ee_01

GATE 2014: General Instructions during Examination

1. Total duration of the GATE examination is 180 minutes. 2. The clock will be set at the server. The countdown timer at the top right corner of

screen will display the remaining time available for you to complete the examination. When the timer reaches zero, the examination will end by itself. You need not terminate the examination or submit your paper.

3. Any useful data required for your paper can be viewed by clicking on the Useful Common Data button that appears on the screen.

4. Use the scribble pad provided to you for any rough work. Submit the scribble pad at the end of the examination.

5. You are allowed to use a non-programmable type calculator, however, sharing of calculators is not allowed.

6. The Question Palette displayed on the right side of screen will show the status of each question using one of the following symbols:

The Marked for Review status for a question simply indicates that you would like to look at that question again. If a question is answered, but marked for review, then the answer will be considered for evaluation unless the status is modified by the candidate.

Navigating to a Question :

7. To answer a question, do the following: a. Click on the question number in the Question Palette to go to that question

directly. b. Select an answer for a multiple choice type question by clicking on the bubble

placed before the 4 choices, namely A, B, C and D. Use the virtual numeric keypad to enter a number as answer for a numerical type question.

c. Click on Save & Next to save your answer for the current question and then go to the next question.

d. Click on Mark for Review & Next to save your answer for the current question and also mark it for review, and then go to the next question.

Caution: Note that your answer for the current question will not be saved, if you navigate to another question directly by clicking on a question number without saving the answer to the previous question.

You can view all the questions by clicking on the Question Paper button. This feature is provided, so that if you want you can just see the entire question paper at a glance.

Answering a Question :

8. Procedure for answering a multiple choice (MCQ) type question: a. Choose one answer from the 4 options (A,B,C,D) given below the question,

click on the bubble placed before the chosen option. b. To deselect your chosen answer, click on the bubble of the chosen option again

or click on the Clear Response button. c. To change your chosen answer, click on the bubble of another option. d. To save your answer, you MUST click on the Save & Next button.

9. Procedure for answering a numerical answer type question: a. To enter a number as your answer, use the virtual numerical keypad. b. A fraction (e.g. -0.3 or -.3) can be entered as an answer with or without '0'

before the decimal point. As many as four decimal points, e.g. 12.5435 or 0.003 or -932.6711 or 12.82 can be entered.

c. To clear your answer, click on the Clear Response button. d. To save your answer, you MUST click on the Save & Next button

10. To mark a question for review, click on the Mark for Review & Next button. If an answer is selected (for MCQ) or entered (for numerical answer type) for a question that is Marked for Review, that answer will be considered in the evaluation unless the status is modified by the candidate.

11. To change your answer to a question that has already been answered, first select that question for answering and then follow the procedure for answering that type of question.

12. Note that ONLY Questions for which answers are saved or marked for review after answering will be considered for evaluation.

Choosing a Section :

13. Sections in this question paper are displayed on the top bar of the screen. Questions in a Section can be viewed by clicking on the name of that Section. The Section you are currently viewing will be highlighted.

14. A checkbox is displayed for every optional Section, if any, in the Question Paper. To select the optional Section for answering, click on the checkbox for that Section.

15. If the checkbox for an optional Section is not selected, the Save & Next button and the Mark for Review & Next button will NOT be enabled for that Section. You will

only be able to see questions in this Section, but you will not be able to answer questions in the Section.

16. After clicking the Save & Next button for the last question in a Section, you will automatically be taken to the first question of the next Section in sequence.

17. You can move the mouse cursor over the name of a Section to view the answering status for that Section.

Changing the Optional Section :

18. After answering the chosen optional Section, partially or completely, you can change the optional Section by selecting the checkbox for a new Section that you want to attempt. A warning message will appear along with a table showing the number of questions answered in each of the previously chosen optional Sections and a checkbox against each of these Sections. Click on a checkbox against a Section that you want to reset and then click on the RESET button. Note that RESETTING a Section will DELETE all the answers for questions in that Section. Hence, if you think that you may want to select this Section again later, you will have to note down your answers for questions in that Section. If you do not want to reset the Section and want to continue answering the previously chosen optional Section, then click on the BACK button.

19. If you deselect the checkbox for an optional Section in the top bar, the following warning message will appear: "Deselecting the checkbox will DELETE all the answers for questions in this Section. Do you want to deselect this Section? If you want to deselect, click on the RESET button. If you do not want to deselect, click on the BACK button.

20. You can shuffle between different Sections or change the optional Sections any number of times.

GATE 2014 Examination

EE: Electrical Engineering Duration: 180 minutes Maximum Marks: 100

Read the following instructions carefully.

1. To login, enter your Registration Number and password provided to you. Kindly go through the various symbols used in the test and understand their meaning before you start the examination.

2. Once you login and after the start of the examination, you can view all the questions in the question paper, by clicking on the View All Questions button in the screen.

3. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject specific GATE paper for 85 marks. Both these sections are compulsory. The GA section consists of 10 questions. Question numbers 1 to 5 are of 1-mark each, while question numbers 6 to 10 are of 2-mark each. The subject specific GATE paper section consists of 55 questions, out of which question numbers 1 to 25 are of 1-mark each, while question numbers 26 to 55 are of 2-mark each.

4. Depending upon the GATE paper, there may be useful common data that may be required for answering the questions. If the paper has such useful data, the same can be viewed by clicking on the Useful Common Data button that appears at the top, right hand side of the screen.

5. The computer allotted to you at the examination center runs specialized software that permits only one answer to be selected for multiple-choice questions using a mouse and to enter a suitable number for the numerical answer type questions using the virtual keyboard and mouse.

6. Your answers shall be updated and saved on a server periodically and also at the end of the examination. The examination will stop automatically at the end of 180 minutes.

7. In each paper a candidate can answer a total of 65 questions carrying 100 marks. 8. The question paper may consist of questions of multiple choice type (MCQ) and numerical answer

type. 9. Multiple choice type questions will have four choices against A, B, C, D, out of which only ONE is the

correct answer. The candidate has to choose the correct answer by clicking on the bubble () placed before the choice.

10. For numerical answer type questions, each question will have a numerical answer and there will not be any choices. For these questions, the answer should be enteredby using the virtual keyboard that appears on the monitor and the mouse.

11. All questions that are not attempted will result in zero marks. However, wrong answers for multiple choice type questions (MCQ) will result in NEGATIVE marks. For all MCQ questions a wrong answer will result in deduction of marks for a 1-mark question and marks for a 2-mark question.

12. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.

13. Non-programmable type Calculator is allowed. Charts, graph sheets, and mathematical tables are NOT allowed in the Examination Hall. You must use the Scribble pad provided to you at the examination centre for all your rough work. The Scribble Pad has to be returned at the end of the examination.

Declaration by the candidate: I have read and understood all the above instructions. I have also read and understood clearly the instructions given on the admit card and shall follow the same. I also understand that in case I am found to violate any of these instructions, my candidature is liable to be cancelled. I also confirm that at the start of the examination all the computer hardware allotted to me are in proper working condition.

GATE 2014 SET- 7 General Aptitude -GA

GA 1/2

Q. 1 Q. 5 carry one mark each.

Q.1 Which of the following options is the closest in meaning to the phrase underlined in the sentence below? It is fascinating to see life forms cope with varied environmental conditions.

(A) adopt to (B) adapt to (C) adept in (D) accept with Q.2 Choose the most appropriate word from the options given below to complete the following

sentence. He could not understand the judges awarding her the first prize, because he thought that her performance was quite __________.

(A) superb (B) medium (C) mediocre (D) exhilarating Q.3 In a press meet on the recent scam, the minister said, "The buck stops here". What did the minister

convey by the statement?

(A) He wants all the money (B) He will return the money (C) He will assume final responsibility (D) He will resist all enquiries

Q.4 If ( + 1/)2 = 98, compute (2 + 1/2).

Q.5 The roots of 2 + + = 0 are real and positive. a, b and c are real. Then 2 + || + = 0

has

(A) no roots (B) 2 real roots (C) 3 real roots (D) 4 real roots

Q. 6 Q. 10 carry two marks each.

Q.6 The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern part of the Western

Ghats in India, is lower than the hilly terrain to its north and south. The exact reasons for the formation of this gap are not clear. It results in the neighbouring regions of Tamil Nadu getting more rainfall from the South West monsoon and the neighbouring regions of Kerala having higher summer temperatures. What can be inferred from this passage?

(A) The Palghat gap is caused by high rainfall and high temperatures in southern Tamil Nadu and Kerala

(B) The regions in Tamil Nadu and Kerala that are near the Palghat Gap are low-lying (C) The low terrain of the Palghat Gap has a significant impact on weather patterns in neighbouring

parts of Tamil Nadu and Kerala (D) Higher summer temperatures result in higher rainfall near the Palghat Gap area

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GATE 2014 SET- 7 General Aptitude -GA

GA 2/2

Q.7 Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely?

(A) Strategies are now available for eliminating psychiatric illnesses (B) Certain psychiatric illnesses have a genetic basis (C) All human diseases can be traced back to genes and how they are expressed (D) In the future, genetics will become the only relevant field for identifying psychiatric illnesses

Q.8 Round-trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In

addition, groups of 4 or more get a discount of 5% on the total fare. If the one way single person fare is Rs 100, a group of 5 tourists purchasing round-trip tickets will be charged Rs _________.

Q.9 In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were

further asked to mention whether they own a car or scooter or both. Their responses are tabulated below. What percent of respondents do not own a scooter?

Men Women

Own vehicle Car 40 34 Scooter 30 20 Both 60 46

Do not own vehicle 20 50

Q.10 When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight

lines to its corners, how many (new) internal planes are created with these lines? _____________

END OF THE QUESTION PAPER

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GATE 2014 SET1 ELECTRICAL EE

EE 1/15

Q. 1 Q. 25 carry one mark each.

Q.1 Given a system of equations: 2 2 5 3

Which of the following is true regarding its solutions

(A) The system has a unique solution for any given and (B) The system will have infinitely many solutions for any given and (C) Whether or not a solution exists depends on the given and (D) The system would have no solution for any values of and

Q.2 Let . The maximum value of the function in the interval (0, ) is

(A) (B) (C) 1 (D) 1 Q.3 The solution for the differential equation

9 ,

with initial conditions 0 1 and | 1, is (A) 1 (B) sin 3 cos 3

(C) sin 3 cos 3 (D) cos 3 Q.4 Let be the Laplace Transform of a signal . Then, 0 is

(A) 0 (B) 3 (C) 5 (D) 21 Q.5 Let be the set of points in the complex plane corresponding to the unit circle. (That is,

|| 1). Consider the function where denotes the complex conjugate of . The maps to which one of the following in the complex plane (A) unit circle (B) horizontal axis line segment from origin to (1, 0) (C) the point (1, 0) (D) the entire horizontal axis

Q.6 The three circuit elements shown in the figure are part of an electric circuit. The total power

absorbed by the three circuit elements in watts is __________.

10 A

100 V

8 A

80 V

15 V

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Q.7 C0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity , the expression for the modified capacitance is

(A) 1 (B) 0

(C) (D) 1

Q.8 A combination of 1 F capacitor with an initial voltage v0 2 V in series with a 100

resistor is connected to a 20 mA ideal dc current source by operating both switches at 0 s as shown. Which of the following graphs shown in the options approximates the voltage v across the current source over the next few seconds?

(A)

(B)

(C)

(D)

Q.9 is nonzero only for , and similarly, is nonzero only for . Let

be convolution of and . Which one of the following statements is TRUE? (A) can be nonzero over an unbounded interval. (B) is nonzero for . (C) is zero outside of . (D) is nonzero for .

(a) (b)

+vs -

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Q.10 For a periodic square wave, which one of the following statements is TRUE?

(A) The Fourier series coefficients do not exist. (B) The Fourier series coefficients exist but the reconstruction converges at no point. (C) The Fourier series coefficients exist and the reconstruction converges at most points. (D) The Fourier series coefficients exist and the reconstruction converges at every point.

Q.11 An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 700 rpm. The frequency of the rotor current of the motor in Hz is ____________.

Q.12 For a specified input voltage and frequency, if the equivalent radius of the core of a transformer is

reduced by half, the factor by which the number of turns in the primary should change to maintain the same no load current is

(A) 1/4 (B) 1/2 (C) 2 (D) 4 Q.13 A star connected 400 V, 50 Hz, 4 pole synchronous machine gave the following open circuit and

short circuit test results: Open circuit test: Voc = 400 V (rms, line-to-line) at field current, If = 2.3 A Short circuit test: Isc = 10 A (rms, phase) at field current, If = 1.5 A The value of per phase synchronous impedance in at rated voltage is __________.

Q.14 The undesirable property of an electrical insulating material is

(A) high dielectric strength (B) high relative permittivity (C) high thermal conductivity (D) high insulation resistivity

Q.15 Three-phase to ground fault takes place at locations F1 and F2 in the system shown in the figure

If the fault takes place at location F1, then the voltage and the current at bus A are VF1 and IF1 respectively. If the fault takes place at location F2, then the voltage and the current at bus A are VF2 and IF2 respectively. The correct statement about voltages and currents during faults at F1 and F2 is

(A) VF1 leads IF1 and VF2 leads IF2 (B) VF1 leads IF1 and VF2 lags IF2 (C) VF1 lags IF1 and VF2 leads IF2 (D) VF1 lags IF1 and VF2 lags IF2

IF1 IF2 B A F1 F2

0BE AE ~ ~VF2 VF1

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Q.16 A 2-bus system and corresponding zero sequence network are shown in the figure.

The transformers T1 and T2 are connected as

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

Q.17 In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero

values. This premature termination of the array indicates the presence of

(A) only one root at the origin (B) imaginary roots (C) only positive real roots (D) only negative real roots

Q.18 The root locus of a unity feedback system is shown in the figure

The closed loop transfer function of the system is

(A) 21)()(

ssK

sRsC

(B) KssK

sRsC

21)()(

(C) KssK

sRsC

21)()(

(D) KssK

sRsC

21)()(

Q.19 Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method.

The first wattmeter reads twice that of the second. Then the load impedance angle in radians is

(A) /12 (B) /8 (C) /6 (D) /3 Q.20 In an oscilloscope screen, linear sweep is applied at the

(A) vertical axis (B) horizontal axis (C) origin (D) both horizontal and vertical axis

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Q.21 A cascade of three identical modulo-5 counters has an overall modulus of

(A) 5 (B) 25 (C) 125 (D) 625 Q.22 In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when

(A) ,

(B)

,

(C)

,

(D)

,

Q.23 The magnitude of the mid-band voltage gain of the circuit shown in figure is (assuming of the

transistor to be 100)

(A) 1 (B) 10 (C) 20 (D) 100 Q.24 The figure shows the circuit of a rectifier fed from a 230-V (rms), 50-Hz sinusoidal voltage source.

If we want to replace the current source with a resistor so that the rms value of the current supplied by the voltage source remains unchanged, the value of the resistance (in ohms) is __________ (Assume diodes to be ideal.)

R1

R4C2 R2

R3

+Vcc

C1

-Vcc

vi

10k

10kC

C

1k

+Vcc

hfe=100

vo

C

10 A

230 V, 50 Hz

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Q.25 Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches can block

voltages of either polarity (applied between terminals a and b) when the active device is in the OFF state?

(A) (i), (ii) and (iii) (B) (ii), (iii) and (iv) (C) (ii) and (iii) (D) (i) and (iv)

Q. 26 Q. 55 carry two marks each.

Q.26 Let : 0, 0, be a function defined by , where represents the integer

part of . (That is, it is the largest integer which is less than or equal to ). The value of the constant term in the Fourier series expansion of is _______

Q.27 A fair coin is tossed n times. The probability that the difference between the number of heads and tails is (n-3) is

(A) 2 (B) 0 (C) 2 (D) 2 Q.28 The line integral of function F = yzi, in the counterclockwise direction, along the circle x2+y2 = 1 at

z = 1 is

(A) -2 (B) - (C) (D) 2 Q.29 An incandescent lamp is marked 40 W, 240V. If resistance at room temperature (26C) is 120 ,

and temperature coefficient of resistance is 4.5x10-3/C, then its ON state filament temperature in C is approximately _______

Q.30 In the figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The

current I is ______

a

b

a a a

bb b

(i)(ii)

(iii) (iv)

R300VI

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Q.31 In an unbalanced three phase system, phase current 190o pu, negative sequence current 4150o pu, zero sequence current 390o pu. The magnitude of phase current in pu is

(A) 1.00 (B) 7.81 (C) 11.53 (D) 13.00 Q.32 The following four vector fields are given in Cartesian co-ordinate system. The vector field which

does not satisfy the property of magnetic flux density is

(A) zyx xzy aaa222 (B) zyx yxz aaa 222

(C) zyx zyx aaa222 (D) zyx yxzxzy aaa 222222

Q.33 The function shown in the figure can be represented as

(A)

2 (B)

2

(C)

(D) 2

2

Q.34 Let be the Z-transform of a causal signal . Then, the values of 2 and 3 are

(A) 0 and 0 (B) 0 and 1 (C) 1 and 0 (D) 1 and 1 Q.35 Let be a continuous time signal and let be its Fourier Transform defined by

Define by

What is the relationship between and ? (A) would always be proportional to . (B) would be proportional to if is an even function. (C) would be proportional to only if is a sinusoidal function. (D) would never be proportional to .

2T

1

0 T

t

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Q.36 The core loss of a single phase, 230/115 V, 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively,

(A) 508 W and 542 W. (B) 468 W and 582 W. (C) 498 W and 552 W. (D) 488 W and 562 W.

Q.37 A 15 kW, 230 V dc shunt motor has armature circuit resistance of 0.4 and field circuit resistance

of 230 . At no load and rated voltage, the motor runs at 1400 rpm and the line current drawn by the motor is 5 A. At full load, the motor draws a line current of 70 A. Neglect armature reaction. The full load speed of the motor in rpm is _________.

Q.38 A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 and reactance of 0.92 . Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is

(A) 1.567 (B) 1.712 (C) 1.948 (D) 2.134

Q.39 A three phase synchronous generator is to be connected to the infinite bus. The lamps are connected

as shown in the figure for the synchronization. The phase sequence of bus voltage is R-Y-B and that of incoming generator voltage is R-Y-B.

It was found that the lamps are becoming dark in the sequence La-Lb-Lc. It means that the phase sequence of incoming generator is

(A) opposite to infinite bus and its frequency is more than infinite bus (B) opposite to infinite bus but its frequency is less than infinite bus (C) same as infinite bus and its frequency is more than infinite bus (D) same as infinite bus and its frequency is less than infinite bus

La

Lb

Lc

R Y B

R'

Y'

B'

Infinite Bus Incoming Generator

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Q.40 A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400V, 50 Hz balanced sources. Both voltage sources S1 and S2 are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure.

The contributions of S1 and S2 in 100 A current supplied at location P respectively, are

(A) 75 A and 25 A (B) 50 A and 50 A (C) 25 A and 75 A (D) 0 A and 100 A Q.41 A two bus power system shown in the figure supplies load of 1.0+j0.5 p.u.

The values of V1 in p.u. and 2 respectively are (A) 0.95 and 6.00 o (B) 1.05 and -5.44o (C) 1.1 and -6.00 o (D) 1.1 and -27.12o

Q.42 The fuel cost functions of two power plants are

BAPgPgCPBAPgPgCP

2310.0:Plant

05.0:Plant

22222

12111

where, Pg1 and Pg2 are the generated powers of two plants, and A and B are the constants. If the two plants optimally share 1000 MW load at incremental fuel cost of 100 Rs/MWh, the ratio of load shared by plants P1 and P2 is

(A) 1:4 (B) 2:3 (C) 3:2 (D) 4:1 Q.43 The overcurrent relays for the line protection and loads connected at the buses are shown in the

figure. The relays are IDMT in nature having the characteristic

1Multiplier Setting PlugSetting Multiplier Time14.0t 02.0op

The maximum and minimum fault currents at bus B are 2000 A and 500 A respectively. Assuming the time multiplier setting and plug setting for relay RB to be 0.1 and 5A respectively, the operating time of RB (in seconds) is ____________

~A B C

100 A 200 A 300 A

RA RB EE

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Q.44 For the given system, it is desired that the system be stable. The minimum value of for this

condition is _______________.

Q.45

The Bode magnitude plot of the transfer function . is shown below: Note that 6 dB/octave = 20 dB/decade. The value of is _____________.

Q.46 A system matrix is given as follows.

0 1 16 11 66 11 5

. The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is _______

.

0.010

6dB/Octave

2 4 8 24 36

6dB/Octave

0dB/Octave

0dB/Octave6dB/Octave

12dB/Octave

dB

(rad/s)

1 1 1 + R(s) C(s)

.

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Q.47 The reading of the voltmeter (rms) in volts, for the circuit shown in the figure is __________

Q.48 The dc current flowing in a circuit is measured by two ammeters, one PMMC and another

electrodynamometer type, connected in series. The PMMC meter contains 100 turns in the coil, the flux density in the air gap is 0.2 Wb/m2, and the area of the coil is 80 mm2. The electrodynamometer ammeter has a change in mutual inductance with respect to deflection of 0.5 mH/deg. The spring constants of both the meters are equal. The value of current, at which the deflections of the two meters are same, is ________

Q.49 Given that the op-amps in the figure are ideal, the output voltage is

(A) (B) 2 (C) /2 (D)

1j 1/j

1j1/j100sin(t)

R = 0.5

V

2R

R

R

R

RV2

V1

V0

RR

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Q.50 Which of the following logic circuits is a realization of the function F whose Karnaugh map is shown in figure

(A)

(B)

(C)

(D)

11

11C

0

1

AB00 01 11 10

A

B

C

A

B

C

C

A

B

B

C

A

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Q.51 In the figure shown, assume the op-amp to be ideal. Which of the alternatives gives the correct Bode plots for the transfer function ?

(A)

(B)

(C)

(D)

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Q.52 An output device is interfaced with 8-bit microprocessor 8085A. The interfacing circuit is shown in figure

The interfacing circuit makes use of 3 Line to 8 Line decoder having 3 enable lines 321 E ,E ,E . The address of the device is (A) 50 H (B) 5000 H (C) A0 H (D) A000 H

Q.53 The figure shows the circuit diagram of a rectifier. The load consists of a resistance 10 and an

inductance 0.05 H connected in series. Assuming ideal thyristor and ideal diode, the thyristor firing angle (in degree) needed to obtain an average load voltage of 70 V is ______

8BDB

AB

8

Output Port

8

Output Device

0

1

2

3

4

5

6

7

3Lx8L DecoderI2 I1I0

IO/M WRBCB

E1

E2E3

A15A14

A13A12

A11

325 sin (314t) V+

-Load

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Q.54 Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit in figure (i) is switched such that the voltage across the diode has the wave shape as shown in figure (ii). The capacitance C is large so that the voltage across it is constant. If switch S and the diode are ideal, the peak to peak ripple (in A) in the inductor current is ______

Figure (i)

Figure (ii)

Q.55 The figure shows one period of the output voltage of an inverter. should be chosen such that

60o 90o. If rms value of the fundamental component is 50 V, then in degree is______________

END OF THE QUESTION PAPER

100 V

S 1 mH

C LoadvD

+

_

vD

t (ms)

100 V

0.05 0.10 0.15 0.2

t(degree)

0 180- 180 180+ 360- 360

100 V

-100 V -100 V -100 V

100 V 100 V

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E 20

14)

Key/Range Marks Key/Range MarksGA 1 B 1 EE 24 23to23 1GA 2 C 1 EE 25 C 1GA 3 C 1 EE 26 0.5to0.5 2GA 4 96to96 1 EE 27 B 2GA 5 D 1 EE 28 B 2GA 6 C 2 EE 29 2470to2471 2GA 7 B 2 EE 30 10to10 2GA 8 850to850 2 EE 31 C 2GA 9 48to48 2 EE 32 C 2GA 10 6to6 2 EE 33 A 2EE 1 B 1 EE 34 B 2EE 2 A 1 EE 35 B 2EE 3 C 1 EE 36 A 2EE 4 B 1 EE 37 1239to1242 2EE 5 C 1 EE 38 C 2EE 6 330to330 1 EE 39 A 2EE 7 A 1 EE 40 D 2EE 8 C 1 EE 41 B 2EE 9 C 1 EE 42 D 2EE 10 C 1 EE 43 0.21to0.23 2EE 11 3.2to3.5 1 EE 44 0.61to0.63 2EE 12 C 1 EE 45 0.7to0.8 2EE 13 14.5to15.5 1 EE 46 2.9to3.1 2EE 14 B 1 EE 47 140to142 2EE 15 C 1 EE 48 3.0to3.4 2EE 16 B 1 EE 49 B 2EE 17 B 1 EE 50 C 2EE 18 C 1 EE 51 A 2EE 19 C 1 EE 52 B 2EE 20 B 1 EE 53 69to70 2EE 21 C 1 EE 54 2.49to2.51 2EE 22 C 1 EE 55 76.5to78.0 2EE 23 D 1

Section Q.No. SECTION7

GATE2014AnswerKeysforEEElectricalEngineering

Section Q.No. SECTION7

General Aptitude

4)

21

z 98z

22

1Z 2. Z

Z

1.

Z98

22

1Z 96

Z

So, answer is 96.

5) 2ax b x c 0

Can be rephrased as

2

a X b X C 0

Because 2x & 2

x will have same value and hence well have 2 positive value of x as

the solution to this equation.

Suppose roots are 1x & 2x both real & positive

Then x = 1x & x = 2x are the roots

& hence 1x x & 2x x

So the equation has 4 real roots.

8) A group of 5 tourists will get a discount of 15% on total fare as no. of people > 4

Total fare of 5 people = 100 5 = 500Rs.

For to & fro journey

Fare = 2 500 = 1000 Rs.

Net fare = 1000 15 % of 1000 = 1000 150 = 850 Rs.

9) The people that do not own a scoter are either the people those own a car or the

people who own neither.

Number of men = 150

Number of women = 150

Total no. of respondents = 300

No. of car owners = 40 + 34 = 74

No. of people who do not own either = 20 + 50 = 70

Total = 70 + 74 = 144

Percentage = 144

100 48%300

10) If we consider a point inside tetrahedron

connected to all corners, we get 4 new lines.

For a new surface we take 2 of these lines & one original edge.

& hence no. of surface = 42

C = 6

Electrical Engineering

1) 1

x 2y 2z b

2

5x y 3z b

Here, we have 3 variables and only two equations which are linearly independent.

So, one variable can be assumed and other two can be obtained from these equations.

Hence, for any value of 1

b & 2

b , there are infinitely many solutions.

2) xf x xe

For maxima , f ' x 0

x x xf ' x xe e 1 x e 0

xe 0 for any value of x

x xf " x e 1 x e

= xx 2 e

At, x = 1, f " x 0 , & hence x = 1 corresponds to a maxima.

1x 1

ef x

3) 2

2d x9x

dt

2

2d x9x 0

dt

The solution of a differential of the form

2

22d x k x 0

dt

Is, x = A cos k.t + B sin kt

solution of this equation is

x = A cos 3t + B sin 3t

x(0) = 1

A cos 0 + B sin 0 = 1 [A=1]

t 0

dx1

dt

3A sin 3t + t = 03B cos 3t =1

3B = 1

B = 13

x = cos 3 t + 1

3 sin 3 t

Shortcut: please satisfy initial conditions in the option given.

4) 2

3s 5X S

s 10s 21

x 0 can be computed by initial value theorem

limx 0 S X S

S

=

lim

S

2

22

53lim3S 5S s10 21SS 10S 21 1

s s

= 3

5) 2

f z zz* z

as z 1

f z 1

So f z corresponds to the point (1, 0)

6) for the 100V source, current goes into positive terminal

So, Power absorbed= 100 10 = 1000 W

for 80 V, current comes out of positive terminal

Power absorbed = 80 8 = 640 W (power delivered)

Current in 15V source = 10 8 = 2A (downwards)

Power absorbed = 15 2 = 30 W (power delivered)

Total power absorbed = 1000 640 30 = 330 W

7) 00

AC

d

A capacitor with half filed dielectric can be considered as a parallel combination of two

capacitors

01

AC

2 d

(for air filled part)

02

rAC

2 d

(for dielectric filled part )

0eq 1 2 rA

C C C 12 d

= 0 0r r

A C11 1

2 d 2

8) Switch across current source is opened at t = 0

c RSWV 0 V 0 V 0

cSV 0 V 0 (Voltage across capacitor is continuous)

3Sw sV 0 2V i R 2 20 10 100 2 2 0V

For t > 0

s

sRsSw

l t1V t i dt V l R

C C

= s tl Rc

So, curve is a straight line passing through origin.

9) If z (t) = x(t) * y(t)

If x(t) exists for x xT t T '

If y(t) exists for y yT t T '

Then z (t) exits for z zT t T '

z x yT T T

z x yT ' T ' T '

Hence (C) is correct

10) Suppose the square wave is as shown

The Fourier series is given by

0n 1,3,5

4Vt sin n t

n

0

2

T

If we observe that function is not differentiable at points where function makes a

transition, also it is not Continuous at those points.

But sine terms are continuous at all points, so Fourier series will not converge.

Hence (C) is correct.

11) Synchronous speed of rotor = 120 f 120 50

750rpmP 8

rotor speed = 700 rpm

slip = rs

s

N N 750 700 50 115N 750 750

rotor frequency = s stator frequency

= 1 50 3.33Hz15

12) No-load current = 2

m

V

X

Where mX = magnetizing reactance

mX wL ,where 02L n A

Where n = no. of turns per unit length

If voltage is constant, mX should be constant for I to be constant.

2n A = constant

2 1nA

where A = 2r

2 21n

r 1 n

r

If radius is halved, n should be doubled.

13) The short circuit characteristics are linear

f

SC I 2.3

10 2.3I 15.33A

1.5

osS

SC

V 400X 15.06

3 I 3 15.33

14) High relative permittivity corresponds to a better conductor as air is an insulator for

which r 1 .

So (B) is correct.

15) If fault occurs at 1

F

A BF

1 2

E 0EI

X 90 X 90

Where 1

X is reactance from source to fault point

2

X is reactance from BE to fault

B BF1 F1V E 0 I . X 90

3X = reactance from BE to point A

B

F12

E 0V

X 90 (ve sign as current flows into fault but

F1I has opposite

direction shown)

= B

2

E90

X

3B BF1

2

XV E 0 E 180

X = 3

2

XE8 1 0

X

So F1I leads

F1V

For fault at 2

F

A 4F2 F2V E I X 90

4

X = reactance from A

E to point A

A AF21 1

E EI 90

X 90 X

A4

F21

XV E 1

x

So F2I lags

F2V .

16) The zero sequence circuit of first transformer,

it is the equivalent circuit of Transformer.

The zero sequence circuit of second transformer

, It is equivalent circuit of transformer.

17) If an entire row of elements is zero, then we form an auxiliary polynomial whose roots

are imaginary and it shows presence of imaginary roots.

18) Rules of root locus

Portion of real axis on whose right side the total number of poles and zeroes are odd,

lies on root locus.

But if we see here, in the diagram, the scale is reversed

as to right S = 1, 0 poles & zeroes lie.

On left of S = 2, 2 poles & zeroes lie.

This indicates presence of positive feedback.

Open loop transfer function =

K

s 1 s 2

Closed loop =

G S K

1 G S s 1 s 2 k

19) Power factor angle

1 2

1 2

P Ptan 3

P P

1 2

P 2P

1 2

1 2

2P P 3 1tan 33 32P P

11

tan63

20) Linear sweep is applied to horizontal axis to sweep the electron wave across the

screen.

21) Cascaded connection means that o/p of one counter acts as a clock to second and

hence the modulos get multiplied

Modulo = 5 5 5 5 = 125

22) This can be simply referred from a text book and hence derivation is avoided here

3 1 2

4 2 1 1 1 2 2

R R C 1 &

R R C R C R C

23) We neglect dc-capacitances (short-circuit) and open circuit C & C (open circuit)

As no information about early effect or dc operating point is given, we neglect 0

r & r

Eq. circuit

ib

S

vi

R

iFE FE0 C Cb

S

vh i R h R

R

0 CFE

Si

R 10h 100 100R 10

24) If rectifier is connected to a resistive load

0

V looks as shown

Then 0 irms rmsV V 230V

0 rms

0 rms

VI

R

For current source at load

LS rmsI I 10A

0rmsV 230

10 R 23R R

25) Switch (i) conducts in reverse direction when active device is OFF.

Same is the case with (iv) switch but switch (ii) & (iii) cannot conduct when active

device is OFF.

Hence (C) is correct.

26) g(x) = x [x] = {x} (fractional part of x)

dc component

= T

0

1g x dx

T

T = Time period of g(x) = 1

= t

0

g x dx 0.5

27) In our case

No. of heads = H

No. of tails = T

H T

n

H T n 3

H2 2n 3

H

32

(not possible as & H should both be integers)

Hence, probability is 0

28) Applying Greens Theorem

2 1F F

F.dl dAx y

1

F yz 2

F 0

F.dl 0 yz dA zdA 1dA dAy

2

F.dl 1 1

29) Resistance in ON state = 2 2V 240

1440P 40

0R T R 1 T

1440 = 120 31 4.5 10 T

T = 3

11

4.5 10 = 2444.44 C

T Troom = T ;

T = 2444.44 + 26 = 2470.44 C

30) I = 300

R

I =

300

I252

22I 25I 300 0 I +50I 600=0

2

2I 60I 10I 600 0

I 60 I 10 0

[I = 10AmP] {I cannot be negative as current cannot enter the source}

31) a a1 a2 a0I I I I

a0 b0 l0I I I 3 90 pu

b2a2 a2b2

I 4 150I I I 4 270

1 120 , where = 1 1200

a1 a2 a0aI I I I 1 90 4 270 3 90

a1I j 4 j 3j 8 j

2 a1b1I I 1 240 8 90

= 8 150

b b1 b2 b0I I I I

= 8 150 4 150 3 90 11.53 154.3 A

32) The property of magnetic flux density is that its divergence is always zero.

.B 0

This is violated in option (C)

As .B 2 x y z 0

Hence (C) is current

33) This wave from can be broken as shown in the

adjoining figure:

The equation of first wave form is

1x t u t u t T

2t T t 2T

x t u t T u t 2TT T

1 2x t x t x t

=

t T t 2T

u t u t T u t T u t 2TT T

34) X(z) = 1

1 Z3

1

Z

n

3n 0,1,2

= 1

Z

n

n 0,3,6....

x n Z

n

n 0,1,2....

X(0) = 1

X(1) = 0 {By comparison}

X(2) = 0

X(3) = 1

35) g(t) = jutF u due

Assume Fourier transform of g(t) to be G(u)

G(u) = jutg t dte

jut jutg t G u dt F u due e

replace u by k in right hand integral

jkt jutg t F k dk G u due e

G u F u ,g t f t iff. G u F u

For that F(u) should be even (F(u)=F(-u))

Which can happen if and only if f(t) is even.

36) Core- loss = hysteresis loss + eddy current loss

ehP P 1050W ..(i)

1.6

mh hP k B f 2 2e mhP k B f

mVB

f

11

230V 13830f 50

mB = constant

h

P f 2eP f

If V = 138 V, f = 30 zH

h h

'h

30P P 0.6P

50 ;

2

e z'e

30P P 0.36 P

50

' 'ehP P 500W

0.6 h

P + 0.36 eP = 500 ..(ii)

Solving (i) & (ii)

h

P = 508 W , eP = 542 W

37) aL fI I I

At no. load, LI = 5A

f

f

230VI 1AR 230

a L fI I I 5 1 4A

a aTbE V I R 230 4 0.4 228.4V

Under full load, LI 70A

aI 70 1 69A

a atbE V I R 230 69 0.4 202.4V

Speed bE

Since 1

V remains constant, f

I remains same & hence = constant

b

N E

b22

1 b1

EN

N E 2

N 202.4 =

1400 228.4

2N 1241rpm

38)

2

2

22 2

2

s

R / s3VT

R / s X

At full load

2

2

s 222

FL

2

FLf1

R

S3VT

RX

S

at maximum torque S = 22

RX

2

max

2s

3V 1T

2X

max

2FL2 2

FL 22

FL

2

T 1 11.948

RT 0.512912X

S RX

S

39) aL corresponds to phase R

b

L corresponds to phase Y

cL corresponds to phase B

A lamp goes off. When voltage of generator &

infinite bus corresponding to that

phase has same magnitude & phase.

If both had same phase sequence, all lamps

would go off at same instant when all phases coincide,

so that cannot be the case.

So, generator & infinite bus have opposite phase sequence.

If frequency is same, then both phases get locked to each other, and no other lamp

goes off.

If frequency of bus is more, magnetic flux of bus rotates wrt generator and after R, B

phase will overlap (120 rotation anticlockwise) so lamp c goes off which is incorrect

as after a bL ,L goes off.

If frequency of generator is more, then generator field moves anti-clockwise wrt bus

field and after R, Y phase will overlap & b

L goes off & hence (A) is correct.

40) Assume point of minimum voltage is A

Assume resistance of entire feeder = R

A 1V 400 I 0.4R .(i)

A 2 2 2V 400 I 0.2R I 200 0.2R I 300 0.2R

= 400 0.6 R2

I + 100R ..(ii)

1 2I I 300 200

1 2I I = 500 .(iii)

From (i) & (ii)

4 001

0.4I R 4 00 2

0.6I R 100R

2 1

0.6I 0.4I 100 (iv)

1 2

0.4I 0.4I 200 .(v)

Contribution of 2

I to P 2I I 200 100A

P1

I to I 0A

41) Current from bus 1 to bus 2

1 2

12

V 0 1I

j0.1

L 2 12

*S V I

=

*

1 22

V 0 11 1 j 0.5

j 0.1

= 1 2

2

V 0 11 1 j 0.5

j 0.1

= 1 2V 1

1 j 0.5j 0.1

= 1 2 1 210 V cos 90 j V sin 90 j 1 j 0.5

1 2 1 2

V sin + j V cos = 0.1 + j 0.05 + j

1 2 1 2V sin + j V cos = 0.1 + 0.05j

tan 2 =

2

0.1 5.44

1.05

221

V 0.1 1.05 1.05

42) 11 g1

g1

dClC 0.1P A

dP

22 g2

g2

dClC 0.2P 3A

dP

Given : g1 g2

P P 100MW

1 2

lC lC 100

g1 g2

0.1P A 0.2P 3A 100

g1

0.1P A 100 ..(i)

g10.2 1000 P 3A 100

g1

0.2P 3A 100 .(ii)

From (i) & (ii)

g1

P = 800MW A = 20

g2

P = 200 MW ; g1

g2

P 8004 : 1

P 200

43) CT ratio = 500 100 : 15

PSM = Maximum fault current 2000

20CT primary current 100

Top =

0.02

0.14 0.1

20 1

= 0.2267 s

44) characteristic equation of the system

3 2S 1 a S a 1 S 1 a s a 0

3 2S 1 a S S 1 0a

3S 1 a Routh criterion

2S (1 + a) 1

1S

2a a 1

1 a

S 1

For system to be stable, (a + 1) > 0 ; a > 1

Also 2a a 1 0

The roots of this equation are

a = 1 1 4 1 5

0.622 2

, 1.62

2a a 1 0 iff a< 1.62 or a > 0.62

If a < 1 .62, (a + 1) < 0 , system is unstable

So, a > 0.62, hence minimum value is 0.62.

45) From Bode plot, corner frequencies

w = 0, w = 2; w = 4; w = 8; w = 24; w = 36

at w = 4, slope goes from 0 dB/octave to 6dB/ octave &

So from expression 1a4

.

Similarly a pole is located at w = 24 1b24

Equation of first segments of bode plot

=

K 1 1K

SS 1 1 1

From the line we see, that gain is 0 dB at w = 0 along the line

Gain = 1 (0 dB)

K 18

K = 8

a 1

24 0.75bK 4 8

46) for calculating the Eigen values of the matrix

I A 0

1 1

6 11 6 0

6 11 5

11 5 66 1 6 5 36 1 66 6 11 0

2 6 55 66 6

30 36 66 6 66 0

23 6 11 6 0

= 1, -2, -3

Ratio = 3

31

Note : this can be conflicting as maximum Eigen value is 1 & minimum is 3.

47) The equivalent impedance of both series combination of inductor & capacitor = 0

& hence current

S100

I 200 sin wt0.5

S1 2

II I

2 (impedance of both equal )

1 1200 0

V I j j 100 90 V2

2 2

200 01 1V I 100 90 Vj j2

V reading = 1 2

V V = (200j)V

rms value = 200

2 = 141.42 V

48) for PMMC meter

2I dN

R d

Since both deflations are the same

NI 2AB I

R

R

dw

d {current is same due to sales connection }

6NAB 100 80 10 0.2I 3.2A

dw 0.5d

49) Voltage at - terminals of OP-Amp are also 1

V & 2

V by virtual ground concept.

1 2V V

I2R

2

V V IR

1 2

2

V VV R

2R

2 13V V2

1

V V IR

= 1 2 1 2

1

V V 3V VV V R

2R 2

If only V is present

1 2

a

3V VVV

2 4

1 2

0

3V V1V 1 V

1 2

If only V is present

2 10

3V V1V V

1 2

1 2 2 1

0

3V V 3V VV

2 2

0 1 2V 2 V V

50) The boolean function from the k map is

F = CA CB

Circuit implementation

Hence (C) is correct

51) The transfer function is

0

i

1V S 1 1sC1 sV S sRC 1R 1sC 1000

Corner frequency = 310 rad/ s

Bode plot

The phase of transfer function goes to 2

as

1tan 1000

52) For output device to be selected, output of decoder should be asserted.

For 2

O to be asserted. 2 1 0I I I 0 1 0

For enable, 1

E 1 12 11

A =1; A =0

15 14 13 12 115

A A A A A 0 1 0 1 0

Minimum address = 5000H

53) Assume firing angle =

0mVV avg 1 cos

2

(Half wave rectifier with FD)

0325

T 1 cos2

1 + cos = 1.3533

Cos = 0.3533

= 69.31

54) when switch S in on

Current in inductor increases from minI to maxI

L 0inV V V

0ini

L V VDTS

in100-DV

i = DTsL

D = ONT 0.05

0.5T 0.1

(from waveform)

S

T = 0.1 ms

3

3

100 1 0.5 0.5 0.1 10i

10

= 2.5A

55) This is not a standard inerter output & hence fundamental needs to be computed.

Since the function is odd function, we need to compute only sine terms.

1 02

V f t sin t dtT

; 0 2 T

Period : 2

V fund = 2

0

2t sin t dt

2

=

180 180

1800

1100sint dt + 100sint dt 100sint dt

180 360 360

180 360180

100sint dt + 100sint dt 100sint dt

= 100

1 cos 1 cos cos cos

1 cos cos cos cos 1

= 100

4 8cos 50V 2

{maximum value}

= 4 8cos 1.57 2

8cos 1.78 cos = 0.2224

77.14

EE01_Que.pdfGeneral InstructionsEE01 2014EE InstructionsGATE 2014 ExaminationEE: Electrical EngineeringRead the following instructions carefully.

EE01 (GATE 2014)GA SET7Q. 1 Q. 5 carry one mark each.Q. 6 Q. 10 carry two marks each.END OF THE QUESTION PAPER

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