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Electrical Engineering Q. No. 1 – 25 Carry One Mark Each In the circuit shown, the diodes are ideal, the inductance is small, and Io  0. Which one of the

1.

following statements is true?

(A) D1 conducts for greater than 180o and D 2 conducts for greater than 180o (B) D 2 conducts for more than 180o and D1 conducts for 180o (C) D1 conducts for 180o and D 2 conducts for 180o . (D) D1 conducts for more than 180o and D 2 conducts for 180o Key: (A) 2.

For a 3-input logic circuit shown below, the output Z can be expressed as P Z

Q R

(A) Q  R

(B) PQ  R

(C) Q  R

(D) P  Q  R

Key: (C) Exp:

 PQ.Q.QR  PQ  Q  QR  Q  QR QR

3.

P

PQ

Z

 Q  QP  Q   A  AB  A  B

Q

PQ.Q.QR

R

QR

An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is 1 4 5 6 (A) (B) (C) (D) 2 9 9 9 49

Key: (A) Exp:

1 2

4R, 5B 59

R

B

5R 5B 1 2

R

5R, 5B B

59

R

49 B

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Here R is red ball, B is black ball

 The probability to get a red ball in the second draw is

4.

1 4 1 5 1     2 9 2 9 2

When a unit ramp input is applied to the unity feedback system having closed loop transfer function

C s

R s



Ks  b  a  0, b  0, K  0  , the steady state error will be s  as  b 2

(A) 0

(B)

a b

(C)

aK b

(D)

aK b

Key: (D) Exp: Given T  s  

C s 

R s

Ct   r t   t 

Apply L.T to above equations

E  s   R  s  1  T  s  ess  C     lt S.E  s   lt .s. s 0

ess 

5.

s 0

 Ks  b   lt 1 s2  s a  K   lt s   a  K  1 1    s 0 s 2  as  b s 2  s 2  as  b  s 0 s s 2  as  b

aK b

A three-phase voltage source inverter with ideal devices operating in 180o conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is Vdc . The peak of the

fundamental component of the phase voltage is V 2Vdc 3Vdc (A) dc (B) (C)    Key: (B) Exp: Fourier series expansion of line to neutral voltage Vao is given by

(D)

4Vdc 



 2Vs    sin  nt  n  6k 1  n  2V for n  1, Vao  s  max value   Vao 

6.



   The figures show diagrammatic representations of vector fields X, Y and Z respectively. Which one of the following choices is true?

   (A) .X  0,   Y  0,   Z  0    (C) .X  0,   Y  0,   Z  0

   (B) .X  0,   Y  0,   Z  0    (D) .X  0,   Y  0,   Z  0

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Key: (C) Exp: for x Divergence not equal to zero    x  0

for y

Divergence  0    t  0 Curl  0 

for z

Divergence  0    z  0 Curl  0 

7.

Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green (vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is ________. Key: 0.9 to 0.9 8.

Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed. (A) uniformly over the entire volume of the sphere (B) uniformly over the outer surface of the sphere (C) concentrated around the centre of the sphere (D) along a straight line passing through the centre of the sphere

Key: Exp:

(B) For a perfect conductor the charge is present only on the surface.

i.e,

Pu  0  inside the conductor E0 

The transfer function C  s  of a compensator is given below.

9.

s  s   1   1   0.1  100  C s   s 1  s  1    10  The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is (A) 0.1    1 (B) 1    10 (C) 10    100 (D)   100 Key: (A)

10.

The figure show the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where  is a complex number with non-zero real and imaginary parts.

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For the given circuit, Ybus and Zbus are bus admittance matrix and bus impedance matrix, respectively, each of size 2  2. Which one of the following statements is true? (A) Both Ybus and Zbus are symmetric (B) Ybus is symmetric and Zbus is unsymmetric (C) Ybus is unsymmetric and Zbus is symmetric (D) Both Ybus and Zbus are unsymmetric Key: (D)  yt  2 a Exp: YBUS     yt   a z BUS  y bus 1

 yt  a*    yt  

11.

A phase-controlled, single-phase, full-bridge converter is supplying a highly inductive DC load. The converter is fed from a 230 V, 50 Hz, AC source. The fundamental frequency in Hz of the voltage ripple on the DC side is (A) 25 (B) 50 (C) 100 (D) 300 Key: (C) Exp: Vo Vm



 

o

2

2  

Q

For one input pulse, Vo has 2 pulses  frequency of Vo ripple = 2f supply  2  50  100Hz 12.

Let x and y be integers satisfying the following equations

2x 2  y 2  34 x  2y  11 The value of  x  y  is ________. Key: 7 to 7 Exp:

Clearly x = 3 and y = 4 satisfies the given two equation

x  y  7 13.

Consider a function f  x, y, z  given by

f  x, y,z    x 2  y2  2z 2  y2  z 2  The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is ________ Key: 40 to 40 f Exp:   y 2  z 2   2x  at x  2, y  1, z  3  1  9  4   40 x





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For the given 2-port network, the value of transfer impedance Z21 in ohms is_______

Key: 3 to 3 Exp:

V Z21  2 I1

I1

I1

I2  0



I1  2I1  3I1 2 V  Z21  2  3 I1 V2  2 

15.

I1

2

2

 2x

2

I1

4

V1



2 



I1

2

2



 V2

2I1



The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals ______. (Give the answer up to one decimal place.)

Key: 99 to 101 Exp:

Before initial charge on the capacitor  0  Vc  0   0V Final voltage Vc     10V To find time constant 

10  10  5 10  10   R eq ; C  5 R eq 

5

5 5

Vc  t   VC      VC  0   VC     e  t   10  10e  t  d  i C  t   C C  2e t 5 dt We know that i r  i C     2e

 5

 5

t 5

Instantaneous power p   i r  10  2e t 5  20e t 5 



0

0

Energy transferred   pdt   20e



t 5

e t 5 dt  20  100 0 1  100J 1 5 0

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The figure below shows the circuit diagram of a controlled rectifier supplied from a 230 V, 50 Hz, 1-phase voltage source and a 10:1 ideal transformer. Assume that all devices are ideal. The firing angles of the thyristors T1 and T2 are 90o and 270o , respectively.

The RMS value of the current through diode D 3 in amperes is ________ Key: 0 to 0 Exp: 0A since D2 is OFF and it will not turn ON for R load. 17.

In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of the Jacobian is 100  100. If there are 20 PV buses in addition to PQ buses and a slack bus, the total number of buses in the system is ________. Key: 61 to 61 Exp: Given the size of bus is 100*100. so [J]= 100 we have formula for [J] = [2n-m-2] 100= [2n-20-2] total no.of buses ,n = 61 18.

A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The speed of the rotor flux in mechanical rad/sec, sensed by a stationary observer, is closest to (A) 1500 (B) 1470 (C) 157 (D) 154

Key:

(C)

Exp:

3 4P 400V

S.C.I.M s  0.02 r  N r

Rotor flux speed is same as stator flux speed. 120  50  1500 4 2N 2  1500 Ws    157.08 rad sec 60 60 Ns 

19.

Two resistors with nominal resistance values R1 and R 2 have additive uncertainties R1 and R 2 , respectively. When these resistances are connected in parallel, the standard

deviation of the error in the equivalent resistance R is 2

 R   R  R1    R 2  (A)    R1   R 2 

2

2

 R   R  R1    R 2  (B)    R 2   R1 

2

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 R   R  (C)    R 2    R 1  R1   R 2 

Key:

(A)

Exp:

R eq 

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2

 R   R  (D)    R1    R 2  R1   R 2 

R1R 2 R1  R 2 2

2

 R  2  R  2    R1    R 2  R1   R 2  OR 2

 R   R  R1    R 2    R1   R 2 

2

The nominal-  circuit of a transmission line is shown in the figure.

20.

Impedance Z  100 80o  and reactance X  3300 . The magnitude of the characteristic impedance of the transmission line, in  , is _______________. (Give the answer up to one decimal place.) Key: 404 to 408 y 1 Exp:  2 x 2 2 y   6.06 10 4 x 3300 z 100 z0    406.2 y 6.06  104 21.

The pole-zero plots of three discrete-time systems P, Q and R on the z-plane are shown below.

Which one of the following is TRUE about the frequency selectivity of these systems? (A) All three are high-pass filters. (B) All three are band-pass filters. (C) All three are low-pass filters. (D) P is a low-pass filter, Q is a band-pass filter and R is a high-pass filter. Key:

(B)

Exp:

  0 rad / samples reprsent lowfrequencies

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   rad / samples reprsent highfrequencies Since zeros are located at   0 rad / samples and   rad / samples they cannot be high pass and low pass filters. Thus they all replresent band pass filters. The mean square value of the given periodic waveform f  t  is_________

22.

Key:

6 to 6

Exp: Mean square value 

f 2 t

Area under the squarred function Period of the function

16

Area  16   0.7  0.3  4  2.7  0.7   16  8  24 volt  second Period  2.7   1.3  4 Mean square value 

23.

4

24 6 4

1.3

0.3

0.7

2.7

A stationary closed Lissajous pattern on an oscilloscope has 3 horizontal tangencies and 2 vertical tangencies for a horizontal input with frequency 3 kHZ. The frequency of the vertical input is (A) 1.5 kHz (B) 2 kHz (C) 3 kHz (D) 4.5 kHz

Key:

(D)

Exp:

3 9 f r   3   4.5 kHz 2 2

n 3

fy fx



nx ny

ny  2

24.

Let y 2  2y  1  x and x  y  5. The value of x  y equals _________. (Give the answer up

to three decimal places) Key: 5.7 to 5.8 Exp: y 2  2y  1  x  x  y  1

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If a synchronous motor is running at a leading power factor, its excitation induced voltage  E f  is

25.

Key: Exp:

(A) equal to terminal voltage Vt

(B) higher than the terminal voltage Vt

(C) less than terminal voltage Vt

(D) dependent upon supply voltage Vt

(B) Higher than the terminal voltage.

Ef

iIa X s V

Ei Ia

Q 

Q. No. 26 – 55 Carry Two Marks Each 26.

Which of the following systems has maximum peak overshoot due to a unit step input? 100 100 (A) 2 (B) 2 s  10s  100 s  15s  100 100 100 (C) 2 (D) 2 s  20s  100 s  5s  100 Key: (C) 

Exp: Peak over shoot  e

12

If   0,peak over shoot 100%  Maximum  In General  If   1 peak over shoot  0%  Minimum  Here which of the following has '  ' value less, the system will have maximum over shoot. Option „A‟, n  10, 2n  10    0.5 Option „B‟ n  10, 2n  15    0.75 Option „C‟ n  10, 2n  5    0.25 Option „D‟ n  10, 2n  20    1 So, option „C‟ is correct (OR) By Inspection, see all options n  cons tan t, 2n varies, so, 2n less means, that system have maximum over shoot. 27.

Consider an overhead transmission line with 3-phase, 50 Hz balanced system with conductors located at the vertices of an equilateral triangle of length Dab  Dbc  Dca  1m as shown in figure below. The resistance of the conductors are neglected. The geometric mean radius (GMR) of each conductor is 0.01m. Neglecting the effect of ground, the magnitude of positive sequence reactance in  / km (rounded off to three decimal places) is ________  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India

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Key: 0.271 to 0.301 Exp: Deq  3 Dab Dbc Dca  1m  GMD

DS  GMR  0.01m Inductance/phase/m  2  107 ln

Dm  1  7  2  107 ln    9.21 10 H DS  0.01 

Inductance/phase/km  9.21104 H Reactance  L  2 50  9.21104  0.2892 / km 28.

Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and 4% respectively from no load to full load. Both the generating units are operating in parallel to share a load of 600 MW. Assuming free governor action, the load shared by the larger unit is _______ MW.

Key: Exp:

395 to 405 Assume No – load speed regulations are equal % Speed Reg A x 4% B G

6% F

E

300MW

C D



400MW

H

Power

Power

From similar triangles method A 

F

6%

 E

x 

 B D

300

x



G 

 C

400

B

4%

 H

BG AB  CH AC

FB AB  ED AD

P1  300 

A 

x 6

P2  400 

x 4

P2  100x

P1  50x

Given that P1  P2  600MW 150x  600 x4

 The load supplied by largest machine is P2=100×4=400MW

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For the network given in figure below, the Thevenin‟s voltage Vab is

29.

(A) -1.5 V Key: Exp:

(B) -0.5 V

(C) 0.5 V

(A) The equivalent CKT is

10

5

(D) 1.5 V Vth

Apply nodal Vth  30 Vth Vth  16   0 15 10 10 2Vth  60  3Vth  3Vth  48  0

30V

a

 

 16V 

10 b

 8Vth  12  Vth  1.5V

30.

The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples uniquely. The required minimum sampling rate is

(A) 1000 samples/s Key: Exp:

(B) 1500 sample/s

(C) 2000 samples/s

(D) 3000samples/s

(B)

500

f cos 1000t 

 Convolution 

x f 

500

f

500

500

f

Consider 1 F cos 1000t      f  500     f  500  2 Input signal to the LTI system is

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1 x  f     f  500     f  500  2 1 1 W  f     f  500     f  500  2 2

If the input signal is defined as w(t) then its Fourier transform can be drawn as follows:  f 

1000

0

f

1000

H f 

sin 150t  1500 sinc 1500t  t  f   H  f   rect    1500 

Given h  f  

 Y  f   W  f  H  f  has a max frequency 750 HZ

∴ Minimum sampling rate = 1500 HZ

750

f

750

A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance R a  0.02.

31.

When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is (A) 34.2 A (B) 30 A (C) 22 A (D) 4.84 A Key: Exp:

(B) Separately excited d.c. motor 2NT 60 2  900  70 E b Ia   6597 60 6597 Ia  ...(1) Eb

Ia

P

V  Ia R a  E b 220 

k

0.02 V  220V

Eb

6597  0.02  E b Eb

220E b   6597  0.02   E 2b By solving above equation

We get E b1  219.39, E b2  0.61 Ia 

V  E b 220  E b 220  219.39   Ia   Ia  30.5 Amps Ra 0.02 0.02

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 

 

A cascade system having the impulse responses h1  n   1, 1 and h 2  n   1,1 is shown in the

32.





figure below, where symbol ↑ denotes the time origin.

The input sequence x  n  for which the cascade system produces an output sequence





y  n   1,2,1, 1, 2, 1 is 

  x  n   1,1,1,1

  (D) x  n   1,2,2,1

(A) x  n   1,2,1,1

(B) x  n   1,1,2,2



(C)







Key: (D) Exp:

Y1  

h(n)  h1[n]*h2[n]  {1, 0, 1} 

Y[n]=h(n)*x(n) Given

y[n]  {1, 2, 1,  1,  2,  1} By observation x[n] should be{1,2,2,1}

2000

1000

For the circuit shown in the figure below, it is given that VCE 

33.

 1000 2000

VCC . The transistor has 2

  29and VBE  0.7V when the B-E junction is forward biased.

RB is R (B) 92

For this circuit, the value of (A) 43 Key:

(C) 121

(D) 129

(D)

Exp: Given VCE 

Vec 10   5V 2 2

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10  1    I B  4R  I B R B  0.7  1    I B .R 10  30I B  4R  I B R B  0.7  30I B  R 9.3  I B 120R  30R  R B  9.3  I B 150R  R B 

...(1)

10  4 R 1    I B  VCE  1    I B  R 10  120RI B  5  30I B .R 5 1  ...(2) 150R 30R Substituting equation (2) in equation (1) 1 9.3  150  R B  30R R RB 279  150  B ;  279  150  129 R R IB 

34.

A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees. Key: 12.5 to 12.9 Pa  Pm  Pe Exp:

 60  0  60mw GH 1000 1   180f 180  50 9 10 t  10cycles   0.25sec 50 5 t  5cycles   0.1sec 50

m

2 Pa t 2 60  0.1 .    2.7 1 m 2 2 9 New ratio,   10  2.7 12.7



35.

For the circuit shown below, assume that the OPAMP is ideal.

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Which one of the following is TRUE? (A) vO  vS (B) vO  1.5vS Key: Exp:

(C) vO  2.5vS

(D) vO  5vS

(C) At node (1)

Vx 

Vs  2R Vs  4R 2

At node (2)

R

Vx Vx  Vy  0 R R 2Vx  Vy ;

R

Vy 

2Vs  Vs 2

R

Vy R

V 

y

 Vx 

V 

y

 Vo 

R R V Vs  Vs  s  Vs  Vo  0; 2 Vs 3Vs   Vo 2 5V Vo  s ; Vo  2.5Vs 2

R

3

Vy

2

At node (3)

36.

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Vx

Vo



0

2R 1

Vx

Vs 2R

The root locus of the feedback control system having the characteristic equation

s  6Ks  2s  5  0 where K  0, enters into the real axis at (A) s  1

(C) s  5

(B) s   5

(D) s  5

Key: (B) Exp: C.E  s2  6ks  2s  5  0 6ks 1 2 0 s  2s  5 6ks G  s   1  2 s  2s  5   s 2  2s  5  1 5 K   s  2   6s 6 s dk  5  0  1  2   0 ds  s 

j 2j

 5

1

2j

s2  5  0  s   5

S=  5 it enters

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For the synchronous sequential circuit shown below, the output Z is zero for the initial conditions QA QBQC  Q'A Q'BQ'C  100.

The minimum number of clock cycles after which the output Z would again become zero is ________ Key: 6 to 6 Exp: Upper part of the circuit is ring counter and lower part of the circuit is Johnson counter as per the connection established. Outputs of the Ring counter and Johnson counter is given to Ex-OR. Gates, whose output is given to the three inputs OR-gate. Ring counter output Ring counter output

Jonson counter

output

QA

QB

QC

Q1A

Q1B

Q1C

Z

1

0

0

1

0

0

0  Inital valume

0

1

0

1

1

0

1

1 CP

0

0

1

1

1

1

1

2 CP

1

0

0

0

1

1

1

3CP

0

1

0

0

0

1

1

4CP

0

0

1

0

0

0

1

5CP

1

0

0

1

0

0

0

6CP

So, output Z will become again 1 after 6 clock pulses.

38.

In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is

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(B) 1 F

(A) 1 nF Key: Exp:

(C) 1 mF

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(D) 10 mF

(D) To get the maximum power the load Ckt must be at resonance i.e. imaginary part of load impedance is zero.

1 R 1  jRC  R jC ZL  jL   jL   jL  1 1  jRC 1  2  R 2  C2 R jC j term  0 R

R 2 C R 2C C  L   5  103  2 2 2 2 2 2 1   R C 1  R C 1  104 C 2 From options C  10mF will satisfy the about equation  L 

39.

In the circuit shown all elements are ideal and the switch S is operated at 10 kHz and 60% duty ratio. The capacitor is large enough so that the ripple across it is negligible and at steady state acquires a voltage as shown. The peak current in amperes drawn from the 50 V DC source is ________. (Give the answer up to one decimal place.)

Key: 39 to 41 Exp: Given is Buckboost converter DVS V0  1 D Given Vs  50V, D  0.6, Vo  75V

Vo Is D 0.6    1.5 Vs Io 1  D 1  0.6 Vo 75  15A R 5 D 3 Is  . Io  15  22.5A 1 D 2 Since capacitor is very large, ic  0 Io 

i L avg  is avg  i o avg I L  Is  Io  22.5 15  37.5A IL 

DVS 0.6  50   5A fL 10,000  0.6 103 

IL 5  37.5   40A 2 2 Peak current drawn from source is 40A  i L peak  IL 

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In the circuit shown in the figure, the diode used is ideal. The input power factor is _______. (Give the answer up to two decimal places.)

Key: 0.70 to 0.71 V Exp: Vor  m 2

Vo

Vm  2VS

Vm Vor Vm Ior   R 2R  0 2 VS Vm 2  VS  2 2 2 Vor Ior Vor PLoad VS 1 IPF       0.707 InputVA VS Ior VS 2 VS 2

41.

3



Consider the system described by the following state space representation    x1  t     0 1   x1  t     0  u  t     0 2   x  t   1   2   x 2  t    x1  t   y  t   1 0    x 2  t    x1  0    1  If u  t  is a unit step input and      , the value of output y  t  at t = 1 sec (rounded  x 2  0   0 

off to three decimal places) is_________ Key: 1.280 to 1.287

0 1  Exp: Given A    0 2  0  B  1  C  1 0  s  2  1 s  2 1    0    0 s   1   0  s 1  1 X  s    SI  A   x  0   Bu  s     2  1 / S    s  2s  0  1  s  s  2   s  

1   s  2   s    1  ; x s  s s  2

y s  

s s  2  1 s2  s  2 

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1 1 1 1 1 1 y s   2    2 s s  s  2  s 4s 2s 4 s  2 3 1 1 3 1 1 y  l   4  t   t4  t   e 2t 4  t     e 2 4 2 4 4 2 4 y 1  1.2838 42.

A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has following equivalent circuit parameters per phase referred to the stator: R1  0.3,R 2  0.3,X1  0.41,X2  0.41. Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3phase AC source is __________.

Key: Exp:

69 to 71 Reactance offered by stator and Rotor will be changes, because of change in frequency. R1

20  0.41  0.13666  60 20 x 2   0.41  0.1366 60 Vph Ist  I ph  Isc  Zeq x1 



0.3

R2 0.3

80 v, 20Hz 3

X1

X2

0.4

0.4

Ist  Isc

80 3 46.18   0.3  0.3  j  0.1366  0.1366   0.65

 71.4 24.46o Amps

A 25 kVA, 400 V,  - connected, 3-phase, cylindrical rotor synchronous generator requires a field current of 5 A to maintain the rated armature current under short-circuit condition. For the same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is _________.

43.

Key:

-15 to -14

Exp:

25kVA, 400V,   connection Voc  360V

25  103  36.084 Amps 3  400 I ph  20.833 Amps IL 

ISC  I rated , If  5A Xs  Zs  E

Voc phase 360   17.28 Isc phase 20.833

 V cos   Ia R a 

2

  Vsin   Ia Xs   2

 400  0.8  0 

2

  400  0.6  20.833  17.28 

2

E ph  341.758 volts 341.758  400  14.6% 400 Hint: Obtained regulation should be negative. % Reg 

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If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA. Y   transformer (the per phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is_____.

44. Key:

37 to 39

Exp:

25KVA, Y  D, 3.3kV   N1 : N 2  5 :1

25  103 3  3.3  103 IL  Iph  4.374 Amps Ist 

Transformer is a constant-Power device

E 2 I2  E1I1 N 2 I2  N1I1  I2 

N1 5 .I1   4.374 N2 1

Iph  I2  21.869 Amps    side IL  3  I2  3  21.869  37.879 Amps

45.

For the balanced Y-Y connected 3-Phase circuit shown in the figure below, the line-line voltage is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading.

The approximate value of the impedance Z is (A) 33  53.1o 

(B) 6053.1o 

(C) 60  53.1o 

(D) 180  53.1o 

Key:

(C)

Exp:

VL  208V, P  432W

cos   0.6 leading P  3 Vph  I ph  cos  2

 208  3  2   0.6 Vph 3  P  3. .cos z ph   60.08 z ph 432 cos 1 0.6  53.1o z ph  60 53.1o z ph 

Vph o

I ph 53.1



Vph Iph

53.1o

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A thin soap bubble of radius R = 1 cm, and thickness a  3.3m  a  R  , is at a potential of 1

46.

V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap 4 in the thin bubble is 4R 2a and that of the drop is r 3 . The potential in volts, of the resulting 3 single spherical drop with respect to the same reference point at infinity is __________. (Give the answer up to two decimal places.)

Key: Exp:

9.50 to 10.50 Charge must be same  4R 2a  P   43 r3  P

r

3

3R 2 a

0.996  103 The potential of thin bubble is 1 V Q 1 4E 0  1  102 Q  40  102 C

Potential of Soap drop Q V 40 r 

40  102 40  0.9966  103

 10.03V 47.

The value of the contour integral in the complex-plane

z 3  2z  3  z  2 dz Along the contour |Z| = 3, taken counter-clockwise is (A) 18i (B) 0 (C) 14i Key: (C)

Exp:

(D) 48i

z = 2 is the singularity lies inside C : z  3   C

z3  2z  3 dz  2i  z3  2z  3  14i z2 z2

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1  x x  0  x x  1 andf  x    2 Let g  x    x  1, x  1 x , x  0

48.

Consider the composition of f and g, i.e.,  f  g  x   f  g  x . The number of discontinuities in  f  g  x  present in the interval  ,0  is: (A) 0

(B) 1

(C) 2

(D) 4

Key:

(A)

Exp:

1  x, x  1 Clearly  fog  x    2 is discontinuous at x  1  ,0   x , x 1

 The number of discontinuities in (fog) (x) present in the interval  , 0  is 0 Alternative Method:

f  x   1  x for x  0 and g  x    x for x < 0

 Both f(x) and g(x) are continuous when x < 0   fog  x  is also continuous for x < 0 (Since the composite function of two continuous functions is continuous)

 The number of discontinuities in the interval  ,0  i.e., x < 0 is „0‟ 49.

A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8  , and the shunt field resistance is 240 . The no load speed, in rpm, is _______________.

Key: 1235 to 1250 Exp:

2A

0.5

2A

0.5

0.8

240

0.8

120

240

120

1.5A

6.5

E b1

N  Eb

No load, N1  ?

Eb2 Full load, N2  1200rpm

 120  1.5  0.8  N1 E b1   N1     1200  1241.82 rpm N 2 E b2 120   6.5  0.8 

50.

A 10 ½ digit timer counter possesses a base clock of frequency 100 MHz. When measuring a particular input, the reading obtained is the same in: (i) Frequency mode of operation with a gating time of one second and (ii) Period mode of operation (in the x 10 ns scale). The frequency of the unknown input (reading obtained) in Hz is _______. Key: 10000 to 10000 51.

A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var Y , where var . denotes the variance, equals  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India

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7 8

(B)

49 64

(C)

7 64

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(D)

105 64

Key: (C) 52.

The figure below shows a half-bridge voltage source inverter supplying an RL-load with  0.3  R  40 and L    H. The desired fundamental frequency of the load voltage is 50 Hz. The    switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index. M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage VDC in volts is

(B) 500

(A) 300 2 Key: (C)

(C) 500 2

(D) 1000 2

0.3  30  z  R L  jx L  40  j30  50 36.86

Exp: x L  L  100  M  0.6 I Load 

PL 1440   6A RL 40

VA o1 VA o1  6 z1 50 VA o1  300V  rms  I Load 

VAo1  300 2 V  max  VAo1  m.vdc 300 2  0.6  vdc  Vdc  500 2 The range of K for which all the roots of the equation s3  3s2  2s  K  0 are in the left half of the complex s-plane is (A) 0 < K < 6 (B) 0 < K < 16 (C) 6< K < 36 (D) 6< K < 16 Key: (A) 53.

Exp: C.E  s3  3s2  2s  k  0 If system to be stable K  06  K 0K6

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The eigen values of the matrix given below are 0 0  0 (A)

1 0 0 1  3 4  (0,-1,-3)

(B) (0,-2,-3)

(C) (0,2,3)

(D) (0,1,3)

Key: (A) Exp:

 1 0 1 0 Characteristic equation is 0  0 3 4  

   4  2  3  0      1   3  0    0, 1, 3 are the eigen values 55.

A 3-phase 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are 0.25  and 3.925  respectively. If the voltage at the generator terminal is 17.87 kV, the power

factor of the load is ________. Key: 0.75 to 0.85 Exp: PS  3MW Z  3.9329 86.35  ph 17.87 E f  17.32 KV Vt   10317.249 V ph 3 Z   0.25  j3.925   ph E f  10.000 V ph V  17.87 KV t

2  90  86.35  3.65 2

V  EV Pog  f t sin     2    t  ra Zs  Zs  10000  10317.249  10317.249  sin    3.65      0.25 3.9329  3.9329    2.3024 2

106 

 I a Zs 

2

 E f 2  Vt 2  2E f Vt cos 

Ia  131.43 A / ph PS  3MW  3  17.87  103  131.43  cos  cos   0.737

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General Aptitude Q. No. 1 – 5 Carry One Mark Each 1.

There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to the West of W, Z is to the East of X and the West of V, W is to the West of Y. Which is the building in the middle? (A) V (B) W (C) X (D) Y Key: (A) Exp:

From the given data, the following is formed

X

Z

V

W

Y

N W

East

West

E S

 The building „V‟ is in the middle 2.

Saturn is _________ to be seen on a clear night with the naked eye. (A) enough bright (B) bright enough (C) as enough bright (D) bright as enough Key: (B) 3.

Choose the option with words that are not synonyms. (A) aversion, dislike (B) luminous, radiant (C) plunder, loot (D) yielding, resistant Key: (D) 4.

There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is (A) 1/5 (B) 7/30 (C) 1/4 (D) 4/15 Key: (D) 3

Exp:

C2  4 C2 3 C2 12 4   10 C2 45 15

5.

A test has twenty questions worth 100 marks in total. There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have? (A) 12 (B) 15 (C) 18 (D) 19 Key: (B) Exp: x  y  20 x  MCQ

3x  11y  100  x  15, y  5

y  essay type

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Q. No. 6 – 10 Carry Two Marks Each 6.

An air pressure contour line joins locations in a region having the same atmospheric pressure. The following is an air pressure contour plot of a geographical region. Contour lines are shown at 0.05 bar intervals in this plot.

If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region, which of the following regions is most likely to have a thunderstorm? (A) P (B) Q (C) R (D) S Key: (C) Exp: Region

Air pressure difference

P

0.95 – 0.90 = 0.05

Q

0.80 – 0.75 = 0.05

R

0.85 – 0.65 = 0.20

S

0.95 – 0.90 = 0.05

In general thunder storms are occurred in a region where suddenly air pressure changes (i.e.,) should rise (or) sudden fall of air pressure. From the given contour map in „R‟ region only more changes in air pressure. So, the possibility of a thunder storms in this region. So option (C) is correct. 7.

There are three boxes. One contains apples, another contains oranges and the last one contains both apples and oranges. All three are known to be incorrectly labelled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes? (A) The box labelled „Apples‟ (B) The box labelled „Apples and Oranges‟ (C) The box labelled „Oranges‟ (D) Cannot be determined Key: (B) Exp:

The person who is opening the boxes, he knew that all 3 are marked wrong. Suppose if 3 boxes are labelled as below.

(1) Apples

(2) Oranges

(3) Apples & Oranges

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If he inspected from Box(1), picked one fruit, found orange, then he don‟t know whether box contains oranges (or) both apples and oranges. Similarly, if he picked one fruit from box(2), found apple then he don‟t know whether box contain apples (or) both apples and oranges. But if he picked one fruit from box(3), i.e., labelled is “apples and oranges‟, if he found apple then he can decide compulsorily that box(3) contains apples and as he knew all boxes are labelled as incorrect, he can tell box(2) contains both apples and oranges, box(1) contain remaining oranges. So, he should open box labelled „Apples and Oranges‟ to determine contents of all the three boxes. “We lived in a culture that denied any merit to literary works, considering them important only when they were handmaidens to something seemingly more urgent – namely ideology. This was a country where all gestures, even the most private, were interpreted in political terms.”

8.

The author‟s belief that ideology is not as important as literature is revealed by the word: (A) „culture‟ (B) „seemingly‟ (C) „urgent‟ (D) „political‟ Key: (B) X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number X3 will have (A) 90 digits (B) 91 digits (C) 92 digits (D) 93 digits Key: (A) 9.

Exp:

X   47........... 30 digits Suppose  47   2  2  2 digits in (47)3 3

Similarly  47   contains 30 + 30 + 30 digits = 90 digits. 3

The number of roots of ex  0.5x 2  2  0 in the range  5,5 is

10.

(A) 0 Key: (C) Exp:

(B) 1

(C) 2

(D) 3

f  x   e x  0.5x 2  2

f  5  10.50; f  4   6.01, f  2   0.135; f  1  1.13; f  0   1, f 1  1.21, f  2   7.38, f  3 , f  4  , f  5  also  ve.

 As there are 2 sign changes from +ve to –ve and –ve to +ve, two roots will be there in the range [-5, 5].

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