General Aptitude -...

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1

General Aptitude

Q. No. 1 – 5 Carry One Mark Each

1. The chairman requested the aggrieved shareholders to ___________him.

(A) bare with (B) bore with (C) bear with (D) bare

Key: (C)

2. Identify the correct spelling out of the given options:

(A) Managable (B) Manageable (C) Manageble (D) Managible

Key: (B)

3. Pick the odd one out in the following

13, 23, 33, 43, 53

(A) 23 (B) 33 (C) 43 (D) 53

Key: (B)

Exp: Given numbers are 13, 23, 33, 43, 53.

All the numbers have second digits as 3.

If We sum of the digits of each number we get 4, 5, 6, 7, 8.

All given numbers are irrational numbers except 33 which is rotation.

So odd one out is 33.

4. R2D2 is a robot. R2D2 can repair aeroplanes. No other robot can repair aeroplanes.

Which of the following can be logically inferred from the above statements?

(A) R2D2 is a robot which can only repair aeroplanes.

(B) R2D2is the only robot which can repair aeroplanes.

(C) R2D2 is a robot which can repair only aeroplanes.

(D) Only R2D2 is a robot.

Key: (B)

5. If 29y 6 3, then y 4y 3 is _________.

(A) 0 (B) 1 3 (C) 1 3 (D) undefined

Key: (C)

Exp: 9Y 6 3

Possibility (A): 9y 9 y 1

Possibility (B): 1

9y 3 y3

When y=1

2 24y 4(1) 3 4 1

y 13 3 3 3




2

When 1

y3

2

14

1 1 4 3 13

3 3 9 9 9 3

Q. No. 6 – 10 Carry Two Mark Each

6. The following graph represents the installed capacity for cement production (in tones) and the

actual production (in tones) of nine cement plants of a cement company Capacity utilization of a

plant is defined as ratio of actual production of cement to installed capacity. A plant with installed

capacity of at least 200 tonnes is called a large plant and a plant with lesser capacity is called a

small plant. The difference between total production of large plants and small plants, in tones is

Key: 120

Exp: Largent plant

Installed Capacity 220 200 250 200

Actual production 160 190 230 190

Plant number 1 4 8 9

Total production of larger plants = 160+190+230+190=770 tonnes

Smaller Plants

Installed Capacity 180 190 160 150 140

Actual production 150 160 120 100 120

Plant number

Total production of smallest plants = 150+160+120+100+120= 650tonnes

Difference = 770-650 = 120 tonnes

300

250

200

150

100

50

0

1 2 3 4 5 6 7 8 9

Plant Number

Capacity

production

tonnes

220

160

180

150

190

160

200

190

160

120

150

100

140

120

250

230

200190

InstalledCapacity Actual Production




3

7. A poll of students appearing for masters in engineering indicated that 60% of the students

believed that mechanical engineering is a profession unsuitable for women. A research study on

women with master or higher degrees in mechanical engineering found that 99% of such women

were successful in their professions.

Which of the following can be logically inferred from the above paragraph?

(A) Many students have isconceptions regarding various engineering disciplines

(B) Men with advanced degrees in mechanical engineering believe women are well suited to be

mechanical engineers.

(C) Mechanical engineering is a profession well suited for women with masters or higher degrees

in mechanical engineering.

(D) The number of women pursuing high degrees in mechanical engineering is small.

Key: (C)

8. Sourya committee had proposed the establishment of Sourya Institutes of Technology (SITs) in

line with Indian Institutes of Technology (IITs) to cater to the technological and industrial needs

of a developing country.

Which of the following can be logically inferred from the above sentence?

Based on the proposal,

(i) In the initial years, SIT students will get degrees from IIT.

(ii) SITs will have a distinct national objective

(iii) SIT like institutions can only be established in consultation with IIT.

(iv) SITs will serve technological needs of a developing country.

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

(C) (ii) and (iv) only (D) (ii) and (iii) only

Key: (C)

9. Shaquille O’ Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free

throws out of 100 attempts on average. What is the probability that he will successfully make

exactly 6 free throws in 10 attempts?

(A) 0.2508 (B) 0.2816 (C) 0.2934 (D) 0.6000

Key: (A)

10. The numeral in the units position of 870 127 424211 146 3 is _____.

Key: (7)




4

Electrical Engineering

Q. No. 1 –25 Carry One Mark Each

1. The output expression for the Karnaugh map shown below is

(A) A B

(B) A C

(C) A C

(D) A C

Key: (B)

Exp:

2. The circuit shown below is an example of a

(A) low pass filter

(B) band pass filter

(C) high pass filter

(D) notch filter

Key: (A)

Exp: 2

21 i 2

22

s

1R

RcsZ R , Z1 1 R CS

Rc

2 2

1 2 1

Z RTF

Z 1 R CS R

WhenS 0 ; 2

1

RTF non zero value

R

When S ; TF 0

Given circuit is an example of low pass filter

BC

00 01 11 10A

0

1

1 0 0 1

1111

2R

C

15V

15V

1RinV

outV

B C B C B C B C

1 0 0

1

1

111

A

A

F A,B,C A C




5

3. The following figure shows the connection of an ideal transformer with primary to secondary

turns ratio of 1:100. The applied primary voltage is 100V (rms), 50Hz, AC. The rms value of the

current I, in ampere, is _____.

Key: (10)

Exp: Where

3secondary

L 22

Z 80 j40Z 10 8 4i

n 100 1

L

100 100 100I 10 36.86

j10 Z J10 8 j4 8 j6

Since all the calculation done with respect to RMS, I also in RMS

4. Consider a causal LTI system characterized by differential equation

dy t 1

y t 3x tdt 6

. The

response of the system to input 1

3x t 3e u t ,

where u(t) denotes the unit step function, is

(A) t

39e u t

(B) t

69e u t

(C) t t

3 69e u t 6e u t

(D) t t

6 354e u t 54e u t

Key: (D)

Exp: 1

S y s 3 s6

3

H s1

s6

Similarly 3

X s1

s3

9 54 54

y s1 1 1 1

s s s s3 6 6 3

Thus t

t 36y t 54 e u t 54e u t 0

LX 10

I~100V

I :100R 80k

CX 40k




6

5. Suppose the maximum frequency in a band-limited signal x(t) is 5kHz. Then, the maximum

frequency in x(t) cos 2000 t , in kHz, is ______.

Key: 6

Exp: Maximum frequency of x t 5kHz

Maximum frequency of cos 2000 t 1kHz

x t cos 2000 t Gives convolution between their respective spectrums in frequency domain

max frequency of x t cos 2000 t 6kHz

6. Consider the function f (z) z z* where z is a complex variable and z* denotes its complex

conjugate. Which one of the following is TRUE?

(A) f(z) is both continuous and analytic

(B) f(z) is both continuous but not analytic

(C) f(z) is not continuous but is analytic

(D) f(z) is neither continuous not analytic

Key: (B)

Exp: f z z z

x iy x iy z is conjugate of z

2x i 0

f z 2x is continues but not analytic, since

C – R equations will not satisfy

7. A 3 3 matrix P is such that, 3P P. Then the eigenvalues of P are

(A) 1, 1, - 1

(B) 1,0.5 j0.866,0.5 j0.866

(C) 1, 0.5 j0.866,0.5 j0.866

(D) 0, 1, - 1

Key: (D)

Exp: If is an Eigen value of p then 3 is an Eigen

value of p3.

3p p

3

3 2 20 1 0 0; 1

0; 1




7

8. The solution of the differential equation, for "t 0, y t 2y ' t y t 0 with initial

conditions y 0 = 0 and y ' 0 1, is u t denotes the unit step function),

(A) tte u t (B) t te te u t

(C) t te te u t (D) te u t

Key: (A)

Exp: The operator form of the of given D.E is

2D 2D 1 y 0

The A.E is 2D 2D 1 0

2

t

1 2

D 1 0 D 1, 1.

y t e C C t

Given y 0 0 & y' 0 1 i.e, t 0; y 0

from (s); 10 C 1C 0 y' 0 1

From (1), t t

1 2 2

dye C C t e C

dt

dy

At t 0; y ' 1dt

2 21 0 C 0 1 C 2 21 C C 1

From (1) t ty(t) e (t) te u(t)

9. The value of the line integral 2 2

c

2xy dx 2x ydy dz along a path joining the origin (0,0,0) and

the point (1,1,1) is

(A) 0 (B) 2 (C) 4 (D) 6

Key: (B)

Exp: Let 2 2f 2xy i 2x y j k

2 2

i j k

curl f f 0yx z

2xy 2x y 1

f is irrotational

Consider a straight line passing through 0,0,0 & 1,1,1

x y z

i.e, t x y z t dx dy dz dt1 1 1

1

2 2 3 3

t 0C

2xy dx 2x ydy dz 2t dt 2t dt dt

1 1

3 4

0t 04t 1 dt t t 2




8

10. Let f(x) be a real, periodic function satisfying f x f x . The general form of its Fourier

series representation would be

(A) 0 kk 1f x a a cos kx

(B) kk 1

f x b sin kx

(C) 0 2kk 1f x a a cos kx

(D) 2k 1k 1

f x a sin 2k 1 x

Key: (B)

Exp: We know that a periodic function f(x) defined in (-c, c) can be represented by the poisoins series

on n

n 1 n 1

a n x n xf x a cos b sin

2 c c

If a periodic function f(x) is odd, its Fourier expression contains only sine terms

11. A resistance and a coil are connected in series and supplied from a single phase, 100V, 50Hz ac

source as shown in the figure below. The rms values of plausible voltage across the resistance

RV and coil CV respectively, in volts, are

(A) 65, 35 (B) 50, 50 (C) 60, 90 (D) 60, 80

Key: (C)

12. The voltage (v) and current (A) across a load are as follows.

v t 100 sin t , Oi t 10sin t 60 2sin 3 t sin 5 t

The average power consumed by the load, in W, is ________.

Key: 250

Exp: The instantaneous power of load is

p t V i t

100sin t 10sin( t 60 100sin t 2sin3 t

100sin t 5sin5 t

T

avg

0

since P P t dt , in the above expression

Only 1st term will result non zero answer

Remaining 2 terms will be 0.

so directly consider

P t 100sin t 10 sin t 60

~SV

RV

CV




9

avg rms rms V IP V I cos 100 10 1000 1

cos 60 250 watt2 22 2

13. A power system with two generators is shown in the figure below. The system (generators, buses

and transmission lines) is protected by six overcurrent relays 1 6R to R . Assuming a mix of

directional and nondirectional relays at appropriate locations, the remote backup relays for R4 are

(A) 1 2R , R (B) 2 6R , R (C) 2 5R , R (D) 1 6R , R

Key: (D)

14. A power system has 100 buses including 10 generator buses. For the load flow analysis using

Newton-Raphson method in polar coordinates, the size of the Jacobian is

(A) 189 189 (B) 100 100 (C) 90 90 (D) 180 180

Key: (A)

Exp: Total no of Buses = 100

generator buses = 10 – 1 = 9 (n)

load buses = 100 -10 = 90(m)

Jacobeam matrix size = 2m n 2m n 2 90 9 2 90 9 189 189

15. The inductance and capacitance of a 400kV. Three-phase. 50Hz lossless transmission line are 1.6

mH/km/phase and 10nF km phase respectively. The sending end voltage is maintained at 400kV.

To maintain a voltage of 400kV at the receiving end, when the line is delivering 300MW load, the

shunt compensation required is

(A) Capacitive (B) Inductive (C) Resistive (D) Zero

Key: (B)

Exp: 3

LX j L j314 1.6 10 j0.5024

C 9

j jX j31847.3376

C 314 10 10

Since C LX X , the shunt compensation is inductive

16. A parallel plate capacitor field with two dielectrics is shown in the figure below. If the electric

field in the region A is 4kV/cm, the electric field in the region B, in kV/cm, is

1S

~

1R

3R

2R

4R

6R5R 2S

~

r

A

1 r

B

4 2cm




10

(A) 1 (B) 2 (C) 4 (D) 16

Key: (C)

Exp: Since the voltage & distance b w the two plates are same for both the regions. The electric field

is same for both the regions V

Ed

Electric field in region B = 4kV cm

17. A 50MVA, 10kV, 50Hz, star-connected, unloaded three-phase alternator has a synchronous

reactance of 1 p.u. and a sub-transient reactance of 0.2 p.u. If a 3-phase short circuit occurs close

to the generator terminals, the ratio of initial and final values of the sinusoidal component of the

short circuit current is _________.

Key: 5

Exp: prefault

SC

VI

X P.U

SC

S

I initial 15

I cfinal 0.2

18. Consider a liner time-invariant system transfer function

1H s

s 1

. If the input is cos(t) and

the steady state output is Acos t , then the value of A is _________.

Key: 0.707

Exp: 1

2

1H tan

1

O1H 45

2

So when input is cost then O/P

O1y t cos t 45

2

So 1

A 0.7072

19. A three-phase diode bridge rectifier is feeding a constant DC current of 100A to a highly inductive

load. If three-phase, 415V, 50Hz AC source is supplying to this bridge rectifier then the rms value

of the current in each diode, in ampere, is _________.

Key: 57.73

Exp: OI 100A

RMS, diode current = 100

57.73A3




11

20. A buck-bost DC-DC converter shown in the figure

below, is used to convert 24 V battery voltage to 36 V

DC voltage to feed a load of 72 W. It is operated at

20kHz with an inductor of 2 mH and output capacitor

of 1000 F. All devices are considered to be ideal.

The peak voltage across the solid-state switch (S), in

volt, is _________.

Key: 60

21. For the network shown in the figure below, the frequency in rad s at which the maximum

phase lag occurs is.______.

Key: 0.316

Exp: The given circuit is standard lag compensator

Whose Transfer function

11

s 1 1 ssG s1 1 10s 1 s

a 1s

So 1, 10 the frequency at which maximum phase lag happen

m

1 10.316 rad sec

10

22. The direction of rotation of a single-phase capacitor run induction motor is reversed by

(A) interchanging the terminals of the AC supply

(B) interchanging the terminals of the capacitor

(C) interchanging the terminals of the auxiliary winding.

(D) interchanging the terminals of both the windings

Key: (C)

23. In the circuit shown below, the voltage and current are ideal. The voltage (Vout) across the current

source, in volts, is

outsV

2

10V

5A

S

2mH

Load

36V24V

9

1

1F

0in




12

(A) 0 (B) 5 (C) 10 (D) 20

Key: (D)

Exp: Writing KVL

O

O

V 10 10 0

V 20V

24. The graph associated with an electrical network has 7 branches and 5 nodes. The number of

independent KCL equations and the number of independent KVL equations, respectively, are

(A) 2 and 5 (B) 5 and 2 (C) 3 and 4 (D) 4 and 3

Key: (D)

Exp: No of branches = 7

Nodes = 5

No of KCL equations = No of Modal equations = n – 1 = 5 – 1 = 4

No of KVL equations = No of Mesh equations = b-(n – 1) = 7-4 = 3

Since no information given regarding how many simple & principal node, if we assume all

principal nodes then the answer for nodal is 5 – 1 = 4

25. The electrodes, whose cross-sectional view is shown in the figure below, are at the same potential

The maximum electric field will be at the point

(A) A (B) B (C) C (D) D

Key: (A)

Exp: At A

Fields are additive 1 2F F

At C

Fields are C

subtractive

1 2F F

At D field is due to one electrode 2F

At B field make an angle

2 2

1 2 1 2F F 2FF cos

So maximum electric field is at ‘A’

A D C

B

OV

10V

10V

5A




13

Q. No. 26 – 50 Carry Two Mark Each

26. The Boolean expression a b c d b c simplifies to

(A) 1 (B) a.b (C) a.b (D) 0

Key: (D)

Exp: F = a b c d b c = a d b b c c = a d 1 1 = 1 = 0

27. For the circuit shown below, taking the opamp is ideal, the output voltage outV in terms of the

input voltages 1 2V , V and 3V is

(A) 1 2 31.8V 7.2V V (B) 1 2 32V 8V 9V

(C) 1 2 37.2V 1.8V V (D) 1 2 38V 2V 9V

Key: (D)

Exp: x 1 x 2 1 2x

V V V V 4V V0; V

1 4 5

x 3 x outV V V V

01 9

1 2

out1 2

3

4V VV4V V 5V 0

5 9

1 2 out

1 2 3

4V V 5V4V V 5V 0

9

1 2 3 1 2 out36V 9V 45V 4V V 5V 0

1 2 3out

40V 10V 45VV

5

out 1 2 3V 8V 2V 9V

9

1

4

2V

1V

3V

outV

1

xV

xV




14

28. Let 1 1 2 2x t X and x t X be two signals whose Fourier Transforms are as shown

in the figure below. In the figure h(t) = 2 t

e

denotes the impulse response.

For the system shown above, the minimum sampling rate required to sample y(t), so that y(t) can

be uniquely reconstructed from its samples, is

(A) 12B (B) 1 22 B B (C) 1 24 B B (D)

Key: (B)

Exp: 1 2y t x t x t * h t

In frequency duration

1 2y X *X H

Max. Frequency of 1 1X B

Max, frequency of 2 2X B

Max frequency of 1 2 1 2X *X B B

Max frequency of H

Thus max, freq of 1 2y B B

Max frequency

Nyquist frequency = 1 22 B B

29. The value of the integral sin 2 t

2 dtt

is equal to

(A) 0 (B) 0.5 (C) 1 (D) 2

Key: (D)

1X

1B1B

2

1B

2

1B2B

2X

2B

1x t

2x t

2 th t e

y t




15

Exp: 0

sin 2 t sin 2 t sin 2 t2 dt 2 2 dt is an even function

t t t

0.t

0

sin 2 t4 e .dt

t

By the defining of L.T; we have

sin 2 t

4L ; where S 0t

4 sin 2 t

L ; wheres 0t

1 14 2 sin at atan ; where s 0 L tan

s t s

Putting s = 0; than

14 4tan 2

2

sin 2 t

2 dt 2t

30. Let y(x) be the solution of the differential equation 2

2

d y dy4 4y 0

dx dx with initial conditions

x 0

dyy 0 0 and 1.

dx

Then the value of y (1) is _________.

Key: 7.398

Exp: The operate form of given D.E is

2D 4D 4 y 0

The A.E is 2D 4D 4 0

2

D 2 0 D 2,2

2

D 2 0 D 2,2

The solution is 2x

1 2y e C C x 1

Given that y 0 0 & y ' 0 1

i.e x 0 y 0 i.e at x 0, y' 1

from (1); 10 1 C 0 from 1 2x 2x

2 1 21 y e C C C x 2c

1C 0 2 21 C 0 C 1

2xfrom 1 y e 0 1.x 2xy xe

2 1 2 2y 1 1e e y 1 e or y 1 7.389




16

31. The line integral of the vector field F = 2 2ˆ ˆ ˆ5xzi 3x 2y j x zk along a path from (0,0,0) to

(1,1,1) parametrized by 2t, t , t is _________.

Key: 4.4167

Exp: 2 2F 5xzi 3x 2y j x z k

2x t; y t ; z t

dx dt d 2tdt; dz dt

The line integral of the vector field is

2 2

C

F.dr 5xzdx 3x 2y dy x z dz

1

2 3 3

t 0

5t dt 10t dt t dt

1

2 3

t 0

5t dt 11t dt

1 13 4

0 0

t t5 11

3 4

20 23 535 11 4.4167

3 4 1̀2 12

32. Let P = 3 1

1 3

. Consider the set S of all vector x

y

such than 2 2a x

a b 1 where pb y

.

Then S is

(A) A circle of radius 10 (B) a circle of radius = 1

10

(C) an ellipse with major axis along 1

1

(D) an ellipse with minor axis along 1

1

Key: (D)

Exp: 3 1

P1 3

a x a 3 1 x

Pb y b 1 3 y

3x y a x 3y b

2 2a b 1

2 2

3x y x 3y 1

2 210x 10y 12xy 1

a 10; b 10; h 6




17

It represents ellipse

The length of semi-axes is 2 4 2ab h r a b r 1 0

4 264r 20r 1 0 2 21 1r or r

4 16

Both 2r values are positive, so it represents ellipse

1 1

r or r2 4

Length of major axis = 2r 1

Length of minor axis = 1 12r 2

24

Equation of the major axis is 2

1

1a x hy 0

r

10 4 x 6y 0 x y 0

Equation of the minor axis is 2

2

1a x hy 0

r

10 16 x 6y 0 y x 0

Major axis exists along y = -x and minor axis exists along y = x.

1

The vector1

Lies on the line y = x

33. Let the probability density function of random variable, X, be given as:

3x 4x

x

3f x e u x ae u x

2

where u(x) is the unit step function. Then the value of ‘a’ and Prob X 0 , respectively, are

(A) 1

2,2

(B) 1

4,2

(C) 1

2,4

(D) 1

4,4

Key: (A)

Exp: we have xf x dx 1

0

x x

0

f x dx f x dx 1

0

4x 3x

0

3a e dx e dx 1

2

u x 1 for x 0

0, other wise




18

u x 1 for x 0

0,otherwise

0

4x 3x

0

3a e dx e dx 1

2

04x 3x

0

e 3 ea 1

4 2 3

1 1

a 0 0 1 14 2

a 1 a 11 a 2

4 2 4 2

Prob 0 0 0

4x 4x

xx 0 f x dx a.e dx a e dx

04xe

2y

2 11 0

4 2

34. The driving point input impedance seen from the source VS of the circuit shown below, in , is

__________.

Key: 20

Exp: The Driving point impedance is nothing but the ratio of voltage to current from the defined port.

In this case it is S

S

V

I

Writing KCL at node x

x xS 1

V VI 4V 0

3 6

Substituting these in Eq(1)

S s S sS s

V 2I V 2II 8I 0

3 6

S S

1 1 2 2V I 1 8

3 6 3 6

S SV 2 1 I 6 4 48 2

S

S

V 6020

I 3

1V

2SI

3 14V

2

4

SV

1V

6

1V

2SI

3 14V

2

4SV

xV

xV

3

1I




19

35. The z-parameters of the two port network shown in the figure are 11 12Z 40 , Z 60 ,

21 22Z 80 , Z 100 . The average power delivered to LR 20 , in watts, is ____.

Key: 35.55

Exp: In the given terminated 2 port network the Z matrix is known and for load of 20 we want to

find power on the load.

→ The get it assuming LR as load let first obtain the thevenin equivalent of 2 port

→ Thevenin equivalent means th thV & R

2

th 2 I 0V V i.e., O.C voltage of port 2

SC 2 2I I V 0 i.e., s.c current of port 2,

thin

sc.

VR

I

→ Evaluation of thV .

The Z matrix equation is

1 1 2V 40I 60I

2 1 2V 80I 100I

In the above two equations if 2I 0 then

1 1V 40I (1)

2 1V 80I (2)

From the input side we can say 1 1V 20 10I

1 120 10I 40I

1

2I A5

Then equation 2 becomes 2

3

2 1

2V 80 I 80 32V

5

th 2so V V 32

Evaluation of ISC

In the Z matrix equation if we put 2V 0 then

1 1 2V 40I 60I …(5)

1 20 80I 100I …(4)

101I

1V

20V Z

2V

LR

2I




20

1 2 1 1 2

10 100I I & V 20 10I 20 I

8 8

Using these 1 1V & I in equation 3

2 2 2

100 40020 I I 60I

8 8

2 2 2160 100I 400I 480I

2 2160 20I I 8A

SC 2I I 8A

inin

sc

V 32R 4

I 8

Now the ckt is from port 2is

2

20 20P I 20

32

204 20

35.55watt

36. In the balanced 3-phase, 50Hz, circuit shown below,

the value of inductance (L) is 10mH. The value of

the capacitance (C) for which all the line current are

zero, in millifarads, is _____.

Key: 3.04

Exp: LI 0 2ph

L Cph

L C

j L j

X .X 3 cZ

j L jX X

3 c

L 1

b c

3

3314

10 10 C

C 3.04 mF

LI

CL

3

L 3 L 3

CC

thV 32

thR 4

LR 20

L

L

C C

L

C




21

37. In the circuit shown below, the initial capacitor voltage is 4V.

Switch S1 is closed at t = 0. The charge in C lost by the

capacitor from t 25 s to t 100 s is _______.

Key: 6.99

Exp: It is given CV 0 4

1

R 5 , C 54f so 40000RC

→ Since it is a source free network we can say

t

C CV t V 0 e ; t 0 40000 t4e

→ We are asked to find the charge last by capacitor

From t 25 s to 100 s

We know in a capacitor Q CV

or Q C V

C CQ C V 25 sec V 100 sec

→ 40000t

CV 4e 640000 25 10

C

t 25 sec

V 4e 1.47

640000 100 10

C

t 100 sec

V 4e 0.073

Q 5 1.47 0.073 6.99 s

38. The single line diagram of a balanced power system is shown in the figure. The voltage magnitude

at the generator internal bus is constant and 1.0 p.u. the p.u. reactances of different components in

the system are also shown in the figure. The infinite bus voltage magnitude is 1.0p.u. A three

phase fault occurs at the middle of line 2.

The ratio of the maximum real power that can be transferred during the pre-fault condition to the

maximum real power that can be transferred under the faulted condition is ______

1S

4V

5 F

5

Generator int ernal bus j0.1

j0.5

Line 1

inf inite bus

j0.5

Line 2

j0.1

j0.2~




22

Key: 2.286

Exp:

ac

0.375 0.125 0.125 0.0729 0.0729 0.375X

0.0729

1.143

EV

Pe max1.143

Pe Prefault 1.1432.286

Pe during fault 20.5

39. The open loop transfer function of a unity feedback control system is given by

K s 1G s , K 0, T 0.

s 1 Ts 1 2s

The closed loop system will be stable if

(A) 4 K 1

0 TK 1

(B)

4 T 20 K

T 2

(C) T 2

0 KT 1

(D)

8 K 10 T

K 1

Key: (C)

Exp: To comment closed 100b system stability we need the characteristic equation. Here it is given that

it is a unity feedback system.

Unity feedback system

So the characteristic equation is

S 1 TS 1 2S K S 1 0

During fault

0.2

0.6

0.1 0.25

~

0.25

0.2

0.35 ~

0.6

0.25

b

C

Before fault

0.1

0.2

0.5

0.1 0.5 ~

0.1 0.5Xeq 0.2 0.5 P.U

2

e

eq

EVP max 2EV

X

0.6b c

x cx b

x e

0.35 0.25

b

0.175 0.125

0.0729

C

a0.375 0.125

0.0729

C

Converting Y

acXC

a




23

2S TS 1 2S KS K 0

2 2 3S 2S TS 2TS KS K 0

3 2S 2T S 2 T S 1 k k 0

3 22 T k 1 ks s s 0

2T 2T 2T

for stability using R t criterion

2 T K 1 K

2T 2T 2T

T 2 K 1 K K 1 1

K T 2

1 1

1̀K T 2

1 T 1 T 2

KK T 2 T 1

40. At no load condition a 3-phase, 50Hz, lossless power transmission line has sending –end and

receiving-end voltage of 400 kV and 420kV respectively. Assuming the velocity of traveling

wave to be the velocity of light, the length of the line, in km, is __________.

Key: 294.84

Exp: 2 2 10

S r

10V V 1

18

2 2 10314 10

400 420 1 294.84km18

41. The power consumption of industry is 500kVA, at 0.8 p.f. lagging. A synchronous motor is added

to raise the power factor of the industry to unity. If the power intake of the motor is 100kW. The

p.f. of the motor is _________.

Key: 0.3162

Exp: 2cos 1

2

1

1

1 2

0

36.86

Pcos

S

P 400; P 100

motor 1 1 1 2 2Q P tan P P tan 400 tan36.86 500 tan 300 kW

motor

m

S 100 j300

cos 0.3162

2 400 100

36.86




24

42. The flux linkage and current (i) relation for an electromagnetic system is i g . When i

= 2A and g air gap length 10cm, the magnitude of mechanical force on the moving part, in

N, is ________.

Key: 186 to 190

43. The starting line current of a 415V. 3-phase, delta connected induction motor is 120A, when the

rated voltage is applied to its stator winding. The starting line current at a reduced voltage of

110V, in ampere is _________.

Key: 31 to 33

44. A single-phase, 2kVA, 100 200V transformer is reconnected as an auto-transformer such that its

kVA rating is maximum. The new rating in kVA, is _______.

Key: 6

Exp:

`

45. A full-bridge converter supplying in RLE load is shown in figure. The firing angle of the bridge

converter is 120O. The supply voltage m t 200 sin 100 t V, R 20 , E 800V. The

inductor L is large enough to make the output current LI a smooth dc current. Switches are

lossless. The real power fed back to the source. In kW is ________.

`

Key: 6

Exp: 0

2Vm 200V cos 2 cos120 200V

Load

L

R 20

E 800V

LI

3T1T

4T 2T

BridgeinV~

20A

100V

200V

300V

30A

10A

max

KVA rating 300 200

6kVA




25

00

E V 800 200I 30A

R 20

fedback 0 0P V I 200 30 6kW .

46. A three – phase Voltage Source Inverter (VSI) as shown in the figure is feeding a delta connected

resistive load of 30 phase . If it is fed from a 600V battery, with 180O conduction of solid-state

devices, the power consumed by the load, in kW, is _________.

Key: 24

Exp: ph

2 2V Vdc 600 200 2 V

3 3

ph

R 30R 10

3 3

2

2

Load

ph

200 23VphP 3 24kW

R 10

47. A DC-DC boost converter, as shown in the figure below, is used to boost 360V to 400V, at a

power of kW. All devices are ideal. Considering continuous inductor current, the rms current in

the solid state switch (S), in ampere, is ______.

Key: 3 to 4

48. A single-phase bi-directional voltage source converter (VSC) is shown in the figure below. All

devices are ideal. It is used to charge a battery at 400V with power of 5kW from a source

sV 220V(rms),50Hz sinusoidal AC mains at unity p.f. If its ac side interfacing inductor is 5mH

and the switches are operated at 20kHz, then the phase shift between AC mains voltage SV

and fundamental AC rms VSC voltage C1V , in degree, is________.

600V

30

30

30

10mH

1mF

Load

400 V

S360V




26

Key: 9.1 to 9.3

49. Consider a linear time invariant system x Ax, with initial condition x(0) at t = 0. Suppose

and are eigenvectors of 2 2 matrix A corresponding to distinct eigenvalues 1 2and

respectively. Then the response x t of the system due to initial condition x (0) = is

(A) 1te (B) 1te (C) 2te (D) 2 2t te e

Key: (A)

Exp: Eigen values are nothing but pole location

Here with respect to the pole is 1

wrt Pole is 2

The section should be of form t2

1te e but we are asking w.r.t

Initial condition x 0 only so the response should be 1te

50. A second-order real system has the following properties:

(a) the damping ratio 0.5 and undamped natural frequency n 10 rad s,

(b) the steady state value of the output, to a unit step input, is 1.02.

The transfer function of the system is

(A) 2

1.02

s 5s 100 (B)

2

102

s 10s 100 (C)

2

100

s 10s 100 (D)

2

102

s 5s 100

Key: (B)

Exp: The standard 2nd

order T/F is 2

n

2 2

n n

Ks 2 s

it is given that n0.5 & 10

2

100G s K

s 10s 100

Now to satisfy the steady state O/P 1.02

2s 0

1 100y t K 1.02 K 1.02

s s 10s 100

2 2

1.02 100 102G s

s 10s 100 s 10s 100

5mH

SXSI

~220VAC

1mH 400V

SI

SV

1 1I XC1V




27

51. Three single-phase transformers are connected to form a delta-star three-phase transformer of

110kV 11kV. The transformer supplies at 11kV a load of 8MW at 0.8 p.f. lagging to a nearby

plant. Neglect the transformer losses. The ratio of phase current in delta side to star side is

(A) 1:10 3 (B) 10 3 :1 (C) 1:10 (D) 3 :10

Key: (A)

Exp: 1 2

11N : N 110: 10 3 :1

3

1 1 2 2I N I N

i 2I 10 3 I .1

1

2

I 1

I 10 3

52. The gain at the breakaway point of the root locus of a unity feedback system with open loop

transfer function

KsG s

s 1 s 4

is

(A) 1 (B) 2 (C) 5 (D) 9

Key: (A)

Exp:

KsG s

s 1 s 4

To find Break away point

We need to find the root of dk

0ds

where

2s 1 s 4 s 5s 4

Ks s

2 2

2

d ds s 5s 4 s 5s 4 s

dk ds ds

ds s

2S 2S 5 S 5S 4 0

2 22S 5S S 5S 4 0

2S 4 0 S 2

From the pole zero plot it is clean that Break away point must be S 2 as it is in between 2

poles

Now to find gain at this point use magnitude condition

s 2

KS1

s 1 s 4

KS1 K 1

1 2

1 4




28

53. Two identical unloaded generators are connected in parallel as shown in the figure. Both the

generators are having positive, negative and zero sequence impedance of j0.4p.u, j0.3p.u and

terminals of the generators, the fault current, in p.u., is ________.

Key: 6

Exp: 0Z 0.15 P.U ; 1

0.4Z 0.2 P.U

2 ; 2

0.3Z 0.15P.U

2

f

O 1 2

3Vprefault 3I 6p.u

Z Z Z 0.15 0.2 0.15

54. An energy meter, having meter constant of 1200 revolutions kWh, makes 20 revolutions in 30

seconds for a constant load. The load, in kW is ______

Key: 2

Exp:

20revolutionsK 1200re v kwh

30P kW hr

3600

P 2kW

55. A rotating conductor of 1m length is placed in a radially

outward (about the z-axis) magnetic flux density (B) of 1 Tesla

as shown in figure below. Conductor is parallel to and at 1m

distance from the z-axis. The speed of the conductor in r.p,m.

required to induce a voltage of 1V across it, should be ______.

Key: 9.55

Exp: Voltage = B

Velocity Voltage 1

1m sB 1 1

i.e, 1m takes

2 2 1 2 m Takes 2 r

2 r for 1 rotation

in 1 minute = 60

2

Velocity = 60

9.55 rpm2

~

~

z

B

1m

1m

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