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.

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. 1 © 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  20kg 4 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) (B) (C) (D)

The industrialist had a personnel jet. I write my experience in my personnel diary. All personnel are being given the day off. 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.

2 © 2015 Kreatryx. All Rights Reserved.

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 (C) 0,1

(B) -1,0 (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 (C) Fibre

(B) Textile (D) Apparel

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

Q5: We ______________________our friend’s birthday and we ____________________how to make it up to him. (A) (B) (C) (D)

Completely forgot-------don’t just know Forgot completely-------don’t just know Completely forgot-------just don’t know Forgot completely-------just don’t know

Correct Answer (C) Solution: We completely forgot our friend’s birthday and we just don’t 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 if (r s)  (p q) (p q) (r s)

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

10

1

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) (B) (C) (D)

Since the report lacked needed information, it was of no use to them. The report was useless to them because there were no needed information in it. Since the report didn’t contain the needed information, it was not real useful to them. 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 QPS  600 . What is the length of PS? (A) (C)

(q r) qr (q2  r 2 )

(B) (D)

qr (q r)

(q r)2 qr


Correct Answer (B)  QPR  2qr Solution: PS  cos   qr  2  QPR  2  QPS  120

PS 

 120   qr  2qr 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) P2 ,R 2 ,S2 , T2 ,U2 (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’.

P2 ,R 2 ,S2 , T2 ,U2 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 (C) 28.5 Correct Answer (A)

(B) 6.74 (D) 45.49

Solution: SN  1sin 450  1

2

 OT

  Vertical distance b/w M & P=  4  2  1  ; Horizontal distance = 1 2 2   2

Distance between M & P =

2

 1    1   5.34km 6     2  2 


Section: Electrical Engineering Q1: The Laplace transform of f(t)  2 t  is s3 2 . The Laplace transform of g(t)  1 t is (A) 3s5 2 2 1 2 (B) s 12 (C) s 32 (D) s

Correct Answer (B) f(t) Solution: g(t)  2t f(t)  F(s) 

f(t)  F(s).ds If t

 s

 g(t) 

 1 3

2

s s



 1  1 2 1 2 .ds   s 2  s 2  1   2  s

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 f1 and f2 respectively. If f1  f2 , the resultant circuit exhibits the characteristics of a (A) Band pass filter (C) All pass filter

(B) band-stop filter (D) High-Q filter

Correct Answer (B) Solution: The filter F1 passes lower frequency component sin the signal upto f1 and stops higher frequency components. The filter F2 passes higher frequency components in the signal from f2 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 ( Zin ) 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




1 L  C b2



1 5000  10 5



5000  10 3 b2

 b  0.5

Input Impedance = =

R

 2.5

a2b2

2.5 a2b2

 2.5  a2b2  1  a  1  2 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  pu.  j4 j5  j8 

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. y10  j1 y 20  j1 y30  j1

Now, if all shunt capacitance are 50 % compensated y10  j0.5 y 20  j0.5 y 30  j0.5


 Y33  y10  y 20  y30   j4  j5  j0.5  Y33   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 t1 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. EV Pe  sin  Xeq As Xeq increases, Pe decreases According to swing equation

H d2  (Pm  Pe ) f dt2 As Pe decreases, rotor accelerates and  increases. Hence, option (A) is correct. 10 © 2015 Kreatryx. All Rights Reserved.

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

d2i dt2

at t= 0

V L V R 0 RV L2

Correct Answer (D) Solution: The current in a RL circuit is given by V i(t)  1  et  R





di V t   e dt L d2i dt2



RV t  e L2

d2i dt2

 t 0

RV L2

Q8: A capacitive voltage divider is used to measure the bus voltage Vbus in a high-voltage 50Hz AC system as shown in figure. The measurement capacitors C1 and C 2 have tolerance of 10% on their nominal capacitance values. If the bus voltage Vbus is 100kV rms, the maximum rms output voltage Vout (in kV), considering the capacitance tolerance, is ______.


Correct Answer 11.956 C1 V Solution: Vout  C1  C2 bus For maximum Vout, C1  1F  10%  1.1F C2  9F  10%  8.1F

Vout 

1.1 V  11.956kV 1.1  8.1 bus

Q9: The Operational amplifier shown in the figure is ideal. The input voltage (in Volt) is Vi  2sin(2  2000 t) . The amplitude of the output voltage Vo (in volts) is ____________.

Correct Answer 1.245 Solution: Reactance of capacitor =

1 1   795.77 C 400  107

Equivalent feedback impedance Z eq 

(R)(  jX C ) R  jXC

 622.67  51.4880 

  Z eq  622.67   2 Vo  Vin   R1  1000   Vo  1.245V

Q10: A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the xaxis as shown in the figure. The value of 0 is 4  107 in SI unit. If a uniform magnetic field 



intensity H  107 z A/m is applied, then the peak value of the induced voltage, Vturn (in volts), is ___________________________.


Correct Answer 248.05



Solution: The area of the ring perpendicular to magnetic field = r2 sin t



  angular speed of rotation Magnetic field density = 0H  4 wb m2 Flux,   B.A  4(  r2sin  t)  42r2 sin t

Vturn 

d  42r2 cos t dt

 2  Peak value = 42r2  4.2 .(1)2 .   60   248.05 V  60 

Q11: Two semi-infinite dielectric regions are separated by a plane boundary at y=0. The dielectric constants of region 1 (y<0) and 2 (y>0) are 2 and 5, respectively. Region 1 has uniform electric field E  3ax  4ay  2az , where ax , ay and az are unit vectors along the x, y and z axes, respectively. The electric field in region 2 is (A) 3ax  1.6ay  2az (B) 1.2ax  4ay  2az (C) 1.2ax  4ay  0.8az (D) 3ax  10ay  0.8az


Correct Answer (A) Solution: Boundary conditions in wave propagation E||,1  E||,2 This means, component of electric filed parallel to boundary of dielectric media remains same. 1E1  2E2

This means, perpendicular component of displacement vector remains same.

E||,2  (3ax  2az )  3ax  2az E ,2 

2 4ay  1.6ay 5

E2  3ax  1.6ay  2az Q12: Nyquist plots of two functions G1 (s) and G2 (s) are shown in figure.

Nyquist plot in the product of G1 (s) and G2 (s) is


Correct Answer (B) Solution: The equation of G1 (j ) will be

G1(j ) 

1  G1(j )  0,at        j  G1(j )   j ,at    0 

The equation of G2 (j ) will be

 G (j )  0,at    0  G2(j )  j  2   G2(j )  j ,at      G1(j ).G(j )  1

So (B) is correct.

Q13: In the following circuit, the transistor is in active mode and VC  2V . To get VC  4V , we R'C ' replace R C with R C ,Then the ratio is ____________. RC

Correct Answer 0.75 Solution: By KVL 10  0.7 9.3 IB   RB RB IC   IB 

10  VC RC

If we replace R C by R'c IB is constant and hence IC is also constant 10  2 10  4  RC R' C

R'C

RC



6 3   0.75 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 VAB  VAC  VBC VAC 

51  2.5v 2

VBC 

51  2.5v 2

VAB  Vth  0v

 Current in 2  resistor = 0 A.

Q16: The open loop control system results in a response of e2t (sin5t  cos5t) for a unit impulse input. The DC gain of the control system is ________. Correct Answer 0.2413 Solution: y(t)  e2t (sin5t  cos5t) Y(s) 

5 (s  2)2  2s



(s  2) (s  2)2  25



(s  7) s2  4s  29

For unit impulse input, X(s)  1 H(s) 

Y(s) (s  7) ;  2 X(s) s  4s  29

DC gain = H(s)

s 0



7  0.2413 29


Q17: Match the following.

 D.ds  Q 2.  f(z).dz  0 3.  (. A).dv   A.ds 4.  (  A).ds   A.dl

P. Stokes’s Theorem

1.

Q. Gauss’s Theorem R. Divergence Theorem S. Cauchy’s Integral Theorem (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: Stoke’s Theorem:

 (  A).ds   A.dl

 D.ds   Divergence Theorem:  (.A).dV   A.ds Cauchy’s Integral Theorem:  f(z).dz  0 Gauss’s Theorem:

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 z0 , then g(z) and h(z) are also differentiable at z0 (B) If g(z) and h(z) are differentiable at , then f(z) is also differentiable at z0 . (C) If f(z) is continuous at z0 , then it is differentiable at z0 . (D) If f(z) is differentiable at z0 , 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 (C) 115 and 2.5

(B) 115 and 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



Elap Ewave



1 2

Elap  115v


In lap connected since no. of parallel paths are now doubled, so current rating is doubled Ilap  2.I wave Elap 

1 E 2 wave

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 TL depends on its rotational speed r (in radian/second), such that TL  2.78  r . Neglecting rotational losses, the steady-state speed (in radian/second ) of the motor, when it drives this load, is______. Correct Answer 179.985 Solution: The variation of motor speed with torque is given as (200  180) 20   200  T  200  T  200  0.04T 500 500 TL  2.78r    200  0.04  2.78  r  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 (C) 5.00 kW and 0.00 kW

(B) 2.50 kW and 2.50 kW (D) 2.96 kW and 2.04 kW

Correct Answer (A) Solution: When power factor is 0.707

cos   0.707    450 P  3VL IL cos   5kW  3VL IL (0.707)  VL IL  4.083kVA


Reading of 1st wattmeter = VL IL cos(30  )  4.083cos(30  45)  3.94kW Reading of 2nd wattmeter= VL IL cos(30  )  4.083cos(30  45)  1.06kW

Q23: In the following circuit, the input voltage Vin 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 D1 gets forward biased. Entire Vin appears across C1 . This gets divided into C1 & R. Vo 

Vin (R) (R  jXC )

100RC  50  R 

50  50.XC 100C

 50XC  Vo  100 sin(100 t)    99.98 sin(100 t  1.145)  50XC  jXC 

Similarly during negative half cycle, Vo  99.98sin(100 t  1.145)

 63.6v  Vo avg  2  99.98 

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) 20 © 2015 Kreatryx. All Rights Reserved.

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, Vg across the magnetizing inductance, is 210 V rms, and the slip, s, is 0.05. The torque (in Nm) produced by the motor is __________.

Correct Answer 401.66 Solution: At s=0.05 Vg  210V I

Vg (1  j0.22)

 205.09  12.400 A

Pg  3  205.092  1  42061.9W Torque 

Pg s



42061.9  401.66 N  m 2  50

Q26: For the system governed by the set of equations: dx1 dx2 =2x1  x2 ,  2x1  u, y  3x1 dt dt The transfer function Y(s) U(s) is given by (A) 3(s 1) (s2  2s 2)

(B) 3(2s 1) (s2  2s 1)

(C) (s 1) (s2  2s 1)

(D) 3(2s 1) (s2  2s 2)


Correct Answer (A) dx1  2x1  1.x2  u Solution: dt dx2  2x1  0.x2  u dt  x1   2 1   x1  1        u  x2   2 0   x2  1 x  y  3 0   1   x2  s  2 1 (sI  A)    s  2 (sI  A)1 

TF  C(sI  A)1B 

TF  TF 

s

1

1 

  (s2  2s  2)  2 s  2

3 0 

3 0 

s 1  1    (s2  2s  2)  2 s  2 1

s  1

  (s2  2s  2) s  4  3(s  1) (s2  2s  2)

Q27: The z-Transform of a sequence x[n] is given as X(z)  2z 4  4 z  3 z2 . If y[n] is the first difference of x[n], then Y(z) is given by (A) 2z  2  8 z  7 z2  3 z3 (B) 2z  2  6 z  1 z2  3 z3 (C) 2z  2  8 z  7 z2  3 z3 (D) 4z  2  8 z  1 z2  3 z3


Correct Answer (A) Solution: X(z)  2 z  4  y[n]  x[n]  x[n 1]

4 3  z z2

Y(z)  X(z) 1  z 1      1  Y(z)   1    2z  4  4  3  2 z z  z   Y(z)  2 z  2  8  7  3 z z2 z3

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: Ef  Vt00  jIaXs 

11000

 j  100  50  8082.9  38.210 3 Suppose when load torque is increased, pf becomes cos  lagging

Ef  Vt  jIa(cos   jsin ) Xs   Vt _ IaXs sin   jIaXs cos  Ef  (Vt  IaXs sin )2  (IaXscos )2 Ef  Vt2  Ia2Xs2  2IaXs Vt sin 

Since excitation is not changed, Ef is constant 8082.9  Vt2  Ia2Xs2  2IaXs Vt sin  2

8082.9 

2

 11000  11000 2 2   50 sin    (120)  (50)  2  120  3  3 

sin   0.1443    8.2980 Ef  Vt0  jIaXs 

11000 3

= 8082.97  47.260 v

 j  120  8.2980  50

  47.260 23 © 2015 Kreatryx. All Rights Reserved.

Q29: Two coins R and S are tossed. The 4 joint events HRHS , TR TS ,HR TS , TRHS 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: P(HRHs )  0.28

P(HR Ts )  0.30 P(TRHs )  0.24

P(TR Ts )  0.18

By total probability theorem

P(HRHs )  P(HR Ts )  P(HR ) P(HR )  0.58 P(HRHs )  P(TR Hs )  P(HS ) P(HS )  0.52 P(HRHs )  P(HR )P(HS )

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= VI cos   30  25  0.8  600w Error in power 

V2 302  0.9w ; R 1000

% error =

0.9  100  0.15% 600


Q31: The volume enclosed by the surface f (x, y) = ex over the triangle bounded by the lines x = y; x = 0; y = 1 in the xy plane is ________. Correct Answer 0.7182 Solution: In this triangle, x varies from 0 to y

z  f(x, y)  ex 1y

Volume 

1

1

ex .dx.dy  (ex  1).dy  e y  y   (e 1)  1  0





00

0

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

Q2 K , where K is given by 4 0 d2

on the charge is (A)

0

1 1  ˆi  ˆj 4 4 1 1  ˆi  ˆj 8 8

(B) (C)

12 2 ˆ 12 2 ˆ i j 8 2 8 2

(D)

Correct Answer (D) Solution: F14 

F12 

F13 

q2 4 0 (2d)2

q2 4 0 (2d)2

(  x)

q2

 x y     2  4 0 (2 2 d)2  2

F1  F12  F13  F14  K

(  y)

12 2 8 2

x

q2

 1  1 1 1   x    y     8 2 4  4 0 (d)2   8 2 4 

12 2 8 2

y

12 2 8 2

i

12 2 8 2

j


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 ______.

Correct Answer 190.98 Solution: Fourier series of square wave 





4V sinnt  n



60 sinnt  n

n1,3,5





n1,3,5





4  15 sinnt n



19.098 sinnt n

n1,3,5



n1,3,5

T  0.4 ms 1 1 f   500 rad/sec T 0.4  103 So we are interested in fundamental 1 1 XC    50 C 5000  4  106 XL  L  5000  102  50

Since XL = XC , circuit is in resonance VC  IfundXC 

19.098  (  j50)sin  t 5

 VC  190.98 sin(  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. 26 © 2015 Kreatryx. All Rights Reserved.

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 L Leq   0.5L 2

  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 ____________.

Correct Answer 1.914 1

Solution: GMR = (D11D12D13 )(D21D22D23 )(D31D32D33 ) 2 1

3 3    0.7788R  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 Eb  Vt  IaRa  100  10  1  90V at N=1000 rpm EbN (Since   cons tant )

E N  b2  2 Eb1 N1

500  45V 1000 TIa (Since   cons tant )

Eb2  90 

TN2  IaN2 2

2 N   500   2    0.25 Ia1  N1   1000 

Ia2

Ia2  10  0.25  2.5A Vt  Eb2  Ia2R a  45  2.5  1  47.5V

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

G(s) 

K 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 28 © 2015 Kreatryx. All Rights Reserved.

Correct Answer (A) Solution: Characteristic equation: 1  G(s)H(s)  0 For unity feedback H(s) = 1 1+G(s)=0  s(s  1)(s  2)  k  0  s3  3s2  2s  k  0  k  (s3  3s2  2s) dk  0  (3s2  6 s  2)  0 ds 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 Vo 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 ( iL ) just continuous is _______.


Correct Answer 2500 Solution: For the boundary between continuous and discontinuous conduction I DVin I0  2  I0  2 R IL 

D(1  D) Vin fL

DVin D(1  D) Vin  R 2fL R

2fL 2  100  103  5  10 3 1000   (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 V1 and V2 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 300 . Therefore phase difference between V1 & V2 is 300 as V2 leads V1 by 300 .


Q41: The incremental costs (in Rupees/MWh) of operating two generating units are functions of their respective powers P1 and P2 in MW, are given by dC1  0.2P1  50 dP1 dC2  0.24P2  40 dP2

Where

20MW  P1  150MW 20MW  P2  150MW.

dC1  76 Rs/MWh and For a certain load demand, P1 and P2 have been chosen such that dP1 dC2 dP2

 68.8 Rs/MWh . If the generations are rescheduled to minimize the total cost, then P2

is________. Correct Answer 136.36 dC1  76  0.2P1  50 Solution: dP1 P1  130MW dC2  68.8  0.24P2  40 dP2 P2  120MW load  P1  P2  250MW

Now Generators are rescheduled for minimum cost  IC1  IC2 0.2P1  50  0.24P2  40 0.2P1  0.24P2  10............(i) P1 P2  250  P1  113.636MW  P2  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 i1 and i2 , if the fault current is 300 A as shown in figure.

(A) i1  1

3 A, i2  0 A (C) i1  0 A, i2  1 3 A

(B) i1  0 A, i2  0 A (D) i1  1 3 A, i2  1

3 A

Correct Answer (A) Solution: The fault lies on C-phase, so there is no fault current in A-phase 5  i2  I  0A 500 A In primary side, Vph,p 33  3 3 Vph,s 11 3 Iph,p 

300 3 3



100 3

A

This current flows through C-line into C-phase and exits through A-line, so that other two phases do not carry any current 100  ia  A 3 i1 

5 1 .i  A 500 a 3

Hence (A) is correct.


Q43: A differential equation

di  0.2i  0 is applicable over -10 < t <10. If i (4)=10, then i(-5) dt

is ___. Correct Answer 1.653 di Solution:  0.2i dt di di  dt  5  dt 0.2i i Integrating both sides 5ln(i)  10  Ke0.8  10 K  10e 0.8 i( 5)  Ke

5

5  Ke 1  10e 1.8

i( 5)  1.653

Q44: The following discrete-time equations result from the numerical integration of the differential equations of un-damped simple harmonic oscillator with state variables x and y. The integration time step is ‘h’. xk 1  xk y  yk  yk and k 1  xk h h Find the discrete-time system, which one of the following statements is TRUE? (A) The system is not stable for h>0. (B) The system is stable for h 

1 . 

(C) The system is stable for 0  h  (D)The system is stable for

1 

1 1 h 2 

Correct Answer (A) Solution: Since x and y are stable variables, transform the equations given as


xk 1  xk  h.yk yk 1  hxk  yk  xk 1   1 h  xk       yk 1   h 1  yk  xk 1  Axk

For TF we calculate (zI  A)1  z  1 h  (zI  A)    z  1  h (zI  A)1 

z  1 h    (z  1)2  h2  h z  1 1

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 Ts . If the voltage VL is as shown, the duty cycle of the switch S is______.

Correct Answer 0.75 Solution: During On-state voltage across inductor = Vin During OFF-state , voltage across inductor =  Vin  Vo   Vin  15V Vin  Vo  45 Vo  60V

This is also verified because it is a step-up chopper. Vin Vo  (1  D) (1  D) 

15 1  60 4

D  3  0.75 4


Q46: The unit step response of a system with the transfer function G(s)  which one of the following waveforms?

Correct Answer (A)  1  2s  Solution: Transfer function, G(s)     1s 

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

Y(s)  G(s).U(s) 



1 s

(1  2s) 1 3   s(1  s) s (1  s)



y(t)  1  3e  t .u(t)


1  2s is given by 1s

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 S2S1S0 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

0

0

0

0

1

0

0

0

1

0

0

0

1

0

1

0

0

1

1

1

0

1

0

0

1

0

1

0

1

0

0

1

1

0

1

0

1

1

1

1

1

0

0

0

1

1

0

0

1

1

1

0

1

0

1

1

0

1

1

1

1

1

0

0

0

1

1

0

1

1

1

1

1

0

1

1

1

1

1

0

f D

1 D

1 1 1 D 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 Vo of the following circuit in steady-state is

(A) Square wave of period 0.55 ms. (C) Square wave of period 0.25 ms.

(B) Triangular wave of period 0.55 ms. (D) Triangular wave of period 0.25 ms.

Correct Answer (A) Solution: When Vo  10v

V 2  5V Voltage of positive terminal V1  o 4 This is UTP of Schmitt Trigger Vc (  )  10 v Vc (0)  5 v Vc (t)  10  ( 5  10)e  t   10  15e  t 

At t  T1

Vc (t)  5 v  5  10  15e  t   5  15e  t  t  T1   ln3  RCln3 T1  0.25ln3  0.275ms Duty cycle of the waveform is 50% due to symmetrical UTP & LTP  T  2T1  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: N1  100,N2  50,N3  50 . If the magnetizing current is neglected, and the currents in two windings are I2 = 2300 and I3 = 21500 A, then what is the value of the current I1 in Ampere? (A) 1900 0 (B) 1270

0 (C) 490 0 (D) 4270

Correct Answer (A) N N 50 50  2300   21500 Solution: I1  2 I2  3 I3  N1 N1 100 100  1300  11500  1900 A

Q50: In the given rectifier, the delay angle of the thyristor T1 measured from the positive going zero crossing of Vs is 300 . If the input voltage Vs is 100sin(100 t) V, the average voltage across R (in Volt) under steady-state is __________.


Correct Answer 61.529 Solution: The output waveforms looks like

 2  1  Vo .d(  t) Average Voltage =  2  0 



   2 1   100 sin t.d(  t)  100 sin t.d(  t)   2    6  







 2  1  100   cos t   100   cos t     6   2 



100 (1  cos30)  (2) 2 



100 3  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 T Solution: max  2.25, smax  0.15 TFL For maximum torque to occur at starting smax  1 R 0.03 smax  2  0.15  R1 X2 X2  0.2

Now smax  1 R 2  R ext

1

X2

 R ext  X2  R 2  0.2  0.03  0.17 e j10t , for t  1  Q52: Consider a signal defined by x(t)   . Its Fourier Transform is 0 , for t  1  

(A)

2sin(   10)   10

(B) 2e j10

sin(  10)   10

(C)

2sin()   10

(D) 2e j10

2sin( ) 

Correct Answer (A) e jt , t  1  Solution: x(t)    0, t  1 

X( ) 



x(t)e  jt 

 1





1

1

e

jt .e  jt .dt

1

e j(10 ) .dt 

1 1 e j(10 t)  1 j(10  ) 



1 e j(10 )  e j(10 )   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 Is  10000 A, then what is the value of Ip in Ampere?

(A) 50300

(B) 50  300

(C) 50 3300

(D) 200300

Correct Answer (A) Solution: Per-phase current on secondary side  Turns ratio of primary to secondary  Iph,s Iph,p



Iph,p 

Vph,p Vph,s



Vph,p Vph,s



100 3

300 A

230 3 2  115 3

2 3

3 100  30  50300 A 2 3


Q54: In the given network V1  10000 V, V2  100  1200 V, V3  100  1200 V. The phasor current i (in ampere) is (A) 173.2  600 0 (B) 173.2120

0 (C) 100.0  60

0 (D) 100.0120

Correct Answer (A) V  V2 Solution: I2  3  173.2A j1 V  V1 I1  3  173.2  1200 A  j1 I  I1  I2  173.2  600 A

Q55: In the following sequential circuit, the initial state (before the first clock pulse) of the circuit is Q1Q0  00 . The state (Q1Q0 ) , immediately after the 333rd clock pulse is (A) 00 (B) 01 (C) 10 (D) 11


Correct Answer (B) Solution: CLK

Q1

Q0

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 ( Q1Q0 ) = (0 0) After 332nd clock pulse, the counter will show ( Q1Q0 ) = (0 1)


GATE 2015 EE Solutions - Session 2

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