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GATEELECTRICAL ENGINEERING

Vol 1 of 4

Second Edition

GATEELECTRICAL ENGINEERING

Vol 1 of 4

RK Kanodia Ashish Murolia

NODIA & COMPANY

GATE Electrical Engineering Vol 1, 2eRK Kanodia & Ashish Murolia

Copyright © By NODIA & COMPANY

Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that NODIA & COMPANY and its author

are supplying information but are not attempting to render engineering or other professional services.

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SYLLABUS

GENERAL ABILITY

Verbal Ability : English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction.

Numerical Ability : Numerical computation, numerical estimation, numerical reasoning and data interpretation.

ENGINEERING MATHEMATICS

Linear Algebra: Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems.

Differential equations: First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method.

Complex variables: Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s and Laurent’ series, Residue theorem, solution integrals.

Probability and Statistics: Sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Discrete and continuous distributions, Poisson,Normal and Binomial distribution, Correlation and regression analysis.

Numerical Methods: Solutions of non-linear algebraic equations, single and multi-step methods for differential equations.

Transform Theory: Fourier transform,Laplace transform, Z-transform.

ELECTRICAL ENGINEERING

Electric Circuits and Fields: Network graph, KCL, KVL, node and mesh analysis, transient response of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts; ideal current and voltage sources, Thevenin’s, Norton’s and Superposition and Maximum Power Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electric field and potential due to point, line, plane and spherical charge distributions; Ampere’s and Biot-Savart’s laws; inductance; dielectrics; capacitance.

Signals and Systems: Representation of continuous and discrete-time signals; shifting and scaling operations; linear, time-invariant and causal systems; Fourier series representation of continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.

Electrical Machines: Single phase transformer – equivalent circuit, phasor diagram, tests, regulation and efficiency; three phase transformers – connections, parallel operation; auto-transformer; energy conversion principles; DC machines – types, windings, generator characteristics, armature reaction and commutation, starting and speed control of motors; three phase induction motors – principles, types, performance characteristics, starting and speed control; single phase induction motors; synchronous machines – performance, regulation and parallel operation of generators, motor starting, characteristics and applications; servo and stepper motors.

Power Systems: Basic power generation concepts; transmission line models and performance; cable performance, insulation; corona and radio interference; distribution systems; per-unit quantities; bus impedance and admittance matrices; load flow; voltage control; power factor correction; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuit breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission and FACTS concepts.

Control Systems: Principles of feedback; transfer function; block diagrams; steady-state errors; Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; state space model; state transition matrix, controllability and observability.

Electrical and Electronic Measurements: Bridges and potentiometers; PMMC, moving iron, dynamometer and induction type instruments; measurement of voltage, current, power, energy and power factor; instrument transformers; digital voltmeters and multimeters; phase, time and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.

Analog and Digital Electronics: Characteristics of diodes, BJT, FET; amplifiers – biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers – characteristics and applications; simple active filters; VCOs and timers; combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators; sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture, programming and interfacing.

Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs, GTOs, MOSFETs and IGBTs – static characteristics and principles of operation; triggering circuits; phase control rectifiers; bridge converters – fully controlled and half controlled; principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

***********

PREFACEThis book doesn’t make promise but provides complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exams as well as fresh set of problems.

The book is categorized into units which are then sub-divided into chapters and the concepts of the problems are addressed in the relevant chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts and techniques which are absolutely necessary. Again time is a critical factor both from the point of view of preparation duration and time taken for solving each problem in the examination. So the problems solving methods is the books are those which take the least distance to the solution.

But however to make a comment that this book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books. But for a wide collection of problems, for a variety of problems and the efficient way of solving them, what one needs to go needs to go through is there in there in the book. Each unit (e.g. Networks) is subdivided into average seven number of chapters on an average each of which contains 40 problems which are selected so as to avoid unnecessary redundancy and highly needed completeness.

I shall appreciate and greatly acknowledge the comments and suggestion from the users of this book.

R. K. KanodiaAshish Murolia

CONTENTS

EF ELECTRIC CIRCUITS & FIELDS

EF 1 Basic Concept EF 3

EF 2 Basic Laws EF 23

EF 3 Graph Theory EF 64

EF 4 Nodal and Loop Analysis EF 82

EF 5 Circuit Theorems EF 105

EF 6 Inductor and Capacitor EF 150

EF 7 First Order RL and RC Circuits EF 190

EF 8 Second Order Circuit EF 249

EF 9 Sinusoidal Steady State Analysis EF 287

EF 10 AC Power Analysis EF 322

EF 11 Three-phase Circuits EF 360

EF 12 Magnetically Coupled Circuits EF 393

EF 13 Frequency Response EF 419

EF 14 Circuit Analysis Using Laplace Transform EF 455

EF 15 Two Port Network EF 500

EF 16 Electric Field EF 546

EF 17 Magnetic Fields EF 562

EF 18 Gate Solved Questions EF 580

EE ELECTRICAL & ELECTRONIC MEASUREMENTS

EE 1 Measurement and Error EE 3

EE 2 Electro-mechanical Instruments EE 28

EE 3 Electronic & Digital Instruments EE 82

EE 4 Instrument Transformer EE 92

EE 5 Measurement of R, L, C and AC Bridges EE 101

EE 6 CRO EE 135

EE 7 Gate Solved Questions EE 151

***********


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GATE Electrical Engineering-2015in 4 Volumes

by R. K. Kanodia & Ashish MuroliaS

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CIRCUIT THEOREMS

EF 5.1 The linear network in the figure contains resistors and dependent sources only. When 10 VVs = , the power supplied by the voltage source is 40 W. What will be the power supplied by the source if 5 VVs = ( in W)?

EF 5.2 In the circuit below, it is given that when 20 VVs = , 200 mAIL = . What values of IL and Vs will be required such that power absorbed by RL is 2.5 W ?

(A) A1 , . V2 5

(B) 0.5 A, 2 V

(C) 0.5 A, 50 V

(D) 2 A, 1.25 V

EF 5.3 For the circuit shown in figure below, some measurements are made and listed in the table.

Which of the following equation is true for IL ?(A) . .I V I0 6 0 4L s s= +

(B) . .I V I0 2 0 3L s s= -(C) . .I V I0 2 0 3L s s= +

(D) . .I V I0 4 0 6L s s= -


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GATE EE vol-1Electric circuit & Field, Electrical & electronic measurement

GATE EE vol-2Analog electronics, Digital electronics, Power electronics

GATE EE vol-3Control systems, Signals & systems

GATE EE vol-4Electrical machines, Power systems

Engineering mathematics, General AptitudeEF 5.4 In the circuit below, the voltage drop across the resistance R2 will be ____ V

EF 5.5 In the circuit below, current I I I I1 2 3= + + , where I1, I2 and I3 are currents due to 60 A, 30 A and 30 V sources acting alone. The values of I1, I2 and I3 are respectively

(A) 8 A, 8 A, 4 A-

(B) 12 A, 12 A, 5 A-(C) 4 A, 4 A, 1 A-

(D) 2 A, 2 A, 4 A-

EF 5.6 In the circuit below, current I is equal to sum of two currents I1 and I2. What are the values of I1 and I2 ?

(A) 6 A, 1 A

(B) 9 A, 6 A

(C) 3 A, 1 A

(D) 3 A, 4 A

EF 5.7 A network consists only of independent current sources and resistors. If the values of all the current sources are doubled, then values of node voltages(A) remains same

(B) will be doubled

(C) will be halved

(D) changes in some other way.


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EF 5.8 Consider a network which consists of resistors and voltage sources only. If the values of all the voltage sources are doubled, then the values of mesh current will be(A) doubled (B) same

(C) halved (D) none of these

EF 5.9 In the circuit below, the 12 V source

(A) absorbs 36 W (B) delivers 4 W

(C) absorbs 100 W (D) delivers 36 W

EF 5.10 The value of current I in the circuit below is equal to ____A.

EF 5.11 Which of the following circuits is equivalent to the circuit shown below ?


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Engineering mathematics, General AptitudeEF 5.12 Consider a dependent current source shown in figure below.

The source transformation of above is given by

EF 5.13 In the given circuit value of current I flowing through 2 W resistance is ___ A.

EF 5.14 In the given circuit, the value of current I will be _____Amps.

EF 5.15 Consider a circuit shown in the figure


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Which of the following circuit is equivalent to the above circuit ?

EF 5.16 For the circuit shown in the figure the Thevenin voltage and resistance seen from the terminal a -b are respectively

(A) 34 V, 0 W (B) V20 , 42 W(C) 14 V, 0 W (D) 14 V- , 24 W

EF 5.17 In the following circuit, Thevenin voltage and resistance across terminal a and b respectively are

(A) 10 V, 18 W (B) 2 V, 18 W(C) 10 V, 18.67 W (D) 2 V, 18.67 W

EF 5.18 What is the value of current I in the circuit shown below (in Amp) ?


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Engineering mathematics, General AptitudeEF 5.19 The value of RTh and VTh such that the circuit of figure (B) is the Thevenin

equivalent circuit of the circuit shown in figure (A), will be equal to

(A) 6RTh W= , 4 VVTh =

(B) 6RTh W= , 28 VVTh =(C) 2RTh W= , 24 VVTh =

(D) 10RTh W= , 14 VVTh =

EF 5.20 What values of RTh and VTh will cause the circuit of figure (B) to be the equivalent circuit of figure (A) ?

(A) 2.4 W, 24 V- (B) 3 W, 16 V

(C) 10 W, 24 V (D) 10 W, 24 V-

Common Data For Q. 21 and 22 :

Consider the two circuits shown in figure (A) and figure (B) below

EF 5.21 The value of Thevenin voltage across terminals a -b of figure (A) and figure (B) respectively are(A) 30 V, V36

(B) 28 V, 12 V-(C) 18 V, 12 V

(D) 30 V, 12 V-


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EF 5.22 The value of Thevenin resistance across terminals a -b of figure (A) and figure (B) respectively are(A) zero, 3 W

(B) 9 W, 16 W(C) 2 W, 3 W

(D) zero, 16 W

EF 5.23 In the given circuit, what is the value of current I (in Amp) through 6 W resistor

EF 5.24 For a network having resistors and independent sources, it is desired to obtain Thevenin equivalent across the load which is in parallel with an ideal current source. Then which of the following statement is true ?(A) Thevenin equivalent circuit is simply that of a voltage source.

(B) Thevenin equivalent circuit consists of a voltage source and a series resistor.

(C) Thevenin equivalent circuit does not exist but the Norton equivalent does exist.

(D) None of these

EF 5.25 Thevenin equivalent circuit of a network consists only of a resistor (Thevenin voltage is zero). Then which of the following elements might be contained in the network ?(A) resistor and independent sources

(B) resistor only

(C) resistor and dependent sources

(D) resistor, independent sources and dependent sources.

EF 5.26 For the given circuit, the Thevenin’s voltage and resistance looking into a -b are

(A) 2 V, 3 W

(B) 2 V, 2 W(C) 6 V, 9 W-

(D) 6 V, 3 W-


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Engineering mathematics, General AptitudeEF 5.27 For the circuit, values of voltage V for different values of R are shown in the table.

The Thevenin voltage and resistance of the unknown circuit are respectively.(A) 14 V, 4 W

(B) 4 V, 1 W(C) 14 V, 6 W

(D) 10 V, 2 W

EF 5.28 For the circuit of figure, some measurements were made at the terminals a -b and given in the table below.

What is the value of IL (in Amps) for 20RL W= ?

EF 5.29 In the circuit shown below, the Norton equivalent current and resistance with respect to terminal a -b is

(A) A617 , 0 W (B) 2 A, 24 W

(C) A67- , 24 W (D) 2 A- , 24 W

EF 5.30 The Norton equivalent circuit for the circuit shown in figure is given by


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EF 5.31 If 5 AI = in the circuit below, then what is the value of voltage source Vs (in volts)?

EF 5.32 What are the values of equivalent Norton current source ( )IN and equivalent resistance ( )RN across the load terminal of the circuit shown in figure ?

IN RN

(A) 10 A 2 W(B) 10 A 9 W(C) 3.33 A 9 W(D) 6.66 A 2 W

EF 5.33 For a network consisting of resistors and independent sources only, it is desired to obtain Thevenin’s or Norton’s equivalent across a load which is in parallel with an ideal voltage sources.Consider the following statements :1. Thevenin equivalent circuit across this terminal does not exist.


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Engineering mathematics, General Aptitude2. Thevenin equivalent circuit exists and it is simply that of a voltage source.

3. The Norton equivalent circuit for this terminal does not exist.

Which of the above statements is/are true ?(A) 1 and 3 (B) 1 only

(C) 2 and 3 (D) 3 only

EF 5.34 For a network consisting of resistors and independent sources only, it is desired to obtain Thevenin’s or Norton’s equivalent across a load which is in series with an ideal current sources.Consider the following statements1. Norton equivalent across this terminal is not feasible.

2. Norton equivalent circuit exists and it is simply that of a current source only.

3. Thevenin’s equivalent circuit across this terminal is not feasible.

Which of the above statements is/are correct ?(A) 1 and 3 (B) 2 and 3

(C) 1 only (D) 3 only

EF 5.35 In the circuit shown below, what is the value of current I (in Amps) ?

EF 5.36 The Norton equivalent circuit of the given network with respect to the terminal a -b , is


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EF 5.37 In the circuit below, if RL is fixed and Rs is variable then for what value of Rs power dissipated in RL will be maximum ?

(A) R RS L= (B) R 0S =(C) /R R 2S L= (D) R R2S L=

EF 5.38 In the circuit shown below the maximum power transferred to RL is Pmax, then

(A) 12RL W= , 12 WPmax =(B) 3RL W= , 96 WPmax =(C) 3RL W= , 48 WPmax =(D) 12RL W= , 24 WPmax =

EF 5.39 In the circuit shown in figure (A) if current 2 AI1 = , then current I2 and I3 in figure (B) and figure (C) respectively are

(A) 2 A, 2 A (B) 2 A- , 2 A

(C) 2 A, 2 A- (D) 2 A- , 2 A-


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Engineering mathematics, General AptitudeEF 5.40 In the circuit of figure (A), if 20 mAI1 = , then what is the value of current I2 in

the circuit of figure (B) ?

(A) 40 mA (B) 20 mA-(C) 20 mA (D) R1, R2 and R3 must be known

EF 5.41 If 2 VV1 = in the circuit of figure (A), then what is the value of V2 in the circuit of figure (B) ( in V)?

EF 5.42 The value of current I in the circuit below is equal to ____mA.

EF 5.43 A simple equivalent circuit of the two-terminal network shown in figure is


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EF 5.44 If V AV BV CI1 2 3= + + in the following circuit, then values of A, B and C respectively are

(A) , ,32

32

31

(B) , ,31

31

3100

(C) , ,21

21

31

(D) , ,31

32

3100

EF 5.45 The V -I relation for the circuit below is plotted in the figure. The maximum power that can be transferred to the load RL will be ______ mW

EF 5.46 For the linear network shown below, V -I characteristic is also given in the figure. The value of Norton equivalent current and resistance respectively are

(A) 3 A, 2 W

(B) 6 W, 2 W(C) 6 A, 0.5 W

(D) 3 A, 0.5 W


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Engineering mathematics, General AptitudeEF 5.47 In the circuit a network and its Thevenin and Norton equivalent are given.

The value of the parameter are

VTh RTh IN RN

(A) 4 V 2 W 2 A 2 W

(B) 4 V 2 W 2 A 3 W

(C) 8 V 1.2 W 330 A 1.2 W

(D) 8 V 5 W 58 A 5 W

EF 5.48 For the following circuit the value of equivalent Norton current IN and resistance RN are

(A) 2 A, 20 W

(B) 2 A, 20- W(C) 0 A, 20 W

(D) 0 A, 20- W

EF 5.49 Consider the network shown below :

The power absorbed by load resistance RL is shown in table :

RL 10 kW 30 kW

P 3.6 mW 4.8 mW

What is the value of RL (in kW ), that would absorb maximum power ?


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EF 5.50 Consider the following circuits shown below

The relation between Ia and Ib is(A) 6I Ib a= + (B) 2I Ib a= +(C) 1.5I Ib a= (D) I Ib a=

Common Data For Q. 51 and 52 :In the following circuit, some measurements were made at the terminals a , b and given in the table below.

EF 5.51 The Thevenin equivalent of the unknown network across terminal a -b is(A) 3 W, 14 V (B) 5 W, 16 V

(C) 16 W, 38 V (D) 10 W, 26 V

EF 5.52 The value of R that will cause I to be 1 A, is _____ W

EF 5.53 In the circuit shown in fig (A) if current . AI 2 51 = then current I2 and I3 in fig (B) and (C) respectively are


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Engineering mathematics, General Aptitude

(A) 5 A, 10 A (B) 5 A- , 10 A

(C) 5 A, 10 A- (D) 5 A- , 10 A-

EF 5.54 The V -I relation of the unknown element X in the given network is V AI B= +. The value of A (in ohm) and B (in volt) respectively are

(A) ,2 20 (B) ,2 8

(C) 0.5, 4 (D) . ,0 5 16

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EF 5.1 Correct option is (B).

For, Vs 10 V= , 40 WP =

So, Is 4 AVP

1040

s= = =

Now, Vsl 5 V= , so 2 AIs =l (From linearity)

New value of the power supplied by source is

Psl 5 2 10 WV Is s #= = =l l

Note: Linearity does not apply to power calculations.

EF 5.2 Correct option is (C).

From linearity, we know that in the circuit IV

L

s ratio remains constant

IV

L

s 200 10

20 1003#

= =-

Let current through load is ILl when the power absorbed is 2.5 W, so

PL ( )I RL L2= l

2.5 ( )I 10L2#= l

ILl 0.5 A=

IV

L

s IV 100

L

s= =ll

So, Vsl 100 100 0.5 50 VIL #= = =l

Thus required values are

ILl 0.5 A= , 50 VVs =l

EF 5.3 Correct option is (D).From linearity,

IL AV BIs s= + , A and B are constants

From the table 2 A B14 6= + ...(1)

6 A B18 2= + ...(2)Solving equation (1) and (2)

.A 0 4= , .B 0 6=-So, IL . .V I0 4 0 6s s= -


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Engineering mathematics, General AptitudeEF 5.4 Correct option is (B).

The circuit has 3 independent sources, so we apply superposition theorem to obtain the voltage drop.Due to 16 V source only : (Open circuit 5 A source and Short circuit 32 V source)Let voltage across R2 due to 16 V source only is V1.

Using voltage division

V1 ( )24 88 16=- + 4 V=-

Due to 5 A source only : (Short circuit both the 16 V and 32 V sources)

Let voltage across R2 due to 5 A source only is V2.

V2 ( || 16 || 16 ) 524 #W W W= 6 5 30 volt#= =Due to 32 V source only : (Short circuit 16 V source and open circuit 5 A source)Let voltage across R2 due to V32 source only is V3


V3 .. (32) 12 V16 9 6

9 6= + =

By superposition, the net voltage across R2 is

V V V V1 2 3= + +

4 30 12=- + + 38 volt=

ALTERNATIVE METHOD :

The problem may be solved by applying a node equation at the top node.

EF 5.5 Correct option is (C).Due to 60 A Source Only : (Open circuit 30 A and short circuit 30 V sources)


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12 || 6W W 4 W=

Using current division

Ia (60) 12 A2 82= + =

Again, Ia will be distributed between parallel combination of 12 W and 6 W

I1 (12) 4 A12 66= + =

Due to 30 A source only : (Open circuit 60 A and short circuit 30 V sources)


Ib 4 (30) 12 A4 6= + =

Ib will be distributed between parallel combination of 12 W and 6 W

I2 (12) 4 A12 66= + =

Due to 30 V Source Only : (Open circuit 60 A and 30 A sources)

Using source transformation


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I3 (5) 1 A12 33=- + =-


Using superposition, I I I1 2= +Let I1 is the current due to 9 A source only. (i.e. short 18 V source)

I1 (9) 3 A6 126= + = (current division)

Let I2 is the current due to 18 V source only (i.e. open 9 A source)

I2 1 A6 1218= + =

So, 3 AI1 = , 1 AI2 =

EF 5.7 Correct option is (B).From superposition theorem, it is known that if all source values are doubled, then node voltages also be doubled.

EF 5.8 Correct option is (A).From the principal of superposition, doubling the values of voltage source doubles the mesh currents.

EF 5.9 Correct option is (D).Using source transformation of 4 A and 6 V source.

Adding parallel current sources


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Source transformation of 5 A source

Applying KVL around the anticlockwise direction

I I5 8 2 12- - + - - 0= I9 3- - 0= I 3 A=-Power absorbed by 12 V source

P12 V I12#= (Passive sign convention)

12 3#= - 36 W=-or, 12 V source supplies 36 W power.

EF 5.10 Correct option is (C).Using source transformation, we can obtain I in following steps.

I 2 A3 46 8

714= +

+ = =

ALTERNATIVE METHOD :Try to solve the problem by obtaining Thevenin equivalent for right half of the circuit.


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We know that source transformation also exists for dependent source, so

Current source values

Is I I2

6 3xx= = (downward)

Rs 2 W=

EF 5.12 Correct option is (C).We know that source transformation is applicable to dependent source also.Values of equivalent voltage source

Vs ( )( )I I4 5 20x x= = Rs 5 W=

EF 5.13 Correct answer is 5.Using super position, we obtain I .Due to 10 V source only : (Open circuit 5 A source)

I1 5 A210= =

Due to 5 A source only : (Short circuit 10 V source)

I2 0= I 5 0 5 AI I1 2= + = + =

ALTERNATIVE METHOD :We can see that voltage source is in parallel with resistor and current source so voltage across parallel branches will be 10 V and 10/2 5 AI = =


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EF 5.14 Correct answer is .0 5- .Applying superposition,Due to 6 V source only : (Open circuit 2 A current source)

I1 0.5 A6 66= + =

Due to 2 A source only : (Short circuit 6 V source)

I2 ( )6 66 2= + - (using current division)

1 A=- I 0.5 1 0.5 AI I1 2= + = - =-


This problem may be solved by using a single KVL equation around the outer

loop.

EF 5.15 Correct option is (C).Combining the parallel resistance and adding the parallel connected current sources.

9 3A A- 6 A= (upward)

3 || 6W W 2 W=


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Engineering mathematics, General AptitudeSource transformation of 6 A source

EF 5.16 Correct option is (D).Thevenin Voltage : (Open Circuit Voltage)The open circuit voltage between a -b can be obtained as

Writing KCL at node a

1V24

10Th - + 0=

10 24VTh - + 0=

or VTh 14 volt=-Thevenin Resistance :To obtain Thevenin’s resistance, we set all independent sources to zero i.e., short circuit all the voltage sources and open circuit all the current sources.

RTh 24 W=

EF 5.17 Correct option is (B).Thevenin Voltage :

Using voltage division V1 (10) 4 volt20 3020= + =

and, V2 (10) 6 volt15 1015= + =

Applying KVL, V V Vab1 2- + 0= 4 6 Vab- + 0= VTh 2 voltVab= =-


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Thevenin Resistance :

Rab [20 || ] [15 || 10 ]30W W W W= +

12 6 18W W W= + = RTh 18Rab W= =

EF 5.18 Correct answer is 1.5 A.Using source transformation of 48 V source and the 24 V source

Using parallel resistances combination

Source transformation of 8 A and 6 A sources


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Engineering mathematics, General AptitudeWriting KVL around anticlock wise direction

I I I12 2 40 4 2 16- - + - - - 0= I12 8- 0=

I 1.5 A812= =

EF 5.19 Option (A) is a correct.Using source transformation of 24 V source

Adding parallel connected sources

So, VTh 4 V= , 6RTh W=

EF 5.20 Correct option is (A).Thevenin Voltage: (Open Circuit Voltage)

VTh ( )6 46 40= + - (using voltage division)

24 volt=-


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RTh 6 || 4 2.46 46 4#W W W= = + =

EF 5.21 Correct option is (B).For the circuit of figure (A)

VTh V Va b= - Va 24 V=

Vb ( 6) 4 V6 36= + - =- (Voltage division)

VTh 24 ( 4) 28 V= - - =For the circuit of figure (B), using source transformation

Combining parallel resistances,

12 || 4W W 3 W=Adding parallel current sources,

8 4- 4 A= (Downward)

VTh 12 V=-


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Engineering mathematics, General AptitudeEF 5.22 Correct option is (C).

For the circuit for fig (A)

RTh 6 || 3 2Rab W W W= = =For the circuit of fig (B), as obtained in previous solution.

RTh 3 W=

EF 5.23 Correct answer is .0 5- .Current I can be easily calculated by Thevenin’s equivalent across 6 W.Thevenin Voltage : (Open Circuit Voltage)

In the bottom mesh I2 1 A=In the bottom left mesh V I12 3Th 2- - + 0= VTh 3 (12)(1) 9 V= - =-Thevenin Resistance :

RTh 12 W= (both 4 W resistors are short circuit)so, circuit becomes as


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I 0.5 ARV

6 12 69

189

Th

Th= + = +- =- =-

Note: The problem can be solved easily by a single node equation. Take the

nodes connecting the top 4 W, 3 V and 4 W as supernode and apply KCL.


The current source connected in parallel with load does not affect Thevenin equivalent circuit. Thus, Thevenin equivalent circuit will contain its usual form of a voltage source in series with a resistor.

EF 5.25 Correct option is (C).The network consists of resistor and dependent sources because if it has independent source then there will be an open circuit Thevenin voltage present.

EF 5.26 Correct option is (D).Thevenin Voltage (Open Circuit Voltage) :

Applying KCL at top middle node

1V V V3

26

Th x Th- + + 0=

V V V32

6 1Th Th Th- + + 0= ( )V VTh x=

V V2 6Th Th- + + 0= VTh 6 volt=Thevenin Resistance :

RTh Short circuit currentOpen circuit voltage

IV

sc

Th= =

To obtain Thevenin resistance, first we find short circuit current through a -b


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Writing KCL at top middle node

V V V V32

6 1 30x x x x- + + + - 0=

V V V2 6 2x x x- + + + 0= or Vx 6 volt=-

Isc 2 AV3

036x= - =- =-

Thevenin’s resistance, RTh 3IV

26

sc

Th W= =- =-

ALTERNATIVE METHOD :Since dependent source is present in the circuit, we put a test source across a -b to obtain Thevenin’s equivalent.

By applying KCL at top middle node

V V V V V32

6 1 3x x x x test- + + + - 0=

V V V V2 6 2 2x x x test- + + + - 0= V V2 test x- 6= ...(1)

We have Itest V V

3test x= -

I3 test V Vtest x= - Vx V I3test test= -Put Vx into equation (1)

( )V V I2 3test test test- - 6= V V I2 3test test test- + 6= Vtest I6 3 test= - ...(2)For Thevenin’s equivalent circuit


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RV V

Th

test Th- Itest=

Vtest V R ITh Th test= + ...(3)Comparing equation (2) and (3)

VTh 6 V= , 3RTh W=-

EF 5.27 Correct option is (D).


V V R RR

ThTh

= +b l

From the table,

6 V R33

ThTh

= +b l ...(1)

8 V R88

ThTh

= +b l ...(2)

Dividing equation (1) and (2), we get

86

( )( )

RR

8 33 8

Th

Th= ++

R6 2 Th+ R8 Th= + RTh 2 W=Substituting RTh into equation (1)

6 V 3 23

Th= +b l or VTh 10 V=

EF 5.28 Correct answer is 4.We find Thevenin equivalent across a -b .

IL R RV

Th L

Th= +From the data given in table

10 RV

2Th

Th= + ...(1)

6 RV

10Th

Th= + ...(2)


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Engineering mathematics, General AptitudeDividing equation (1) and (2), we get

610 R

R210

Th

Th= ++

R10 20Th + R6 60Th= + R4 Th 40 10RTh& W= =Substituting RTh into equation (1)

10 V10 2

Th= +

VTh 10(12) 120 V= =

For 20RL W= , IL R RV

Th L

Th= +

A10 20120 4= + =

EF 5.29 Correct option is (C).Norton Current : (Short Circuit Current)The Norton equivalent current is equal to the short-circuit current that would flow when the load replaced by a short circuit as shown below

Applying KCL at node a

I I 2N 1+ + 0=

Since I1 A240 20

65= - =-

So, I 65 2N - + 0=

IN A67=-

Norton Resistance :Set all independent sources to zero (i.e. open circuit current sources and short circuit voltage sources) to obtain Norton’s equivalent resistance RN .

RN 24 W=

EF 5.30 Correct option is (C).Using source transformation of 1 A source


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Again, source transformation of 2 V source

Adding parallel current sources

ALTERNATIVE METHOD :Try to solve the problem using superposition method.

EF 5.31 Correct answer is 56 V.6 W and 3 W resistors are in parallel, which is equivalent to 2 W.

Using source transformation of 6 A source

Source transform of 4 A source


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Adding series resistors and sources on the left

Source transformation of 48 V source

Source transformation of 4 A3 source.

I V19 9

12 72 s= ++ +

Vs (28 ) 12 72I#= - - (28 5) 12 72#= - - 56 V=

EF 5.32 Correct option is (C).Short circuit current across terminal a -b is


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For simplicity circuit can be redrawn as

IN ( )3 63 10= + 3.33 A= (Current division)

Norton’s equivalent resistance

RN 6 3 9 W= + =


The voltage across load terminal is simply Vs and it is independent of any other current or voltage. So, Thevenin equivalent is V VTh s= and R 0Th = (Voltage source is ideal). Norton equivalent does not exist because of parallel connected voltage source.


The output current from the network is equal to the series connected current source only, so I IN s= . Thus, effect of all other component in the network does not change IN .In this case Thevenin’s equivalent is not feasible because of the series connected current source.


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Engineering mathematics, General AptitudeEF 5.35 Correct answer is 4.

We solve this problem using linearity and assumption that 1 AI = .

V1 I I4 2= + (Using KVL)

6 V= I2 I I1= + (Using KCL)

V I41= + 1 2.5 A4

6= + =

V2 I V4 2 1= + (Using KVL)

4(2.5) 6 16 V= + = I Is 3+ I2= (Using KCL)

I V4 12s

2- + I2=

Is 2.5 3.5 A1616= + =

When 3.5 AIs = , I 1 A=

But 14 AIs = , so I .. 14 4 A3 51

#= =

EF 5.36 Correct option is (C).Norton Current : (Short Circuit Current)

Using source transformation

Nodal equation at top center node

( )

I60 24

3 30 6

N- + +

- - + 0=

4 1 IN- + + 0=


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IN 3 A=Norton Resistance :

RN 6 || (3 3) 6 || 6 3Rab W= = + = =So, Norton equivalent will be


Voltage V V R RR

ss L

L= +b l

Power absorbed by RL , PL ( )

( )RV

R RV R

L s L

s L2

2

2

= =+

From above expression, it is known that power is maximum when 0Rs =

NOTE :Do not get confused with maximum power transfer theorem. According to maximum power transfer theorem if RL is variable and Rs is fixed then power dissipated by RL is maximum when R RL s= .

EF 5.38 Correct option is (C).We solve this problem using maximum power transfer theorem. First, obtain Thevenin equivalent across RL .Thevenin Voltage : (Open circuit voltage)


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Engineering mathematics, General AptitudeUsing source transformation

Using nodal analysis V V6

242 4

24Th Th- + +- 0=

2V 48Th - 0 24 VVTh&= =


RTh 6 || 6 3W W W= =Circuit becomes as

For maximum power transfer

RL 3RTh W= =Value of maximum power

Pmax ( )

RV4 L

Th2

=

4 3( )

48 W24 2

#= =

EF 5.39 Correct option is (D).This can be solved by reciprocity theorem. But we have to take care that the polarity of voltage source have the same correspondence with branch current in each of the circuit.In figure (B) and figure (C), polarity of voltage source is reversed with respect to direction of branch current so

IV

1

1 IV

IV

2

2

3

3=- =-

I2 2 AI3= =-


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EF 5.40 Correct option is (C).According to reciprocity theorem in any linear bilateral network when a single voltage source Va in branch a produces a current Ib in branches b, then if the voltage source Va is removed(i.e. branch a is short circuited) and inserted in branch b, then it will produce a current Ib in branch a.

So, I2 20 mAI1= =

EF 5.41 Correct option is (A).According to reciprocity theorem in any linear bilateral network when a single current source Ia in branch a produces a voltage Vb in branches b, then if the current source Ia is removed(i.e. branch a is open circuited) and inserted in branch b, then it will produce a voltage Vb in branch a.

So, V2 2 volt=

EF 5.42 Correct option is (A).We use Millman’s theorem to obtain equivalent resistance and voltage across a -b

Vab 2401

2001

8001

24096

20040

80080

=+ +

- + + -

28.8 V5144=- =-

The equivalent resistance

Rab 1 96

2401

2001

8001 W =

+ +=

Now, the circuit is reduced as

I . 100 mA96 19228 8= + =

EF 5.43 Correct option is (B).Thevenin Voltage: (Open Circuit Voltage):The open circuit voltage will be equal to V , i.e. V VTh =Thevenin Resistance:Set all independent sources to zero i.e. open circuit the current source and short circuit the voltage source as shown in figure


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Open circuit voltage V1=

EF 5.44 Correct option is (B).V is obtained using super position.Due to source V1 only : (Open circuit source I3 and short circuit source V2)

V ( )V100 5050

1= + V31

1= (using voltage division)

so, A 31=

Due to source V2 only : (Open circuit source I3 and short circuit source V1)

V ( )V100 5050

2= + V31

2= (Using voltage division)

So, B 31=

Due to source I3 only : (short circuit sources V1 and V2)

V [ || | | ]I 100 100 1003= I 3100

3= b l

So, C 3100=

EF 5.45 Correct answer is 4.Redrawing the circuit in Thevenin equivalent form


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I RV V

Th

Th= -

or, V R I VTh Th=- + (General form)From the given graph

V I4 8=- +So, by comparing RTh 4 kW= , 8 VVTh =For maximum power transfer R RL Th=Maximum power absorbed by RL

Pmax ( )

4 mWRV4 4 4

8Th

Th2 2

#= = =

EF 5.46 Correct option is (C).The circuit with Norton equivalent

So, I IN + RV

N=

I RV I

NN= - (General form)

From the given graph, the equation of line

I 2 6V= -Comparing with general form

R1N

2= or 0.5RN W=

IN 6 A=

EF 5.47 Correct option is (D).Thevenin voltage: (Open circuit voltage)

VTh 4 4 4 8 V2 2#= + = + =^ h


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Engineering mathematics, General AptitudeThevenin Resistance:

RTh 2 3 5 RNW= + = =Norton Current:

IN ARV

58

Th

Th= =

EF 5.48 Correct option is (C).Norton current, 0IN = because there is no independent source present in the circuit. To obtain Norton resistance we put a 1 A test source across the load terminal as shown in figure.

Norton or Thevenin resistance

RN V1test=

Writing KVL in the left mesh

I I I20 10 1 301 1 1+ - -^ h 0= I I I20 10 30 101 1 1- - + 0= or I1 0.5 A=Writing KVL in the right mesh

V I5 1 30test 1- -^ h 0= 0.V 5 30 5test - - ^ h 0= V 5 15test - - 0=

RN 20V1test W= =

EF 5.49 Correct answer is 30.

For 10 kRL W= , Vab1 610k 3.6m#= = V

For 30 kRL W= , Vab2 1230k 4.8m#= = V

Vab1 R V1010 6

ThTh= + = ...(1)

Vab2 12R V3030

ThTh= + = ...(2)

Dividing equation (1) and (2), we get RTh 30 kW= . Maximum power will be transferred when 30 kR RL Th W= = .


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EF 5.50 Correct option is (C).In circuit (b) transforming the 3 A source in to 18 V source all source are 1.5 times of that in circuit (a) as shown in figure.

Using principal of linearity, 1.5I Ib a=


I R RV

Th

Th= +

From the table, 2 RV

3 Th

Th= + ...(1)

1.6 RV

5 Th

Th= + ...(2)

Dividing equation (1) and (2), we get

.1 62 R

R35

Th

Th= ++

R6 2 Th+ . R8 1 6 Th= + . R0 4 Th 2= RTh 5 W=Substituting RTh into equation (1)

2 V3 5

Th= + or

VTh 2(8) 16 V= =

EF 5.52 Correct option is (D).

We have, I R RV

Th

Th= +

16 VVTh = , 5RTh W=

I R516 1= + =

16 R5= + or R 11 W=


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It can be solved by reciprocity theorem. Polarity of voltage source should have same correspondence with branch current in each of the circuit. Polarity of voltage source and current direction are shown below

So, IV

1

1 IV

IV

2

2

3

3=- =

.2 510 I I

20 402 3

=- =

I2 5 A=- I3 10 A=

EF 5.54 Correct option is (A).To obtain V -I equation we find the Thevenin equivalent across the terminal at which X is connected.Thevenin Voltage : (Open Circuit Voltage)

V1 6 1 6 V#= = V V12 1 3+ - 0= (KVL in outer mesh)

V3 12 6 18 V= + = V V VTh 2 3- - 0= (KVL in Bottom right mesh)

VTh V V2 3= + ( 2 1 2 )VV2 #= = 2 18 20 V= + =Thevenin Resistance :

RTh 1 1 2 W= + =


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Now, the circuit becomes as

I RV V

Th

Th= -

V R I VTh Th= +so A 2RTh W= = B 20 VVTh= =


In the mesh ABCDEA, we have KVL equation as

( ) ( )V I I1 2 1 6 12- + - + - 0= V I2 20= +So, A 2= , B 2=

*************

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