Vpm Classes _free Solved Expected Paper _ Gate 2013 _ Electrical Engg

7/30/2019 Vpm Classes _free Solved Expected Paper _ Gate 2013 _ Electrical Engg

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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.

Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05

Page 1

For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exams

GATE - ELECTRICAL ENGINEERING

1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714

Web Site www.vpmclasses.com [email protected]

MOCK TEST PAPER

Pattern of questions : MCQs

There are a total of 65 questions carrying 100 marks.

Questions (56-65) belongs to general aptitude (GA).

Questions (56-60) will carry 1-mark each, and

question (61-65) will carry 2-marks each

l

l

l

Total marks : 100

Duration of test : 3 Hours

l

l

l

Questions (1-25) will carry 1-mark each and

questions (26-55) will carry 2-marks each.

For Q.1-25 and Q.56-60 1/3 mark will be deducted

for each wrong answer.For Q.26-51 and Q. 61-65 2/3

mark will be deducted for each wrong answer.

The question pairs (Q.52, Q.53) and (Q.54, Q.55)

are linked questions.For Q.52 &54 2/3 mark will be

deducted. There is no negative marking for Q.53 &Q.55.

l

Q.48-51 are common data questions.

If first question is attempted wrongly then answer of

second question will not be evaluated.

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1 In the matrix equation Px = q, which of the following is a necessary condition for the existence of at

least one solution for the unknown vector x?

(A) Augmented matrix [Pq] must have the same rank as matrix P

(B) Vector q must have only non - zero elements

(C) Matrix P must be singular

(D) Matrix P must be square

2. An arbitrary vector X is an eigen vector of the matrix A =

1 0 0

0 a 0

0 0 b

, if (a, b) =

(A) (0, 0)

(B) (1, 1)

(C) (0, 1)

(D) (1, 2)

3. The integration of logx.dx has the value

(A) (x log x 1)

(B) log x x

(C) x (log x 1)

(D) None of these

4. If f(x) = | x |, then the interval [1, 1], f(x) is

(A) Satisfied all the conditions of Rolles Theorem

(B) Satisfied all the conditions of Mean Value Theorem

(C) Does not satisfied the conditions of Mean Value Theorem

(D) None of these

5. Differential equation,

2

2

d x dx10 25x 0

dt dt

will have a solution of the form

(A)C1

+ C2

t)e5t

(B) C1 e2t

(C) C1

e5t + C2

e5t


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(D) C1 e5t + C2

e2t

Where C1

and C2

are constants.

6. For the differential equationdy

5y 0dt

with y(0) = 1, then general solution is

(A) e5t

(B) e5t

(C) 5e5t

(D) 5te

7. For | z | = 1, where C is the circle, is

(A)

(B)

(C)

(D) None of these

8. If A and B are independent and P (C)= 0, t hen A, B and C are independent

(A) True

(B) False

(C) Both (a) and (b)

(D) None of these

9. Following are the value of a function

y(x): y(1) = 5, y(0), y (1) = 8

dy

dxat x = 0 as per Newtons central difference scheme is

(A) 0

(B) 1.5

(C) 2.0

(D) 3.0

10. For any two events A and B

(A) P(B) = P(A B) + P( A B)

(B) P(A B) = P(A) +P(B) P(A B)


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(C) P(A/B) P(A).

(D) All of these

11. A unit step current to a network consisting of only passive elements. The voltage across the current

source observed is v(t) = (1 + e-t). The simplest possible network will consist of the elements

(A) 1 resistor and 2 capacitors

(B) 1 resistor and 2 inductors

(C) 2 resistors and 1 capacitor

(D) 2 resistor and 1 inductor

12. The energy stored in the magnetic field at a solenoid 30 cm long and 3 cm diameter wound with 1000

turns of wire carrying a current at 10 A, is

(A) 0.015 Joule

(B) 0.15 Joules

(C) 0.5 Joule

(D) 1.15 Joules

13. The current wave from in a pure resistor at 10 is shown in the given figure. Power dissipated in the

resistor is

(A) 7.29 W

(B) 52.4 W

(C) 135 W

(D) 270 W

14. A series RLC circuit consisting of R = 10 ohms, XL

= 20 ohms and XC

= 20 ohms is connected across

an ac supply of 100 V (rms). The magnitude and phase angle (with reference to supply voltage) of the

voltage across the inductive coil are respectively,

(A) 100 V, 90

(B) 100 V, - 90

(C) 200 V, 90

(D) 200V, 90


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15. When a unit impulse voltage is applied to an inductor of 1 H, the energy supplied by the source is

(A)

(B) 1 J

(C)1

J2

(D) 0

16. The resonant frequency of the given series circuit is

(A)1

Hz2 3

(B)1

Hz4 3

(C)1

Hz4 2

(D)1

Hz2

17. The final value of L-1

4 3 2

2s 1

s 8s 16s s

is

(A) Infinity

(B) 2

(C) 1

(D) Zero

18. If (t) denotes a unit impulse, then the laplace transform of 22

d d t

dtwill be

(A) 1

(B) s2


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(C) s

(D) s-2

19. The unit step response of a system is given by (1 e) u (t)

(A)e u (t)

(B) e u (t)

(C) 1

e u t

(D) e u (t)

20. If f(s) = / (s2 + 2) , then the value of dtLt

f(t)

(A) Cannot be determined(B) Is zero

(C) Is unity

(D) None of these

21. The function f(t) has a Fourier transform g(w). The Fourier transform

j tf t g t g t e dt,is

(A) 1

f

2

(B) 1

f2

(C) 2 f

(D) None of these

22. If f (t) and f (t) satisfy the Dicrichlets conditions, then f(t) can be expanded in a Fourier series

containing

(A) Only sine terms

(B) Only cosine terms

(C) Cosine terms and constant terms

(D) Sine terms and a constant term


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23. A single - phase starter winding when excited with ac voltage produces

(A) A single rotating field rotating as synchronous speed.

(B) Two rotating fields rotating at synchronous speed in opposite directions.

(C) Two rotating fields rotating in the same direction but at different speeds

(D) Two rotating fields rotating in opposite directions but with difference speeds.

24. The mmf of a sinusoidally distributed current sheet is

(A) Triangular shifted from the current by 90

(B) Triangular is phase with the current wave

(C) Sinusoidal shifted from the current wave by 90

(D) Sinusoidal in phase with the current wave

25. In a synchronous machine, the voltage induced by armature reaction flux acts like the voltage drop in

(A) Inductive reactance

(B) Resistance

(C) Capacitive reactance

(D) Inductive impedance

26. Synchronous reactance is

(A) The difference of armature leakage reactance and reactance equivalent of armature reaction

(B) The same as armature leakage of armature reaction

(C) The reactance equivalent of armature reaction

(D) The sum of armature leakage reactance and reactance equivalent of armature reaction

27. A synchronous motor is floating on infinite mains at no load. If its excitation is now increased, it will

draw

(A) Unity power factor current

(B) Zero power factors lagging current

(C) Zero power factor loading current

(C) No current

28. Voltage regulation of a synchronous generator calculated by synchronous impedance method is

(A) Higher than actual because of saturation of magnetic circuit

(B) Lower than actual because of saturation of magnetic circuit

(C) Nearly accurate because it takes account of magnetic circuit.

(D) Nearly accurate because the generator is normally operated with an unsaturated magnetic circuit.


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29. A two conductor 1 - line operates at 50 Hz. The diameter of each conductor is 4 cm and is spaced 6

m apart. What is the capacitive susceptance to neutral per km?

(A) 1.42 10-9 s/km

(B) 3.25 10-8 s/km

(C) 4.8 10-9 s/km

(D) 3.06 10-9 s/km

30. A two conductor 1 - line operates at 50 Hz. The diameter of each conductor is 20 mm and the

spacing between the conductors is 3m. The height of conductor above the ground is 6 m. The

capacitance of the line to neutral will be

(A) 9.7 pF/m

(B) 10.2 pF/m

(C) 8.7 F/km

(D) 2.4 F/m

31. A three phase line has conductor of 5 mm diameter placed at the corner of an equilateral triangle of

1.5 m side. The capacitive reactance to neutral per phase per km will be

(A) 1.25 106 /km

(B) 8.68 105/km

(C) 4.25m/ km(D) 3.66 105 /km

32. Three insulating materials with same maximum working stress and permittivities 2.5, 3.0, 4.0, are

used in a single core cable. The location of the materials with respect to the core of the cable will be

(A) 2.5, 3.0, 4.0

(B) 3.0, 2.5, 4.0

(C) 4.0, 3.0, 2.5

(D) 4.0, 2.5, 3.0

33. The incremental generating costs of two generating units are given by

IC1

= 0.1 X + 20 Rs /MWhr.

IC2

= 0.15 Y + 18 Rs/ MWhr.


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Where X and Y are power (in MW) generated by the two units. For a total demand of 300 MW. The

values (in MW) of X and Y will be respectively

(A) 172 and 128

(B) 128 and 172

(C) 175 and 125

(D) 200 and 100

34. For protection of rotating machines against lighting surges a combination of

(A) Lighting conductor and capacitor are used

(B) Lighting conductor and lighting arrester are used

(C) Lighting arrester and capacitor are used

(D) Lighting arrester alone is used

35. For the system shown, the transfer function C(s) / R(s) is equal to

(A)2

10

s s 10

(B) 2 10s 11s 10

(C)2

10

s 9s 10

(D)2

10

s 2s 10

36. In the following block diagram 1 2 1

10 10G : G ;H s 3

s s 1

and H

2= 1.

The overall transfer function C/R is given by


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(A)2

10

11s 31s 10

(B)2

100

11s 31s 100

(C)2

100

11s 31s 10

(D)2

100

11s 31s

37. Consider a unity feedback control system with open -loop transfer function

K

G(s)s s 1

The steady state error of the system due to a unit step input is

(A) Zero

(B) K

(C) 1 / K

(D)

38. For a gain constant K, the phase - lead compositor

(A) Reduces the slope of the magnitude curve in the entire range of frequency domain

(B) Decreases the grain cross - over frequency

(C) Reduce the phase margin

(D) Reduce the resonance MP

39. A control system is as shown in the given figure. The maximum value of gain K for which the system

is stable is

(A) 3

(B) 3

(C) 4

(D) 5


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40. Straight line asymptotic Bode magnitude plot for a certain system is shown in the given figure. What

will be its transfer function?

(A)4s

s s1 1

4 10

(B)2

s4 1

4

s1

10

(C)0.25

s s1 1

4 10

(D)2

0.25

s s1 1

4 10

41. A compensated wattmeter has its reading corrected for error due to the

(A) Frequency

(B) Friction

(C) Power consumed in current coil

(D) Power consumed in potential coil

42. The sensitivity of an instrument is the

(A) Smallest increment in the input that can be detected with certainty

(B) Largest input change to which the instrument fails to respond

(C) Ratio of the change in the magnitude of the output to the corresponding change in the magnitude

of the input

(D) Closeness of the output values for repeated applications of a constant input.


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43. If a dc voltmeter is made from an ammeter having a full - scale deflection of 100 micro amperes, then

its sensitivity (in k - ohm. (V) Will be

(A) 1

(B) 10

(C) 100

(D) 1000

44. What of the following are advantages of instrument transformers?

(A) The readings of instruments used in conjunction with them do not depend upon their resistance,

inductance etc.

(B) The readings of instruments transformers have been standardized and the rating of instruments

used in conjunction used with them also get standardized. Therefore, there is reduction of cost

and ease in replacements.

(C) The metering circuit is electrically isolated from the power circuit thereby providing safety to

operating personnel.

(D) All of these

45. The limiting error of measurement of power consumed by and the current passing through a

resistance are 1.5 % and 1 % respectively. Then the limiting error of measurement of resistance

will be.

(A) 0.5 %

(B) 1.0 %

(C) 2.5 %

(D) 3 .5 %

46 The braking torque provided by a permanent magnet in a single phase energy meter can be changed

by

(A) Providing a magnetic shunt and changing its position

(B) Changing the distance of the permanent magnet from the center of the revolving disc


(D) None of these


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47. The power in d.c. circuit is measured with the help of ammeter and a voltmeter. The voltmeter is

connected on the load side. The power indicated by the produced to reading of two instruments (VI) is

(A) The power consumed in the load

(B) The sum of power consumed by load and the voltmeter

(C) The sum of power consumed by load and the ammeter

(D) None of these

Linked Answer Q. 48-49

The first and the second stage of a two stage RC coupled amplifier have the lower cut off frequencies

to be 100 Hz and 200 Hz respectively. Their upper cuts off frequencies are 140 kHz and 100 kHz

respectively.

48. Overall lowed cut-off frequency will be

(A) 212Hz

(B) 232 Hz

(C) 238 Hz

(D) 242 Hz

49. Overall 3 dB band-width of the amplifier will be

(A) 74.236 kHz

(B) 74.362 kHz

(C) 74.6 kHz

(D) None of these

Linked Answer Q. 50-51

The input voltage Viin the circuit shown in the given figure is a 1 kHz since-wave of 1V amplitude.

Assume ideal operational amplifiers with 15 V DC supply.


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50. The peak value of V1

will be

(A) 1.06 V

(B) 1.33 V

(C) 2.33 V

(D) 3.33 V

51. The average value of Vo

will be

(A) 1.06 V

(B) 1.33 V

(C) 2.33 V

(D) 3.33 V

Common Data for Q (52-53)

A separately excited DC motor runs at 1500 rpm under no - load with 200 V applied to the armature.

The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of

5 Nm, is 1400 rpm as shown in the figure. The rotational losses and armature reaction are neglected.

52. The armature resistance of the motor is

(A) 2

(B) 3.4

(C) 4.4

(D) 7.7

53. For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage to be applied is

(A) 125.5 V

(B) 193.3 V

(C) 200 V

(D) 241.7 V


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Common Data Question 54 55

Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50W and

a resistive load is shown in the figure.

l

1

4

Load

|V(z)|

54. Value of the load resistance is

(A) 50

(B) 200

(C) 12.5

(D) 0

55. Reflection coefficient is

(A) 0.6

(B) 1

(C) 0.6

(D) 0

56. No doubt, it was our own government but it was being run on borrowed ideas, using

_________solutions.

(A) Worn out

(B) Second hand

(C) Impractical

(D) Appropriate


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The question below consists of pair of related words followed by four pairs of words. Select the pair that

best expresses the relation in the original pair:

57. Ratify: Approval:

(A) Mutate: change

(B) Pacify: conquest

(C) Duel: combat

(D) Appeal: authority

58. A car goes 35 km in 1 hour, next 270 km in 3 hrs. and next 80 km in1

22

hrs.Find the average speed

of the car

(A) 59.23 km/h.(B) 61.5 km/h

(C) 80 km/h

(D) None of these

59. Some critics believe that Satyajit Ray never quite came back to the great beginning he made in this

path breaking film Pather Panchali. ______have endured decades of well-traveled bad prints to

become a signpost in cinematic history.

(A) The bizarre history of its misty origins

(B) Its haunting images

(C) Its compelling munificence

(D) The breathtaking awe it inspires

Choose the most appropriate word from the options given below that is most nearly opposite in

meaning to the given word:

60. Valedictory

(A) Sad

(B) Collegiate

(C) Derivative

(D) Generosity

Each of the 11 letters A, H, I M, O,T, U V, W, X and Z appears same when looked at in mirror. They

are called symmetric letters. Other letters in the alphabet are asymmetric letters.


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61. If the area of a given square ABCD is 3 find the total area of the entire figure?

(A) 452

(B) 45

(C) 48

(D) 31

62. The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the amino

terminal portions of both the H and L chains.

(A) H chain(B) L chain


(D) None of these

63. In a certain code Language

134 means good and tasty

478 means see good picture

729 means picture are faint

Which number has been used here for faint?

(A) 9

(B) 2

(C) Data are inadequate

(D) 253

64. A bag contains an equal number of one rupee, 50 paisa and 25 paisa coins. If the value of money in

the bag is Rs. 35, find the total number of coins of each type?

(A) 7

(B) 40

(C) 30

(D) 20


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Profit to sale-Table for three companies A, B and C for 1996-97

Companies 1996 1997Total units 300000 400000

Shares A 5% 25%

B 60% 40%

C 35% 35%

Price A 10% 8%

(per unit) B 7% 14%

(in rupees) C 9% 10%

Profit A 3% 1%

(per unit) B 0.5 5%

(in rupees) C 2% 2.5

65. What is the increase in the total profits of company B in 1997?

(A) 800%

(B) 900%

(C) 750%

(D) 789%


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ANSWER KEY

Que Ans Que Ans Que Ans Que Ans Que Ans

1 A 16 B 31 D 46 C 61 B

2 B 17 C 32 C 47 D 62 C

3 C 18 A 33 A 48 C 63 C

4 C 19 B 34 C 49 B 64 D

5 A 20 B 35 B 50 D 65 D6 B 21 C 36 B 51 A

7 A 22 A 37 A 52 B

8 A 23 B 38 A 53 B

9 B 24 C 39 D 54 B

10 D 25 A 40 D 55 C

11 D 26 D 41 D 56 B

12 B 27 C 42 D 57 C

13 D 28 A 43 B 58 A

14 D 29 D 44 D 59 B

15 C 30 A 45 C 60 D

Hints and Solution

1(A) According to Rouches theorem, the system is consistent if and only if the coefficient matrix and the

augmented matrix K are of the same rank, otherwise the system is inconsistent.

2.(B) Since the matrix is triangular, the eigen values are , a, b.

If (X1, X

2, X

3) is an arbitrary eigen vector, say corresponding to 1, then


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1 1

2 2

3 3

1 0 0 x x

0 a 0 x 1 x

0 0 b x x

X2, X3 being not zero, we have, X1 = X1 ; a X2 = X2 which gives

a = 1

and bX = X3

which gives b = 1

(a, b) = (1, 1).

3.(C)d

logx.dx logx.x x. (logx)dxdx

= x log x 1. dx

= x log x x

= x (log x 1)

4.(C) Since f(x) = | x | is continuous is [1, 1] be it is not diff erentiable at x = 0 (1, 1)

5.(A)2

2

d x dx10 25x ]0

dt dt

(D2 + 10D + 25) X = 0

(D + 5)2 = 0

D = -5, 5

Hence solution is, (C1

+ C2t)e5t

6.(B) Given:dy

5y 0dt

dy

5 dty

Integrating, we get

loge

y = c 5t

When t = 0, y = 1.

loge1 = c 5 0

c = 0


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logey = 5t

y = e5t

7.(A) Poles of f(z) =2

z 3

z 2z 5

are given by

z2 + 2z + 5 = 0

z =2 4i

1 2i2

Since, both poles lie outside the circle | z | = 1, therefore f(z) is analytic inside the circle

2

z 3dz 0

z 2z 5

8. (A) P(C) = 0

C =

P(A B C) = P (A B )

= P() = 0

P(A) P(B) P(C) = 0 .....Since P(C) = 0

P(A B C) = P (A) P (B) P (C)

Hence A, B, C are independent.

9. (B) 2 1

2 1at x 0

dy y y

dx x x

=y(1) y( 1) 8 5

1.51 ( 1) 2

10.(D) (a) B = (A B) ( A B)

P(B) = P(A B) + P( A B)

A B and A B are mutually exclusive.


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P ( A B) = P(B) P (A B).

(B) P (A B) = P[A ( A B)]

= P (A) + P( A B)

= P (A) + P(B) P(A B)

(C) P (A/B) = P(A) P (B/A)

P (A) as P(B/A) 1.

11. (D) z(s) =

V s

I s

=1 1

s1s s

=s

11

s

12. (B) 6 7 42

0

10 4 10 9 10N A 4L

l 0.3

=2 59 10

0.3

Energy = 21

LI2

= 0.148 or 0.15 Joules

13. (D)

323 32 2rms

0 0

1 9 1 tI t .dt 9 27A

3 3 3 3

Power = I2R = 27 10 = 270 W

14. (D) Since the reactance are canceling each other, the circuit is purely resistive and the current is phase

with the applied voltage. The voltage across the inductor leads the current through it by 90

Since I = 100 0 / {10 + j (20 - 20) } = 10 0

LV = jL I = 20 10 90 = 200 90


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15. (C) Current that flows is given by,

V(s) = sL1I (s)

I = s I (s) I (s) =1

s= u (t) = 1 A.

Energy supplied =1 1

1 1 J.2 2

16. (B) Leq

= L1

+ L2

+ 2M = 6H

At resonance, Leq

=1

C

or =eq

1

L .C

=1

12

f =1

Hz.4 3

17.(C)

4 3 2s s 0 s 0

s 2s 1Limf t LimsF s Lim

s 8s 16s s

3 2s 0

2s 1Lim s 8s 16s s

18.(A) Since (t) = 1

19.(B) Impulse response = td

1dt

te

20.(B) 22 2f s 0s

Lt f t Lt sF(s) 0s

s s

21.(C) Since g() = j t1

g e dt2


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22.(A) Condition f (t) = -f(-t) implies that it has rotational symmetry.

23.(B) a single - phase starter winding when excited with ac voltage produces two rotating fields rotating at

synchronous speed in opposite directions.

24. (C) The mmf of a sinusoidally distributed current sheet is sinusoidal shifted from the current wave by 90.

25.(A) In a synchronous machine, the voltage induced by armature reaction flux acts like the voltage drop in

Inductive reactance

26. (D) Synchronous reactance is the sum of armature leakage reactance and reactance equivalent of

armature reaction

27.(C) A synchronous motor is floating on infinite mains at no load. If its excitation is now increased, it will

draw no current

28. (A) Voltage regulation of a synchronous generator calculated by synchronous impedance method is

Higher than actual because of saturation of magnetic circuit.

29.(D) Here, D = 6m, r = 2 cm = 2 10-2 M

Cn

=

0

9

2

2 16In D / r

18 10 In2 10

=9

1

318 10 In

0.01

= 9.74 10-9 F/km

Capacitive susceptance to neutral

bc

= n1

2 F.C

Xc

= 2 50 9.74 10-9 s/km

= 3.06 10-9 s / km


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30.(A) Cn

= 0

m

s

2

D HIn In

r H

Here, r = 10 mm = 10 10-3 m,

D = 3 m, h = 6 m,

H1

= H2

= 2h = 12 m

H12

= H21

= (D2

+ 4 h2)1/2

= (32 + 4 62)1/2 = 153

Hm

= (H12

. H21

)1/2 = H12

= 153 m

Hs= (H

1. H

2)1/2 = (2h. 2h)1/2 = 2h = 12m

Cn

= 0

3

2

3 153In In10 10 12

=9

2

1

3 1218 10 In

10 153

= 9.7 10-12 F/m = 9.7 pF/m

31.(D) Here, d = 5 mm, r = 2.5 mm = 2.5 10-3 m,

D = 1.5 m

Deq = (DAB. DBC. DCA)1/3

= (1.5 1.5 1.5)1/3 = 1.5 m

Cn

= 0

eq

2

DIn

r

=9

3

1

1.518 10 In

2.5 10

=

39

1

1.5 1018 10 In

1.5

= 8.68 10-12 F/ m

= 8.68 10-9 F/km


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Page 26

= 8.68 10-3 F/km

Capacitive reactance,

c 9n1 1

X2 f C 2 50 8.68 10

= 3.66 105/km

32.(C) When all the three materials are subjected to the same maximum stress,

max

1 2 2 3 2

g2 r 2 r 2 r

or 1

r = 2

r1

= 3

r2

Since r < r1

< r2,

Therefore 1

> 2>

3

Thus, the dielectric material with highest permittivity should be placed near the conductor and other

layer in the descending order.

33.(A) For most economic load sharing

I C1

= IC2

We have X + Y = 300 MW

and 0.1 + 20 = 0.15 Y + 18

= 0.15 (300 - X) + 18

= 45 - 0.15 + 18

025 X = 43or X = 172 MW

Y = 300 - 172 = 128 MW


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Page 27

34.(C) The rotating machine should be protected against major and minor insulation. Major insulation is the

insulation between the winding and minor insulation means inter- turn insulation. The major isnulation

is determined by line to ground voltage across the terminal of the machine whereas minor insulation

is determined by the rate of rise of voltage. Hence to protect the rotating machine against surges

requires limiting the surge voltage magnitude at the machine terminals and sloping the wave front of

the incoming surge. To protect the major insulation, a special lighting arrester and to protect the minor

insulation a capacitor are required and connected as depicted in the give figure.

35.(B)

2

10

10s(s 1)G s

10 s 11s1 s

s s 1

2

2

2

10C s 10s 11s10R s s 11s 10

1s 11s

36.(B) Successive block diagram reduction can be


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Page 28

1 2

2 1

1 2 2

2 1

G G

C s 1 G H

G G HR s1

1 G H

= 1 2

2 1 1 2 2

G G

1 G H G G H

10 10.

C s s s 1

s 3R s 10 101 10 . .1

s 1 s s 1

=2

100

11s 31s 100

37.(A) Unity feedback control system is shown as below :

Steady state error due to unit step.

Since it is a type 1 system, hence steady as below:

Error of system due to unit step input is zero.

38.(A) Lead compensator increases the gain - crossover frequency. It also increases the phase margin. The

high frequency end of the log magnitude plot has been raised by a dB gain of 20 log 1( ).2

39.(D) The characteristics equation is

1 + GH = 0

or 1 +3 2

K0

s 3s 2s 1

or3 2

3 2

s 3s 2s K0

s 3s 2s 1

Hurwitz criterion can be applied.

s3 1 25 K

3

and (1+ K), both should be positive


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Page 29

s2 3 1+Ks1 6 1 K3

0

K should be < - 1 and less than 5

s 1 K

40.(D) Initially plot has a slope of + 20 d/B decade. So there must be zero at origin.

At = 4, slope changes to 0 dB/ decade. so pole at = 4.

Again slope at = 10 changes to - 40 dB/decade.

So there are two poles at = 10.

Now, 20 lop |K| = 0 at = 4.

Which gives, K = 0.25.

41. (D) A compensated wattmeter has its reading corrected for error due to power consumed in potential coil.

42. (D) Voltage across R1, V

1= 35 - 5 = 30 V

Current in the circuit,

I =5 5

A600 1200 400

600 1200

Rx =30 400

5

= 2.4k


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Page 30

43.(B) Sensitivity =Total of resistance of the inst

ohm./VFull scale current figure

=6

110K ohm / V

10 10

44.(D) All are correct option.

45.(C) M1

= 20 0.001 = 0.02,

Relative error at 1 A =0.02

2%1

M2

= 10 0.002 = 0.02,

Relative error =0.02

2%1

M3

= 5 0.005 = 0.025,


2.5%1

M4

= 1 0.01 = 0.01,


1%1

46. (C) The braking torque provided by a permanent magnet in a single phase energy meter can be changedby Providing a magnetic shunt and changing its position & Changing the distance of the permanent

magnet from the center of the revolving disc

47.(D) All of these are incorrect.

48. (C) Given,

fL1

= 100 Hz and fL2

= 200 Hz

fH1

= 140 KHz and fH2

= 100 KHz

Let fL

= Overall lower cut off frequency,

and fH

= Overall upper cut-off frequency


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Page 31

Then

1 2

2 2

H H

H H

1 1 1

2f f

1 1f f

or 2 = 1 +

1 2 1 2

42HH2 2 2 2

H H H H

f 1 1f

f . f f f

or fH

4 + 2 21 2H H

(f f ) fH

2 2 21 2H H

f . f 0

or fH

4 = (1402 + 1002) fH

2 (14000)2 = 0

fH

2 = 29600 2 829600 4 (1.96 10 )

2

= 5572

fH = 74.6 kHz

1 2

2 2

L L

L L

1 1 1

2f f1 1

f f

or 2 = 1 + 1 2 2 21 2

2 2L L

L L4 2L L

f f 1(f f ) 0

f f

or2 2 2 24 2

L L L L L L1 2 1 2f f f f (f f ) 0

or fL4 f

L2 (10000 + 40000) 1002 2002 = 0

or fL2 = 50000

2 2 250000 4 (100 200 )

2

= 57015

fL= 238 Hz

49. (B) 3 dB bandwidth = 74.6 kHz 238 Hz

= 74.362 kHz


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Page 32

50. (D) & 51. (A)

Given, Vi= sin t

= 6283 rad/sec

First op amp works as half-wave rectifier, with amplification factor

=10 5

10 5

= 3.33

V1

= 3.33 sin t, < t < 2

= 0, 0 < t <

Second op-amp network as average defector,

Vo

=3.33

= 1.06 V

Peak value of V1

= 3.33 V

Average value of Vo= 1.06 V

51.(A)

1 2 1 1x y x x x x z

0 0 1 0 1 0 10 1 1 1 1 1 1

1 0 1 1 1 1 1

1 1 0 1 0 1 1


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Page 33

Then x y z

0 0 1

0 1 1

1 0 1

1 1 1

52. (B)

Given:

No load speed

N0 = 1500 rpm, V = 200V,

T = 5 Nm, N = 1400

Emf at no load,1b

E V 200

Now b11

2 b2

EN

N E

2b2 b11

N 1400E .E 200

N 1500

= 186.67 V

But B a

m

E IT


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Page 34

m2 N

60

a60* 186.67* I

5

2 * 1400

Ia

= 3.926amp

2 2b a a

V E I R

22

ba

a

V ER

I

=200 186.67

3.926

= 3.395 =3.4

53. (B) Given: T = 2.5 Nm, 1400 rpm

Now

2.5 =

Ia

= 1.963 amp

V = Eb

+ Ra

Ia

= 186.66 + 1.9634 3.4

= 193.34 V

54 & 55. (B, C)

Here vmax = 4, and Vmin = 1

VSWR = 4 = S

Reflection coefficient,

S 1 4 1 3K 0.6

S 1 4 1 5

Now, R 0

R 0

Z ZK

Z Z

3ZR + 150 = 5Z R 250

ZR = 200

56.(B) No doubt, it was our own government but it was being run on borrowed ideas, using second hand

solutions.


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Page 35

57.(C) Ratify: Approval:: duel: combat

58.(A) Total distance covered

= 35 + 270 + 80 = 385 km.

= 1 + 3 +1

22

hrs.

= 13/2 hrs

Average speed =D 385 2

T 13

= 59.23 km/hr.

59.(B) some critics believe that Satyajit Ray never quite came back to the great beginning he made in this

path breaking film Pather Panchali. Its haunting images have endured decades of well-traveled bad

prints to become a signpost in cinematic history.

60(D) Generosity is nearly opposite to Valedictory

61.(B) Count the number of squares in the figure and multiply it by 3 i.e. 45.

62(C) The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the Aminoterminal portions of both the H and L chains.

63.(C) 4 = good 7 = picture and 2 and 9 = are and faint respectively.

64.(D)X X X 7X

35 or1 2 4 4

= 35 or 7X

= 35 4 or X = 20

65(D) The increase in the total profits of company B in 1997 is 789%.


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Page 36

Please find below in Table the details of theCourse forGATE 2013Electrical Engineering offered by VPM CLASSES:

TABLE - GATE 2013 ENGINEERING

GATE 2013

CORRESPONDENCE COURSE

(withGATE Aptitude study material) (withoutGATE Aptitude study material)Course A Course B

Rs.7,100/- Rs.6,300/-

6 volumes of theory 6 volumes of theory

1 volume of theory covering various aspects of

GATE General Aptitude

36 Topic-wise Unit test papers(UTPs)covering the GATE syllabus

12 Topic-wise Unit test papers covering the

GATE General Aptitude syllabus

6 Volume Test Papers (VTPs) for better

practice and revision of your syllabus

36 Topic-wise Unit test papers(UTPs)

covering the GATE syllabus

12 Full length test papers (on GATE pattern)

6 Volume Test Papers (VTPs) for better

practice and revision of your syllabus

Previous 2 year solved question bank(2011-

2012)

12 Full length test papers (on GATE pattern) Hints and solutions with all test papers

Previous 2 year solved question bank(2011-

2012)

Hints and solutions with all test papers

NOTE: All prices include service tax.


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Page 37

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For the course you want to get enrolled with us, please prepare a Demand Draft (D.D.) of

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