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GATEELECTRICAL ENGINEERING

Vol 2 of 4

Second Edition

GATEELECTRICAL ENGINEERING

Vol 2 of 4

RK Kanodia Ashish Murolia

NODIA & COMPANY

GATE Electrical Engineering Vol 2, 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

AE ANALOG ELECTRONICS

AE 1 Diode Circuits AE 3

AE 2 BJT Biasing AE 39

AE 3 BJT Amplifiers AE 79

AE 4 FET Biasing AE 109

AE 5 FET Amplifiers AE 134

AE 6 Output Stages and Power Amplifiers AE 155

AE 7 Op-Amp Characteristics and Basic Circuits AE 177

AE 8 Op-Amp Application AE 210

AE 9 Active Filters AE 250

AE 10 Gate Solved Questions AE 281

DE DIGITAL ELECTRONICS

DE 1 Number System and Codes DE 3

DE 2 Boolean Algebra and Logic Simplification DE 28

DE 3 The K-Map DE 75

DE 4 Combinational Circuits DE 101

DE 5 Sequential Circuits DE 140

DE 6 Digital Systems DE 168

DE 7 Logic Families DE 191

DE 8 Microprocessor DE 223

DE 9 Gate Solved Questions DE 249

PE POWER ELECTRONICS

PE 1 Power Semiconductor Devices PE 3

PE 2 Diode Circuits and Rectifiers PE 16

PE 3 Thyristor PE 29

PE 4 Phase Controlled Converters PE 48

PE 5 Choppers PE 76

PE 6 Inverters PE 94

PE 7 AC and DC Drives PE 114

PE 8 Gate Solved Questions PE 125

***********

PE 9 Phase Controlled Converters PE 1EF 9 Phase Controlled Converters EF 1

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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 Aptitude

PE 1PHASE CONTROLLED CONVERTERS

Common Data For Q. 1 to 3:A single phase 230 V, 50 Hz ac source is feeding a fully controlled bridge converter shown in the figure. The firing angle is 30c.

PE 1.1 The dc output voltage will be(A) 126.8 V (B) 96.6 V

(C) 179.3 V (D) 63.4 V

PE 1.2 If a freewheeling diode is connected across the load, then what is the value of dc output voltage ?(A) 193.2 V (B) 136.6 V

(C) 386.4 V (D) 273.2 V

PE 1.3 When the thyristor Th3 gets open circuited, the value of dc output current flowing through a load of 10 W is ___ A.

PE 1.4 In single-phase to single-phase cyclo converter, if 1a and 2a are the trigger angles of positive converter and negative converter, then(A) 1 2 2a a+ = p (B) 1 2a a p+ =(C) 1 2 2

3a a+ = p (D) 21 2a a p+ =

PE 1.5 A three-phase, half-wave controlled converter is fed from a 380 V (line), 50 Hz ac supply and is operating at a firing angle of 45c. The thyristors have a forward voltage-drop of 1.2 V. What will be the approximate average load voltage ? ( in V)

PE 1.6 In the given circuit, the thyristor is fired at an angle /4p in every positive half-cycle of the input ac voltage. The average power across the load will be ___ kW.


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GATE Electrical Engineering-2015in 4 Volumes

by R. K. Kanodia & Ashish MuroliaS

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PE 1.7 A line commutated ac to dc converter is shown in the figure. It operates from a three phase, 50 Hz, 580 V (line to line) supply. The load current I0 is ripple free and constant at 3464 A. For an average output voltage of 648 V, the delay angle a is ___ degree.

PE 1.8 In a 3-phase to 1-phase cyclo converter employing 6-pulse bridge circuit, if the input voltage is 200 V per phase the fundamental rms value of output voltage will be

(A) 600p V (B) 300 3 V

(C) 300 3p V (D) 300

p V

PE 1.9 A single-phase half controlled bridge rectifier is operated from a source sinV t100 314s = . The average power drawn by a resistive load of 10 ohms at a

firing angle 45ca = is(A) 295.5 W

(B) 500 W

(C) 267 W

(D) 454.5 W

PE 1.10 In a fully-controlled converter the load voltage is controlled by which of the following quantity ?(A) extension angle (B) firing angle

(C) conduction angel (D) none

PE 1.11 The fully controlled bridge converter shown in the figure is fed from a single-phase source. The peak value of input voltage is Vm , What will be the average output dc voltage Vdc for a firing of 30c ?

(A) . V0 6 m (B) . V077 m

(C) . V0 155 m (D) . V0 424 m


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Engineering mathematics, General AptitudePE 1.12 When the firing angle a of a single phase fully controlled rectifier feeding constant

d.c. current into the load is 30c, what is the displacement factor of the rectifier ?(A) 1 (B) 0.5

(C) 3 (D) 23

PE 1.13 A single-phase ac voltage regulator is fed from a 50 Hz supply system. If it supplies a load comprising a resistance of 2 W connected in series with an inductance of 6.36 mH, then the range of firing angle ‘a ’ providing controlled voltage would be(A) 0 180< <c ca (B) 45 180< <c ca (C) 90 180< <c ca (D) 0 45< <c ca

PE 1.14 In a single phase full wave controlled bridge rectifier, minimum output voltage and maximum output voltage are obtained at which conduction angles ?(A) ,0 180c c respectively

(B) ,180 0c c respectively

(C) ,0 0c c respectively

(D) ,180 180c c respectively

PE 1.15 A half-controlled bridge converter is operating from an r.m.s input voltage of 120 V. Neglecting the voltage drops, what are the mean load voltage at a firing delay angle of 0c and 180c, respectively ?

(A) 120 2 2#p V and 0 (B) 0 and 120 2 2#

p V

(C) 120 2p V and 0 (D) 0 and 120 2

p V

PE 1.16 In the single phase voltage controller circuit shown in the figure, for what range of triggering angle ( )a , the input voltage ( )V0 is not controllable ?

(A) 0 45< <c ca (B) 45 135< <c ca(C) 90 180< <c ca (D) 135 180< <c ca

PE 1.17 A single-phase full-bridge converter with a free-wheeling diode feeds an inductive load. The load resistance is .15 53 W and it has a large inductance providing constant and ripple free d.c. current. Input to converter is from an ideal 230 V, 50 Hz single phase source. For a firing delay angle of 60c, the average value of diode current is ___ A.

PE 1.18 A single phase fully controlled bridge converter feeds an inductive load. Assume the load current to be constant and the firing angle is 45c, the input harmonic factor is ___


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PE 1.19 The most suitable solid state converter for controlling the speed of the three-phase cage motor as 25 Hz is(A) Cyclo converter

(B) Current source inverter

(C) Voltage source inverter

(D) load commutated inverter

PE 1.20 The fully controlled thyristor converter in the figure is fed from a single-phase source. When the firing angle is 0c, the dc output voltage of the converter is 300 V. What will be the output voltage for a firing angle of 60c, assuming continuous conduction? (in V)

PE 1.21 A single-phase half controlled converter shown in the figure feeding power to highly inductive load. The converter is operating at a firing angle of 60c.

If the firing pulses are suddenly removed, the steady state voltage ( )V0 waveform of the converter will become


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PE 1.22 A three pulse converter is feeding a purely resistive load. What is the value of firing delay angle a , which dictates the boundary between continuous and discontinuous mode of current conduction ?(A) 0ca =

(B) 30ca =(C) 60ca =

(D) 150ca =

PE 1.23 A single phase fully controlled bridge converter supplies a load drawing constant and ripple free load current, if the triggering angle is 30c, the input power factor will be ___

PE 1.24 A 3-phase cycloconverter is used to obtain a variable frequency single-phase a.c. output. The single phase a.c. load is 220 V, 60 A at a power factor of 0.6 lagging. The rms value of input voltage per phase required is ____ V.

PE 1.25 The total harmonic distortion (THD) of ac supply input current of rectifiers is maximum for(A) single-phase diode rectifier with dc inductive filter

(B) 3-phase diode rectifier with dc inductive filter

(C) 3-phase thyristor with inductive filter

(D) Single-phase diode rectifier with capacitive filter

PE 1.26 A six pulse thyristor rectifier bridge is connected to a balanced 50 Hz three phase ac source. Assuming that the dc output current of the rectifier is constant, the lowest frequency harmonic component in the ac source line current is ___ Hz.

PE 1.27 A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.


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Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of 10I0 = A will be ___ degree

PE 1.28 A three phase fully controlled bridge converter is feeding a load drawing a constant and ripple free load current of 10 A at a firing angle of 30c. The approximate Total harmonic Distortion (%THD) and the rms value of fundamental component of input current will respectively be(A) 31% and 6.8 A

(B) 31% and 7.8 A

(C) 66% and 6.8 A

(D) 66% and 7.8 A

PE 1.29 An AC voltage-regulator using back-to-back connected SCRs is feeding an RL load. The SCR firing angle <a f (f is power factor angle of the load). If SCRs are fired using short-duration gate pulses, the output load-voltage waveform will be(A) Full-wave rectified

(B) Sinusoidal

(C) Symmetrical chopped a.c. voltage

(D) half-wave rectified

PE 1.30 A single-phase fully controlled thyristor bridge ac-dc converter is operating at a firing angle of 25c and an overlap angle of 10c with constant dc output current of 20 A. The fundamental power factor (displacement factor) at input ac mains is(A) 0.78 (B) 0.827

(C) 0.866 (D) 0.9

PE 1.31 A single phase full-wave half-controlled bridge converter feeds an inductive load. The two SCRs in the converter are connected to a common DC bus. The converter has to have a freewheeling diode.(A) because the converter inherently does not provide for free-wheeling

(B) because the converter does not provide for free-wheeling for high values of triggering angles

(C) or else the free-wheeling action of the converter will cause shorting of the AC supply

(D) or else if a gate pulse to one of the SCRs is missed, it will subsequently cause a high load current in the other SCR.


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Engineering mathematics, General AptitudePE 1.32 In the circuit shown in Figure, L is large and the average value of I0 is 100 A.

Then which of the following is true for the circuit ?

(A) The thyristor is gated in the positive half cycle of ( )e t at a delay angle a equal to .167 9c.

(B) The thyristor is gated in the negative half cycle of ( )e t at a delay angle a equal to 122. 52 c.

(C) The thyristor is gated in the positive half cycle of ( )e t at a delay angle a equal to 122. 52 c.

(D) The thyristor is gated in the negative half cycle of ( )e t at a delay angle a equal to .167 9c.

PE 1.33 When a line commutated converter operates in the inverter mode(A) it draws both real and reactive power from the A.C. supply.

(B) it delivers both real and reactive power to the A.C. supply

(C) it delivers real power to the A.C. supply

(D) it draws reactive power from the A.C. supply.

PE 1.34 In a 3-phase controlled bridge rectifier, with an increase of overlap angle, the output dc voltage.(A) decreases (B) increases

(C) does not change (D) depends upon load inductance

PE 1.35 For a single phase a.c. voltage controller feeding a resistive load, what is the power factor?(A) Unity for all values of firing angle

(B) sin121 2

/1 2

pp a a- +^ h: D& 0

(C) sin121 2

/1 2

pp a a+ +^ h; E& 0

(D) sin121 2

/1 2

pp a a- -^ h; E& 0

where a is firing angle measured from voltage zero.

PE 1.36 In a dual converter, the circulating current(A) allows smooth reversal of load current, but increases the response time

(B) does not allow smooth reversal of load current, but reduces the response time

(C) allows smooth reversal of load current with improved speed of response

(D) flows only if there is no interconnecting inductor.


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PE 1.37 A PWM switching scheme is used with a three phase inverter to(A) reduce the total harmonic distortion with modest filtering.

(B) minimize the load on the DC side

(C) increase the life of the batteries

(D) reduce low order harmonics and increase high order harmonics

PE 1.38 A half controlled bridge converter feeds a resistive load of 10 W with ripple free current. If the input voltage is 240 V, 50 Hz and the triggering angle is 60c then the value of rms input current is ___ A.

PE 1.39 A three phase fully controlled bridge converter is fed from a 400 V (line to line) ac source. A resistive load of 100 W draws 400 W of power form the converter, the input power factor will be ___

PE 1.40 A single-phase half-wave controlled converter is fed from a sinusoidal source. If the average output voltage is 25% of the maximum possible average output voltage for a purely resistive load, then firing angle is(A) /4p (B) /2p(C) /3p (D) /6p

PE 1.41 What is the power factor of a single phase a.c. regulator feeding a resistive load?(A) Per unit power 2

^ h (B) Per unit power /1 2^ h

(C) 2

Per unit power 2^ h

(D) 2Per unit power /1 2^ h

PE 1.42 A single-phase half-controlled bridge rectifier is feeding a load drawing a constant and ripple free load current at a firing angle /6a p= . The harmonic factor(HF) of input current and the input power factor respectively are(A) 30.80%, 0.922 (B) 4.72%, 0.6

(C) 60%, 0.827 (D) 96.6%, 0.477

PE 1.43 A full-wave controlled bridge rectifier is fed by an ac source of 230 V rms, 50 Hz . The value of load resistance is 15 ohm. For a delay angle of 30c the input power factor is(A) 0.840 (B) 0.70

(C) 0.985 (D) 0.492

PE 1.44 In the continuous conduction mode the output voltage waveform does not depend on(A) firing angle (B) conduction angle

(C) supply (D) load

PE 1.45 The rectification efficiency of a single phase half-wave controlled rectifier having a resistive load and the delay angle of /2p is(A) 24.28% (B) 45.04%

(C) 20.28% (D) 26.30%


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Engineering mathematics, General AptitudePE 1.46 A single phase ac voltage controller is controlling current in a purely inductive

load. If the firing angle of the SCR is a , What will be the conduction angle of the SCR?(A) p (B) p a-^ h

(C) 2p a-^ h (D) 2p

PE 1.47 For a single phase half-controlled bridge converter having highly inductive load, the delay angle is /2p . The harmonic factor will be _____ %

PE 1.48 In the circuit shown in the figure, the SCRs are triggered at 30c delay. The current through 100 W resistor is ____A

PE 1.49 A three phase half wave controlled rectifier circuit is shown in the figure. It is operated from 3-f star connected, supply transformer with a line to line ac supply voltage of 440 volts rms, at 50 Hz. The thyristor are triggered at a delay angle of 30ca = . Assume continuous ripple free current.

The average output current is _____A.

PE 1.50 In the circuit shown in figure, a battery of 6 V is charged by a 1-f one pulse thyristor controlled rectifier. A resistance R is to be inserted in series with the battery to limit the charging current to 4 A. The value of R is ___ W .


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PE 1.51 A single-phase, 230 V, 50 Hz ac mains fed step down transformer (4:1) is supplying power to a half-wave uncontrolled ac-dc converter used for charging a battery(12 V dc) with the series current limiting resistor being 19.04 W. The charging current is ____ A.

PE 1.52 An integral cycle AC voltage controller is feeding a purely resistive circuit from a single phase ac voltage source. The current waveform consists alternately burst of N -complete cycle of conduction following by M -complete cycles of extinction. The rms value of the load voltage equals the rms value of the supply voltage for:(A) N M= (B) N 0=(C) N M 0= = (D) M 0=

PE 1.53 A 3-phase fully controlled bridge converter with free wheeling diode is fed from 400 V, 50 Hz AC source and is operating at a firing angle of 60c. The load current is assumed constant at 10 A due to high load inductance. The input displacement factor (IDF) and the input power factor (IPF) of the converter will be(A) 0.867; 0.828IDF IPF= = (B) 0.867; 0.552IDF IPF= =(C) 0. 5; 0.47895IDF IPF= = (D) 0.5; 0.318IDF IPF= =

PE 1.54 A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 W,then each thyristor will be reverse biased for a period of ___ degree.

PE 1.55 A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of .20 W as shown in figure. The SCRs are triggered by a constant dc signal. If SCR2 gets open circuited, what will be the average charging current ? (in A)

PE 1.56 Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle a in every positive half cycle of the input voltage. If the peak value of the instantaneous output voltage equals 230 V, the firing angle a is close to

(A) 45c (B) 135c

(C) 90c (D) 83.6c


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Engineering mathematics, General AptitudePE 1.57 A single-phase ac regulator fed from 50 Hz supply feeds a load having 4 W

resistance and 12 : 73 mH inductance. The control range of firing angle will be(A) 0° to 180°

(B) 45° to 180°

(C) 90° to 180°

(D) 0° to 45°

PE 1.58 In the single phase diode bridge rectifier shown in figure, the load resistor is 50R W= . The source voltage is ( )sinV t200 w= , where 2 50#w p= radians per

second. The power dissipated in the load resistor R is

(A) 3200p W (B) 400

p W

(C) 400 W (D) 800 W

PE 1.59 A half-wave thyristor converter supplies a purely inductive load as shown in figure. If the triggering angle of the thyristor is 120c, the extinction angle will be ___ degree.

PE 1.60 A single phase half wave rectifier circuit is shown in the figure. The thyristor is fired at 30c in each positive half cycle. The values of average load voltage and the rms load voltage will respectively be

(A) 475.2 V, 190.9 V

(B) 237.64 V, 194.2 V

(C) 118.8 V, 197.1 V

(D) 237.6 V, 197.1 V

PE 1.61 A dc battery of 50 V is charged through a 10 W resistor as shown in the figure. Assume that the thyristor is continuously fired. The average value of charging current is ___ A.


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PE 1.62 A bridge converter is fed from a source sinV V ts m w= as shown in the following figure. What will be the output voltage for a firing angle of a ? Assume continuous conduction.

(A) cosV2 m

p a (B) cosV2 1m

p a+^ h

(C) cosV2 3m

p a+^ h (D) cosVm

p a

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SOLUTIONS

PE 1.1 Correct option is (C).DC output voltage

Vdc cosV2 m

p a=

( )

30 179.3cos V2 2 230

cp= =

PE 1.2 Correct option is (A).When free wheeling diode is present, Th1 and Th2 will conduct from a to p while Th3 and Th4 will conduct for p a+ to 2p.

Vdc (230)( )

(1 )sin cosd1 22 230

p q p a= = +a

p#

( )

(1 30 ) 193.185cos V2 230

cp= + =

PE 1.3 Correct answer is 9.7 A.When Th3 gets open circuited, the circuit will work as a half wave rectifier, the output dc voltage

Vdc (230)sin d21 2p q q=

a

p#

( )

( ) ( )cos cos2

2301

2230 1 30c

pa

p= + = +

96.6 V=

Average dc output current

Idc . 9.7 A10

96 6= =

PE 1.4 Correct option is (B).In practice, the firing angle pa of positive group cannot be reduced to zero, for this firing angle corresponding to 180 180n p ca a= - = for negative group, because of commutation overlap and thyristor turn off time problems. But

180p n ca a+ = .

PE 1.5 Correct answer is 180.2 V.

Here Vm 310.3 V3

380 2= =

Let the thyristor voltage drop is ( )Vt , then average dc voltage

Vdc cosV V23 3

m tp a= -

45 1.2cos2 3

3 3 380 2# cp

= -

180.2 V=


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PE 1.6 Correct answer is 1.1 kW.

RMS load voltage

V ( )rmsdc sinV 4 82

m21

pp a

pa= - +: D

220 1 .sin

V44

82 04 88

21

pp p

p

p=

-+ == G

Average power across the load

Pac ( . )

. kWRV

10104 88

1 1( )rmsdc2 2

= = =

PE 1.7 Correct answer is 34.18°.

Here 580V VLine =Average output voltage

Vdc cos cosV V3 3 3 2inem Lp a p a= =

648 cos3 2 580# #p a=

or a 34.18c=

PE 1.8 Correct option is (A).Given 3 phase 6 pulse bridge, thus m 6= and VV 200ph =The fundamental r.m.s output voltage

Vor sinVm

mph pp

= a ak k9 C

sin2006

6600

pp

p= =b al k: D

PE 1.9 Correct option is (D).

Vs ,sin t100 314= R 10 W= , 45ca =RMS load voltage

V ( )rmsdc sinV 2 42 /

m

1 2

pp a

pa= - +: D

V ( )rmsdc 100 67.42sin

V24

42

/1 2

pp p

p

p=

-+ == G

The average power delivered to the load is

Pac RV ( )rmsdc

2

=

( . )

454.5 W1067 42 2

= =

PE 1.10 Correct option is (B).The average value of dc voltage i.e. load voltage can be varied by controlling the phase angle(a) of firing pulses.

PE 1.11 Correct option is (C).The average output dc voltage


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Engineering mathematics, General Aptitude Vdc ( )sinV td t1

mp w w=a

p a+

#

( )cosV tm

p w= - ap a+

[ ( )]cos cosVm

p a p a= - +

cosV2 m

p a=

Given 30ca = , the average dc output voltage

Vdc cosV2 m

p a=

( )cosV2 30m cp=

0.155132Vm=

PE 1.12 Correct option is (D).For a 1-f full converter the displacement factor is

DF cos cos30ca= = 23=

PE 1.13 Correct option is (B).We have R 2L W = , . mHL 6 36= , Hzf 50= ,

. .Tan 22 50 6 36 10 44 971

3# # #p =-

-b l

mina 45cf = =

PE 1.14 Correct option is (B).For a single phase fully controlled bridge rectifier, the average output voltage is given by

V0 cosV 1m

p a= +^ h

Output voltage is minimum for 180ca = and maximum for 0ca = .

PE 1.15 Correct option is (A).For half-controlled bridge rectifier, average output voltage

V0 cosV 1m

p a= +6 @

For 0ca = , V0 cos120 2 1 0 120 2 2#cp p= + =6 @

For 180ca = , V0 cos120 2 1 180 0cp + =6 @

PE 1.16 Correct option is (A).

R jXL+ j50 50= +

tanf RL

5050 1w= = = or f 45c=

so, firing angle ‘a ’ must be higher the 45c, Thus for 0 45< < ca , V0 is uncontrollable.


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PE 1.17 Correct answer is 3.33 A.Average output current

I0 cosRV 1m

p a= +^ h

. cos15 53230 2 1 60#

cp +^ h

Average current through diode

IFW ( )I d t131 10

/o

0

3#p w= =

p#

3.33 A=

PE 1.18 Correct answer is 0.482.

Input harmonic factor .8 1 0 482/2 1 2p= - =: D

PE 1.19 Correct option is (A).Speed can be controlled by changing the frequency. Cycloconverter directly converts ac power at input frequency to a different frequency.

PE 1.20 Correct answer is 150 V.Given fully-controlled thyristor converter, when firing angle 0a = , dc output voltage Vdc0 300 V= If a 60c= , then ?Vdc =For fully-controlled converter

Vdc0 cosV2 2 dc1

p a=

Since a 0= , 300Vdc0 = V

300 0cosV2 2 dc1 cp=

Vdc1 2 2300p=

At a 60c= , Vdc2 cos2 22 2300 60# cp

p=

300 15021 V#= =

PE 1.21 Correct option is (A).Output of this

Here the inductor makes T1 and T3 in ON because current passing through T1 and T3 is more than the holding current.


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Engineering mathematics, General AptitudePE 1.22 Correct option is (D).

PE 1.23 Correct option is 0.78.Given 30ca = , in a 1-f fully bridge converterwe know that,

Power factor cosDistortion factor# a= D.f. (Distortion factor) / 0.9I Is s(fundamental)= = power factor . cos0 9 30# c= .0 78=

PE 1.24 Correct answer is 266 V.We have VV 220or =For 3-phase to single phase cyclo converter

Vor sinr Vm

mph pp

= a ak k9 C

where Vph = perphase input voltage m 3= for 3-phase pulse drive.

cosr mma = is voltage reduction factor

Thus 220 sinV13

3ph pp

= b al k: D

Vph .266 02=

PE 1.25 Correct option is (D).Single phase diode rectifier with capacitive filter has maximum THD.

PE 1.26 Correct answer is 250 Hz.For six pulse thyristor rectifier bridge the lowest frequency component in AC source line current is of 250 Hz.

PE 1.27 Correct answer is 129°.Here for continuous conduction mode, by Kirchoff’s voltage law, average load current

V I2 150a- + 0=

Ia V

2150= +

10I1` = A, So V 130=- V

cosV2 m

p a 130=-

cos2 2 230# #p a 130c=-

a 129c=


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PE 1.28 Correct option is (B).

Total rms current Ia 10 8.1632 A#= =

Fundamental current Ia1 0.78 10 7.8 A#= =

THD 1 1DF2= -

where DF .. 0.955I

I0 816 100 78 10

a

a1

##= = =

THD . 3 %0 9551 1 1

2= - =b l

PE 1.29 Correct option is (D).In the case of RL load, the output voltage can be controlled for a in the range of f p- . If <a f and triggered with short duration pulse, then one of the thyristor will not come in to conduction. The resultant circuit act as half wave rectifier.

PE 1.30 Correct option is (A).

Firing angle a 25c=Overlap angle m 10c=

so, I0 [ ( )]cos cosLsVm

w a a m= - +

20 [ ( )]cos cosLs2 50230 2 25 25 10#

c c cp= - +

Ls 0.0045 H=

V0 cosV LsI2 m 0

pa

pw= -

. .. .cos

3 142 230 2 25

3 142 3 14 50 4 5 10 203

# # # # # #c= --

.187 73 9= - .178 74c=

Displacement factor V IV I

s s

0 0=

.230 20

178 25 20##= .0 78=

PE 1.31 Correct option is (B).Single phase full wave half controlled bridge converter feeds an Inductive load. The two SCRs in the converter are connected to a common dc bus. The converter has to have free wheeling diode because the converter does not provide for free wheeling for high values of triggering angles.

PE 1.32 Correct option is (C).Assuming continuous conduction. We have

so V0 I R E0= + 100 0.01 20 21#= + =For a single-phase half-wave converter, average output voltage is

V0 ( )cosV

21 21m

pa= = +

( )cos1 a+ 122.252 200

21 2 &## cp a= =


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Engineering mathematics, General AptitudeFrom the output waveform given below, we observed that the thyristor is gated in positive half cycle.

PE 1.33 Correct option is (C).In the inverting mode a line commutated converted operates for phase angles 90c to 180c. When the dc voltage is negative power flow is from dc to ac and the converter functions as inverter. As dc power is fed back, it is real power.

PE 1.34 Correct option is (A).For a 3-phase fully-controlled converter, output dc voltage is given as

Vdc ( )cosV L I

3 3 30mph sdp a m p

w= + +

Where m is the overlap angle. So when the overlap angle is increased, the cosine term in the above expression decreases and the output dc voltage also decreases.


P.F VVor

S=

r.m.s value Vor sinV td t1 /

m2 2

1 2

p w w=a

p< F#

sinV21 2

/S

1 2

pp a a= - +^ h: D

PE 1.36 Correct option is (C).The circulating current helps in maintaining continuous conduction of both the converters irrespective of load and the time response to change the operation from one quadrant to other is faster.

PE 1.37 Correct option is (D).In a three-phase inverter, the supply current consists of one pulse per half-cycle and the lowest order harmonic is third. It is difficult to eliminate the lowest order harmonic current. The lowest order harmonics can be reduced if the supply current has more than one pulse per half-cycle. In PWM lowest order harmonic can be eliminated and higher order harmonics can be increased.

PE 1.38 Correct answer is 13.23 A.

Vdc (1 )cosVm

p a= +

( )

(1 60 )cos2 240

cp= +


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162.03 V=

Load current I . 16.203 ARV

10162 03

L

dc= = =

RMS input current

Is .I 1 16 203 1 60. .0 5 0 5cpa

p= - = -a bk l

13.23 A=

PE 1.39 Correct answer is 0.5.

Load Current

IL 2 A100400 .0 5

= =b l

In a three-phase fully controlled bridge converter input rms current Is or the current in each supply phase exists for 120c in every 180c.Therefore rms value of input current

Is 1.15 A1802 120 .0 5#= =b l

Input apparent power 400 1.15 796.72 VA3 # #= = . cos796 72 q 400=Power factor cos q 0.5 lagging=

PE 1.40 Correct option is (C).Average output voltage

Vdc (1 )cosV2

m

p a= +

The maximum output voltage is obtained when 0a =

Vdcmax Vm

p=

Given Vdc 25% 0.25V Vm m

p p= =b l

So 0.25Vm

p (1 )cosV2

m

p a= +

The Firing angle is

a 60c=


In a AC voltage controller P.f VVor

S=

Per unit power //

Rated powerPresent power

V RV Ror

S2

2

= =^ h

Thus p.f VV Per unit poweror

S= =

PE 1.42 Correct option is (A).Supply rms current

Irms I 1 /dc

1 2

pa= -a k


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/0.91I I1

6 /

dc dc

1 2

pp= - =c m

Now, the rms value of the supply fundamental component of input current.

Irms1 /2cosI2 2 dc

p a= 15 0.869cosI I2 2dc dcp= =

Harmonic factor (HF) on input current

HF II 1

1

/

rms

rms2 1 2

= -b l; E .. 30.80 %0 869

0 91 1/2 1 2

= - =b l; E

Input power factor /2 0.922 ( )cos laggingII 1

rms

rms a= =

PE 1.43 Correct option is (C).The rms load voltage,

Vrms sinV 2 4

2 /

m

1 2

pp a

pa= - +: D

230sin

2 26

462 /1 2

pp p

p

p#=

-+> H

226.713 V=

Input power factor VVrms

s= .

230226 713=

cosf 0.985 lag=

PE 1.44 Correct option is (D).

PE 1.45 Correct option is (C).Average load voltage is given by

V ( )0 av ( )cosV2 1m

p a= +

.cosV V2 1 2 0 159mmp

p= + =a k

Average load current

I ( )0 av .R

VR

V0 159( ) m0 av= =RMS load voltage

V0( )rms sinV 4 82

m21

pp a

pa= - +: D

/

0.353sin

V V42

8

2 2m m

21

#

pp p

p

p

= - + =a k

> H

RMS load current

I0( )rms .R

VR

V0 3530( )rms m= =

To obtain rectification efficiency

h PP

V IV I

( ) ( )

0( ) 0( )

rms rmsac

dc

0 0

av av= =

. .. .

.V R

VV R

V

0 353 0 3530 159 0 159

0 2028m

m

mm

#

#= = or 20.28%


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PE 1.46 Correct option is (A).In AC voltage controller feeding purely inductive load, then each thyristor conducts for 2a . The range of firing angle is 90° to 180°. If 90ca = , then each SCR conducts for 180°.

PE 1.47 Correct answer is 0.4334.Let the average load current is Idc

Fundamental RMS current

Irms I 1 /dc

1 2

pa= -a k

The fundamental component of RMS current

Irms1 cosI2 22

dc

pa=

The harmonic factor (HF) is given as,

HF II 1rms

rms

12

2

= -

Putting values in above equation,

HF cosI

I

82

1dc

dc

2

22

2

pa

pp a

=

-

-a k

( )

cos8 2

12 a

p p a= - -

For /2a p= , HF ( / )

.cos8 4

21 1 23 1

2 pp p p= - - = -

0.4834=

PE 1.48 Correct is 184.4 V.This is a fully controlled bridge. The average value of output voltage.

V0( )av ( )cosV 1m

p a= +

( )cos230 2 1 30cp= + 184.8 V=

This voltage is applied to the load. The equivalent circuit is shown in the figure

Applying KVL to above circuit,

V0( )av 50I R0( )av= + 184.8 100 50I0( )av #= +

I0(av) 1.348 A=


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Engineering mathematics, General AptitudePE 1.49 Correct answer is 12.86 A.

The average output voltage for continuous ripple free output current is,

V ( )av0 cosV2

3 3 m

p a=

Here Vm is peak value of supply phase voltage. We have

Vline rms( ) 440 V=

V ( )rmsph 254 VV3 3

440line= = =

Vm V2 ( )rmsph=

. V2 254 359 26#= =

V ( )av0 . 30 257.3cos V23 3 359 26# cp= =

Average output current I ( )av0

I ( )av0 . 12.86 ARV

20257 3( )av0= = =

PE 1.50 Correct answer is .2 544 W .

Let the supply is sinV V tS m w= and battery emf is E . For the circuit voltage equation is

sinV tm w E I R0= +

or, I0 sin

RV t Em w= -

Since the SCR is turn on when sinV Em 1q = and is turned off when sinV Em 2q =, where 2 1q p q= - .

1q .sin sinVE

306 11 53

m

1 1 c= = =- -b bl l

The battery charging requires only the average current I0 given by:

I0 ( ) ( )sinR V t E d t21

m

1

1

p w w= -q

p q-

> H#

[2 ( 2 )]cosR V E21

m 1 1p q p q= - -

4 Amp 2 2 30 11.53 6 1802 11.53cosR2

1p p p

# # #= - -b l; E

4 Amp [ . . ]R21 83 13 19 172p= -

[ . ]R21 63 95p=

or R [ . ]2 41 63 95#p W=

.2 544 W=


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PE 1.51 Correct answer is .1 059 /AW .

Vs .4230 57 5= =

Here charging current I= sinVm q 12= 1q 8.486 0.148 radian= = Vm .81 317= V

e 12 V=There is no power consumption in battery due to ac current, so average value of charging current.

Iav(charging) . [ ( )]cosV2 19 041 2 2m 1 1#p q e p q= - -

. [ ( )]cosV2 19 041 2 12 2m 1 1#

# #p q p q= - -

1.059 /AW=

PE 1.52 Correct option is (D).For N -on cycles, M -off cycles

rms value of output voltage N mN Vs= +c m

If M 0= , V Vor s=

PE 1.53 Correct option is (C).Given that

400 V, 50 Hz AC source, a 60c= , 10I AL =so,

Input displacement factor .cos 0 5a= =and, input power factor cosD.F.# a=

distortion factor II

s

s(fundamental)=

/

sin

10 2 32

4 10 60

### c

p=

0.955=so, input power factor . .0 955 0 5#=

.0 478=

PE 1.54 Correct answer is 125°.Let we have

Rsolar 0.5 W= , 20I A0 =so Vs 350 20 0.5 340 V#= - =

340 cos3 440 2# #p a=

cosa 55c=So each thyristor will reverse biased for 180 55c c- 125c= .


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Engineering mathematics, General AptitudePE 1.55 Correct answer is 11.9 A.

In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier.

So

Iavg Average value of current=

( )sinR V t E d21

m1

1

p w q= -q

p q-#

a I0(avg) ( )cosR V E21 2 2m 1p q p q= - -6 @

[ ( ) ( )]cos2 21 2 230 2 200 2 1#

# #p q p q= - -

1q sin VE

m

1= -b l

38 0.66sin230 2

200 Rad1

#c= = =-

c m

I (0 avg) [2 230 38 200( 2 0.66)]cos2 21 2#

# #cp p= - -

11.9 A=


We know that Vrms 230 V=

so, Vm 230 2 V#=If whether a 90c1

Then Vpeak sinV 230m a= =

sin230 2 a 230=

sina 2

1=

angle a 135c=

PE 1.57 Correct option is (B).For controlling the load, Minimum value of firing angle a = load phase angle f .

Thus f tan RL1 w = -


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.tan 42 50 12 73 101

3# # #p = -

-

c m

.tan 0 99931= -^ h

.44 97 45c c.=The maximum possible value of a is 180°.Firing angle control range is 45° to 180°.

PE 1.58 Correct option is (C).

Given that, V sin t200 w= f 50 Hz=Power dispatched in the load resistor R ?=First we have to calculate output of rectifier.

( ) ( )sin cos

sin

V t d t t d t

t t

1 200 2002

1 2

20021

22 200

21

2200

/ /

/ /

02

0

1 2

0

1 2

0

1 2 1 2

rms

#

p w wp

w w

pw w

pp

= = -

= - = =

p p

p

b

b

l

l

: ;

; :

D E

E D

# #

Power dissipated to resistor

PR RV0

2rms=

^ h

40050200 2

W2

= =e o

PE 1.59 Correct answer is 120°.Given a half wave Thyristor converter supplies a purely inductive load where triggering angle is a 120c=

First we have to draw its output characteristics as shown below


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Engineering mathematics, General AptitudeOutput is given by

i0 ( ) ( )sin sin expZV t Z

VLRm mw f a f w a= - - - - -b l ...(1)

We know at extinction angle i.e. tw b= , i 00 =from equation (1), at ( )tw b=

0 ( ) ( )sin sinZV

ZV em m cb f a f= - - -

or ( )sin b f- ( )sin a f= -or b f- a f= -or b 120ca= =

PE 1.60 Correct option is (C).Peak value of secondary voltage

Vm 4002800 V= =

and a 30c=Average dc voltage is given by

Vdc (1 )cosV2

m

p a= + (1 30 ) 118.8cos V2400 cp= + =

RMS voltage

Vrms sinV 4 8

2 /

m

1 2

pp a

pa= - +b l

400 197.1sin V430

860 /1 2c c

pp

p= - + =b l

PE 1.61 Correct answer is 1.09 A.SCR will conduct when the instantaneous value of ac voltage is more than 50 V or

sin t100 w 50=

or tw 6p= and 6

5p

i sin sint t10100 50 10 5w w= - = -

Average current ( ) ( )sin t d t21 10 5

/

/

6

5 6

p w w= -p

p

#

cos t t21 10 5

//6

5 6

p w w= - -pp

cos cos21 10 6

5 10 6 5 65 5 6# #p

p p p p= - + - +b l

1.09 A=

PE 1.62 Correct option is (C).In positive half cycle Th1 and D2 conduct from a to p. During negative half cycle D3 and D4 are forward biased and conduct from p to 2p. From t 0w = to

tw a= , Th1 is off but D2 is forward biased. D4 continues to conduct during this interval because it was conducting prior to t 0w = i.e. during previous negative half cycle. Therefore from 0 to a , D2 and D4 conduct, the load is short circuited and load voltage is zero.


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The input output voltage waveforms are shown as below

Output voltage

V0 ( ) (sin sinV td t V td t21

m m

2

p w w w w= +a

p

a

p

> H##

(3 )cosV2

m

p a= +

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

GATE ELECTRICAL ENGINEERING Vol 2 of 4 · GATE ELECTRICAL ENGINEERING Vol 2 of 4 ... KCL, KVL, node...

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Transcript of GATE ELECTRICAL ENGINEERING Vol 2 of 4 · GATE ELECTRICAL ENGINEERING Vol 2 of 4 ... KCL, KVL, node...