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.

1 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.


Section: General Aptitude

Q1: Didnt you buy _____________ when you went shopping?

(A) any paper (B) much paper (C) no paper (D) a few paper

Correct Answer (A)

Solution: Didnt 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.


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

1 1P P P P4 52,14 3,13 4,12 5,11

P P P P P16 2,14 3,13 4,12 5,11

1 1 1 1 1 0.20

520 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

20x 40

100

x = 200

Total no. of females across all department

4 200 1605

Female in civil department = 30

160 48100

Female in mechanical department = 10

160 16100

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.

Answered Answered NotQ. No Marks

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 (D) 8.795

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

Average marks = 299

6.79544


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 Pexactly one exactly two exactly three

PCM PCM PCM PCM PCM PCM PCM 0.75 -------------(i)

Probability of passing in at least two

P Pexactly 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)

1PCM 0.110


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

271.35

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

12 X s 0

Y s?

X s

(A) 21

1

G

1 G 1 G (B)

22

2

G

1 G 1 G (C)

1 2

1G

1 G G (D)

1 2

2G

1 G G

Correct Answer (B)

Solution: Forward path, P G1 2

Loops, L G1 1

L G G2 1 2

By Masons Gain Formulas

Gain

P G 11 1 2

1 G G G1 L L1 1 21 2

=

G

2

1 G 1 G1 2


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 bA

c d

Trace (A) = sum of diagonal elements = a + d = -6

Det (A) = ad bc

a + d = - 6 => d = (- 6 - a)


Det (A) = - a (a + 6) bc

For maximum det (A)

d

det A 0da

= -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 0.1 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 F 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 (D) 77.3

Correct Answer (B)

Solution: Induced emf in a dc generator

NZ P A60

E Ng

E Ng2 2 2

E Ng1 1 1

Now since field current & speed are halved

2 0.5

1

&

N2 0.5

N1


Eg2

0.5 0.5 0.25Eg1

E 200 0.25 50Vg2

For charging RC circuit

t RC

V E 1 ec g

t RC

25 50 1 e

t RC

e 0.5

t RC ln 0.5 69.3

Q6: The voltages developed across the 3 and 2 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

62A

3

Current through 2 resistor

2

1A2

Sum of currents into N2

should be zero

I 1 2A1

I 1A1

Power delivered by 5V source = 5 x 1 = 5W


Q7: For the given circuit, the Thevenin equivalent is to be determined. The Thevenin Voltage

ThV (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 i1

_________(i)

By KVL in Loop 2

1 i 20i 2 i i 01

23i 2i1

__________(ii)

From (i) & (ii)

2i 2i 41

2i 23i 41

25 i = 4

i 0.16A, i 1.84A1

V V 2 i i 2 1.84 0.16TH AB 1

= 3.36 V


Q8: A steady current I is flowing in the x direction through each of two infinitely long wires at

Ly

2 as shown in the figure. The permeability of the medium is

0 . The B field at (0, L, 0) is

(A) 04 I

Z3 L

(B) 0

4 IZ

3 L

(C) 0 (D) 0

3 IZ

4 L

Correct Answer (A)

Solution: Since direction of current in both wires is in x direction.

So magnetic field at y = L due to both wires will be in z direction as per Right hand thumb rule.

By Amperes Circuital law

B d l I1 0

LB 2 I21 0 2 I

0B1 2 L

L LB : radius L 32 22

By Amperes Circuital law

B .d l I2 0

3LB 2 I

2 02

2 I0B

2 6 L

2 I

0 1 1B 0,L,0 B B2 61 2 L

4 I

0 B 0,L,0 z3 L


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 Ma 1

= 800 480 = 320 mH

L L Mb 2

= 600 480 = 120 mH

L M 480mHc

L L L || Leq a b c

480 120320 416mH

600

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 8V0 in

V E 8 50I 1A0 R 3

Charging current = 1 A

Q11: When the Wheatstone bridge shown in the figure is used to find the value of resistor XR ,

the galvanometer G indicates zero current when 1 2

R 50 , R 65 and 3

R 100 . If 3

R is

known with 5% tolerance on its normal value of 100 , what is the range of XR 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 R2 3R

x R1

R 100 5% 95 1053

If R 953

65 95R 123.5

x 50

R 1053

65 105R 136.5

x 50

R 123.5, 136.5x


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 : 1

V =500 kV, 2

V =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) 1 2

V 500 kV, V 485 kV and I = 1.5 kA

(B) 1 2

V 485 kV, V 500 kV and I =1.5 kA

(C) 1 2

V 500 kV, V 485 kV and I = 1.5kA

(D) 1 2

V 500 kV, V 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 V1 2I

R

In case of option (A), V 500kV,V 485kV1 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 500kV1 2

485 500I 1.5kA

R

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

Slope 20 20

20dB dec10

log1000

So, a pole exists at w = 10.

After w = 1k, slope again becomes zero

So a zero exists at w = 1000

Transfer function =

K s 1000

s 10

At s = 0 , Gain = 20 dB

At s = 0 , TF = 1000

K 100k10

Gain = 20 log (100k) = 20 dB

100 k = 10

K = 0.1

s 1000TF

10 s 10


Q14: In the 4 x 1 multiplexer, the output F is given by F = A B. find the required input 3 2 1 0' I I I I '

.

(A) 1010

(B) 0110

(C) 1000

(D) 1110

Correct Answer (B)

Solution: For Y A B

A B Y

0 0 0 I0

0 1 1 I1

1 0 1 I2

1 1 0 I3

I I I I 01100 1 2 3

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 function2

1f r

r , where r is the distance from the origin and r 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 1 12.f r fr f sin f2 r r sin r r sinr

1f r

2r

r

1f , f 0, f 0

2r

1 2 1.f r 022 r rr

Q17: In the given circuit the silicon transistor has 75 and a collector C

V =9 V. Then the ratio

of BR and CR is _____________.


Correct Answer 105.133

Solution: I I 75Ic B B

By KVL

15 I I R I R 0.7B BC C B 14.3 75R I I R

CB B B

14.3 I 76R RB C B __________(i)

15 I I R VC B C C

I I R 6C B C 76I R 6

B C __________(ii)

Divide (i) By (ii)

76R R14.3 C B

6 76RC

RB 1.3833

76RC

RB 105.133

RC

Q18: A random variable X has probability density function f(x) as given below:

a bx for 0 < x < 1f x

0 otherwise

If the expected value E[X] = 2/3, then Pr [X < 0.5] is ______________.

Correct Answer 0.25

Solution: Total Probability = 1

11 21 bxf x dx a bx dx ax

0 20 0

ba 12

____________(i)


11 12 3x xE X xf x dx x a bx dx a b

2 30 0 0

2 a b2 33

____________(ii)

Solving (i) & (ii)

a = 0, b = 2

f(x) = 2x

0.50.52P x 0.5 2x dx x

0 0

= 0.25

Q19: A moving average function is given by t

t T

1y t u d

T . If the input u is a sinusoidal

signal of frequency 1

Hz,2T

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

Correct Answer 90

Solution: Assume u Asinwt

2 2w

TT 2T

tu Asin

T

t t1 t A T t

y Asin dt cosT T T T t Tt T

A t ty cos cos

T T

1A t 2A t 0cos sin 90T T

Y lags u by 090


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 0.4, 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 ____________.


Solution: Emf induced in loop d

dt

4BA 0.25sinwt 10 5 10

412.5sinwt 10 wb

4e 12.5 10 wcoswt

412.5 10 10 coswt

0.3927coswt


Power absorbed

2 2e .3927rms 0.1927WR 2 0.4

Q22: Consider the circuit shown in the figure. In this circuit R = 1 k and C = 1 F. The input

voltage is sinusoidal with a frequency of 50 Hz, represented as a phasor with magnitude i

V and

phase angle 0 radian as shown in the figure. The output voltages is represented as a phasor with

magnitude 0

V and phase angle radian. What is the value of the output phase angle (in

radian) relative to the phase angle of the input voltage?

(A) 0 (B) (C) 2

(D) 2

Correct Answer (D)

Solution: R

V V01 1 1

CS

SCR V1

RV 1 RC V02 S 21R

CS

SCR V2

V V V0 01 02

SCR V V SCR V1 2 i

V V V 01 2 i

C 1 F R 1k s jw j 100

0V 0.1 V 900 i

0V 0.1 V , 900 i

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

Solution:

0.1R 0.222m

sh 450

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)

Solution:

L1jwL wcjwC CZ

1 2j(wL ) j w LC 1wC

wjjwL cZ

2 2 1w LC 1 wLC

Which

wj

c1w , ZLC 21 w

LC

(inductive)

When

wj1 cw , Z

2LC 1wLC

(capacitive)

At 1

wLC

, 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 1 resistance connected across port 2. When 3 V is applied to port 1, the current (in Ampere)

through a 2 resistance connected across Port 2 is ____________.


Solution: Using Transmission Parameters

V AV BI1 2 2

When part 2 is short circuited

V 02

V BI1 2

10 B 4

B 2.5

When a resistance of 1 is connected at part 2

V 1 I I2 2 2

5 A B I A B 1.252

A + B = 4

A = 1.5


When a resistance of 2 is connected

V 2I2 2

3 2AI BI2 2

3 3 3I 0.5454A2 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.


Q28: The maximum value of a such that the matrix

3 0 2

1 1 0

0 a 2

has three linearly

independent real eigenvectors is

(A) 2

3 3 (B)

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

0 a 2

3 26 11 2a 6 0

3 22a 6 11 6

_________(i)

To maximum a we can maximum 2a

d 2a0

d

23 12 11 0

32

3

Substituting 3

23

in equation (i), we get

1a

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 5, 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

2H d

P Pm e2f dt

In case of bolted three phase fault

P 0e

P 1pum

22 d1

250 dt

2t

25 rad0 2

2t25 180

0 2 (deg)

20.025 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 in

V 100VDC , L = 9.55 mH, C = 63.66 F , R = 5, the

amplitude of the fundamental component in the output voltage 0

V (in Volt) under steady-state

is __________________.

Correct Answer 0.23

Solution: This is an example of multiple pulse modulation

No. of pulses 350 10fc 500

2f 2 50

Pulse width, Vr2d 1

Vc

Vr

Vc

modulation index = 0.7

2d 0.3


2d d 0.3 0.15

N 501 500N 1

31.6792 10

Output fourier series 8V nds sin n sin sin nwtn 2

n 1,3,5

Amplitude of fundamental

8V

s dsin sin2

8 100 0.33sin 1.6792 10 sin2 2

= 0.3135V

3X wL 100 9.55 10 3L

1 1X 50

C 6wc 100 63.66 10

1R

jwcZ 2.989 0.179 j

eq 1Rjwc

Zeq

V V 0.23V0 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.


(A) 1 2

1 2

G G

1 HG 1 HG

(B) 1 2

1 2

G G

1 HG 1 HG

(C) 1 2

1 2

G G

1 H G G

(D)

1 2

1 2

G G

1 H G G

Correct Answer (C)

Solution: X S X S HY s1

X S X S HY s2

Y S G X s G X s1 1 2 2

G X s HY S X s G G1 1 2

G GY s 1 2

X s 1 H G G1 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 8A0

Solving with AC source

w = 1

X wL 1L

V(t) = 12 sin t


V wL1i t tan

RZ

12 0sin t 452 2I 1

12 0sin t 452

By superposition theorem

aci I i

DC

12

802 sin t 45

RMS value of i

2I2 mI0 2

2

1 1228 64 36 10A2 2

Q33: In the signal flow diagram given in the figure 1

u and 2

u are possible whereas 1

y and 2

y

are possible outputs. When would the SISO system derived from this diagram be controllable

and observable?

(A) when 1

u is the only input and 1

y is the only output.

(B) when 2



(C) when 1



(D) when 2



Correct Answer (B)

Solution: From the block diagram

1X1 u 5x 2xs 1 1 2

sX 5x 2x u1 1 2 1

x ' 5x 2x u1 1 2 1

________(i)


1x u u 2x xs2 1 2 1 2

sx 2x x u u2 1 2 1 2

x ' 2x x u u2 1 2 1 2

___________(ii)

x x5 2 1 01 1 u u1 2x 2 1 x 1 1

2 2

y x1 1

y x x2 1 2

x x1 1y 1 0 ; y 1 1

1 2x x2 2

Case 1 u1

is input, y1

is output

5 2 1

A B C 1 02 1 1

5 2 1 3AB

2 1 1 3

1 3

Q B ABC 1 3

det Q 0c (not controllable)

Case 2 u2

is input, y1

is output

5 2 0

A B C 1 02 1 1

2 0 2AB ; Q B AB

c1 1 1

det Q 2c (controllable)

5 2

CA 1 0 5 22 1

C 1 0Q

0 CA 5 2

det Q 20 (Observable)


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 1. 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 0a

V Et b

E 200Vb at N = 1000 rpm

Since motor is separately excited, flux can be assumed constant.

E Nb

E N 500b2 2

E N 1000b1 1

E 100Vb2

For rated torque, let armature current = Ia

Since = constant

T Ia

At half rated torque, armature current = Ia

2

At N = 520 rpm

E Nb3 3

E Nb1 1

E 104Vb3

IaV E I R E R

t b2 a a b3 a2

Ia100 I R 104 R

a a a2

Ia4 R

a2

8I 8ARa

a

Full load armature current = 8A


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 . 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 ____________.


Solution: 0.01Wb

N = 1000 rpm

Z = 666

A = 2

P = 4

Induced emf NZ P

Eb 60 A

0.01 1000 666 4Eb 60 2

6660

2 222V60

V E 230 222t bI 29.9625Aa R 0.267

a

Developed Power = E I 6651.685Wb a

Shaft power = Developed power Rotational loss

= 6651.685 600

= 6051.685W

P 6051.685shT 57.789N msh 2W

100060

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 2 1 0

Q Q Q 000 . This state 2 1 0

Q Q Q 000 will repeat after ________

number of cycles of the clock CLK.

Correct Answer 6

Solution:

Q Q Q2 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: n nx n 0.25 u n 0.5 u n 1

x n x n1 2

nx n 0.25 u n1

ROC Z 0.251

nx n 0.5 u n 12

ROC Z 0.52

So, ROC of x n ROC ROC1 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 2

2

d y dy5 6y 0

dtdt is such that y(0) = 2

and 3

1 3ey 1

e

. The value of

dy0

dt is _____________.

Correct Answer -3

Solution: 2d y 5dy

6y 02 dtdt

By method of differential operator

2D 5D 6 y 0 D 3 D 2 y 0


Characteristics roots = -3, -2

3t 2ty C e C e1 2

y 0 C C 21 2

_______(i)

3 2y 1 C e C e1 2

C C e 3e 11 23 3e e

C 3 & C 12 1

3t 2ty t e 3e

dy t 3t 2t3e 6edt

dy

0 3 6 3dt

dy

0 3dt

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 _____.


Solution: Capacitance of air capacitor = A1 C0 15mm

Capacitance of glass capacitor = A

4 4C0 15mm

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 V1 g 1 a


15V 3.75KV4g

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) = 16

P(A wins on third trial) = 5 5 16 6 6

25 1

6 6

Because A must lose on first trial & B must lose on second trial.

P(A wins on 5th trial) 45 1

6 6

P(A wins) 2 41 5 51 1 .............

6 6 6 66

1 1 1 36

256 6 11136

611

[Since this is an infinite GP]

Q42: The circuit shown is meant to supply a resistive load LR 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 0

V (in Volt) across LR is ________________.


Correct Answer 7

Solution: When S1 is ON, V 10V0

When S2 is ON, V 5V0

Output voltage wave form looks like

10 0.2 5 0.5

V0 avg 0.5

10 2 5 37V

5

Q43: The signum function is given by

x; x 0

xsgn 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 1

T and 2

T each rated at 500 kVA are operated in parallel.

Percentage impedances of 1

T and 2

T 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 2

T (in KVA) is ______________.


Solution:

*Z1S S

2 L Z Z1 2

0S 1000 36.86L KVA

*1 j6

S 1000 36.862 1.8 j10.8

0555.85 36.86 KVA

So contribution of T2

is 555.55 KVA at a power factor of 0.8 lag.

Q45: In the given circuit the parameter k is positive, and the power dissipated in the 2 resistor

is 12.5W. The value of k is _______________.


Correct Answer 0.5

Solution: Current through V02 resistor

2

By KCL V0 KV 5

2 0 ______________(i)

2V0P 12.5W

2 2

2V 250

V 5V0

From (i)

2.5 KV 50

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) LV 0 and C dc1

V V1

(B) L dcV V2

and

C dc

1V V

1

(C) LV 0 and C dcV V1

(D) L dcV V2

and

C dcV V

1


Correct Answer (A)

Solution: This is an example of boost converter or step up chopper

ViV V

0 c1

By volt sec balance, average voltage across inductor

V 0L

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) = M (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.

21 1 1 1 1C P 0.01P 30P 10 ; 100 MW P 150MW

22 2 2 2 2C P 0.05P 10P 10 ; 100 MW P 180MW

The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is ___________.


Correct Answer 20

Solution: Allocating minimum power to both units

P 100MW1

P 100MW2

P P 200MW1 2

dC

11C 0.02P 301 1dP

1

dC

21C 0.01P 102 2dP

2

At P 100MW, 1C 32Rs /MWhr1 1

At P 100MW, 1C 20Rs /MWhr2 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 i

V =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 voltage0

V is

(A) 1001 (B) 101 (C) 11 (D) 1


Correct Answer (A)

Solution: V 0

V A V V0

V A 0 V0 V0V

1000

V V V V 1000s s 0IR R 1000

Vd d 0I C V V C V

C 0 0dt dt 1000

610 d

1001V03 dt10

dV

6 01.001 10dt

Due to infinite input resistance

I IR C

dV3 6 6 010 V 10 V 1.001 10

s 0 dt

V 1mVs

dv6 6 6010 10 V 1.001 10

0 dt

dV01 V 1.001 0

0 dt

dV01.001 V 1

0dt

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 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

Solution: Current in load resistance 600

50A12

Since there is no losses so input power = output power

200 I 600 50p

I 150Ap

V 200inZ 1.33in 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

Solution: N N

s rslip 0.04N

s

Nr1 0.04

Ns

N 0.96Nr s

Speed of 5th harmonic = N

s5

(opposite mdr to fundamental)

Ns N 0.2 0.965 rslip 5.8N 0.2s

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 (B) 1000 0000 (C) 0000 0001 (D) 0000 0011

Correct Answer (A)

Solution: Resolution = 5 5

0.01968 2552 1

Output after 1st clock pulse = 10000000 = 128 (decimal)

5128 2.5098V255

V V0 a , so next bit will be set in next clock cycle

Output after 2nd clock pulse = 11000000 = 192 (decimal)

5192 3.7647V255

V V0 a , so this bit is reset and next bit will be set

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 0

V be maximum?

(A) 1 x

(B) (1 + x)

(C) 1 / 1 x

(D) 1 x

Correct Answer (A)

Solution: V V V0 Q S

R 1 x R

E ER pRR 1 x pR

R 1 xR RE

R pR xR R pR


1 x 1V E0 1 x p 1 p

For maximum V0

dV0 0

dp

dV 1 x 10 E 02 2dp

1 x p 1 p

2 21 x 1 p 1 x p

22

1 x 1 p x 1 p

2 2x 1 p x 2x 1 p

2x 1 p 2 1 p x 0

2p 2p 1 2 2p x 0

2p 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 & I2 4

are equal & opposite phasors as consistant with circuit so, there can be no fault

between I & I2 4

.

But I & I1 3

are not equal & opposite, so fault lies between I & I1 3

but since I1

is more than I3

so fault is closer to V1

than V2

Hence, fault is located at Q.

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