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General Aptitude Q. No. 1 – 5 Carry One Mark Each 1. Key: 2. Key: 3.

Key: Exp:

4.

The chairman requested the aggrieved shareholders to ___________him. (A) bare with (B) bore with (C) bear with (D) bare (C) Identify the correct spelling out of the given options: (A) Managable (B) Manageable (C) Manageble (B) Pick the odd one out in the following 13, 23, 33, 43, 53 (A) 23 (B) 33 (C) 43 (B) 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.

(D) Managible

(D) 53

Key:

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. (B)

5.

If 9y  6  3, then y2  4y 3 is _________. (A) 0

Key: Exp:

(B) 1 3

(C) 1 3

(D) undefined

(C) 9Y  6  3 Possibility (A): 9y  9  y  1 Possibility (B): 9y  3  y 

1 3

When y=1 4y 2 4(1) 3  4 1 y2  1    3 3 3 3  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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1 3 1 4  2 1  3   1  4  3   1    3 9 9 9 3  3

When y 

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

300 250

Installed Capacity

Actual Pr oduction 250 230

220

190

180

200

160

Capacity production 150  tonnes  100

200

200 190

190 150

160

160

150

140

120

100

5

6

120

50 0

1

2

3

4

7

8

9

Plant Number

Key: Exp:

120 Largent plant Installed Capacity Actual production Plant number

220 160 1

200 190 4

250 230 8

200 190 9

Total production of larger plants = 160+190+230+190=770 tonnes Smaller Plants Installed Capacity Actual production Plant number

180 150

190 160

160 120

150 100

140 120

Total production of smallest plants = 150+160+120+100+120= 650tonnes Difference = 770-650 = 120 tonnes  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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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.

Key:

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 (A)

10.

The numeral in the units position of 211870  146127  3424 is _____.

Key:

(7)

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Electrical Engineering Q. No. 1 –25 Carry One Mark Each 1.

The output expression for the Karnaugh map shown below is (A) A  B

BC 00

A

(B) A  C

0

(C) A  C

1

11

10

0

0

1

1

1

1

01

1 1

(D) A  C Key: Exp:

(B) A

BC

BC

1

0

0

1

1

1

BC

BC F  A, B, C   A  C

1 1

A

R2

2.

The circuit shown below is an example of a C

(A) low pass filter

15V

(B) band pass filter Vin

(C) high pass filter



Vout



(D) notch filter Key:

R1

15V

(A) 1 R2 cs  Z1  R i , Z2  1 1  R 2 CS R2  cs Z R2 TF   2   Z1 1  R 2CS R1 R2 

Exp:

When S  0 ; TF  

R2  non zero value  R1

When S  ; TF  0 Given circuit is an example of low pass filter

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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 _____. XL  10

(10)

Exp:

Where ZL 

I

R  80k

XC  40k

I

100V ~

Key:

I :100

 80  j40  3   10    8  4i   2  100 1 

Zsec ondary n2

100 100 100    10 36.86 j10  ZL J10  8  j4 8  j6

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

4.

dy  t 

Consider a causal LTI system characterized by differential equation

dt



1 y  t   3x  t  . The 6

1  3

response of the system to input x  t   3e u  t  , where u(t) denotes the unit step function, is (A) 9e (C) 9e Key: Exp:



t 3

t  3

ut

(B) 9e t  6

u  t   6e u  t 



t 6

ut

(D) 54e



t 6



t 3

u  t   54e u  t 

(D) 1   S   y s   3  s  6  3  H s  1  s   6 

Similarly X  s  

 y s 

3  1 s    3

9 54 54    1  1   1   1   s   s    s    s   6  6  3  3 

Thus y  t   54 e

t

6

u  t   54e t 3 u  t   0

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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

Key:

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 (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 A 3  3 matrix P is such that, P3  P. Then the eigenvalues of P are

7.

(A) 1, 1, - 1 (B) 1,0.5  j0.866,0.5  j0.866 (C) 1, 0.5  j0.866,0.5  j0.866 Key: Exp:

(D) 0, 1, - 1 (D) If  is an Eigen value of p then 3 is an Eigen value of p3.  p3  p

 3  

 3    0    2  1  0    0;  2  1

   0;    1

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The solution of the differential equation, for t  0, y"  t   2y '  t    y  t   0 with initial

8.

conditions y  0  = 0 and y '  0   1, is u  t  denotes the unit step function), (A) te  t u  t 

(B) e t  te t u  t 





(C) e t  te t u  t 

(D) e t u  t 

Key: (A) Exp: The operator form of the of given D.E is

D2  2D  1 y  0 The A.E is D2  2D  1  0   D  1  0  D  1,  1. 2

 y  t   e  t  C1  C2 t 

Given y  0   0 & y '  0   1 i.e, t  0; y  0 from (s); 0  C1  C1  0  y '  0   1

dy  e  t  C1  C2 t   e  t C2  dt dy At t  0;  y' 1 dt

From (1),

1  0  C2  0  1C2   1  C2  C2  1

From (1) y(t)  e t (t)  te t u(t) 9.

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

The value of the line integral

2

c

the point (1,1,1) is (A) 0 Key: (B) Exp:

(B) 2

(C) 4

(D) 6

Let f  2xy 2 i  2x 2 y j  k

i   curl f    f  x 2xy 2

j 

y

2x 2 y

k  0 z 1

 f is irrotational

 Consider a straight line passing through  0,0,0  & 1,1,1 i.e,

x y z    t  x  y  z  t  dx  dy  dz  dt 1 1 1

 2xy dx  2x 2

C

2

ydy  dz 



1

t 0

2t 3 dt  2t 3 dt  dt  

1

t 0

 4t

3

 1 dt   t 4  t   2 1

0

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Let f(x) be a real, periodic function satisfying f   x   f  x  . The general form of its Fourier

10.

series representation would be (A) f  x   a 0   k 1 a k cos  kx 

(B) f  x    k 1 bk sin  kx 



(C) f  x   a 0  k 1 a 2k cos  kx  

Key: Exp:



(D) f  x    k 1 a 2k 1 sin  2k  1x 

(B) We know that a periodic function f(x) defined in (-c, c) can be represented by the poisoins series   a nx nx f  x   o   a n cos   bn sin 2 n 1 c c n 1 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

 VR  and coil  VC 

respectively, in volts, are VR ~

VC

VS

Key:

(A) 65, 35 (C)

(B) 50, 50

12.

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



(C) 60, 90

(D) 60, 80



v  t  100 sin   t  , i  t   10sin t  60O  2sin 3t   sin 5t 

Key: Exp:

The average power consumed by the load, in W, is ________. 250 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

 since Pavg 

 P  t  dt , in the above expression 0

st

Only 1 term will result non zero answer Remaining 2 terms will be 0.  so directly consider P  t   100sin t  10 sin t  60  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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Pavg  Vrms I rms cos  V  I  

13.

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100 10 1000 1 cos  60   250 watt 2 2 2 2

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 R1 to R 6 . Assuming a mix of directional and nondirectional relays at appropriate locations, the remote backup relays for R 4 are R1

R2

S1 ~

(A) R1 , R 2

R6

R5 R3

~

R4

(B) R 2 , R 6

S2

(C) R 2 , R 5

(D) R1 , R 6

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: Exp:

(A) 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.

Key: Exp:

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 (B) XL  jL  j314 1.6 103  j0.5024 XC 

j j    j31847.3376 C 314 10 109

Since XC  XL , 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 A r  1

B r  4

2cm

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Key:

(A) 1 (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 E 

(C) 4

(D) 16

V d

Electric field in region B = 4kV cm 17.

Key: Exp:

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 _________. 5 Vprefault ISC  X  P.U  ISC  initial  IScfinal



1 5 0.2

Consider a liner time-invariant system transfer function H  s  

18.

Key: Exp:

1 . If the input is cos(t) and  s  1

the steady state output is Acos  t   , then the value of A is _________. 0.707 1 H     tan 1  2  1 1 H    45O 2 So when input is cost then O/P 1 yt  cos t  45O 2 1  0.707 So A  2





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: IO  100A RMS, diode current =

100  57.73A 3

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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.

 24V

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Load 36V

S



2mH



The peak voltage across the solid-state switch (S), in volt, is _________. Key: 60 For the network shown in the figure below, the frequency  in rad s  at which the maximum

21.

9

phase lag occurs is.______.

1

in

0

1F

Key: Exp:

0.316 The given circuit is standard lag compensator Whose Transfer function 1 1 s  s  1  1  s G s 1 1  10s 1  s a 1 s So   1,   10 the frequency at which maximum phase lag happen 1 1 m    0.316 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 2 10V

 

5A



Vouts 

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(C) 10

(D) 20

Key: (D) Exp:

10V 

Writing KVL VO  10  10  0

10V

 

5A

VO  20V



VO 

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

D

C

B

(A) A Key:

(A)

Exp:

At A

(B) B

(C) C

(D) D

Fields are additive F1  F2 At C C  Fields are subtractive F1  F2

At D field is due to one electrode F2 At B field make an angle

F12  F22  2F1F2 cos  So maximum electric field is at ‘A’  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

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Q. No. 26 – 50 Carry Two Mark Each





26.

The Boolean expression a  b  c  d   b  c  simplifies to

Key:

(A) 1 (D)

Exp:

F

27.

For the circuit shown below, taking the opamp is ideal, the output voltage Vout in terms of the

=

(C) a.b

(B) a.b

a  b  c  d  b  c



= a d  bb  cc

input voltages V1 , V2 and V3 is

1

V3 V1

1

 

(D) 0



= a  d 1  1 = 1 = 0

9

Vx Vx





Vout

4 V2

Key: Exp:

(A) 1.8V1  7.2V2  V3

(B) 2V1  8V2  9V3

(C) 7.2V1  1.8V2  V3

(D) 8V1  2V2  9V3

(D) Vx  V1 Vx  V2 4V  V2   0; Vx  1 1 4 5

 Vx  V3  

Vx  Vout 0 1 9  4V1  V2   V out  4V1  V2   V  5 0 3 5 9 4V  V  5V  4V1  V2  5V3    1 92 out   0 36V1  9V2  45V3  4V1  V2  5Vout  0 40V1  10V2  45V3 5  8V1  2V2  9V3

Vout 

Vout

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Let x1  t   X1   and x 2  t   X 2   be two signals whose Fourier Transforms are as shown

28.

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

2 t

denotes the impulse response.

X1  

B1

 B1 2

X 2  

B1 2





 B2

B1

B2

x1  t  h t  e

2 t

yt

x2  t 

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) 2B1

(B) 2  B1  B2 

Key:

(B)

Exp:

y  t    x1  t  x 2  t  * h  t 

(C) 4  B1  B2 

(D) 

In frequency duration y    X1   *X2   H  

Max. Frequency of X1    B1 Max, frequency of X 2    B2 Max frequency of X1   *X 2     B1  B2  Max frequency of H     Thus max, freq of y     B1  B2  Max frequency

 Nyquist frequency = 2  B1  B2    sin 2t  The value of the integral 2  dt is equal to   t  (A) 0 (B) 0.5 (C) 1

29.

Key:

(D) 2

(D)

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

Exp:



sin 2 t  sin 2 t  2  dt  dt  2  2  t  t   0 

 4  e0.t 0

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 sin 2 t   t is an even function   

sin 2 t .dt t

By the defining of L.T; we have

 sin 2t   4L   ; where S  0  t  

4 4  sin 2t  1  2  L  ; wheres  0  tan   ; where s  0    t   s 

  sin at  1  a   L  t   tan  s       

Putting s = 0; than



4 4 tan 1        2 2   

2

30.

 sin 2t  dt  2 t  

 

Let y(x) be the solution of the differential equation

y  0   0 and Key: Exp:

dy dx

d2 y dy 4  4y  0 with initial conditions 2 dx dx

 1. Then the value of y (1) is _________. x 0

7.398 The operate form of given D.E is

D2  4D  4 y  0 The A.E is D2  4D  4  0   D  2  0  D  2,2 2

  D  2  0  D  2,2 2

 The solution is y  e 2x  C1  C2 x   1 Given that

 from 1 y  e

y  0  0

&

y '0  1

i.e x  0  y  0

i.e at x  0, y'  1

 from (1); 0  1  C1  0

from 1  y1  e2x C2    C1  C2 x  2c 2x

 C1  0

 1  C2  0  C2  1

2x

 0  1.x   y  xe

 y 1  1e 21  e 2  y 1  e 2

2x

 or  y 1  7.389

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

The line integral of the vector field F = 5xziˆ  3x 2  2y ˆj  x 2 zkˆ along a path from (0,0,0) to

31.





(1,1,1) parametrized by t, t 2 , t is _________. Key: Exp:

4.4167

F  5xzi  3x 2  2y  j  x 2 z  k x  t; y  t 2 ; z  t

 dx  dt d  2t dt;  dz  dt  The line integral of the vector field is

 F.dr   5xzdx  3x

2

 2y dy   x 2 z  dz

C

1



 5t

2

dt  10t 3dt  t 3dt

2

dt  11t 3dt

t 0 1



 5t

t 0

1

1

 t3   t4   5    11    3 0  4 0

20  23 53  5  11    4.4167 3 4 `12 12

32.

Key: Exp:

a x 3 1  x Let P =  . Consider the set S of all vector   such than a 2  b 2  1 where    p   .  b  y 1 3  y Then S is 1 (A) A circle of radius 10 (B) a circle of radius = 10 1 1 (C) an ellipse with major axis along   (D) an ellipse with minor axis along   1 1 (D) 3 1  P   1 3 a   x   a  3 1  x   b   P  y    b   1 3  y  3x  y  a  x  3y  b         

a 2  b2  1   3x  y    x  3y   1 2

2

 10x 2  10y2 12xy  1

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 It represents ellipse



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

The length of semi-axes is ab  h 2 r 4   a  b  r 2  1  0

1 1  or  r 2  4 16 2 Both r values are positive, so it represents ellipse 1 1  r   or  r  2 4 Length of major axis = 2r  1  64r 4  20r 2  1  0  r 2 

1 Length of minor axis = 2r  2    1 2 4  1  Equation of the major axis is  a  2  x  hy  0 r1  

 10  4  x  6y  0  x  y  0  1 Equation of the minor axis is  a  2  x  hy  0 r2  

  10  16  x  6y  0  y  x  0

Major axis exists along y = -x and minor axis exists along y = x. 1  The vector   Lies on the line y = x 1 33.

Let the probability density function of random variable, X, be given as: 3 f x  x   e 3x u  x   ae 4x u   x  2 where u(x) is the unit step function. Then the value of ‘a’ and Prob X  0 , respectively, are (A) 2,

Key:

1 2

(B) 4,

1 2

(C) 2,

1 4

(D) 4,

1 4

(A) 

Exp:

we have

 f  x dx  1 x





0



 f  x  dx   f  x  dx  1 x

x





0

0





0

4x  a e dx 

3

2 e

3x

dx  1

 u  x   1 for x  0  0, other wise

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u   x   1 for x  0 0,otherwise 

 ae



4x



0



0

 e4x  3 3  e3x  dx   e 3x dx  1  a      1 2  4   2  3  0 0

a 1 a 1 1  1  a   0  0  1  1   1    a  2 4 2 4 2 4  2 Prob  x  0  

0

 f x  x  dx 



0

4x  a.e dx  a



0

e

4x

dx



0

 e4x  2 1  2    1  0  2  y   4 The driving point input impedance seen from the source V S of the circuit shown below, in  , is __________.

34.



IS  VS 

Key: Exp:

V1

2



2 3

4V1

4

20 The Driving point impedance is nothing but the ratio of voltage to current from the defined port. V In this case it is S Vx 2 V1  IS   Writing KCL at node x V V  IS  x  4V1  x  0 3 6 Substituting these in Eq(1) V  2Is V  2Is  IS  S  8Is  S 0 3 6

2 2 1 1   VS     IS 1   8   3 6 3 6 

IS  VS 

2 3

4

4V1 Vx 3

V1 6

I1

 VS  2  1  IS  6  4  48  2 



VS 60   20 IS 3

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The z-parameters of the two port network shown in the figure are Z11  40, Z12  60 ,

35.

Z21  80, Z22  100 . The average power delivered to R L  20, in watts, is ____. I2

10 I1  20V  

Key: Exp:

V1



 Z

V2



RL



35.55 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 R L as load let first obtain the thevenin equivalent of 2 port → Thevenin equivalent means Vth & R th Vth  V2 I

2 0

i.e., O.C voltage of port 2

ISC   I2 V2  0 i.e., s.c current of port 2, R in 

Vth Isc.

→ Evaluation of Vth . The Z matrix equation is V1  40I1  60I2

V2  80I1  100I2 In the above two equations if I2  0 then

V1  40I1

(1)

V2  80I1

(2)

From the input side we can say  V1  20 10I1 

 20  10I1  40I1 2  I1  2 A Then equation 2 becomes 5 3 2  V2  80 I1  80   32 V 5 so Vth  V2  32 Evaluation of ISC In the Z matrix equation if we put V2  0 then

V1  40I1  60I2

…(5)

0  80I1  100I2

…(4)

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10 100 I 2 & V1  20  10I1  20  I2 8 8 Using these V1 & I1 in equation 3 I1  

100 400 I2   I 2  60I 2 8 8  160  100I2   400I2  480I2 20 

 160  20I2  I2  8A ISC   I2  8A  R in 

R th  4

Vin 32   4 Isc 8

 Now the ckt is from port 2is

R L  20

Vth  32

P20   I20  20 2

 32    20  35.55watt  4  20  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 _____.

C

L

C

L

C

L

Key: Exp:

IL

3.04 IL  0  2ph    jL   j     X L .XC 3  c  Zph   jL j X L  XC  3 c L 1  b c 3 314  10  103  C C  3.04 mF

L 3

C

L3

L3

C

C

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S1

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 _______. 4V

5

5F

Key:

6.99

Exp:

It is given VC 0  4

 

1  40000 RC Since it is a source free network we can say VC  t   VC 0 e t  ; t  0  4e40000 t

R  5, C  54f so →

 

→

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 

Q  C VC  25 sec   VC 100 sec  → VC  4e40000t 6

 VC

 4e40000 2510  1.47

 VC

 4e 40000 10010  0.073

t  25 sec 6

t 100  sec

 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 ______ Generator int ernal bus

inf inite bus

j0.1 j0.5 Line 1

j0.2 ~

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2.286 During fault

Before fault

0.6

0.5

0.1

0.2

0.2

~

~

0.5

0.1

Xeq  0.2 

0.1  0.25

0.1  0.5  0.5 P.U 2

0.6 0.2 ~

EV Pe  max    2EV Xeq

0.35

Converting Y   c

0.175

xe

0.35

0.125

C

b

xc

C

0.25

b

0.6 b

xb

 0.25

a

0.375

0.125

a C

Xac

C



0.0729

0.25

0.0729

0.375  0.125  0.125  0.0729  0.0729  0.375  1.143 0.0729 Pe  Prefault  EV 1.143 Pe  max      2.286 Pe during fault 1.143   20.5

X ac 

39.

The open loop transfer function of a unity feedback control system is given by K  s  1 G s  , K  0, T  0. s 1  Ts 1  2s  The closed loop system will be stable if (A) 0  T  (C) 0  K 

Key: Exp:

4  K  1 K 1 T2 T 1

(B) 0  K  (D) 0  T 

4  T  2 T2 8  K  1

K 1 (C) 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

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 S  TS2  1  2S  KS  K  0  S  2S2  TS2  2TS3  KS  K  0

 S3  2T   S2  2  T   S 1  k   k  0

k  2  T  2  k 1  s3   0 s    s 2T  2T   2T   for stability using R  t  criterion

 2  T   K 1 K      2T   2T  2T K 1 1    T  2   K  1  K  K T2 1 1 1  T 1  T  2   `1      K     K T2 K T2  T 1 

40.

Key:

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 __________. 294.84

 2  2  1010  Exp: VS  Vr 1   18    3142   2  1010  400  420 1      294.84km 18   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 2 400 Exp: cos 2  1 100

2  0 1  36.86

36.86

P S P1  400; P2  100

cos 1 

Qmotor  P1 tan 1   P1  P2  tan 2  400 tan36.86  500 tan   300 kW

Smotor  100  j300 cos m  0.3162

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The flux linkage    and current (i) relation for an electromagnetic system is  

42.

 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.

Key:

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 _________. 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 20A Exp:   ` max 30A

100V 

10A  200V 

45.

300V

KVA rating  300  200  6kVA



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 IL a smooth dc current. Switches are lossless. The real power fed back to the source. In kW is ________. Load

IL

T1 ~ Vin

L

T3 R  20

Bridge

` T4

T2



E  800V



Key: 6 Exp: V0 

2Vm 200 cos   2  cos120   200V  

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E  V0 800  200   30A R 20 Pfedback  V0 I0   200  30  6kW . I0 

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

46.

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

30

30 600V

Key:

24

Exp: Vph 

R ph 47.

30

2 2 Vdc   600  200 2 V 3 3



200 2 R 30 3Vph 2    10  PLoad   3 3 3 R ph 10



2

 24kW

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 ______. 10mH

Load

S

360V

1mF

 400 V



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 Vs  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  VS  and fundamental AC rms VSC voltage  VC1  , in degree, is________.

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5mH

 IS

220VAC ~

1mH

XS

 400V

IS

VS



VC1

I1X1

Key:

9.1 to 9.3

49.

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

 2  2  matrix A corresponding to distinct eigenvalues 1 and 2 respectively. Then the response x  t  of the system due to initial condition x (0) =  is and  are eigenvectors of (A) e1t  Key: Exp:

(B) e1t 

(C) e2 t 

(D) e2 t   e2 t 

(A) 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 e1t   e

2t

but we are asking w.r.t

Initial condition x  0    only so the response should be e1t 50.

A second-order real system has the following properties: (a) the damping ratio   0.5 and undamped natural frequency n  10 rad s,

Key: Exp:

(b) the steady state value of the output, to a unit step input, is 1.02. The transfer function of the system is 1.02 102 100 (A) 2 (B) 2 (C) 2 s  5s  100 s  10s  100 s  10s  100 (B)   n 2 The standard 2nd order T/F is  K 2 2   s  2n s  n 

(D)

102 s  5s  100 2

it is given that   0.5 & n 10 G s  K

100 s  10s  100 2

Now to satisfy the steady state O/P 1.02 y     t

s 0

G s 

 1 100  2  K  1.02  K  1.02 s  s  10s  100 

1.02  100 102  2 s  10s  100 s  10s  100 2

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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

Key:

(A)

Exp:

N1 : N2  110 :

(B) 10 3 :1

(C) 1:10

(D)

3 :10

11  10 3 :1 3

I1N1  I2 N2





Ii 10 3  I 2 .1

I1 1  I2 10 3

52.

The gain at the breakaway point of the root locus of a unity feedback system with open loop Ks transfer function G  s   is  s  1  s  4 

Key:

(A) 1 (A)

Exp:

G s

(B) 2

(C) 5

(D) 9

Ks  s  1  s  4 

To find Break away point

 1

dk  0 where We need to find the root of ds  s  1  s  4     s 2  5s  4  K   s s  

 4

d  d 2 s s  5s  4    s 2  5s  4   s   dk  ds  ds   ds  s2   

 S  2S  5  S2  5S  4   0  2S2  5S  S2  5S  4  0

 S2  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



KS KS 1  1 K 1  s  1 s  4 s2 1 2 

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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

0.4 0.3 Z2   0.15P.U  0.2 P.U ; 2 2 3Vprefault 3 If    6p.u ZO  Z1  Z2 0.15  0.2  0.15

Exp: Z0  0.15 P.U ; Z1 

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 20revolutions Exp: K  1200re v kwh   30  P  kW     hr  3600   P  2kW 54.

z

55.

Key: Exp:

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 ______. 9.55 Voltage =  B  Velocity  

B

1m 1m

Voltage 1   1m s B 11

i.e, 1m takes  2  21  2 m Takes 2r

 2r for 1 rotation  in 1 minute =

Velocity =

60 2

60  9.55 rpm 2

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