GATE 2015 EE Solutions - Session 1

GATE 2015 (EE) Solutions Session 1 (7th Feb)

Prepared by

Ankit Goyal AIR 1 GATE 2014 (EE) © 2015 Kreatryx. All Rights Reserved.

Paper Analysis by Ankit Goyal This paper had even distribution of Numerical Aptitude and Verbal Aptitude in the Aptitude Section. The questions except were one or two were of standard level. So it was expected that a student can secure 8-10 marks easily in this section. In 1 mark Technical Section 15 questions at-least were straight forward and rest of the questions were either lengthy or non-trivial. Basic understanding was required to attempt core questions as mostly the language had to be decoded and once that was done question was straight forward. The surprising thing to me was that there were a few straight forward theory questions and that too from the topics which most students neglect. In 2 marks Technical Section 20 questions were based on straight concepts but many might have committed errors due to exam pressure or not reading the question properly. Machines had mostly DC Machines which had questions that looked tough on the outlook but once you go through the question statement the solution was easy. The level of Power Systems was a bit difficult as questions were mostly thinking based and non-trivial. What I felt while solving the paper was that there were some questions that were out of my league also (cannot say about you guys) so those questions even I would have probably left. The paper was balanced among different subjects as you can see the pie-chart below so core subjects were not as dominant as they were expected to be. But definitely core had some of the most amazing questions that I have come across.

© 2015 Kreatryx. All Rights Reserved. 1 © 2015 Kreatryx. All Rights Reserved.

Section: General Aptitude Q1: Didn’t you buy _____________ when you went shopping? (A) any paper

(B) much paper

(C) no paper

(D) a few paper

Correct Answer (A) Solution: Didn’t you buy any paper when you went shopping. Q2: Based on the given statements, select the most appropriate option to solve the given question. If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building? Statements: (I) Each step is 3/ 4 foot high. (II)

Each step is 1 foot wide.

(A) Statements I alone is sufficient but statement II alone is not sufficient. (B) Statement II alone is sufficient, but statement I alone is not sufficient. (C) Both statements together are sufficient but neither statement alone is sufficient. (D) Statement I and II together are not sufficient. Correct Answer (A) Solution: Since two floors in a building are separated by vertical distance & thus the height of each stair is required to calculate no. of steps and width of steps do not matter. Q3: Which of the following options is the closest in meaning to the sentence below? She enjoyed herself immensely at the party. (A) She had a terrible time at the party (B) She had a horrible time at the party. (C) She had a terrific time at the party. (D) She had a terrifying time at the party Correct Answer (C) Solution: Since, she enjoyed immensely at the party so she had a terrific time at the party.

2 © 2015 Kreatryx. All Rights Reserved.

Q4: Which one of the following combinations is incorrect? (A) Acquiescence – Submission (B) Wheedle – Roundabout (C) Flippancy – Lightness (D) Profligate – Extravagant Correct Answer (B) Solution: Wheedle: use of endearments or flattery to persuade someone Roundabout: not direct Q5: Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals 16? (A) 0.20

(B) 0.25

(C) 0.30

(D) 0.33

Correct Answer (A) Solution: To make the sum equal to 16, the following combinations can be considered. 2 + 14 ; 3 + 13 ; 4 + 12 ; 5 + 11 P  1 4  15  P P P 2,14 3,13 4,12 5,11 P P P P P 16 2,14 3,13 4,12 5,11 1 1 1 1      1 5  0.20 20 20 20 20 Q6: The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each departments is 5 : 4. There are 40 males in Electrical Engineering. What is the difference between the numbers of female student in the Civil departments and the female students in the Mechanical departments?


Correct Answer 32 Solution: Males in electrical engineering = 40 Let total no. of males across all departments = x 20  x  40 100

x = 200 Total no. of females across all department

 4 5  200  160 Female in civil department =

30  160  48 100

Female in mechanical department =

10  160  16 100

Difference = 48 – 16 = 32 Q7: The number of students in a class who have answered correctly, wrongly or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking. Q. No Marks

Answered Answered

Not

Correctly

Wrongly

Attempted

1

2

21

17

6

2

3

15

27

2

3

1

11

29

4

4

2

23

18

3

5

5

31

12

1

What is the average of the marks obtained by the class in the examination? (A) 2.290

(B) 2.970

(C) 6.795

Correct Answer (C) Solution: Total marks obtained by the class   2  21  3  15  1  11  2  23  5  31  = 299 Total no. of students in class = 44 299 Average marks =  6.795 44


(D) 8.795

Q8: The probabilities that a student passes in Mathematics, physics and Chemistry are m, p, and c respectively. Of these subject, the student has 75% chance of passing in a least one, a 50 % chance of passing at least two a 40% chance of passing exactly two, following relations are drawn in m, p, c: (I) p + m + c = 27/20 (II) p + m + c = 13/20 (III) (p)x(m)x(c) = 1/20 (A) Only relation I is true. (B) Only relation II is true. (C) Relations II and III are true. (D) Relation I and III are true. Correct Answer (D) Solution: If p denotes probability of passing in physics then p denotes probability of failing. Similarity, m & c denote probability of failing in maths & chemistry respectively. Probability of passing in at least one P P P exactly one exactly two exactly three



 



 PCM  PCM  PCM  PCM  PCM  PCM  PCM  0.75

-------------(i)

Probability of passing in at least two

P P exactly two exactly three





 PCM  PCM  PCM  PCM  0.5

-------------(ii)

Probability of passing in exactly two





 PCM  PCM  PCM  0.4

-------------(iii)

From (ii) & (iii)

PCM  0.1  1 10


So relation (iii) is true Now P  C  M  PMC  PCM  PCM  2 PCM  PCM  PCM  3 PCM

-------------(iv)

[By venn diagram] From (i) & (ii) PMC  PCM  PCM  0.25

Substituting into (iv) P  C  M  0.25  2  0.4  3  0.1

 1.35 

27 20

Hence relation (i) is also true Q9: Select the alternative meaning of the underlined part of the sentence. The chain snatchers took to their heels when the police part arrived. (A) took shelter in think jungle (B) open indiscriminate fire (C) took to flight (D) unconditionally surrendered Correct Answer (C) Solution: Took to their heels means to run away and hence option (C) is most appropriate. Q10: The given statements is followed by some courses of action. Assuming the statements to be true, decide the correct option. Statement: There has been a significant drop in the water level in the lakes supplying water to the city. Courses of action: (I)

The water supply authority should impose a partial cut in supply to tackle the situation.

(II)

The government should appeal to all the residents thought mass media for minimal use of water.

(III)

The government should ban the water supply in lower area.


(A) Statements I and II follow. (B) Statements I and III follow. (C) Statements II and III follow. (D) All statements follow. Correct Answer (A) Solution: Statements I & II are most appropriate course of action for the problem at hand.

Section: Electrical Engineering Q.1 For the signal-flow graph shown in the figure. Which one of the following expressions is equal to the transfer function

Y s

?

X2  s  X

1  s  0

(A)

G1

1  G2 1  G1 

(B)

G2

1  G2 1  G2 

(C)

G1

1  G1 G2

Correct Answer (B) Solution: Forward path, P  G 1 2 Loops, L  G 1 1 L  G G 2 1 2 By Mason’s Gain Formulas P  G 1 1 1 2  Gain   1  G  G G  1 L L 1 2 1 2  1 G 2 = 1G 1G 1 2










(D)

G2

1  G1 G2

Q2: The primary mmf is least affected by the secondary terminal conditions in a (A) power transformer

(B) potential transformer

(C) current transformer

(D) distribution transformer

Correct Answer (D) Solution: Distribution transformer encounter varying load so it should be least affected by secondary terminal conditions. Q3: If a continuous function f(x) does not have a root in the interval [a, b], then which one of the following statements is TRUE ? (A) f(a). f(b) = 0

(B) f(a). f(b) < 0

(C) f(a). f(b) > 0

(D) f(a). f(b) ≤ 0

Correct Answer (C) Solution: If the sign of function is opposite at two points then there exist a root of function between those two points. f  a  f b   0 c   a,b  such that f(c) = 0

So, it there is no between a & b then sign of f(a) & f(b) should be same & hence f(a) f(b) > 0

Q4: If the sum of the diagonal elements of a 2 x 2 matrix is – 6, then the maximum possible value of determinant of the matrix is ____________. Correct Answer 9 Solution: Assuming 2 x 2 matrix as a b  A   c d Trace (A) = sum of diagonal elements = a + d = -6 Det (A) = ad – bc a+d=-6

=> d = (- 6 - a) 8 © 2015 Kreatryx. All Rights Reserved.

Det (A) = - a (a + 6) – bc For maximum det (A) d  det  A    0 da

= -2a – 6 = 0 a=-3 => d = - 3 Det (A) = 9 – bc For maximum value bc = 0 Det (A) = 9 Q5: A separately excited DC generator has an armature resistance of 0.1Ω and negligible armature inductance. At rated field current and rated rotor speed, its open –circuit voltage is 200 V. when this generator is operated at half the rated speed, with half the rated field current, an uncharged 1000 μF capacitor is suddenly connected across the armature terminals. Assume that the speed remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected will the voltages across it reach 25 V? (A) 62.25

(B) 69.3

(C) 73.25

Correct Answer (B) Solution: Induced emf in a dc generator NZ P A 60 E  N g 

 

E g2 2N2  E  N g1 1 1

Now since field current & speed are halved

 N 2  0.5 & 2  0.5  N 1 1 9 © 2015 Kreatryx. All Rights Reserved.

(D) 77.3

E g2  0.5  0.5  0.25 E g1 E  200  0.25  50V g2

For charging RC circuit

V  E  1  e c g

t RC   

t RC  25  50  1  e   

e

t RC

 0.5

t  RC ln 0.5  69.3

Q6: The voltages developed across the 3Ω and 2Ω resistors shown in the figure are 6V and 2V respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5 V voltage source? (A) 5

(B) 7

(C) 10

(D) 14

Correct Answer (A) Solution: Current through 3 resistor 

6  2A 3

Current through 2 resistor 

2  1A 2

Sum of currents into N should be zero 2

 I  1  2A 1  I  1A 1 Power delivered by 5V source = 5 x 1 = 5W 10 © 2015 Kreatryx. All Rights Reserved.

Q7: For the given circuit, the Thevenin equivalent is to be determined. The Thevenin Voltage VTh (in Volt), seen from terminal AB is ____________.

Correct Answer 3.36 Solution: Since the terminals A & B are extended beyond so it is already opened. By KVL in Loop – 1

2 i i 1

_________(i)

By KVL in Loop – 2





1  i  20i  2 i  i  0 1

23i  2i 1

__________(ii)

From (i) & (ii)

2i  2i  4 1 2i  23i  4 1 25 i = 4

i  0.16A, i  1.84A 1





V V  2 i  i  2  1.84  0.16  TH AB 1

= 3.36 V


Q8: A steady current I is flowing in the – x direction through each of two infinitely long wires at L y   as shown in the figure. The permeability of the medium is 0 . The B –field at (0, L, 0) is 2

(A) 

40 I 3L

Zˆ

(B) 

40 I 3L

Zˆ

(C) 0

(D) 

30 I 4L

Zˆ

Correct Answer (A) Solution: Since direction of current in both wires is in  xˆ direction. So magnetic field at y = L due to both wires will be in  zˆ direction as per Right hand thumb rule. By Ampere’s Circuital law

 B1d l  0 I

 

B  2 L 2   I 1 0 2 I B  0 1 2L B : radius  L  L 2  3 L 2 2 By Ampere’s Circuital law

 B2 .d l  0 I

 3L  B  2     I 2 0 2

2 I B  0 2 6L

2 I B  0,L,0   B  B  0 1 2  1 6   1 2 L  4 I B  0,L,0    0 zˆ 3L 12 © 2015 Kreatryx. All Rights Reserved.

Q9: The self-inductance of the primary winding of a single phase, 50 Hz, transformer is 800 mH, and that of the secondary winding is 600 mH. The mutual inductance between these two windings is 480 mH. The secondary winding of this transformer is short circuited and the primary winding is connected to a 50 Hz, single phase, sinusoidal voltage source. The current flowing in both the windings is less than respective rated currents. The resistance of both windings can be neglected. In this condition, what is the effective inductance (in mH) seen by the source? (A) 416

(B) 440

(C) 200

(D) 920

Correct Answer (A) Solution: L  L M a 1 = 800 – 480 = 320 mH

L  L M b 2 = 600 – 480 = 120 mH

L  M  480mH c L

eq



 L  L || L a b c

 320 



 480  120  600

 416mH

Q10: In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is _____________.


Correct Answer 1 Solution: This is an example of Buck – converter or step – Down chopper V  DV  0.4  20  8V 0 in V E 8 5 I  0   1A 0 R 3 Charging current = 1 A

Q11: When the Wheatstone bridge shown in the figure is used to find the value of resistor R X , the galvanometer G indicates zero current when R1  50 , R 2  65  and R3  100  . If R 3 is known with ± 5% tolerance on its normal value of 100 Ω , what is the range of R X in ohms ?

(A) [123.50, 136.50]

(B) [125.89, 134.12]

(C) [117.00, 143.00]

(D) [120.25, 139.75]

Correct Answer (A) Solution: For bridge balancing R R R  2 3 x R 1

R  100  5%  95  105 3 If R  95 3 R  x

65  95  123.5 50

R  105 3 R  x

65  105  136.5 50

R  123.5, 136.5 x


Q12: Consider a HVDC link which uses thyristor based line- commutated converter as shown in the figure. For a power flow of 750 MW from system 1 to system 3, the voltages at the two ends, and the current are given by : V1 =500 kV, V2 =485 kV and I = 1.5 kA. If the direction of power flow is to be reversed (that is from system 2 to system 1) without changing the electrical connections, then which one of the following combinations is feasible?

(A) V1  500 kV, V2  485 kV and I = 1.5 kA (B) V1  485 kV, V2  500 kV and I =1.5 kA (C) V1  500 kV, V2  485 kV and I = – 1.5kA (D)

V1  500 kV, V2  485 kV and I = – 1.5 kA

Correct Answer (B) Solution: For reversing the power flow (P = VI) either the direction of current at the polarity of voltage must be reversed but not both. As per the directions shown in figure.

V V 2 I 1 R In case of option (A), V  500kV, V  485kV 1 2 = I = - 1.5 kA So direction of current is also reversed & hence power flow is not reversed. In case of option (B), V  485kV, V  500kV 1 2

I

485   500  R

 1.5kA

So, direction of current is same but polarity of voltage is reversed so power flow is reversed.


Q13: A bode magnitude plot for the transfer function G(s) of a plant is shown in the figure. Which one of the following transfer functions best describes the plant?

(A)

1000  s  10  s  1000

(B)

10  s  10  s  s  1000 

(C)

s  1000 10s  s  10 

(D)

s  1000 10  s  10 

Correct Answer (D) Solution: Since, the bode plot has zero slope near w = 0 so there is no pole at origin. Between w = 10 & w = 1000 20   20   20 dB dec Slope   10  log    1000  So, a pole exists at w = 10. After w = 1k, slope again becomes zero So a zero exists at w = 1000 K  s  1000  Transfer function =  s  10  At s = 0 , Gain = 20 dB 1000 At s = 0 , TF = K   100k 10 Gain = 20 log (100k) = 20 dB 100 k = 10 K = 0.1  s  1000  TF  10  s  10 


Q14: In the 4 x 1 multiplexer, the output F is given by F = A  B. find the required input ' I3 I2 I1 I0 ' . (A) 1010 (B) 0110 (C) 1000 (D) 1110 Correct Answer (B) Solution: For Y  A  B

A B Y 0 0 0I

0

0 1 1I 1 1 0 1I 2 1 1 0I 3 I I I I   0110  0123

Q15: The impulse response g(t) of a system, G is as shown in figure (a). What is the maximum value attained by the impulse response of two cascaded block of G as shown in figure (b)?

(A)

2 3

(B)

3 4

(C)

4 5


(D) 1

Correct Answer (D) Solution: The impulse response of cascaded block ht  g t * g t 

*

=>

So maximum value of cascaded response = 1 Q16: Consider a function f 

1 r2

rˆ , where r is the distance from the origin and rˆ is the vector in

the redial direction. The divergence of this function over a sphere of radius R, which includes the origin, is (A) 0

(B) 2

(C) 4

(D) R

Correct Answer (A) Solution: Divergence in spherical co – ordinates. 1  2 1  1  .f  r fr  f sin   f  2 r sin  r r sin    r r

 

f 

1 rˆ r2

.f 

 fr 





1 , r2

f  0, 

 

f 0 

1  2 1  r  2   0 r  r2 r 

Q17: In the given circuit the silicon transistor has   75 and a collector VC =9 V. Then the ratio of RB and RC is _____________.


Correct Answer 105.133 Solution: I  I  75I c B B By KVL





15  I  I R  I R  0.7 C B C B B

14.3  75R I  I R CB B B



14.3  I 76R  R B C B





__________(i)



15  I  I R  V C B C C

 IC  IB RC  6

76I R  6 B C

__________(ii)

Divide (i) By (ii) 14.3 76RC  RB  6 76R C

R B  1.3833 76R C R B  105.133 R C

Q18: A random variable X has probability density function f(x) as given below: a  bx f x    0

for 0 < x < 1 otherwise

If the expected value E[X] = 2/3, then Pr [X < 0.5] is ______________. Correct Answer 0.25 Solution: Total Probability = 1

1 0 f  x  dx 

a  b2  1

1  bx2    a  bx  dx  ax  2   0 0

1

____________(i)


1

1 E  X    xf  x  dx   x  a  bx  dx  0 0 2 a  2  b3 3

1 3  x2  x a 2  b 3    0

____________(ii)

Solving (i) & (ii) a = 0, b = 2 f(x) = 2x

P  x  0.5 

0.5



0

2x dx   x2   

0.5 0

= 0.25

Q19: A moving average function is given by y  t   signal of frequency

1 t u    d . If the input u is a sinusoidal T t  T

1 Hz, then in steady state, the output y will lag u (in degree) by ___________. 2T

Correct Answer 90 Solution: Assume u  A sinwt 2 2  w   T T 2T  t  u  A sin   T 1 y T y



t



tT

t A T  t   t   A sin   dt     cos    T  T  T t  T

A t   t  cos      cos    T  T 

1A  t  2A  t  cos    sin   900    T T  Y lags u by 900


Q20:

Base load power plants are P : wind farms Q : run-of-river plants R : nuclear power plants S : diesel power plants

(A) P, Q and S only

(B) P, R and S only

(C) P, Q and R only

(D) Q and R only

Correct Answer (D) Solution: Wind farms and diesel power plants are peak load power plants. Run – of – river & nuclear power plants are base load power plants. Q21: Consider a one – turn rectangular loop of wire placed in a uniform magnetic field as shown in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the loop is 0.4Ω, and its inductance is negligible. The magnetic flux density (in Tesla) is a function of time, and is given by B(t) = 0.25 sin t ,where   2  50 radian/ second. The power absorbed (in Watt) by the loop from the magnetic field is ____________.

Correct Answer 0.1927 Solution: Emf induced in loop 

d dt

  BA  0.25sinwt  10  5 104  12.5sinwt  104 wb

e  12.5  104  w cos wt  12.5  104  10  cos wt  0.3927 cos wt


e2 .39272 Power absorbed  rms   0.1927W R 2  0.4

Q22: Consider the circuit shown in the figure. In this circuit R = 1 kΩ and C = 1 μF. The input voltage is sinusoidal with a frequency of 50 Hz, represented as a phasor with magnitude Vi and phase angle 0 radian as shown in the figure. The output voltages is represented as a phasor with magnitude V0 and phase angle  radian. What is the value of the output phase angle  (in radian) relative to the phase angle of the input voltage?

(C) 

(B) 

(A) 0

2

Correct Answer (D)  R     SCR V Solution: V  V  01 1 1  1  CS  R  SCR V V  1  RC   V 02  S  R  1  2 2 CS  

V V V 0 01 02  SCR  V  V   SCR V  1 2  i C  1F R  1k

 V  V  V 0  2 i   1 s  jw  j 100  

V  0.1V   900 0 i V  0.1V 0 i

,

  900



2


(D)   2

Q23: Of the four characteristics given below. Which are the major requirements for an instrumentation amplifier ? P . High common mode rejection ratio Q . High input impedance R . High linearity S. High output impendence (A) P, Q and R only (B) P and R only (C) P, Q and S only (D) Q, R and S only Correct Answer (A) Solution: Instrumentation amplifier should have low output impedance but rest of requirements all correct. Q24: A(0-50A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0 – 500A) is ____________. Correct Answer 0.22 0.1 R   0.222m sh 450 Solution: Q25: An inductor is connected in parallel with a capacitor as shown in the figure. As the frequency of current I is increased, the impedance (Z) of the network varies as

(A)

(B)

(C)

(D)


Correct Answer (B)

L  wc jwC  C Solution: Z  j(wL  1 ) j  w2LC  1 wC   w j c  jwL Z  w2LC  1 w2  1 LC jwL  1

 



Which w  1

When w 

1

At w 

LC

LC

1 LC

,

,



Z

Z





jwc

1

2 LC  w

jwc

w2  1 LC





(inductive)

(capacitive)

, impedance goes to infinity.

So, option (B) is correct.

Q26:

In a linear two-port network, when 10 V applied to port 1, current of 4 A flows through

port 2 when it is short- circuited. When 5 V is applied to port 1, a current of 1.25 A flows though a 1Ω resistance connected across port 2. When 3 V is applied to port 1, the current (in Ampere) through a 2 Ω resistance connected across Port 2 is ____________. Correct Answer 0.5454 Solution: Using Transmission Parameters V  AV  BI 1 2 2 When part 2 is short – circuited V 0 2 V  BI 1 2 10  B  4

B  2.5

When a resistance of 1 is connected at part – 2 V  1I  I 2 2 2 5   A  B  I   A  B  1.25 2 A+B=4 A = 1.5 24 © 2015 Kreatryx. All Rights Reserved.

When a resistance of 2 is connected V  2I 2 2 3  2AI  BI  2  2

3 3 3 I     0.5454A 2  2A  B  2  1.5  2.5 5.5

Q27:

The open loop poles of a third order unity feedback system are at 0, –1, –2,. Let the

frequency corresponding to the points where the root locus of the system transits to instable region be K. Now suppose we introduce a zero in the open loop transfer function at – 3, while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits to unstable region? (A) It corresponds to a frequency greater than K (B) It corresponds to a frequency less than K (C) It corresponds to a frequency K (D) Root locus of modified system never transits to unstable region Correct Answer (D) Solution: Open – loop transfer function K S  S  1  S  2  The root locus for this TF looks like as shown in first figure

Now a zero is added at s = 3, open loop transfer function becomes K  s  3  s  s  1  s  2  So, in second figure, root locus never enters right half plan which is the unstable region.


  3 0 2    The maximum value of “a” such that the matrix  1 1 0  has three linearly 0 a 2  

Q28:

independent real eigenvectors is

2

(A)

(B)

3 3

1 3 3

(C)

12 3 3 3

(D)

1 3 3 3

Correct Answer (B) Solution: Characteristics equation det  A  I   0 3

0

2

1

 1

0

0

a

2

0

 3  62  11  2a  6  0  2a    3  62  11  6   

_________(i)

To maximum ‘a’ we can maximum ‘2a’ d  2a 0 d 32  12  11  0

  2 

3 3

Substituting   2  a

3 in equation (i), we get 3

1 3 3

Q29:

A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating

steady state at synchronous speed, and producing 1 pu of real power. The initial value of the rotor angle  is 5º, when a bolted there phase to ground short circuit fault occurs terminal of the generator. Assuming the inputs mechanical power to remain at 1 pu, the value degree 0.02 second after the fault is ____________.


Correct Answer 5.9 Solution: Swing equation H d2   P P m e f dt2 In case of bolted three – phase fault P 0 e P  1pu m 2 2 d  1 50 dt2 t2      25  rad  0 2 t2     25  180  (deg) 0 2 2  0.02   5  25  180     2 





  5.9

Q30: The single-phase full-bridge voltage inverter (VSI), shown in figure, has an output frequency of 50 Hz, It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For Vin  100VDC , L = 9.55 mH, C = 63.66 F , R = 5Ω, the amplitude of the fundamental component in the output voltage V0 (in Volt) under steady-state is __________________.

Correct Answer 0.23 Solution: This is an example of multiple pulse modulation 50  103  500 No. of pulses  fc 2f  2  50  V  Pulse width, 2d   1  r  V   c V r  modulation index = 0.7 V c 2d  0.3






  2d d   0.3 0.15    N  1 N 501 500



 1.6792  103 

Output fourier series 





n  1,3,5

8V s sin n sin  nd  sin nwt   2   n  

Amplitude of fundamental 8V s sin  sin d  2  8  100  0.3   sin 1.6792  10 3   sin     22

 







= 0.3135V X  wL  100  9.55  10 3  3 L 1 1 X    50 C wc 100  63.66  10 6

R  1 jwc Z    2.989  0.179 j  eq R  1 jwc

 Z  eq  V V   0.23V 0 s  Z  jXL   eq 

Amplitude of fundamental output voltage = 0.23V

Q31: Find the transfer function

Y s X s

of the system given below.


G1

(A)

(C)

1  HG1



G2

(B)

1  HG2

G1  G2

1  HG1

(D)

1  H  G1  G2 

G1



G2

1  HG2

G1  G2

1  H  G1  G2 

Correct Answer (C) Solution: X S   X S   HY  s  1 X S   X S   HY  s  2 Y S   G X  s   G X  s  1 1 2 2  G  X  s   HY S    X  s  G  G  1 2  1 G G Y s 1 2  X s 1  H G  G 1 2



Q32:



The circuit shown in the figure has two sources connected in series. The instantaneous

voltage of the AC source (in Volt) is given by v(t) = 12 sin t. If the circuit is in steady state. The the rms value of the current (in Ampere) flowing in the circuit is ____________.

Correct Answer 10 Solution: Solving with DC source In steady state, inductor is short – circuited I  8A 0 Solving with AC source w=1 X  wL  1 L V(t) = 12 sin t


i t  



 wL  V   tan1   Z  R  12

I2  12

12 2



sin t  450



sin t  450





By superposition theorem

iI i DC ac 8

12



2 sin t  450



I2 2 RMS value of i  I  m 0 2 2 1  12   82    64  36  10A 2  2 

Q33: In the signal flow diagram given in the figure u1 and u2 are possible whereas y1 and y 2 are possible outputs. When would the SISO system derived from this diagram be controllable and observable? (A) when u1 is the only input and y1 is the only output. (B) when u2 is the only input and y1 is the only output. (C) when u1 is the only input and y 2 is the only output. (D) when u2 is the only input and y 2 is the only output. Correct Answer (B) Solution: From the block diagram X1  1 u  5x  2x  s  1 1 2 

sX  5x  2x  u 1 1 2 1  x '  5x  2x  u 1 1 2 1

________(i) 30 © 2015 Kreatryx. All Rights Reserved.

x  1 u  u  2x  x  s  1 2 2 1 2 

sx  2x  x  u  u 2 1 2 1 2

x '  2x  x  u  u 2 1 2 1 2

___________(ii)

 x  5 2  x  1 0  1  1         u1    u2  x  2 1   x  1  1  2  2

y x 1 1 y  x x 2 1 2 x  x  1   y  1 0  ; y  1  1  1  1 x  2 x   2  2

Case – 1 u is input, y is output 1 1 5 2 1 A B    C  1 0  2 1  1 5 2 1 3 AB      2 1  1 3 1 3 Q  B AB    C 1 3

 c  0

det Q

(not controllable)

Case – 2 u is input, y is output 2 1 5 2 0  A B    C  1 0  2 1  1 

 2 0 2 AB    ; Q  B AB     c 1 1 1 

 c  2

det Q

(controllable)

5 2 CA  1 0    5  2 2 1 

 C  1 0  Q   det Q  2 (Observable) 0 0 CA  5 2

 


Q34: A separately excited DC motor run at 1000 rpm on no load when its armature terminals are connected to a 200V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1Ω. The no-load armature current is negligible. With the motor developing its full load torque. The armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational losses, the full armature current (in Ampere) is ________________. Correct Answer 8 Solution: Under no load, I  0 a V  E t b E  200V at N = 1000 rpm b Since motor is separately excited, flux can be assumed constant. E  N b E N b2  2  500 E N 1000 b1 1 E  100V b2 For rated torque, let armature current = I a Since  = constant TI a I At half rated torque, armature current = a 2 At N = 520 rpm E N b3  3 E N b1 1 E  104V b3 I V E I R E  aR t b2 a a b3 2 a I 100  I R  104  a R a a 2 a I 4 aR 2 a I  8 R  8A a a Full load armature current = 8A 32 © 2015 Kreatryx. All Rights Reserved.

Q35: A DC motor has the following specifications: 10 hp, 37.5A, 230 V; flux/pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature resistance = 0.267 Ω. The armature reaction is negligible and rotational losses are 600 W. the motor operates from a 230 V DC supply. If the motor runs at 1000 rpm, the output torque produced (in Nm) is ____________. Correct Answer 57.789 Solution:   0.01W b N = 1000 rpm Z = 666 A=2 P=4

NZ  P  Induced emf E    b 60  A 

E  b

0.01  1000  666  4    60 2

6660  2  222V 60 V E b  230  222  29.9625A I  t a R 0.267 a 

Developed Power = E I  6651.685W ba Shaft power = Developed power – Rotational loss = 6651.685 – 600 = 6051.685W

P 6051.685 T  sh   57.789N  m sh W 2  1000 60

Q36: Determine the correctness or otherwise of the followings Assertion [a] and Reason [r] Assertion: Fast decoupled load flow method. Gives approximate load flow solution because it uses several assumptions. Reason: Accuracy depends on the power mismatch vector tolerance.


(A) Both [a] and [r] are true [r] is the correct reason for [a] (B) Both [a] and [r] are true [r] is not the correct reason for [a]. (C) Both [a] and [r] are false. (D) [a] is false and [r] is true. Correct Answer (B) Solution: FDLF method assume real power primarily depends on voltage angles and reactive power depends on voltage magnitudes and due to these assumptions the result is approximate. Also, lower the tolerance of power mismatch vector higher is the accuracy. So both statements are correct but reason does not explain assertion. Q37: The figure shown a digital circuit constructed using negative edge triggered J-K flips. Assume a starting state of Q2 Q1 Q0  000 . This state Q2 Q1 Q0  000 will repeat after ________ number of cycles of the clock CLK.

Correct Answer 6 Solution: Q Q Q 2 1 0 0 0 0 CLK1 0 0 1 CLK2 0 1 0 CLK3 0 1 1 CLK4 1 0 0 CLK5 1 0 1 CLK6 0 0 0 So, original state is reached after 6 clock cycles.


Q38: Consider a discrete time signal given by n

n

x n   0.25 un  0.5 u n  1

The region of convergence of its Z-transform would be (A) The region inside the circles of radius 0.5 an centered at origin (B) The region outside the circle of radius 0.25 and centered at origin (C) The annular region between the two circles, both centered at origin and having radii 0.25 and 0.5 (D) The entire Z plane. Correct Answer (C)

Solution: x n   0.25 n u n   0.5 n u  n  1

 x n  x n 1 2 x n   0.25 n u n 1 ROC  Z  0.25 1 x n   0.5 n u  n  1 2 ROC  Z  0.5 2

So, ROC of x n  ROC  ROC 1 2 Hence ROC is annular region between two circles both centered at origin and hawing radius 0.25 & 0.5 Q39: A solution of the ordinary differential equation and y 1   

1  3e 3

e

. The value of

d2 y dt

2

5

dy  6y  0 is such that y(0) = 2 dt

dy  0  is _____________. dt

Correct Answer -3 d2 y 5dy   6y  0 Solution: dt dt2 By method of differential operator D2  5D  6 y  0





D  3 D  2  y  0 35 © 2015 Kreatryx. All Rights Reserved.

Characteristics roots = -3, -2 y  C e3t  C e2t 1 2

y 0  C  C  2 1 2 y 1   C e3  C e2 1 2 C  C e 3e  1 2   1 3 e e3 C  3 & C  1 2 1  3t y  t   e  3e2t dy  t 

_______(i)

 3e3t  6e2t

dt dy  0   3  6  3 dt dy  0   3 dt

Q40: A parallel plate capacitor is partially filled with glass of dielectric constant 4.0 as shown below. The dielectric strengths of air and glass are 30 kV/cm and 300 kV/cm respectively. The maximum voltage (in kilovolts), which can be applied across the capacitor without any breakdown is _____.

Correct Answer 18.75





A Solution: Capacitance of air capacitor =   1  C 0 1 5mm A Capacitance of glass capacitor =   4   4C 0 1 5mm If air reaches dielectric strength 30 KV/cm Voltage across air capacitor = 30 x 0.5 = 15 kV Voltage across glass capacitor = 4C  V  C  V 1 g 1 a


V  15  3.75KV 4 g Total voltage = Va + Vg = 18.75 KV If glass reaches dielectric strength of 300 KV/cm, the air will exceed dielectric strength & dielectric breakdown will occur.

Q41: Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is (A) 5/ 11

(B) 1/2

(C) 7/13

(D) 6/11

Correct Answer (D) Solution: A can only win on an odd number of trial P(A wins on first trial) = 1 6 P(A wins on third trial) = 5  5  1 6 6 6 2  56  16

 

Because A must lose on first trial & B must lose on second trial. 4 1 P(A wins on 5th trial)  5 6  6

 

4 1 5 2 1  6  6  5 6  1 6  ............. 6  1 36 1 1     6  1  25  6 11 36  

P(A wins) 

 6 11

 

 

[Since this is an infinite GP]

Q42: The circuit shown is meant to supply a resistive load R L from separate DC voltage sources. The switches S1 and S2 are controlled so that only of them is ON at any instant. S1 is turned on for 0.2 ms and S2 is turned on for 0.3 ms in a 0.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the outputs voltage V0 (in Volt) across R L is ________________. 37 © 2015 Kreatryx. All Rights Reserved.

Correct Answer 7 Solution: When S1 is ON, V  10V 0 When S2 is ON, V  5V 0 Output voltage wave form looks like 10  0.2  5  0.5 V  0  avg 0.5



10  2  5  3  7V 5

Q43: The signum function is given by

x  ;x  0 sgn  x    x  0; x  0  The Fourier series expansion of sgn(cos(t)) has (A) Only sine terms with harmonics. (B) Only cosine terms with all harmonics. (C) Only sine terms terms with even numbered harmonics. (D) Only cosine terms with odd numbered harmonics.


Correct Answer (D) Solution: The wave form of sgn (cos (t)) will look like

Since this function is even, so it contains only cosine terms in fourier series. The function is also half wave symmetric so only has odd harmonics in fourier series.

Q44: Two single-phase transformers T1 and T2 each rated at 500 kVA are operated in parallel. Percentage impedances of T1 and T2 are (1 + j6) and (0.8 + j4.8), respectively. To share a load of 1000 kVA at 0.8 lagging power factor, the contributions of T2 (in KVA) is ______________. Correct Answer 555.55  Z  1  Solution: S  S  2 LZ Z  2  1 0 S  100036.86 KVA L

*

 1  j6  S  100036.86   2  1.8  j10.8 

*

 555.8536.860 KVA

So contribution of T is 555.55 KVA at a power factor of 0.8 lag. 2

Q45: In the given circuit the parameter k is positive, and the power dissipated in the 2 Ω resistor is 12.5W. The value of k is _______________. 39 © 2015 Kreatryx. All Rights Reserved.

Correct Answer 0.5

V Solution: Current through 2 resistor  0 2 V By KCL 0 2  KV  5 ______________(i) 0 V2 P  12.5W  0 2 2 V2  25 0 V  5V 0

From (i) 2.5  KV  5 0 K  0.5

Q46: A self-commutating switch SW, operated at duty cycle  is used to control the load voltage as shown in the figure. Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are

(A) VL  0 and VC 

1 V 1   dc

(B) VL 

 1 Vdc and VC  V 2 1   dc

(C) VL  0 and VC 

 V 1   dc

(D) VL 

  Vdc and VC  V 2 1   dc


Correct Answer (A) Solution: This is an example of boost converter or step – up chopper V V  i  V 0 1  c By volt – sec balance, average voltage across inductor V 0 L Q47: The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at (2 – j3). List all the poles and zeroes. (A) Poles at (2  j3), no zeroes. (B) Poles at (  2 – j3), one zero at origin. (C) Poles at (2 – j3), ( –2 + j3), zeroes at (–2 – j3), (2 + j3). (D) Poles at (2  j3), zeroes at (–2  j3) Correct Answer (D) Solution: Since the magnitude response is flat, it is an all – pass filter so poles & zeros are symmetric about y – axis Complex poles and zeroes occur in conjugate pairs, so poles at  2  j3 Zeroes at  2  j3

Q48:

f(A, B, C, D) = πM (0, 1, 3, 4, 5, 7, 9, 11, 12, 13, 14, 15,) is a maxterm representation of a

Boolean function f(A, B, C, D) where A is the MSB and D is the LSB. The equivalent minimized representation of this function is





(A) A  C  D A  B  D



(B) ACD  ABD (C) ACD  ABCD  ABCD (D)

B  C  D  A  B  C  D  A  B  C  D 


Correct Answer (C) Solution: f  A,B, C,D   M  0,1,3, 4,5, 7, 9,11,12,13,14,15 

m2,6,8,10

f  ACD  ABD

f  ACD  ABCD  ABCD f  A,B, C,D   M  0,1,3, 4,5, 7, 9,11,12,13,14,15 

f  D  A  B  A  C So, (C) is correct Q49: Consider the economic dispatch problem for power plant having two generating units. The fuel costs in Rs./Mwh along with the generation limits for the two unity are given below. C1 P1   0.01P12  30P1  10 ; 100 MW  P1  150MW C2 P2   0.05P22  10P2  10 ; 100 MW  P2  180MW

The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ___________. 42 © 2015 Kreatryx. All Rights Reserved.

Correct Answer 20 Solution: Allocating minimum power to both units P  100MW 1 P  100MW 2 P  P  200MW 1 2 dC 1C  1  0.02P  30 1 dP 1 1 dC 2  0.01P  10 1C  2 dP 2 2 At P  100MW, 1C  32Rs / MWhr 1 1









At P  100MW, 1C  20Rs / MWhr 2 2 So, incremental cost of plant = 20 Rs/MWh as when additional load is applied, it will be taken by plant – 2.

Q50: The op-amp shown in the figure has a finite gain A = 1000 and an infinite input resistance. A step-voltage Vi =1 mV is applied at the time t = 0 as shown. Assuming that the operational amplifier is not saturated the time constant (in millisecond) of the output voltage V0 is

(A) 1001

(B) 101

(C) 11


(D) 1

Correct Answer (A) Solution: V  0 

  V  A 0  V  0  V  A V V 0  

V 0 V   1000 I  R

 Vs  V    Vs  V0 1000  R

1000

 d d V I C V  V  C  0  V  C 0 0 dt  dt  1000  









106 d  1001V   0  3 dt 10

  1.001  106

dV

 dt0

Due to infinite input resistance I I R C

dV 103 V  106 V  1.001  106 0 s 0 dt V  1mV s dv 106  106 V  1.001 0  106 0 dt dV 1  V  1.001 0  0 0 dt dV 1.001 0  V  1 0 dt Time constant   1.001sec   1001msec


Q51:

A 200/400V, 50 Hz, two-winding transformer is rated at 20 kVA. Its windings are

connected as an auto-transformer of rating 200/600 V. A resistive load of 12 Ω is connected to the high voltage (600V) side of the auto-transformer. The value of equivalent load resistance (in ohm) as seen from low voltage side is _____________. Correct Answer 1.33

600  50A 12 Since there is no losses so input power = output power 200  I  600  50 p

Solution: Current in load resistance 

I  150A p

Z

V 200  in   1.33 in I 150 m

Q52: A 3 – phase 50 Hz square wave (6-step) VSI feeds a 3 – phase, 4 pole induction motor. The VSI line voltage has a dominant 5th harmonic component. If the operating slip of the motor with respect to fundamental component voltages is 0.04, the slip of the motor with respect to 5th harmonic component of voltage is __________________.

Correct Answer 5.8 N N r  0.04 Solution: slip  s N s N 1  r  0.04 N s N  0.96N r s N Speed of 5th harmonic = s (opposite drm to fundamental) 5 N s N 5 r  0.2  0.96  5.8 slip  N 0.2 s 5


Q53: An 8-bit, unipolar successive Approximation Register type ADC is used to convert 3.5 V to digital equivalent output. The reference voltages is +5 V. the output of the ADC. At the end 3rd clock pulse after start of conversion, is (A) 1010 0000 Correct Answer (A) Solution: Resolution =

(B) 1000 0000

(C) 0000 0001

(D) 0000 0011

5 5   0.0196 28  1 255

Output after 1st clock pulse = 10000000 = 128 (decimal)  128  5 255  2.5098V V  V , so next bit will be set in next clock cycle 0 a Output after 2nd clock pulse = 11000000 = 192 (decimal)  192  5 255  3.7647V V  V , so this bit is reset and next bit will be set 0 a Output after 3rd clock pulse = 10100000

Q54: An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the magnitude of V0 be maximum? (A)

1  x 

(B) (1 + x) (C) 1 / (D)

1  x 

1  x 

Correct Answer (A) Solution: V  V  V 0 Q S

 R 1  x    R  E  E   R 1  x   pR   R  pR   R 1  xR  R   E    R  pR  xR R  pR 


 1x 1  V  E   0 1  x  p 1  p 

For maximum V 0

dV 0 0 dp

  dV 1  0  E   1  x   0  2 2 dp  1  x  p  1  p  

1  x 1  p 2  1  x  p 2 2

1  x 1  p2   x  1  p 2 x 1  p   x2  2x 1  p  2 x 1  p   2 1  p   x  0

 p2  2p  1  2  2p  x  0     p2  1  x p  1x

Q55: A sustained three-phase fault occurs in the power system shown in the figure. The current and Voltage Phasors during the fault (on common reference), after the natural transients have died down are also shown. Where is the fault located?

(A) Location P

(B) Location Q

(C) Location R


(D) Location S

Correct Answer (B) Solution: Fault current must always lag voltage due to inductive reactance of the system & if any currents lead the voltage, that is only possible if direction of current is assumed opposite to assumed direction. Since I & I are equal & opposite phasors as consistant with circuit so, there can be no fault 2 4 between I & I . 2 4 But I & I are not equal & opposite, so fault lies between I & I but since I is more than I 1 1 3 3 1 3 so fault is closer to V than V 1 2 Hence, fault is located at Q.


GATE 2015 EE Solutions - Session 1

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