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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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gineering,V
olume-2
GATE EE vol-1
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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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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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 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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olume-2
GATE EE vol-1
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 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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olume-2
GATE EE vol-1
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 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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SampleChapterofGATEE
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olume-2
GATE EE vol-1
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 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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SampleChapterofGATEE
lectricalEn
gineering,V
olume-2
GATE EE vol-1
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 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=
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 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.
PE 1.11 Correct option is (C).
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
=
PE 1.13 Correct option is (B).
We have R 2L W= , . mHL 636= , Hzf 50= ,
. .T an2
2 50 636 10 449713
# # #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 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 @
PE 1.16 Correct option is (A).
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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PE 1.17 Correct answer is 3.33 A.
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=
PE 1.18 Correct answer is 0.482.
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#= =
PE 1.21 Correct option is (A).
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 1.22 Correct option is (D).
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=
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 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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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.. 0.955
II
0816 10078 10
a
a1
#
#= = =
THD.
3 %0955
1 1 12
= - =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
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From 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
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.
PE 1.34 Correct option is (A).
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.
PE 1.35 Correct option is (B).
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
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 operationfrom 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 )cosVmp
a= +
( )
(1 60 )cos2 240
cp
= +
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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=
PE 1.39 Correct answer is 0.5.
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=
PE 1.40 Correct option is (C).
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=
PE 1.41 Correct option is (B).
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= =
PE 1.43 Correct option is (C).
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=
PE 1.44 Correct option is (D).
PE 1.45 Correct option is (C).
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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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 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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PE 1.49 Correct answer is 12.86 A.
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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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=
PE 1.52 Correct option is (D).
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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PE 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 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=
PE 1.56 Correct option is (B).
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=
PE 1.57 Correct option is (B).
For controlling the load,
Minimum value of firing angle a = load phase angle f.
Thus f tanR
L1w= -
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PE 1 Phase Controlled Converters PE 34EF 1 Phase Controlled Converters EF 34
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Sample
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GATEElectricalEngine
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.
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
1 200 200
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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SampleChapterofGATEE
lectricalEn
gineering,V
olume-2
GATE EE vol-1
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
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= =
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 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=
PE 1.62 Correct option is (C).
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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PE 1 Phase Controlled Converters PE 36EF 1 Phase Controlled Converters EF 36
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mple
Chapterof
GATEElectricalEngine
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= +
************