the Eigen signal Z(t) = sin t

For sint, y(t) sin(t-τ)h(τ)dτ

=sin t cos τh(τ)dτ cos t sin τh(τ)dτ

Thus y(t)=sin t cos τh(τ)dτ

Eigen value ‘ γ ’is same for both the Eigen functions sint & cost

36. The current state QA QB of a two JK flip-flop system is 00. Assume that the clock rise-time is

much smaller than the delay of the JK flip-flop. The next state of the system is

(A) 00 (B) 01 (C) 11 (D) 10

Key: (C)

Exp:

A A

B B A

J = K =1

J = K =Q

It is given initially A BQ Q = 0

Since it is a synchronous counter, when clock is applied both flipflop will change there state

simultaneously based on JK FF state table

A A A AJ =1, K =1 , Q =0 Q = 1

B B B BJ =1, K =1 , Q =0 Q = 1

So next state A BQ Q is 11

5V

J

K

CLK

VQ

AQ

J

K

OQ

Logic 1

AJ

AK

BK

BJ

AQ

BQ

AQ

BQ




20

37. A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is IN0 V 3 Volts.

The expression for the LSB of the output 0B as a Boolean function of

2 1 0X ,X , and X is

(A) 0 2 1X X X

(B) 0 2 1X X X

(C) 0 2 1X X X (D) 0 2 1X X X

Key: (A)

38. Two electric charges q and -2q are placed at (0,0) and (6,0) on the x-y plane. The equation of the

zero equipotential curve in the x-y plane is

(A) x 2 (B) y 2 (C) 2 2x y 2 (D) 2 2x 2 y 16

Key: (D)

Exp: The potential due to Q 4-2Q at (x, y) is2 2 2 2

q 2q

k x + y k (x-6) + y

If potential at (x, y) = 0 where 1

k=4πε

2 2 2 2

q 2q0

k x + y k (x-6) + y

2 2 2 2(x-6) + y 2 x + y 0

2 2 2 2(x-6) + y 4(x + y ) 2 2 2 2x +36 -12x + y = 4x 4y

2 2x y - 4x-12 = 0

Option:D

2 2(x+2) + y 16

2 2x +4x +4+ y = 16

2 2x y + 4x-12 = 0

3V

100

200

200

100

2X

1X

0X

Digital

circuit

1B

0B

INV




21

39. In the circuit shown, switch S2 has been closed for a long time. A time t = 0 switch S1 is closed At

t 0 , the rate of change of current through the inductor, in amperes per second, is _______.

Key: 2

Exp: At -t = 0 Network is in steady state with 1S opens 2S (closed) So we can say

-

L

3i (0 )= 1.5A

2

At

+t = 0 indicator behaves as ideal current source of 1.5A if we draw the network at +t = 0 ,

both switch closed

Writing Nodes equation at LV (0 ) node

L

1 1 3 3V (0 )= 1.5

1 2 1 2

LV (0 ) = 2

+di(0 )L 2

dt

+

Ldi (0 )2A/sec

dt

40. A three-phase cable is supplying 800kW and 600kVAr to an inductive load, It is intended to

supply an additional resistive load of 100kW through the same cable without increasing the heat

dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the

load. Given the line voltage is 3.3kV, 50Hz, the capacitance per phase of the bank, expressed in

microfarads, is _________.

Key: 47 to 49

41. A 30MVA, 3-phase, 50Hz 13.8kV, star-connected synchronous generator has positive, negative

and zero sequence reactance, 15%,15% and 5% respectively. A reactance nX is connected

between the neutral of the generator and ground. A double line to ground fault takes place

involving phases ‘b’ and ‘c’, with a fault impedance of j0.1 p.u. The value of nX in p.u. that will

limit the positive sequence generator current to 4270 A is ________.

Key: 1.108

Exp: 1X 0.15 P.U

1S1 2S

2

3V3V 1H

+at t = 0

1.5

+

LV (0 )

2 1

3V

3V




22

6

base 3

30 10I 1255.109

3 13.8 10

2X 0.15P.U

0X 0.05P.U

fX 0.1P.U

actual

base

I 4270I P.U 3.4 P.U

I 1255.109

a1

1 2 O f n

EaI

Z Z || Z 3z 3z

2 0 f n

13.4

0.15 Z ||Z 3z 3z

2 f nZ || 20 3z 3z 0.144

2 f

1 1 1

Z 20 3z 3zln 0.144

f n20 3z 3z 3.675

n3z 3.675 0.05 3 0.1

nZ 1.108P.U

42. If the star side of the star-delta transformer shown in the figure is excited by a negative sequence

voltage, then

(A) VAB leads Vab by 60O

(B) VAB lags Vab by 60

O

(C) VAB leads Vab by 30O

(D) VAB lags Vab by 30

O

Key: (D)

43. A single-phase thyristor-bridge rectifier is fed from a 230V, 50Hz, single-phase, AC mains. If it is

delivering a constant DC current 10A, at firing angle of 30O, then value of the power factor at AC

mains is

(A) 0.87 (B) 0.9 (C) 0.78 (D) 0.45

Key: (C)

Exp: IPF = 0.9cos 0.9 cos30 0.78

A

C

B N

a

c

b




23

44. The switches T1 and T2 are shown in Figure (a).

They are switched in a complementary fashion with sinusoidal pulse width modulation technique.

The modulating voltage m t 0.8sin 200 t V and the triangular carrier voltage c are as

shown in Figure (b). The carrier frequency is 5kHz. The peak value of the 100Hz component of

the load current Li , in ampere, is ______.

Key: 10

Exp: m

a

Vm 0.8

Vcarrier

0l

VdsV max ma. 0.8 250 200V

2

2 22L 16 12 20

L

200I max 10A

20

45. The voltage s across and the current gl through a semiconductor switch during a turn-ON

transition are shown in figure. The energy dissipated during the turn – ON transition, in mJ, is

_________.

0.8

C

1

t

mV

b

dcV 2 250V

Li

T1

LX 16 at

100Hz

R 12

dcV 2 250V

T2

a

sv

0

si

0

T1 1 s T2 1 s

100A

50A

t

t

600V




24

Key: 75

Exp: 61E Pdt V t tdt 600 150 10

2

+ 61

100 600 102

75mJ

46. A single-phase 400V, 50Hz transformer has an iron loss of 5000W at the condition. When

operated at 200V, 25Hz, the iron loss is 2000W. When operated at 416V, 52Hz, the value of the

hysteresis loss divided by the eddy current loss is

Key: 1.4423

Exp: Since V 400 200

8f 50 2.5

is constant, Bm is constant

2

h e 1 2Pcore P P k f k f

h 1P k f 2

1 2 1 25000 50k 50 . k 50k 2500k

2

e 2P K f 2

1 1 2 1 22000 25k 25k 25 k 25k 625k

1 2 2

1 2 1

k 50k 100 k 0.8

k 25k 80 k 60

∴ Pcore = 260f 0.8 f

hP 60f

2

eP 0.8f

When hf 52Hz, P 60 52 3120

2

eP 0.8 52 2163.2

h

e

P 31201.4423

P 2163.2

47. ADC shunt generator delivers 45 A at a terminal voltage of 220V. The armature and the shunt

field resistance are 0.01 and 44 respectively. The stray losses are 375W. The percentage

efficiency of the DC generator is ______.

Key: 86.84

Exp: outP 220 45 990W

f

220I 5A

44

a L fI I I 45 5 50A

Armature losses = 2 2

a aI r 50 0.01 25W

Field losses = 2 2

f fI R 5 44 1100W

Stray losses = 375W

Total losses = 1500W

out

1losses 1 P

fI

LI 45A

ar 0.01 220V

44




25

1

1500 1 9900 86.84%

48. A three-phase, 50Hz salient-pole synchronous motor has a per-phase direct-axis reactance dX

of 0.8pu and a per-phase quadrature-axis reactance qX of 0.6pu . Resistance of the machine is

negligible. It is drawing full-load current at 0.8pf (leading)j. When the terminal voltage is 1pu,

per-phase induced voltage, in pu, is ______.

Key: 1.608

Exp: dX 0.8P.U P.f 0.8 leading Ra 0

qX 0.6PU Vt 1P.U

cos 0.8 36..86

aI 136.86

a a a

t a a a

I q cos I R sintan

V I q sin I R cos

a

t

I q cos 1 0.6 0.80.3529

V Ia qsin 1 1 0.6 0.6

O19.44

36.86 19.44 56.3

d aI I sin 1 sin56.4 0.832

q aI I cos 1 0.5547 0.5547

q aEf Vcos Id d I R 0

dVcos I d 1.cos19.44 0.832 0.8 1.608pu

FE 1.608PU

49. A single-phase, 22kVA, 2200V/220, 50Hz, distribution transformer is to be connected as an auto-

transformer to get an output voltage of 2420 V. Its maximum kVA rating as an auto-transformer is

(A) 22 (B) 24.2 (C) 242 (D) 2420

Key: (C)

Exp: Rated LV current = 22kVA

100A220

max

2200 10KVA 242kVA

1000

Or

2420 100

242kVA1000

50. A single-phase full-bridge voltage source inverter (VSI) is fed from a 300 V battery. A pulse of

120O duration is used to trigger the appropriate devices in each half-cycle. The rms value of the

fundamental component of the output voltage, in volts, is

(A) 234 (B) 245 (C) 300 (D) 331

Key: (A)

100A

220V

10A

2420V

110A

2200V2200VLOV




26

Exp: O

n 1,3,s

4Vs nV t cos sin n t

n 6

S

01

4V1V cos 6 233.9 234V

2

51. A single-phase transmission line has two conductors each of 10mm radius. These are fixed a

center-to-center distance of 1m in a horizontal plane. This is now converted to a three-phase

transmission line by introducing a third conductor of the same radius. This conductor is fixed at an

equal distance D from the two single-phase conductors. The three-phase line is fully transposed.

The positive sequence inductance per phase of the three-phase system is to be 5% more than that

of the inductance per conductor of the single-phase system. The distance D, in meters, is _______.

Key: 1.439

Exp: For single phase

m

1

DL 0.2 n

Ds

s

10.2 n

D

For three phase

23DM D

L3 0.2 n 0.2 nDS Ds

3L 1.05L1

2/3

S

D 10.2 1.05 0.2 n

D DS

D 1.439 mts

52. In the circuit shown below, the supply voltage is 10sin 1000t volts. The peak value of the steady

state current through the 1 resistor, in amperes, is _________.

Key: 1

Exp: W =1000, the various impedance at this frequency are

4

240 F1

4mH

5

500mH

2 F

~

10sin 1000t

1m

D D

1m

23

mD 3 D.D.1 D




27

3

250μf 4mH 6

-jZ Z ( j 1000 4 10 )

1000 250 10

( j4) (j4) open circuit

3

24f 500mH 6

-jZ Z ( j 1000 500 10 )

1000 250 10

( j500) (j500) open circuit

Since both LC pair parallel combination becomes

open then the circuit can be redrawn as

1Ω

10sin1000tI sin1000t

4+1+5

So peak value of 1ΩI 1A

53. A dc voltage with ripple is given by t 100 sin t 5sin 3 t volts. Measurements of

this voltage t , made by moving-coil and moving-iron voltmeters, show readings of 1V and

2V

respectively. The value of 2 1V V , in volts, is _______.

Key: 0.312

Exp: 1V 100V

2 2

2

2

10 5V 100 100.312V

2 2

2 1V V 0.312V

54. The circuit below is excited by a sinusoidal source. The value of R, in , for which the

admittance of the circuit becomes a pure conductance at all frequencies is ________.

Key: 14.14

Exp: Admittance becomes pure conductance means the

imaginary part of Y must be zero which imply

resonance condition.

Let first get Y expression interms of L,C

then by equalising imaginary part we will

100 F R

0.02H R

~

C

R

R

L

4

5 1

10sin1000tV




28

get the answer.

1 1

1R+j LR-

Y

C

eq22 2

2

jR+

R-j L C m [ ] 0R +( L) 1

R +C

g Y

22 2

2

1

L C

R +( L) 1R +

C

Cross multiplying

2

2 2 21 1 1( L)R L R +( L)

C C C

2 1 1 1R L- L- 0

C C

C

2 L 1R L- 0

C

C

Now by looking into above equation we can say that if

2 L

RC

then it will have no depending on frequency

for resonance2 L

RC

So 6

L 0.02R 10 2 14.14

C 100 10

55. In the circuit shown below, the node voltage VA is _________V.

Key: 11.42

Exp: All the branch currents are expressed interval of AV now writing KCL at node A

A A 1 AV V 10I V 105 0

5 5 5

5 5A

A 1I 5

5

110I

5

10V




29

A 1

1 1 1V 2I 5 1

5 5 10

AA

V 102 1V 2 6

5 10 10

A

2 1 2V 6 2

5 10 10

A

7V 8

10

A

80V 11.42V

7

AV -10

10

A 1V 10I

5

AV

5

1I

10

110I

5A

5 5

AV

5

5

Documents

EE gate 2016 set 1