GATE 2015 EE Solutions - Session 2

GATE 2015 (EE)

Solutions

Session 2 (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 also had even distribution of Numerical and Verbal Ability but some questions in

this section were really outstanding. Still I believe if carefully attempted 8-10 marks can be

scored in this section.

In 1 mark Technical Section 15 questions were very basic and required a straight and calm

head. the other questions did involve some tricks or were based on the concepts that students

mostly leave. Like I felt the questions based on Electric and Magnetic Fields required

understanding of the concepts studied in 12th and many graduate students do forget those

concepts. So they might have left those questions. But most of the questions tested the very

basic understanding of the concepts that students mostly lack in lieu of solving more formula

based questions. Machines had mostly DC Machines questions which could be done if you

read the question carefully and did not have any hidden tricks in them.

In 2 mark Technical Section again 20 questions were very basic and could be done if you do it

carefully and devote proper time to them. Again I liked the questions from Electric and

Magnetic Fields here and one or two question were like that did not click even which chapter

they were from. So I must say IIT Kanpur did keep the level of GATE exam pretty good and did

not fall into mediocrity of formula based questions. An unusual thing that I noticed was the

dominance of DC Machines in the paper though it is not the case mostly.

2015 Kreatryx. All Rights Reserved.


Section: General Aptitude

Q1: Based on the given statements, select the most appropriate option to solve the given

question.

What will be the total weight of 10 poles each of same weight?

Statements:

(I) One fourth of the weight of a pole is 5Kg.

(II) The total weight of these poles is 160Kg more than the total weight of two poles.

(A) Statement I alone is not sufficient.

(B) Statement II alone is not sufficient.

(C) Either I or II alone is sufficient

(D) Both statements I and II together are not sufficient.

Correct Answer (C)

Solution: Suppose weight of each pole is x

By statement -1:

x

5 x 20kg4

Weight of 10 poles =10x=200kg

By Statement-2:

10x=2x+160

X= 20kg

Weight of 10 poles =10x = 200kg

Both statements alone are sufficient.

Q2: Choose the statement where underlined word is used correctly

(A) The industrialist had a personnel jet.

(B) I write my experience in my personnel diary.

(C) All personnel are being given the day off.

(D) Being religious is a personnel aspect.

Correct Answer (C)

Solution: Personnel refers to people employed in an organization and hence (c) is most

appropriate choice.


Q3: Consider a function f(x) 1 x on 1 x 1 . The value of x at which the function attains

a maximum, and the maximum value of the function are:

(A) 0,-1 (B) -1,0

(C) 0,1 (D) -1,2

Correct Answer (C)

Solution: The graph of f(x) 1 x looks like as shown below

So minimum occurs at x = 0 , f(x) = 1

Q4: A generic term that includes various items of clothing such as a skirt, a pair of trousers

and a shirt is

(A) Fabric (B) Textile

(C) Fibre (D) Apparel

Correct Answer (D)

Solution: Apparel refers to formal or casual clothing and so (D) is most appropriate choice

Q5: We ______________________our friends birthday and we ____________________how to make it

up to him.

(A) Completely forgot-------dont just know

(B) Forgot completely-------dont just know

(C) Completely forgot-------just dont know

(D) Forgot completely-------just dont know

Correct Answer (C)

Solution: We completely forgot our friends birthday and we just dont know how to make it

up to him.


Q6: If p, q, r, s are distinct integers such that:

f (p, q, r, s)=max(p, q, r, s)

g (p, q, r, s)=min(p, q, r, s)

h (p, q, r, s)=remainder of

(p q)

(r s) if (p q) (r s) or remainder of

(r s)

(p q) if (r s) (p q)

Also a function fgh (p, q, r, s) = f (p, q, r, s) g (p, q, r, s) h (p, q, r, s)

Also the same operations are valid with two variable functions of the form f(p, q).

What is the value of fg( h(2, 5, 7, 3), 4, 6, 8) ?

Correct Answer 8

Solution: h (2, 5, 7, 3) = remainder of 7 3 2 5

= remainder of 21 110

fg(1,4,6,8) f(1,4,6,8).g(1,4,6,8)

= 8 1 =8

Q7: Out of the following four sentences, select the most suitable sentence with respect to

grammar and usage

(A) Since the report lacked needed information, it was of no use to them.

(B) The report was useless to them because there were no needed information in it.

(C) Since the report didnt contain the needed information, it was not real useful to them.

(D) Since the report lacked needed information, it would not had been useful to them.

Correct Answer (A)

Solution: Sentence (A) seems most grammatically appropriate.

Q8: In a triangle PQR, PS is the angle bisector of QPR and 0QPS 60 . What is the length

of PS?

(A) (q r)

qr

(B)

qr

(q r)

(C) 2 2(q r ) (D) 2(q r)

qr


Correct Answer (B)

Solution: 2qr QPR

PS cosq r 2

QPR 2 QPS 120

2qr 120 qrPS cos

q r 2 q r

Q9: If the letters P, R, S, T, U is an arithmetic sequence, which of the following are also in

arithmetic sequence?

(I) 2P, 2R, 2S, 2T, 2U

(II) P-3, R-3, S-3, T-3, U-3

(III) 2 2 2 2 2P ,R ,S ,T ,U

(A) I only

(B) I and II

(C) II and III

(D) I and III

Correct Answer (B)

Solution: If P, R, S, T, U are in arithmetic sequence with a common difference d

Then 2P, 2R, 2S, 2T, 2U will be an arithmetic sequence with common difference 2d

P-3, R-3, S-3, T-3, U-3 will be an arithmetic sequence with common difference d.

2 2 2 2 2P ,R ,S ,T ,U Will not be an arithmetic sequence.

Q10: Four branches of a company are located at M, N, O, and P. M is north of N at a distance

of 4 Km; P is south of O at a distance of 2Km; N is southeast of O by 1Km. What is the distance

between M and P in Km?

(A) 5.34 (B) 6.74

(C) 28.5 (D) 45.49

Correct Answer (A)

Solution: 0 1SN 1sin45 OT

2

Vertical distance b/w M & P= 14 22

; Horizontal distance = 1

2

Distance between M & P = 2 2

1 16 5.34km2 2


Section: Electrical Engineering

Q1: The Laplace transform of f(t) 2 t is 3 2s . The Laplace transform of g(t) 1 t is

(A) 5 23s 2

(B) 1 2s

(C) 1 2s

(D) 3 2s

Correct Answer (B)

Solution: f(t)

g(t)2t

If

s

13 12

2 2

ss

f(t) F(s)

f(t)F(s).ds

t

1 1 sg(t) s .ds s

12 22

Q2: The filters F1 and F2 having characteristics as shown in Figure (a) and (b) are connected as

shown in Figure (c).


The cut-off frequencies of F1 and F2 are 1f and 2f respectively. If 1 2f f , the resultant circuit

exhibits the characteristics of a

(A) Band pass filter (B) band-stop filter

(C) All pass filter (D) High-Q filter

Correct Answer (B)

Solution: The filter 1F passes lower frequency component sin the signal upto 1f and stops

higher frequency components.

The filter 2F passes higher frequency components in the signal from 2f and stops lower

frequency components

The characteristics look like

This is the characteristic of Band-Stop filter.

Q3: Find the transformer ratios a and b such that the impedance ( inZ ) is resistive and equals

2.5 when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s.

(A) a =0.5 , b=2.0 (B) a =2.0 , b=0.5

(C) a =1.0 , b=1.0 (D) a =4.0 , b=0.5

Correct Answer (B)

Solution: Since impedance is resistive


2

3

5 2

1 L

C b

1 5000 10

5000 10 b

b 0.5

Input Impedance = 2 2

R2.5

a b

= 2 22 2

2.5 12.5 a b 1 a 2b

a b

Q4: A 3-bus power system network consists of 3 transmission lines. The bus admittance matrix

of the uncompensated system is

j6 j3 j4

j3 j7 j5

j4 j5 j8

pu.

If the shunt capacitance of all transmission line is 50% compensated, the imaginary part of the

3rd row 3rd column element (in pu) of the bus admittance matrix after compensation is

(A) -j7.0

(B) j8.5

(C) j7.5

(D) j9.0

Correct Answer (B)

Solution: If we take the sum of all elements in a row, we obtain the admittance between bus

& ground.

10

20

30

y j1

y j1

y j1

Now, if all shunt capacitance are 50 % compensated

10

20

30

y j0.5

y j0.5

y j0.5


33 10 20 30

33

Y y y y j4 j5 j0.5

Y j8.5

Q5: We have a set of 3 linear equations in 3 unknowns. X Y means X and Y are equivalent

statements and X Y means X and Y are not equivalent statements.

P: There is a unique solution.

Q: The equations are linearly independent.

R: All eigenvalues of the coefficient matrix are nonzero.

S: The determinant of the coefficient matrix is nonzero.

Which one of the following is TRUE?

(A) P Q R S

(B) P R Q S

(C) P Q R S

(D) P Q R S

Correct Answer (A)

Solution: For a unique solution, the rank of coefficient matrix must be equal to order of matrix.

This implies determinant of coefficient matrix is non-zero.

Determinant= Product of eigen values

So if determinant is non-zero, all the eigen values must also be non-zero.

Q6: The synchronous generator shown in the figure is supplying active power to an infinite

bus via two short, lossless transmission lines, and is initially in steady state. The mechanical

power input to the generator and the voltage magnitude E are constant. If one line is tripped

at time 1t by opening the circuit breakers at the two ends (although there is no fault), then it

is seen that the generator undergoes a stable transient. Which one of the following waveforms

of the rotor angle shows the transient correctly?


Correct Answer (A)

Solution: When one of the line is opened, the equivalent reactance of the system is increased,

which leads to reduction in electrical power output of system.

eeq

EVP sin

X

As eqX increases, eP decreases

According to swing equation

2

m e2

H d(P P )

f dt

As eP decreases, rotor accelerates and increases.

Hence, option (A) is correct.


Q7: A series RL circuit is excited at t=0 by closing a switch as shown in figure. Assuming zero

initial conditions, the value of 2

2

d i

dt at t= 0

(A) V

L

(B) V

R

(C) 0

(D)

2

RV

L

Correct Answer (D)

Solution: The current in a RL circuit is given by

t

t

2 t

2 2

2

2 2t 0

Vi(t) 1 e

R

di Ve

dt L

d i RVe

dt L

d i RV

dt L

Q8: A capacitive voltage divider is used to measure the bus voltage busV in a high-voltage

50Hz AC system as shown in figure. The measurement capacitors 1C and 2C have tolerance

of 10% on their nominal capacitance values. If the bus voltage busV is 100kV rms, the

maximum rms output voltage outV (in kV), considering the capacitance tolerance, is ______.


Correct Answer 11.956

Solution: 1out bus1 2

CV V

C C

For maximum out,V 1C 1 F 10% 1.1 F

2C 9 F 10% 8.1 F

out bus1.1

V V 11.956kV1.1 8.1

Q9: The Operational amplifier shown in the figure is ideal. The input voltage (in Volt) is

iV 2sin(2 2000t) . The amplitude of the output voltage oV (in volts) is ____________.


Solution: Reactance of capacitor =7

1 1795.77

C 400 10

Equivalent feedback impedance

0Ceq

C

eqo in

1

o

(R)( jX )Z 622.67 51.488

R jX

Z 622.67V V 2

R 1000

V 1.245V

Q10: A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-

axis as shown in the figure. The value of 0 is 74 10 in SI unit. If a uniform magnetic field

intensity 7H 10 z

A/m is applied, then the peak value of the induced voltage, turnV (in

volts), is ___________________________.



Solution: The area of the ring perpendicular to magnetic field = 2r sin t angular speed of rotation

Magnetic field density = 0H 4 2wb

m

Flux, 2 2 2B.A 4 ( r sin t) 4 r sin t

2 2turn

dV 4 r cos t

dt

Peak value = 2 24 r 2 22

4. .(1) . 60 248.0560

V

Q11: Two semi-infinite dielectric regions are separated by a plane boundary at y=0. The

dielectric constants of region 1 (y0) are 2 and 5, respectively. Region 1 has

uniform electric field x y zE 3a 4a 2a , where xa , ya and za are unit vectors along the x, y

and z axes, respectively. The electric field in region 2 is

(A) x y z3a 1.6a 2a

(B) x y z 1.2a 4a 2a

(C) x y z1.2a 4a 0.8a

(D) x y z3a 10a 0.8a


Correct Answer (A)

Solution: Boundary conditions in wave propagation

||,1 ||,2E E

This means, component of electric filed parallel to boundary of dielectric media remains

same.

1 1 2 2E E

This means, perpendicular component of displacement vector remains same.

x z x z||,2

y y,2

x y z2

E (3a 2a ) 3a 2a

2E 4a 1.6a

5

E 3a 1.6a 2a

Q12: Nyquist plots of two functions 1G (s) and 2G (s) are shown in figure.

Nyquist plot in the product of 1G (s) and 2G (s) is


Correct Answer (B)

Solution: The equation of 1G (j ) will be

111

G (j ) 0,at1G (j )

G (j ) j ,at 0j

The equation of 2G (j ) will be

222

G (j ) 0,at 0G (j ) j

G (j ) j ,at

1G (j ).G(j ) 1

So (B) is correct.

Q13: In the following circuit, the transistor is in active mode and CV 2V . To get CV 4V , we

replace CR with'CR ,Then the ratio

'C

C

R

R is ____________.

Correct Answer 0.75

Solution: By KVL

BB B

10 0.7 9.3I

R R

CC B

C

10 VI I

R

If we replace CR by 'cR

BI is constant and hence CI is also constant

'C C

'C

C

10 2 10 4

R R

R 6 30.75

R 8 4


Q14: When a bipolar junction transistor is operating in the saturation mode, which one of the

following statements is TRUE about the state of its collector-base (CB) and the base-emitter

(BE) junctions?

(A) The CB junction is forward biased and the BE junction is reverse biased.

(B) The CB junction is reverse biased and the BE junction is forward biased.

(C) Both the CB and BE junctions are forward biased.

(D) Both the CB and BE junctions are reverse biased.

Correct Answer (C)

Solution: In saturation mode, both the junctions in a transistor are forward biased

Q15: The current i (in Amp) in the 2 resistor of the given network is__________.

Correct Answer 0

Solution: Finding Thevenin equivalent across 2 resistor

AB AC BC

AC

BC

AB th

V V V

5 1V 2.5v

2

5 1V 2.5v

2

V V 0v

Current in 2 resistor = 0 A.

Q16: The open loop control system results in a response of 2te (sin5t cos5t) for a unit

impulse input. The DC gain of the control system is ________.


Solution: 2ty(t) e (sin5t cos5t)

2 2 2

5 (s 2) (s 7)Y(s)

(s 2) 2s (s 2) 25 s 4s 29

For unit impulse input, X(s) 1

2

Y(s) (s 7)H(s)

X(s) s 4s 29

; DC gain =

s 0

7H(s) 0.2413

29


Q17: Match the following.

P. Stokess Theorem 1. D.ds Q

Q. Gausss Theorem 2. f(z).dz 0

R. Divergence Theorem 3. ( .A).dv A.ds

S. Cauchys Integral Theorem 4. ( A).ds A.dl (A) (B) (C) (D)

P-2 P-4 P-4 P-3

Q-1 Q-1 Q-3 Q-4

R-4 R-3 R-1 R-2

S-3 S-2 S-2 S-1

Correct Answer (B)

Solution: Stokes Theorem: ( A).ds A.dl

Gausss Theorem: D.ds

Divergence Theorem: ( .A).dV A.ds

Cauchys Integral Theorem: f(z).dz 0

Q18: Match the following.

Instrument type Used for

P. Permanent magnet moving coil 1. DC only

Q. Moving iron connected through current transformer 2. AC only

R. Rectifier 3. AC and DC

S. Electrodynamometer

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

P-1 P-1 P-1 P-3

Q-2 Q-3 Q-2 Q-1

R-1 R-1 R-3 R-2

S-3 S-2 S-3 S-1


Correct Answer (C)

Solution: PMMC instrument can only measure DC quantities. Moving Iron in conjunction

with current Transformer can only be used for AC as Transformer will not respond to DC.

Rectifier type instruments can be used with AC as well as DC.

Electrodynamometer Type Instrument can also be used for AC & DC.

Q19: Given f(z) g(z) h(z) , where f, g, h are complex valued functions of a complex variable

z. Which one of the following statements is TRUE?

(A) If f(z) is differentiable at 0z , then g(z) and h(z) are also differentiable at 0z

(B) If g(z) and h(z) are differentiable at , then f(z) is also differentiable at 0z .

(C) If f(z) is continuous at 0z , then it is differentiable at 0z .

(D) If f(z) is differentiable at 0z , then so are its real and imaginary parts.

Correct Answer (B)

Solution: f(z) g(z) h(z)

If g(z) and h(z) are differentiable then their sum is also differentiable.

Q20: A 4-pole, separately excited, wave wound DC machine with negligible armature

resistance is rated for 230V and 5kW at a speed of 1200 rpm. If the same armature coils are

reconnected to form a lap winding, what is rated voltage (in volts) and power (in kW)

respectively at 1200 rpm of the reconnected machine if the field circuit is left unchanged?

(A) 230 and 5 (B) 115 and 5

(C) 115 and 2.5 (D) 230 and 2.5

Correct Answer (B)

Solution: Induced E.M.F.=( NZ) P

60 A

For lap connected, P = A = 4

For wave connected, A = 2

lap

wave

E 1

E 2

lapE 115v


In lap connected since no. of parallel paths are now doubled, so current rating is doubled

lap wave

lap wave

I 2.I

1E E

2

P E.I Cons tant

So, Power rating = 5 kW.

Q21: A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is

200 radian/sec. At its rated torque of 500 Nm, its speed is 180 radian/sec. The motor is used

to directly drive a load whose load torque LT depends on its rotational speed r (in

radian/second), such that LT 2.78 r . Neglecting rotational losses, the steady-state speed

(in radian/second ) of the motor, when it drives this load, is______.


Solution: The variation of motor speed with torque is given as

(200 180) 20

200 T 200 T 200 0.04T500 500

L rT 2.78

r

200 0.04 2.78

179.985rad / sec

Q22: A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced

supply. The power consumed by the load is 5kW. The power is measured by the two-wattmeter

method. The reading of the two watt meters are

(A) 3.94 kW and 1.06 kW (B) 2.50 kW and 2.50 kW

(C) 5.00 kW and 0.00 kW (D) 2.96 kW and 2.04 kW

Correct Answer (A)

Solution: When power factor is 0.707

0cos 0.707 45

L L

L L

L L

P 3V I cos

5kW 3V I (0.707)

V I 4.083kVA


Reading of 1st wattmeter = L LV I cos(30 ) 4.083cos(30 45) 3.94kW

Reading of 2nd wattmeter= L LV I cos(30 ) 4.083cos(30 45) 1.06kW

Q23: In the following circuit, the input voltage inV is 100sin(100 t) . For 100 RC 50 , the

average voltage across R (in Volts) under steady-state is nearest to

(A) 100

(B) 31.8

(C) 200

(D) 63.6

Correct Answer (D)

Solution: During positive half cycle 1D gets forward biased.

Entire inV appears across 1C .

This gets divided into 1C & R.

ino

C

C

Co

C C

V (R)V

(R jX )

50100 RC 50 R 50.X

100 C

50XV 100sin(100 t) 99.98sin(100 t 1.145)

50X jX

Similarly during negative half cycle,

oV 99.98sin(100 t 1.145)

o avg2 99.98

V 63.6v

Q24: Consider the following sum of products expression, F ABC ABC ABC ABC ABC

The equivalent Product of Sums expression is

(A) F (A B C)(A B C)(A B C)

(B) F (A B C)(A B C)(A B C)

(C) F (A B C)(A B C)(A B C)

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


Correct Answer (A)

Solution: F ABC ABC ABC ABC ABC

F m(0,1,3,5,7) M(2,4,6)

F (A B C)(A B C)(A B C)

Q25: The figure shows the per-phase equivalent circuit of a two-pole three-phase induction

motor operating at 50 Hz. The air-gap voltage, gV across the magnetizing inductance, is 210

V rms, and the slip, s, is 0.05. The torque (in Nm) produced by the motor is __________.


Solution: At s=0.05

g

g 0

2g

g

s

V 210V

VI 205.09 12.40 A

(1 j0.22)

P 3 205.09 1 42061.9W

P 42061.9Torque 401.66 N m

2 50

Q26: For the system governed by the set of equations:

1 21 2 1 1dx dx

=2x x , 2x u, y 3xdt dt

The transfer function Y(s) U(s) is given by

(A) 23(s 1) (s 2s 2) (B) 23(2s 1) (s 2s 1)

(C) 2(s 1) (s 2s 1) (D) 23(2s 1) (s 2s 2)


Correct Answer (A)

Solution: 1 1 2dx

2x 1.x udt

21 2

1 1

2 2

1

2

1

2

1

2

dx2x 0.x u

dt

x x2 1 1u

x x2 0 1

xy 3 0

x

s 2 1(sI A)

2 s

s 11(sI A)

2 s 2(s 2s 2)

3 0 s 1 1TF C(sI A) B

2 s 2 1(s 2s 2)

2

2

3 0 s 1TF

s 4(s 2s 2)

3(s 1)TF

(s 2s 2)

Q27: The z-Transform of a sequence x[n] is given as 2X(z) 2z 4 4 z 3 z . If y[n] is the

first difference of x[n], then Y(z) is given by

(A) 2 32z 2 8 z 7 z 3 z

(B) 2 32z 2 6 z 1 z 3 z

(C) 2 32z 2 8 z 7 z 3 z

(D) 2 34z 2 8 z 1 z 3 z


Correct Answer (A)

Solution: 2

4 3X(z) 2z 4

z z

1

2

2 3

y[n] x[n] x[n 1]

Y(z) X(z) 1 z

1 34Y(z) 1 2z 4zz z

8 37Y(z) 2z 2z z z

Q28: A three-phase, 11 kV, 50 Hz, 2pole, star connected, and cylindrical rotor synchronous

motor is connected to an 11kV, 50 Hz source. Its synchronous reactance is 50 per phase,

and its stator resistance is negligible. The motor has a constant field excitation. At a particular

load torque, its stator current is 100A at unity power factor. If the load torque is increased so

that the stator current is 120 a, then the load angle (in degrees) at this load is __________.

Correct Answer -47.26

Solution: 0 0f t a s11000

E V 0 jI X j 100 50 8082.9 38.213

Suppose when load torque is increased, pf becomes cos lagging

f t a sE V jI (cos jsin )X

t a s a sV _ I X sin jI X cos

2 2f t a s a s

2 2 2f t a s a s t

E (V I X sin ) (I X cos )

E V I X 2I X V sin

Since excitation is not changed, fE is constant

2 2 2t a s a s t

22 2 2

0

8082.9 V I X 2I X V sin

11000 110008082.9 (120) (50) 2 120 50sin

3 3

sin 0.1443 8.298

0f t a s

11000E V 0 jI X j 120 8.298 50

3

= 08082.97 47.26 v 047.26


Q29: Two coins R and S are tossed. The 4 joint events R S R S R S R SH H ,T T ,H T ,T H have

probabilities 0.28, 0.18, 0.30, 0.24, respectively, where H represents head and T represents tail.

Which one of the following is TRUE?

(A) The coin tosses are independent.

(B) R is fair, S is not.

(C) S is fair, R is not.

(D)The coin tosses are dependent.

Correct Answer (D)

Solution: R sP(H H ) 0.28 R sP(H T ) 0.30

R sP(T T ) 0.18 R sP(T H ) 0.24

By total probability theorem

R s R s R

R

P(H H ) P(H T ) P(H )

P(H ) 0.58

R s R s S

S

R s R S

P(H H ) P(T H ) P(H )

P(H ) 0.52

P(H H ) P(H )P(H )

So, coins are not independent, so coin tosses are dependent.

Q30: The coils of a wattmeter have resistance 0.01 and 1000 ; their inductances may be

neglected. The wattmeter is connected as shown in the figure, to measure the power

consumed by a load, which draws 25 A at power factor 0.8. The voltage across the load

terminals is 30 V. The percentage error on the wattmeter reading is __________.

Correct Answer 0.15

Solution: Load power= VIcos 30 25 0.8 600w

Error in power2 2V 30

0.9wR 1000

; % error = 0.9

100 0.15%600


Q31: The volume enclosed by the surface f (x, y) = xe over the triangle bounded by the lines

x = y; x = 0; y = 1 in the xy plane is ________.


Solution: In this triangle, x varies from 0 to y

xz f(x,y) e

y1 1 1x x y

00 0 0

Volume e .dx.dy (e 1).dy e y (e 1) 1

Volume= e-2 = 0.7182

Q32: Two semi-infinite conducting sheets are placed at right angles to each other as shown in

the figure. A point charge of +Q is placed at a distance of d from both sheets. The net force

on the charge is 2

20

Q K,

4 d where K is given by

(A) 0

(B) 1 1 i j4 4

(C) 1 1 i j8 8

(D)

1 2 2 1 2 2 i j

8 2 8 2

Correct Answer (D)

Solution: 2

14 20

qF ( y)

4 (2d)

2

12 20

qF ( x)

4 (2d)

2

13 20

2

1 12 13 14 20

q x yF

2 24 (2 2 d)

q 1 1 1 1F F F F x y

4 48 2 8 24 (d)

1 2 2 1 2 2 1 2 2 1 2 2K x y i j

8 2 8 2 8 2 8 2


Q33: A symmetrical square wave of 50% duty cycle has amplitude of 15V and time period

of 0.4 ms. This square wave is applied across a series RLC circuit with R 5 ,L 10mH, and

C 4 F . The amplitude of the 5000 rad/s component of the capacitor voltage (in Volt) is ______.


Solution: Fourier series of square wave

n 1,3,5 n 1,3,5

n 1,3,5 n 1,3,5

4V 4 15sinn t sinn t

n n

60 19.098sinn t sinn t

n n

T 0.4 ms

3

1 1f 500

T 0.4 10

rad/sec

So we are interested in fundamental

C 6

2L

1 1X 50

C 5000 4 10

X L 5000 10 50

Since LX = CX , circuit is in resonance

C fund C

C

19.098V I X ( j50)sin t

5

V 190.98sin( t 90)

Amplitude 190.98V

Q34: For linear time invariant systems, that are Bounded Input Bounded Output stable, which

one of the following statements is TRUE?

(A) The impulse response will be integrable, but may not be absolute integrable.

(B) The unit impulse response will have finite support.

(C) The unit step response will be absolutely integrable.

(D)The unit step response will be bounded.


Correct Answer (D)

Solution: Since system is BIBO stable and unit step function is balanced, so the unit step

response should also be bounded.

Q35: Two identical coils each having inductance L is placed together on the same core. If an

overall inductance of L is obtained by interconnecting these two coils, the minimum value

of is _____.

Correct Answer 0.5

Solution: Since no mutual coupling is given, we assume M = 0, for minimum equivalent

inductance, the coils must be connected in parallel

eqL

L 0.5L2

0.5

Q36: A composite conductor consists of three conductors of radius R each. The conductors

are arranged as shown below. The geometric mean radius (GMR) (in cm) of the composite

conductor is kR. The value of k is ____________.


Solution: GMR = 1

211 12 13 21 22 23 31 32 33(D D D )(D D D )(D D D )

1

3 30.7788R 3R 3R 1.914R

K=1.914


Q37: With an armature voltage of 100 V and rated field winding voltage, the speed of a

separately excited DC motor driving a fan is 1000rpm, and its armature current is 10 A. The

armature resistance is 1 . The load torque of the fan load is proportional to the square of

the rotor speed. Neglecting rotational losses, the value of the armature voltage (in Volt) which

will reduce the rotor speed to 500rpm is ___________________.

Correct Answer 47.5

Solution: Since the motor is separately excited, flux can be assumed constant

b t a aE V I R 100 10 1 90V at N=1000 rpm

bE N (Since cons tant )

b2 2

b1 1

E N

E N

b2500

E 90 45V1000

aT I (Since cons tant )

2

2a

2 2a2 2

a1 1

a2

t b2 a2 a

T N

I N

I N 5000.25

I N 1000

I 10 0.25 2.5A

V E I R 45 2.5 1 47.5V

Q38: An open loop transfer function G(s) of a system is

K

G(s)s(s 1)(s 2)

For a unity feedback system, the breakaway point of the root loci on the real axis occurs at,

(A) -0.42

(B) -1.58

(C) -0.42 and -1.58

(D) none of the above


Correct Answer (A)

Solution: Characteristic equation: 1 G(s)H(s) 0

For unity feedback H(s) = 1

1+G(s)=0

3 2

3 2

s(s 1)(s 2) k 0

s 3s 2s k 0

k (s 3s 2s)

2dk 0 (3s 6s 2) 0ds

s 0.42, 1.577

The root locus looks like

So break-away point lies between 0 & -1

Hence s=-0.42 is break-away point.

Q39: A buck converter feeding a variable load is shown in the figure. The switching frequency

of the switch is 100 kHz and the duty ratio is 0.6. The output voltage oV is 36 V. Assume that

all the components are ideal, and that the output voltage is ripple-free. The value of R(in Ohm)

that will make the inductor current ( Li ) just continuous is _______.


Correct Answer 2500

Solution: For the boundary between continuous and discontinuous conduction

2 in0 0

inL

in in

3 3

I DVI I

2 R

D(1 D)VI

fL

DV D(1 D)V

R 2fL

2fL 2 100 10 5 10 1000R

(1 D) 0.4 0.4

R 2500

Q40: The three-phase transformers are realized using single-phase transformers as shown in

the figure.

The phase difference (in degree) between voltages 1V and 2V is ___________.

Correct Answer -30

Solution: The first transformer is a Dd0 connection, so phase difference primary and

secondary is 00 .

The second transformer is Dy11 connection, so secondary leads primary by 030 .

Therefore phase difference between 1 2V & V is 030 as 2V leads 1V by

030 .


Q41: The incremental costs (in Rupees/MWh) of operating two generating units are functions

of their respective powers 1P and 2P in MW, are given by

11

1

22

2

dC0.2P 50

dP

dC0.24P 40

dP

Where 1

2

20MW P 150MW

20MW P 150MW.

For a certain load demand, 1P and 2P have been chosen such that 1

1

dC76

dP Rs/MWh and

2

2

dC68.8

dPRs/MWh . If the generations are rescheduled to minimize the total cost, then 2P

is________.


Solution: 1 11

dC76 0.2P 50

dP

1

22

2

2

1 2

P 130MW

dC68.8 0.24P 40

dP

P 120MW

load P P 250MW

Now Generators are rescheduled for minimum cost

1 2

1 2

1 2

1 2

1

2

IC IC

0.2P 50 0.24P 40

0.2P 0.24P 10............(i)

P P 250

P 113.636MW

P 136.36MW


Q42: A 3-phase transformer rated for 33 kV/11kV is connected in delta/star as shown in figure.

The current transformer (CTs) on low and high voltage sides has a ratio of 500/5. Find the

currents 1i and 2i , if the fault current is 300 A as shown in figure.

(A) 1i 1 3 A, 2i 0 A (B) 1i 0 A, 2i 0 A

(C) 1i 0 A, 2i 1 3 A (D) 1i 1 3 A, 2i 1 3 A

Correct Answer (A)

Solution: The fault lies on C-phase, so there is no fault current in A-phase

2 A5

i I 0A500

In primary side,

ph,p

ph,s

ph,p

V 333 3

V 11 3

300 100I A

3 3 3

This current flows through C-line into C-phase and exits through A-line, so that other two

phases do not carry any current

a

1 a

100i A

3

5 1i .i A

500 3

Hence (A) is correct.


Q43: A differential equation di

0.2i 0dt

is applicable over -10 < t 0.

(B) The system is stable for 1

h

.

(C) The system is stable for 1

0 h

(D)The system is stable for 1 1

h2

Correct Answer (A)

Solution: Since x and y are stable variables, transform the equations given as


k 1 k k

k 1 k k

k 1 k

k 1 k

k 1 k

x x h.y

y hx y

x x1 h

y yh 1

x Ax

For TF we calculate 1(zI A)

1

2 2

z 1 h(zI A)

h z 1

z 1 h1(zI A)

h z 1(z 1) h

Poles lie at z 1 ih which are outside unit circle h 0 & hence system is unstable

for h>0.

Q45: For switching converter shown in the following figure, assume steady-state operation.

Also assume that the components are ideal, the inductor current is always positive and

continuous and switching period is sT . If the voltage LV is as shown, the duty cycle of the

switch S is______.

Correct Answer 0.75

Solution: During On-state voltage across inductor = inV

During OFF-state , voltage across inductor = in oV V

in

in o

o

V 15V

V V 45

V 60V

This is also verified because it is a step-up chopper.

ino

VV

(1 D)

15 1(1 D)

60 4

3D 0.754


Q46: The unit step response of a system with the transfer function1 2s

G(s)1 s

is given by

which one of the following waveforms?

Correct Answer (A)

Solution: Transfer function, 1 2s

G(s)1 s

Laplace transform of u(t)=U(s)=1

s

t

(1 2s) 1 3Y(s) G(s).U(s)

s(1 s) s (1 s)

y(t) 1 3e .u(t)


Q47: A Boolean function f(A, B, C, D) = (1,5,12,15) is to be implemented using 8 1

multiplexer (A is MSB). The inputs ABC are connected to the select inputs 2 1 0S S S of the

multiplexer respectively.

Which one of the following options gives the correct inputs to pins 0,1,2,3,4,5,6,7 in order?

(A) D,0,D,0,0,0,D,D

(B) D,1,D,1,1,1,D,D

(C) D,1,D,1,1,1,D,D

(D)D,0,D,0,0,0,D,D

Correct Answer (B)

Solution: f(A,B,C,D) (1,5,12,15) (0,2,3,4,6,7,8,9,10,11,13,14) Truth table looks like

A B C D f f

0 0 0 0 1

0 0 0 1 0 D

0 0 1 0 1

0 0 1 1 1 1

0 1 0 0 1

0 1 0 1 0 D

0 1 1 0 1

0 1 1 1 1 1

1 0 0 0 1

1 0 0 1 1 1

1 0 1 0 1

1 0 1 1 1 1

1 1 0 0 0

1 1 0 1 1 D

1 1 1 0 1

1 1 1 1 0 D

Inputs are D,1,D,1,1,1,D,D


Q48: The saturation voltage of the ideal op-amp shown below is 10 V. The output voltage

oV of the following circuit in steady-state is

(A) Square wave of period 0.55 ms. (B) Triangular wave of period 0.55 ms.

(C) Square wave of period 0.25 ms. (D) Triangular wave of period 0.25 ms.

Correct Answer (A)

Solution: When oV 10v

Voltage of positive terminal o1V 2

V 5V4

This is UTP of Schmitt Trigger

c

c

t tc

V ( ) 10 v

V (0) 5v

V (t) 10 ( 5 10)e 10 15e

At 1t T

c

t

t

1

1

V (t) 5v

5 10 15e

5 15e

t T ln3 RCln3

T 0.25ln3 0.275ms

Duty cycle of the waveform is 50% due to symmetrical UTP & LTP

1T 2T 0.55ms


Q49: A three-winding transformer is connected to an AC voltage source as shown in the figure.

The number of turns are as follows: 1 2 3N 100,N 50,N 50 . If the magnetizing current is

neglected, and the currents in two windings are 2I =02 30 and 3I =

02 150 A, then what

is the value of the current 1I in Ampere?

(A) 01 90

(B) 01 270

(C) 04 90

(D) 04 270

Correct Answer (A)

Solution: 0 032

1 2 31 1

NN 50 50I I I 2 30 2 150

N N 100 100

0 0

0

1 30 1 150

1 90 A

Q50: In the given rectifier, the delay angle of the thyristor 1T measured from the positive

going zero crossing of sV is 030 . If the input voltage sV is 100sin(100 t) V, the average

voltage across R (in Volt) under steady-state is __________.



Solution: The output waveforms looks like

Average Voltage =

2

o

0

1V .d( t)

2

2

6

2

6

1100sin t.d( t) 100sin t.d( t)

2

1100 cos t 100 cos t

2

100(1 cos30) (2)

2

1003 cos30 61.529V

2

Q51: A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated

voltage and frequency. The stator resistance, magnetizing reactance, and core loss are

negligible. The maximum torque produced by the rotor is 225% of full load torque and it

occurs at 15% slip. The actual rotor resistance is 0.03 / phase . The value of external

resistance (in Ohm) which must be inserted in a rotor phase if the maximum torque is to occur

at start is______.


Correct Answer 0.17

Solution: max

FL

T2.25,

T maxs 0.15

For maximum torque to occur at starting

max

2max

1 2

2

s 1

R 0.03s 0.15

R X

X 0.2

Now maxs 1

2 ext

2

ext 2 2

R R1

X

R X R 0.2 0.03 0.17

Q52: Consider a signal defined by

j10te , for t 1x(t)

0 , for t 1 . Its Fourier Transform is

(A) 2sin( 10)

10

(B) j10

sin( 10)2e

10

(C) 2sin( )

10

(D) j10

2sin( )2e

Correct Answer (A)

Solution: j te , t 1

x(t)0, t 1

1j t j t j t

1

X( ) x(t)e e .e .dt

1 1j(10 ) j(10 t)

11

j(10 ) j(10 )

1e .dt e

j(10 )

1e e

j(10 )

2sin(10 ) 2sin( 10)

(10 ) ( 10)


Q53: A balanced (positive sequence) three-phase AC voltage source is connected to a

balanced, star connected load through a star-delta transformer as shown in the figure. The

line-to-line voltage rating is 230 V on the star side, and 115 V on the delta side. If the

magnetizing current is neglected and 0sI 100 0 A, then what is the value of pI in Ampere?

(A) 050 30 (B) 050 30

(C) 050 3 30 (D) 0200 30

Correct Answer (A)

Solution: Per-phase current on secondary side 0100

30 A3

Turns ratio of primary to secondary ph,p

ph,s

V 230 3 2

V 115 3

ph,s ph,p

ph,p ph,s

0ph,p

I V 2

I V 3

3 100I 30 50 30 A

2 3


Q54: In the given network 01V 100 0 V, 0

2V 100 120 V, 0

3V 100 120 V. The

phasor current i (in ampere) is

(A) 0173.2 60

(B) 0173.2 120

(C) 0100.0 60

(D) 0100.0 120

Correct Answer (A)

Solution: 3 22V V

I 173.2Aj1

03 11

V VI 173.2 120 A

j1

01 2I I I 173.2 60 A

Q55: In the following sequential circuit, the initial state (before the first clock pulse) of the

circuit is 1 0Q Q 00 . The state 1 0(Q Q ) , immediately after the 333rd clock pulse is

(A) 00

(B) 01

(C) 10

(D) 11


Correct Answer (B)

Solution: CLK 1Q 0Q

0 0 0

1 0 1

2 1 1

3 1 0

4 0 0

After 332nd clock pulse, the counter will again reach initial state ( 1 0Q Q ) = (0 0)

After 332nd clock pulse, the counter will show ( 1 0Q Q ) = (0 1)

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GATE 2015 EE Solutions - Session 2