GATE ELECTRICAL ENGINEERING Vol 1 of 4

Second Edition GATE ELECTRICAL ENGINEERING Vol 1 of 4 RK Kanodia Ashish Murolia NODIA & COMPANY

GATE Electrical Engineering Vol 1, 2e RK 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. MRP 590.00 NODIA & COMPANY B 8, Dhanshree Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039 Ph : +91 141 2101150, email : enquiry@nodia.co.in Printed by Nodia and Company, Jaipur

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; autotransformer; 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 overcurrent, 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. ***********

PREFACE This 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. Kanodia Ashish 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 ***********

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 105 EF 5.1 EF 5 CIRCUIT THEOREMS The linear network in the figure contains resistors and dependent sources only. When V s = 10 V, the power supplied by the voltage source is 40 W. What will be the power supplied by the source if V s = 5V ( in W)? EF 5.2 In the circuit below, it is given that when V s = 20 V, I L = 200 ma. What values of I L and V s will be required such that power absorbed by R L is 2.5 W? EF 5.3 (A) 1 A, 25. V (B) 0.5 A, 2V (C) 0.5 A, 50 V (D) 2A, 1.25 V For the circuit shown in figure below, some measurements are made and listed in the table. Which of the following equation is true for I L? (A) IL = 06. Vs+ 04. Is (B) I = 02. V -03. I L s s (C) I = 02. V + 03. I (D) I = 04. V -06. I L s s L s s Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude In the circuit below, the voltage drop across the resistance R 2 will be V EF 106 Circuit Theorems EF 5 EF 5.4 EF 5.5 In the circuit below, current I = I1+ I2+ I3, where I 1, I 2 and I 3 are currents due to 60 A, 30 A and 30 V sources acting alone. The values of I 1, I 2 and I 3 are respectively EF 5.6 EF 5.7 (A) 8A, 8A, (B) 12 A, 12 A, (C) 4A, 4A, (D) 2A, 2A, - 4A -5A - 1A -4A In the circuit below, current I is equal to sum of two currents I 1 and I 2. What are the values of I 1 and I 2? (A) 6A, 1A (B) 9A, 6A (C) 3A, 1A (D) 3A, 4A 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. Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 107 EF 5.8 EF 5.9 EF 5.10 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 In the circuit below, the 12 V source (A) absorbs 36 W (C) absorbs 100 W (B) delivers 4 W (D) delivers 36 W 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? Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 109 EF 5.16 EF 5.17 Which of the following circuit is equivalent to the above circuit? 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) 20 V, 24 W (C) 14 V, 0 W (D) - 14 V, 24 W In the following circuit, Thevenin voltage and resistance across terminal a and b respectively are (A) 10 V, 18 W (C) 10 V, 18.67 W (B) 2V, 18 W (D) 2V, 18.67 W EF 5.18 What is the value of current I in the circuit shown below (in Amp)? Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude The value of R Th and V Th such that the circuit of figure (B) is the Thevenin equivalent circuit of the circuit shown in figure (A), will be equal to EF 110 Circuit Theorems EF 5 EF 5.19 EF 5.20 EF 5.21 (A) R Th = 6 W, V Th = 4V (B) = 6 W, V Th = 28 V R Th (C) R Th = 2 W, V Th = 24 V (D) = 10 W, V Th = 14 V R Th What values of R Th and V Th 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 The value of Thevenin voltage across terminals a-b of figure (A) and figure (B) respectively are (A) 30 V, 36 V (B) 28 V, -12 V (C) 18 V, 12 V (D) 30 V, -12 V Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 111 EF 5.22 EF 5.23 EF 5.24 EF 5.25 EF 5.26 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 In the given circuit, what is the value of current I (in Amp) through 6 W resistor 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 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. For the given circuit, the Thevenin s voltage and resistance looking into a-b are (A) 2V, 3 W (B) 2V, 2 W (C) 6V, (D) 6V, - 9 W -3 W Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude For the circuit, values of voltage V for different values of R are shown in the table. EF 112 Circuit Theorems EF 5 EF 5.27 EF 5.28 EF 5.29 EF 5.30 The Thevenin voltage and resistance of the unknown circuit are respectively. (A) 14 V, 4 W (B) 4V, 1 W (C) 14 V, 6 W (D) 10 V, 2 W 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 I L (in Amps) for = 20 W? R L In the circuit shown below, the Norton equivalent current and resistance with respect to terminal a-b is (A) 17 A, 0 W (B) 2A, 24 W 6 (C) - A, 24 W (D) - 2A, 24 W 6 7 The Norton equivalent circuit for the circuit shown in figure is given by Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 113 EF 5.31 If I = 5A in the circuit below, then what is the value of voltage source V s (in volts)? EF 5.32 What are the values of equivalent Norton current source ( I N ) and equivalent resistance ( R N ) across the load terminal of the circuit shown in figure? EF 5.33 I N R N (A) 10 A 2 W (B) 10 A 9 W (C) 3.33 A 9 W (D) 6.66 A 2 W 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. Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude 2. 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 114 Circuit Theorems EF 5 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 statements 1. 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 Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 115 EF 5.37 EF 5.38 In the circuit below, if R L is fixed and R s is variable then for what value of R s power dissipated in R L will be maximum? (A) R S = R (B) R 0 L S = (C) R = R / 2 (D) R = 2R S L In the circuit shown below the maximum power transferred to R L is P max, then (A) = 12 W, P max = 12 W R L (B) = 3 W, P max = 96 W R L (C) = 3 W, P max = 48 W R L (D) = 12 W, P max = 24 W R L EF 5.39 In the circuit shown in figure (A) if current I 1 = 2A, then current I 2 and I 3 in figure (B) and figure (C) respectively are (A) 2A, 2A (B) - 2A, 2A (C) 2A, - 2A (D) - 2A, -2A S L Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 117 EF 5.44 If V = AV1+ BV2+ CI3 in the following circuit, then values of A, B and C respectively are EF 5.45 EF 5.46 2 2 (A) 3, 3, 1 1 (B),, 3 3 1 1 (C) 2, 2, 1 2 (D),, 3 3 1 3 100 3 1 3 100 3 The V -I relation for the circuit below is plotted in the figure. The maximum power that can be transferred to the load R L will be mw 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) 3A, 2 W (B) 6 W, 2 W (C) 6A, 0.5 W (D) 3A, 0.5 W Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude In the circuit a network and its Thevenin and Norton equivalent are given. EF 118 Circuit Theorems EF 5 EF 5.47 EF 5.48 The value of the parameter are V Th R Th I N R N (A) 4 V 2 W 2 A 2 W (B) 4 V 2 W 2 A 3 W 30 (C) 8 V 1.2 W 3 A 1.2 W 8 (D) 8 V 5 W 5 A 5 W For the following circuit the value of equivalent Norton current I N and resistance R N 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 R L is shown in table : R L 10 kw 30 kw P 3.6 mw 4.8 mw What is the value of R L (in kw ), that would absorb maximum power? Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 119 EF 5.50 EF 5.51 EF 5.52 Consider the following circuits shown below The relation between I a and I b is (A) I = I + 6 (B) I = I + 2 b a (C) I = 1.5I (D) I = I b 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. 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 b b (D) 10 W, 26 V The value of R that will cause I to be 1A, is W EF 5.53 In the circuit shown in fig (A) if current I1 = 25. A then current I 2 and I 3 in fig (B) and (C) respectively are a a Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 120 Circuit Theorems EF 5 (A) 5A, 10 A (B) - 5A, 10 A (C) 5A, - 10 A (D) - 5A, -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) 220, (B) 28, (C) 0.5, 4 (D) 05., 16 ************* Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 121 EF 5.1 EF 5.2 EF 5.3 Correct option is (B). SOLUTION For, V s = 10 V, P = 40 W So, I s = 4A V P = 10 = s Now, V s l = 5V, so I s l = 2A (From linearity) New value of the power supplied by source is P s l = VI ll= 5 2 = 10 W s s # Note: Linearity does not apply to power calculations. Correct option is (C). From linearity, we know that in the circuit Vs ratio remains constant I Vs = 20 I -3 = 100 L 200 # 10 Let current through load is I L l when the power absorbed is 2.5 W, so 2 P L = ( I l) R 2.5 = ( I l) 10 L 2 # I L l = 0.5 A Vs Vs = l = 100 IL I l So, V s l = 100I L l= 100 # 0.5 = 50 V Thus required values are I L l = 0.5 A, V s l= 50 V L L Correct option is (D). From linearity, I L = AVs+ BIs, A and B are constants From the table 2 = 14A+ 6B...(1) 6 = 18A+ 2B...(2) Solving equation (1) and (2) A = 04., B =-06. So, I L = 04. V -06. I s s L L Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 GATE EE vol-4 Control systems, Signals & systems Electrical machines, Power systems Engineering mathematics, General Aptitude 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 R 2 due to 16 V source only is V 1. EF 122 Circuit Theorems EF 5 EF 5.4 EF 5.5 Using voltage division V 1 =- 8 ( 16) =-4V 24 + 8 Due to 5A source only : (Short circuit both the 16 V and 32 V sources) Let voltage across R 2 due to 5A source only is V 2. V 2 = ( 24 W 16 W 16 W) # 5 = 6 # 5 = 30 volt Due to 32 V source only : (Short circuit 16 V source and open circuit 5 A source) Let voltage across R 2 due to 32 V source only is V 3 Using voltage division V 3 = 96. (32) = 12 V 16 + 9. 6 By superposition, the net voltage across R 2 is V = V1+ V2+ V3 =- 4+ 30+ 12 = 38 volt ALTERNATIVE METHOD : The problem may be solved by applying a node equation at the top node. Correct option is (C). Due to 60 A Source Only : (Open circuit 30 A and short circuit 30 V sources) Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 123 12 W 6 W = 4 W Using current division I a = 2 (60) = 12 A 2 + 8 Again, I a will be distributed between parallel combination of 12 W and 6 W I 1 = 6 (12) = 4 A 12 + 6 Due to 30 A source only : (Open circuit 60 A and short circuit 30 V sources) Using current division I 4 b = (30) = 12 A 4 + 6 I b will be distributed between parallel combination of 12 W and 6 W I 2 = 6 (12) = 4 A 12 + 6 Due to 30 V Source Only : (Open circuit 60 A and 30 A sources) Using source transformation Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 124 Circuit Theorems EF 5 EF 5.6 EF 5.7 EF 5.8 EF 5.9 Using current division I 3 =- 3 (5) =-1 A 12 + 3 Correct option is (C). Using superposition, I = I1+ I2 Let I 1 is the current due to 9A source only. (i.e. short 18 V source) I 1 = 6 (9) = 3 A (current division) 6 + 12 Let I 2 is the current due to 18 V source only (i.e. open 9A source) I 2 = 18 = 1A 6 + 12 So, I 1 = 3A, I 2 = 1A Correct option is (B). From superposition theorem, it is known that if all source values are doubled, then node voltages also be doubled. Correct option is (A). From the principal of superposition, doubling the values of voltage source doubles the mesh currents. Correct option is (D). Using source transformation of 4A and 6V source. Adding parallel current sources Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 125 EF 5.10 Source transformation of 5A source Applying KVL around the anticlockwise direction -5- I+ 8-2I- 12 = 0-9- 3I = 0 I =-3A Power absorbed by 12 V source P 12 V = 12 # I (Passive sign convention) = 12 #- 3 =-36 W or, 12 V source supplies 36 W power. Correct option is (C). Using source transformation, we can obtain I in following steps. I = 6+ 8 = 14 = 2A 3 + 4 7 ALTERNATIVE METHOD : Try to solve the problem by obtaining Thevenin equivalent for right half of the circuit. Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 126 Circuit Theorems EF 5 EF 5.11 EF 5.12 GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude Correct option is (B). We know that source transformation also exists for dependent source, so Current source values I 6Ix s = = 3Ix (downward) 2 = 2 W R s Correct option is (C). We know that source transformation is applicable to dependent source also. Values of equivalent voltage source V s = ( 4I )( 5) = 20I R s x = 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) I 1 = 10 = 5A 2 Due to 5 A source only : (Short circuit 10 V source) I 2 = 0 I = I + I = 5+ 0 = 5A 1 2 x 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 I = 10/2 = 5 A Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 127 EF 5.14 Correct answer is - 05.. Applying superposition, Due to 6V source only : (Open circuit 2A current source) EF 5.15 I 1 = 6 = 0.5 A 6 + 6 Due to 2 A source only : (Short circuit 6 V source) I 2 = 6 (- 2) (using current division) 6 + 6 =-1A ALTERNATIVE METHOD : I = I + I = 0.5-1 =-0.5 A 1 2 This problem may be solved by using a single KVL equation around the outer loop. Correct option is (C). Combining the parallel resistance and adding the parallel connected current sources. 9A- 3A = 6A (upward) 3 W 6W = 2 W Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 128 Circuit Theorems EF 5 EF 5.16 EF 5.17 Source transformation of 6 A source 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 VTh - 10 + 1 = 0 24 GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude V Th - 10 + 24 = 0 or V Th =-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. R Th = 24 W Correct option is (B). Thevenin Voltage : Using voltage division V 1 = 20 (10) = 4 volt 20 + 30 and, V 2 = 15 (10) = 6 volt 15 + 10 Applying KVL, V 1 - V 2 + V ab = 0 4-6+ V ab = 0 V Th = =-2 volt V ab Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 129 Thevenin Resistance : R ab = [20 W 30 W] + [15 W 10 W] = 12 W+ 6 W = 18 W R Th = = 18 W R ab 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 8A and 6A sources Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 131 EF 5.21 Thevenin Resistance : Correct option is (B). For the circuit of figure (A) R Th = 6 W 4 W = 6# 4 = 2.4 W 6 + 4 V Th = Va-Vb V a = 24 V V b = 6 (- 6) =- 4V (Voltage division) 6 + 3 V Th = 24 -(- 4) = 28 V For the circuit of figure (B), using source transformation Combining parallel resistances, 12 W 4 W = 3 W Adding parallel current sources, 8-4 = 4A (Downward) V Th =-12 V Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 132 Circuit Theorems EF 5 EF 5.22 Correct option is (C). For the circuit for fig (A) R Th = R ab = 6 W 3W = 2W For the circuit of fig (B), as obtained in previous solution. R Th = 3 W GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 5.23 Correct answer is - 05.. Current I can be easily calculated by Thevenin s equivalent across 6 W. Thevenin Voltage : (Open Circuit Voltage) In the bottom mesh I 2 = 1A In the bottom left mesh -V - 12I + 3 = 0 Thevenin Resistance : Th 2 V Th = 3 -(12)(1) =-9 V R Th = 12 W (both 4 W resistors are short circuit) so, circuit becomes as Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 133 EF 5.24 EF 5.25 EF 5.26 I VTh = 9 0.5 A R + 6 = 12 - =- + 6 18 9 =- Th Note: The problem can be solved easily by a single node equation. Take the nodes connecting the top 4 W, 3V and 4 W as supernode and apply KCL. Correct option is (B). 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. 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. Correct option is (D). Thevenin Voltage (Open Circuit Voltage) : Applying KCL at top middle node VTh - V V 1 3 2 x Th + + = 0 6 VTh - V V 3 2 Th Th + + 1 = 0 ( VTh = Vx) 6-2VTh + VTh + 6 = 0 V Th = 6 volt Thevenin Resistance : Open circuit voltage R VTh Th = = Short circuit current Isc To obtain Thevenin resistance, first we find short circuit current through a-b Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 134 Circuit Theorems EF 5 Writing KCL at top middle node Vx- V V V 3 2 x x x + + 1 + 6 3-0 = 0-2Vx+ Vx+ 6+ 2Vx = 0 or V x =-6 volt I Vx sc = - 0 =- 6 =-2A 3 3 Thevenin s resistance, R VTh Th = =- 3 I 2 6 =- W sc 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 Vx- V V V V 3 2 x x x test + + 1 + 6 - = 0 3 We have - 2V + V + 6+ 2V - 2V = 0 x x x test I test = V test - V 3 3I test = V -V test V x = Vtest -3Itest Put V x into equation (1) 2V -( V - 3I ) = 6 test test test 2V - V + 3I = 6 test test test 2V test - V x = 6...(1) x x V test For Thevenin s equivalent circuit = 6-3I test...(2) Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 135 EF 5.27 V test - V R Th Th = I test V test = VTh + RThItest...(3) Comparing equation (2) and (3) = 6V, =-3 W V Th Correct option is (D). R Th Using voltage division V = V R Th b R + R From the table, 6 = V Th Th l b3 + 3 R l...(1) Th 8 = V 8 8 Th b + R l...(2) Th Dividing equation (1) and (2), we get 8 6 38 ( + RTh) = 83 ( + R ) 6+ 2R Th = + R 8 Th R Th = 2 W Substituting R Th into equation (1) 6 = V 3 Th b 3 + 2 l or V Th = 10 V EF 5.28 Correct answer is 4. We find Thevenin equivalent across a-b. Th I VTh L = RTh + RL From the data given in table 10 VTh = RTh + 2 6 VTh = R + 10 Th...(1)...(2) Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 136 Circuit Theorems EF 5 EF 5.29 EF 5.30 Dividing equation (1) and (2), we get 10 RTh = + 10 6 R + 2 Th 10RTh + 20 = 6RTh + 60 4R Th = 40 & R Th = 10 W Substituting R Th into equation (1) 10 VTh = 10 + 2 V Th = 10(12) = 120 V For R L = 20 W, I VTh L = RTh + RL = 120 = 4 A 10 + 20 GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude 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 IN + I1 + 2 = 0 Since I 1 = 0-20 =- 5 A 24 6 So, I 5 N - + 2 = 0 6 I N =- 7 A 6 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 R N. R N = 24 W Correct option is (C). Using source transformation of 1A source Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 137 Again, source transformation of 2V 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 Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 138 Circuit Theorems EF 5 EF 5.32 Adding series resistors and sources on the left Source transformation of 48 V source Source transformation of 4 A source. 3 I = 12 + V 19 72 + s + 9 V s = (28 # I) -12-72 = (28 # 5) -12-72 = 56 V Correct option is (C). Short circuit current across terminal a-b is Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 139 EF 5.33 EF 5.34 For simplicity circuit can be redrawn as I N = 3 ( 10) = 3.33 A (Current division) 3 + 6 Norton s equivalent resistance Correct option is (C). R N = 6+ 3 = 9W The voltage across load terminal is simply V s and it is independent of any other current or voltage. So, Thevenin equivalent is VTh = Vs and RTh = 0 (Voltage source is ideal). Norton equivalent does not exist because of parallel connected voltage source. Correct option is (B). The output current from the network is equal to the series connected current source only, so IN = Is. Thus, effect of all other component in the network does not change I N. In this case Thevenin s equivalent is not feasible because of the series connected current source. Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 140 Circuit Theorems EF 5 EF 5.35 Correct answer is 4. We solve this problem using linearity and assumption that EF 5.36 I s I GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude I = 1A. V 1 = 4I+ 2I (Using KVL) = 6V I 2 = I1 + I (Using KCL) = V I 4 1 + = 6 + 1 = 2.5A 4 V 2 = 4I 2 + V 1 (Using KVL) = 4(2.5) + 6 = 16 V + I = I 2 (Using KCL) s 3 V2 - = I 4 + 12 2 I s = 16 + 2.5 = 3.5 A 16 When I s = 3.5 A, I = 1A But I s = 14 A, so I =. 14 4 A 35. # = Correct option is (C). Norton Current : (Short Circuit Current) Using source transformation Nodal equation at top center node 0-24 0-( - 6) + + IN = 0 6 3+ 3-4+ 1+ I N = 0 Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 141 EF 5.37 EF 5.38 Norton Resistance : So, Norton equivalent will be I N = 3A R N = = 6 (3+ 3) = 6 6= 3W R ab Correct option is (B). Voltage V V RL = s b R + R 2 2 ( V) Power absorbed by R L, P Vs L = = R 2 L ( Rs+ RL) From above expression, it is known that power is maximum when R s = 0 NOTE : Do not get confused with maximum power transfer theorem. According to maximum power transfer theorem if R L is variable and R s is fixed then power dissipated by R L is maximum when R = R. L s Correct option is (C). We solve this problem using maximum power transfer theorem. First, obtain Thevenin equivalent across R L. Thevenin Voltage : (Open circuit voltage) s L l Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 142 Circuit Theorems EF 5 EF 5.39 Using source transformation Using nodal analysis V Thevenin Resistance : Th - 6 24 + V Th - 24 = 0 2+ 4 R Th = 6 W 6W = 3W Circuit becomes as For maximum power transfer R L = R Th = 3 W Value of maximum power ( VTh) 2 P max = 4R L ( 24) 2 = = 48 W 4# 3 GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude 2VTh - 48 = 0 & V Th = 24V 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 V1 V2 V3 =- =- I I I 1 I 2 2 3 = =-2A I 3 Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 143 EF 5.40 EF 5.41 EF 5.42 EF 5.43 Correct option is (C). According to reciprocity theorem in any linear bilateral network when a single voltage source V a in branch a produces a current I b in branches b, then if the voltage source V a is removed(i.e. branch a is short circuited) and inserted in branch b, then it will produce a current I b in branch a. So, I 2 = I 1 = 20 ma Correct option is (A). According to reciprocity theorem in any linear bilateral network when a single current source I a in branch a produces a voltage V b in branches b, then if the current source I a is removed(i.e. branch a is open circuited) and inserted in branch b, then it will produce a voltage V b in branch a. So, V 2 = 2 volt Correct option is (A). We use Millman s theorem to obtain equivalent resistance and voltage across a-b 96 40 80-240 + 200 + 800 V ab = - 1 1 1 240 + 200 + 800 =- 144 =-28.8 V 5 The equivalent resistance R 1 ab = 1 1 1 = 96 W + + 240 Now, the circuit is reduced as I 200 800 = 28. 8 = 100 ma 96 + 192 Correct option is (B). Thevenin Voltage: (Open Circuit Voltage): The open circuit voltage will be equal to V, i.e. VTh = V 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 Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude EF 144 Circuit Theorems EF 5 EF 5.44 Open circuit voltage = V 1 Correct option is (B). V is obtained using super position. Due to source V 1 only : (Open circuit source I 3 and short circuit source V 2 ) V = 50 ( V1) = V 100 + 50 1 (using voltage division) 3 1 so, A = 1 3 Due to source V 2 only : (Open circuit source I 3 and short circuit source V 1 ) V = 50 ( V2) = V 100 + 50 1 (Using voltage division) 3 2 So, B = 1 3 Due to source I 3 only : (short circuit sources V 1 and V 2 ) So, V = I3[ 100 100 100] = I 100 3b 3 l C = 100 3 EF 5.45 Correct answer is 4. Redrawing the circuit in Thevenin equivalent form Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 145 EF 5.46 EF 5.47 I VTh = - V R Th or, V =- RThI+ VTh (General form) From the given graph V =- 4I + 8 So, by comparing R Th = 4kW, V Th = 8V For maximum power transfer RL = RTh Maximum power absorbed by R L P VTh 2 2 () 8 max = = = 4mW 4RTh 4# 4 Correct option is (C). The circuit with Norton equivalent So, IN + I = R V N I = R V - IN (General form) N From the given graph, the equation of line I = 2V -6 Comparing with general form R 1 N I N = 2 or = 0.5 W = 6A R N Correct option is (D). Thevenin voltage: (Open circuit voltage) V Th = 4+ ^2# 2h= 4+ 4 = 8V Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 146 Circuit Theorems EF 5 EF 5.48 Thevenin Resistance: R Th = 2+ 3 = 5W = R N Norton Current: I VTh N = = 8 A R 5 Th GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude Correct option is (C). Norton current, I N = 0 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 R N = V 1 test Writing KVL in the left mesh 20I1+ 10^1 -I1h-30I1 = 0 20I1-10I1-30I1+ 10 = 0 or I 1 = 0.5 A Writing KVL in the right mesh Vtest -51 ^ h -30I1 = 0 Vtest -5-30^0. 5h = 0 Vtest -5-15 = 0 R N = V 20 1 test = W EF 5.49 Correct answer is 30. For R L = 10 kw, V ab1 = 10k # 3.6m = 6 V For R L = 30 kw, V ab2 = 30k # 4.8m = 12 V V ab1 = 10 10 R V Th = 6...(1) + Th V ab2 = 12 30 30 R V Th =...(2) + Th Dividing equation (1) and (2), we get R Th = 30 kw. Maximum power will be transferred when R = R = 30 k. L Th W Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 147 EF 5.50 EF 5.51 EF 5.52 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, I = 1.5I Correct option is (B). b a I VTh = R + RTh From the table, 2 VTh =...(1) 3 + RTh 1.6 VTh =...(2) 5 + RTh Dividing equation (1) and (2), we get 16 2 5 RTh = +. 3 + R Th 6+ 2R Th = 8+ 1. 6R Th 04. R Th = 2 R Th = 5 W Substituting R Th into equation (1) 2 VTh = or 3 + 5 V Th = 2(8) = 16 V Correct option is (D). We have, I VTh = R + R Th V Th = 16 V, = 5 W I R Th = 5 16 = 1 + R 16 = 5 + R or R = 11 W Buy Online:

GATE EE vol-1 Electric circuit & Field, Electrical & electronic measurement GATE EE vol-2 Analog electronics, Digital electronics, Power electronics GATE EE vol-3 Control systems, Signals & systems EF 148 Circuit Theorems EF 5 EF 5.53 EF 5.54 Correct option is (B). GATE EE vol-4 Electrical machines, Power systems Engineering mathematics, General Aptitude 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, V1 V2 V3 =- = I1 I2 I3 10 =- 20 = 40 25. I I I 2 I 3 =-5A = 10 A 2 3 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) V 1 = 6# 1 = 6V + V - V = 0 (KVL in outer mesh) V 3 = 12 + 6 = 18 V 12 1 3 VTh -V2- V3 = 0 (KVL in Bottom right mesh) V Th = V2+ V3 ( V 2 = 2# 1 = 2 V) = 2+ 18 = 20V Thevenin Resistance : R Th = 1+ 1 = 2W Buy Online:

GATE Electrical Engineering-2015 in 4 Volumes by R. K. Kanodia & Ashish Murolia EF 5 Circuit Theorems EF 149 Now, the circuit becomes as I = V - V R Th Th V = R I+ V so A = = 2 W B Th R Th Th = = 20 V V Th ALTERNATIVE METHOD : In the mesh ABCDEA, we have KVL equation as V- 1( I+ 2) - 1( I+ 6) - 12 = 0 V = 2I + 20 So, A = 2, B = 2 ************* Buy Online: