gate ee vol-2

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GATE

ELECTRICAL ENGINEERINGVol 2 of 4


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Second Edition

GATEELECTRICAL ENGINEERING

Vol 2 of 4

RK Kanodia

Ashish Murolia

NODIA & COMPANY


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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 ofuse 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 anddata 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 andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series.Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gaussand Greens theorems.

Differential equations: First order equation (linear and nonlinear), Higher order lineardifferential equations with constant coefficients, Method of variation of parameters, Cauchysand Eulers equations, Initial and boundary value problems, Partial Differential Equations andvariable separable method.

Complex variables: Analytic functions, Cauchys integral theorem and integral formula,Taylors and Laurent series, Residue theorem, solution integrals.

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

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

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

ELECTRICAL ENGINEERING

Electric Circuits and Fields:Network graph, KCL, K VL, node and mesh analysis, transientresponse of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts;ideal current and voltage sources, Thevenins, Nortons and Superposition and MaximumPower Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electricfield and potential due to point, line, plane and spherical charge distributions; Amperes andBiot-Savarts laws; inductance; dielectrics; capacitance.


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Signals and Systems: Representation of continuous and discrete-time signals; shifting andscaling operations; linear, time-invariant and causal systems; Fourier series representation ofcontinuous 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, generatorcharacteristics, armature reaction and commutation, starting and speed control of motors;three phase induction motors principles, types, performance characteristics, starting andspeed control; single phase induction motors; synchronous machines performance, regulationand parallel operation of generators, motor starting, characteristics and applications; servo andstepper 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 factorcorrection; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuitbreakers; system stability concepts, swing curves and equal area criterion; HVDC transmissionand 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; statespace 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, energyand power factor; instrument transformers; digital voltmeters and multimeters; phase, timeand frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.

Analog and Digital Electronics:Characteristics of diodes, BJ T, FET; amplifiers biasing,equivalent circuit and frequency response; oscillators and feedback amplifiers; operationalamplifiers 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; triggeringcircuits; phase control rectifiers; bridge converters fully controlled and half controlled;principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

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PREFACE

This book doesnt make promise but provides complete satisfaction to the readers. The

market scenario is confusing and readers dont 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. T he 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 GAT E preparation will bean 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


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CONTENTS

AE ANALOG ELECTRONICS

AE 1 Diode Circuits AE 3

AE 2 BJ T Biasing AE 39

AE 3 BJ T 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 210AE 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

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

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SampleChapterofGATEE

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Electric circuit & Field, Electrical & electronic measurementGATE EE vol-2

Analog electronics, Digital electronics, Power electronics

GATE EE vol-3

Control systems, Signals & systems

GATE EE vol-4

Electrical machines, Power systemsEngineering 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. T he 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 dcoutput 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 10Wis _ _ _ 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 23a 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. T he average power across the load will be _ _ _ kW.


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GATE Electrical Engineering-2015in 4 Volumes

by R. K. Kanodia & Ashish Murolia

Sample

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PE 1.7 A line commutated ac to dc converter is shown in the figure. I t operates from a

three phase, 50 Hz, 580 V (line to line) supply. The load current I0is ripple free

and constant at 3464 A. For an average output voltage of 648 V, the delay angle

ais _ _ _ degree.

PE 1.8

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

will be

(A) 600p

V (B) 300 3V

(C) 300 3p

V (D) 300p

V

PE 1.9 A single-phase half controlled bridge rectifier is operated from a source

sinV t100 314s= . T he 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 Vdcfor a firing of 30c?

(A) . V06 m (B) . V077 m

(C) . V0155 m (D) . V0424 m


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PE 1.12 When the firing angleaof 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) 2

3

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 2W 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<


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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 of60c, 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 ismaximum 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 ofinput 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


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PE 1.32 In the circuit shown in Figure, L is large and the average value of I0is 100A.

Then which of the following is true for the circuit ?

(A) T he thyristor is gated in the positive half cycle of ( )e t at a delay angle a

equal to .1679c.

(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 .1679c.

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) sin1 21 2/1 2

pp a a- +^ h: D& 0

(C) sin121 2

/1 2

pp a a+ +^ h; E& 0

(D) sin1 21 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 10Wwith ripple free

current. I f the input voltage is 240 V, 50 Hz and the triggering angle is 60cthen

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 100Wdraws 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)

2Per unit power

2

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

^ hPE 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 30cthe inputpower 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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PE 1.46 A single phase ac voltage controller is controlling current in a purely inductive

load. I f 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 100Wresistor is _ _ _ _ A

PE 1.49 A three phase half wave controlled rectifier circuit is shown in the figure. It

is operated from 3-fstar connected, supply transformer with a line to line acsupply 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-fone pulse

thyristor controlled rectifier. A resistance Ris to be inserted in series with the

battery to limit the charging current to 4 A. The value of Ris _ _ _ 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 currentis 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. I f the solar cell resistance is0.5W,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 Was shown in figure. The SCRs are triggered by

a constant dc signal. I f 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 ain every positive half cycle of the input voltage. If the peak value

of the instantaneous output voltage equals 230 V, the firing angleais close to

(A) 45c (B) 135c

(C) 90c (D) 83.6c


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PE 1.57 A single-phase ac regulator fed from 50 Hz supply feeds a load having 4Wresistance 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= . T he source voltage is ( )sinV t200 w= , where 2 50#w p= radians per

second. T he power dissipated in the load resistor Ris

(A) 3200p

W (B) 400p

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 30cin 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 10Wresistor 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 ofa? Assume continuous

conduction.

(A) cosV2 mp

a (B) cosV2

1mp

a+^ h(C) cosV

23m

p a+^ h (D) cosVmp a

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SOLUTIONS

PE 1.1 Correct option is (C).

DC output voltage

Vdc cosV2 mp

a=

( )30 179.3cos V

2 2 230c

p= =

PE 1.2 Correct option is (A).

When free wheeling diode is present,T h1andT h2will conduct from ato pwhile

T h3and T h4will 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.

WhenTh3gets 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= + = +

96.6V=

Average dc output current

Idc. 9.7A

10966

= =

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 310.3V3

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

= -

180.2V=


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PE 1.6 Correct answer is 1.1 kW.

RMS load voltage

V ( )rmsdc sin

V4 8

2m

21

pp a

pa

= -

+: D

220 1 .sin

V4

48

2 048821

p

p p

p

p

=-

+ == GAverage power across the load

Pac( . )

. kWR

V

1010488

11( )rmsdc

2 2

= = =

PE 1.7 Correct answer is 34.18.

Here 580V VLine =

Average output voltage

Vdc cos cosV V3 3 3 2 inem Lpa

pa= =

648 cos3 2 580# #p

a=

or a 34.18c=


Given 3 phase 6 pulse bridge, thus m 6= and VV 200ph=

The fundamental r.m.s output voltage

Vor sinV m

mph p p= a ak k9 C sin200

66

600p

pp

= =b al k: DPE 1.9 Correct option is (D).

Vs ,sin t100 314= R 10W= , 45ca =

RMS load voltage

V ( )rmsdc sin

V2 4

2 /m

1 2

pp a

pa

= -

+: D

V ( )rmsdc 100 67.42sin V

24

42

/1 2

pp p

pp= - + == G

The average power delivered to the load is

Pac RV ( )rmsdc

2

=

( . )454.5W

106742 2

= =

PE 1.10 Correct option is (B).

The average value of dc voltage i.e. load voltage can be varied by controlling thephase angle(a) of firing pulses.


The average output dc voltage


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Vdc ( )sinV td t 1

mp w w=

a

p a+

#

( )cosV tmp

w= - ap a+

[ ( )]cos cosVmp

a p a= - +

cosV2 mp

a=

Given 30ca = , the average dc output voltage

Vdc cosV2 mp

a=

( )cosV2 30m cp

=

0.155132Vm=

PE 1.12 Correct option is (D).

For a 1-ffull converter the displacement factor is

DF cos cos30ca= = 23

=


We have R 2L W= , . mHL 636= , Hzf 50= ,

. .T an2

2 50 636 10 449713

# # #p =--b l

mina 45cf= =


For a single phase fully controlled bridge rectifier, the average output voltage is

given by

V0 cosV 1mp

a= +^ hOutput 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 1mp

a= +6 @For 0ca = , V0 cos

120 2 1 0 120 2 2#cp p

= + =6 @For 180ca = , V0 cos

120 2 1 180 0cp

+ =6 @


R j X L+ j50 50= +

tanfR

L5050 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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Sample

Chapterof


ering,

Volum

e-2


Average output current

I0 cosRV 1mp

a= +^ h

.cos

1553230 2 1 60#

cp

+^ hAverage current through diode

IFW ( )I d t1

31 10

/

o0

3

#p w= =

p

#

3.33A=


Input harmonic factor .

8

1 0482/2 1 2p

= - =

: DPE 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

voltageVdc0 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 c

p=

Vdc12 2300p

=

At a 60c= , Vdc2 cos2 2

2 2300 60# cp

p=

300 15021 V#= =


Output of this

Here the inductor makes T1and T3in ON because current passing through T1

andT3is more than the holding current.


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PE 1.23 Correct option is 0.78.

Given 30ca = , in a 1-ffully bridge converter

we know that,

Power factor cosDistortionfactor # a=

D.f. (Distortion factor) / 0.9I Is s(fundamental)= =

power factor . cos09 30# c= .078=

PE 1.24 Correct answer is 266 V.

We have VV 220or=

For 3-phase to single phase cyclo converter

V

or sinr V

m

mph p

p

= a ak k9 CwhereVph= 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 .26602=


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 K irchoffs voltage law, average load

current

V I2 150a- + 0=

IaV

2150

= +

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

cosV2 mp a 130=-

cos2 2 230# #p

a 130c=-

a 129c=


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Sample

Chapterof


ering,

Volum

e-2


Total rms current Ia 10 8.1632 A#= =

Fundamental current Ia1 0.78 10 7.8 A#= =

THD 1 1DF2

= -

where DF.. 0.955

II

0816 10078 10

a

a1

#

#= = =

THD.

3 %0955

1 1 12

= - =b l


In the case of RL load, the output voltage can be controlled for a in the range of

f p- . If


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From the output waveform given below, we observed that the thyristor is gated

in positive half cycle.


In the inverting mode a line commutated converted operates for phase angles

90cto180c. When the dc voltage is negative power flow is from dc to ac and theconverter functions as inverter. As dc power is fed back, it is real power.


For a 3-phase fully-controlled converter, output dc voltage is given as

Vdc ( )cosV L

I3 3 30mph s

dpa m

pw

= + +

Where mis 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.FVVor

S

=

r.m.s valueVor sinV td t 1

/

m2 2

1 2

pw w=

a

p< F# sinV

21 2

/S

1 2

pp a a= - +^ h: D


The circulating current helps in maintaining continuous conduction of both the

converters irrespective of load and the time response to change the operationfrom one quadrant to other is faster.


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.


Vdc (1 )cosVmp

a= +

( )

(1 60 )cos2 240

cp

= +


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Sample

Chapterof


ering,

Volum

e-2

162.03V=

Load current I . 16.203ARV

1016203

L

dc= = =

RMS input current

Is .I 1 16203 1 60. .05 05c

pa

p= - = -a bk l

13.23A=


Load Current

IL 2A100400 .05

= =b lIn a three-phase fully controlled bridge converter input rms current Isor the

current in each supply phase exists for 120cin every 180c.Therefore rms value of input current

Is 1.15A1802 120 .05#= =b l

Input apparent power 400 1.15 796.72VA3 # #= =

. cos79672 q 400=

Power factor cosq 0.5lagging=


Average output voltage Vdc (1 )cos

V2

m

pa= +

The maximum output voltage is obtained when 0a=

VdcmaxVmp

=

Given Vdc 25% 0.25V Vm mp p

= =b lSo 0.25Vm

p (1 )cosV

2m

pa= +

The Firing angle is

a 60c=


In a AC voltage controller P.fVVor

S

=

Per unit power/

/RatedpowerPresent power

V R

V Ror

S2

2

= = h

Thus p.fVV Per unit poweror

S

= =

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

Ir ms I 1/

dc

1 2

pa

= -a k


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/

0.91I I16 /

dc dc

1 2

pp

= - =c mNow, the rms value of the supply fundamental component of input current.

Ir ms1 / 2cos

I2 2 dcp a= 15 0.869cosI I

2 2dc dc p= =

Harmonic factor (HF) on input current

HFII 1

1

/

r ms

r ms2 1 2

= -b l; E .. 30.80%0869091 1/2 1 2

= - =b l; EInput power factor / 2 0.922( )cos lagging

II 1

r ms

r ms a= =


The rms load voltage,

Vr ms sinV 2 42/

m

1 2

pp a pa= - +: D 230

sin2

26

462 /1 2

p

p p

p

p

#=-

+> H 226.713V=

Input power factorV

Vrmss

= .230

226713=

cosf 0.985lag=



Average load voltage is given by

V ( )0 av ( )cosV2

1mp

a= +

.cosV V2

12

0159m mpp

= + =a kAverage load current

I ( )0 av .RV

RV0159( ) m0 av= =

RMS load voltage

V0( )rms sin

V4 8

2m

21

pp a

pa

= -

+: D

/

0.353sin

V V4

28

22

m m

21

#

pp p

p

p

= -

+ =a k> H

RMS load current

I0( )rms .

R

V

RV03530( )rms m= =

To obtain rectification efficiency

hPP

V I

V I

( ) ( )

0( ) 0( )

rms rmsac

dc

0 0

av av= =

. .

. .

.V

RV

VR

V

0353 0353

0159 0159

02028m

m

mm

#

#= = or 20.28%


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Sample

Chapterof


ering,

Volum

e-2


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.


Let the average load current is IdcFundamental RMS current

Ir ms I 1/

dc

1 2

pa= -a k

The fundamental component of RMS current

Ir ms1 cosI2 2

2dc

pa

=

The harmonic factor (HF) is given as,

H F I

I 1r ms

r ms

12

2

= -

Putting values in above equation,

H F cosI

I

82

1dc

dc

2

22

2

pa

pp a

=

-

-a k

( )

cos82

12 a

p p a=

--

For / 2a p= , H F ( / )

.cos8

4

21 123 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

= + 184.8V=

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) 1.348A=


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

V ( )av0 cosV

23 3 m

p a=

HereVmis peak value of supply phase voltage. We have

Vl ine rms( ) 440V=

V ( )rmsph 254VV

3 3440l ine= = =

Vm V2 ( )rmsph=

. V2 254 35926#= =

V ( )av0 . 30 257.3cos V

23 3 35926# c

p= =

Average output current I ( )av0

I ( )av0 . 12.86A

R

V

202573( )av0

= = =

PE 1.50 Correct answer is .2544W.

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, I0sin

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 1153

m

1 1c= = =- -

b bl lThe battery charging requires only the average current I0given by: I0 ( ) ( )sinR

V t E d t 2

1m

1

1

p w w= -

q

p q-> H# [2 ( 2 )]cos

R V E

21

m 1 1p q p q= - -

4 Amp 2 2 30 11.53 6180

2 11.53cosR2

1p

p p# # #= - -b l; E

4 Amp [ . . ]R2

1 8313 19172p

= -

[ . ]R2

1 6395p

=

or R [ . ]2 4

1 6395#p

W=

.2544W=


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Sample

Chapterof


ering,

Volum

e-2

PE 1.51 Correct answer is .1 059 / AW .

Vs .4230 575= =

Here charging current I=

sinVm q 12=

1q 8.486 0.148 radian= =

Vm .81317= V

e 12 V=

There is no power consumption in battery due to ac current, so average value of

charging current.

Iav(charging) .[ ( )]cosV

2 19041 2 2m 1 1

#p q e p q= - -

.

[ ( )]cosV

2 1904

1 2 12 2m 1 1#

# #

p

q p q= - -

1.059 /AW=


For N -on cycles, M -off cycles

rms value of output voltageN m

NVs= +

c mIf 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 05a= =

and, input power factor cosD.F. # a=

distortion factorI

I

s

s(fundamental)=

/

sin

10 2 3

24 10 60

#

#

# cp

=

0.955=

so, input power factor . .0955 05#=

.0478=

PE 1.54 Correct answer is 125.

Let we have

Rsolar 0.5W= , 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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In this circuitry if SCR gets open circuited, than circuit behaves like a half wave

rectifier.

So

Iavg Averagevalueof current=

( )sinR

V t E d 2

1m

1

1

p w q= -

q

p q-

#

a I0(avg) ( )cosR V E

21 2 2m 1p

q p q= - -6 @

[ ( ) ( )]cos2 2

1

2 230 2 200 2 1# # #p q p q= - -

1q sin VE

m

1= - b l38 0.66sin

230 2200 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=

sina2

1=

angle a 135c=


For controlling the load,

Minimum value of firing angle a = load phase angle f.

Thus f tanR

L1w= -


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Sample

Chapterof


ering,

Volum

e-2

.tan4

2 50 1273 101 3# # #p= --c m

.tan 099931= - ^ h .4497 45c c.=

The maximum possible value of a is 180.

Firing angle control range is 45 to 180.


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

1 200 200

2

1 2

20021

22 200

21

2200

/ /

/ /

02

0

1 2

0

1 2

0

1 2 1 2

rms

#

p

w w

p

ww

pw

w

pp

= = -

= - = =

p p

p bb ll: ;; :D EE D# #

Power dissipated to resistor

PR R

V02

rms= h

40050

200 2W

2

= =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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Output is given by

i0 ( ) ( )sin sin expZV

tZ

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 sinZ

VZ

Vem m cb f a f= - - -

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

or b f- a f= -

or b 120ca= =


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 V

2400

cp

= + =

RMS voltage

Vr mssin

V4 8

2 /m

1 2

pp a

pa= - +b l

400 197.1sin V4

30860 /1 2c c

pp

p=

-+ =

b lPE 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

= and65p

i sin sint t10

100 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= - -

p

p

cos cos21 10

65 10

65

65 5

6# #pp p p p

= - + - +b l 1.09A=


In positive half cycle T h1and D2conduct from a to p. During negative halfcycle D3and D4are forward biased and conduct from pto 2p. From t 0w = to

tw a= , T h1is off but D2is forward biased. D4continues 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 , D2and D4conduct, the load is short circuited

and load voltage is zero.


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mple

Chapterof


ering,

Volum

e-2

The input output voltage waveforms are shown as below

Output voltage

V0 ( ) (sin sinV td t V t d t 21

m m

2

p w w w w= +

a

p

a

p> H## (3 )cosV

2m

p a= +

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

gate ee vol-2

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