GATE Electrical 2015 Solved Paper by Kanodia PDF
GATE Electrical 2015 Solved Paper by Kanodia PDF
GATE Electrical 2015 Solved Paper by Kanodia PDF
ELECTRICAL ENGINEERING
RK Kanodia
Ashish Murolia
www.nodia.co.in
MRP Free
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 Greens theorems.
Differential equations: First order equation (linear and nonlinear), Higher order linear
differential equations with constant coefficients, Method of variation of parameters, Cauchys
and Eulers equations, Initial and boundary value problems, Partial Differential Equations and
variable 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 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, Thevenins, Nortons 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; Amperes and
Biot-Savarts laws; inductance; dielectrics; capacitance.
Signals and Systems: Representation of continuous and discrete-time signals; shifting and
scaling operations; linear, time-invariant and causal systems; Fourier series representation of
continuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.
Electrical Machines: Single phase transformer equivalent circuit, phasor diagram, tests,
regulation and efficiency; three phase transformers connections, parallel operation; autotransformer; energy conversion principles; DC machines types, windings, generator
characteristics, armature reaction and commutation, starting and speed control of motors;
three phase induction motors principles, types, performance characteristics, starting and
speed control; single phase induction motors; synchronous machines performance, regulation
and parallel operation of generators, motor starting, characteristics and applications; servo and
stepper motors.
Power Systems: Basic power generation concepts; transmission line models and performance;
cable performance, insulation; corona and radio interference; distribution systems; per-unit
quantities; bus impedance and admittance matrices; load flow; voltage control; power factor
correction; economic operation; symmetrical components; fault analysis; principles of overcurrent, differential and distance protection; solid state relays and digital protection; circuit
breakers; system stability concepts, swing curves and equal area criterion; HVDC transmission
and FACTS concepts.
Control Systems: Principles of feedback; transfer function; block diagrams; steady-state errors;
Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; state
space model; state transition matrix, controllability and observability.
Electrical and Electronic Measurements: Bridges and potentiometers; PMMC, moving iron,
dynamometer and induction type instruments; measurement of voltage, current, power, energy
and power factor; instrument transformers; digital voltmeters and multimeters; phase, time
and frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.
Analog and Digital Electronics: Characteristics of diodes, BJT, FET; amplifiers biasing,
equivalent circuit and frequency response; oscillators and feedback amplifiers; operational
amplifiers characteristics and applications; simple active filters; VCOs and timers;
combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators;
sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture,
programming and interfacing.
Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs,
GTOs, MOSFETs and IGBTs static characteristics and principles of operation; triggering
circuits; phase control rectifiers; bridge converters fully controlled and half controlled;
principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.
***********
PREFACE
This book 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. The aim of the book is
to avoid the unnecessary elaboration and highlights only those concepts and techniques
which are absolutely necessary. Again time is a critical factor both from the point of view
of preparation duration and time taken for solving each problem in the examination. So
the problems solving methods is the books are those which take the least distance to the
solution.
But however to make a comment that this book is absolute for GATE preparation will be
an inappropriate one. The theory for the preparation of the examination should be followed
from the standard books. But for a wide collection of problems, for a variety of problems
and the efficient way of solving them, what one needs to go needs to go through is there
in there in the book. Each unit (e.g. Networks) is subdivided into average seven number of
chapters on an average each of which contains 40 problems which are selected so as to avoid
unnecessary redundancy and highly needed completeness.
I shall appreciate and greatly acknowledge the comments and suggestion from the users of
this book.
R. K. Kanodia
Ashish Murolia
CONTENTS
SP 1
SP 1 - 83
SP 2
SP 84 - 124
SP 3
Electrical Machines
SP 125 - 200
SP 4
Power Systems
SP 201 - 277
SP 5
Control Systems
SP 278 - 343
SP 6
SP 7
Analog Electronics
SP 390 - 451
SP 8
Digital Electronics
SP 452 - 493
SP 8
Power Electronics
SP 494 - 551
SP 9
Engineering Mathematics
SP 552 - 595
SP 9
General Aptitude
SP 596 - 623
***********
www.gatehelp.com
Chapter 1
Page 1
CHAPTER 1
ELECTRIC CIRCUITS AND FIELDS
ONE MARK
EE SP 1.1
The three circuit elements shown in the figure are part of an electric circuit. The
total power absorbed by the three circuit elements in watts is _____.
EE SP 1.2
nodia
(A) C 0 ^1 + e r h
2
(C) C 0 e r
2
EE SP 1.3
(B) ^C 0 + e r h
(D) C 0 ^1 + e r h
www.nodia.co.in
www.gatehelp.com
Page 2
EE SP 1.4
Chapter 1
nodia
The undesirable property of an electrical insulating material is
(A) high dielectric strength
(B) high relative permittivity
(C) high thermal conductivity
(D) high insulation resistivity
YEAR 2014 EE01
EE SP 1.5
EE SP 1.6
TWO MARKS.
In the figure, the value of resistor R is ^25 + I/2h ohms , where I is the current
in amperes. The current I is _____.
The following four vector fields are given in Cartesian co-ordinate system. The
vector field which does not satisfy the property of magnetic flux density is
(A) y2 ax + z2 ay + x2 az
(B) z2 ax + x2 ay + y2 az
(C) x2 ax + y2 ay + z2 az
(D) y2 z2 ax + x2 z2 ay + x2 y2 az
YEAR 2014 EE02
EE SP 1.7
ONE MARK
Two identical coupled inductors are connected in series. The measured inductances
for the two possible series connections are 380 mH and 240 mH . Their mutual
inductance in mH is _____.
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.8
Page 3
The switch SW shown in the circuit is kept at position 1 for a long duration.
At t = 0+ , the switch is moved to position 2. Assuming Vo2 > Vo1 , the voltage
vc ^ t h across the capacitor is
EE SP 1.9
nodia
If the potential difference between one of the plates and the nearest surface of
dielectric interface is 2 Volts, then the ratio e 1 : e 2 is
(A) 1 : 4
(B) 2 : 3
(C) 3 : 2
(D) 4 : 1
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Page 4
EE SP 1.11
Chapter 1
nodia
EE SP 1.12
EE SP 1.13
ONE MARK.
EE SP 1.14
(D) - 10 3 90c V
4
2
(A) s +3 3s + 1
s + 2s
2
(C) 4 s +2 1
s +s +1
4
2
(B) s +22s + 4
s +2
3
(D) 4 s +2 1
s +s +1
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.15
EE SP 1.17
EE SP 1.18
Page 5
TWO MARKS
nodia
(A) 2 2 k
(B) - 2k
(C) 2k
(D) - 2 2 k
www.nodia.co.in
www.gatehelp.com
Page 6
EE SP 1.19
Chapter 1
nodia
YEAR 2013
EE SP 1.20
ONE MARK
Consider a delta connection of resistors and its equivalent star connection as shown
below. If all elements of the delta connection are scaled by a factor k , k > 0 , the
elements of the corresponding star equivalent will be scaled by a factor of
(A) k2
(B) k
(C) 1/k
(D)
EE SP 1.21
The flux density at a point in space is given by Bv = 4xavx + 2kyavy + 8avz Wb/m2 .
The value of constant k must be equal to
(A) - 2
(B) - 0.5
(C) + 0.5
(D) + 2
EE SP 1.22
EE SP 1.23
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.24
(A) 0.5s + 1
s+1
(C) s + 2
s+1
V2 ^s h
of the circuit shown below is
V1 ^s h
(B) 3s + 6
s+2
(D) s + 1
s+2
nodia
YEAR 2013
EE SP 1.25
EE SP 1.26
TWO MARKS
A dielectric slab with 500 mm # 500 mm cross-section is 0.4 m long. The slab
is subjected to a uniform electric field of Ev = 6avx + 8avy kV/mm . The relative
permittivity of the dielectric material is equal to 2. The value of constant e0 is
8.85 # 10-12 F/m . The energy stored in the dielectric in Joules is
(B) 8.85 # 10-5
(A) 8.85 # 10-11
(D) 885
(C) 88.5
Page 7
In the circuit shown below, if the source voltage VS = 100+53.13c V then the
Thevenins equivalent voltage in Volts as seen by the load resistance RL is
www.nodia.co.in
www.gatehelp.com
Page 8
(A) 100+90c
(C) 800+90c
(B) 800+0c
(D) 100+60c
YEAR 2012
EE SP 1.28
ONE MARK
nodia
(A) 50 W
(C) 5 kW
EE SP 1.29
(B) 100 W
(D) 10.1 kW
(B) - 1 A
1+j
2 A
1+j
(C) 1 A
1+j
(A)
EE SP 1.30
EE SP 1.31
Chapter 1
(D) 0 A
(s2 + 9) (s + 2)
(s + 1) (s + 3) (s + 4)
is excited by sin (wt). The steady-state output of the system is zero at
(A) w = 1 rad/s
(B) w = 2 rad/s
(C) w = 3 rad/s
(D) w = 4 rad/s
A system with transfer function G (s) =
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.32
Page 9
In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12
V before the ideal switch S is closed at t = 0. The current i (t) for all t is
(A) zero
EE SP 1.33
nodia
If VA - VB = 6 V then VC - VD is
(A) - 5 V
(C) 3 V
EE SP 1.34
TWO MARKS
(B) 2 V
(D) 6 V
Assuming both the voltage sources are in phase, the value of R for which
maximum power is transferred from circuit A to circuit B is
(A) 0.8 W
(C) 2 W
(B) 1.4 W
(D) 2.8 W
www.nodia.co.in
www.gatehelp.com
Page 10
Chapter 1
EE SP 1.35
EE SP 1.36
(D) 9 V
nodia
In the circuit shown, the three voltmeter readings are V1 = 220 V, V2 = 122 V,
V3 = 136 V .
EE SP 1.37
EE SP 1.38
(B) 0.50
(D) 0.60
EE SP 1.39
ONE MARK
The r.m.s value of the current i (t) in the circuit shown below is
(A) 1 A
2
(C) 1 A
(B) 1 A
2
(D) 2 A
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.40
The voltage applied to a circuit is 100 2 cos (100pt) volts and the circuit draws
a current of 10 2 sin (100pt + p/4) amperes. Taking the voltage as the reference
phasor, the phasor representation of the current in amperes is
(A) 10 2 - p/4
(B) 10 - p/4
(C) 10 + p/4
EE SP 1.41
(D) 10 2 + p/4
In the circuit given below, the value of R required for the transfer of maximum
power to the load having a resistance of 3 W is
nodia
(B) 3 W
(D) infinity
(A) zero
(C) 6 W
YEAR 2011
EE SP 1.42
EE SP 1.43
Page 11
TWO MARKS
EE SP 1.45
www.nodia.co.in
www.gatehelp.com
Page 12
Chapter 1
EE SP 1.46
nodia
(C)
2W
(D) 2 W
EE SP 1.47
(B) - j
(C) + j
(D) + j2A
1
2
1
2
YEAR 2010
EE SP 1.48
The switch in the circuit has been closed for a long time. It is opened at t = 0.
At t = 0+ , the current through the 1 mF capacitor is
(A) 0 A
(C) 1.25 A
EE SP 1.49
ONE MARK
(B) 1 A
(D) 5 A
(A) 25 A
(C) 100 A
(B) 50 A
(C) 200 A
www.nodia.co.in
www.gatehelp.com
Chapter 1
YEAR 2010
EE SP 1.50
TWO MARKS
(A) 4 W
(C) 8 W
EE SP 1.51
(B) 6 W
(D) 18 W
nodia
The two-port network P shown in the figure has ports 1 and 2, denoted by
terminals (a,b) and (c,d) respectively. It has an impedance matrix Z with
parameters denoted by Zij . A 1 W resistor is connected in series with the network
at port 1 as shown in the figure. The impedance matrix of the modified two-port
network (shown as a dashed box ) is
Z11 + 1
(A) e
Z21
Z11 + 1
(C) e
Z21
Z12 + 1
Z22 + 1o
Z12
Z22 o
Z11 + 1 Z12
(B) e
Z21 Z22 + 1o
Z11 + 1 Z12
(D) e
Z21 + 1 Z22 o
YEAR 2009
EE SP 1.52
ONE MARK
(A) 0 mA
(C) 2 mA
EE SP 1.53
Page 13
(B) 1 mA
(D) 6 mA
How many 200 W/220 V incandescent lamps connected in series would consume
the same total power as a single 100 W/220 V incandescent lamp ?
(A) not possible
(B) 4
(C) 3
(D) 2
www.nodia.co.in
www.gatehelp.com
Page 14
YEAR 2009
EE SP 1.54
EE SP 1.55
TWO MARKS
In the figure shown, all elements used are ideal. For time t < 0, S1 remained closed
and S2 open. At t = 0, S1 is opened and S2 is closed. If the voltage Vc2 across the
capacitor C2 at t = 0 is zero, the voltage across the capacitor combination at
t = 0+ will be
(A) 1 V
(B) 2 V
(C) 1.5 V
(D) 3 V
nodia
(A) 2 mF
(C) 200 mF
EE SP 1.56
Chapter 1
(B) 100 mF
(D) 4 mF
For the circuit shown, find out the current flowing through the 2 W resistance.
Also identify the changes to be made to double the current through the 2 W
resistance.
(A) (5 A; PutVS = 30 V)
(C) (5 A; Put IS = 10 A)
(B) (2 A; PutVS = 8 V)
(D) (7 A; Put IS = 12 A)
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 15
EE SP 1.57
For the circuit given above, the Thevenins resistance across the terminals A and B is
(A) 0.5 kW
(B) 0.2 kW
(C) 1 kW
(D) 0.11 kW
EE SP 1.58
For the circuit given above, the Thevenins voltage across the terminals A and B is
(A) 1.25 V
(B) 0.25 V
(C) 1 V
(D) 0.5 V
YEAR 2008
EE SP 1.59
ONE MARK
nodia
(A) 3
(C) 5
EE SP 1.60
(B) 4
(D) 6
EE SP 1.61
(A) 1/9 s
(C) 4 s
EE SP 1.62
TWO MARKS
(B) 1/4 s
(D) 9 s
www.nodia.co.in
www.gatehelp.com
Page 16
EE SP 1.63
(A) 1 rad/s
(B) 2 rad/s
(C) 3 rad/s
(D) 4 rad/s
Chapter 1
Assuming ideal elements in the circuit shown below, the voltage Vab will be
(A) - 3 V
(C) 3 V
(B) 0 V
nodia
(D) 5 V
The current i (t) sketched in the figure flows through a initially uncharged 0.3
nF capacitor.
EE SP 1.64
EE SP 1.65
The capacitor charged upto 5 ms, as per the current profile given in the figure,
is connected across an inductor of 0.6 mH. Then the value of voltage across the
capacitor after 1 ms will approximately be
(A) 18.8 V
(B) 23.5 V
(C) - 23.5 V
(D) - 30.6 V
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.66
EE SP 1.67
Page 17
In the circuit shown in the figure, the value of the current i will be given by
(A) 0.31 A
(B) 1.25 A
(C) 1.75 A
(D) 2.5 A
nodia
(D) 22.5 mC
EE SP 1.68
EE SP 1.69
A capacitor consists of two metal plates each 500 # 500 mm2 and spaced 6 mm
apart. The space between the metal plates is filled with a glass plate of 4 mm
thickness and a layer of paper of 2 mm thickness. The relative primitivities of
the glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the
capacitance will be (Given that e0 = 8.85 # 10 - 12 F/m )
(A) 983.3 pF
(B) 1475 pF
(C) 637.7 pF
(D) 9956.25 pF
EE SP 1.70
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 18
YEAR 2007
EE SP 1.71
Chapter 1
TWO MARKS
The state equation for the current I1 in the network shown below in terms of the
voltage VX and the independent source V , is given by
EE SP 1.72
EE SP 1.73
nodia
The R-L-C series circuit shown in figure is supplied from a variable frequency
voltage source. The admittance - locus of the R-L-C network at terminals AB for
increasing frequency w is
In the circuit shown in figure. Switch SW1 is initially closed and SW2 is open.
The inductor L carries a current of 10 A and the capacitor charged to 10 V with
polarities as indicated. SW2 is closed at t = 0 and SW1 is opened at t = 0 . The
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 19
EE SP 1.74
EE SP 1.75
EE SP 1.76
(A) 55 A, 4.5 V
(B) 5.5 A, 45 V
(C) 45 A, 5.5 A
(D) 4.5 A, 55 V
nodia
In the figure given below all phasors are with reference to the potential at point
''O'' . The locus of voltage phasor VYX as R is varied from zero to infinity is shown
by
The matrix A given below in the node incidence matrix of a network. The columns
correspond to branches of the network while the rows correspond to nodes.
Let V = [V1V2 .....V6]T denote the vector of branch voltages while I = [i1 i2 .....i6]T
that of branch currents. The vector E = [e1 e2 e3 e4]T denotes the vector of node
voltages relative to a common ground.
www.nodia.co.in
www.gatehelp.com
Page 20
Chapter 1
R1 1 1 0 0 0 V
W
S
S 0 -1 0 -1 1 0 W
S- 1 0 0 0 - 1 - 1W
W
S
S 0 0 -1 1 0 1 W
X
T
Which of the following statement is true ?
(A) The equations V1 - V2 + V3 = 0,V3 + V4 - V5 = 0 are KVL equations for the
network for some loops
(B) The equations V1 - V3 - V6 = 0,V4 + V5 - V6 = 0 are KVL equations for the
network for some loops
(C) E = AV
(D) AV = 0 are KVI equations for the network
EE SP 1.77
nodia
A solid sphere made of insulating material has a radius R and has a total charge
Q distributed uniformly in its volume. What is the magnitude of the electric field
intensity, E , at a distance r (0 < r < R) inside the sphere ?
1 Qr
4pe0 R3
Q
(C) 1 2
4pe0 r
3 Qr
4pe0 R3
QR
(D) 1
4pe0 r3
(A)
(B)
An inductor designed with 400 turns coil wound on an iron core of 16 cm2 cross
sectional area and with a cut of an air gap length of 1 mm. The coil is connected
to a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and
leakage inductance, (m0 = 4p # 10 - 7 H/M)
EE SP 1.78
EE SP 1.79
(B) 9.04 A
(D) 2.28 A
The average force on the core to reduce the air gap will be
(A) 832.29 N
(B) 1666.22 N
(C) 3332.47 N
(D) 6664.84 N
YEAR 2006
EE SP 1.80
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 1
(A)
2 +0 V, (1 + 2j) W
(C) 2+45% V, (1 + j) W
Page 21
2 +45% V, (1 + j) W
YEAR 2006
EE SP 1.81
TWO MARKS
nodia
(A) 1 V, 3, 10 W
(C) 1 V, 0, 3
EE SP 1.82
(B) 1 V, 0, 10 W
(D) 10 V, 3, 10 W
In the circuit shown in the figure, the current source I = 1 A , the voltage source
V = 5 V, R1 = R2 = R3 = 1 W, L1 = L2 = L3 = 1 H, C1 = C2 = 1 F
The currents (in A) through R3 and through the voltage source V respectively
will be
(A) 1, 4
(B) 5, 1
(C) 5, 2
(D) 5, 4
EE SP 1.83
The parameter type and the matrix representation of the relevant two port
parameters that describe the circuit shown are
0 0
(A) z parameters, = G
0 0
0 0
(C) h parameters, = G
0 0
1 0
(B) h parameters, = G
0 1
1 0
(D) z parameters, = G
0 1
www.nodia.co.in
www.gatehelp.com
Page 22
EE SP 1.84
Chapter 1
nodia
EE SP 1.85
EE SP 1.86
EE SP 1.87
Which of the following statement holds for the divergence of electric and magnetic
flux densities ?
(A) Both are zero
(B) These are zero for static densities but non zero for time varying densities.
(C) It is zero for the electric flux density
(D) It is zero for the magnetic flux density
www.nodia.co.in
www.gatehelp.com
Chapter 1
YEAR 2005
EE SP 1.88
ONE MARK
(A) 2.5 W
(C) 7.5 W
EE SP 1.89
Page 23
(B) 5.0 W
(D) 10.0 W
nodia
The RMS value of the voltage u (t)= 3 + 4 cos (3t) is
(A) 17 V
(B) 5 V
(C) 7 V
(D) (3 + 2 2 ) V
EE SP 1.90
EE SP 1.91
For the two port network shown in the figure the Z -matrix is given by
Z1
Z1 + Z2
(A) =
Z1 + Z2
Z2 G
Z1
Z1
(B) =
Z1 + Z2 Z2 G
Z1
Z2
(C) =
Z2 Z1 + Z2 G
Z1 Z1
(D) =
Z1 Z1 + Z2 G
In the figure given, for the initial capacitor voltage is zero. The switch is closed
at t = 0 . The final steady-state voltage across the capacitor is
(A) 20 V
(C) 5 V
EE SP 1.92
(B) 10 V
(D) 0 V
www.nodia.co.in
www.gatehelp.com
Page 24
Chapter 1
YEAR 2005
TWO MARKS
If, at t = 0+ , the voltage across the coil is 120 V, the value of resistance R is
nodia
(A) 0 W
(C) 40 W
EE SP 1.94
EE SP 1.95
EE SP 1.96
(B) 20 W
(D) 60 W
For the value as obtained in (a), the time taken for 95% of the stored energy to
be dissipated is close to
(A) 0.10 sec
(B) 0.15 sec
(C) 0.50 sec
(D) 1.0 sec
The RL circuit of the figure is fed from a constant magnitude, variable frequency
sinusoidal voltage source Vin . At 100 Hz, the Rand L elements each have a
voltage drop mRMS .If the frequency of the source is changed to 50 Hz, then new
voltage drop across R is
(A)
5
8
uRMS
(B)
2
3
uRMS
(C)
8
5
uRMS
(D)
3
2
uRMS
For the three-phase circuit shown in the figure the ratio of the currents IR: IY : IB
is given by
www.nodia.co.in
www.gatehelp.com
Chapter 1
(A) 1 : 1 :
(C) 1 : 1 : 0
EE SP 1.97
Page 25
(B) 1 : 1 : 2
(D) 1 : 1 : 3/2
The circuit shown in the figure is in steady state, when the switch is closed at
t = 0 .Assuming that the inductance is ideal, the current through the inductor at
t = 0+ equals
nodia
(A) 0 A
(B) 0.5 A
(C) 1 A
(D) 2 A
EE SP 1.98
www.nodia.co.in
www.gatehelp.com
Page 26
EE SP 1.99
Chapter 1
In the given figure, the Thevenins equivalent pair (voltage, impedance), as seen
at the terminals P-Q, is given by
(A) (2 V, 5 W)
(B) (2 V, 7.5 W)
(C) (4 V, 5 W)
nodia
(D) (4 V, 7.5 W)
YEAR 2004
EE SP 1.100
(A) 125.00 mH
(C) 2.0 mF
EE SP 1.101
(B) 304.20 mF
(D) 0.05 mF
A parallel plate capacitor is shown in figure. It is made two square metal plates
of 400 mm side. The 14 mm space between the plates is filled with two layers of
dielectrics of er = 4 , 6 mm thick and er = 2 , 8 mm thick. Neglecting fringing of
fields at the edge the capacitance is
(A) 1298 pF
(C) 354 pF
EE SP 1.102
ONE MARK
(B) 944 pF
(D) 257 pF
The inductance of a long solenoid of length 1000 mm wound uniformly with 3000
turns on a cylindrical paper tube of 60 mm diameter is
(A) 3.2 mH
(B) 3.2 mH
(D) 3.2 H
(C) 32.0 mH
www.nodia.co.in
www.gatehelp.com
Chapter 1
YEAR 2004
EE SP 1.103
EE SP 1.104
(A) 12 V
(B) 24 V
(C) 30 V
(D) 44 V
nodia
(B) 5, 2.5, 5
(D) 2.5, 5, 2.5
EE SP 1.106
TWO MARKS
(A) 2.5, 5, 5
(C) 5, 5, 2.5
EE SP 1.105
Page 27
(B) 5 - j18
(D) 5 - j12
(A) 10
(C) 30
(B) 20
(D) 40
www.nodia.co.in
www.gatehelp.com
Page 28
EE SP 1.107
EE SP 1.108
EE SP 1.109
In figure, the capacitor initially has a charge of 10 Coulomb. The current in the
circuit one second after the switch S is closed will be
(A) 14.7 A
(B) 18.5 A
(C) 40.0 A
(D) 50.0 A
The rms value of the current in a wire which carries a d.c. current of 10 A and a
sinusoidal alternating current of peak value 20 A is
(A) 10 A
(B) 14.14 A
(C) 15 A
(D) 17.32 A
nodia
0.9 0.2
The Z-matrix of a 2-port network as given by =
0.2 0.6G
The element Y22 of the corresponding Y-matrix of the same network is given by
(A) 1.2
(B) 0.4
(C) - 0.4
(D) 1.8
YEAR 2003
EE SP 1.110
ONE MARK
(A) 144 J
(C) 132 J
EE SP 1.111
Chapter 1
(B) 98 J
(D) 168 J
www.nodia.co.in
www.gatehelp.com
Chapter 1
(A) 3 - 8 cos 2t
(C) 16 sin 2t
EE SP 1.112
(B) 60+30%
nodia
(C) 70+30%
(D) 34.4+65%
Two conductors are carrying forward and return current of +I and - I as shown
in figure. The magnetic field intensity H at point P is
(A) I Y
pd
(C) I Y
2pd
EE SP 1.114
(B) 32 sin 2t
(D) 16 cos 2t
(A) 56.66+45%
EE SP 1.113
Page 29
(B) I X
pd
(D) I X
2pd
(A) 2L H/m
(C) L/2 H/m
www.nodia.co.in
www.gatehelp.com
Page 30
YEAR 2003
EE SP 1.115
EE SP 1.116
Chapter 1
TWO MARKS
In the circuit of figure, the magnitudes of VL and VC are twice that of VR . Given
that f = 50 Hz , the inductance of the coil is
(A) 2.14 mH
(B) 5.30 H
(C) 31.8 mH
(D) 1.32 H
nodia
In figure, the potential difference between points P and Q is
(A) 12 V
(B) 10 V
(C) - 6 V
(D) 8 V
EE SP 1.117
Two ac sources feed a common variable resistive load as shown in figure. Under
the maximum power transfer condition, the power absorbed by the load resistance
RL is
(A) 2200 W
(B) 1250 W
(C) 1000 W
(D) 625 W
www.nodia.co.in
www.gatehelp.com
Chapter 1
EE SP 1.118
Page 31
(A) 10 W
(B) 18 W
(C) 24 W
nodia
(D) 12 W
EE SP 1.119
In the circuit shown in figure, the switch S is closed at time (t = 0). The voltage
across the inductance at t = 0+ , is
(A) 2 V
(B) 4 V
(C) - 6 V
(D) 8 V
EE SP 1.120
(A) 0.125
(B) 0.167
(C) 0.625
(D) 0.25
www.nodia.co.in
www.gatehelp.com
Page 32
EE SP 1.121
(A) 0.22 kV
(C) - 2.24 kV
EE SP 1.122
EE SP 1.123
Chapter 1
(B) - 225 V
(D) 15 V
A parallel plate capacitor has an electrode area of 100 mm2, with spacing of
0.1 mm between the electrodes. The dielectric between the plates is air with
a permittivity of 8.85 # 10 - 12 F/m. The charge on the capacitor is 100 V. The
stored energy in the capacitor is
(A) 8.85 pJ
(B) 440 pJ
(C) 22.1 nJ
(D) 44.3 nJ
nodia
(A) 52 V
(C) 67 V
(B) 60 V
(D) 33 V
***********
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 33
SOLUTION
SOL 1.1
nodia
From the given circuit, we have the following observations for the connected
batteries:
100 V battery: As the current flows through the battery from positive to negative
terminal, so it absorbs power.
80 V battery: As the current flows through the battery from negative to positive
terminal, so it delivers power.
15 V battery: As the current flows through the battery negative to positive
terminal, so it delivers power.
Thus, the net power absorbed by circuit elements is obtained as
Pnet = Power absorbed by 100 V battery
[Power emitted by 80 V battery + Power emitted by 15 V battery]
= 100 # 10 - 6^80 # 8h + ^15 # 2h@
= 1000 - 640 - 30
= 330 W
SOL 1.2
(a)
(b)
...(1)
www.nodia.co.in
www.gatehelp.com
Page 34
C1 =
Chapter 1
eo
d
A
2
eo e r
d
Therefore, we get the net capacitance as
and
C2 =
A
2
C net = C1 + C2
Ae o /2 Ae o e r /2 Ae o Ae o e r
+
=
+
= Ae o ^1 + e r h
d
d
2d
2d
2d
Thus, from equations (1) and (2), we get
C net = Co ^1 + e r h
2
=
SOL 1.3
...(2)
nodia
Correct answer is (C)
We have the given circuit as
In the given circuit, the switches operates as shown by the respective arrows. So,
at t = 0 ,
Switch 1 (switch across Vs ) changes it state from short circuit to open circuit.
Switch 2 changes it state from open circuit to short circuit.
Therefore, we have the equivalent circuit at t = 0- as (given Vc ^0 h =- 2 V )
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 35
nodia
- 2 + 20 # 10
s
s
Thus, we obtain
or
-3
1
20 # 10-3 V
100
+
#
s
s
s # 1006
3
Vs =- 2 + 20 #2 10 + 2
s
s
s
= 22 # 10 4
s
So,
Vs ^ t h = 2 # 10 4 t
i.e. the plot of source voltage is a straight line passing through origin, as shown
below.
SOL 1.4
SOL 1.5
Correct answer is 10 A.
We redraw the given circuit as
www.nodia.co.in
www.gatehelp.com
Page 36
Chapter 1
2
300 = 25I + I
2
I 2 + 50I - 600 = 0
I 2 + 60I - 10I - 600 = 0
I ^I + 60h - 10 ^I + 60h = 0
^I - 10h^I + 60h = 0
So,
I = 10 A , - 60 A
Now, for the two obtained values of current, we get
R = b 25 + 10 l = 30
2
R = b 25 - 60 l =- 5
2
negative)
nodia
Thus, the current through the circuit is
I = 10 A
SOL 1.6
D:B = 0
i.e. the divergence of magnetic field is zero. Now, we check this property for each
of the given vectors.
Option (A):
So,
Option (B):
So,
Option (C):
So,
B = y 2 a x + z 2 ay + x 2 a z
D : B = 2 y2 + 2 z2 + 2 x2 = 0
2x
2y
2z
B = z 2 a x + x 2 ay + y 2 a z
D : B = 2 z2 + 2 x2 + 2 y2 = 0
2x
2y
2z
B = x 2 a x + y 2 ay + z 2 a z
D : B = 2 x2 + 2 y2 + 2 z2
2x
2y
2z
= 2x + 2y + 2z ! 0
Option (D):
B = y 2 z 2 a x + x 2 z 2 ay + x 2 y 2 a z
D : B = 2 y2 z2 + 2 x2 z2 + 2 x2 y2 = 0
2x
2y
2z
Thus, the vector given in option (C) does not satisfy the property of magnetic
flux density.
So,
SOL 1.7
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 37
Equivalent inductance
Leq = L1 + L2 + 2M
380 = L + L + 2M
or
L + M = 190
(2) Series opposing connection
...(1)
Equivalent inductance
Leq = L1 + L2 - 2M
240 = L + L - 2M
or
L - M = 120
Substituting eq (2) from eq (1), we get
nodia
...(2)
2L = 70
or
SOL 1.8
L = 35 mH
Therefore
vC ^0-h =- Vo1
Step 2 : For t > 0 , the switch moves to position 2. Again in steady state capacitor
acts as an open circuit.
vC ^3h = Vo2
Time constant
t = RTh C
RTh is the Thevenin resistance across the capacitor terminal after t = 0 . So
t = 2RC
Now, at any time t > 0 , capacitor voltage is given by
vC ^ t h = vC ^3h + 6vC ^0 h - vC ^3h@e-t/t
www.nodia.co.in
www.gatehelp.com
Page 38
Chapter 1
or,
or,
or,
SOL 1.9
nodia
^Vs = 10 Vh
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 39
superposition to obtain vC ^ t h.
First consider source v ^ t h = 20 sin 10t . Open circuit current source and redrawing
circuit in domain with w 1 = 10 rad/ sec .
1
j10
(Using voltage division)
VC1 =
# 20
1+ 1
j10
1
=
20
1 + j10 #
VC1 = 20 - tan-1 10
101
A1 = VC1 = 20 = 1.99 A
101
Now, consider current source i ^ t h = 10 sin 5t . Short circuit voltage source and
redrawing circuit in frequency domain with w 2 = 5 rad/ sec .
nodia
IC2 =
1 10 0c = j5
10 0c
1 + j5 #
1+ 1
j5
VC2 = IC2 # 1
j5
10 0c
VC2 =
1 + j5
VC2 = 10 - tan-5
26
A2 = VC2 = 10 = 1.96 A
26
SOL 1.11
www.nodia.co.in
www.gatehelp.com
Page 40
Chapter 1
H By wire on x -axis = I k
2pd
=4# 1 k
1
2p
H by wire on y -axis = I ^- k h
2pd
= 2 # 1 ^- k h
2
2p
H on (2, 1, 0) = 4 k - 1 k
2p
2p
H = 3 k A/m
2p
SOL 1.12
nodia
...(1)
d : ^ f v h = x2 y + y2 z + z2 x
where f , v are scalar and vector field respectively. From the properties of vector
field, we have
d : ^ f v h = v : ^df h - f ^d : v h
or
Again, we have
So,
v : ^df h = d : ^ f v h + f ^d : v h
...(2)
v = yi + zj + xk
2y 2z 2x
d:v =
+
+
2x 2y 2z
=0
Therefore, we get
f ^d : v h = 0
Substituting it in equation (2), we get
v : ^df h = d : ^ f v h
Thus,
SOL 1.13
v : ^df h = x2 y + y2 z + z2 x
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 41
nodia
The capacitor and inductor can be replaced with its equivalent reactance (in
frequency domain) as shown below.
Solving the above circuit, we get the equivalent driving point impedance as
Z ^s h = s + 1 || bs + 1 l
s
s
2
= s + 1 || b s + 1 l
s
s
or
1 s2 + 1
# s
= s+ s
1 + s2 + 1
s
s
2
+
s
1
= s+ 3
s + 2s
4
2
Z ^s h = s +3 3s + 1
s + 2s
www.nodia.co.in
www.gatehelp.com
Page 42
SOL 1.15
Chapter 1
Q = D : ds
Since, the electric potential is defined as
Q
v =
4pe 0 r
Q = v # 4pe 0 r
= 4pe 0 r
So,
# D : ds = 4pe r
Hence,
SOL 1.16
(since, v = 1 volt)
Correct answer is 2.
To obtain the Norton equivalent along X -Y terminal, we redraw the given circuit
as
nodia
10I1 - 5Isc = 5
2I1 = Isc + 1
...(1)
...(2)
...(3)
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 43
or
or
SOL 1.17
Correct answer is 3 W.
To obtain the required unknown, we redraw the given circuit as
nodia
Applying KCL at node Vx , we have
Vx - 1 + Vx = 2
1
1
2Vx - 1 = 2
or
Vx = 3
2
Therefore, the power delivered by current source is obtained as
= Voltage across current source # current through
or
current source
= 3#2 = 3W
2
SOL 1.18
Due to presence of conducting plane, we can assume an image charge for the
system as shown below
www.nodia.co.in
www.gatehelp.com
Page 44
Chapter 1
So, we can observe from the figure that electric field vector along x -y plane
is cancelled, i.e. zero. Therefore, the electric field vector acts along negative z
-direction and given as
"
...(1)
E = d E1 + E2 n^- kth
2
2
where E1 and E2 are the field intensity due to the charge Q and - Q at point
( 2 , 2 , 0). So, we obtain
Q
32pe 0 2
E1 = 1 2 = 1
4pe 0 r
4pe 0 ( ( 2 ) 2 + ( 2 ) 2 + (2) 2) 2
= 2
Similarly, we get, E2 = 2
Substituting these values in equation (1), we obtain
nodia
"
2 +
2
=- 2kt
E =e
SOL 1.19
2 - kt
o^ h
2
V s2 RL
^^Rs + RL h2 + X s2h
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 45
dP = 0
dRL
or
or
R s2 + R L2 + 2Rs RL + X s2 - 2Rs RL - 2R L2 = 0
R s2 + X s2 = R L2
or
or
RL = R s2 + X s2
i.e. to maximise the power transferred to pure resistive load (RL ), its value must
be equal to the magnitude of Zs of internal circuit.
SOL 1.20
nodia
cos 31c =
or
50
2500 + b103 L -
1
103 C l
...(2)
Rb Ra
Ra Rc
Rb Rc
; RB =
and RA =
Ra + Rb + Rc
Ra + Rb + Rc
Ra + Rb + Rc
www.nodia.co.in
www.gatehelp.com
Page 46
SOL 1.22
Chapter 1
or,
SOL 1.23
nodia
We obtain the power delivered by load as
SOL 1.24
SOL 1.25
2
2
+ XTh
RL = RTh
is the equivalent thevinin (input) impedance of the circuit.
i.e.,
RL =
42 + 32 = 5 W
So,
SOL 1.26
10 4 + 10 4
V2 ^s h
s+1
= 4 s
4 = s+2
V1 ^s h
10 + 10 4 + 10
s
s
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 47
#W
v
nodia
Consider that the voltage across the three capacitors C1 , C 2 and C 3 are V1 , V2 and
V3 respectively. So, we can write
V2 = C 3
....(1)
V3
C2
Since, Voltage is inversely proportional to capacitance
Now, given that
C1 = 10 mF ; ^V1hmax = 10V
C2 = 5 mF ; ^V2hmax = 5 V
C 3 = 2 mF ; ^V3hmax = 2V
So, from Eq (1) we have
V2 = 2
5
V3
for
We obtain,
^V3hmax = 2
V2 = 2 # 2 = 0.8 volt < 5
5
i.e.,
V2 < ^V2hmax
Hence, this is the voltage at C2 . Therefore,
V3 = 2 volt
V2 = 0.8 volt
and
V1 = V2 + V3 = 2.8 volt
Now, equivalent capacitance across the terminal is
Ceq = C 2 C 3 + C1
C2 + C3
= 5 # 2 + 10 = 80 mF
5+2
7
Equivalent voltage is (max. value)
Vmax = V1 = 2.8
So, charge stored in the effective capacitance is
Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC
7
www.nodia.co.in
www.gatehelp.com
Page 48
SOL 1.28
Chapter 1
nodia
SOL 1.29
1
V1 + 1 0c - 1 + j + 1
=
I1 =
j1
j1
j
=
= 1 A
(1 + j) j 1 + j
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.30
Page 49
G (jw) = 0
-w 2 + 9 = 0
w = 3 rad/s
SOL 1.31
nodia
ZTh = Vtest
Itest
...(i)
Ib =- Vtest =-Vtest
9k + 1k
10k
But
Z = 4 - j3
Z = 5 - 36.86cW
I = 5 100c A
Average power delivered :
Pavg. = 1 I 2 Z cos q
2
= 1 # 25 # 5 cos 36.86c = 50 W
2
www.nodia.co.in
www.gatehelp.com
Page 50
Chapter 1
Alternate Method:
Z = (4 - j3) W
I = 5 cos (100pt + 100) A
2
Pavg = 1 Re $ I Z .
2
= 1 # Re "(5) 2 # (4 - j3),
2
= 1 # 100 = 50 W
2
SOL 1.33
nodia
vc (0) /s
v (0)
= c
1 + 1
1 + 1
C1 C 2
C1 s C 2 s
= b C1 C2 l (12 V)
C1 + C 2
I (s) =
vC (0) = 12 V
= 12Ceq
Taking inverse Laplace transform for the current in time domain,
i (t) = 12Ceq d (t)
SOL 1.34
(Impulse)
VA - VB = 6 V
So current in the branch,
IAB = 6 = 3 A
2
We can see, that the circuit is a one port circuit looking from terminal BD as
shown below
For a one port network current entering one terminal, equals the current leaving
the second terminal. Thus the outgoing current from A to B will be equal to the
incoming current from D to C as shown
i.e.
IDC = IAB = 3 A
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 51
= 2+3 = 5A
VCD = 1 # (- I1) =- 5 V
So,
SOL 1.35
nodia
Correct option is (A).
We obtain Thevenin equivalent of circuit B .
Thevenin Impedance :
ZTh = R
Thevenin Voltage :
VTh = 3 0c V
Now, circuit becomes as
I1 = 10 - 3
2+R
Power transfer from circuit A to B
P = (I 12) 2 R + 3I1
Current in the circuit,
www.nodia.co.in
www.gatehelp.com
Page 52
Chapter 1
2
= :10 - 3D R + 3 :10 - 3D = 49R 2 + 21
2+R
2+R
(2 + R)
(2 + R)
49R + 21 (2 + R)
=
= 42 + 70R2
2
(2 + R)
(2 + R)
2
dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0
dR
(2 + R) 4
(2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0
nodia
Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let
Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance
is RTh .
IL =
VTh,10 V
RTh + RL
For RL = 1 W , IL = 3 A
3=
VTh,10 V
RTh + 1
...(i)
2=
VTh,10 V
RTh + 2.5
...(ii)
For RL = 2.5 W , IL = 2 A
Substituting RTh
3RTh + 3 = 2RTh + 5
RTh = 2 W
into equation (i)
VTh,10 V = 3 (2 + 1) = 9 V
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 53
Note that it is a non reciprocal two port network. Thevenin voltage seen at port
B depends on the voltage connected at port A. Therefore we took subscript
VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input
port A. If the voltage connected to port A is different, then Thevenin voltage
will be different. However, Thevenins resistance remains same.
Now, the circuit is
nodia
For RL = 7 W ,
SOL 1.37
IL =
VTh,10 V
= 9 = 1A
2 + RL 2 + 7
VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V
3
3
3 3
This is a linear network, so VTh at port B can be written as
VTh = V1 a + b
where V1 is the input applied at port A.
We have V1 = 10 V , VTh,10 V = 9 V
...(i)
9 = 10a + b
When V1 = 6 V , VTh, 6 V = 9 V
...(ii)
7 = 6a + b
Solving (i) and (ii)
a = 0.5 , b = 4
Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit
voltage at port B will be
So,
VTh, V = 0.5V1 + 4
For
V1 = 8 V
1
www.nodia.co.in
www.gatehelp.com
Page 54
Chapter 1
220 =
nodia
Power absorbed in RL
SOL 1.40
Since the capacitor and inductive reactances are equal in magnitude, the net
impedance of that branch will become zero.
Equivalent circuit
Current
rms value of current
SOL 1.41
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 55
In phasor form,
SOL 1.42
nodia
Power transferred to the load
2
10
RL
l
Rth + RL
For maximum power transfer Rth , should be minimum.
Rth = 6R = 0
6+R
P = I 2 RL = b
R =0
Note: Since load resistance is constant so we choose a minimum value of Rth
SOL 1.43
2
(5 103) 2
Power loss = V rated = #
= 20 W
Rp
1.25 # 106
For an parallel combination of resistance and capacitor
1
1
=
tan d =
wC p R p 2p # 50 # 1.25 # 0.102
= 1 = 0.025
40
SOL 1.44
Q = CV = e0 er A V = (e0 er A)V
d
d
C = e0 er A
d
Q = Q max
We have e0 = 8.85 # 10 F/cm , er = 2.26 , A = 20 # 40 cm2
-14
www.nodia.co.in
www.gatehelp.com
Page 56
Chapter 1
V = 50 103 kV/cm
#
d
Maximum electrical charge on the capacitor
V = V
when
b d l = 50 kV/cm
d
max
Thus,
SOL 1.45
nodia
Input power factor
I1 (rms)
cos f1
Irms
I1 (rms) is rms values of fundamental component of current and Irms is the rms value
of converter current.
10
pf =
cos p/3 = 0.44
2
10 + 52 + 22
pf =
SOL 1.46
SOL 1.47
SOL 1.48
2 cos p/4 = 1 W
2 p/ 4 -
2 - p/ 4
= 2 2 j sin p/4 = 2j
SOL 1.49
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 57
For capacitor at t = 0+
vc (0+) = vc (0) = 4 V
v (0+)
current in 4 W resistor at t = 0+ , i1 = c
=1A
4
nodia
so current in capacitor at t = 0+ , ic (0+) = i1 = 1 A
SOL 1.50
i = 100 = 50 A
1+1
SOL 1.51
Current in R W resistor is
i = 2-1 = 1 A
Voltage across 12 W resistor is
VA = 1 # 12 = 12 V
So,
i = VA - 6 = 12 - 6 = 6 W
1
R
www.nodia.co.in
www.gatehelp.com
Page 58
SOL 1.52
Chapter 1
V l1 = Zl11 I l1 + Zl12 I l2
V1 = Z11 I1 + Z12 I2
V l2 = Zl21 I l1 + Zl22 I l2
V2 = Z21 I1 + Z22 I2
Here, I1 = I l1, I2 = I l2
When R = 1 W is connected
V l1 = V1 + I l1 # 1 = V1 + I1
V l1
V l1
So,
Zl11
Zl12
Similarly for output port
V l2
So, Zl21 = Z21 , Zl22 = Z22
nodia
= Z11 + 1
= Z12
Z11 + 1 Z12
Z =>
Z21 Z22H
Z-matrix is
SOL 1.53
= Z11 I1 + Z12 I2 + I1
= (Z11 + 1) I1 + Z12 I2
In the bridge
R1 R 4 = R 2 R 3 = 1
So it is a balanced bridge
I = 0 mA
SOL 1.54
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.55
Page 59
SOL 1.56
nodia
Applying KVL in the input loop
v1 - i1 (1 + 1) # 103 - 1 (i1 + 49i1) = 0
jwC
Input impedance,
Equivalent capacitance,
v1 = 2 # 103 i1 + 1 50i1
jw C
1
Z1 = v1 = 2 # 103 +
i1
jw (C/50)
100 nF
Ceq = C =
= 2 nF
50
50
SOL 1.57
SOL 1.58
www.nodia.co.in
www.gatehelp.com
Page 60
Chapter 1
VP - 5 + VP + VS = I
S
2
2
1
VP - 5 + VP + 2VS = 2IS
2VP + 2VS = 2Is + 5
VP + VS = IS + 2.5
&
So,
VP - VS = 3VS
VP = 4VS
4VS + VS = IS + 2.5
5VS = IS + 2.5
VS = 0.2IS + 0.5
For Thevenin equivalent circuit
nodia
VS = IS Rth + Vth
By comparing (2) and (3),
Thevenin resistance Rth = 0.2 kW
SOL 1.59
SOL 1.60
...(1)
...(2)
...(3)
Vth = 0.5 V
SOL 1.61
R = 2.38 W
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 61
nodia
Ceq = 2 F
3
Equivalent Resistance
Req = 3 + 3 = 6 W
Time constant
SOL 1.63
R
1 - jwCR
R
= jwL +
# 1 - jwCR
1
+
w
j
CR
+R
R (1 - jwCR)
= j wL +
1 + w2 C2 R2
jwL (1 + w2 C2 R2) + R - jwCR2
=
1 + w2 C2 R2
j [wL (1 + w2 C2 R2) - wCR2]
R
=
+
1 + w2 C2 R2
1 + w2 C2 R2
For resonance Im (Z) = 0
So,
wL (1 + w2 C2 R2) = wCR2
L = 0.1 H, C = 1 F, R = 1 W
So,
w # 0.1 [1 + w2 (1) (1)] = w (1) (1) 2
Z = jw L +
1 + w2 = 10
&
SOL 1.64
w=
9 = 3 rad/sec
Vab - 2i + 5 = 0
Vab = 2 # 1 - 5 =- 3 Volt
www.nodia.co.in
www.gatehelp.com
Page 62
SOL 1.65
Chapter 1
Q =
# i (t) dt
nodia
Q =Area OABCDO
=Area (OAD)+Area(AEB)+Area(EBCD)
= 1#2#4+1#2#3+3#2
2
2
= 4 + 3 + 6 = 13 nC
SOL 1.66
SOL 1.67
vc (t) = Vo cos wo t
1
wo = 1 =
-9
LC
0.3 # 10 # 0.6 # 10- 3
= 2.35 # 106 rad/sec
vc (t) = 43.33 cos (2.35 # 106 # 1 # 10- 6)
= 43.33 # (- 0.70) =- 30.44 V
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 63
Vb = 1.25 V
i = Vb = 1.25 A
1
Current
SOL 1.68
SOL 1.69
nodia
- 12
-6
# 500 # 10
C1 = e0 er1 A = 8.85 # 10 # 8 # 500
d1
4 # 10- 3
= 442.5 # 10- 11 F
- 12
-6
# 500 # 10
C2 = e0 er2 A = 8.85 # 10 # 2 # 500
3
d2
2 # 10
= 221.25 # 10- 11 F
- 11
- 11
Ceq = 442.5 # 10 - 11 # 221.25 # 10- 11
442.5 # 10 + 221.25 # 10
= 147.6 # 10- 11
- 1476 pF
SOL 1.70
l = 300 mm
n = 300
A = 300 mm2
m n2 A
L = 0
l
4p # 10- 7 # (300) 2 # 300 # 10- 6
(300 # 10- 3)
= 113.04 mH
=
SOL 1.71
www.nodia.co.in
www.gatehelp.com
Page 64
SOL 1.72
Chapter 1
...(1)
In second loop
- 5I2 + 0.2Vx + 0.5 dI1 = 0
dt
I2 = 0.04Vx + 0.1 dI1
dt
...(2)
SOL 1.73
nodia
Correct option is (A).
Impedance of the given network is
Z = R + j b wL - 1 l
wC
1
AdmittanceY = 1 =
Z
R + j b wL - 1 l
wC
R - j ^wL - w1C h
R - j ^wL - w1C h
1
=
=
#
R + j ^wL - w1C h R - j ^wL - w1C h
R2 + ^wL - w1C h2
j ^wL - w1C h
R
= 2
R + ^wL - w1C h2 R2 + ^wL - w1C h2
= Re (Y) + Im (Y)
Varying frequency for Re (Y) and Im (Y) we can obtain the admittance-locus.
SOL 1.74
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 65
Applying KCL
vL (0+) vL (0+) - vc (0+)
+
= iL (0+) = 10
10
10
2vL (0+) - 10 = 100
Voltage across inductor at t = 0+
vL (0+) = 100 + 10 = 55 V
2
nodia
So, current in capacitor at t = 0+
vL (0+) - vc (0+)
iC (0 ) =
10
= 55 - 10 = 4.5 A
10
+
SOL 1.75
Applying KVL
2
= VNL
3 - 2 # I NL
2
2
3 - 2I NL
= I NL
2
3I NL
= 3 & INL = 1 A
VNL = (1) 2 = 1 V
SOL 1.76
VX = V+0c
www.nodia.co.in
www.gatehelp.com
Page 66
Chapter 1
Vy - 2V+0c
+ (Vy) jwC = 0
R
Vy (1 + jwCR) = 2V+0c
Vy = 2V+0c
1 + jwCR
VYX = VX - VY = V -
2V
1 + jwCR
VYX = V - 2V =- V
R " 0,
R " 3,
VYX = V - 0 = V
So power dissipated in the non-linear resistance
P = VNL INL = 1 # 1 = 1 W
SOL 1.77
nodia
E = AV
V
R
S1 1 1 0 0 0W
S 0 -1 0 -1 1 0 W
T
T
8e1 e2 e 3 e 4B = S- 1 0 0 0 - 1 - 1W8V1 V2 -- V6B
W
S
S 0 0 -1 1 0 1 W
X
R V RT
V
Se1W S V1 + V2 + V3 W
Se2W S- V2 - V4 + V5W
Se W = S- V - V - V W which is true.
5
6W
S 3W S 1
Se 4W S- V3 + V4 + V6W
T X T
X
SOL 1.78
# D : ds = Q enclosed
3
Q
4 pr3 = Qr
#
3
4
3
R3
3 pR
www.nodia.co.in
www.gatehelp.com
Chapter 1
So,
or
SOL 1.79
Qr3
R3
Qr3
Q r
D # 4pr2 = 3 =
4pe0 R3
R
# D : ds
Page 67
a D = e0 E
m0 N2 A
l
SOL 1.80
SOL 1.81
nodia
Correct option is (A).
Energy stored is inductor
E = 1 LI2
2
= 1 # 321.6 # 10- 3 # (2.28) 2
2
Force required to reduce the air gap of length 1 mm is
F = E = 0.835- 3 = 835 N
l
1 # 10
Correct option is (D).
Thevenin voltage:
Vth = I (R + ZL + ZC )
= 1+0c [1 + 2j - j]
= 1 (1 + j) =
2 +45% V
Thevenin impedance:
Zth = R + ZL + ZC = 1 + 2j - j = (1 + j) W
www.nodia.co.in
www.gatehelp.com
Page 68
SOL 1.82
Chapter 1
Output voltage
vo = Avi = 106 # 1 mV = 1 V
Input impedance
Zi = vi = vi = 3
0
ii
Output impedance
Zo = vo = Avi = Ro = 10 W
io
io
SOL 1.83
nodia
Voltage across R 3 is
5 = I1 R 3
5 = I1 (1)
I1 = 5 A
By applying KCL, current through voltage source
(current through R 3 )
1 + I 2 = I1
I2 = 5 - 1 = 4 A
SOL 1.84
SOL 1.85
www.nodia.co.in
www.gatehelp.com
Chapter 1
Current
IR = V1 +0c
R1 + R 2
Voltage
V2 = IR R2 + j (VL - VC )
V2 = V1 +0c R2
R1 + R 2
Page 69
VC =
So phasor diagram is
SOL 1.86
nodia
Correct option is (B).
This is a second order LC circuit shown below
dvc (t)
dt
# vc (t) dt
V (s)
IL (s) = 1
L s
For t > 0 , applying KCL(in s-domain)
IC (s) + IL (s) = 0
V (s)
=0
CsV (s) - V (0) + 1
L s
1
2
:s + LCs D V (s) = Vo
V (s) = Vo 2 s 2 ,
s + w0
Taking inverse Laplace transformation
v (t) = Vo cos wo t ,
a w20 = 1
LC
t>0
www.nodia.co.in
www.gatehelp.com
Page 70
SOL 1.87
Chapter 1
2.3 # 103 =
R = 23 W (Resistance of heater)
Now it is connected with a square wave source of 400 V peak to peak
Power dissipated is
2
P = V rms ,
R
=
SOL 1.88
(200) 2
= 1.739 kW
23
nodia
4: B = 0
I =
Or
SOL 1.90
100
=8 A
R + (10 || 10)
(given)
100 = 8
R+5
R = 60 = 7.5 W
8
SOL 1.91
SOL 1.89
32 +
(4) 2
2
9 + 8 = 17 V
...(1)
...(2)
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.92
Page 71
nodia
= 10 V
SOL 1.93
SOL 1.94
120 = 2 A
20 + 40
At t = 0 , when switch is moved to position 1,inductor current does not change
simultaneously so
iL (0-) =
iL (0+) = iL (0-)=2 A
Voltage across inductor at t = 0+
vL (0+) = 120 V
By applying KVL in loop
120 = 2 (40 + R + 20)
120 = 120 + R
R = 0W
www.nodia.co.in
www.gatehelp.com
Page 72
SOL 1.95
Chapter 1
60
So,
SOL 1.96
nodia
Correct option is (C).
At f1 = 100 Hz, voltage drop across R and L is mRMS
mRMS = Vin .R
R + jw1 L
V (jw L)
= in 1
R + jw1 L
So,
R = w1 L
At f2 = 50 Hz, voltage drop across R
mlRMS = Vin .R
R + jw2 L
mRMS
R + jw2 L
=
=
R + jw1 L
mlRMS
=
=
=
mlRMS =
SOL 1.97
w12 L2 + w22 L2 ,
w12 L2 + w12 L2
R2 + w22 L2
R2 + w12 L2
R = w1 L
f 12 + f 22
w12 + w22 =
2f 12
2w12
(100) 2 + (50) 2
= 5
8
2 (100) 2
8m
5 RMS
= I R2 + I y2 + IR Iy
I R = Iy
I B2 = I R2 + I R2 + I R2
= 3I R2
IB = 3 IR
=
3 Iy
IR: Iy: IB = 1: 1: 3
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.98
Page 73
iL (0-) = 10 = 1 A
10
Inductor current does not change simultaneously so at t = 0 when switch is
closed current remains same
nodia
iL (0+) = iL (0-)=1 A
SOL 1.99
SOL 1.100
Nodal analysis at P
Vth - 4 + Vth = 0
10
10
2Vth - 4 = 0
Vth = 2 V
Thevenin resistance:
Rth = 10 W || 10 W = 5 W
SOL 1.101
www.nodia.co.in
www.gatehelp.com
Page 74
Chapter 1
1 + j wC = 0
j wL
1 - w2 LC = 0
1 (resonant frequency)
LC
1
1
C = 2 =
= 0.05 m F
2
4 # p # (500) 2 # 2
wL
w=
SOL 1.102
nodia
Similarly
- 12
16 # 10- 2 = 94.4 10- 11 F
= 8.85 # 10 # 4 #
#
-3
6 # 10
- 11
- 11
= 25.74 # 10- 11 - 257 pF
Ceq = 94.4 # 10 # 35.4 #-10
11
(94.4 + 35.4) # 10
SOL 1.103
SOL 1.104
Voltage
So,
VA = (2 + 1) # 6 = 18 Volt
2 = E - VA
6
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 75
2 = E - 18
6
E = 12 + 18 = 30 V
SOL 1.105
SOL 1.106
given as
= 10 # 10 = 2.5 W
20 + 10 + 10
= 20 # 10 = 5 W
20 + 10 + 10
= 20 # 10 = 5 W
20 + 10 + 10
nodia
Correct option is (D).
For parallel circuit
I = E = EYeq
Zeq
= 0.5 - j1.2
So, current I = 10 (0.5 - j1.2) = (5 - j12) A
SOL 1.107
100
(10 || R)
10 + (10 || R) #
100
10R
=
f 10 + 10R pb 10 + R l
10 + R
= 1000R = 50R
100 + 20R
5+R
Current in R W resistor
2 = VA
R
2 = 50R
R (5 + R)
Voltage
or
SOL 1.108
VA =
R = 20 W
www.nodia.co.in
www.gatehelp.com
Page 76
Chapter 1
nodia
Current in the circuit
= 0.5 # 80e- t
= 40e- t
At t = 1 sec,
SOL 1.109
I = 10 + 20 sin wt
102 +
Irms =
(20) 2
2
= 100 + 200
=
300 = 17.32 A
SOL 1.110
SOL 1.111
EL =
# Pdt
0
Where power
P = VI = I bL dI l
dt
t
So,
EL =
dt
# LIb dI
dt l
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 77
For0 # t # 4 sec
4
dt
# Ib dI
dt l
EL = 2
a dI = 3, 0 # t # 2
,
#
#
* dt
0
2
= 0, 2 < t < 4
2
= 6 # I.dt =6(area under the curve i (t) - t )
2
=2
I (3) dt + 2
I (0) dt
= 6 # 1 # 2 # 6 = 36 J
2
Energy absorbed by 1 W resistor is
t
ER =
I2 Rdt =
# (3t) 2 # 1dt +
0
# (6) 2 dt
2
I = 3t, 0 # t # 2
)
= 6A 2 # t # 4
nodia
3 2
= 9 # :t D + 36[t]2 = 24 + 72 =96 J
3 0
Total energy absorbed in 4 sec
E = EL + ER = 36 + 96 = 132 J
SOL 1.112
iL = iC + 1 + 2
iL = iC + 3
iC =- C dvc =- 1 d [4 sin 2t]
dt
dt
so
Voltage across inductor
=- 8 cos 2t
iL =- 8 cos 2t + 3
www.nodia.co.in
www.gatehelp.com
Page 78
Chapter 1
Given that
= 10 c
1-
3j
m = 5 (1 -
3 j)
3j
m = 5 (1 +
3 j)
Z2 = 10+60c
= 10 c
1+
Z 3 = 50+53.13c
3 + 4j
= 50 b
= 10 (3 + 4j)
5 l
5 (1 - 3j) 5 (1 + 3 j)
5 (1 - 3 j) + 5 (1 + 3 j)
25 (1 + 3)
= 10 (3 + 4j) +
10
Zth = 10 (3 + 4j) +
So,
nodia
= 30 + 40j + 10 = 40 + 40j
Zth = 40 2 +45c W
SOL 1.114
SOL 1.115
Correct option is ( ).
SOL 1.116
VL = wL = 2VR
wL = 2 # 5
2 # p # 50 # L = 10
L = 10 = 31.8 mH
314
VR = 5 V, at resonance
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.117
Page 79
nodia
VPQ = VP - VQ
= 24 - 54 =- 6 V
5
5
SOL 1.118
Thevenin voltage
Vth - 110+0c + Vth - 90+0c 0
=
6 + 8j
6 + 8j
2Vth - 200+0c = 0
Vth = 100+0c V
Thevenin impedance
32 + 42 = 5 W
www.nodia.co.in
www.gatehelp.com
Page 80
Chapter 1
Power in load
2
P = ieff
RL
2
100
P =
5
3 + 4j + 5 #
(100) 2
5
80 #
= 625 Watt
SOL 1.119
nodia
Correct option is (D).
Applying mesh analysis in the circuit
I1 = 10 A, I2 =- 5 A
Current in 2 W resistor
I2W = I1 - (- I2)
= 10 - (- 5) = 15 A
So, voltage
VA = 15 # 2 = 30 Volt
Now we can easily find out current in all branches as following
Current in resistor R is 5 A
5 = 100 - 40
R
R = 60 = 12 W
5
www.nodia.co.in
www.gatehelp.com
Chapter 1
SOL 1.120
Page 81
nodia
Simplified circuit
At node A
E A - E1 + E A - E 2 + E A = 0
2
2
4
5EA = 2E1 + 2E2
...(1)
www.nodia.co.in
www.gatehelp.com
Page 82
Chapter 1
Similarly
E1 - E A + E1 = 0
2
2
2E1 = EA
...(2)
nodia
9
10- 9 - 9 # 109 # 1 # 10- 9
= 9 # 10 # 1 #
-3
40 # 10
20 # 10- 3
= 9 # 103 : 1 - 1 D =- 225 Volt
40 20
SOL 1.123
- 12
-6
# 10
C = e0 A = 8.85 # 10 # 100
3
d
0.1 # 10
= 8.85 # 10- 12 F
E = 1 # 8.85 # 10- 12 # (100) 2
2
= 44.3 nJ
SOL 1.124
www.nodia.co.in
www.gatehelp.com
Chapter 1
Page 83
************
nodia
www.nodia.co.in
www.gatehelp.com
Page 84
Chapter 2
CHAPTER 2
SIGNALS AND SYSTEMS
2014 EE01
EE SP 2.1
EE SP 2.2
ONE MARK
Let X ^s h = 2 3s + 5
be the Laplace Transform of a signal x ^ t h. Then, x ^0+h
s + 10s + 21
is
(A) 0
(B) 3
(C) 5
(D) 21
nodia
For a periodic square wave, which one of the following statements is TRUE ?
(A) The Fourier series coefficients do not exist.
(B) The Fourier series coefficients exist but the reconstruction converges at no
point.
(C) The Fourier series coefficients exist and the reconstruction converges at
most points.
(D) The Fourier series coefficients exist and the reconstruction converges at
every point.
2014 EE02
EE SP 2.3
TWO MARKS
Let f ^ t h be a continuous time signal and let F ^wh be its Fourier Transform
defined by
F ^w h =
Define g ^ t h by
#
g^t h = #
-3
3
-3
f ^ t h e-jwt dt
F ^u h e-jut du
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 85
^t - T h
^t - 2T h
u ^t - 2T h
u ^t - T h T
T
(B) u ^ t h + t u ^t - T h - t u ^t - 2T h
T
T
^t - T h
^t - 2T h
(C) u ^ t h - u ^t - T h +
u^t h u^t h
T
T
^t - T h
^t - 2T h
(D) u ^ t h +
u ^t - 2T h
u ^t - T h - 2
T
T
(A) u ^ t h - u ^t - T h +
2014 EE02
EE SP 2.5
EE SP 2.6
ONE MARK
nodia
EE SP 2.7
EE SP 2.8
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Page 86
Chapter 2
2014 EE03
EE SP 2.9
ONE MARK
nodia
(D) imaginary and even function of w
EE SP 2.10
A signal is represented by
1
t <1
x^t h = *
0
t >1
The Fourier transform of the convolved signal y ^ t h = x ^2t h * x ^t/2h is
EE SP 2.11
(B) 42 sin a w k
2
w
(D) 42 sin2 w
w
For the signal f ^ t h = 3 sin 8pt + 6 sin 12pt + sin 14pt , the minimum sampling
frequency (in Hz) satisfying the Nyquist criterion is _____.
2014 EE03
EE SP 2.12
TWO MARKS
A continuous-time LTI system with system function H ^wh has the following polezero plot. For this system, which of the alternatives is TRUE ?
(A) H ^0 h
(B) H ^wh
(C) H ^0 h
(D) H ^wh
> ^w h ; w > 0
has multiple maxima, at w 1 and w 2
< H ^w h ; w > 0
= constant; - 3 < w < 3
www.nodia.co.in
www.gatehelp.com
Chapter 2
EE SP 2.13
EE SP 2.14
Page 87
(D) 90 Hz
nodia
YEAR 2013
EE SP 2.15
ONE MARK
(D) 20 kHz
EE SP 2.16
EE SP 2.17
EE SP 2.18
For a periodic signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the
fundamental frequency in rad/s
(A) 100
(B) 300
(C) 500
(D) 1500
Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected in cascade.
Then the overall impulse response of the cascaded system is given by
(A) product of h1 ^ t h and h2 ^ t h
(B) sum of h1 ^ t h and h2 ^ t h
(C) convolution of h1 ^ t h and h2 ^ t h
(D) subtraction of h2 ^ t h from h1 ^ t h
Which one of the following statements is NOT TRUE for a continuous time
causal and stable LTI system?
(A) All the poles of the system must lie on the left side of the jw axis
(B) Zeros of the system can lie anywhere in the s-plane
(C) All the poles must lie within s = 1
(D) All the roots of the characteristic equation must be located on the left side
of the jw axis.
EE SP 2.19
www.nodia.co.in
www.gatehelp.com
Page 88
output is
2
(A) t u ^ t h
2
(C)
Chapter 2
(B)
^t - 1h2
u ^t - 1h
2
t ^t - 1h
u ^t - 1h
2
2
(D) t - 1 u ^t - 1h
2
YEAR 2013
EE SP 2.20
TWO MARKS
(D) 3
nodia
YEAR 2012
EE SP 2.21
EE SP 2.22
ONE MARK
If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC) of its z
-transform in the z -plane will be
(A) 1 < z < 3
(B) 1 < z < 1
3
3
2
(C) 1 < z < 3
(D) 1 < z
2
3
TWO MARKS
EE SP 2.23
Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and
g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals
(A) 0
(B) 1/2
(C) 1
(D) 3/2
EE SP 2.24
The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w . The value
of h (0) is
(A) 1/4
(B) 1/2
(C) 1
(D) 2
EE SP 2.25
The input x (t) and output y (t) of a system are related as y (t) =
. The system is
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 89
YEAR 2011
EE SP 2.26
ONE MARK
of
n=1
the periodic signal shown below will contain the following nonzero terms
nodia
(A) a 0 and bn, n = 1, 3, 5, ...3
(B) a 0 and an, n = 1, 2, 3, ...3
(C) a 0 an and bn, n = 1, 2, 3, ...3
(D) a 0 and an n = 1, 3, 5, ...3
EE SP 2.27
Given two continuous time signals x (t) = e-t and y (t) = e-2t which exist for
t > 0 , the convolution z (t) = x (t) * y (t) is
(A) e-t - e-2t
(B) e-3t
(C) e+t
(D) e-t + e-2t
YEAR 2011
TWO MARKS
EE SP 2.28
Let the Laplace transform of a function f (t) which exists for t > 0 be F1 (s)
and the Laplace transform of its delayed version f (t - t) be F2 (s). Let F1 * (s)
be the complex conjugate of F1 (s) with the Laplace variable set s = s + jw . If
F (s) F1 * (s)
, then the inverse Laplace transform of G (s) is an ideal
G (s) = 2
F1 (s) 2
(A) impulse d (t)
(B) delayed impulse d (t - t)
(C) step function u (t)
(D) delayed step function u (t - t)
EE SP 2.29
The response h (t) of a linear time invariant system to an impulse d (t), under
initially relaxed condition is h (t) = e-t + e-2t . The response of this system for a
www.nodia.co.in
www.gatehelp.com
Page 90
Chapter 2
ONE MARK
For the system 2/ (s + 1), the approximate time taken for a step response to
reach 98% of the final value is
(A) 1 s
(B) 2 s
nodia
(C) 4 s
(D) 8 s
EE SP 2.31
EE SP 2.32
5t
The second harmonic component of the periodic waveform given in the figure
has an amplitude of
(A) 0
(B) 1
(C) 2/p
(D)
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 91
YEAR 2010
EE SP 2.34
TWO MARKS
(A) 2
(B) 2p
(D) 4p
nodia
(C) 4
EE SP 2.35
#- 3
Given the finite length input x [n] and the corresponding finite length output y [n]
of an LTI system as shown below, the impulse response h [n] of the system is
EE SP 2.36
EE SP 2.37
(B) g (t) = f` t - 3j
2
(D) g (t) = f` t - 3 j
2 2
(B) 1 (e - 5s - e - 3s)
s
- 3s
(C) e (1 - e - 2s)
s
www.nodia.co.in
www.gatehelp.com
Page 92
Chapter 2
YEAR 2009
EE SP 2.38
ONE MARK
A Linear Time Invariant system with an impulse response h (t) produces output
y (t) when input x (t) is applied. When the input x (t - t) is applied to a system
with impulse response h (t - t), the output will be
(A) y (t)
(B) y (2 (t - t))
(C) y (t - t)
(D) y (t - 2t)
YEAR 2009
EE SP 2.39
TWO MARKS
nodia
A cascade of three Linear Time Invariant systems is causal and unstable. From
this, we conclude that
(A) each system in the cascade is individually causal and unstable
(B) at least on system is unstable and at least one system is causal
(C) at least one system is causal and all systems are unstable
(D) the majority are unstable and the majority are causal
EE SP 2.40
EE SP 2.41
- ji
- ji
YEAR 2008
EE SP 2.42
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 93
A signal e - at sin (wt) is the input to a real Linear Time Invariant system. Given K
and f are constants, the output of the system will be of the form Ke - bt sin (vt + f)
where
(A) b need not be equal to a but v equal to w
(B) v need not be equal to w but b equal to a
(C) b equal to a and v equal to w
(D) b need not be equal to a and v need not be equal to w
nodia
YEAR 2008
EE SP 2.44
TWO MARKS
z
Given X (z) =
with z > a , the residue of X (z) zn - 1 at z = a for n $ 0
2
(z - a)
will be
(A) an - 1
(B) an
(D) nan - 1
(C) nan
EE SP 2.45
A system with x (t) and output y (t) is defined by the input-output relation :
y (t) =
- 2t
#- 3x (t) dt
A signal x (t) = sinc (at) where a is a real constant ^sinc (x) = px h is the input
to a Linear Time Invariant system whose impulse response h (t) = sinc (bt), where
b is a real constant. If min (a, b) denotes the minimum of a and b and similarly,
max (a, b) denotes the maximum of a and b, and K is a constant, which one of
the following statements is true about the output of the system ?
(A) It will be of the form Ksinc (gt) where g = min (a, b)
(B) It will be of the form Ksinc (gt) where g = max (a, b)
(C) It will be of the form Ksinc (at)
(D) It can not be a sinc type of signal
sin (px)
EE SP 2.47
Let x (t) be a periodic signal with time period T , Let y (t) = x (t - t0) + x (t + t0)
for some t0 . The Fourier Series coefficients of y (t) are denoted by bk . If bk = 0 for
all odd k , then t0 can be equal to
(A) T/8
(B) T/4
(C) T/2
(D) 2T
EE SP 2.48
www.nodia.co.in
www.gatehelp.com
Page 94
Chapter 2
input to such a system, the output is zero. Further, the Region of convergence
(ROC) of ^1 - 12 z - 1h H(z) is the entire Z-plane (except z = 0 ). It can then be
inferred that H (z) can have a minimum of
(A) one pole and one zero
(B) one pole and two zeros
(C) two poles and one zero
D) two poles and two zeros
EE SP 2.49
1
2
sin (px)
EE SP 2.50
nodia
Given a sequence x [n], to generate the sequence y [n] = x [3 - 4n], which one of
the following procedures would be correct ?
(A) First delay x (n) by 3 samples to generate z1 [n], then pick every 4th sample
of z1 [n] to generate z2 [n], and than finally time reverse z2 [n] to obtain y [n].
(B) First advance x [n] by 3 samples to generate z1 [n], then pick every 4th
sample of z1 [n] to generate z2 [n], and then finally time reverse z2 [n] to
obtain y [n]
(C) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n]
to obtain v2 [n], and finally advance v2 [n] by 3 samples to obtain y [n]
(D) First pick every fourth sample of x [n] to generate v1 [n], time-reverse v1 [n]
to obtain v2 [n], and finally delay v2 [n] by 3 samples to obtain y [n]
YEAR 2007
EE SP 2.51
ONE MARK
Let a signal a1 sin (w1 t + f) be applied to a stable linear time variant system. Let
the corresponding steady state output be represented as a2 F (w2 t + f2). Then
which of the following statement is true?
(A) F is not necessarily a Sine or Cosine function but must be periodic
with w1 = w2 .
(B) F must be a Sine or Cosine function with a1 = a2
(C) F must be a Sine function with w1 = w2 and f1 = f2
(D) F must be a Sine or Cosine function with w1 = w2
EE SP 2.52
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 95
nodia
YEAR 2007
EE SP 2.53
TWO MARKS
EE SP 2.54
Consider the discrete-time system shown in the figure where the impulse response
of G (z) is g (0) = 0, g (1) = g (2) = 1, g (3) = g (4) = g = 0
www.nodia.co.in
www.gatehelp.com
Page 96
Chapter 2
nodia
For a distortion free output signal wave form, G (s) must
(A) provides zero phase shift for all frequency
(B) provides constant phase shift for all frequency
EE SP 2.56
(C) a = b(1/3)
EE SP 2.57
If u (t), r (t) denote the unit step and unit ramp functions respectively and
u (t) * r (t) their convolution, then the function u (t + 1) * r (t - 2) is given by
(A) 12 (t - 1) u (t - 1)
(B) 12 (t - 1) u (t - 2)
(C)
EE SP 2.58
1
2
(t - 1) 2 u (t - 1)
EE SP 2.59
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 2
EE SP 2.60
Page 97
x (t) is a real valued function of a real variable with period T . Its trigonometric
Fourier Series expansion contains no terms of frequency w = 2p (2k) /T; k = 1, 2g
Also, no sine terms are present. Then x (t) satisfies the equation
(A) x (t) =- x (t - T)
(B) x (t) = x (T - t) =- x (- t)
(C) x (t) = x (T - t) =- x (t - T/2)
(D) x (t) = x (t - T) = x (t - T/2)
EE SP 2.61
A discrete real all pass system has a pole at z = 2+30% : it, therefore
(A) also has a pole at 12 +30%
(B) has a constant phase response over the z -plane: arg H (z) = constant
constant
nodia
(C) is stable only if it is anti-causal
(D) has a constant phase response over the unit circle: arg H (eiW) = constant
YEAR 2006
EE SP 2.62
EE SP 2.63
TWO MARKS
denotes a
/
(B) y2 [n] ) Y2 (z) = / n3= 0 (5n - n) z - (2n + 1)
(C) y3 [n] ) Y3 (z) = / n3=- 3 2 - n z - n
(D) y4 [n] ) Y4 (z) = 2z - 4 + 3z - 2 + 1
EE SP 2.64
EE SP 2.65
#- 3 x (t') dt'
(A) has no finite singularities in its double sided Laplace Transform Y (s)
(B) produces a bounded output for every causal bounded input
www.nodia.co.in
www.gatehelp.com
Page 98
Chapter 2
For the triangular wave from shown in the figure, the RMS value of the voltage
is equal to
nodia
(A)
1
6
(C) 1
3
EE SP 2.67
TWO MARKS
(B)
(D)
1
3
2
3
2
The Laplace transform of a function f (t) is F (s) = 5s 2+ 23s + 6 as t " 3, f (t)
s (s + 2s + 2)
approaches
(A) 3
(B) 5
(D) 3
(C) 17
2
EE SP 2.68
EE SP 2.69
If u (t) is the unit step and d (t) is the unit impulse function, the inverse z
-transform of F (z) = z +1 1 for k > 0 is
(A) (- 1) k d (k)
(B) d (k) - (- 1) k
(C) (- 1) k u (k)
(D) u (k) - (- 1) k
YEAR 2004
TWO MARKS
EE SP 2.70
The rms value of the resultant current in a wire which carries a dc current of 10
A and a sinusoidal alternating current of peak value 20 is
(A) 14.1 A
(B) 17.3 A
(C) 22.4 A
(D) 30.0 A
EE SP 2.71
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 99
(A) 2 6 A
(B) 6 2 A
(C)
(D) 1.5 A
4/3 A
***********
nodia
www.nodia.co.in
www.gatehelp.com
Page 100
Chapter 2
SOLUTION
SOL 2.1
3s + 5
s2 + 10s + 21
x ^ t h = lim sX ^s h
s"3
So, we obtain
nodia
s"3
= lim s d
s"3
= lim s
s"3
SOL 2.2
3s + 5
n
s2 + 10s + 21
s _3 + s5 i
s 2 `1 + 10s + 21
s j
2
3+ 5
3
=3
1 + 10 + 21
3 3
For the square wave, we observe that the Fourier series coefficients exist and the
reconstruction converges at most point.
SOL 2.3
# f ^t he
-jwt
dt
g^t h =
# F^u he
-jut
du
...(1)
-3
# F ^w h e
jwt
dw
-3
# F^u he
-jut
du
-3
...(2)
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 101
nodia
The function represented in the waveform can be resolved in the following four
waveforms; where u (t) is the unit step function.
x ^ t h = k cos 3t , k = 1
www.nodia.co.in
www.gatehelp.com
Page 102
Chapter 2
y ^ t h = A sin ^3t + ah
We can see that output differs only in phase. In Laplace domain
Y ^s h = X ^s h H ^s h
Y ^s h = X ^s h H ^s h
For w = 3 ,
SOL 2.6
nodia
Impulse response
Laplace transform,
h ^ t h = e-5t u ^ t h
H ^s h = 1
s+5
y ^ t h = e-3t u ^ t h - e-5t u ^ t h
Laplace transform,
Y (s) = 1 + 1
s+3 s+5
2
=
^s + 3h^s + 5h
We know that,
Y (s) = X ^s h H ^s h
Y ^s h
So,
X ^s h =
H ^s h
Substituting Y ^s h and H ^s h, we get
X ^s h = 2
s+3
Taking inverse Laplace, transform
Output,
x ^ t h = 2e-3t u ^ t h
SOL 2.7
where
z
zI - A = >
0
So,
z
a a-1
H->
H
0
a+1 a
z - a -a + 1
H
=>
-a - 1 z - a
For a = 1,
Characteristic equation
z-1 0
H
zI - A = >
-2 z - 1
zI - A = 0
^z - 1h2 = 0
z = 1 ! j0
The roots of characteristic equation gives the system poles.
www.nodia.co.in
www.gatehelp.com
Chapter 2
SOL 2.8
Page 103
x ^ t h = 2 + 5 sin ^100pt h
x ^nTs h = 2 + 5 sin ^100p # nTs h
Ts " Sampling period, Ts = 1
400
x ^nTs h = 2 + 5 sin b100p # n # 1 l
400
x ^nTs h = 2 + 5 sin a np k
4
Transfer function
H ^z h =
N
Y ^z h
= 1 c 1 - z-1 m
N 1-z
X ^z h
N-1
z-n
= 1
N n=0
1
0 # n # N-1
N
=*
0
otherwise
400
=
=8
f = 100p = 50 Hz
50
2p
1
0#n#7
= *8
0
otherwise
= x ^n h * h ^n h
5
5
= )2, 2 + 2 , ^2 + 5h, c 2 + 2 m, 2 .......3
nodia
h ^n h
or
N=
fs
f
h ^n h
So
y ^n h
x ^n h
y ^n h =
N-1
/ x^k hh^n - k h
n=0
The integration sum of sine term will be zero and for constant term
y ^n h =
/ 2 # 18 = 2
n=0
SOL 2.9
www.nodia.co.in
www.gatehelp.com
Page 104
Chapter 2
1, t < 1
x (t ) = *
0, t > 1
The waveform for the given signal is drawn as
nodia
So, comparing the two waveforms, we get
x (t ) = rect a t k
2
Now, the Fourier transform pair for rectangular function is
rect c t m * t sin C c wt m
t
2p
so,
Similarly,
sin ^ w2 h
w
rect (t ) * sin C c m = w
2p
2
and
2 sin (2w)
w
Thus, we obtain the fourier transform of signal y (t ) as
F {y (t )} = F {x (2t) * x (t/2)}
= F {x (2t)} .F {x (t/2)}
sin ^ w2 h 2 sin (2w)
= w
w
2
and
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 105
w max = 14p
2pf max = 14p
or
f max = 7
Thus, the sampling frequency is
fs > 2 # 7
or
fs > 14
i.e. minimum sampling frequency is 14 Hz.
SOL 2.12
nodia
...(1)
| P1 | = | Z2 |
and
| P2 | = | Z1 |
Thus, the numerator and denominator terms of equation are cancelled. Hence,
we get
| H (w) | = K
SOL 2.13
www.nodia.co.in
www.gatehelp.com
Page 106
Chapter 2
it is clear that only lower component (fs - fx ) passes through the filter, i.e.
fS - fx < 25
and
fS - fx = 20
50 - fx = 20
fx = 50 - 20 = 30 Hz
SOL 2.14
nodia
...(1)
Y (w) = jwX (w)
where X (w) and Y (w) are the Fourier transform of x (t ) and y (t ) respectively.
Now, we have the function x (t ), which is differentiable, non-constant, even
Then
function. So, its frequency response will be real, i.e. X (w) is real
Hence, equation (1) gives the result that Y (w) is imaginary.
SOL 2.15
fm = 5 kHz
According to the Nyquist sampling theorem, the sampling frequency must be
greater than the Nyquist frequency which is given as
fN = 2fm = 2 # 5 = 10 kHz
So, the sampling frequency fs must satisfy
fs $ fN
fs $ 10 kHz
only the option (A) does not satisfy the condition therefore, 5 kHz is not a
valid sampling frequency.
SOL 2.16
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 107
T0 = 2p
100
or,
SOL 2.18
SOL 2.19
nodia
h^t h = t u^t h
Its Laplace transform is
H ^s h = 12
s
Hence, the overall response at the output is
-s
Y ^s h = X ^s h H ^s h = e 3
s
Its inverse Laplace transform is
^t - 1h2
y^t h =
u ^t - 1h
2
SOL 2.20
www.nodia.co.in
www.gatehelp.com
Page 108
Chapter 2
y ^2 h = u ^2 - 1h + u ^2 - 3h = 1
At t = 2
SOL 2.21
nodia
X 6z @ =
-
1 n -n
1 -n -n
b 3 l z u [n ] +
b 3 l z u [- n - 1]
n =- 3
n =- 3
-1
1 n -n
1 n -n
1 -n -n
1 n -n
b 2 l z u [n ] =
b3l z +
b3l z b2l z
n =- 3
n=0
n =- 3
n=0
/
3
/ b 31z l + / b 13 z l
n=0
14
42
4
43
I
m=1
1 44 2
44 3
II
/ b 21z l
Taking m =- n
n=0
14
42
4
43
III
1 < 1 or z > 1
3
3z
Series II converges if 1 z < 1 or z < 3
3
Series III converges if 1 < 1 or z > 1
2z
2
Region of convergence of X (z) will be intersection of above three
So,
ROC : 1 < z < 3
2
Series I converges if
SOL 2.22
f (t)
F (s)
dF (s)
L
tf (t)
ds
L [tf (t)] = - d ; 2 1
ds s + s + 1E
= 2 2s + 1 2
(s + s + 1)
So,
SOL 2.23
y [n] =
/ h [n] g [n - k]
k =- 3
/ h [n] g [n - k]
k=0
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 109
For n = 1,
g [- 1] = g [- 2] = ....0
...(i)
1
h [1] = b 1 l = 1
2
2
1 = g [1] + g [0]
From equation (i),
nodia
g [1] = 1 - 1 = 0
So,
SOL 2.24
g [1] = 1 - g [0]
y [0] 1
= =1
g [0] =
h [0] 1
x (t) = d (t)
y (t) =
www.nodia.co.in
www.gatehelp.com
Page 110
Chapter 2
# d (t - t ) cos (3t) dt
0
-3
Delayed output
y (t, t 0) ! y (t - t 0)
System is not time invariant.
Stability :
Consider a bounded input x (t) = cos 3t
y (t) =
-3
cos2 3t =
t
= 1 1dt - 1
2 -3
2
As t " 3, y (t) " 3 (unbounded)
System is not stable.
SOL 2.26
1 - cos 6t
2
-3
# cos 6t dt
nodia
-3
f (t) = a 0 +
n=1
1
s+1
y (t) = e-2t
Y (s) = 1
s+2
Convolution in time domain is equivalent to multiplication in frequency domain.
z (t) = x (t) ) y (t)
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 111
f (t)
F1 (s)
f (t - t)
nodia
g (t) = L - 1 [e-st] = d (t - t)
SOL 2.29
or
Output,
= 1; 1 + 1 E
s s+1 s+2
By partial fraction
Y (s) = 3 - 1 - b 1 l 1
2s s + 1
s+2 2
Taking inverse Laplace
e-2t u (t)
y (t) = 3 u (t) - e-t u (t) 2
2
= u (t) 61.5 - e-t - 0.5e-2t@
SOL 2.30
H (s) =
Step input
Output
www.nodia.co.in
www.gatehelp.com
Page 112
Chapter 2
=2- 2
s (s + 1)
Taking inverse Laplace transform
y (t) = (2 - 2e- t) u (t)
Final value of y (t),
yss (t) = lim y (t) = 2
t"3
Let time taken for step response to reach 98% of its final value is ts .
So,
2 - 2e- ts = 2 # 0.98
0.02 = e- ts
ts = ln 50
= 3.91 sec.
SOL 2.31
nodia
Correct option is (D).
Period of x (t),
T = 2p = 2 p
0.8 p
w
= 2.5 sec
SOL 2.32
5t
#- 3x (t) dt,
t>0
Causality :
y (t) depends on x (5t), t > 0 system is non-causal.
For example t = 2
y (2) depends on x (10) (future value of input)
Linearity :
Output is integration of input which is a linear function, so system is linear.
SOL 2.33
n=1
So,
T/2
T
= 2=
(1) sin nw0 t dt +
(- 1) sin nw0 t dt G
T 0
T/2
T/2
T
= 2 =c cos nw0 t m - c cos nw0 t m G
- nw0 0
- nw0 T/2
T
= 2 6(1 - cos np) + (cos 2np - cos np)@
nw0 T
= 2 61 - (- 1) n @
np
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 113
4 , n odd
bn = * np
0 , n even
So only odd harmonic will be present in x (t)
For second harmonic component (n = 2) amplitude is zero.
SOL 2.34
#- 3
SOL 2.35
#- 3 x2 (t) dt
3
X (w) 2 dw = 2p # 2 = 4p
nodia
Correct option is (C).
Given sequences
y [n] = {a, - a + b, - b + c, - c + d, - d}
Comparing we have
a =1
-a + b = 0 & b = a = 1
-b + c = 0 & c = b = 1
-c + d = 0 & d = c = 1
So,
SOL 2.36
h [n] = {1, 1, 1, 1}
-
www.nodia.co.in
www.gatehelp.com
Page 114
Chapter 2
g (t) = g1 (t - 3) = f` t - 3 j
2
Shift g1 (t) by 3,
nodia
g (t) = f` t - 3 j
2 2
SOL 2.37
g (t) = u (t - 3) - u (t - 5)
By shifting property we can write Laplace transform of g (t)
G (s) = 1 e - 3s - 1 e - 5s
s
s
- 3s
= e (1 - e - 2s)
s
SOL 2.38
x (t - t)
e - st X (s)
h (t - t)
e- st H (s)
(shifting property)
SOL 2.39
Assume
H1 (z) = z2 + z1 + 1 (non-causal)
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 115
H2 (z) = z3 + z2 + 1 (non-causal)
Overall response of the system
H (z) = H1 (z) H2 (z) H3 (z)
H (z) = (z2 + z1 + 1) (z3 + z2 + 1) H3 (z)
To make H (z) causal we have to take H3 (z) also causal.
Let
H3 (z) = z - 6 + z - 4 + 1
= (z2 + z1 + 1) (z3 + z2 + 1) (z - 6 + z - 4 + 1)
H (z) " causal
Similarly to make H (z) unstable atleast one of the system should be unstable.
SOL 2.40
nodia
x (t) =
/ ak e j2pkt/T
k =- 3
/ ak e jkw t
0
k =- 3
a 2p = w0
T
Im [x (t)] = 2 (constant)
SOL 2.41
SOL 2.42
www.nodia.co.in
www.gatehelp.com
Page 116
Chapter 2
Since the given system is LTI, So principal of Superposition holds due to linearity.
For causal system h (t) = 0 , t < 0
Both statement are correct.
SOL 2.43
nodia
Correct option is (D).
z
, z >a
(z - a) 2
at z = a is
= d (z - a) 2 X (z) zn - 1 z = a
dz
z
= d (z - a) 2
zn - 1
dz
(z - a) 2
z=a
n
d
=
z
dz z = a
X (z) =
Given that
Residue of X (z) zn - 1
= nzn - 1 z = a
= nan - 1
SOL 2.45
- 2t
#- 3x (t) dt
Causality :
Since y (t) depends on x (- 2t), So it is non-causal.
Time-variance :
y (t) =
- 2t
#- 3x (t - t0) dt =Y y (t - t0)
So this is time-variant.
Stability :
Output y (t) is unbounded for an bounded input.
For example
Let
x (t) = e - t (bounded)
y (t) =
SOL 2.46
- 2t
- t - 2t
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 117
nodia
Y (jw) = K rect ` w j
2g
So,
g = min (a, b)
y (t) = K sinc (gt)
Where
And
SOL 2.47
For k = 1,
SOL 2.48
SOL 2.49
So,
Similarly
x (t) = rect `t - 1 j
2
1, - 1 # t - 1 # 1 or 0 # t # 1
2
2 2
x (t) = *
0, elsewhere
x (- t) = rect`- t - 1 j
2
1, - 1 # - t - 1 # 1 or - 1 # t # 0
2
2 2
x (- t) = *
0, elsewhere
www.nodia.co.in
www.gatehelp.com
Page 118
F [x (t) + x (- t)] =
=
Chapter 2
- jw t
- jw t
= ;e E +;e E
- jw 0
- jw - 1
= 1 (1 - e - jw) + 1 (e jw - 1)
jw
jw
- jw /2
jw /2
=e
(e jw/2 - e - jw/2) + e
(e jw/2 - e - jw/2)
jw
jw
SOL 2.50
nodia
Correct option is (B).
In option (A)
z1 [n] = x [n - 3]
z2 [n] = z1 [4n] = x [4n - 3]
y [n] = z2 [- n]
= x [- 4n - 3] =
Y x [3 - 4n]
In option (B)
z1 [n] = x [n + 3]
z2 [n] = z1 [4n] = x [4n + 3]
y [n] = z2 [- n] = x [- 4n + 3]
In option (C)
v1 [n] = x [4n]
v2 [n] = v1 [- n] = x [- 4n]
y [n] = v2 [n + 3]
= x [- 4 (n + 3)] =
Y x [3 - 4n]
In option (D)
v1 [n] = x [4n]
v2 [n] = v1 [- n] = x [- 4n]
y [n] = v2 [n - 3] = x [- 4 (n - 3)] =
Y x [3 - 4n]
SOL 2.51
SOL 2.52
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 119
1
= 1 kHz
fs = 1 =
Ts
1 m sec
nodia
SOL 2.53
www.nodia.co.in
www.gatehelp.com
Page 120
Chapter 2
G (z)
H (z) =
1 - KG (z)
g [n] = d [n - 1] + d [n - 2]
G (z) = z - 1 + z - 2
(z - 1 + z - 2)
= 2 z+1
-1
-2
z - Kz - K
1 - K (z + z )
For system to be stable poles should lie inside unit circle.
So
H (z) =
z #1
2
z = K ! K + 4K # 1 K ! K2 + 4K # 2
2
K 2 + 4K # 2 - K
K2 + 4K # 4 - 4K + K2
nodia
8K # 4
K # 1/2
SOL 2.55
H (w) = Ke - jwt
Amplitude response H (w) = K
Phase response,
qn (w) =- wtd
For distortion less output, phase response should be proportional to frequency.
So,
SOL 2.56
SOL 2.57
and
So
(Time-shifting property)
L [u (t + 1)] = es c 12 m
s
L [r (t)] = 1/s2
(Time-shifting property)
L [r (t - 2) = e - 2s c 12 m
s
H (s) = ;es ` 1 jE;e - 2s c 12 mE
s
s
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 121
H (s) = e - s c 13 m
s
Taking inverse Laplace transform
h (t) = 1 (t - 1) 2 u (t - 1)
2
SOL 2.58
a x [n - 1]
nodia
z - 1 x (z)
H (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2)
Output of the system for input u [n] = d [n - 1] is ,
y (z) = H (z) U (z)
U [n ]
U (z) = z - 1
So
Y (z) = z - 1 (1 - 3z - 1) (1 + 2z - 2) z - 1
= z - 2 (1 - 3z - 1 + 2z - 2 - 6z - 3)
= z - 2 - 3z - 3 + 2z - 4 - 6z - 5
Taking inverse z-transform on both sides we have output.
y [n] = d [n - 2] - 3d [n - 3] + 2d [n - 4] - 6d [n - 5]
SOL 2.59
#- t
E =
g (t) 2 dt < 3
SOL 2.60
x (t) = A0 +
n=1
#0
= 2=
T0
T0
#0
T0 /2
Where t = T - t & dt =- dt
= 2;
T0
= 2;
T0
= 2;
T0
T0 /2
#T
TO
T0
T0
= 2 ;T0
bn = 0 if
T0
T0
x (t) = x (T - t)
www.nodia.co.in
www.gatehelp.com
Page 122
Chapter 2
( 3 + j)
= ( 3 + j)
2
nodia
system is stable if z < 1 and for this it is anti-causal.
SOL 2.62
So, a = 1
y [n] = {- a, 2a - b, 2b - c, 2c}
y [n] = {- 1, 3, - 1, - 2}
-
www.nodia.co.in
www.gatehelp.com
Chapter 2
Page 123
2a - b = 3 & b =- 1
2a - c =- 1 & c =- 1
Impulse response h [n] = "1, - 1, - 1,
SOL 2.63
Correct option is ( ).
SOL 2.64
SOL 2.65
nodia
Correct option is (B).
y (t) =
Given
# x (t') dt'
-3
# x (t') dt'
is always bounded.
-3
SOL 2.66
1
T
Vrms =
Where
#0
V2 (t) dt
2
T
` T j t, 0 # t # 2
V (t) = *
So
1
T
#0
T <t#T
2
0,
V 2 (t) dt = 1 =
T
#0
T/2
= 1 $ 42
T T
2t 2
` T j dt +
#0
T /2
#T/2 (0) dt G
t2 dt
T/2
3
= 43 ; t E
T 3 0
3
= 43 # T = 1
24
6
T
Vrms =
SOL 2.67
1 V
6
t"3
s"0
= lim s
s"0
SOL 2.68
(5s2 + 23s + 6)
= 6 =3
2
2
s (s + 2s + 2)
www.nodia.co.in
www.gatehelp.com
Page 124
Chapter 2
nodia
F (z) =
Z-transform
f (k) = d (k) - (- 1) k
Z
1
(- 1) k
1 + z- 1
so,
Thus
SOL 2.70
1 = 1- z = 1- 1
z+1
z+1
1 + z- 1
I = 10 + 20 sin wt
Irms =
SOL 2.71
(10) 2 +
(20) 2
= 17.32 A
2
Irms =
So
1
T
#0
I2 dt = 1 =
T
#0
T /2
- 12t 2
` T j dt +
#T/2 (6) 2 dt G
T/2
t3
= 1 e 144
+ 36 6t @TT/2 o
;
T T2 3 E0
T3 + 36 T
= 1 ; 144
c
` 2 jE
2
T T 24 m
= 1 [6T + 18T] = 24
T
Irms =
24 = 2 6 A
***********
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 125
CHAPTER 3
ELECTRIC MACHINES
ONE MARK
EE SP 3.1
EE SP 3.2
For a specified input voltage and frequency, if the equivalent radius of the core of
a transformer is reduced by half, the factor by which the number of turns in the
primary should change to maintain the same no load current is
(A) 1/4
(B) 1/2
nodia
(C) 2
EE SP 3.3
(D) 4
A star connected 400 V, 50 Hz, 4 pole synchronous machine gave the following
open circuit and short circuit test results :
Open circuit test : Voc = 400 V (rms, line-to-line) at field current, I f = 2.3 A
Short circuit test : Isc = 10 A (rms, phase) at field current, I f = 1.5 A
The value of per phase synchronous impedance in W at rated voltage is_____.
YEAR 2014 EE01
TWO MARKS
EE SP 3.4
EE SP 3.5
A 15 kW, 230 V dc shunt motor has armature circuit resistance of 0.4 W and
field circuit resistance of 230 W . At no load and rated voltage, the motor runs at
1400 rpm and the line current drawn by the motor is 5 A. At full load, the motor
draws a line current of 70 A. Neglect armature reaction. The full load speed of
the motor in rpm is_____.
EE SP 3.6
A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 W and
reactance of 0.92 W . Neglect the voltage drop in stator and assume that the rotor
resistance is constant. Given that the full load slip is 3%, the ratio of maximum
torque to full load torque is
(A) 1.567
(B) 1.712
(C) 1.948
(D) 2.134
www.nodia.co.in
www.gatehelp.com
Page 126
Electric Machines
Chapter 3
EE SP 3.8
ONE MARK
nodia
EE SP 3.10
EE SP 3.11
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 127
TWO MARKS
EE SP 3.12
A 250 V dc shunt machine has armature circuit resistance of 0.6 W and field
circuit resistance of 125 W . The machine is connected to 250 V supply mains. The
motor is operated as a generator and then as a motor separately. The line current
of the machine in both the cases is 50 A. The ratio of the speed as a generator
to the speed as a motor is _____.
EE SP 3.13
nodia
If the rotor winding is open circuited and the system is made to run at rotational
speed f r with the help of prime-mover in anti-clockwise direction, then the
frequency of voltage across slip rings is f 1 and frequency of voltage across
commutator brushes is f 2 . The values of f 1 and f 2 respectively are
(B) f - f r and f
(A) f + f r and f
(C) f - f r and f + f r
(D) f + f r and f - f r
EE SP 3.14
EE SP 3.15
For a single phase, two winding transformer, the supply frequency and voltage
are both increased by 10%. The percentage changes in the hysteresis loss and
eddy current loss, respectively, are
(A) 10 and 21
(B) - 10 and 21
(D) - 21 and 10
(C) 21 and 10
YEAR 2014 EE03
EE SP 3.16
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 128
EE SP 3.17
Electric Machines
Chapter 3
A single phase induction motor is provided with capacitor and centrifugal switch
in series with auxiliary winding. The switch is expected to operate at a speed of
0.7 Ns, but due to malfunctioning the switch fails to operate. The torque-speed
characteristic of the motor is represented by
nodia
EE SP 3.18
EE SP 3.19
The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The
armature resistance drop and the brush drop are neglected. The field current is
kept constant at rated value. The torque of the motor in Nm for an armature
current of 8 A is ________.
The torque speed characteristics of motor ^TM h and load ^TL h for two cases are
shown in the figures (a) and (b). The load torque is equal to motor torque at
points P , Q , R and S .
(B) P and S
(D) Q and S
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 129
TWO MARKS
nodia
EE SP 3.21
EE SP 3.22
A separately excited 300 V DC shunt motor under no load runs at 900 rpm
drawing an armature current of 2 A. The armature resistance is 0.5 W and
leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then,
the speed in rpm is _____.
EE SP 3.23
The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.
www.nodia.co.in
www.gatehelp.com
Page 130
Electric Machines
Chapter 3
ONE MARK
nodia
(B) flux that links both stator and rotor windings
(C) flux that links none of the windings
(D) flux that links the stator winding or the rotor winding but not both
EE SP 3.26
EE SP 3.27
TWO MARKS
EE SP 3.28
EE SP 3.29
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 131
VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX.
Then, VYZ1 /VWX1 , VWX2 /VYZ2 are respectively,
nodia
YEAR 2012
EE SP 3.30
ONE MARK
TWO MARKS
EE SP 3.31
A 220 V, 15 kW, 100 rpm shunt motor with armature resistance of 0.25 W, has
a rated line current of 68 A and a rated field current of 2.2 A. The change in
field flux required to obtain a speed of 1600 rpm while drawing a line current of
52.8 A and a field current of 1.8 A is
(A) 18.18% increase
(B) 18.18% decrease
(C) 36.36% increase
(D) 36.36% decrease
EE SP 3.32
The locked rotor current in a 3-phase, star connected 15 kW, 4 pole, 230 V, 50 Hz
induction motor at rated conditions is 50 A. Neglecting losses and magnetizing
current, the approximate locked rotor line current drawn when the motor is
connected to a 236 V, 57 Hz supply is
(A) 58.5 A
(B) 45.0 A
(C) 42.7 A
(D) 55.6 A
EE SP 3.33
www.nodia.co.in
www.gatehelp.com
Page 132
Electric Machines
(C) 1 A, 110 W
Chapter 3
(D) 1 A, 220 W
YEAR 2011
EE SP 3.34
ONE MARK
EE SP 3.35
nodia
(B) First increases and then decreases steeply
EE SP 3.36
A single phase air core transformer, fed from a rated sinusoidal supply, is operating
at no load. The steady state magnetizing current drawn by the transformer from
the supply will have the waveform
YEAR 2011
EE SP 3.37
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 133
nodia
YEAR 2010
EE SP 3.39
(A) 1.41 A
(C) 2.24 A
EE SP 3.40
ONE MARK
(B) 2 A
(D) 3 A
EE SP 3.41
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Page 134
Electric Machines
Chapter 3
nodia
(A) (3 + j0) W
(C) (0.866 + j0.5) W
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 135
EE SP 3.43
EE SP 3.44
nodia
For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage
to be applied is
(A) 125.5 V
(B) 193.3 V
(C) 200 V
(D) 241.7 V
YEAR 2009
EE SP 3.45
ONE MARK
EE SP 3.46
The single phase, 50 Hz iron core transformer in the circuit has both the vertical
arms of cross sectional area 20 cm2 and both the horizontal arms of cross
sectional area 10 cm2 . If the two windings shown were wound instead on opposite
horizontal arms, the mutual inductance will
(A) double
(C) be halved
EE SP 3.47
A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives
a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the
load(dotted curve) are shown. Of the two equilibrium points A and B, which of
the following options correctly describes the stability of A and B ?
www.nodia.co.in
www.gatehelp.com
Page 136
Electric Machines
Chapter 3
YEAR 2009
EE SP 3.48
TWO MARKS
nodia
(A)
(B)
(C)
(D)
rotates clockwise
rotates anti-clockwise
does not rotate
rotates momentarily and comes to a halt
www.nodia.co.in
www.gatehelp.com
Chapter 3
EE SP 3.49
EE SP 3.50
Electric Machines
Page 137
nodia
The peak voltage across A and B, with S open is
(B) 800 V
(A) 400 V
p
(D) 800 V
(C) 4000 V
p
p
If the wave form of i (t) is changed to i (t) = 10 sin (100pt) A, the peak voltage
across A and B with S closed is
(A) 400 V
(B) 240 V
(C) 320 V
(D) 160 V
The star-delta transformer shown above is excited on the star side with balanced,
4-wire, 3-phase, sinusoidal voltage supply of rated magnitude. The transformer is
under no load condition
EE SP 3.51
www.nodia.co.in
www.gatehelp.com
Page 138
EE SP 3.52
Electric Machines
Chapter 3
With S2 closed and S1 open, the current waveform in the delta winding will be
(A) a sinusoid at fundamental frequency
(B) flat-topped with third harmonic
(C) only third-harmonic
(D) none of these
EE SP 3.53
nodia
Statement for Linked Answer Questions 54 and 55
The figure above shows coils-1 and 2, with dot markings as shown, having 4000
and 6000 turns respectively. Both the coils have a rated current of 25 A. Coil-1
is excited with single phase, 400 V, 50 Hz supply.
EE SP 3.54
400
The coils are to be connected to obtain a single-phase, 1000
V,
auto-transformer to drive a load of 10 kVA. Which of the options given should
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 139
nodia
YEAR 2008
EE SP 3.56
ONE MARK
EE SP 3.57
EE SP 3.59
www.nodia.co.in
www.gatehelp.com
Page 140
Electric Machines
Chapter 3
The Wattmeters used in open circuit test and short circuit test of the transformer
will respectively be
(A) W1 and W2
(B) W2 and W4
(C) W1 and W4
(D) W2 and W3
YEAR 2008
EE SP 3.60
EE SP 3.61
nodia
EE SP 3.62
TWO MARKS
(D) 26.21 kW
The induced emf (ers) in the secondary winding as a function of time will be of
the form
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 141
nodia
The motor is coupled to a 220 V, separately excited d.c generator feeding power
to fixed resistance of 10 W. Two-wattmeter method is used to measure the
input power to induction motor. The variable resistance is adjusted such the
motor runs at 1410 rpm and the following readings were recorded W1 = 1800 W,
W2 =- 200 W.
EE SP 3.63
The speed of rotation of stator magnetic field with respect to rotor structure will
be
(A) 90 rpm in the direction of rotation
(B) 90 rpm in the opposite direction of rotation
(C) 1500 rpm in the direction of rotation
(D) 1500 rpm in the opposite direction of rotation
EE SP 3.64
Neglecting all losses of both the machines, the dc generator power output and
the current through resistance (Rex) will respectively be
(A) 96 W, 3.10 A
(B) 120 W, 3.46 A
(C) 1504 W, 12.26 A
(D) 1880 W, 13.71 A
EE SP 3.65
A 400 V, 50 Hz, 4-pole, 1400 rpm, star connected squirrel cage induction motor
has the following parameters referred to the stator:
R'r = 1.0 W, Xs = X'r = 1.5 W
Neglect stator resistance and core and rotational losses of the motor. The motor
is controlled from a 3-phase voltage source inverter with constant V/f control.
The stator line-to-line voltage(rms) and frequency to obtain the maximum torque
at starting will be :
(A) 20.6 V, 2.7 Hz
(B) 133.3 V, 16.7 Hz
(C) 266.6 V, 33.3 Hz
(D) 323.3 V, 40.3 Hz
www.nodia.co.in
www.gatehelp.com
Page 142
Electric Machines
Chapter 3
80 W.
EE SP 3.66
The net voltage across the armature resistance at the time of plugging will be
(A) 6 V
(B) 234 V
(C) 240 V
(D) 474 V
EE SP 3.67
The external resistance to be added in the armature circuit to limit the armature
current to 125% of its rated value is
(A) 31.1 W
(B) 31.9 W
(C) 15.1 W
(D) 15.9 W
nodia
EE SP 3.68
EE SP 3.69
ONE MARK
EE SP 3.70
EE SP 3.71
The dc motor, which can provide zero speed regulation at full load without any
controller is
(A) series
(B) shunt
(C) cumulative compound
(D) differential compound
EE SP 3.72
The electromagnetic torque Te of a drive and its connected load torque TL are
as shown below. Out of the operating points A, B, C and D, the stable ones are
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 143
(A) A, C, D
(B) B, C
(C) A, D
(D) B, C, D
nodia
YEAR 2007
EE SP 3.73
TWO MARKS
EE SP 3.74
EE SP 3.75
EE SP 3.76
EE SP 3.77
A three-phase squirrel cage induction motor has a starting torque of 150% and
a maximum torque of 300% with respect to rated torque at rated voltage and
www.nodia.co.in
www.gatehelp.com
Page 144
Electric Machines
Chapter 3
rated frequency. Neglect the stator resistance and rotational losses. The value of
slip for maximum torque is
(A) 13.48%
(B) 16.42%
(C) 18.92%
(D) 26.79%
EE SP 3.79
EE SP 3.80
If an auto transformer is used for reduced voltage starting to provide 1.5 per unit
starting torque, the auto transformer ratio(%) should be
(A) 57.77 %
(B) 72.56 %
(C) 78.25 %
(D) 81.33 %
nodia
If a star-delta starter is used to start this induction motor, the per unit starting
torque will be
(A) 0.607
(B) 0.816
(C) 1.225
(D) 1.616
If a starting torque of 0.5 per unit is required then the per unit starting current
should be
(A) 4.65
(B) 3.75
(C) 3.16
(D) 2.13
YEAR 2006
ONE MARK
EE SP 3.81
EE SP 3.82
For a single phase capacitor start induction motor which of the following
statements is valid ?
(A) The capacitor is used for power factor improvement
(B) The direction of rotation can be changed by reversing the main winding
terminals
(C) The direction of rotation cannot be changed
(D) The direction of rotation can be changed by interchanging the supply
terminals
EE SP 3.83
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 145
TWO MARKS
nodia
(C) 0.96
EE SP 3.85
(D) 1.06
EE SP 3.86
Two transformers are to be operated in parallel such that they share load in
proportion to their kVA ratings. The rating of the first transformer is 500
kVA ratings. The rating of the first transformer is 500 kVA and its pu leakage
impedance is 0.05 pu. If the rating of second transformer is 250 kVA, its pu
leakage impedance is
(A) 0.20
(B) 0.10
(C) 0.05
(D) 0.025
EE SP 3.87
EE SP 3.88
www.nodia.co.in
www.gatehelp.com
Page 146
Electric Machines
Chapter 3
EE SP 3.89
A 3-phase, 10 kW, 400 V, 4-pole, 50Hz, star connected induction motor draws 20
A on full load. Its no load and blocked rotor test data are given below.
No Load Test :
400 V
6A
1002 W
Blocked Rotor Test :
90 V
15 A
762 W
Neglecting copper loss in no load test and core loss in blocked rotor test, estimate
motors full load efficiency
(A) 76%
(B) 81%
(C) 82.4%
(D) 85%
EE SP 3.90
nodia
(C) 51%
(D) 50%
EE SP 3.91
The line-to-line induced emf(in volts), for a three phase star connection is
approximately
(A) 808
(B) 888
(C) 1400
(D) 1538
EE SP 3.92
The line-to-line induced emf(in volts), for a three phase connection is approximately
(A) 1143
(B) 1332
(C) 1617
(D) 1791
EE SP 3.93
The fifth harmonic component of phase emf(in volts), for a three phase star
connection is
(A) 0
(B) 269
(C) 281
(D) 808
The iron loss (Pi) and copper loss (Pc) in kW, under full load operation are
(B) Pc = 6.59, Pi = 9.21
(A) Pc = 4.12, Pi = 8.51
(C) Pc = 8.51, Pi = 4.12
(D) Pc = 12.72, Pi = 3.07
EE SP 3.95
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 147
(A) 95.1
(B) 96.2
(C) 96.4
(D) 98.1
YEAR 2005
EE SP 3.96
EE SP 3.97
The equivalent circuit of a transformer has leakage reactances X1, X'2 and
magnetizing reactance XM . Their magnitudes satisfy
(A) X1 >> X'2 >> XM
nodia
(C) Delta-Delta
EE SP 3.98
(D) Delta-Zigzag
On the torque/speed curve of the induction motor shown in the figure four points
of operation are marked as W, X, Y and Z. Which one of them represents the
operation at a slip greater than 1 ?
(A) W
(C) Y
EE SP 3.99
ONE MARK
(B) X
(D) Z
For an induction motor, operation at a slip s , the ration of gross power output
to air gap power is equal to
(A) (1 - s) 2
(B) (1 - s)
(C)
(1 - s)
(D) (1 -
s)
YEAR 2005
EE SP 3.100
TWO MARKS
Two magnetic poles revolve around a stationary armature carrying two coil
(c1 - c1l , c2 - c2l ) as shown in the figure. Consider the instant when the poles are
in a position as shown. Identify the correct statement regarding the polarity of
the induced emf at this instant in coil sides c1 and c2 .
www.nodia.co.in
www.gatehelp.com
Page 148
Electric Machines
Chapter 3
(A) 9 in c1 , no emf in c2
(B) 7 in c1 , no emf in c2
(C) 9 in c2 , no emf in c1
(D) 7 in c2 , no emf in c1
EE SP 3.101
nodia
EE SP 3.102
EE SP 3.103
List-II
Performance Variables
Proportional to
P. Armature emf (E )
Q. Developed torque (T )
2. f and w only
R Developed power (P )
3. f and Ia only
4. Ia and w only
5. Ia only
Codes:
(A)
(B)
P
3
2
Q
3
5
R
1
4
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
(C)
(D)
EE SP 3.104
3
2
5
3
Page 149
4
1
EE SP 3.105
nodia
(C) 2.40
EE SP 3.106
EE SP 3.107
(D) 6.00
In a single phase induction motor driving a fan load, the reason for having a high
resistance rotor is to achieve
(A) low starting torque
(B) quick acceleration
(C) high efficiency
(D) reduced size
EE SP 3.109
(B) 18.3c
(D) 33.0c
www.nodia.co.in
www.gatehelp.com
Page 150
Electric Machines
Chapter 3
YEAR 2004
EE SP 3.110
ONE MARK
A 500 kVA, 3-phase transformer has iron losses of 300 W and full load copper
losses of 600 W. The percentage load at which the transformer is expected to
have maximum efficiency is
(A) 50.0%
(B) 70.7%
(C) 141.4%
EE SP 3.111
EE SP 3.112
(D) 200.0%
For a given stepper motor, the following torque has the highest numerical value
(A) Detent torque
(B) Pull-in torque
(C) Pull-out torque
(D) Holding torque
nodia
The following motor definitely has a permanent magnet rotor
(A) DC commutator motor
(B) Brushless dc motor
(C) Stepper motor
(D) Reluctance motor
EE SP 3.113
The type of single-phase induction motor having the highest power factor at full
load is
(A) shaded pole type
(B) split-phase type
(C) capacitor-start type
EE SP 3.114
EE SP 3.115
EE SP 3.116
TWO MARKS
The synchronous speed for the seventh space harmonic mmf wave of a 3-phase,
8-pole, 50 Hz induction machine is
(A) 107.14 rpm in forward direction
(B) 107.14 rpm in reverse direction
(C) 5250 rpm in forward direction
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 151
A rotating electrical machine its self-inductances of both the stator and the rotor
windings, independent of the rotor position will be definitely not develop
(A) starting torque
(B) synchronizing torque
(C) hysteresis torque
(D) reluctance torque
EE SP 3.118
nodia
(C) 59.2%
EE SP 3.119
(D) 88.8%
A 50 kVA, 3300/230 V single-phase transformer is connected as an autotransformer shown in figure. The nominal rating of the auto- transformer will be
EE SP 3.120
EE SP 3.121
EE SP 3.122
www.nodia.co.in
www.gatehelp.com
Page 152
Electric Machines
Chapter 3
A 400 V, 15 kW, 4-pole, 50Hz, Y-connected induction motor has full load slip of
4%. The output torque of the machine at full load is
(A) 1.66 Nm
(B) 95.50 Nm
(C) 99.47 Nm
(D) 624.73 Nm
EE SP 3.124
EE SP 3.125
EE SP 3.127
is
nodia
EE SP 3.126
rate
(D) 768 V
ONE MARK
EE SP 3.128
A simple phase transformer has a maximum efficiency of 90% at full load and
unity power factor. Efficiency at half load at the same power factor is
(A) 86.7%
(B) 88.26%
(C) 88.9%
(D) 87.8%
EE SP 3.129
Group-I lists different applications and Group-II lists the motors for these
applications. Match the application with the most suitable motor and choose the
right combination among the choices given thereafter
Group-I
Group-II
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 153
P.
Food mixer
R.
3. Universal motor
S.
Escalator
Codes:
(A)
(B)
(C)
(D)
EE SP 3.130
P
3
1
3
3
Q
6
3
1
2
R
4
2
2
1
S
5
4
4
4
nodia
(C) the field current has to be increased and fuel input left unaltered
(D) the field current has to be reduced and fuel input left unaltered
EE SP 3.131
Curves X and Y in figure denote open circuit and full-load zero power factor(zpf)
characteristics of a synchronous generator. Q is a point on the zpf characteristics
at 1.0 p.u. voltage. The vertical distance PQ in figure gives the voltage drop
across
www.nodia.co.in
www.gatehelp.com
Page 154
Electric Machines
YEAR 2003
EE SP 3.133
Chapter 3
TWO MARKS
Figure shows an ideal single-phase transformer. The primary and secondary coils
are wound on the core as shown. Turns ratio N1 /N2 = 2 .The correct phasors of
voltages E1, E2 , currents I1, I2 and core flux F are as shown in
nodia
EE SP 3.134
EE SP 3.135
www.nodia.co.in
www.gatehelp.com
Chapter 3
EE SP 3.136
Electric Machines
Page 155
(A) 240 V
(B) 480 V
(C) 415 V
(D) 0 V
nodia
(A) (- 10 + j10) A
(C) (10 + j10) A
EE SP 3.137
EE SP 3.138
(B) (- 10 - j10) A
(D) (10 - j10) A
Synchronous Machines
Induction Machines
(A) P,S
(B) Q,U
(C) P,S
(D) R,S
Q,T
P,T
R,U
Q,U
R,U
R,S
Q,T
P,T
When stator and rotor windings of a 2-pole rotating electrical machine are excited,
each would produce a sinusoidal mmf distribution in the airgap with peal values
Fs and Fr respectively. The rotor mmf lags stator mmf by a space angle d at any
instant as shown in figure. Thus, half of stator and rotor surfaces will form one
pole with the other half forming the second pole. Further, the direction of torque
www.nodia.co.in
www.gatehelp.com
Page 156
Electric Machines
Chapter 3
The following table gives four set of statement as regards poles and torque. Select
the correct set corresponding to the mmf axes as shown in figure.
EE SP 3.139
nodia
Stator Surface
ABC forms
Stator Surface
CDA forms
Rotor Surface
abc forms
Rotor surface
cda forms
Torque
is
South Pole
North Pole
North Pole
North Pole
South Pole
South Pole
South Pole
South Pole
North Pole
North Pole
South Pole
North Pole
Clockwise
Counter
Clockwise
Counter
Clockwise
Clockwise
EE SP 3.140
EE SP 3.141
A single-phase induction motor with only the main winding excited would exhibit
the following response at synchronous speed
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 157
A dc series motor driving and electric train faces a constant power load. It is
running at rated speed and rated voltage. If the speed has to be brought down to
0.25 p.u. the supply voltage has to be approximately brought down to
(A) 0.75 p.u
(B) 0.5 p.u
(C) 0.25 p.u
nodia
***********
www.nodia.co.in
www.gatehelp.com
Page 158
Electric Machines
Chapter 3
SOLUTION
SOL 3.1
120f
= 120 # 50 = 750 rpm
8
P
nodia
Frequency of rotor current = s # f
= 1 # 50 = 3.33 Hz
15
SOL 3.2
f = B#A
Now, the given radius is reduced by half, i.e.
r " r/2
So, the area will be reduced by 1/4, i.e.
A " A/4
To maintain no load current constant [in turn emf (E ) to be constant], number
of turn should be increased by 4 times.
SOL 3.3
Ph
Ph
Since Isc and I f are linearly related, so we get the short circuit phase current at
I f = 2.3 A as
Iscl = 10 # 2.3 = 46 A
1.5
3
I = 2.3 A
Thus, per phase synchronous impedance is
V
Z ph = oc
Iscl I = 2.3 A
Ph
Ph
Ph
400/ 3
= 400 #
46
46/3
3 = 15.06 W
www.nodia.co.in
www.gatehelp.com
Chapter 3
SOL 3.4
Electric Machines
Page 159
...(i)
Iron loss = Pi = ki V 2
Hysteresis loss = PH = KH V 1.6 f - 0.6
Substituting it in equation (i), we get
^230h1.6
Ki ^230h2 + KH
= 1050
^50h0.6
or
Ki ^52900h + KH ^574.62h = 1050
Again, for V = 138 V and f = 30 Hz , we have
and
nodia
or
or
...(ii)
or
KP ^19044h + KH ^344.77h = 500
Solving equations (i) and (ii), we get
and
For 230 V, 50 Hz; we have
...(ii)
KP = 0.01024
KH = 0.855
= 541.69 W
SOL 3.5
www.nodia.co.in
www.gatehelp.com
Page 160
Electric Machines
Chapter 3
V = E f 1 + Ia # Rs
230 = E f 1 + 4 # 0.4
or
At full load,
E f 1 = 228.4 V
I = 70 A
So,
Ia = 70 - 1 = 69 A
and
E f 2 = 230 - 69 # 0.4 = 202.4 V
as
emf \ flux # speed
For DC shunt motor flux ^fh is constant. So,
Ef \ w
Ef 1
or
= w1
w2
Ef 2
228.4 = 1400
or
w2
202.4
Thus,
w 2 = 202.4 # 1400 = 1241.1 rpm
228.4
SOL 3.6
nodia
Correct option is (C).
For the induction motor, we have
s f R2
R + ^s f X2h2
or
Tmax ^maxm torqueh \ 1
2X2
Tf
2s R X
So, we get
= 2 f 2 2 2
Tmax
R 2 + ^s f X2h
Tf
2
or
=
Tmax
s max + s f
sf
s max
Now, the slip at full load is 3% , i.e.
Tf (Full load torque) \
2
2
...(i)
s f = 0.003
and
s max = R2 = 0.1 = 0.108
X2 0.92
Substituting these values in equation (i), we get
Tf
2
=
0.108 + 0.03
Tmax
0.03
0.108
Tmax = 3.623 + 0.276
or
2
Tf
Tmax = 1.948
or
Tf
i.e. the ratio of maximum torque to full load torque is 1.948.
SOL 3.7
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 161
VTh = b N2 l Vs
N1
VTh = 2 sin ^wt h = 2 sin ^wt h
1
2
RTh = b 2 l RP
1
2
= b 2 l RP = 4 # 1 = 4 W
1
SOL 3.8
nodia
Vi = V1 = 1000 V
Vo = V1 + V2 = 1000 + 100 = 1100 V
Rating of auto transformer
S 0a = V0 # I2a
= 1100 # 500 = 550 kVA
SOL 3.9
^I2a = I2h
www.nodia.co.in
www.gatehelp.com
Page 162
Electric Machines
Chapter 3
As the load decreases, the speed increases. Relative speed between stator rmf
and rotor speed reduced which decrease the slip consequently emf induced in the
rotor and current decreases. Hence frequency decreases.
SOL 3.10
So
f 1, f 2 < 50 Hz
but
f2 > f1
Correct answer is 7.
Let complex power (kVA) is given as
S = P + jQ
P " Real power ^W h
Q " Reactive Power ^VR h
P = S cos q
Q = S sin q
or
Q = P sin q = P tan q
cos q
Initially when the real power is 12 MW and power factor is 0.6, the reactive
power is
Real power
Reactive power
nodia
Q1 = P1 tan q 1
Q2 = P1 tan q 2
Q2 = 12 # tan ^36.86ch
= 12 # 3 = 9 MAR
4
From the power triangle (before and after capacitor is added)
(P1 is same)
www.nodia.co.in
www.gatehelp.com
Chapter 3
SOL 3.11
Electric Machines
Page 163
SOL 3.12
IL = 50 A
Ia = IL + Isn = 50 + 2 = 52 A
Load current
Armature current
IlL = 50 A
Ila = I lL - I lsn = 50 - 2 = 48 A
Load current
nodia
Induced Emf
Eg = 250 + 52 ^0.6h = 281.2 V
When the machine is running as a motor
Field current
Ilsn = 250 = 2 A
125
Induced Emf
Em = 250 - 48 ^0.6h = 221.2 V
For DC machine, f is constant, so
Ng
= Nm
Em
Eg
Ng
E
Speed Ratio
= g = 281.2 = 1.27
Nm
Em 221.2
SOL 3.13
SOL 3.14
Distribution factor
No. of turns,
www.nodia.co.in
www.gatehelp.com
Page 164
Electric Machines
Chapter 3
r = 20c
180 = 3
20 # 3
sin ^nr/2h
sin ^3 # 20/2h
=
= 0.9597
kd 2 =
n sin ^r/2h
3 # sin ^20/2h
1
N2 = no. of slots # no. of conductor #
2
no. of phase
= 180 # 6 # 1
2
3
n =
Distribution factor
no. of turns,
= 180
The rms value of generator voltage
nodia
V1 = 4.44 kd 1 ffN1
V2 = 4.44kd2 ffN2
V1 = kd 1 # N1 = 0.6398 # 540
0.9597 # 180
V2
kd 2 # N2
-2
SOL 3.15
n
Ph = kn fB max
Where, the kn is the proportionality constant dependent on the characteristics
and volume of material, n is known as steinmetz exponents.
B max \ V
f
Change in voltage and frequency is same, so B max is constant.
So,
Ph \ f
Change in voltage and frequency is same, so B max is constant.
So,
Ph \ f
Therefore percentage change in hysteris loss will be same as that of change in
frequency
TPh = Tf = 10%
Eddy current loss is given by
Pe = ke ^B max f d h2
Where
d =
B max =
f =
ke =
Again,
So,
lamination thickness
maximum flux density
frequency
Proportionality constant
(Constant)
B max \ V
f
Pe \ f 2
2
Pe1 = f 1
Pe2
f 22
f1
f
=
= 1
1.1
f2
f + 0.1f
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 165
Pe1 = 1 2
Pe2 b 1.1 l
So
or
= 21%
SOL 3.16
SOL 3.17
nodia
But, the switch does not operate due to malfunctioning. So, there will be no
discontinuity in the characteristic curve. Thus, we have the modified characteristic
curve of the motor as
SOL 3.18
www.nodia.co.in
www.gatehelp.com
Page 166
Electric Machines
Chapter 3
Given
No load speed,
Armature current,
N = 1400 rpm
Ia = 8 A
V = 230 V
So, we obtain the torque of the motor as
t = V Ia = 2pV Ia
w
60 # N
Excitation voltage,
= 2p230 # 8
60 # 1400
= 9.55 # 230 # 8
1400
= 12.55 N-m
SOL 3.19
SOL 3.20
nodia
where
PC = PH f + Pe f 2
PC = total core losses of the transformer
f = frequency
PH = Hysteresis losses constant
...(i)
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
PH = 10 ;
Page 167
Pe = 1
10
From equation (i), we have the hysteresis and eddy current losses as
Pe =
PH f = 10 # 25 = 250 W
Pe f 2 = 1 # ^25h2 = 62.5 W
10
Thus, we get
SOL 3.21
PH f + Pe f 2
S
S
Hysteresis loss Eddy current loss
XS = 0.8 p.u.
P = 1 p.u.
nodia
cos f = 1
Terminal voltage,
Vt = 1.1 p.u.
Since, power factor is unity, so phase angle of armature current is also zero. So,
we get
1
=
= 1
Ia = P
Vt cos f (1.1) (1) 1.1
or
Ia = 1 < 0c
1.1
Again, we define the armature current as
E +d - Vt +0
Ia = f
jXS
E f +d = Ia (jXS ) + Vt < 0
= 1 (j 0.8) + 1.1+0
1.1+0
= 1 0.8+90c + 1.1+0
1.1+0
= 1.1+0 + 0.8 +90c
1.1
or
E f cos d + jE f sin d = 1.1 + j 0.8
1.1
Now, we have the power equation
E V
P = f t sin d
XS
E (1.1)
1= f
sin d
0.8
E f sin d = 0.8
1.1
So, from equation (i), we get
So,
E f cos d = 1.1
Dividing equation (ii) by equation (iii), we get
E f sin d
0.8
=
1.1 # 1.1
E f cos d
tan d = 0.8
1.21
0.8
d = tan-1 b 1.21 l = 33.47c
SOL 3.22
...(i)
...(ii)
...(iii)
www.nodia.co.in
www.gatehelp.com
Page 168
Electric Machines
Armature resistance,
ra = 0.5 W
Armature current,
Leakage inductance,
So, we draw the circuit as
Ia = 2 A
L = 0.01 H
Chapter 3
nodia
E no load (1) = V - Ia Ra
E1 = 300 - 2 ^0.5h
or
= 300 - 1 = 299 V
Again, when the load is applied, we have
Armature resistance,
ra = 0.5 W
Armature current,
So, we draw the circuit as
Ia = 15 A
Therefore, we obtain
E load (2) = V - Ia Ra
E2 = 300 - 15 # 0.5
= 300 - 7.5 = 292.5
or
Since, we know that
E\N
So,
or
or
SOL 3.23
E1 = N1
N2
E2
299 = 900
292.5
N2
N2 = 880 rpm
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 169
nodia
= 40.858+ - 27.13c
This is the current through secondary winding of transformer. So, the current
through primary winding is given by
I1 = N 2
I 2 N1
where N1 = number of turns in primary winding
N2 = Number of turns in secondary winding.
From the given circuit, we have
N1 = 2
1
N2
I1 = 1
So,
2
I2
Therefore,
I1 = I2 = 40.858+ - 27.13c
2
2
= 20.43+27.13c
Since, it is required to improve the input power factor to unity. So, the input
current should be red. Therefore, the imaginary part of I1 should pass through
reactance X . Thus, current through X is
IX = Im {I1}
= 20.43 sin (27.13c)
= 9.31
Hence, the reactance is
X = 220 = 220
9.31
IX
= 23.618 W
SOL 3.24
www.nodia.co.in
www.gatehelp.com
Page 170
Electric Machines
Chapter 3
...(ii)
nodia
= j 200 # 0.934 = j 186.8 W
| Zl | = 186.8 W
Primary leakage reactance
Ll1 = L1 - 2M = 45 - 2 (20) = 5 mH
Self inductance of secondary winding refers to primary side
N1 2
V1 2
Ll2 = b N2 l L2 = bV2 l L 2
220 2
= b 110 l # 30
So,
Ll2 = 120 mH
Leakage inductance of secondary referred to primary side
Ll2 = Ll2 - 2M
= 120 - 2 (20) = 80 mH
Magnetizing inductance, Lm = 2M = 2 (20) = 40 mm
SOL 3.25
SOL 3.26
SOL 3.27
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 171
measured as
copper loss at 90% load = 81 watt
or,
Wcu ^at 90%h = 81 watt
Now, we have the copper loss for x load as
2
Wcu ^x h = Wcu ^at full loadh # c Ix m
IFL
2
I
or,
Wcu ^at full loadh = Wcu ^at x% loadh # c FL m
I
x
= ^81h #
1
^0.9h2
= 100 watt
For maximum efficiency we must have
copper lossed = no-load loss
= 64 watt
Consider at x% load copper loss is 64 watt. So,
^x2h Wcu ^at full loadh = 64
64 = 0.8
or,
x =
100
nodia
x = 80% load
or
SOL 3.28
f = 50 Hz
frequency of source,
no. of poles
P =4
rotating speed
N = 1440 rpm
Now, the synchronous speed is determined as
120f
NS =
P
= 120 ^50h = 1500 rpm
4
So, the slip in the motor is
S = NS - N = 1500 - 1440 = 0.04
1500
NS
Now, the electrical frequency of the induced negative sequence current in rotor
is obtained as
f0 = ^2 - 5h f
where f is stator frequency given as f = 50 Hz .
Therefore,
f0 = ^2 - 0.04h 50 = 98 Hz
SOL 3.29
www.nodia.co.in
www.gatehelp.com
Page 172
Electric Machines
Chapter 3
Since,
So,
VYZ = VWX
or,
VYZ
= 100
100
VWX
V
= 100 V ; WX = 100
100
VYZ
VWX = 100 V ;
at
nodia
1
at
VWZ
SOL 3.30
S = ns - n
ns
Slip is given as
where,
ns = synchronous speed
n = rotor speed
Thus, slip depend on synchronous speed and the rotor speed. Also, torque
increases with increasing slip up to a maximum value and then decreases. Slip
does not depend on core/loss component.
SOL 3.31
Now,
Armature current,
E1
E2
203.55
207.45
f 2 = 0.6369f1
f1 - f2
100
f1 #
f - 0.6369f1
= 1
# 100
f1
% reduce in flux =
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 173
- 36.3%
SOL 3.32
I2l = 45.0 A
SOL 3.33
nodia
a2 = 2a1
Length
l2 =
2 l1
N 2 ma
L =
l
Magnetizing inductance,
N = no. of turns
m = length of flux path
a = cross section area
l = length
L\a
l
L1 = a1 : l 2
a 2 l1
L2
L 1 = a1 :
2a1
L2
2 l1
l1
L 2 = 2 L1
Thus, magnetizing reactance of second transformer is
Magnetizing current
2 times of first.
Xm2 =
2 Xm1
Im = V
Xm
Im1 = V1 : Xm2 = V1
2 Xm1
b 2V1 lc Xm1 m
Im2 V2 Xm1
(V2 = 2V1)
Im2 = 2 Im1
Thus, magnetizing current of second transformer
Im2 = 2 # 0.5 = 0.707 A
Since voltage of second transformer is twice that of first and current is
that of first, so power will be 2 2 times of first transformer.
2 times
P2 = 2 2 # 55 = 155.6 W
SOL 3.34
www.nodia.co.in
www.gatehelp.com
Page 174
Electric Machines
Chapter 3
Ia = V - Eb
Ra
at the time of starting, Eb = 0 . If the full supply voltage is applied to the motor,
it will draw a large current due to low armature resistance.
A variable resistance should be connected in series with the armature resistance
to limit the starting current.
A 4-point starter is used to start and control speed of a dc shut motor.
SOL 3.35
SOL 3.36
SOL 3.37
nodia
Initially
Eb = V - Ia Ra = 220 - 1 # 10 = 210 V
Now the flux is reduced by 10% keeping the torque to be constant, so the current
will be
1
Ia f1 = Ia f2
f
Ia = Ia 1 = 10 # 1 = 11.11 A
0.9
f2
1
&
Eb \ Nf
Nf
Eb
= 2 2 = 0.9
Eb
N1 f 1
2
` f2 = 0.9f1
N1 = N 2
Eb = V - Ia (Ra + R)
2
Option ( ) is correct.
The steady state speed of magnetic field
ns = 120 # 50 = 1000 rpm
6
S = 0.05
nr = (1 - S) ns
= 0.95 # 1000 = 950 rpm
In the steady state both the rotor and stator magnetic fields rotate in synchronism
, so the speed of rotor field with respect to stator field would be zero.
Speed of rotor which respect to stator field
Speed of rotor
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 175
nodia
IP =
i 02 + i 22
= 1+4 =
SOL 3.40
SOL 3.41
5 = 2.24 Amp
Option ( ) is correct.
than
Ia = 1.963 A
V = Eb + Ia Ra
= 186.6 + 1.963 # 3.4 = 193.34 V
SOL 3.43
Correct option is ( ).
SOL 3.44
www.nodia.co.in
www.gatehelp.com
Page 176
Electric Machines
Chapter 3
Eb1 = Va = 200 V
E
N \ Eb & N1 = b
N2 Eb
Eb = b N2 l Eb = 1400 # 200 = 186.67 V
1500
N1
T = Eb ^Ia /wh & 186.67 # 60 Ia = 5
2p # 1400
1
Ia = 3.926 A
V = Eb + Ia Ra
Ra = Va - Eb = 200 - 186.67 = 3.4 W
3.926
Ia
SOL 3.45
nodia
So,
SOL 3.46
SOL 3.47
Correct option is ( ).
SOL 3.49
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 177
Option ( ) is correct.
SOL 3.51
SOL 3.52
SOL 3.53
SOL 3.54
VAlBl
nodia
N1
N2
I
V
= 4000
= 6000
= 25 A
= 400 V , f = 50 Hz
Coil are to be connected to obtain a single Phase, 400 V auto transfer to drive
1000
Load 10 kVA. Connected A & D common B
www.nodia.co.in
www.gatehelp.com
Page 178
SOL 3.55
Electric Machines
Chapter 3
SOL 3.56
SOL 3.57
SOL 3.58
nodia
Correct option is (B).
Transformer connection will be represented by Y d1.
SOL 3.59
SOL 3.60
SOL 3.61
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 179
3 # 400 # 50 # 0.8
= 27.71 kW
=- N2
df
dt
nodia
During - 0 < t < 1,
E1 =- (100)
df
=- 12 V
dt
E2 = 2E1 = 24 V
df
During time 1 < t < 2 ,
= 0 , then E1 = E2 = 0
dt
df
During 2 < t < 2.5 ,
=- 24 V
E1 =- (100)
dt
E2 =- 0 - 48 V
Then
SOL 3.63
So,
SOL 3.65
I R = 1504
I = 1504 = 12.26 A
10
www.nodia.co.in
www.gatehelp.com
Page 180
Electric Machines
Chapter 3
motor.
R = 1.00 W, Xs = Xlr = 1.5 W
Rlr
So, for max. torque slip Sm =
Xsm + Xlrm
For starting torque Sm = 1
Then
Xsm + Xlrm = Rlr
2pfm Ls + 0.2pfm Llr = 1
Frequency at max. torque
1
fm =
2p (Ls + Llr )
Xs
= 1.5
Ls =
2p # 50 2p # 50
Llr = 1.5
2p # 50
1
fm =
= 50 = 16.7 Hz
1.5 + 1.5
3
50
50
In const V/f control method
nodia
V1 = 400 = 8
50
f1
V2 = 8
f1
So
SOL 3.66
V2 = f2 # 8
= 16.7 # 8 = 133.3 V
SOL 3.67
SOL 3.68
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 181
V = 1+0c p.u.
Ia = 0.6+0c p.u.
Zs = Ra + jXs = 0 + j1 = 1+90c p.u.
V = E+d + Ia Zs = 1+0c - 0.6+0c # 1+90c
E+d = 1.166+ - 30.96c p.u.
Excitation voltage = 1.17 p.u.
Load angle (d) = 30.96c(lagging)
SOL 3.69
Correct option is ( ).
SOL 3.70
SOL 3.71
nodia
The shunt motor provides speed regulation at full load without any controller.
SOL 3.72
SOL 3.73
SOL 3.74
SOL 3.75
www.nodia.co.in
www.gatehelp.com
Page 182
Electric Machines
Chapter 3
50 # 1 # 1
50 # Wcu + Wi
So
Wcu + Wi = 2.631
Reconfigured as a 500 V/750 V auto-transformer
Efficiency
95% =
nodia
auto-transformer efficiency
h=
150
= 98.276%
150 + 2.631
SOL 3.76
SOL 3.77
SOL 3.78
= 3TFL
=1
2
= 22S max 2
S max + 1
...(1)
...(2)
=1
2
=0
= 26.786%
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 183
ISt = 7I Fl
S Fl = 5%
TSt = ISt 2 x2 S
bTFl l # # Fl
TFl
1.5 = (7) 2 # x2 # 0.05
x = 78.252%
SOL 3.79
SOL 3.80
nodia
Correct option is (C).
Given starting torque is 0.5 p.u.
TSt = Isc 2 S
So,
b I Fl l # Fl
TFl
2
0.5 = b Isc l # 0.05
I Fl
Per unit starting current
Isc =
0.5 = 3.16 A
0.05
I Fl
SOL 3.81
SOL 3.82
The two windings are displaced 90c in space. The direction of rotation can be
changed by reversing the main winding terminals.
SOL 3.83
www.nodia.co.in
www.gatehelp.com
Page 184
Electric Machines
Chapter 3
nodia
for motor
So
SOL 3.85
SOL 3.86
SOL 3.87
= 200 + 20 # 0.2
Eg = 204 volt
Em = V - Ia Ra
= 200 - 20 # 0.2
Em = 196 volt
N
Eg
f
= g # g
Nm
Em
fm
204 = Ng
1
196
Nm # 1.1
Nm =
196
= 0.87
204 # 1.1
Ng
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
So
Page 185
2
T \ cV m S
f
S2 = V1 2 f2 T2
S1 bV2 l # f1 # T1
Given
T1 = T2
Then
2
S2 = 0.04 # b 400 l # 30
50
240
S2 = 0.066
Nr = Ns (1 - S)
120f
P
Nr = 120 # 30 ^1 - 0.066h
4
Nr =
So
nodia
= 840.6 rpm
SOL 3.88
P = 4 , V = 400 V , f = 50 Hz
r1 = 1.0 W , r2l= 0.5 W
X1 = Xl2 = 1.2 W, Xm = 35 W
So, Speed of motor is
120f
Ns =
= 120 # 50 = 1500 rpm
4
P
Torque
SOL 3.89
V2 rl2
Tst = 180 #
2pNs
(r1 + rl2) 2 + X 2
400 2 0.5
c 3m#
180
=
2 # 3.14 # 1500 # (1.5) 2 + (2.4) 2
= 63.58 Nm
SOL 3.90
www.nodia.co.in
www.gatehelp.com
Page 186
Electric Machines
Chapter 3
5 # 103
= 7.22
3 # 400 # 1
So,
E =
(E2) 2 - V 2
nodia
(236) 2 - c 400 m = 48.932
3
Ia = 48.932 = 4.8932
10
4
.
8932
Load (%) =
= 67.83 %
7.22
=
SOL 3.91
E ph = 4.44 ffTph KW
Slot/Pole/ph = 48 = 4
4#3
Slot/Pole = 48 = 12
4
Slot angle = 180 = 15c
12
sin (4 # 15/2)
Kd =
= 0.957
4 sin (15/2)
K p = cos a = cos 18c = 0.951
2
In double layer wdg
No. of coil = No of slots
No. of turns/ph = 48 # 10 = 160
3
Then
3 # 808
EL = 1400 V (approximate)
SOL 3.92
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 187
SOL 3.94
nodia
Correct option is (C).
Given that: A 300 kVA transformer
Efficiency at full load is 95% and 0.8 p.f. lagging
96% efficiency at half load and unity power factor
So
For Ist condition for full load
kVA # 0.8
95% =
kVA # 0.8 + Wcu + Wi
...(1)
...(2)
Wcu + Wi = 12.63
0.25Wcu + 0.96Wi = 6.25
Wcu = 8.51, Wi = 4.118
SOL 3.96
kVA # 0.5
kVA # 0.5 + Wcu + Wi
X # p.f. # kVA
X # kVA + Wi + Wcu # X2
4.118 = 0.6956
8.51
0.6956 # 1 # 300
h% =
0.6956 # 300 + 4.118 + 8.51 # (0.6956) 2
h = 96.20%
X =
www.nodia.co.in
www.gatehelp.com
Page 188
Electric Machines
Chapter 3
SOL 3.97
SOL 3.98
nodia
When the speed of the motor is in forward direction then slip varies from 0 to 1
but when speed of motor is in reverse direction or negative then slip is greater
then 1. So at point W slip is greater than 1.
SOL 3.99
SOL 3.100
SOL 3.101
Pnew = 50 = 25 kW
2
At 1.5 time the rated speed by field control
P = constant
P = 50 kW
So
SOL 3.102
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 189
In synchronous machine, when the armature terminal are shorted the field current
should first be decreased to zero before started the alternator.
In open circuit the synchronous machine runs at rated synchronous speed. The
field current is gradually increased in steps.
The short circuit ratio is the ratio of field current required to produced the rated
voltage on open to the rated armature current.
SOL 3.103
In DC motor,
or
E = Kfwn
So armature emf E depends upon f and w only and torque developed depends
upon
PZfIa
T =
2pA
So, torque(T ) is depends of f and Ia and developed power(P ) is depend of flux
f, speed w and armature current Ia .
SOL 3.104
SOL 3.105
nodia
Correct option is ( ).
SOL 3.106
...(2)
From equation (1) and (2) the high resistance of rotor then the motor achieves
quick acceleration and torque of starting is increase.
SOL 3.107
www.nodia.co.in
www.gatehelp.com
Page 190
Electric Machines
Chapter 3
value, the magnetic flux is maintained almost constant at the rated value which
keeps maximum torque constant.
SOL 3.108
So
I =
3 VL IL
1000 = 87.47 A
3 # 6. 6
nodia
IX = 87.47 # 20 = 1.75 kV
2
2
= c 6.5 m + (1.75) 2
E ph
3
So,
6.5 2
2
c 3 m + (1.75)
= 4.2 kV
E ph =
E ph
EL = 3 E ph
EL = 1.732 # 4.2 = 7.26 kV
SOL 3.109
a Star connection
az = tan- 1 c
SOL 3.110
3 # 1.75 = 24.6c
m
6.6
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
So,
X =
Wi =
Wc
Page 191
300 = 0.707
600
SOL 3.112
SOL 3.113
SOL 3.114
SOL 3.115
nodia
circuit
the
field
energy
is
equal
to
the
Wf = W f' = 1 Li2 = 1 yi = 1 y2
2
2
2L
Wf = field energy
W f' = co energy
SOL 3.116
www.nodia.co.in
www.gatehelp.com
Page 192
Electric Machines
Chapter 3
SOL 3.118
nodia
Ra = 0.8 W
At no load condition
P
P
Efficiency
h
SOL 3.119
Vin = 3300 V
Vout = 3300 + 230 = 3530 V
Output current I2 and output voltage 230 V
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
Page 193
So
3
I2 = 50 # 10 = 217.4 A
230
When the output voltage is Vout then kVA rating of auto transformer will be
I2 = 3530 # 217.4
= 767.42 kVA
SOL 3.120
nodia
Z primary = Z pu # Z Base
= (0.02 + j0.07) (3630) = 72.6 + j254.1
SOL 3.121
Zm = Rm + Xm = 6.0 + j4.0 W
ZA = RA + XA = 8.0 + j6.0 W
Phase angle of main winding
+I m = + - Z m
=- + (6 + j4) =- +33.7c
So angle of the auxiliary winding when the capacitor is in series.
+IA =- + (8 + j6) + 1
jwC
j
= + (8 + j6) wC
a = +I A - +I m
So
90 =- tan >f
-1
6- 1
wC p - (- 33.7)H
8
1 = 18
wC
w = 2pf
www.nodia.co.in
www.gatehelp.com
Page 194
Electric Machines
So
C =
Chapter 3
1
1
=
18 # 2 # 3.14 # 50
18 # 2pf
= 176.8 mF
SOL 3.122
synchronous
reactance
of
nodia
both alternator voltage are in phase
So,
E f1 = 3300
3
E f2 = 3200
3
Synchronizing current or circulating current
EC
=
TS1 + TS2
Reactance of both alternator are same
E - Ef 2
So
= f1
TS 1 + TS 2
= 1 b 3300 - 3200 l = 16.98 A
3 1.7 + 1.7
SOL 3.123
SOL 3.124
www.nodia.co.in
www.gatehelp.com
Chapter 3
Electric Machines
and
SOL 3.125
Page 195
rev/min = 30 rpm
nodia
Eg = 384 volt
SOL 3.126
Input power =
I2 =
SOL 3.127
3 V2 I2 = 11.8 kW
11.8 kW
= 21.29 A
3 # 400 # 0.8
3 VL IL = 500 MW
500 # 106
IL =
3 # 21.5 # 103 # 0.85
= 15.79 # 103
IL = 15.79 kA
SOL 3.128
www.nodia.co.in
www.gatehelp.com
Page 196
Electric Machines
so,
Chapter 3
Pi = 0.0555 MVA
At half load, load factor is
L.F = 1 = .5
2
0.5 # 1
So,
h=
# 100 = 87.8%
0.5 # 0.0555 # (0.5) 2 + 0.0555
SOL 3.129
SOL 3.130
SOL 3.131
nodia
Given open circuit and full-load zero power factor of a synchronous generator. At
point Q the zero power factor at 1.0 pu voltage. The voltage drop at point PQ is
across synchronous reactance.
SOL 3.132
SOL 3.133
Option ( ) is correct.
www.nodia.co.in
www.gatehelp.com
Chapter 3
SOL 3.134
Electric Machines
Page 197
nodia
Eb = 0.98
The DC shunt motor is mechanically coupled by the generator so the emf
induced by motor and generator is equal
Eg = Eb
so voltage generated by the generator is
SOL 3.135
SOL 3.136
R
V1
V1
V2
V2
and
V1
V3
N1 : N2: N 3 is 4: 2: 1
= 10 W
= 400 V
= N1 = 4
N2 2
= 2V1 = 200 V
4
= N1 = 4
N3 1
V3 = 100 V
so current in secondary winding
I2 = V2 = 200 = 20 A
10
R
The current in third winding when the capacitor is connected
www.nodia.co.in
www.gatehelp.com
Page 198
Electric Machines
Chapter 3
I 3 = V3 = 100 = j40
- jXc - j2.5
When the secondary winding current I2 is referred to primary side i.e I 1'
I 1' = N2 = 2
So
I2
N1 4
I 1' = 20 = 10 A
2
and winding third current I 3 is referred to Primary side i.e I 1'' . I 3 flows to opposite
to I1
I 1'' = N 3 = 1
So
N1 4
- I3
so
I 1'' =- j10
So total current in primary winding is
nodia
I1 = I 1'' + I 2'' = 10 - j10 A
SOL 3.137
T
U
Rotor has salient pole and slip rings and stator is cylindrical
Both stator and rotor have poly-phase windings
So
DC motor/machines:
The stator winding is connected to dc supply and rotor winding flows ac current.
Stator is made of salient pole and Commutator is connected to the rotor so rotor
winding is supply ac power.
Induction machines:
In induction motor the ac supply is connected to stator winding and rotor and
stator are made of poly-phase windings.
Synchronous machines:
In this type machines the stator is connected to ac supply but rotor winding is
excited by dc supply. The rotor is made of both salient pole and slip rings and
stator is made of cylindrical.
SOL 3.138
www.nodia.co.in
www.gatehelp.com
Chapter 3
SOL 3.139
Electric Machines
Page 199
SOL 3.140
nodia
SOL 3.141
SOL 3.142
...(1)
...(2)
www.nodia.co.in
www.gatehelp.com
Page 200
Electric Machines
Chapter 3
nodia
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 201
CHAPTER 4
POWER SYSTEMS
ONE MARK
nodia
If the fault takes place at location F1 , then the voltage and the current at bus
A are VF1 and IF1 respectively. If the fault takes place at location F2 , then the
voltage and the current at bus A and VF2 and IF2 respectively. The correct
statement about voltages and currents during faults at F1 and F2 is
(B) VF1 leads IF1 and VF2 lags IF2
(A) VF1 leads IF1 and VF2 leads IF2
(C) VF1 lags IF1 and VF2 leads IF2
(D) VF1 lags IF1 and VF2 lags IF2
EE SP 4.2
A 2-bus system and corresponding zero sequence network are shown in the figure.
www.nodia.co.in
www.gatehelp.com
Page 202
Power Systems
Chapter 4
TWO MARKS
EE SP 4.4
nodia
It was found that the lamps are becoming dark in the sequence La - Lb - Lc . It
means that the phase sequence of incoming generator is
(A) opposite to infinite bus and its frequency is more than infinite bus
(B) opposite to infinite bus but its frequency is less than infinite bus
(C) same as infinite bus and its frequency is more than infinite bus
(D) same as infinite bus and its frequency is less than infinite bus
EE SP 4.5
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 203
(B) 50 A and 50 A
(C) 25 A and 75 A
(D) 0 A and 100 A
EE SP 4.6
A two bus power system shown in the figure supplies load of 1.0 + j0.5 p.u.
nodia
The values of V1 in p.u. and d 2 respectively are
(A) 0.95 and 6.00c
EE SP 4.7
C1 = 0.05Pg 12 + APg1 + B
C2 = 0.10Pg 22 + 3APg2 + 2B
Plant P1 :
Plant P2 :
where, Pg1 and Pg2 are the generated powers of two plants, and A and B are the
constants. If the two plants optimally share 1000 MW load at incremental fuel
cost of 100 Rs/MWh, the ratio of load shared by plants P1 and P2 is
(A) 1 : 4
(B) 2 : 3
(C) 3 : 2
(D) 4 : 1
EE SP 4.8
The over current relays for the line protection and loads connected at the buses
are shown in the figure.
www.nodia.co.in
www.gatehelp.com
Page 204
Power Systems
Chapter 4
ONE MARK
EE SP 4.9
EE SP 4.10
nodia
YEAR 2014 EE02
EE SP 4.11
EE SP 4.12
TWO MARKS
There are two generators in a power system. No-load frequencies of the generators
are 51.5 Hz and 51 Hz, respectively, and both are having droop constant of
1 Hz/MW . Total load in the system is 2.5 MW. Assuming that the generators
are operating under their respective droop characteristics, the frequency of the
power system in Hz in the steady state is ______.
EE SP 4.13
EE SP 4.14
A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The
positive, negative, and the zero sequence reactances of the generator are 0.2 pu,
0.2 pu and 0.05 pu, respectively, at the machine base quantities. If a bolted single
phase to ground fault occurs at the terminal of the unloaded generator, the fault
current in amperes immediately after the fault is_____.
YEAR 2014 EE03
EE SP 4.15
ONE MARK
(D) g = c/l
www.nodia.co.in
www.gatehelp.com
Chapter 4
EE SP 4.16
Power Systems
Page 205
EE SP 4.17
A 183 bus power system has 150 PQ buses and 32 PV buses. In the general
case, to obtain the load flow solution using Newton-Raphson method in polar
coordinates, the minimum number of simultaneous equations to be solved is
_____.
YEAR 2014 EE03
EE SP 4.18
EE SP 4.19
EE SP 4.20
TWO MARKS
nodia
For a 400 km long transmission line, the series impedance is ^0.0 + j0.5h W/km
and the shunt admittance is ^0.0 + j5.0h mmho/km . The magnitude of the series
impedance (in W ) of the equivalent p circuit of the transmission line is ____.
TWO MARKS
For a power system network with n nodes, Z 33 of its bus impedance matrix is
j0.5 per unit. The voltage at node 3 is 1.3 - 10c per unit. If a capacitor having
reactance of - j3.5 per unit is now added to the network between node 3 and the
reference node, the current drawn by the capacitor per unit is
(A) 0.325 - 100c
(B) 0.325 80c
(C) 0.371 - 100c
www.nodia.co.in
www.gatehelp.com
Page 206
Power Systems
Chapter 4
EE SP 4.22
EE SP 4.23
nodia
The voltage phase angles in rad at buses 2 and 3 are
(B) q2 = 0 , q3 =- 0.1
(A) q2 =- 0.1, q3 =- 0.2
(C) q2 = 0.1, q3 = 0.1
(D) q2 = 0.1, q3 = 0.2
If the base impedance and the line-to line base voltage are 100 ohms and 100 kV
respectively, then the real power in MW delivered by the generator connected at
the slack bus is
(A) - 10
(B) 0
(D) 20
(C) 10
YEAR 2012
ONE MARK
EE SP 4.24
EE SP 4.25
A two-phase load draws the following phase currents : i1 (t) = Im sin (wt - f 1),
i2 (t) = Im cos (wt - f 2). These currents are balanced if f 1 is equal to.
(B) f 2
(A) - f 2
(C) (p/2 - f 2)
(D) (p/2 + f 2)
EE SP 4.26
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 207
nodia
The sequence components of the fault current are as follows : I positive = j 1.5 pu,
I negative =- j 0.5 pu, I zero =- j1 pu . The type of fault in the system is
(A) LG
(B) LL
(C) LLG
(D) LLLG
YEAR 2012
EE SP 4.28
For the system below, SD1 and SD2 are complex power demands at bus 1 and bus
2 respectively. If V2 = 1 pu , the VAR rating of the capacitor (QG2) connected
at bus 2 is
(A) 0.2 pu
(C) 0.312 pu
EE SP 4.29
TWO MARKS
(B) 0.268 pu
(D) 0.4 pu
EE SP 4.30
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 208
EE SP 4.31
EE SP 4.32
Power Systems
Chapter 4
nodia
A negative sequence relay is commonly used to protect
(A) an alternator
(B) an transformer
(C) a transmission line
(D) a bus bar
For enhancing the power transmission in along EHV transmission line, the most
preferred method is to connect a
(A) Series inductive compensator in the line
(B) Shunt inductive compensator at the receiving end
(C) Series capacitive compensator in the line
(D) Shunt capacitive compensator at the sending end
YEAR 2011
EE SP 4.33
TWO MARKS
A load center of 120 MW derives power from two power stations connected by
220 kV transmission lines of 25 km and 75 km as shown in the figure below. The
three generators G1, G2 and G 3 are of 100 MW capacity each and have identical
fuel cost characteristics. The minimum loss generation schedule for supplying the
120 MW load is
P1 = 80 MW + losses
(A) P2 = 20 MW
P3 = 20 MW
P1 = 60 MW
(B) P2 = 30 MW + losses
P3 = 30 MW
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
P1 = 40 MW
(C) P2 = 40 MW
P3 = 40 MW + losses
EE SP 4.34
P1 = 30 MW + losses
(D) P2 = 45 MW
P3 = 45 MW
The direct axis and quadrature axis reactances of a salient pole alternator are
1.2 p.u and 1.0 p.u respectively. The armature resistance is negligible. If this
alternator is delivering rated kVA at upf and at rated voltage then its power
angle is
(A) 30c
(B) 45c
(C) 60c
EE SP 4.35
Page 209
(D) 90c
nodia
A three bus network is shown in the figure below indicating the p.u. impedance
of each element.
Two generator units G1 and G2 are connected by 15 kV line with a bus at the
mid-point as shown below
G1 = 250 MVA , 15 kV, positive sequence reactance XG = 25% on its own base
G2 = 100 MVA , 15 kV, positive sequence reactance XG = 10% on its own base L1
and L2 = 10 km , positive sequence reactance XL = 0.225 W/km
1
www.nodia.co.in
www.gatehelp.com
Page 210
Power Systems
Chapter 4
nodia
EE SP 4.37
EE SP 4.38
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 211
nodia
(A) - 1 pu
(C) 2 pu
EE SP 4.40
(B) 1 pu
(D) 3 pu
YEAR 2010
TWO MARKS
EE SP 4.41
EE SP 4.42
(A) 0.1875 A
(C) 0.375 A
(B) 0.2 A
(D) 60 kA
www.nodia.co.in
www.gatehelp.com
Page 212
EE SP 4.43
EE SP 4.44
Power Systems
The zero-sequence circuit of the three phase transformer shown in the figure is
nodia
Chapter 4
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 213
nodia
(A) 2.0 A
(C) 2.7 A
EE SP 4.47
(B) 2.4 A
(D) 3.5 A
For the power system shown in the figure below, the specifications of the
components are the following :
G1 : 25 kV, 100 MVA, X = 9%
G2 : 25 kV, 100 MVA, X = 9%
T1 : 25 kV/220 kV, 90 MVA, X = 12%
T2 : 220 kV/25 kV, 90 MVA, X = 12%
Line 1: 200 kV, X = 150 ohms
Choose 25 kV as the base voltage at the generator G1 , and 200 MVA as the MVA
base. The impedance diagram is
www.nodia.co.in
www.gatehelp.com
Page 214
Power Systems
Chapter 4
nodia
YEAR 2009
ONE MARK
EE SP 4.48
EE SP 4.49
For a fixed value of complex power flow in a transmission line having a sending
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 215
TWO MARKS
For the Y-bus matrix of a 4-bus system given in per unit, the buses having shunt
elements are
R- 5 2 2.5 0 V
W
S
S 2 - 10 2.5 4 W
YBUS = j S
2.5 2.5 - 9 4 W
W
S
S0
4
4 - 8W
X
T
(A) 3 and 4
(B) 2 and 3
nodia
(C) 1 and 2
EE SP 4.51
(D) 1, 2 and 4
Match the items in List-I (To) with the items in the List-II (Use) and select the
correct answer using the codes given below the lists.
List-I
List-II
a.
1.
shunt reactor
b.
2.
shunt capacitor
c.
3.
series capacitor
4.
series reactor
Match the items in List-I (Type of transmission line) with the items in List-II
(Type of distance relay preferred) and select the correct answer using the codes
given below the lists.
List-I
List-II
a.
Short Line
1.
Ohm Relay
b.
Medium Line
2.
Reactance Relay
c.
Long Line
3.
Mho Relay
Three generators are feeding a load of 100 MW. The details of the generators are
Rating
(MW)
Efficiency
(%)
Regulation (Pu.)
( on 100 MVA base)
www.nodia.co.in
www.gatehelp.com
Page 216
Power Systems
Chapter 4
Generator-1
100
20
0.02
Generator-2
100
30
0.04
Generator-3
100
40
0.03
In the event of increased load power demand, which of the following will happen
?
(A) All the generator will share equal power
(B) Generator-3 will share more power compared to Generator-1
(C) Generator-1 will share more power compared to Generator-2
(D) Generator-2 will share more power compared to Generator-3
EE SP 4.54
A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated
p.f = 0.9, has a moment of inertia of 27.5 # 103 kg-m2 .The inertia constant (H
) will be
(A) 2.44 s
(B) 2.71 s
nodia
(C) 4.88 s
(D) 5.42 s
YEAR 2008
EE SP 4.55
ONE MARK
A two machine power system is shown below. The Transmission line XY has
positive sequence impedance of Z1 W and zero sequence impedance of Z0 W
An a phase to ground fault with zero fault impedance occurs at the centre of the
transmission line. Bus voltage at X and line current from X to F for the phase
a, are given by Va Volts and Ia amperes, respectively. Then, the impedance
measured by the ground distance relay located at the terminal X of line XY will
be given by
(B) ^Z0 /2h W
(A) ^Z1 /2h W
(D) ^Va /Ia h W
(C) (Z0 + Z1) /2 W
EE SP 4.56
EE SP 4.57
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 217
Voltage drop across the transmission line is given by the following equation :
VR V
V R
R
S3 Va W SZs Zm Zm WSIa W
S3 Vb W = SZm Zs Zm WSIb W
SS3 V WW SSZ Z Z WWSSI WW
c
s
m
m
c
X
X
T
T
T
Shunt capacitance of the X line can be neglect. If the has positive sequence
impedance of 15 W and zero sequence impedance of 48 W, then the values of Zs
and Zm will be
(A) Zs = 31.5 W; Zm = 16.5 W
(B) Zs = 26 W; Zm = 11 W
nodia
(D) Zs = 11 W; Zm = 26 W
YEAR 2008
EE SP 4.58
TWO MARKS
EE SP 4.59
A loss less transmission line having Surge Impedance Loading (SIL) of 2280 MW
is provided with a uniformly distributed series capacitive compensation of 30%.
Then, SIL of the compensated transmission line will be
(A) 1835 MW
(B) 2280 MW
(C) 2725 MW
(D) 3257 MW
EE SP 4.60
A loss less power system has to serve a load of 250 MW. There are tow generation
(G 1 and G 2 ) in the system with cost curves C1 and C2 respectively defined as
follows ;
2
C1 (PG1) = PG1 + 0.055 # PG1
2
C2 (PG2) = 3PG2 + 0.03 # PG2
Where PG1 and PG2 are the MW injections from generator G 1 and G 2 respectively.
Thus, the minimum cost dispatch will be
www.nodia.co.in
www.gatehelp.com
Page 218
Power Systems
Chapter 4
A loss less single machine infinite bus power system is shown below :
The synchronous generator transfers 1.0 per unit of power to the infinite bus. The
critical clearing time of circuit breaker is 0.28 s. If another identical synchronous
generator is connected in parallel to the existing generator and each generator is
scheduled to supply 0.5 per unit of power, then the critical clearing time of the
circuit breaker will
(A) reduce to 0.14 s
(B) reduce but will be more than 0.14 s
nodia
(C) remain constant at 0.28 s
(D) increase beyond 0.28 s
EE SP 4.62
Single line diagram of a 4-bus single source distribution system is shown below.
Branches e1, e2, e3 and e4 have equal impedances. The load current values indicated
in the figure are in per unit.
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 219
EE SP 4.63
EE SP 4.64
nodia
The instant (t0) of the fault will be
(A) 4.682 ms
(B) 9.667 ms
(C) 14.667 ms
(D) 19.667 ms
EE SP 4.65
Instead of the three phase fault, if a single line to ground fault occurs on phase
a at point F with zero fault impedance, then the rms of the ac component of
fault current (Ix) for phase a will be
(A) 4.97 p.u
(B) 7.0 p.u
(C) 14.93 p.u
(D) 29.85 p.u
YEAR 2007
EE SP 4.66
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 220
Power Systems
Chapter 4
nodia
(D) Star-Zigzag (q = 30%)
EE SP 4.67
The incremental cost curves in Rs/MWhr for two generators supplying a common
load of 700 MW are shown in the figures. The maximum and minimum generation
limits are also indicated. The optimum generation schedule is :
Two regional systems, each having several synchronous generators and loads are
inter connected by an ac line and a HVDC link as shown in the figure. Which of
the following statements is true in the steady state :
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 221
nodia
(A) Point X
(C) Point Z
(B) Point Y
(D) Point W
YEAR 2007
EE SP 4.70
TWO MARKS
The figure below shows a three phase self-commutated voltage source converter
connected to a power system. The converters dc bus capacitor is marked as C
in the figure. The circuit in initially operating in steady state with d = 0 and the
capacitor dc voltage is equal to Vdc0 . You may neglect all losses and harmonics.
What action should be taken to increase the capacitor dc voltage slowly to a new
steady state value.
The total reactance and total suspectance of a lossless overhead EHV line,
operating at 50 Hz, are given by 0.045 pu and 1.2 pu respectively. If the velocity
of wave propagation is 3 # 105 km/s, then the approximate length of the line is
(B) 172 km
(A) 122 km
(C) 222 km
(D) 272 km
www.nodia.co.in
www.gatehelp.com
Page 222
EE SP 4.72
EE SP 4.73
Power Systems
Consider the protection system shown in the figure below. The circuit breakers
numbered from 1 to 7 are of identical type. A single line to ground fault with zero
fault impedance occurs at the midpoint of the line (at point F), but circuit breaker
4 fails to operate (Stuck breaker). If the relays are coordinated correctly, a
valid sequence of circuit breaker operation is
nodia
(A) 1, 2, 6, 7, 3, 5
(B) 1, 2, 5, 5, 7, 3
(C) 5, 6, 7, 3, 1, 2
(D) 5, 1, 2, 3, 6, 7
A three phase balanced star connected voltage source with frequency w rad/s
is connected to a star connected balanced load which is purely inductive. The
instantaneous line currents and phase to neutral voltages are denoted by (ia, ib, ic)
and (Van, Vbn, Vcn) respectively, and their rms values are denoted by V and I .
R
V
1
- 13 W RSia VW
S 0
3
1 WS W
If R = 8Van Vbn Vcn B S- 13
0
i , then the magnitude of
3 W b
S 1
SSi WW
1
0 W c
S 3 - 3
T
XT X
of R is
(A) 3VI
(C) 0.7VI
EE SP 4.74
Chapter 4
(B) VI
(D) 0
(A) 0.87
(C) 0.67
(B) 0.74
(D) 0.54
www.nodia.co.in
www.gatehelp.com
Chapter 4
EE SP 4.75
EE SP 4.76
Power Systems
Page 223
nodia
The bus voltage phase angular difference between generator bus X and generator
bus Y after interconnection is
(A) 10c
(B) 25c
www.nodia.co.in
www.gatehelp.com
Page 224
Power Systems
(C) - 30c
EE SP 4.77
Chapter 4
(D) 30c
A 230 V (Phase), 50 Hz, three-phase, 4-wire, system has a phase sequence ABC.
A unity power-factor load of 4 kW is connected between phase A and neutral N.
It is desired to achieve zero neutral current through the use of a pure inductor and
a pure capacitor in the other two phases. The value of inductor and capacitor is
(A) 72.95 mH in phase C and 139.02 mF in Phase B
(B) 72.95 mH in Phase B and 139.02 mF in Phase C
(C) 42.12 mH in Phase C and 240.79 mF in Phase B
(D) 42.12 mH in Phase B and 240.79 mF in Phase C
EE SP 4.78
nodia
YEAR 2006
EE SP 4.79
ONE MARK
The concept of an electrically short, medium and long line is primarily based on
the
(A) nominal voltage of the line
(B) physical length of the line
(C) wavelength of the line
EE SP 4.80
Keeping in view the cost and overall effectiveness, the following circuit breaker
is best suited for capacitor bank switching
(A) vacuum
(B) air blast
(C) SF6
(D) oil
EE SP 4.81
EE SP 4.82
An HVDC link consist of rectifier, inverter transmission line and other equipments.
Which one of the following is true for this link ?
(A) The transmission line produces/ supplies reactive power
(B) The rectifier consumes reactive power and the inverter supplies reactive
power from/ to the respective connected AC systems
(C) Rectifier supplies reactive power and the inverted consumers reactive power
to/ from the respective connected AC systems
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 225
(D) Both the converters (rectifier and inverter) consume reactive power from
the respective connected AC systems
YEAR 2006
TWO MARKS
EE SP 4.83
EE SP 4.84
A single phase transmission line and a telephone line are both symmetrically
strung one below the other, in horizontal configurations, on a common tower,
The shortest and longest distances between the phase and telephone conductors
are 2.5 m and 3 m respectively.
The voltage (volt/km) induced in the telephone circuit, due to 50 Hz current of
100 amps in the power circuit is
(A) 4.81
(B) 3.56
(C) 2.29
(D) 1.27
EE SP 4.85
nodia
EE SP 4.86
EE SP 4.87
www.nodia.co.in
www.gatehelp.com
Page 226
Power Systems
Chapter 4
EE SP 4.89
nodia
For a power system the admittance and impedance matrices for the fault studies
are as follows.
V
V
R
R
S- j8.75 j1.25 j2.50 W
Sj0.16 j0.08 j0.12 W
Z bus = Sj0.08 j0.24 j0.16 W
Ybus = S j1.25 - j6.25 j2.50 W
SS j2.50 - j2.50 - j5.00 WW
SSj0.12 j0.16 j0.34 WW
X
X
T
T
The pre-fault voltages are 1.0 pu. at all the buses. The system was unloaded
prior to the fault. A solid 3-phase fault takes place at bus 2.
EE SP 4.90
The post fault voltages at buses 1 and 3 in per unit respectively are
(A) 0.24, 0.63
(B) 0.31, 0.76
(C) 0.33, 0.67
(D) 0.67, 0.33
EE SP 4.91
The per unit fault feeds from generators connected to buses 1 and 2 respectively
are
(A) 1.20, 2.51
(B) 1.55, 2.61
(C) 1.66, 2.50
(D) 5.00, 2.50
EE SP 4.92
A 400 V, 50 Hz, three phase balanced source supplies power to a star connected
load whose rating is 12 3 kVA, 0.8 pf (lag). The rating (in kVAR) of the delta
connected (capacitive) reactive power bank necessary to bring the pf to unity is
(A) 28. 78
(B) 21.60
(C) 16.60
(D) 12.47
YEAR 2005
EE SP 4.93
ONE MARK
The p.u. parameter for a 500 MVA machine on its own base are:
inertia, M = 20 p.u. ; reactance, X = 2 p.u.
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 227
The p.u. values of inertia and reactance on 100 MVA common base, respectively, are
(A) 4, 0.4
(B) 100, 10
(C) 4, 10
(D) 100, 0.4
EE SP 4.94
EE SP 4.95
EE SP 4.96
nodia
High Voltage DC (HVDC) transmission is mainly used for
(A) bulk power transmission over very long distances
(C) inter-connecting two systems with same nominal frequency
(C) eliminating reactive power requirement in the operation
(D) minimizing harmonics at the converter stations
YEAR 2005
TWO MARKS
EE SP 4.97
EE SP 4.98
EE SP 4.99
The network shown in the given figure has impedances in p.u. as indicated. The
diagonal element Y22 of the bus admittance matrix YBUS of the network is
(A) - j19.8
(C) + j0.2
EE SP 4.100
(B) + j20.0
(D) - j19.95
www.nodia.co.in
www.gatehelp.com
Page 228
Power Systems
Chapter 4
and G2 as shown in the figure. The fuel cost characteristic of the generating
stations are given by
F1 = a + bP1 + cP12 Rs/hour
F2 = a + bP2 + 2cP22 Rs/ hour
EE SP 4.101
nodia
Two networks are connected in cascade as shown in the figure. With usual
notations the equivalent A, B, C and D constants are obtained. Given that,
C = 0.025+45c, the value of Z2 is
(A) 10+30c W
(C) 1 W
EE SP 4.102
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
(C) 5.5 W
EE SP 4.104
Page 229
(D) 12.1 W
EE SP 4.105
ONE MARK
nodia
(C) pulsating with zero average
EE SP 4.106
EE SP 4.107
(A) RYB
(B) RBY
(C) BRY
(D) YBR
EE SP 4.108
In the thermal power plants, the pressure in the working fluid cycle is developed by
(A) condenser
(B) super heater
(C) feed water pump
(D) turbine
EE SP 4.109
For harnessing low variable waterheads, the suitable hydraulic turbine with high
percentage of reaction and runner adjustable vanes is
(A) Kaplan
(B) Francis
(C) Pelton
(D) Impeller
www.nodia.co.in
www.gatehelp.com
Page 230
EE SP 4.110
Power Systems
Chapter 4
The transmission line distance protection relay having the property of being
inherently directional is
(A) impedance relay
(B) MHO relay
(C) OHM relay
(D) reactance relay
YEAR 2004
EE SP 4.111
nodia
(C) 2085 MW
EE SP 4.112
EE SP 4.113
TWO MARKS
(D) 2606 MW
A 110 kV, single core coaxial, XLPE insulated power cable delivering power at
50 Hz, has a capacitance of 125 nF/km. If the dielectric loss tangent of XLPE is
2 # 10 - 4 , then dielectric power loss in this cable in W/km is
(A) 5.0
(B) 31.7
(C) 37.8
(D) 189.0
EE SP 4.114
EE SP 4.115
pu.
[equivalent
to
A
new
generator
having
Eg = 1.4+30c
(1.212 + j0.70) pu] and synchronous reactance 'XS ' of 1.0 pu on the system base,
is to be connected to a bus having voltage Vt , in the existing power system. This
existing power system can be represented by Thevenins voltage Eth = 0.9+0c pu
in series with Thevenins impedance Zth = 0.25+90c pu. The magnitude of the
bus voltage Vt of the system in pu will be
(A) 0.990
(B) 0.973
(C) 0.963
(D) 0.900
www.nodia.co.in
www.gatehelp.com
Chapter 4
EE SP 4.116
Power Systems
EE SP 4.117
Page 231
(D) 38.45 kA
nodia
(A) 5.78+ - 30c
(C) 6.33+90c
(B) 5.78+90c
(D) 10.00+ - 30c
EE SP 4.118
EE SP 4.119
A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavoltamperes at 0.8 power factor. Suddenly a fault occurs reducing in electric power
output by 40%. Neglect losses and assume constant power input to the shaft.
The accelerating torque in the generator in MNm at the time of fault will be
(A) 1.528
(B) 1.018
(C) 0.848
(D) 0.509
EE SP 4.120
www.nodia.co.in
www.gatehelp.com
Page 232
Power Systems
YEAR 2003
EE SP 4.121
Chapter 4
ONE MARK
Bundled conductors are mainly used in high voltage overhead transmission lines
to
(A) reduces transmission line losses
(B) increase mechanical strength of the line
(C) reduce corona
(D) reduce sag
EE SP 4.122
EE SP 4.123
A power system consist of 300 buses out of which 20 buses are generator bus,
25 buses are the ones with reactive power support and 15 buses are the ones
with fixed shunt capacitors. All the other buses are load buses. It is proposed to
perform a load flow analysis in the system using Newton-Raphson method. The
size of the Newton Raphson Jacobian matrix is
(A) 553 # 553
(B) 540 # 540
(C) 555 # 555
(D) 554 # 554
nodia
EE SP 4.124
EE SP 4.125
The interrupting time of a circuit breaker is the period between the instant of
(A) initiation of short circuit and the arc extinction on an opening operation
(B) energizing of the trip circuit and the arc extinction on an opening
operation
(C) initiation of short circuit and the parting of primary arc contacts
(D) energizing of the trip circuit and the parting of primary arc contacts
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
YEAR 2003
Page 233
TWO MARKS
EE SP 4.126
EE SP 4.127
EE SP 4.128
nodia
A dc distribution system is shown in figure with load current as marked. The two
ends of the feeder are fed by voltage sources such that VP - VQ = 3 V. The value
of the voltage VP for a minimum voltage of 220 V at any point along the feeder is
(A) 225.89 V
(B) 222.89 V
(C) 220.0 V
(D) 228.58 V
EE SP 4.129
EE SP 4.130
www.nodia.co.in
www.gatehelp.com
Page 234
Power Systems
Chapter 4
EE SP 4.132
nodia
EE SP 4.133
Incremental fuel costs (in some appropriate unit) for a power plant consisting of
three generating units are
IC1 = 20 + 0.3P1, IC2 = 30 + 0.4P2, IC3 = 30
Where P1 is the power in MW generated by unit i for i = 1, 2 and 3. Assume
that all the three units are operating all the time. Minimum and maximum loads
on each unit are 50 MW and 300 MW respectively. If the plant is operating on
economic load dispatch to supply the total power demand of 700 MW, the power
generated by each unit is
(A) P1 = 242.86 MW; P2 = 157.14 MW; and P3 = 300 MW
(B) P1 = 157.14 MW; P2 = 242.86 MW; and P3 = 300 MW
(C) P1 = 300 MW; P2 = 300 MW; and P3 = 100 MW
(D) P1 = 233.3 MW; P2 = 233.3 MW; and P3 = 233.4 MW
EE SP 4.134
A list of relays and the power system components protected by the relays are
given in List-I and List-II respectively. Choose the correct match from the four
choices given below:
List-I
P.
List-II
Distance relay
1.
Transformers
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 235
Q.
2.
Turbines
R.
Differential relay
3.
Busbars
S.
Buchholz relay
4.
Shunt capacitors
5.
Alternators
6.
Transmission lines
Codes:
(A)
(B)
(C)
(D)
EE SP 4.135
EE SP 4.136
P
6
4
5
6
Q
5
3
2
4
R
3
2
1
5
S
1
1
6
3
nodia
A generator delivers power of 1.0 p.u. to an infinite bus through a purely reactive
network. The maximum power that could be delivered by the generator is 2.0
p.u. A three-phase fault occurs at the terminals of the generator which reduces
the generator output to zero. The fault is cleared after tc second. The original
network is then restored. The maximum swing of the rotor angle is found to be
dmax = 110 electrical degree. Then the rotor angle in electrical degrees at t = tc is
(A) 55
(B) 70
(C) 69.14
(D) 72.4
A three-phase alternator generating unbalanced voltages is connected to an
unbalanced load through a 3-phase transmission line as shown in figure. The neutral
of the alternator and the star point of the load are solidly grounded. The phase
voltages of the alternator are Ea = 10+0c V, Eb = 10+ - 90c V, Ec = 10+120c V
. The positive-sequence component of the load current is
www.nodia.co.in
www.gatehelp.com
Page 236
Power Systems
Chapter 4
SOLUTION
SOL 4.1
nodia
SOL 4.2
Here, on T1 side series switch are closed hence star-connection and grounded due
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 237
to 3XGn . On T2 side, shunt switches are closed hence one delta and one star with
ground.
SOL 4.3
T1
T2
nodia
Ia1 = 1 - 90c pu ;
Ib2 = 4 - 150c pu ;
Ic0 = 3 90c pu
The phase current is given by
Ia = Ia1 + Ia2 + Ia0
...(i)
Now, we have
Ib2 = aIa2
So, for the given system
or
Also, we have
Ia2 = 4 - 270c
Thus, we obtain
Ib = Ib1 + Ib2 + Ib0
= 8 150c + 4 - 150c + 3 90c
= 11.53 154.3c pu
SOL 4.4
www.nodia.co.in
www.gatehelp.com
Page 238
Power Systems
Chapter 4
If all the bulbs glow dark simultaneously, then all the gear and infinite bus
are in same phase sequence. But, in the given problem, they are coming in
sequence (not simultaneously) La then Lb , and then Lc . So, phase sequence of
the generator is opposite to infinite bus system and frequency is high (as dark
period is for more time).
SOL 4.5
nodia
Correct option is (D).
We have the distribution feeder as
Let current I be supplied by source ^s1h, and r be the resistance per unit length.
Applying KVL, we get
400 - ^r # 400h I - ^200r h^I - 200h - ^200r h^I - 300h - ^200r h^I - 500h = 400
or
400Ir + 200Ir - 40000r + 200Ir - 60000r + 200Ir - 100000r = 0
or
1000Ir = 200000r
So,
I = 200 A
So, we redraw the current in distribution system as
Thus, the contribution of source ^s1h in 100 A load at P is 0 A , and the contribution
of s2 in 100 A load at P is 100 A.
SOL 4.6
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 239
nodia
or
d 2 =- 5.65c
V1 - cos d 2 = 0.11 cos 116.56c
or
Also,
V1 = 1.05
or
SOL 4.7
2
2
...(i)
...(ii)
...(iii)
...(iv)
...(v)
Pg = 800 MW
and
Pg = 200 MW
Therefore, the ratio of load shared by plants P1 and P2 is
Pg
= 800 = 4
200 1
Pg
1
SOL 4.8
www.nodia.co.in
www.gatehelp.com
Page 240
Power Systems
Chapter 4
...(i)
SOL 4.9
nodia
Correct answer is 118.80 .
SOL 4.11
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 241
d = 30.504c
Reactive power is given by
2
E V
Q = f cos d - V
Xs
Xs
2
= 1.3 # 1 cos ^30.504ch - 1
1. 1
1. 1
nodia
...(i)
51.5 - f = P1
Now in steady state total load is 2.5 MW , so load by second generator is ^2.5 - P1h
. For G2 we write
51 - f
=1
2.5 - P1
51 - f = 2.5 - P1
Adding eq (i) and (ii), we get
...(ii)
51.5 - f + 51 - f = P1 + 2.5 - P1
102.5 - 2f = 2.5
2f = 100
f = 50 Hz
SOL 4.13
SOL 4.14
www.nodia.co.in
www.gatehelp.com
Page 242
Power Systems
Chapter 4
100 # 106
= 2309.40 A
3 # 25 # 103
So actual value of fault current
Ibase =
where
nodia
or
SOL 4.16
SOL 4.17
SOL 4.18
SOL 4.19
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 243
load is
Q # 0.25 kVAR
Since, we have Q1 = 1 kVAR , therefore, capacitor will supply
QC = .75 kVAR
i.e. capacitor must supply .75 kVAR .
SOL 4.20
SOL 4.21
SOL 4.22
nodia
SOL 4.23
i.e.,
q2 = 0
and
q3 =- 0.1 rad
www.nodia.co.in
www.gatehelp.com
Page 244
Power Systems
Chapter 4
= 10 # 106 watt = 10 MW
SOL 4.24
SOL 4.25
nodia
We know that,
So, i1 (t) can be written as
I 2 = Im f 2
Current are balanced if
I1 + I 2 = 0
Im f1 + 90c + Im f 2 = 0
Im cos ^f1 + 90ch + jIm sin ^f1 + 90ch + cos f 2 + j sin f 2 = 0
Im 8cos ^f1 + 90ch + j sin ^f1 + 90chB + Im 6cos f 2 + j sin f 2@ = 0
Im 8cos ^f1 + 90ch + cos f 2B + jIm 8sin f 2 + sin ^f1 + 90chB = 0
cos ^f1 + 90ch + cos f2 = 0
cos ^f1 + 90ch =- cos f2 = cos ^p + f2h
f1 + 90c = p + f2
f1 = p + f2
2
or,
SOL 4.26
2
G1
PL = 0.5P
2PL = 0.5 (2P ) = P
G
G
2PG
1
L1 =
1 - PG
1
So,
1
=1
2
1 - PL
2PG
2PL
ca 2PG = 0 m
2
C1 # L1 = C 2 # L 2
where C1 and C2 are the incremental fuel cost of plant G1 and G2 .
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
(10000) b
So,
Page 245
1
= 12500 # 1
1 - PG l
4 = 1-P
G
5
PG = 1 pu
5
PG = 1 # 100 = 20 MW
5
2
PL = 0.5 b 1 l = 1 pu
5
50
PL = 1 # 100 = 2 MW
50
2
It is an 100 MVA, so
Loss
or
PL = PG + PG - PL
40 = 20 + P2 - 2
Total power,
nodia
PG = 22 MW
2
SOL 4.27
SOL 4.28
V2 = 1 - 30c V
1 0c - 1 30c
I12 = V1 - V2 =
j 0.5
Z
Current,
= (1 - j 0.288) pu
Current in SD is I2 ,
2
SD = V2 I2)
1 = 1 - 30c I2)
2
I2 = 1 - 30c pu
IG = I2 - I12 = 1 - 30c - (1 - j 0.288)
Current in QG ,
2
= 0.268 - 120c
VAR rating of capacitor,
QC = V2 VG = 1 # 0.268 = 0.268 pu
SOL 4.29
www.nodia.co.in
www.gatehelp.com
Page 246
Power Systems
Chapter 4
Total reactance,
X = j1 + j 0.5 = j1.5 pu
Critical angle is given as,
dcr = cos-1 [(p - 2d0) sin d0 - cos d0]
d0 " steady state torque angle.
Steady state power is given as
Pm = Pmax sin d0
E V
Pmax =
X
E V
Pm =
sin d0
X
(1.5) (1)
sin d0
0.5 =
1.5
where,
nodia
So,
...(i)
Pm = 0.5 pu
sin d0 = 0.5
d0 = 30c
d0 = 30c # p = 0.523
180c
In radian,
SOL 4.30
SOL 4.31
SOL 4.32
SOL 4.33
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 247
nodia
= 1 & d = 45c
SOL 4.35
Here
For generator G2
XL = XL = 0.225 # 10 = 2.25 W
2
www.nodia.co.in
www.gatehelp.com
Page 248
Power Systems
Chapter 4
SOL 4.37
nodia
SOL 4.38
SOL 4.39
I >0
VAB > 0 since it is Rectifier O/P
VCD > 0 since it is Inverter I/P
VAB > VCD , Than current will flow in given direction.
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 249
Applying KVL
V + VL = 0
VL =- V
VL =- 1 pu
SOL 4.40
nodia
P + jQ = VI)
Real power
Reactive power
SOL 4.41
SOL 4.42
= I 2 - I1
www.nodia.co.in
www.gatehelp.com
Page 250
Power Systems
Chapter 4
nodia
No option seems to be appropriate but (C) is the nearest.
SOL 4.44
SOL 4.45
11 (6C)
e1 = 3
= 11 # 6 = 3.46 kV
11
6C + 5C
3
e2 = 11 # 5
11
3
= 2.89 kV
SOL 4.46
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 251
w = 2pf = 314
Changing current IC = V = V (wC)
XC
3
= 11 # 10 # 314 # 1 # 10- 6
3
= 2 Amp
SOL 4.47
nodia
The Impedance diagram is being given by as
SOL 4.48
Correct option is ( ).
SOL 4.49
So
SOL 4.50
www.nodia.co.in
www.gatehelp.com
Page 252
Power Systems
We know
Here
Chapter 4
V
R
Sy11 y12 y13 y14W
Sy21 y22 y23 y24W
YBus = S
W
Sy 31 y 32 y 33 y 34W
Sy 41 y 42 y 43 y 44W
X
T
y11 = y10 + y12 + y13 + y14 =- 5j
y22 = y20 + y21 + y23 + y24 =- 10j
y 33 = y 30 + y 31 + y 32 + y 34 =- 9j
y 44 = y 40 + y 41 + y 42 + y 43 =- 8j
y12 = y21 =- y12 = 2j
y13 = y 31 =- y13 = 2.5j
y14 = y 41 =- y14 = 0j
y23 = y 32 =- y23 = 2.5j
y24 = y 42 =- y24 = 4j
y10 = y11 - y12 - y13 - y14 =- 5j + 2j + 2.5j + 0j =- 0.5j
nodia
So
From figure. it is cleared that branch (1) & (2) behaves like shunt element.
SOL 4.51
SOL 4.52
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 253
SOL 4.54
nodia
= 1357.07 MJ
Stored K.E
Inertia constant (H) =
Rating of Generator (MVA)
H = 1357.07
555.56
= 2.44 sec
SOL 4.55
SOL 4.56
SOL 4.57
...(1)
www.nodia.co.in
www.gatehelp.com
Page 254
Power Systems
Chapter 4
...(2)
Correct option is ( ).
SOL 4.59
SOL 4.60
SIL = 2280 MW
nodia
Correct option is (C).
Given
and
dC1 = 1 + 0.11P
G1
dPG1
dC2 = 3 + 0.06P
and
G2
dPG2
Since the system is loss-less
dC1 = dC2
Therefore
dPG2
dPG1
So from equations (3a) and (3b)
We have
0.11PG1 - 0.06PG2 = 2
Now solving equation (1) and (4), we get
...(1)
...(2)
...(3a)
...(3b)
...(4)
PG1 = 100 MW
PG2 = 150 MW
SOL 4.61
SOL 4.62
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 255
0 = Ae- (R/L) t +
0
2 Vm cos (wt - a)
0
Z
nodia
Ae- (R/L) t =- 2 Vm cos (wt 0 - a)
Z
Maximum value of the dc offset current
0
(wt 0 - a) = 0
t0 = a
w
or
...(1)
Z = 0.004 + j0.04
and
Z = Z +a = 0.0401995+84.29c
a = 84.29cor 1.471 rad.
From equation (1)
t0 =
SOL 4.64
Z = 0.0201+84.29c
2
Z1 (Positive sequence) = Z = 0.0201+84.29c
2
also Z1 = Z2 = Z 0 (for 3-f fault)
`
1+0c
I f (pu) = 1+0c =
Z1
0.0201+84.29c
So magnitude
If
(p.u.)
= 49.8
www.nodia.co.in
www.gatehelp.com
Page 256
Power Systems
I f = 49.8 #
` Fault current
SOL 4.65
Chapter 4
100
= 7.18 kA
3 # 400
nodia
Z1 = Z = 0.0201+84.29c
2
and
Then
and
So
Z2 = Z1 = 0.0201+84.29c
Z 0 = 3Z1 = 0.0603+84.29c
Ia /3 = Ia1 = Ia2 = Ia0
Ia1 (pu) = 1.0+0c
Z1 + Z 2 + Z 0
1 .0
Ia1 =
= 9.95 pu
(0.0201 + 0.0201 + 0.0603)
SOL 4.66
SOL 4.67
PA + PB = 700 MW
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 257
SOL 4.68
SOL 4.69
nodia
www.nodia.co.in
www.gatehelp.com
Page 258
SOL 4.71
Power Systems
Chapter 4
Reactance of line
Suspectance of Line
SOL 4.72
SOL 4.73
nodia
Solving we get
V
R
1
- 13 W RSiaVW
S 0
3
1 W
R = [Van Vbn Vcn] S- 13
0
SibW
3
WS W
S 1
1
0 W SicW
S 3 - 3
XT X
T
Here
Since
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
P2 max = EX ,
X2
Page 259
To find maximum value of X for which system does not loose synchronism
P2 = Pm (shown in above figure)
EV sin d = P
`
m
2
X2
as Pm = 1 pu, E = 1.0 pu,V = 1.0 pu
1.0 # 1.0 sin 130c = 1
X2
&
&
&
SOL 4.75
X2 = 0.77
(0.1 + X) = 0.77
X = 0.67
nodia
Correct option is (B).
Given that
FP = KAFS
Rf V
Rf V
S aW
S pW
where, Phase component FP = SfbW, sequence component FS = SfnW
SSf WW
SSf WW
c
o
T
X
T
X
R 1 1 1V
S
W
and
A = Sa2 a 1W
SS a a2 1WW
VP = TKAVS X
`
3
IP = KAIS
and
VS = Zl [IS ]
R0.5 0 0 V
S
W
where
Zl = S 0 0.5 0 W
SS 0 0 2.0WW
T
X
We have to find out Z if VP = ZIP
From equation (2) and (3)
VP = KAZl [IS ]
-1
VP = KAZlb A l I p
K
VP = AZlA- 1 I p
R 1 1 1V
S
W
A = Sa2 a 1W
SS a a2 1WW
T
X
Adj A
-1
A =
A
R
2V
S1 a a W
Adj A = S1 a2 a W
S
W
S1 1 1 W
T
X
1
A =
3
R
2V
S1 a a W
A- 1 = 1 S1 a2 a W
3S
W
S1 1 1 W
T
X
...(1)
...(2)
...(3)
...(4)
...(5)
www.nodia.co.in
www.gatehelp.com
Page 260
Power Systems
Chapter 4
SOL 4.76
R 1 0.5 0.5V
R 1 1 1VR0.5 0 0VR1 a a2V
W
S
W
W
S
W
S
S
Vp = 1 Sa2 a 1WS 0 0.5 0WS1 a2 a W I p = S0.5 1 0.5W I p ...(6)
3S
W
S
SS0.5 0.5 1 WW
S a a2 1WWSS 0 0 2WWS1 1 1 W
XT
XT
T
X
T and (6)
X
Comparing of equation (5)
R 1 0.5 0.5V
S
W
Z = S0.5 1 0.5W
SS0.5 0.5 1 WW
T
X
Correct option is (A).
Given that the first two power system are not connected and separately loaded.
Now these are connected by short transmission line.
as P1 = P2 = Q1 = Q2 = 0
So here no energy transfer. The bus bar voltage and phase angle of each system
should be same than angle difference is
nodia
q = 30c - 20c = 10c
SOL 4.77
a
IN = 0 = IA + IB + IC
Network and its Phasor is being as
...(1)
IA =-c IB #
IA =
Now
and
a Ib - Ic
3 +I
3
C #
2
2 m
3 IB = 3 IC
IB - IC = 17.39 - 10 Amp
3
XC = V = 230 - 23 W
10
IC
XC = 1
2pfC
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
&
&
So
SOL 4.78
Page 261
1 =
1
= 139.02 mF
2p # 50 # 23
2pfXC
XL = V = 230 - 23 W = 2pfL
10
IL
23
= 72.95 mH
L = XL =
2p # 100
2pf
C =
L = 72.95 mH in phase B
C = 139.02 mF in phase C
nodia
System frequency is = 50 - 0.58 = 49.42 Hz
SOL 4.79
SOL 4.80
SOL 4.81
SOL 4.82
SOL 4.83
www.nodia.co.in
www.gatehelp.com
Page 262
Power Systems
Chapter 4
240 - 220
0
%V.R. = .94
# 100
220
%V.R. = 16
SOL 4.84
Option ( ) is correct.
SOL 4.85
nodia
From figure we conclude that positive sequence line voltage leads phase voltage
by 30c
VAN1 = X+q1 - 30c
VAN2 = 4+q2 + 30c
SOL 4.86
SOL 4.87
SOL 4.88
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 263
= 1 - 1.1 # 1 # 1 = 0.56 pu
2
1/0.8
SOL 4.89
SOL 4.90
nodia
Correct option is (D).
Vi c = Prefault voltage
Vi (f) = Vi c (0) - Zir I (f),
V1 (f) = Vi c - Z12 I f = 1+0c - j0.08 (- j4) = 1 - 0.32
V1 (f) = 0.68 pu
V3 (f) = V3 c - Z 32 I f = 1+0c - j0.16 (- j4) = 1 - 0.64
V3 (f) = 0.36 pu
SOL 4.91
Option ( ) is correct.
SOL 4.92
SOL 4.93
www.nodia.co.in
www.gatehelp.com
Page 264
Power Systems
Chapter 4
SOL 4.94
nodia
Correct option is (D).
800 kV has Power transfer capacity = P
At 400 kV Power transfer capacity = ?
We know Power transfer capacity
P = EV sin d
X
P \ V2
So if V is half than Power transfer capacity is 1 of previous value.
4
SOL 4.95
SOL 4.96
SOL 4.97
SOL 4.98
X 0 = XS + 2Xm
= 0.4 + 2 (0.1) = 0.6 W/km
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 265
nodia
Correct option is (D).
Y22 = ?
I1 = V1 Y11 + (V1 - V2) Y12
SOL 4.100
...(1)
...(2)
P1 = 200 MW , P2 = 100 MW
SOL 4.101
www.nodia.co.in
www.gatehelp.com
Page 266
Power Systems
or
SOL 4.102
Chapter 4
1
= 40+ - 45c
Z2 = 1 =
C
0.025+45c
nodia
Steady state stability power limit = ?
Pm1 = EV = 1 # 1 = 6.25
X
0.12 + X
2
1
= 6.25
0.12 + 0.5X
&
X = 0.008 pu
If one of double circuit tripped than
Pm2 = EV
X
1
= 1#1 =
0.12 + 0.08
0.12 + X
Pm2 = 1 = 5 pu
0.2
SOL 4.103
SOL 4.104
220
3
4000
3 # 220
5000
3 # 220
If
5000
=
3
3 3 # 220
220
Vph
3
= 5000
Ia1
220 # 3
220
220 = 29.04 W
#
X1 + X 2 + X 0 =
3 # 5000
X1 + X 2 + X 0 =
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 267
X1 = X2 = 12.1 W
X 0 = 29.04 - 12.1 - 12.1 = 4.84 W
SOL 4.105
nodia
SOL 4.106
SOL 4.107
SOL 4.109
SOL 4.110
SOL 4.111
L =
C
11 # 10- 3
11.68 # 10- 9
= 306.88 W
Ideal power transfer capability
www.nodia.co.in
www.gatehelp.com
Page 268
Power Systems
Chapter 4
2
(800) 2
P =V =
= 2085 MW
306.88
Z0
SOL 4.112
nodia
km
SOL 4.113
SOL 4.114
...(1)
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 269
QR = PR tan f
= PR tan (cos- 1 f) = 50 tan (cos- 1 0.9)
= 24.21 MW
VS VR
A VR 2
sin (b - d) sin (b - a)
QR =
B
B
VS # 220
0.936 # (220) 2
=
sin (76.4c - d) sin 75.6c
142
142
...(2)
(24.21) 142 + 0.936 # 220 # 0.9685 = VS sin (76.4c - d)
220
from equation (1) & (2)
Same as
VS
= (215) 2 + (83.46) 2
VS =
SOL 4.115
53190.5716 = 230.63 kV
nodia
Correct option is (B).
I =
www.nodia.co.in
www.gatehelp.com
Page 270
Power Systems
Chapter 4
110 # 106
3 # 11 # 103
IB = 5773.67 Amp
Symmetrical RMS current = IB # Isc
= 5773.67 # 5.26 = 30369.50 Amp
& Irms = 30.37 kA
IB =
SOL 4.117
SOL 4.118
nodia
SOL 4.119
Where
www.nodia.co.in
www.gatehelp.com
Chapter 4
SOL 4.120
Power Systems
Page 271
P =?
a N = 120 f
P
P = 120 f = 120 # 50 = 24
250
N
No. of Poles
P = 24 Poles
SOL 4.121
SOL 4.122
nodia
Correct option is (B).
= 2 (n - ns) # 2 (n - ns)
= 2 (300 - 30) # 2 (300 - 30)
= 540 # 540
SOL 4.123
SOL 4.124
from eq(1),
P
E
V
Xd
...(1)
www.nodia.co.in
www.gatehelp.com
Page 272
Power Systems
Chapter 4
SOL 4.126
nodia
Receiving end voltage(VR) = sending end voltage(VS )
ohmic value of reactor = ?
We know
VS = AVR + BIR
VS = VR
VR = AVR + BIR
VR (1 - A) = BIR
VR = B = 200+90c
IR
1-A
1 - 0.9+0c
VR = 2000+90c
IR
The ohmic value of reactor = 2000 W
SOL 4.127
L;
C
0.4 # 10 = 28.284
=
0.5 # 10- 6
Surge impedance of overhead transmission line
Z2 = Z3 = L ;
L = 1.5 mm/km, C = 0.015 mF/km
C
1.5 # 10- 5 = 316.23
0.015 # 10- 6
Now the magnitude of voltage at junction due to surge is being given by as
V = 20 kV
Vl = 2 # V # Z2
Z 2 + Z1
Z2 = Z 3 =
3
= 2 # 20 # 10 # 316.23
316 + 28.284
= 36.72 kV
SOL 4.128
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
Page 273
nodia
= 220 + Line voltage
= 220 + 8.58
= 228.58 V
SOL 4.129
www.nodia.co.in
www.gatehelp.com
Page 274
Power Systems
Chapter 4
oa = I l cos f2 = I cos f1
In figure
I l cos 25.84c = 32
I l # 0.9 = 32
Il = 35.55
ac = 24 Amp.
ab = I l sin f2 = 35.55 sin 25.84c
ab = 15.49 Amp
nodia
Correct option is (D).
3
3
(11 # 10 )
= 11 # 10 # 11 # 10 cos (11.537c) 0.242
0.242
6
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
V
R
Sj0.2860W
Sj0.3408W
1
=
W8j0.2860 j0.3408 j0.2586 j0.2414B
S
6j (0.3408) + j0.2@ Sj0.2586W
Sj0.2414W
X
T
Given that we are required to change only Z22, Z23
j2 (0.3408) 2
So in equation (1)
= j0.2147
Zl22 =
j (0.5408)
j2 (0.3408) (0.2586)
= j0.16296
Zl23 =
0.5408
Page 275
...(1)
nodia
= j0.2586 - j0.16296 = j0.0956
SOL 4.132
Ea
Z 0 + Z 1 + Z 2 + 3Z n
0.1
=
j0.2 + j0.2 + j0.34 + j0.15
Ia1 =
=- j1.12 pu
generator MVA
IB =
3 generator kV
20 # 106
=
= 1750 Amp
3 # 6.6 # 103
Fault current
I f = (3Ia) IB
= 3 (- j1.12) (1750) =- j5897.6 Amp
Neutral Voltage
and
Vn = I f Zn
Zn = ZB # Z pu
(6.6) 2
=
0.05 = 0.1089 W
20 #
Vn = 5897.6 # 0.1089 = 642.2 V
SOL 4.133
www.nodia.co.in
www.gatehelp.com
Page 276
Power Systems
Chapter 4
0.3P1 - 0.4P2 = 10
P1 + P2 + P3 = 700
P1 + P2 + 300 = 700
...(1)
P1 + P2 = 400
From equation (1) and (2)
...(2)
P1 = 242.8 MW
P2 = 157.14 MW
SOL 4.134
SOL 4.135
nodia
Correct option is (C).
#d
dm
...(1)
www.nodia.co.in
www.gatehelp.com
Chapter 4
Power Systems
We know
Page 277
nodia
www.nodia.co.in
www.gatehelp.com
Page 278
Control Systems
Chapter 5
CHAPTER 5
CONTROL SYSTEMS
ONE MARK
nodia
(B) imaginary roots
EE SP 5.2
TWO MARKS
EE SP 5.3
For the given system, it is desired that the system be stable. The minimum value
of a for this condition is ______.
EE SP 5.4
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 279
nodia
The value of a is _____
bK
EE SP 5.5
EE SP 5.6
ONE MARK
4
The closed-loop transfer function of a system is T ^s h = 2
. The
+
s
0
.
4s + 4 h
^
steady state error due to unit step input is _____.
et tet
H
(D) >
0 et
TWO MARKS
EE SP 5.7
EE SP 5.8
5 ^s + 4h
s ^s + 0.25h^s2 + 4s + 25h
The values of the constant gain term and the highest corner frequency of the
Bode plot respectively are
(A) 3.2, 5.0
(B) 16.0, 4.0
(C) 3.2, 4.0
G ^s h =
www.nodia.co.in
www.gatehelp.com
Page 280
Control Systems
Chapter 5
nodia
(D) Not controllable and not observable
YEAR 2014 EE03
EE SP 5.10
ONE MARK
The signal flow graph of a system is shown below. U ^s h is the input and C ^s h is
the output.
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 281
(D) It is not possible to say whether or not the system is stable from the
information given
YEAR 2014 EE03
EE SP 5.12
TWO MARKS
nodia
(C) 1/s
EE SP 5.13
(D)
-s
s3 + s2 - s - 2
If u is unit step input, then the steady state error of the system is
(A) 0
(B) 1/2
(C) 2/3
(D) 1
EE SP 5.14
The maximum phase angle f m and the corresponding gain Gm respectively, are
(A) - 30c and 1.73 dB
(B) - 30c and 4.77 dB
(C) + 30c and 4.77 dB
(D) + 30c and 1.73 dB
YEAR 2013
EE SP 5.15
ONE MARK
Assuming zero initial condition, the response y ^ t h of the system given below to
a unit step input u ^ t h is
www.nodia.co.in
www.gatehelp.com
Page 282
Control Systems
(A) u ^ t h
2
(C) t u ^ t h
2
EE SP 5.16
Chapter 5
(B) tu ^ t h
(D) e-t u ^ t h
nodia
EE SP 5.17
Y ^s h
The signal flow graph for a system is given below. The transfer function
U ^s h
for this system is
s+1
5s2 + 6s + 2
(C) 2 s + 1
s + 4s + 2
(A)
EE SP 5.18
TWO MARKS
s+1
s 2 + 6s + 2
(D) 2 1
5s + 6s + 2
(B)
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 283
(A) 1
(B) 5
(C) 10
(D) 100
EE SP 5.19
nodia
The response y ^ t h to the unit step input is
(A) 1 - 1 e-2t
(B) 1 - 1 e-2t - 1 e-t
2 2
2
2
(C) e-2t - e-t
EE SP 5.20
(D) 1 - e-t
The system is
(A) controllable but not observable
(B) not controllable but observable
(C) both controllable and observable
TWO MARKS
EE SP 5.21
EE SP 5.22
www.nodia.co.in
www.gatehelp.com
Page 284
Control Systems
Chapter 5
EE SP 5.24
nodia
The phase of the above lead compensator is maximum at
(A) 2 rad/s
(B)
3 rad/s
(C)
6 rad/s
(D) 1/ 3 rad/s
YEAR 2011
EE SP 5.25
The frequency response of a linear system G (jw) is provided in the tubular form
below
G (jw)
1.3
+G (jw) - 130c
(A) 6 dB and 30c
(C) - 6 dB and 30c
EE SP 5.26
ONE MARK
1.2
1.0
0.8
0.5
0.3
- 140c
- 150c
- 160c
- 180c
- 200c
The steady state error of a unity feedback linear system for a unit step input is
0.1. The steady state error of the same system, for a pulse input r (t) having a
magnitude of 10 and a duration of one second, as shown in the figure is
(A) 0
(C) 1
(B) 0.1
(D) 10
www.nodia.co.in
www.gatehelp.com
Chapter 5
EE SP 5.27
Control Systems
Page 285
EE SP 5.28
TWO MARKS
The open loop transfer function G (s) of a unity feedback control system is given
as
K bs + 2 l
3
G (s) = 2
s (s + 2)
From the root locus, at can be inferred that when K tends to positive infinity,
(A) Three roots with nearly equal real parts exist on the left half of the s -plane
(B) One real root is found on the right half of the s -plane
nodia
(C) The root loci cross the jw axis for a finite value of K; K ! 0
(D) Three real roots are found on the right half of the s -plane
EE SP 5.29
YEAR 2010
EE SP 5.30
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Page 286
EE SP 5.31
Control Systems
Chapter 5
The
characteristic
equation
of
a
closed-loop
system
is
s (s + 1) (s + 3) k (s + 2) = 0, k > 0 .Which of the following statements is true ?
(A) Its root are always real
(B) It cannot have a breakaway point in the range - 1 < Re [s] < 0
(C) Two of its roots tend to infinity along the asymptotes Re [s] =- 1
(D) It may have complex roots in the right half plane.
EE SP 5.32
1
plotted in the
s (s + 1) (s + 2)
complex G (jw) plane (for 0 < w < 3) is
nodia
YEAR 2009
EE SP 5.33
EE SP 5.34
ONE MARK
The measurement system shown in the figure uses three sub-systems in cascade
whose gains are specified as G1, G2, 1/G3 . The relative small errors associated with
each respective subsystem G1, G2 and G3 are e1, e2 and e3 . The error associated
with the output is :
(A) e1 + e2 + 1
e3
(B) e1 e2
e3
(C) e1 + e2 - e3
(D) e1 + e2 + e3
The first two rows of Rouths tabulation of a third order equation are as follows.
s3 2 2
s2 4 4
This means there are
(A) Two roots at s = ! j and one root in right half s -plane
(B) Two roots at s = ! j2 and one root in left half s -plane
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 287
The polar plot of an open loop stable system is shown below. The closed loop
system is
nodia
(A) always stable
EE SP 5.36
10 (s + 5)
s (s + 2) (s + 25)
100 (s + 5)
(C)
s (s + 2) (s + 25)
(A)
1000 (s + 5)
s (s + 2) (s + 25)
80 (s + 5)
(D) 2
s (s + 2) (s + 25)
(B)
YEAR 2009
EE SP 5.37
TWO MARKS
The unit-step response of a unity feed back system with open loop transfer
function G (s) = K/ ((s + 1) (s + 2)) is shown in the figure. The value of K is
(A) 0.5
(C) 4
(B) 2
(D) 6
www.nodia.co.in
www.gatehelp.com
Page 288
EE SP 5.38
Control Systems
Chapter 5
The open loop transfer function of a unity feed back system is given by
G (s) = (e - 0.1s) /s . The gain margin of the is system is
(A) 11.95 dB
(B) 17.67 dB
(C) 21.33 dB
(D) 23.9 dB
EE SP 5.39
EE SP 5.40
nodia
The system transfer function is
(A) 2 s + 2
s + 5s - 6
(C) 2 2s + 5
s + 5s + 6
s+3
s + 5s + 6
(D) 2 2s - 5
s + 5s - 6
(B)
e3t e - 2t - e - 3t
(D) =
G
0
e - 2t
YEAR 2008
EE SP 5.41
ONE MARK
(D) e - t u (t)
YEAR 2008
EE SP 5.42
TWO MARK
www.nodia.co.in
www.gatehelp.com
Chapter 5
EE SP 5.43
Control Systems
Page 289
EE SP 5.44
nodia
This transfer function has
(A) Three poles and one zero
(B) Two poles and one zero
(C) Two poles and two zero
(D) One pole and two zeros
EE SP 5.45
www.nodia.co.in
www.gatehelp.com
Page 290
Control Systems
Chapter 5
EE SP 5.48
A unity feedback is provided to the above system G (s) to make it a closed loop
system as shown in figure.
nodia
For a unit step input r (t), the steady state error in the input will be
(A) 0
(B) 1
(C) 2
(D) 3
YEAR 2007
EE SP 5.49
ONE MARK
(A) Stable
(B) Unstable
(C) Conditionally stable
(D) Stable for input u1 , but unstable for input u2
YEAR 2007
EE SP 5.50
TWO MARKS
If x = Re [G (jw)], and y = Im [G (jw)] then for w " 0+ , the Nyquist plot for
G (s) = 1/s (s + 1) (s + 2) is
(A) x = 0
(B) x =- 3/4
(C) x = y - 1/6
(D) x = y/ 3
www.nodia.co.in
www.gatehelp.com
Chapter 5
EE SP 5.51
Control Systems
Page 291
EE SP 5.52
nodia
If the loop gain K of a negative feed back system having a loop transfer function
K (s + 3) / (s + 8) 2 is to be adjusted to induce a sustained oscillation then
(A) The frequency of this oscillation must be 4 3 rad/s
EE SP 5.53
with
(A) X = c0 s + c1, Y = 1/ (s2 + a0 s + a1), Z = b0 s + b1
(B) X = 1, Y = (c0 s + c1) / (s2 + a0 s + a1), Z = b0 s + b1
(C) X = c1 s + c0, Y = (b1 s + b0) / (s2 + a1 s + a0), Z = 1
(D) X = c1 s + c0, Y = 1/ (s2 + a1 s + a), Z = b1 s + b0
EE SP 5.54
Consider the feedback system shown below which is subjected to a unit step input.
The system is stable and has following parameters Kp = 4, Ki = 10, w = 500 and
www.nodia.co.in
www.gatehelp.com
Page 292
Control Systems
Chapter 5
(A) 1
(B) 0.25
(C) 0.1
(D) 0
nodia
R-L-C circuit shown in figure
EE SP 5.55
EE SP 5.56
ONE MARK
3 (s - 2)
,
4s2 - 2s + 1
the matrix A in the state space form Xo = AX + Bu is equal to
V
V
R
R
S0 1 0 W
S1 0 0 W
(B) S 0 0 1 W
(A) S 0 1 0 W
SS- 1 2 - 4 WW
SS- 1 2 - 4 WW
XV
TR
V X
RT
0
1
1
0
0
0
W
W
S
S
(C) S3 - 2 1 W
(D) S 0 0 1 W
SS1 - 2 4 WW
SS- 1 2 - 4 WW
X in 4 Volumes by RK Kanodia
X
T
T
Shop GATE Electrical
at maximum
discount at
EE SP 5.57
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 293
YEAR 2006
EE SP 5.58
TWO MARKS
nodia
(A) (1) only
(C) all, except (3)
EE SP 5.59
www.nodia.co.in
www.gatehelp.com
Page 294
EE SP 5.60
Control Systems
Chapter 5
nodia
YEAR 2005
ONE MARK
EE SP 5.61
A system with zero initial conditions has the closed loop transfer function.
s2 + 4
T (s) =
(s + 1) (s + 4)
The system output is zero at the frequency
(A) 0.5 rad/sec
(B) 1 rad/sec
(C) 2 rad/sec
(D) 4 rad/sec
EE SP 5.62
Figure shows the root locus plot (location of poles not given) of a third order
system whose open loop transfer function is
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
(A) K3
s
(C)
EE SP 5.63
Page 295
K
s (s + 1)
K
(D)
2
s (s - 1)
(B)
K
s (s + 1)
2
nodia
The gain margin of a unity feed back control system with the open loop transfer
(s + 1)
function G (s) =
is
s2
(A) 0
(B) 1
2
(C) 2
(D) 3
YEAR 2005
EE SP 5.64
TWO MARKS
(B) K > 1
(D) K < - 1
When subject to a unit step input, the closed loop control system shown in the
figure will have a steady state error of
(A) - 1.0
(C) 0
EE SP 5.66
(B) - 0.5
(D) 0.5
In the G (s) H (s)-plane, the Nyquist plot of the loop transfer function
G (s) H (s) = pes passes through the negative real axis at the point
(A) (- 0.25, j0)
(B) (- 0.5, j0)
(C) 0
(D) 0.5
- 0.25s
EE SP 5.67
If the compensated system shown in the figure has a phase margin of 60c at the
www.nodia.co.in
www.gatehelp.com
Page 296
Control Systems
Chapter 5
(A) 0.366
(B) 0.732
(C) 1.366
(D) 2.738
EE SP 5.68
nodia
The state transition matrix
- 3t
1
1
)
3 (1 - e
(A) =
G
- 3t
0
e
1
(C) >
0
EE SP 5.69
1
3
(e3 - t - e- 3t)
H
e- 3t
(e- t - e- 3t)
H
e- t
1
(B) >
0
1
3
1
(D) >
0
(1 - e- t)
H
e- t
1 - e-t
(B) X (t) = = - 3t G
3e
t - e 3t
(C) X (t) = = - 3t G
3e
t - e - 3t
(D) X (t) = = - t G
e
YEAR 2004
ONE MARK
EE SP 5.70
The Nyquist plot of loop transfer function G (s) H (s) of a closed loop control
system passes through the point (- 1, j 0) in the G (s) H (s)plane. The phase
margin of the system is
(A) 0c
(B) 45c
(C) 90c
(D) 180c
EE SP 5.71
5
s (s + 3s + 2)
where F (s) is the Laplace transform of the of the function f (t). The initial value
of f (t) is equal to
(A) 5
(B) 25
(C)
EE SP 5.72
5
3
(D) 0
For a tachometer, if q (t) is the rotor displacement in radians, e (t) is the output
voltage and Kt is the tachometer constant in V/rad/sec, then the transfer
E (s)
will be
function,
Q (s)
(A) Kt s2
(B) Kt s
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
(C) Kt s
Page 297
(D) Kt
YEAR 2004
EE SP 5.73
For the equation, s3 - 4s2 + s + 6 = 0 the number of roots in the left half of s
-plane will be
(A) Zero
(B) One
(C) Two
EE SP 5.74
EE SP 5.76
C (s)
is equal to
R (s)
nodia
2
(B) s + s2 + 1
s
(D) 2 1
s +s+1
o = AX where
The state variable description of a linear autonomous system is, X
X is the two dimensional state vector and A is the system matrix given by
0 2
A = = G. The roots of the characteristic equation are
2 0
(A) - 2 and + 2
(B) - j2 and + j2
(C) - 2 and - 2
(D) + 2 and + 2
The block diagram of a closed loop control system is given by figure. The values
of K and P such that the system has a damping ratio of 0.7 and an undamped
natural frequency wn of 5 rad/sec, are respectively equal to
(D) Three
2
(A) s +2 1
s
2
(C) s + s + 1
s
EE SP 5.75
TWO MARKS
The unit impulse response of a second order under-damped system starting from
rest is given by c (t) = 12.5e - 6t sin 8t, t $ 0 . The steady-state value of the unit
step response of the system is equal to
(A) 0
(B) 0.25
(C) 0.5
(D) 1.0
www.nodia.co.in
www.gatehelp.com
Page 298
EE SP 5.78
Control Systems
In the system shown in figure, the input x (t) = sin t . In the steady-state, the
response y (t) will be
1 sin (t - 45c)
2
(C) sin (t - 45c)
(A)
EE SP 5.79
Chapter 5
1 sin (t + 45c)
2
(D) sin (t + 45c)
(B)
The open loop transfer function of a unity feedback control system is given as
1.
G (s) = as +
2
s
The value of a to give a phase margin of 45c is equal to
(B) 0.441
(A) 0.141
(C) 0.841
(D) 1.141
nodia
YEAR 2003
EE SP 5.80
EE SP 5.81
ONE MARK
EE SP 5.82
2
A second order system starts with an initial condition of = G without any external
3
e - 2t 0
input. The state transition matrix for the system is given by =
G. The state
0 e-t
of the system at the end of 1 second is given by
0.271
(A) =
1.100G
0.135
(B) =
0.368G
0.271
(C) =
0.736G
0.135
(D) =
1.100 G
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
YEAR 2003
EE SP 5.83
Page 299
TWO MARKS
EE SP 5.84
The block diagram shown in figure gives a unity feedback closed loop control
system. The steady state error in the response of the above system to unit step
input is
nodia
(A) 25%
(B) 0.75 %
(C) 6%
(D) 33%
EE SP 5.85
The roots of the closed loop characteristic equation of the system shown above
(Q-5.55)
(A) - 1 and - 15
(B) 6 and 10
(C) - 4 and - 15
(D)- 6 and - 10
EE SP 5.86
www.nodia.co.in
www.gatehelp.com
Page 300
Control Systems
EE SP 5.87
Chapter 5
- K - BJ
(B) = LJ
G
0
1
1
0
(D) =- B - K G
J
LJ
2
The loop gain GH of a closed loop system is given by the following expression
K
The value of K for which the system just becomes unstable is
s (s + 2) (s + 4)
(A) K = 6
(B) K = 8
(C) K = 48
(D) K = 96
EE SP 5.88
nodia
The asymptotic Bode plot of the transfer function K/ [1 + (s/a)] is given in figure.
The error in phase angle and dB gain at a frequency of w = 0.5a are respectively
(B) 5.7c, 3 dB
(D) 5.7c, 0.97 dB
The block diagram of a control system is shown in figure. The transfer function
G (s) = Y (s) /U (s) of the system is
1
18^1 + 12s h^1 + s3 h
1
(C)
27^1 + 12s h^1 + s9 h
(A)
1
27^1 + s6 h^1 + s9 h
1
(D)
27^1 + s9 h^1 + s3 h
(B)
***********
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 301
SOLUTION
SOL 5.1
nodia
Option (D):
Again, the system has only negative real roots, as shown below. So, the root
location diagram does not satisfy the symmetrical condition.
SOL 5.2
www.nodia.co.in
www.gatehelp.com
Page 302
Control Systems
Chapter 5
As the root locus have poles s =- 1, - 2 and root lies in even multiple of poles,
so it is converse of the main transfer function. Hence, gain should be negative, i.e.
-K
G ^s h H ^s h =
^s + 1h^s + 2h
This is open loop transfer function and closed loop transfer function is given by
G ^s h H ^s h
C ^s h
=
1 + G ^s h H ^s h
R ^s h
-K
^s + 1h^s + 2h
=
-K
1+
^s + 1h^s + 2h
-K
=
^s + 1h^s + 2h - K
SOL 5.3
nodia
Correct answer is 0.618.
The Block diagram of given system is
^s + ah
s + ^1 + ah s + ^a - 1h s + ^1 - ah
So, we obtain the character equation as
1 + G ^s h H ^s h = 0
G ^s h H ^s h =
or
or
or
or
For
^s + ah
=0
s + ^1 + ah s + ^a - 1h s + ^1 - ah
s3 + ^1 + ah s2 + ^a - 1h s + ^1 - ah + ^s + ah = 0
s3 + ^1 + ah s2 + ^a - 1 + 1h s + 1 - a + a = 0
s3 + ^1 + ah s2 + as + 1 = 0
the characteristic equation, we form the Rouths array as
1
a
s3
2
1+a
1
s
1
1
+
a
a
^
h
0
s1
1+a
0
1
s
1+
1+a > 0
a >-1
a ^1 + ah - 1
>0
1+a
or
a ^1 + ah - 1 > 0
Solving the inequality, we obtain the roots
a = -1 2
5 , -1 +
2
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 303
nodia
Also, from the given transfer function, we have
K ^1 + 0.55h^1 + as h
G ^s h =
s a1 + s k^1 + bs ha1 + s k
8
36
K ^1 - s/2h^1 + s/ a1 h
=
s a1 + s kb1 + s la1 + s k
8
36
1/b
The first slope - 6 dB/octave is due to one pole that is 1/s
Then, slope 0 dB/octave is due to addition of a zero in T.F. ^1 + s/2h.
Again, + 6 dB/octave slope is due to one zero at corner frequency wC = 4 .
Comparing it to the transfer function, we get
^1 + as h = ^1 + s/4h
or
a = 1/4
Similarly, at wC = 24 , there is an addition of a pole ^- 6 dB/octaveh. So, we get
^1 + bs h = ^1 + s/24h
or
b = 1
24
From the shown Bode plot, we observe that if we extended the slope - 6 dB/octave
, it meets the frequency axis at wC = 8 . So, we have
0 = 20 log KX
s w =8
or
1=K
8
C
or
K =8
Therefore, we obtain the desired value as
a = 1/4 = 24 = 0.75
1
4#8
bK
24 # 8
www.nodia.co.in
www.gatehelp.com
Page 304
SOL 5.5
Control Systems
Chapter 5
Correct answer is 0.
Closed loop T.F.
T ^s h =
G ^s h
1 + G ^s h
1 + G ^s h
G ^s h
1+ 1
G ^s h
1
G ^s h
4
s2 + 0.4s + 4
4
s2 + 0.4s + 4
2
= s + 0.4s + 4
4
2
= s + 0.4s + 4
4
2
= s + 0.4s + 4 - 4
4
4
G ^s h =
s ^s + 0.4h
K p = lim G ^s h
s"0
nodia
Error constant,
4
=3
s ^s + 0.4h
ess = 1 = 0
1 + Kp
= lim
s"0
SOL 5.6
1 0 x1
1
xo1
> o H = >1 1H>x H + >1H U
x2
2
Y = AX + BU
f ^ t h = L-1
So,
^s - 1h2
s-1
e 0
f ^ t h = > t tH
te e
SOL 5.7
Correct answer is 5.
G ^s h =
K
s ^s + 2h^s2 + 2s + 2h
G ^s h
Closed loop T.F. =
1 + G ^s h
K
s ^s + 2h^s2 + 2s + 2h
=
K
1+
s ^s + 2h^s2 + 2s + 2h
Closed loop T.F. =
K
s ^s + 2h^s2 + 2s + 2h + K
Characteristics equation
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 305
s ^s + 2h^s2 + 2s + 2h + K = 0
^s2 + 2s h^s2 + 2s + 2h + K = 0
s 4 + 4s3 + 4s2 + 4s + K = 0
Routh array
s4
s2
s1
20 - 4k
5
s0
nodia
For marginally stable
20 - 4k = 0
k =5
or
SOL 5.8
5 ^s + 4h
^s + 0.25h^s2 + 4s + 25h
^1 + s/4h
= 5#4
0.25 # 25
s
4s s2
a1 + 0.25 kb1 + 25 + 25 l
^1 + s/4h
= 3.2 #
s
4s s2
a1 + 0.25 kb1 + 25 + 25 l
G ^s h =
K = 3.2
wH = 5
or,
So,
CM = 6Q PQ@
-1 1 0
1
H> H = > H
PQ = >
0 -3 1
-3
0 1
H
CM = >
1 -3
det ^CM h =- 1 ! 0
So, the system is controllable.
For observalibility, we check observalibility matrix
C
OM = > H
CA
or
R
OM = > H
RP
-1 1
H = 80 - 3B
RP = 80 1B>
0 -3
www.nodia.co.in
www.gatehelp.com
Page 306
Control Systems
So,
Chapter 5
0 1
OM = >
0 - 3H
det ^OM h = 0
So, the system is not observable.
SOL 5.10
nodia
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 307
h1 = b 1
and
h 0 = b 0 - b 1 a1
Substituting it in equation (1), we get
T.F. =
or
SOL 5.11
G ^s h =
b 0 + b1 s
s + a1 s + a 0
2
C ^s h
= 2 b 0 + b1 s
V ^s h
s + a1 s + a 0
G ^s h H ^s h < 1
So, we may easily conclude that the Nyquist-plot intersect the negative real axis
between 0 and - 1 point, i.e. the Nyquist plot does not enclose the point ^- 1, 0h
or in other words number of encirclements is zero.
nodia
N =0
From Nyquist criterion, we have the relation
...(1)
...(2)
N = P-Z
where P is the number of open loop poles in right-half of s -plane and Z is the
number of closed loop poles (roots of characteristic equation) in right-half of s
-plane. Since, for a stable closed loop system, we must have
Z =0
Thus, from equations (1), (2), and (3), we get
...(3)
P =0
i.e. no poles of open loop system lies in right-half of s -plane, or in other words,
all the poles lies in left-half of s -plane.
Hence, the given system is stable if all poles of G ^s h H ^s h are in left half of the
s -plane.
SOL 5.12
www.nodia.co.in
www.gatehelp.com
Page 308
Control Systems
Chapter 5
G ^s h
1 + G ^s h
Again, we obtain the closed loop transfer function for the dotted box as
sG ^s h
^s + 1h G ^s h + 1
So, the block diagram reduces to
nodia
Thus, we have the overall transfer function as
G ^s h
^s + 1h G ^s h + 1
T.F. 2
^s + s h G ^s h + s + G ^s h
s 6^s + 1h G ^s h + 1@
sG ^s h
= 2
^s + s + 1h G ^s h + s
ALTERNATIVE METHOD :
For the given block diagram, we draw the signal flow graph as
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 309
L13 = G ^s h s ^- 1h
and
=- sG ^s h
So, the transfer function is obtained as
pT
T.F. = 1 1
T
=
G ^s h^1 h
G ^s h
1 - ;- G ^s h - sG ^s hE
s
sG ^s h
s + G ^s h^s2 + s + 1h
For any value of G ^s h given in the options, the desired system transfer function
can not be achieved.
T.F. =
SOL 5.13
nodia
Correct option is (A).
We have the state space equation,
0 1 x1
0
xo1
> H =>
H> H + > H u
o
1
- 1 - 1 x2
x2
x1
y = 81 0B> H
x2
From the state-space equation, we obtain the matrices
0 1
H
A =>
-1 -1
0
B => H
1
C = 81 0B
So, we obtain
S 0
-1
H->
H=>
H
6SI - A@ = >
0 S
-1 -1
1 S+1
Therefore,
S+1 1
1
>
H
S2 + S + 1 - 1 S
Hence, we obtain the transfer function of the system as
T.F. = C [SI - A] -1 B
S+1 1 0
>
H> H
= 81 0B 2 1
S + S + 1 -1 S 1
1
81 0B> H
= 2 1
S
S +S+1
= 2 1
S +S+1
This is the closed loop transfer function,
So we get
G (S )
= 2 1
1 + G (S )
S +S+1
G (S ) = 2 1
S +S
Thus, the steady state error for unit step input is
6SI - A@-1 =
www.nodia.co.in
www.gatehelp.com
Page 310
Control Systems
Chapter 5
1
1 + lim G (S )
S"0
1
=
1 + lim 2 1
d"0 d + d
= 1 =0
1+3
eSS =
SOL 5.14
nodia
Since, slope at 1/3 is increasing and slope at 1 is decreasing. So, we can write
frequency response function as
1
b 3 + jw l
G ^ jwh =
^1 + jwh
jw
1
3 c1 +
1/3 m
or
G ^ jwh =
^1 + jwh
a
^1 + jw/a h
or
G ^ jwh = b
^1 + jw/b h
Therefore, the maximum phase angle is
f m = 90c - 2 tan-1 _ a/b i
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
u^t h =
dy ^ t h
dt
U ^s h = SY ^s h - y ^0 h
U ^s h = s c 12 m - y ^0 h
s
1
U ^s h =
s
&
&
or,
Page 311
^y ^0 h = 0h
SOL 5.16
nodia
Correct option is (B).
From the given plot, we obtain the slope as
Slope =
20 log G2 - 20 log G1
log w2 - log w1
20 log G2 =- 8 dB
20 log G1 = 32 dB
w1 = 1 rad/s
w2 = 10 rad/s
and
or,
k = 10 = 39.8
Hence, the Transfer function is
G ^s h = k2 = 392.8
s
s
SOL 5.17
20
www.nodia.co.in
www.gatehelp.com
Page 312
Control Systems
Chapter 5
L1 = ^- 4h^1 h =- 4
L2 = ^- 4h^s-1h = 4s-1
L 3 = ^- 2h^s-1h^s-1h =- 2s-2
L 4 = ^- 2h^s-1h^1 h =- 2s-1
As all the loop L1, L2, L 3 and L 4 are touching to each other so,
D = 1 - ^L1 + L2 + L 3 + L 4h
= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h
= 5 + 6s-1 + 2s-2
From Masons gain formulae
Y ^s h
s-2 + s-1
s+1
= SPk Dk =
2
-1
-2 =
D
U ^s h
5s + 6s + 2
5 + 6s + 2s
SOL 5.18
nodia
Correct option is (C).
Given, open loop transfer function
G ^s h = 10Ka = Ka 1
1 + 10s s + 10
By taking inverse Laplace transform, we have
g ^ t h = e- t
1
10
Comparing with standard form of transfer function, Ae-t/t , we get the open
loop time constant,
tol = 10
Now, we obtain the closed loop transfer function for the given system as
G ^s h
10Ka
=
H ^s h =
1 + G ^s h 1 + 10s + 10Ka
Ka
=
s + ^Ka + 101 h
By taking inverse laplace transform, we get
h ^ t h = ka .e-^k + ht
So, the time constant of closed loop system is obtained as
tcl = 1 1
ka + 10
or,
tcl = 1
ka
a
1
10
(approximately)
Now, given that ka reduces open loop time constant by a factor of 100. i.e.,
tcl = tol
100
1 = 10
or,
100
ka
or,
SOL 5.19
ka = 10
....(1)
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 313
x1
y = 61 0@> H
x2
and
....(2)
nodia
....(3)
y = x1
Taking Laplace transform both the sides,
Y = XL
or,
or,
1
s ^s + 2h
Y = 1 ;1 - 1 E
2 s s+2
Y =
SOL 5.20
^for t > 0h
1
B = > H; C = 61 0@
1
www.nodia.co.in
www.gatehelp.com
Page 314
SOL 5.21
Control Systems
Chapter 5
SOL 5.22
nodia
Correct option is (A).
Y (s) =
K (s + 1)
[R (s) - Y (s)]
s + as2 + 2s + 1
3
K (s + 1)
s + as2 + 2s + 1E
K (s + 1)
R (s)
= 3
s + as2 + 2s + 1
Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s)
Transfer Function,
Y (s)
K (s + 1)
H (s) =
=
R (s) s3 + as2 + s (2 + k) + (1 + k)
Routh Table :
Y (s) ;1 +
For oscillation,
a (2 + K) - (1 + K)
=0
a
a = K+1
K+2
Auxiliary equation
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 315
A (s) = as2 + (k + 1) = 0
s2 =- k + 1
a
s2 = - k + 1 (k + 2) =- (k + 2)
(k + 1)
s = j k+2
jw = j k + 2
w =
(Oscillation frequency)
k =2
a = 2 + 1 = 3 = 0.75
2+2 4
and
SOL 5.23
k+2 = 2
nodia
jw + a
GC (s) = s + a =
s+b
jw + b
Phase lead angle,
f = tan-1 a w k - tan-1 a w k
a
b
Jw - wN
-1 K a
b O = tan-1 w (b - a)
= tan
c ab + w 2 m
2
O
KK
1+ w O
ab P
L
For phase-lead compensation f > 0
b-a > 0
b >a
Note: For phase lead compensator zero is nearer to the origin as compared to
pole, so option (C) can not be true.
SOL 5.24
SOL 5.25
2 rad/ sec
www.nodia.co.in
www.gatehelp.com
Page 316
Control Systems
Chapter 5
frequency ( wg ). Gain cross over frequency is the frequency at which gain is unity.
From the table it is clear that G (jwg) = 1, at which phase angle is - 150c
fPM = 180c + +G (jwg)
= 180 - 150 = 30c
SOL 5.26
nodia
1 + G (0) = 10
Given input
or
G (0) = 9
r (t) = 10 [m (t) - m (t - 1)]
R (s) = 10 :1 - 1 e-sD
s s
-s
= 10 :1 - e D
s
SOL 5.28
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 317
(2q + 1) 180c
, q = 0, 1
n-m
(2 # 0 + 1) 180c
(1)
= 90c
(3 - 1)
(2 # 1 + 1) 180c
(2)
= 270c
(3 - 1)
(4) The two asymptotes intersect on real axis at centroid
/ Poles - / Zeroes
x =
n-m
- 2 - b- 2 l
3
=
=- 2
3-1
3
(5) Between two open-loop poles s = 0 and s =- 2 there exist a break away
point.
s2 (s + 2)
K =2
bs + 3 l
dK = 0
ds
nodia
s =0
Root locus is shown in the figure
Three roots with nearly equal parts exist on the left half of s -plane.
SOL 5.29
1
s (s + 1 + K )
Y (s)
=
1
R (s) 1 +
s (s + 1 + K )
1
= 2
s + s (1 + K ) + 1
This is a second order system transfer function, characteristic equation is
www.nodia.co.in
www.gatehelp.com
Page 318
Control Systems
Chapter 5
s2 + s (1 + K) + 1 = 0
Comparing with standard form
We get
s2 + 2xwn s + wn2 = 0
x = 1+K
2
M p = e- px/
Peak overshoot,
1 - x2
nodia
A - lI = 0
-1 2
l 0
A - lI = >
->
H
0 2
0 lH
-1 - l 2
=>
0
2 - lH
A - lI = (- 1 - l) (2 - l) - 2 # 0 = 0
l1, l2 =- 1, 2
&
Since eigen values of the system are of opposite signs, so it is unstable
Controllability :
-1 2
0
, B=> H
A =>
H
0 2
1
2
AB = > H
2
0 2
[B: AB] = > H
1 2
Y 0
6B: AB@ =
So it is controllable.
SOL 5.31
3+K
s2
2K
s1
s0
4(3 + K) - 2K (1)
4
12 + 2K
4
K>0
>0
2K
There is no sign change in the first column of routh table, so no root is lying in
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 319
4.
Breakaway point lies between open loop poles of the system. Here
breakaway point lies in the range - 1 < Re [s] < 0 .
Asymptotes meet on real axis at a point C , given by
nodia
5.
C =
n-m
(0 - 1 - 3) - (- 2)
=
3-1
=- 1
As no. of poles is 3, so two root loci branches terminates at infinity along
asymptotes Re (s) =- 1
SOL 5.32
1
s (s + 1) (s + 2)
1
G (jw) =
jw (jw + 1) (jw + 2)
1
G (jw) =
2
w w + 1 w2+ 4
+G (jw) =- 90c - tan- 1 (w) - tan- 1 (w/2)
G (s) =
In nyquist plot
For
w = 0, G (jw) = 3
+G (jw) =- 90c
For w = 3, G (jw) = 0
+G (jw) =- 90c - 90c - 90c
=- 270c
Intersection at real axis
1
G (jw) =
jw (jw + 1) (jw + 2)
1
=
jw (- w2 + j3w + 2)
- 3w2 - jw (2 - w2)
1
=
#
- 3w2 + jw (2 - w2) - 3w2 - jw (2 - w2)
- 3w2 - jw (2 - w2)
=
9w4 + w2 (2 - w2) 2
2
jw (2 - w2)
= 4 -23w
2 2 4
9w + w (2 - w )
9w + w2 (2 - w2) 2
www.nodia.co.in
www.gatehelp.com
Page 320
Control Systems
Chapter 5
Im [G (jw)] = 0
w (2 - w2)
=0
9w4 + w2 (2 - w2)
2 - w2 = 0 & w = 2 rad/sec
2 rad/sec, magnitude response is
1
G (jw) at w = 2 =
2 2+1 2+4
=1<3
6 4
At real axis
So,
At w =
SOL 5.33
SOL 5.34
nodia
Correct option is (D).
Given Rouths tabulation.
s3
s2
s1
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 321
nodia
Transfer function
K (s + 5)
s (s + 2) (s + 25)
Constant term can be obtained as.
T (s) =
T (jw) at w = 0.1 = 80
So,
80 = 20 log
K (5)
(0.1) 2 # 50
K = 1000
therefore, the transfer function is
1000 (s + 5)
T (s) = 2
s (s + 2) (s + 25)
SOL 5.37
www.nodia.co.in
www.gatehelp.com
Page 322
Control Systems
Chapter 5
2 = 0.5 + 0.25K
K = 1.5 = 6
0.25
SOL 5.38
- j0.1w
jw
Phase cross over frequency can be calculated as,
+G (jwp) =- 180c
180
b- 0.1wp # p l - 90c =- 180c
0.1wp # 180c = 90c
p
0.1wp = 90c # p
180c
nodia
wp = 15.7 rad/sec
1
G (jwp) o
= 20 log = 11 G
^ 15.7 h
= 20 log 15.7
= 23.9 dB
= 20 log e
SOL 5.39
...(1)
...(2)
Y (s) = X1 (s)
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
So,
Y (s) = U (s)
Page 323
(2s + 5)
(s + 2) (s + 3)
nodia
f (t) = L- 1 6F (s)@
F (s) = (sI - A) - 1
s 0
-3 1
(sI - A) = > H - >
0 s
0 - 2H
s + 3 -1
(sI - A) = >
0 s + 2H
(sI - A) - 1 =
So
SOL 5.41
s+2 1
1
> 0 s + 3H
(s + 3) (s + 2)
V
R
1
W
S 1
(s + 3) (s + 3) (s + 2)W
S
-1
F (s) = (sI - A) = S
W
1
S 0
(s + 2) W
X
T
e- 3t e- 2t - e- 3t
-1
f (t) = L [F (s)] = >
H
0
e- 2t
1
s+1
G (s) =
Input
www.nodia.co.in
www.gatehelp.com
Page 324
Control Systems
Chapter 5
R (s) = L [d (t - 1)] = e- s
Output is given by
Y (s) = R (s) G (s)
-s
= 2 e
s + 3s + 2
Steady state value of output
lim y (t) = lim sY (s)
t"3
s"0
se- s
s " 0 s + 3s + 2
=0
= lim
SOL 5.43
nodia
qC = tan- 1 (w) - tan- 1 a w k
10
Jw - w N
-1 K
10 O
= tan
2
O
KK
w
1+ O
10
P
L
= tan- 1 c 9w 2 m > 0 (Phase lead)
10 + w
1
SOL 5.44
SOL 5.45
30
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
s2
13
s1
(13 # 30) - K
13
K
s0
Page 325
SOL 5.46
nodia
100
s + 20s + 100
Characteristic equation of the system is given by
H (s)) =
s2 + 20s + 100 = 0
wn2 = 100 & wn = 10 rad/sec.
2xwn = 20
or
x = 20 = 1
2 # 10
(x = 1) so system is critically damped.
SOL 5.47
www.nodia.co.in
www.gatehelp.com
Page 326
Control Systems
Chapter 5
V
R
1 W
S1
s s (s + 2)W 0
G (s) = C [sI - A] - 1 B = 81 0BSS
1 W>1H
0
S (s + 2) W
TR 1 V X
W
S
s (s + 2)W
S
= 81 0BS 1 W
S (s + 2) W
X
1T
=
s (s + 2)
SOL 5.48
nodia
Here
So,
SOL 5.49
System response is
(s - 1)
(s + 2)
H1 (s) =
(s - 1) 1
1+
(s + 2) (s - 1)
(s - 1)
(s + 3)
Poles of the system is lying at s =- 3 (negative s -plane) so this is stable.
For input u2 the system is (u1 = 0)
=
System response is
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 327
1
(s - 1)
H2 (s) =
(s - 1)
1+ 1
(s - 1) (s + 2)
(s + 2)
=
(s - 1) (s + 3)
One pole of the system is lying in right half of s -plane, so the system is unstable.
SOL 5.50
nodia
By simplifying
- jw
1 - jw
2 - jw
1
1
G (jw) = c 1 #
#
#
m
c
m
c
jw - jw 1 + jw 1 - jw 2 + jw
2 - jw m
jw 1 - jw 2 - j w
= c- 2 mc
w 1 + w2 mc 4 + w2 m
- jw (2 - w2 - j3w)
= 2
w (1 + w2) (4 + w2)
jw (w2 - 2)
- 3w2
= 2
2
2 + 2
w (1 + w ) (4 + w ) w (1 + w2) (4 + w2)
G (jw) = x + iy
x = Re [G (jw)] w " 0 = - 3 =- 3
1#4
4
+
SOL 5.51
wp
9k
J w + wp N
p
9 O
90c = tan KK
2 O
K 1 - wp O
9 P
L
w2p
1=0
9
-1
So,
wp = 3 rad/sec.
(wg) compensated = 3 rad/sec.
www.nodia.co.in
www.gatehelp.com
Page 328
Control Systems
Chapter 5
nodia
900 GC (jwg)
wg
wg2 + 1 wg2 + 81
=1
GC (jwg) = 3 9 + 1 9 + 81
900
= 3 # 30 = 1
900
10
in dB GC (wg) = 20 log b 1 l
10
=- 20 dB (attenuation)
SOL 5.52
64 + 3K
1
16 + K
64 + 3K
s2 + 64 + 3 # (- 16) = 0
s2 + 64 - 48 = 0
s2 =- 16 & jw = 4j
w = 4 rad/sec
SOL 5.53
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 329
c 0 P + c1 P
s
s2
T.F =
1 + a1 + a20 - Pb2 0 - Pb1
s
s
s
s
(c 0 + c1 s) P
(s2 + a1 s + a 0) - P (b 0 + sb1)
(c 0 + c1 s) P
2
^s + a1 s + a 0h
=
P (b + sb1)
1- 2 0
s + a1 s + a 0
from the given reduced form transfer function is given by
T.F = XYP
1 - YPZ
by comparing above two we have
=
nodia
X = (c 0 + c1 s)
1
Y = 2
s + a1 s + a 0
Z = (b 0 + sb1)
SOL 5.54
w2
G (s) = b Ki + K p le 2
s
s + 2xws + w2 o
H (s) = 1 (Unity feed back)
So,
SOL 5.55
R
V
sb 1 l
S
W
s
S
Wb Ki l
Z = lim
2
s"0S
K
w
W s
i
S1 + b s + K p l (s2 + 2xws + w2) W
T
X
K
i
= Ki = 1
= lim
2
s"0
>s + (Ki + K p s) 2 w
H Ki
2
(s + 2xws + w )
www.nodia.co.in
www.gatehelp.com
Page 330
Control Systems
% peak overshoot = e
SOL 5.57
C
L
- px
1 - x2
# 100
nodia
=e
SOL 5.56
Chapter 5
- p # 0.5
1 - (0.5) 2
# 100 = 16%
Correct option is ( ).
sn + an - 1 sn - 1 + ... + a1 s + a 0 = 0
in its state variable representation matrix A is given as
V
R
1
0 g
0 W
S 0
S 0
0
1 g
0 W
A =S
W
Sh h h h h W
S- a 0 - a1 - a2 g - an - 1W
X
T
Characteristic equation of the system is
SOL 5.58
SOL 5.59
4s2 - 2s + 1 = 0
So, a2 = 4, a1 =- 2, a 0 = 1
R 0
1
0 VW RS 0 1 0 VW
S
A =S 0
0
1 W=S 0 0 1 W
SS- a - a - a WW SS- 1 2 - 4WW
0
1
2
T
X T
X
Correct option is (A).
In the given options only in option (A) the nyquist plot does not enclose the unit
circle (- 1, j0), So this is stable.
Correct option is (A).
Given function is,
10 4 (1 + jw)
H (jw) =
(10 + jw) (100 + jw) 2
Function can be rewritten as,
10 4 (1 + jw)
H (jw) =
2
10 91 + j w C 10 4 91 + j w C
10
100
0.1 (1 + jw)
=
w
w 2
a1 + j 10 ka1 + j 100 k
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 331
The system is type 0, So, initial slope of the bode plot is 0 dB/decade.
Corner frequencies are
w1 = 1 rad/sec
w 2 = 10 rad/sec
w 3 = 100 rad/sec
As the initial slope of bode plot is 0 dB/decade and corner frequency w1 = 1 rad/
sec, the Slope after w = 1 rad/sec or log w = 0 is(0 + 20) =+ 20 dB/dec.
After corner frequency w2 = 10 rad/sec or log w2 = 1, the Slope is (+ 20 - 20) = 0
dB/dec.
Similarly after w3 = 100 rad/sec or log w = 2 , the slope of plot is (0 - 20 # 2) =- 40
dB/dec.
SOL 5.60
nodia
Correct option is (B).
Given characteristic equation.
(s2 - 4) (s + 1) + K (s - 1) = 0
K (s - 1)
or
1+ 2
=0
(s - 4) (s + 1)
So, the open loop transfer function for the system.
K (s - 1)
,
no. of poles n = 3
G (s) =
(s - 2) (s + 2) (s + 1)
no of zeroes m = 1
Steps for plotting the root-locus
(1) Root loci starts at s = 2, s =- 1, s =- 2
(2) n > m , therefore, number of branches of root locus b = 3
(3) Angle of asymptotes is given by
(2q + 1) 180c
, q = 0, 1
n-m
(2 # 0 + 1) 180c
(1)
= 90c
(3 - 1)
(2 # 1 + 1) 180c
(2)
= 270c
(3 - 1)
(4) The two asymptotes intersect on real axis at
/ Poles - / Zeroes
x =
n-m
(- 1 - 2 + 2) - (1)
=
=- 1
3-1
(5) Between two open-loop poles s =- 1 and s =- 2 there exist a break away
point.
(s2 - 4) (s + 1)
K =(s - 1)
dK = 0
ds
s =- 1.5
SOL 5.61
www.nodia.co.in
www.gatehelp.com
Page 332
Control Systems
T (jw) =
Chapter 5
(jw) 2 + 4
(jw + 1) (jw + 4)
4 - w2
=0
^ jw + 1h (jw + 4)
4 - w2 = 0
w2 = 4
& w = 2 rad/sec
SOL 5.62
SOL 5.63
nodia
Correct option is (A).
Open loop transfer function is.
(s + 1)
G (s) =
s2
jw + 1
G (jw) =
- w2
Phase crossover frequency can be calculated as.
+G (jwp) =- 180c
tan- 1 (wp) =- 180c
wp = 0
Gain margin of the system is.
1
G.M =
G (jwp)
=
SOL 5.64
1
w +1
w2p
2
p
w2p
=0
w2p + 1
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 333
K <1
SOL 5.65
nodia
s"0
= R (s) ; 6
- 2 - Y (s) ; 6 E
s (s + 2) s + 2 E
s (s + 2)
6
Y (s) ;1 +
= R (s) ; 6 - 2s E
E
s (s + 2)
s (s + 2)
(6 - 2s)
Y (s) = R (s) 2
(s + 2s + 6)
(6 - 2s)
So,
R (s)
E (s) = R (s) - 2
(s + 2s + 6)
2
= R (s) ; 2 s + 4s E
s + 2s + 6
For unit step input R (s) = 1
s
Steady state error ess = lim sE (s)
s"0
(s2 + 4s)
ess = lim =s 1 2
=0
s (s + 2s + 6)G
s"0
SOL 5.66
www.nodia.co.in
www.gatehelp.com
Page 334
Control Systems
Chapter 5
nodia
= 1.366
SOL 5.68
SOL 5.69
1
3
(1 - e-3t)
H
e-3t
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 335
nodia
t - e- 3t
X (t) = > - 3t H
3e
SOL 5.70
Option () is correct
Phase margin of a system is the amount of additional phase lag required to bring
the system to the point of instability or (- 1, j0)
So here phase margin = 0c
SOL 5.71
www.nodia.co.in
www.gatehelp.com
Page 336
SOL 5.72
Control Systems
Chapter 5
SOL 5.73
nodia
Correct option is (B).
Given characteristic equation,
s3 - 4s2 + s + 6 = 0
Applying Rouths method,
s3
s2
1
-4
s1
- 4 - 6 = 2.5
-4
s0
There are two sign changes in the first column, so no. of right half poles is 2.
No. of roots in left half of s -plane
SOL 5.74
= (3 - 2) = 1
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 337
nodia
s (s + 2) + K (1 + sP) = 0
s2 + s (2 + KP) + K = 0
SOL 5.77
t"3
s"0
s"0
100
1
= lim s ; 2
s"0
s + 12s + 100 E s
= 1.0
SOL 5.78
Amplitude response
H (jw) =
w
w+1
www.nodia.co.in
www.gatehelp.com
Page 338
Control Systems
Chapter 5
nodia
G (jwg) = 1
a2 wg2 + 1
=1
- wg2
a2 wg2 + 1 = wg4
wg4 - a2 wg2 - 1 = 0
Phase margin of the system is
...(1)
...(2)
t"3
s"0
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 339
24
= lim s ;
s"0
s (s + 2) (s + 5) (s + 1)E
= 24 = 2.4
2#5
SOL 5.81
nodia
qh (w) = tan- 1 a w k - tan- 1 a w k
a
b
Jw - wN
-1 K a
b O = tan- 1 w (b - a)
= tan
; ab + w2 E
2
O
KK
1+ w O
ab P
L
qh should be positive for phase lead compensation
w (b - a)
So,
>0
qh (w) = tan- 1 ;
ab + w2 E
b >a
SOL 5.82
SOL 5.83
www.nodia.co.in
www.gatehelp.com
Page 340
Control Systems
Chapter 5
SOL 5.84
t1 = 3 sec
nodia
So, time constants )
t2 = 6 sec
So
(unity feedback)
s ^ s1 h
ess = lim
45
s"0
1+
(s + 15) (s + 1)
= 15
= 15
60
15 + 45
%ess = 15 # 100
60
= 25%
SOL 5.85
So,
H (s) = 1
(unity feedback)
G (s) = b 3 lb 15 l
s + 15 s + 1
1 + b 3 lb 15 l = 0
s + 15 s + 1
(s + 15) (s + 1) + 45 = 0
s2 + 16s + 60 = 0
(s + 6) (s + 10) = 0
s =- 6, - 10
SOL 5.86
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 341
d 2 w =- b dw - K 2 w + K V
LJ a
J dt LJ
dt 2
Here state variables are defined as,
dw = x
1
dt
w = x2
So state equation is
2
xo1 =- B x1 - K x2 + K Va
J
LJ
LJ
xo2 = dw = x1
dt
In matrix form
K/LJ
xo1
- B/J - K 2 /LJ x1
>o H = >
>x H + > 0 H Va
H
x2
1
0
2
R 2 V
Sd w
W
S dt2 W = P >dwH + QVa
dt
S dw W
S dt W
T
X
nodia
So matrix P is
- B/J - K 2 /LJ
> 1
H
0
SOL 5.87
8
K
s1
6
K - 48
6
s0
1 + w2
a
(0.5a) 2 1/2
= 20 log K - 20 log ;1 +
E
a2
= 20 log K - 0.96
www.nodia.co.in
www.gatehelp.com
Page 342
Control Systems
Chapter 5
=- 26.56c
Phase angle calculated from asymptotic plot at (w = 0.5a) is - 22.5c
Error in phase angle
=- 22.5 - (- 26.56c) = 4.9c
SOL 5.89
nodia
Correct option is (B).
Given block diagram
^ s1 h
= 1
s+3
1 + ^ s1 h3
1
bs l
G2 =
= 1
1
s + 12
1 + b l 12
s
Further reducing the block diagram.
Where
G1 =
2G 1 G 2
1 + (2G1 G2) 9
(2) b 1 lb 1 l
s + 3 s + 12
=
1 + (2) b 1 lb 1 l (9)
s + 3 s + 12
Y (s) =
www.nodia.co.in
www.gatehelp.com
Chapter 5
Control Systems
Page 343
2
(s + 3) (s + 12) + 18
2
= 2
s + 15s + 54
2
1
=
=
s
(s + 9) (s + 6)
27 a1 + ka1 + s k
9
6
=
***********
nodia
www.nodia.co.in
www.gatehelp.com
Page 344
Chapter 6
EE 6
ELECTRONICS AND ELECTRICAL MESUREMENTS
EE SP 6.2
ONE MARK
nodia
In an oscilloscope screen, linear sweep is applied at the
(A) vertical axis
(B) horizontal axis
(C) origin
TWO MARK
EE SP 6.3
EE SP 6.4
The reading of the voltmeter (rms) in volts, for the circuit shown in the figure
is _________
EE SP 6.5
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 345
EE SP 6.7
ONE MARK
The saw-tooth voltage waveform shown in the figure is fed to a moving iron
voltmeter. Its reading would be close to __________
nodia
EE SP 6.8
TWO MARKS
The total power dissipated in the circuit, shown in the figure, is 1 kW.
The voltmeter, across the load, reads 200 V. The value of XL is _____.
EE SP 6.9
EE SP 6.10
(D) 2%
Two ammeters X and Y have resistances of 1.2 W and 1.5 W respectively and
they give full-scale deflection with 150 mA and 250 mA respectively. The ranges
have been extended by connecting shunts so as to give full scale deflection with
15 A. The ammeters along with shunts are connected in parallel and then placed
in a circuit in which the total current flowing is 15 A. The current in amperes
indicated in ammeter X is_____.
YEAR 2014 EE03
EE SP 6.11
ONE MARK
An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150
www.nodia.co.in
www.gatehelp.com
Page 346
Chapter 6
V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150.
If the wattmeter is used with 150 V voltage setting and 10 A current setting, the
multiplying factor of the wattmeter is _______.
EE SP 6.12
The two signals S1 and S2, shown in figure, are applied to Y and X deflection
plates of an oscilloscope.
nodia
The waveform displayed on the screen is
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 347
TWO MARKS
EE SP 6.13
EE SP 6.14
In the bridge circuit shown, the capacitors are loss free. At balance, the value of
capacitance C1 in microfarad is ______.
nodia
YEAR 2013
EE SP 6.15
ONE MARK
Three moving iron type voltmeters are connected as shown below. Voltmeter
readings are V , V1 and V2 as indicated. The correct relation among the voltmeter
readings is
(A) V = V1 + V2
2
2
(C) V = V1 V2
EE SP 6.16
(B) V = V1 + V2
(D) V = V2 - V1
The input impedance of the permanent magnet moving coil (PMMC) voltmeter
is infinite. Assuming that the diode shown in the figure below is ideal, the
reading of the voltmeter in Volts is
www.nodia.co.in
www.gatehelp.com
Page 348
(A) 4.46
(B) 3.15
(C) 2.23
(D) 0
nodia
YEAR 2013
EE SP 6.17
EE SP 6.18
TWO MARKS
(A) 56.02
(C) 29.85
(B) 40.83
(D) 10.02
YEAR 2012
EE SP 6.19
Chapter 6
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 6
(A) 4 V
(C) 8 V
EE SP 6.20
(B) 5 V
(D) 10 V
nodia
The bridge method commonly used for finding mutual inductance is
(A) Heaviside Campbell bridge
(B) Schering bridge
(C) De Sauty bridge
EE SP 6.21
For the circuit shown in the figure, the voltage and current expressions are
v (t) = E1 sin (wt) + E 3 sin (3wt) and
i (t) = I1 sin (wt - f1) + I 3 sin (3wt - f3) + I5 sin (5wt)
The average power measured by the wattmeter is
(A) 1 E1 I1 cos f1
2
(C) 1 [E1 I1 cos f1 + E 3 I 3 cos f3]
2
YEAR 2012
EE SP 6.22
TWO MARKS
EE SP 6.23
Page 349
ONE MARK
A dual trace oscilloscope is set to operate in the ALTernate mode. The control
input of the multiplexer used in the y -circuit is fed with a signal having a
www.nodia.co.in
www.gatehelp.com
Page 350
Chapter 6
frequency equal to
(A) the highest frequency that the multiplexer can operate properly
(B) twice the frequency of the time base (sweep) oscillator
(C) the frequency of the time base (sweep) oscillator
(D) haif the frequency of the time base (sweep) oscillator
EE SP 6.24
EE SP 6.25
nodia
The bridge circuit shown in the figure below is used for the measurement of an
unknown element ZX . The bridge circuit is best suited when ZX is a
YEAR 2011
EE SP 6.26
TWO MARKS
A 4 12 digit DMM has the error specification as: 0.2% of reading + 10 counts. If
a dc voltage of 100 V is read on its 200 V full scale, the maximum error that can
be expected in the reading is
(A) ! 0.1%
(B) ! 0.2%
(C) ! 0.3%
(D) ! 0.4%
YEAR 2010
EE SP 6.27
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 351
nodia
An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 W. In
order to change the range to 0-25 A, we need to add a resistance of
(A) 0.8 W in series with the meter
(B) 1.0 W in series with the meter
(C) 0.04 W in parallel with the meter
(D) 0.05 W in parallel with the meter
EE SP 6.29
As shown in the figure, a negative feedback system has an amplifier of gain 100
with ! 10% tolerance in the forward path, and an attenuator of value 9/100 in
the feedback path. The overall system gain is approximately :
(B) 10 ! 2%
(D) 10 ! 10%
(A) 10 ! 1%
(C) 10 ! 5%
YEAR 2010
EE SP 6.30
TWO MARKS
The Maxwells bridge shown in the figure is at balance. The parameters of the
inductive coil are.
(A) R = R2 R 3 /R 4, L = C 4 R2 R 3
www.nodia.co.in
www.gatehelp.com
Page 352
Chapter 6
(B) L = R2 R 3 /R 4, R = C 4 R2 R 3
(C) R = R 4 /R2 R 3, L = 1/ (C 4 R2 R 3)
(D) L = R 4 /R2 R 3, R = 1/ (C 4 R2 R 3)
YEAR 2009
EE SP 6.31
ONE MARK
EE SP 6.32
The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y
mode, the screen shows a figure which changes from ellipse to circle and back
to ellipse with its major axis changing orientation slowly and repeatedly. The
following inference can be made from this.
(A) The signals are not sinusoidal
nodia
(B) The amplitudes of the signals are very close but not equal
(C) The signals are sinusoidal with their frequencies very close but not equal
(D) There is a constant but small phase difference between the signals
YEAR 2009
EE SP 6.33
The figure shows a three-phase delta connected load supplied from a 400V, 50
Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a
wattmeter are connected to the load as shown, with the coil polarities suitably
selected to ensure a positive deflection. The wattmeter reading will be
(A) 0
(C) 800 Watt
EE SP 6.34
TWO MARKS
EE SP 6.35
ONE MARK
Two 8-bit ADCs, one of single slope integrating type and other of successive
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 353
TWO MARKS
Two sinusoidal signals p (w1, t) = A sin w1 t and q (w2 t) are applied to X and Y
inputs of a dual channel CRO. The Lissajous figure displayed on the screen
shown below :
nodia
The signal q (w2 t) will be represented as
(A) q (w2 t) = A sin w2 t, w2 = 2w1
(B) q (w2 t) = A sin w2 t, w2 = w1 /2
(C) q (w2 t) = A cos w2 t, w2 = 2w1
(D) q (w2 t) = A cos w2 t, w2 = w1 /2
EE SP 6.37
If the bridge is balanced for oscillator frequency f = 2 kHz, then the impedance
Z will be
(A) (260 + j0) W
(B) (0 + j200) W
(C) (260 - j200) W
(D) (260 + j200) W
YEAR 2007
EE SP 6.38
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 354
Chapter 6
B and C in the circuit of the adjacent figure. Vin is a square wave of a suitable
low frequency. The display on Ch1 and Ch2 are as shown on the right. Then the
Signal and Ground probes S1, G1 and S2, G2 of Ch1 and Ch2 respectively are
connected to points :
nodia
(A) A, B, C, A
(B) A, B, C, B
(C) C, B, A, B
(D) B, A, B, C
YEAR 2007
EE SP 6.39
TWO MARKS
A bridge circuit is shown in the figure below. Which one of the sequence given
below is most suitable for balancing the bridge ?
YEAR 2006
EE SP 6.40
ONE MARK
The time/div and voltage/div axes of an oscilloscope have been erased. A student
connects a 1 kHz, 5 V p-p square wave calibration pulse to channel-1 of the scope
and observes the screen to be as shown in the upper trace of the figure. An
unknown signal is connected to channel-2(lower trace) of the scope. It the time/
div and V/div on both channels are the same, the amplitude (p-p) and period of
the unknown signal are respectively
www.nodia.co.in
www.gatehelp.com
Chapter 6
EE SP 6.41
(A) 5 V, 1 ms
(B) 5 V, 2 ms
(C) 7.5 V, 2 ms
(D) 10 V, 1 ms
Page 355
nodia
(A) 0 W
(C) 50 W
(B) 25 W
(D) 100 W
YEAR 2006
TWO MARKS
EE SP 6.42
EE SP 6.43
EE SP 6.44
A 200/1 Current transformer (CT) is wound with 200 turns on the secondary
on a toroidal core. When it carries a current of 160 A on the primary, the ratio
and phase errors of the CT are found to be - 0.5% and 30 minutes respectively.
If the number of secondary turns is reduced by 1 new ratio-error(%) and phase-
www.nodia.co.in
www.gatehelp.com
Page 356
Chapter 6
(B) - 0.5, 35
(D) - 1.0, 25
R1 and R4 are the opposite arms of a Wheatstone bridge as are R3 and R2 . The
source voltage is applied across R1 and R3 . Under balanced conditions which one
of the following is true
(B) R1 = R2 R3 /R4
(A) R1 = R3 R4 /R2
(D) R1 = R2 + R3 + R4
(C) R1 = R2 R4 /R3
YEAR 2005
EE SP 6.46
EE SP 6.47
nodia
The Q-meter works on the principle of
(A) mutual inductance
EE SP 6.48
ONE MARK
3 /2) V
(D) ( 17 /2) V
TWO MARKS
EE SP 6.49
The simultaneous application of signals x (t) and y (t) to the horizontal and
vertical plates, respectively, of an oscilloscope, produces a vertical figure-of-8
display. If P and Q are constants and x (t) = P sin (4t + 30c), then y (t) is equal to
(A) Q sin (4t - 30c)
(B) Q sin (2t + 15c)
(C) Q sin (8t + 60c)
(D) Q sin (4t + 30c)
EE SP 6.50
EE SP 6.51
The set-up in the figure is used to measure resistance R .The ammeter and
voltmeter resistances are 0.01W and 2000 W, respectively. Their readings are 2 A
and 180 V, respectively, giving a measured resistances of 90 W The percentage
error in the measurement is
www.nodia.co.in
www.gatehelp.com
Chapter 6
EE SP 6.52
(A) 2.25%
(B) 2.35%
(C) 4.5%
(D) 4.71%
Page 357
A 1000 V DC supply has two 1-core cables as its positive and negative leads
: their insulation resistances to earth are 4 MW and 6 MW, respectively, as
shown in the figure. A voltmeter with resistance 50 kW is used to measure the
insulation of the cable. When connected between the positive core and earth,
then voltmeter reads
nodia
(A) 8 V
(C) 24 V
EE SP 6.53
(B) 16 V
(D) 40 V
Two wattmeters, which are connected to measure the total power on a three-phase
system supplying a balanced load, read 10.5 kW and - 2.5 kW, respectively. The
total power and the power factor, respectively, are
(A) 13.0 kW, 0.334
(B) 13.0 kW, 0.684
(C) 8.0 kW, 0.52
(D) 8.0 kW, 0.334
YEAR 2004
ONE MARK
EE SP 6.54
EE SP 6.55
The circuit in figure is used to measure the power consumed by the load. The
current coil and the voltage coil of the wattmeter have 0.02 W and 1000W
resistances respectively. The measured power compared to the load power will be
www.nodia.co.in
www.gatehelp.com
Page 358
EE SP 6.56
nodia
(C) 11
(D) 10
YEAR 2004
EE SP 6.57
Chapter 6
TWO MARKS
(A) 3.53 V
(C) 4.54 V
(B) 4.37 V
(D) 5.00 V
EE SP 6.58
A moving coil of a meter has 100 turns, and a length and depth of 10 mm and
20 mm respectively. It is positioned in a uniform radial flux density of 200 mT.
The coil carries a current of 50 mA. The torque on the coil is
(A) 200 mNm
(B) 100 mNm
(C) 2 mNm
(D) 1 mNm
EE SP 6.59
A dc A-h meter is rated for 15 A, 250 V. The meter constant is 14.4 A-sec/rev.
The meter constant at rated voltage may be expressed as
(A) 3750 rev/kWh
(B) 3600 rev/kWh
(C) 1000 rev/kWh
(D) 960 rev/kWh
EE SP 6.60
A moving iron ammeter produces a full scale torque of 240 mNm with a deflection
of 120c at a current of 10 A . The rate of change of self induction (mH/radian)
of the instrument at full scale is
(A) 2.0 mH/radian
(B) 4.8 mH/radian
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 359
nodia
(A) - 795 W
(C) + 597 W
EE SP 6.62
EE SP 6.63
(B) - 597 W
(D) + 795 W
A 50 Hz, bar primary CT has a secondary with 500 turns. The secondary supplies
5 A current into a purely resistive burden of 1 W. The magnetizing ampere-turns
is 200. The phase angle between the primary and second current is
(A) 4.6c
(B) 85.4c
(C) 94.6c
(D) 175.4c
The core flux in the CT of Prob Q.44, under the given operating conditions is
(A) 0
(B) 45.0 mWb
(C) 22.5 mWb
(D) 100.0 mWb
YEAR 2003
ONE MARK
EE SP 6.64
EE SP 6.65
The effect of stray magnetic field on the actuating torque of a portable instrument
is maximum when the operating field of the instrument and the stray fields are
(A) perpendicular
(B) parallel
(C) inclined at 60%
EE SP 6.66
A reading of 120 is obtained when standard inductor was connected in the circuit
of a Q-meter and the variable capacitor is adjusted to value of 300 pF. A lossless
capacitor of unknown value Cx is then connected in parallel with the variable
www.nodia.co.in
www.gatehelp.com
Page 360
Chapter 6
capacitor and the same reading was obtained when the variable capacitor is
readjusted to a value of 200 pF. The value of Cx in pF is
(A) 100
(B) 200
(C) 300
(D) 500
YEAR 2003
EE SP 6.67
EE SP 6.68
TWO MARKS
The simplified block diagram of a 10-bit A/D converter of dual slope integrator
type is shown in figure. The 10-bit counter at the output is clocked by a 1 MHz
clock. Assuming negligible timing overhead for the control logic, the maximum
frequency of the analog signal that can be converted using this A/D converter is
approximately
nodia
(A) 2 kHz
(B) 1 kHz
(C) 500 Hz
(D) 250 Hz
List-II
1. Wheatstone Bridge
2. Kelvin Double Bridge
Codes :
(A) P=2, Q=3, R=6, S=5
(B) P=2, Q=6, R=4, S=5
(C) P=2, Q= 3, R=5, S=4
(D) P=1, Q=3, R=2, S=6
EE SP 6.69
www.nodia.co.in
www.gatehelp.com
Chapter 6
(A) 63.56 W
(C) 89.93 W
EE SP 6.70
EE SP 6.71
Page 361
(B) 69.93 W
(D) 141.3 kW
A wattmeter reads 400 W when its current coil is connected in the R-phase and
its pressure coil is connected between this phase and the neutral of a symmetrical
3-phase system supplying a balanced star connected 0.8 p.f. inductive load. This
phase sequence is RYB. What will be the reading of this wattmeter if its pressure
coil alone is reconnected between the B and Y phases, all other connections
remaining as before ?
(A) 400.0
(B) 519.6
(C) 300.0
(D) 692.8
nodia
EE SP 6.72
EE SP 6.73
EE SP 6.74
Group-II represents the figures obtained on a CRO screen when the voltage
signals Vx = Vxm sin wt and Vy = Vym sin (wt + F) are given to its X and Y plates
respectively and F is changed. Choose the correct value of F from Group-I to
match with the corresponding figure of Group-II.
www.nodia.co.in
www.gatehelp.com
Page 362
Group-I
Chapter 6
Group-II
P. F = 0
Q. F = p/2
nodia
S. F = 3p/2
Codes :
(A) P=1, Q= 3, R=6, S=5
(B) P=2, Q= 6, R=4, S=5
(C) P=2, Q= 3, R=5, S=4
(D) P=1, Q=5, R=6, S=4
***********
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 363
SOLUTIONS
SOL 6.1
SOL 6.2
SOL 6.3
nodia
at t = 26cC ,
RQ = R 61 + a ^Q2 - Q1h@
Substituting the given values, we get
1440 = 120 81 + 4.5 # 10-3 ^Q2 - 26hB
12 = 1 + 4.5 # 10-3 ^Q2 - 26h
or
or
or
Thus,
SOL 6.4
11
4.5 # 10-3
Q2 - 26 = 2.444.44
Q2 = 2470.44cC
Q2 - 26 =
www.nodia.co.in
www.gatehelp.com
Page 364
Chapter 6
nodia
z1 = z 4 = j 1 W
z2 = z3 = 1 W
j
Substituting these values in equation (1), we get
j#j = 1#1
j
j
j2 = 12
j
- 1 =- 1
Hence, it is a balanced Wheatstone bridge, and reading of voltmeter (rms) is
and
V = 100 # 2
= 141.42 V
SOL 6.5
...(1)
...(2)
At balance, we have
Tc = Td
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 365
...(3)
I1 = I2 = I (series)
where
Td = I 2 dM
dq
So,
...(4)
nodia
or
Thus,
SOL 6.6
=
=
=
=
=
=
SOL 6.7
1
20 # 10-3
25 # 106
20 # 10-3
20 # 10-3
20 # 10-3
2
^5 # 103 # t h dt
t2 dt
-3
25 # 10 t3 20 # 10
: D
20 # 10-3 3 0
3
^20 # 10-3h
25 # 106
#
3
20 # 10-3
6
25 # 10 # 400 # 10-6
3
100 = 57.74 A
3
6
3 ^W1 - W2h
G
W1 + W2
www.nodia.co.in
www.gatehelp.com
Page 366
Chapter 6
3 ^250 - 100h
250 + 100
f = tan-1 ^0.7423h
f = 36.586c
Power factor = cos f
SOL 6.8
nodia
VL = 200 V
VL = I R2 + X L2
200 = 10 ^9.96h2 + X L2
^20h2 = ^9.96h2 # X L2
XL = 400 - 99.20
XL = 17.34 W
SOL 6.9
R1 + R2 = Rsum
...(1)
Req = TR1 ^%h + TR2 ^%h - TRsum ^%h
TRsum = ^300 ! 1%h + ^200 ! 1%h
= ;300 ! b 300 # 1 lE + ;200 ! b 200 # 1 lE
100
100
= ^300 ! 3h + ^200 ! 2h
= 500 ! 5
TRsum ^%h = b 5 # 100 l
500
= 1%
www.nodia.co.in
www.gatehelp.com
Chapter 6
SOL 6.10
Page 367
TReq ^%h = 1% + 1% - 1%
TReq ^%h = 1%
nodia
ImX RmX = ^I - ImX h RsnX
0.15 # 1.2 = ^15 - 0.15h RsnX
RsnX = 0.18 = 0.0121 W
14.85
For ammeter Y , Let shunt resistor be RMY
Correct answer is 2.
In low power factor (LPF) wattmeter, the multiplying factor is given by
power setting
M.F. =
power for maximum difflection
www.nodia.co.in
www.gatehelp.com
Page 368
Chapter 6
SOL 6.12
nodia
www.nodia.co.in
www.gatehelp.com
Chapter 6
SOL 6.13
Page 369
nodia
6p
1 + sin wt h2 d ^wt h +
Vrms = = 1 ;
12p 0 ^
== 1 ;
12p 0
6p
6p
12p
6p
12p
SOL 6.14
1/2
1 - cos 2wt d wt +
= = 1 ;6p + 6p +
^ h
12p
2
0
1/2
= : 1 612p + 6p@D
12p
1/2
= b 18 l = 3
12
2
1/2
6p
12p
6p
1
1
3
-6 = 105 # 10 #
w
j
#C
jw # 0.1 # 10
C = 0.3 mF
www.nodia.co.in
www.gatehelp.com
Page 370
Chapter 6
For an ideal voltmeter interval resistance is always zero. So we can apply the
KVL along the two voltmeters as
V - V1 - V2 = 0
V = V1 + V2
or
SOL 6.16
nodia
SOL 6.17
SOL 6.18
SOL 6.19
10
= 1 ; tdt +
20 0
20
v (t) dt
20
12
# (- 5) dt + # 5dtE
10
12
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 371
2 10
20
= 1 c :t D - 5 6t @12
+ 5 6t @12
m
10
20 2 0
SOL 6.21
nodia
#
...(1)
2p
(2)
1
2p
(3)
1
2p
(4)
1
2p
2p
2p
Result (3) and (4) implies that power is transferred between same harmonics of
voltages and currents. Thus integration of equation (1) gives.
P = 1 E1 I1 cos f + 1 E 3 I 3 cos f3
2
2
SOL 6.22
Here
www.nodia.co.in
www.gatehelp.com
Page 372
Chapter 6
V = R m + Rs = 1 + Rs
Vm
Rm
Rm
Rs1 = 20 kW , Vm1 = 440 V
V = 1 + 20k
440
Rm
Here when,
So,
...(1)
When,
V = 480 V , Rm = 220 kW
Rs3 = 40 kW , Vm3 = ?
480 = 1 + 40 k & V - 406 V
m2
Vm3
220 k
So when
SOL 6.23
...(2)
nodia
SOL 6.24
SOL 6.25
Z1 = R1 || jwC1
www.nodia.co.in
www.gatehelp.com
Chapter 6
so admittance
Page 373
Y1 = 1 = 1 + jwC1
Z1 R 1
Z2 = R2 and Z 4 = R 4
Let
ZX = RX + jwLX (Unknown impedance)
For current balance condition of the Bridge
Z 2 Z 4 = Z X Z1 = Z X
Y1
Let
ZX = Z2 Z 4 Y1
RX + jwLX = R2 R 4 b 1 + jwC1 l
R1
Equating imaginary and real parts
RX = R2 R 4 and LX = R2 R 4 C1
R1
Quality factor of inductance which is being measured
Q = wL X = wR 1 C 1
RX
From above equation we can see that for measuring high values of Q we need
a large value of resistance R 4 which is not suitable. This bridge is used for
measuring low Q coils.
Note: We can observe directly that this is a maxwells bridge which is suitable
for low values of Q (i.e. Q < 10 )
SOL 6.26
nodia
Percentage error
SOL 6.27
www.nodia.co.in
www.gatehelp.com
Page 374
Chapter 6
Current in shunt Il = IR - I fs = 25 - 5 = 20 A
20 # Rsh = 5 # 0.2
Rsh = 1 = .05 W
20
SOL 6.29
nodia
error 3 g = 10.091 - 10 - 0.1
100 - 10%
90
Similarly
g =
=
9
1 + (100 - 10%)
1 + 90 # 9
100
100
= 9.89
error 3 g = 9.89 - 10 -- 0.1
So gain g = 10 ! 0.1 = 10 ! 1%
SOL 6.30
jR2 R 3
wC 4
jR2 R 3
wC 4
L = R2 R3 C 4
SOL 6.31
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 375
SOL 6.32
SOL 6.33
Power
nodia
= 1600+240c
= 1600 # 1 = 800 Watt
2
SOL 6.34
SOL 6.35
SOL 6.36
Frequency ratio
w2 = w1 /2
Since the Lissajous figures are ellipse, so there is a phase difference of 90c exists
between vertical and horizontal inputs.
www.nodia.co.in
www.gatehelp.com
Page 376
So
SOL 6.37
Chapter 6
q (w2 t) = A cos w2 t, w2 = w1 /2
ZBC =
ZAD
To balance the bridge
nodia
500Z = 130000
Z = (260 + j0) W
SOL 6.38
SOL 6.39
(R1 + jX1) (R 4 - jX 4) = R2 R 3
(R1 R 4 + X1 X 4) + j (X1 R 4 - R1 X 4) = R2 R 3
comparing real and imaginary parts on both sides of equations
...(1)
R 1 R 4 + X1 X 4 = R 2 R 3
...(2)
X 1 R 4 - R 1 X 4 = 0 & X1 = R 1
X4 R4
from eq(1) and (2) it is clear that for balancing the bridge first balance R 4 and
then R1 .
SOL 6.40
SOL 6.41
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 377
nodia
I PMMC =- 8 A
rms meter reads rms value so
Irms =
(- 8) 2 +
(6 2 ) 2
2
64 + 36 = 10 A
SOL 6.43
xy
z
So
dy
= ! dx !
! dz
x
y
z
= ! 0.5% reading
= ! 1% full scale = ! 1 # 100 = ! 1
100
= ! 1 # 100 = ! 5% reading
20
= 1.5% reading
= ! 0.5% ! 5% ! 1.5%
= ! 7%
SOL 6.44
SOL 6.45
www.nodia.co.in
www.gatehelp.com
Page 378
Chapter 6
nodia
R 2 R 3 + R 3 R 4 = R1 R 4 + R 3 R 4
R 1 = R 2 R 3 /R 4
SOL 6.46
At resonance VC = VL
and I = V
R
Quality factor Q = wL = 1
R
wCR
= wL # I = VL = VC
R#I
E
E
Thus, we can obtain Q
SOL 6.47
SOL 6.48
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 379
n = 8 bit
SOL 6.49
SOL 6.50
nodia
Correct option is (C).
The configuration is shown below
www.nodia.co.in
www.gatehelp.com
Page 380
I + IV = 2 amp
I = 2 - .09 = 1.91 V
R = E = 180 = 94.24 W
1.91
I
R 0 = 180 = 90 W
2
% error = R - R 0 # 100 = 94.24 - 90 # 100 = 4.71%
90
R0
So
Ideally
SOL 6.52
Chapter 6
nodia
Voltmeter reading
1000
(50 kW z 4 MW)
6 MW + 50 kW z 4 MW l
= 1000 # .049 = 8.10 V
6 + .049
V =b
SOL 6.53
SOL 6.54
SOL 6.55
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 381
Current in CC is 20 A
20 = I1 b
1000
1000 + 0.02 l
I1 = 20.0004 A - 20 A
200 = V1 - .02 # 20 = 200.40
Power measured Pm = V1 I1 = 20 (200.40) = 4008 W
Load power
PL = 20 # 200 = 4000 W
% Change = Pm - PL = 4008 - 4000 # 100
4000
PL
nodia
= 0.2% more
SOL 6.56
We have to obtain n = I
I1
I1 = Rsh = 100 = 1
I2
Rm 1000 10
I1 + I 2 = I
I1 + 10I1 = I
11I1 = I
n = I = 11
I1
SOL 6.57
1
1 =
jw C
2p # 100 # 103 # 10 # 10- 12
writing node equation at P
Rectance
Xc =
www.nodia.co.in
www.gatehelp.com
Page 382
Chapter 6
VP - 10 + V 1 + 1 - j = 0
Pb
100
100 500 159 l
10 - VP = VP (1.2 - j0.628)
10 = (2.2 - j0.628) VP
VP 10 = 4.38 V
2.28
SOL 6.58
nodia
So,
SOL 6.59
N = 100
I = 50 mA
B = 200 mTA " Area,
A = 10 mm # 20 mm
SOL 6.60
SOL 6.61
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 383
SOL 6.62
nodia
Correct option is (A).
For small values of phase angle
IP = nf , f " Phase angle (radians)
IS
n " turns ratio
Magnetizing ampere-turns
= 200
So primary current IP = 200 # 1 = 200 amp
Turns ratio n = 500
So
SOL 6.63
SOL 6.64
www.nodia.co.in
www.gatehelp.com
Page 384
Chapter 6
SOL 6.65
SOL 6.66
nodia
Let
C1 = 300 pF
1
wC 1 R
Now when Cx is connected in parallel with variable resistor C1 ' = 200 pF
1
Q = 120 =
w (C1 ' + Cx ) R
Q = 120 =
So
SOL 6.67
C 1 = C1 ' + C x
300 = 200 + Cx
Cx = 100 pF
so
Tm = 2n TC
fm = 1 " maximum frquency of input
Tm
fC = 1 " clock frequency
TC
f
fm = Cn , n = 10
2
6
= 10 = 1 kHz (approax)
1024
SOL 6.68
SOL 6.69
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 385
Im = 1.57 mA
Full scale ac current
(Irms) fs = 1.57 = 1.11 mA
2
nodia
100
= Rs + 100
(1.11 mA)
100 # 900 = Rs + 100
Rs = 89.9 kW
SOL 6.70
Reading of wattmeter
W1 = IP VP cos q1 , cos q1 = 0.8 & q1 = 36.86c
400 = IL VL cos q1
3
...(1)
400 = IL VL # 0.8
3
Now when pressure coil is connected between B and Y-phases, the circuit is
www.nodia.co.in
www.gatehelp.com
Page 386
Chapter 6
phasor diagram
nodia
Angle
Now wattmeter reading
From equation (1)
so
SOL 6.71
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 387
nodia
Total primary current (IT ) =
IT =
=
I p2 + I m2
SOL 6.73
SOL 6.74
www.nodia.co.in
www.gatehelp.com
Page 388
Chapter 6
Vx = Vxm sin wt
Vy = Vym sin (wt + 0c) = sin wt
Draw Vx and Vy as shown below
For f = 0c,
nodia
Divide both Vy and Vx equal parts and match the corresponding points on the
screen.
Similarly for f = 90c
Vx = Vxm sin wt
Vy = Vym sin (wt + 90c)
Similarly for f = 3p
2
www.nodia.co.in
www.gatehelp.com
Chapter 6
Page 389
nodia
we can also obtain for 0 < f < 3p
2
***********
www.nodia.co.in
www.gatehelp.com
Page 390
Analog Electronics
Chapter 7
CHAPTER 7
ANALOG ELECTRONICS
ONE MARK
In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when
nodia
1
(A) R 3 = R1 , w =
R 4 R2
R1 C1 R 2 C 2
R
C
1
(B) 2 = 2 , w =
R1 C1
R1 C1 R 2 C 2
1
(C) R 3 = R1 + C2 , w =
R 4 R 2 C1
R1 C1 R 2 C 2
1
(D) R 3 + R1 = C 2 , w =
R 4 R 2 C1
R1 C1 R 2 C 2
EE SP 7.2
The magnitude of the mid-band voltage gain of the circuit shown in figure is
(assuming h fe of the transistor to be 100)
(A) 1
(B) 10
(C) 20
(D) 100
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 391
TWO MARKS
EE SP 7.3
Given that the op-amps in the figure are ideal, the output voltage V0 is
EE SP 7.4
nodia
(A) ^V1 - V2h
(C) ^V1 - V2h /2
In the figure shown, assume the op-amp to be ideal. Which of the alternatives
v i ^w h
?
gives the correct Bode plote for the transfer function
v i ^w h
www.nodia.co.in
www.gatehelp.com
Page 392
Analog Electronics
nodia
YEAR 2014 EE02
Chapter 7
ONE MARK
EE SP 7.5
The transistor in the given circuit should always be in active region. Take
VCE^sat h = 0.2 V , VBE = 0.7 V . The maximum value of Rc in W which can be used,
is _____.
EE SP 7.6
The sinusoidal ac source in the figure has an rms value of 202 V . Considering all
possible values of RL , the minimum value of Rs in W to avoid burnout of the
Zener diode is _____.
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 393
TWO MARKS
Assuming the diodes to be ideal in the figure, for the output to be clipped, the
input voltage vi must be outside the range
(A) - 1 V to - 2 V
(C) + 1 V to - 2 V
EE SP 7.8
(B) - 2 V to - 4 V
(D) + 2 V to - 4 V
nodia
EE SP 7.9
An oscillator circuit using ideal op-amp and diodes is shown in the figure.
The time duration for + ve part of the cycle is Tt1 and for - ve part is Tt2 . The
Tt1 - Tt 2
RC
value of e
will be ______.
ONE MAERK
An operational amplifier circuit is shown in the figure. The output of the circuit
for a given input vi is
(A) -b R2 l vi
(B) -b1 + R2 l vi
R1
R1
(C) b1 + R2 l vi
(D) + Vsat or - Vsat
R1
www.nodia.co.in
www.gatehelp.com
Page 394
Analog Electronics
nodia
YEAR 2014 EE03
EE SP 7.11
TWO MARKS
YEAR 2013
EE SP 7.12
Chapter 7
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 395
nodia
(D) The input impedance decreases and output impedance increases
EE SP 7.13
In the circuit shown below what is the output voltage ^Vouth if a silicon transistor
Q and an ideal op-amp are used?
(A) - 15 V
(C) + 0.7 V
(B) - 0.7 V
(D) + 15 V
YEAR 2013
EE SP 7.14
TWO MARKS
In the circuit shown below the op-amps are ideal. Then, Vout in Volts is
(A) 4
(B) 6
www.nodia.co.in
www.gatehelp.com
Page 396
Analog Electronics
(C) 8
EE SP 7.15
(D) 10
A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the
voltage measured across WX in Volts, is
nodia
(A) sin wt
(C) ^sin wt - sin wt h /2
EE SP 7.16
Chapter 7
In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA
. To maintain 5 V across RL , the minimum value of RL in W and the minimum
power rating of the Zener diode in mW, respectively, are
YEAR 2012
EE SP 7.17
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 397
(A) 10 mA
(B) 9.3 mA
(C) 6.67 mA
(D) 6.2 mA
YEAR 2012
TWO MARKS
EE SP 7.18
EE SP 7.19
nodia
(A) Av . 200
(B) Av . 100
(C) Av . 20
(D) Av . 10
1
rad/s
(R1 + R2) C
(B) high pass filter with f3dB = 1 rad/s
R1 C
(C) low pass filter with f3dB = 1 rad/s
R1 C
1
(D) high pass filter with f3dB =
rad/s
(R1 + R2) C
(A) low pass filter with f3dB =
YEAR 2011
EE SP 7.20
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 398
Analog Electronics
Chapter 7
nodia
YEAR 2011
EE SP 7.22
TWO MARKS
The transistor used in the circuit shown below has a b of 30 and ICBO is negligible
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 399
If the forward voltage drop of diode is 0.7 V, then the current through collector
will be
(A) 168 mA
(B) 108 mA
(C) 20.54 mA
(D) 5.36 mA
EE SP 7.23
nodia
A clipper circuit is shown below.
www.nodia.co.in
www.gatehelp.com
Page 400
Analog Electronics
Chapter 7
YEAR 2010
EE SP 7.24
ONE MARK
(A) 4 V
nodia
(B) 6 V
(C) 7.5 V
(D) 12.12 V
EE SP 7.25
Assuming that the diodes in the given circuit are ideal, the voltage V0 is
(A) 4 V
(C) 7.5 V
(B) 5 V
(D) 12.12 V
YEAR 2010
EE SP 7.26
TWO MARKS
The transistor circuit shown uses a silicon transistor with VBE = 0.7, IC . IE and
a dc current gain of 100. The value of V0 is
(A) 4.65 V
(B) 5 V
(C) 6.3 V
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 401
(D) 7.23 V
YEAR 2009
ONE MARK
EE SP 7.27
EE SP 7.28
nodia
(A) Current-Current feedback
(B) Voltage-Voltage feedback
(C) Current-Voltage feedback
(D) Voltage-Current feedback
YEAR 2009
TWO MARKS
Transformer and emitter follower can both be used for impedance matching at
the output of an audio amplifier. The basic relationship between the input power
Pin and output power Pout in both the cases is
(A) Pin = Pout for both transformer and emitter follower
(B) Pin > Pout for both transformer and emitter follower
(C) Pin < Pout for transformer and Pin = Pout for emitter follower
(D) Pin = Pout for transformer and Pin < Pout for emitter follower
www.nodia.co.in
www.gatehelp.com
Page 402
EE SP 7.30
Analog Electronics
Chapter 7
An ideal op-amp circuit and its input wave form as shown in the figures. The
output waveform of this circuit will be
nodia
YEAR 2008
EE SP 7.31
ONE MARK
The equivalent circuits of a diode, during forward biased and reverse biased
conditions, are shown in the figure.
(1)
(2)
If such a diode is used in clipper circuit of figure given above, the output voltage
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 403
nodia
YEAR 2008
EE SP 7.32
TWO MARKS
Two perfectly matched silicon transistor are connected as shown in the figure
assuming the b of the transistors to be very high and the forward voltage drop
in diodes to be 0.7 V, the value of current I is
(A) 0 mA
(C) 4.3 mA
EE SP 7.33
(B) 3.6 mA
(D) 5.7 mA
In the voltage doubler circuit shown in the figure, the switch S is closed at
t = 0 . Assuming diodes D1 and D2 to be ideal, load resistance to be infinite and
initial capacitor voltages to be zero. The steady state voltage across capacitor C1
and C2 will be
www.nodia.co.in
www.gatehelp.com
Page 404
EE SP 7.34
Analog Electronics
Chapter 7
The block diagrams of two of half wave rectifiers are shown in the figure. The
transfer characteristics of the rectifiers are also shown within the block.
It is desired to make full wave rectifier using above two half-wave rectifiers. The
resultants circuit will be
nodia
Statement for Linked Answer Questions 35 and 36
A general filter circuit is shown in the figure :
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
EE SP 7.35
EE SP 7.36
Page 405
nodia
YEAR 2007
EE SP 7.37
ONE MARK
The common emitter forward current gain of the transistor shown is bF = 100
www.nodia.co.in
www.gatehelp.com
Page 406
EE SP 7.39
Analog Electronics
Chapter 7
(A) 0.6 W
(B) 2.4 W
(C) 4.2 W
(D) 5.4 W
nodia
rV
R1 < R2
r < R2
(B) a voltage source with voltage
V
R1
r < R2 V
(C) a current source with current c
R1 + R2 m r
(D) a current source with current c R2 mV
R1 + R2 r
(A) a voltage source with voltage
YEAR 2007
EE SP 7.40
TWO MARKS
The input signal Vin shown in the figure is a 1 kHz square wave voltage that
alternates between + 7 V and - 7 V with a 50% duty cycle. Both transistor
have the same current gain which is large. The circuit delivers power to the load
resistor RL . What is the efficiency of this circuit for the given input ? choose the
closest answer.
(A) 46%
(C) 63%
(B) 55%
(D) 92%
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
YEAR 2006
EE SP 7.41
Page 407
ONE MARK
For a given sinusoidal input voltage, the voltage waveform at point P of the
clamper circuit shown in figure will be
nodia
EE SP 7.42
What are the states of the three ideal diodes of the circuit shown in figure ?
(A) D1 ON, D2 OFF, D3 OFF
(B) D1 OFF, D2 ON, D3 OFF
(C) D1 ON, D2 OFF, D3 ON
(D) D1 OFF, D2 ON, D3 ON
YEAR 2006
EE SP 7.43
TWO MARKS
Assuming the diodes D1 and D2 of the circuit shown in figure to be ideal ones,
the transfer characteristics of the circuit will be
www.nodia.co.in
www.gatehelp.com
Page 408
EE SP 7.44
Analog Electronics
Chapter 7
nodia
www.nodia.co.in
www.gatehelp.com
Chapter 7
EE SP 7.45
Analog Electronics
Page 409
Consider the circuit shown in figure. If the b of the transistor is 30 and ICBO is
20 mA and the input voltage is + 5 V , the transistor would be operating in
nodia
(A) saturation region
(B) active region
EE SP 7.46
A TTL NOT gate circuit is shown in figure. Assuming VBE = 0.7 V of both the
transistors, if Vi = 3.0 V, then the states of the two transistors will be
ONE MARK
Assume that D1 and D2 in figure are ideal diodes. The value of current is
www.nodia.co.in
www.gatehelp.com
Page 410
EE SP 7.48
Analog Electronics
Chapter 7
(A) 0 mA
(B) 0.5 mA
(C) 1 mA
(D) 2 mA
Assume that the N-channel MOSFET shown in the figure is ideal, and that its
threshold voltage is + 1.0 V the voltage Vab between nodes a and b is
nodia
(A) 5 V
(C) 1 V
(B) 2 V
(D) 0 V
YEAR 2005
EE SP 7.49
The common emitter amplifier shown in the figure is biased using a 1 mA ideal
current source. The approximate base current value is
(A) 0 mA
(C) 100 mA
EE SP 7.50
TWO MARKS
(B) 10 mA
(D) 1000 mA
www.nodia.co.in
www.gatehelp.com
Chapter 7
EE SP 7.51
Analog Electronics
nodia
(A) Infinity
(C) 100 kW
EE SP 7.52
Page 411
(B) 1 MW
(D) 40 kW
In the given figure, if the input is a sinusoidal signal, the output will appear as
shown
www.nodia.co.in
www.gatehelp.com
Page 412
Analog Electronics
Chapter 7
EE SP 7.53
EE SP 7.54
nodia
The transconductance of the MOSFET is
(A) 0.75 ms
(B) 1 ms
(C) 2 ms
(D) 10 ms
(B) - 7.5
(D) - 10
YEAR 2004
EE SP 7.55
(A) 33 mA
(C) 2 mA
EE SP 7.56
ONE MARK
(B) 3.3 mA
(D) 0 mA
Two perfectly matched silicon transistor are connected as shown in figure. The
value of the current I is
www.nodia.co.in
www.gatehelp.com
Chapter 7
EE SP 7.57
Analog Electronics
Page 413
(A) 0 mA
(B) 2.3 mA
(C) 4.3 mA
(D) 7.3 mA
nodia
The feedback used in the circuit shown in figure can be classified as
YEAR 2004
EE SP 7.58
Assuming that the diodes are ideal in figure, the current in diode D1 is
(A) 9 mA
(C) 0 mA
EE SP 7.59
TWO MARKS
(B) 5 mA
(D) - 3 mA
www.nodia.co.in
www.gatehelp.com
Page 414
Analog Electronics
(A) 10.0 kW
(C) 5.0 kW
EE SP 7.60
nodia
(B) 470 W
(D) 1200 W
In the active filter circuit shown in figure, if Q = 1, a pair of poles will be realized
with w0 equal to
(B) 8.3 kW
(D) 2.5 kW
The value of R for which the PMOS transistor in figure will be biased in linear
region is
(A) 220 W
(C) 680 W
EE SP 7.61
Chapter 7
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
(A) + 100 kW
(C) + 1 MW
Page 415
(B) - 100 kW
(D) - 1 MW
YEAR 2003
EE SP 7.63
ONE MARK
nodia
The variation of drain current with gate-to-source voltage (ID - VGS characteristic)
of a MOSFET is shown in figure. The MOSFET is
In the circuit of figure, assume that the transistor has hfe = 99 and VBE = 0.7 V.
The value of collector current IC of the transistor is approximately
(A) [3.3/3.3] mA
(B) [3.3/(3.3+3.3)] mA
(C) [3.3/.33] mA
(D) [3.3(33+3.3)] mA
EE SP 7.65
For the circuit of figure with an ideal operational amplifier, the maximum phase
shift of the output vout with reference to the input vin is
www.nodia.co.in
www.gatehelp.com
Page 416
Analog Electronics
(A) 0c
(C) + 90c
Chapter 7
(B) - 90c
(D) !180c
YEAR 2003
EE SP 7.66
nodia
For the n-channel enhancement MOSFET shown in figure, the threshold voltage
Vth = 2 V. The drain current ID of the MOSFET is 4 mA when the drain resistance
RD is 1 kW.If the value of RD is increased to 4 kW, drain current ID will become
(A) 2.8 mA
(C) 1.4 mA
EE SP 7.67
TWO MARKS
(B) 2.0 mA
(D) 1.0 mA
Assuming the operational amplifier to be ideal, the gain vout /vin for the circuit
shown in figure is
(A) - 1
(B) - 20
(C) - 100
(D) - 120
EE SP 7.68
A voltage signal 10 sin wt is applied to the circuit with ideal diodes, as shown in
figure, The maximum, and minimum values of the output waveform Vout of the
circuit are respectively
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 417
(A) + 10 V and - 10 V
(B) + 4 V and - 4 V
(C) + 7 V and - 4 V
(D) + 4 V and - 7 V
nodia
************
www.nodia.co.in
www.gatehelp.com
Page 418
Analog Electronics
Chapter 7
SOLUTIONS
SOL 7.1
nodia
At balance, we have
z1 z 4 = z 2 z 3
j
R2
c R 1 - w C 1 m R 4 = c 1 + j wC 2 R 2 m R 3
R 3 = R 1 + C 2 + j wC R - 1
or
b 2 1 wC 1 R 2 l
R4
R 2 C1
Comparing real and imaginary parts, we get
R 3 = R1 + C 2
R4
R 2 C1
1
and
wC 2 R 1 - 1
=0 &w =
wC 1 R 2
R1 C1 R 2 C 2
or
SOL 7.2
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 419
= 100 # 10 # 10
= 100
0 + 10 # 103
i.e. the mid-band gain is 100.
SOL 7.3
nodia
Node voltage Vc is connected to inverting mode of op-amp. So, we have the
output due to Vc as
R
Vo+ = f V1 =- R Vc
R1
R
Again, node voltage Vd is connected to the non-inverting mode of op-amp. So, we
have the output due to Vd as
R R
Vo- =+c1 + f m f V2 = b1 + R l R Vd
R1 2R1
R 2R
Therefore, the net output of the op-amp is
Vo = Vo+ + Vo=- Vc # b R l + Vd R b1 + R l
R
2R
R
=- Vc + Vd # 2 = Vd - Vc
2
Applying KVL in the op-amp circuit, we have
- Vc + IR + I ^2Rh + IR + Vd = 0
or
Vd - Vc =- 4IR
or
Vo =- 4IR
Also, we have
- Va + 2IR + Vb = 0
or
2IR = Va - Vb
From equations (3) and (4), we get
...(1)
...(3)
Vo =- 2 ^Va - Vb h
www.nodia.co.in
www.gatehelp.com
Page 420
Analog Electronics
Chapter 7
...(4)
= 2 ^Vb - Va h
Since, the voltage at inverting and non-inverting terminals are same for ideal opamp. So, from the op-amp circuit, we have
V1 = Vb and V2 = Va
Substituting it in equation (4), we obtain
Vo = 2 ^V1 - V2h
SOL 7.4
nodia
Let the voltage at inverting terminal of op-amp be X . So, we have
X ^s h - Vo ^s h
=0
Rf
or
X ^s h = Vo ^s h
...(1)
www.nodia.co.in
www.gatehelp.com
Chapter 7
SOL 7.5
Analog Electronics
Page 421
Given:
VCE^sath = 0.2
VBE = 0.7
nodia
b = 100
Writing KVL around the input loop
5 - I B R S - 0. 7 = 0
IB = 5 - 0.7 = 4.3 3 = 2.15 mA
2 kW
2 # 10
IC = bIB = 100 # 2.15 mA = 0.215 A
Writing KVL around the output loop
SOL 7.6
5 - IC RC - VCE = 0
VCE = 5 - 0.215RC
VCE > VCE^sath
I S = IZ + I L
www.nodia.co.in
www.gatehelp.com
Page 422
Analog Electronics
Chapter 7
IZ = I S - I L
IZ max = IS max - IL min
Because when IL will be minimum IZ will be maximum.
We have
IS max = Vo - VZ = 20 - 5
Rs min
Rs min
IL min = 0
^1/4h
= 1 A
IZ max = PZ =
5
20
VZ
Substituting these values in eq (1), we get
1 = 20 - 5
20
Rs min
...(1)
(for RL " 3)
Rs min = 15 # 20 = 300 W
SOL 7.7
nodia
Correct option is (B).
Assume when both the diodes will be OFF.
In that case
vo = vi
2
For proper clipping operation, both the diode can not be forward biased together.
When both diodes are off, we have
vo = vi
2
Let v p1 , vn1 are voltage of anode and cathode of diode D1 and v p2 and vn2 are
voltage of anode and cathode of D2 . So
vo = v p1 = vi
2
vo = vn1 = vi
2
To make both diodes forward biased
v p1 > - 1 V and vn2 < - 2 V
vi > - 1 V and vi < - 2 V
or
2
2
So
clipping operation.
SOL 7.8
SOL 7.9
www.nodia.co.in
www.gatehelp.com
Chapter 7
SOL 7.10
Analog Electronics
Page 423
nodia
For
1.
2.
3.
SOL 7.11
www.nodia.co.in
www.gatehelp.com
Page 424
Analog Electronics
Chapter 7
nodia
Case II : when vi < 0
For this case, the diode is open and the output of first op-amp is
Vo1 =- R vi =- vi
R
Therefore, the output of second op-amp is
vo =- R vo1 = vi
R
Thus, we have the output voltage
0
vi > 0
vo = )
vi
vi < 0
Hence, the transfer characteristic of op-amp circuit is
SOL 7.12
www.nodia.co.in
www.gatehelp.com
Chapter 7
SOL 7.13
Analog Electronics
Page 425
VC = 0
VB = 0
IC = 5 mA
i.e.,the base collector junction is reverse biased (zero voltage) therefore, the
collector current (IC ) can have a value only if base-emitter is forward biased.
Hence,
nodia
&
&
SOL 7.14
Vout = 2V1
and, as the I/P current in Op-amp is always zero therefore, there will be no
voltage drop across 1 KW in II op-amp
i.e.,
V2 = 1 V
V2 - ^- 2h
Therefore, V1 - V2 =
1
1
or,
or,
So,
V1 - 1 = 1 + 2
V1 = 4
Vout = 2V1 = 8 volt
www.nodia.co.in
www.gatehelp.com
Page 426
SOL 7.15
Analog Electronics
Chapter 7
nodia
Correct option is (D).
Given, the input voltage VYZ = 100 sin wt
VYZ > 0
i.e., VY is a higher voltage than VZ
So, the diode will be in cutoff region. Therefore, there will no voltage difference
between X and W node.
i.e.,
VWX = 0
Now, for - ve half cycle all the four diodes will active and so, X and W
terminal is short circuited
i.e.,
VWX = 0
Hence,
VWX = 0 for all t
SOL 7.16
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 427
Is = IZ + I L
(1)
or,
IZ = Is - I L
Since, voltage across zener diode is 5 V so, current through 100 W resistor is
obtained as
Is = 10 - 5 = 0.05 A
100
Therefore, the load current is given by
IL = 5
RL
Since, for proper operation, we must have IZ $ Iknes
So, from Eq. (1), we write
0.05 A - 5 $ 10 mA
RL
50 mA - 5 $ 10 mA
RL
40 mA $ 5
RL
40 # 10-3 $ 5
RL
1
# RL
5
40 # 10-3
5
# RL
40 # 10-3
or,
125 W # RL
Therefore, minimum value of RL = 125 W
Now, we know that power rating of Zener diode is given by
nodia
PR = VZ IZ^maxh
IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent
through zener diode flows when load current is zero. i.e.,
IZ^maxh = Is = 10 - 5 = 0.05
100
Therefore,
SOL 7.17
So,
SOL 7.18
PR = 5 # 0.05 W = 250 mW
10 - (v - 0.7) # 2 - v = 0
v = 11.4 = 3.8 V > 0.7
3
i = v - 0.7 = 3.8 - 0.7 = 6.2 mA
500
500
(Assumption is true)
www.nodia.co.in
www.gatehelp.com
Page 428
Analog Electronics
Chapter 7
nodia
This is a shunt-shunt feedback amplifier.
Given parameters,
rp = VT = 25 mV = 2.5 kW
IB
0.01 mA
b
100
= 0.04 s
gm = =
rp 2.5 # 1000
Writing KCL at output node
v0 + g v + v0 - vp = 0
m p
RC
RF
v 0 : 1 + 1 D + v p :gm - 1 D = 0
RC RF
RF
Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s
v 0 (9.33 # 10-5) + v p (0.04) = 0 or v 0 =- 428.72Vp
Writing KCL at input node
vi = v p + v p + v p - vo
Rs
Rs rp
RF
= v p : 1 + 1 + 1 D - v 0 = v p (5.1 # 10-4) - v 0
RF
Rs rp RF
RF
Substituting Vp from equation (1)
vi = - 5.1 # 10-4 v - v 0
0
428.72
Rs
RF
vi
=- 1.16 # 10-6 v 0 - 1 # 10-5 v 0 Rs = 10 kW
10 # 103
...(1)
...(2)
...(1)
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 429
vi
=- 1.116 # 10-5
10 # 103
1
- 8.96
Av = v 0 =
vi
10 # 103 # 1.116 # 10-5
SOL 7.19
nodia
0 - Vi (jw) 0 - Vo (jw)
+
=0
1
R2
jwC + R1
- V (jw)
Vo (jw)
= 1 i
R2
jwC + R1
V (jw) R2
Vo (jw) =- i
R1 - j jw1C
1 " 3, so V = 0
o
wC
1 " 0, so V (jw) =- R2 V (jw)
At w " 3 (higher frequencies),
o
R1 i
wC
The filter passes high frequencies so it is a high pass filter.
H (jw) = Vo = - R2
Vi
R1 - j 1
wC
R
R
2
2
=
H (3) =
R1
R1
At w " 0 (Low frequencies),
So,
SOL 7.20
www.nodia.co.in
www.gatehelp.com
Page 430
Analog Electronics
Chapter 7
nodia
So, it will act as a Band pass filter.
SOL 7.21
We can see that both BE and BC Junction are forwarded biased. So the BJT is
operating in saturation.
Collector current
IC = 12 - 0.2 = 5.36 mA
2.2k
Y bIB
Note:- In saturation mode IC SOL 7.22
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 431
nodia
Since
Threshold value
SOL 7.23
The output
vo =- 0.7 V
2. When - 0.7 1 vi # 5.7 , both zener and diode D will be off. The circuit is
www.nodia.co.in
www.gatehelp.com
Page 432
Analog Electronics
Chapter 7
vo = 5 + 0.7 = 5.7 V
SOL 7.24
nodia
4 + 2 - vo = 0 or vo = 6 volt
SOL 7.25
We can observe that diode D2 is always off, whether D1 ,is on or off. So equivalent
circuit is.
10
10 = 5 volt
10 + 10 #
...(1)
IC - IE = bIB
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 433
V0 = 100I
B
100
V0
IB =
10 # 103
So,
&
SOL 7.27
V0
- 0.7 - V0 = 0
10 # 103
9.3 - 2V0 = 0
V0 = 9.3 = 4.65 A
2
nodia
Feedback samples output voltage and adds a negative feedback voltage (vfb) to
input.
So, it is a voltage-voltage feedback.
SOL 7.28
www.nodia.co.in
www.gatehelp.com
Page 434
SOL 7.29
Analog Electronics
Chapter 7
SOL 7.30
nodia
Similarly
3VTH - 6 = 0
VTH = 2 V (Upper threshold)
VTL - (- 3) VTL
+
=0
2
1
3VTL + 3 = 0
VTL =- 1 V (Lower threshold)
Vin < 2 Volt, V0 =+ 6 Volt
Vin > 2 Volt, V0 =- 3 Volt
Vin < - 1 Volt V0 =+ 6 Volt
Vin > - 1 Volt V0 =- 3 Volt
For
Output waveform
SOL 7.31
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 435
V0 = 10 sin wt # 10
10 + 10
Output voltage
= 5 sin wt
Due to resistor divider, voltage across diode VD < 0 (always). So it in reverse bias
for given input.
nodia
V0 = 5 sin wt
Output,
SOL 7.32
This is a current mirror circuit. Since b is high so IC1 = IC2, IB1 = IB2
VB = (- 5 + 0.7)
=- 4.3 volt
Diode D1 is forward biased.
So, current I is,
SOL 7.33
I = IC2 = IC1
0 - (- 4.3)
=
= 4.3 mA
1
www.nodia.co.in
www.gatehelp.com
Page 436
Analog Electronics
Chapter 7
Option () is correct.
Let input Vin is a sine wave shown below
nodia
To get output V0
K - gain of op-amp
V0 = K (- VP + VQ)
So, P should connected at inverting terminal of op-amp and Q with non-inverting
terminal.
www.nodia.co.in
www.gatehelp.com
Chapter 7
SOL 7.35
Analog Electronics
Page 437
1 "3
wC
nodia
a R1 = R 2 = R A
Similarly,
vA - vi + vA - 0 = 0
R3
R4
2vA = vin ,
a R 3 = R 4 = RB
Output,
vo = vi
So it will pass high frequency signal.
This is a high pass filter.
SOL 7.36
www.nodia.co.in
www.gatehelp.com
Page 438
Analog Electronics
Chapter 7
SOL 7.37
nodia
IC (sat) = 5 mA
IC (sat)
IB(sat) =
= 5 = .050 mA
b
100
IB 1 IB(sat), so transistor is in forward active region.
SOL 7.38
IC - IE
VCE - 0
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 439
PT = VCE # IC = 4 # 0.6
= 2.4 W
SOL 7.39
` IC - IE = 0.6 amp
nodia
Since op-amp is ideal v+ = v- = v1
Writing node equation.
v1 - v + v1 - 0 = 0
R1
R2
v1 c R1 + R2 m = V
R1
R1 R2
v1 = V c
R2
R1 + R2 m
iL = i1 = v1
r
= V c R2 m
r R1 + R2
SOL 7.40
Option () is correct.
This is a class-B amplifier whose efficiency is given as
h = p VP
4 VCC
where VP " peak value of input signal
VCC " supply voltage
here VP = 7 volt, VCC = 10 volt
so,
h = p # 7 # 100
10
4
= 54.95% - 55%
SOL 7.41
www.nodia.co.in
www.gatehelp.com
Page 440
Analog Electronics
Chapter 7
nodia
Since there is no feed back to the op-amp and op-amp has a high open loop gain
so it goes in saturation. Input is applied at inverting terminal so.
VP =- VCC =- 12 V
In negative half cycle of input, diode D1 is in forward bias and equivalent
circuit is shown below.
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 441
SOL 7.43
nodia
Correct option is (A).
In the circuit when Vi < 10 V, both D1 and D2 are off.
So equivalent circuit is,
Output,
Vo = 10 volt
Output,
Vo = Vi
Transfer characteristic of the circuit is
www.nodia.co.in
www.gatehelp.com
Page 442
SOL 7.44
Analog Electronics
Chapter 7
nodia
writing node equation at positive terminal of op-amp
Vth - 12 + Vth - 0 = 0
10
10
Vth = 6 volt (Positive threshold)
So, the capacitor will charge upto 6 volt.
When output V0 =- 12 V, the equivalent circuit is.
node equation
Vth + 12 + Vth - 0 = 0
2
10
5 Vth + 60 + Vth = 0
Vth =- 10 volt (negative threshold)
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 443
SOL 7.45
nodia
Vth - Vi + Vth - (- 12) = 0
15
100
20Vth - 20Vi + 3Vth + 36
23Vth
Vth
Thevenin resistance
Rth
=0
= 20 # 5 - 36 , Vi = 5 V
= 2.78 V
= 15 KW || 100 KW
= 13.04 KW
So the circuit is
www.nodia.co.in
www.gatehelp.com
Page 444
Analog Electronics
Chapter 7
So,
SOL 7.47
nodia
Voltage Vn is given by
Vn = 1 # 2 = 2 Volt
Vp = 0
Vn > Vp (so diode is in reverse bias, assumption is true)
Current through D2 is ID2 = 0
SOL 7.48
SOL 7.49
SOL 7.50
RF = 1 MW
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
R1 =- 1 # 10
Av
Av =- 10 to - 25 so value of R1
Page 445
6
R1 = 10 = 100 kW
10
for Av =- 10
6
R1' = 10 = 40 kW
25
for Av =- 25
SOL 7.52
SOL 7.53
SOL 7.54
nodia
vo =- gm vgs RD
vgs = vin
vo =- gm RD vin
Av = vo =- gm RD
vi
So,
Voltage gain
www.nodia.co.in
www.gatehelp.com
Page 446
Analog Electronics
Chapter 7
In the circuit
V1 = 3.5 V (given)
Current in zener is.
IZ = V1 - VZ
RZ
= 3.5 - 3.33 = 2 mA
0.1 # 10
SOL 7.56
nodia
a IB1 = IB2
a IC1 = IC2, IC2 = bIB2
IR can be calculate as
IR = - 5 + 03.7 =- 4.3 mA
1 # 10
So,
SOL 7.57
I =
4.3
- 4.3 mA
2
1
+
`
100 j
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 447
Here the feedback circuit samples the output voltage and produces a feed back
current Ifb which is in shunt with input signal. So this is a shunt-shunt feedback
configuration.
SOL 7.58
nodia
Applying node equation
Vp - 5 Vp + 8
+
=0
1
1
2Vp =- 3
Vp =- 1.5
Vn = 0
Vp < Vn (so the assumption is true and D1 is in reverse bias) and current in D1
ID1 = 0 mA
SOL 7.59
Here
Input resistance
www.nodia.co.in
www.gatehelp.com
Page 448
SOL 7.60
Analog Electronics
Chapter 7
= 3 Volt
So,
VSD < 3
VS - VD < 3
4 - ID R < 3
1 < ID R
ID R > 1,
R > 1000 W
SOL 7.61
SOL 7.62
nodia
ID = 1 mA
Option () is correct.
Option (B is correct.
If op-amp is ideal, no current will enter in op-amp. So current ix is
v - vy
...(1)
ix = x
1 # 106
(ideal op-amp)
v+ = v- = vx
vx - vy
+ vx - 0 3 = 0
3
100 # 10
10 # 10
vx - vy + 10vx = 0
11vx = vy
...(2)
SOL 7.64
a IE = (hfe + 1) IB
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 449
IC = hfe IB
=
SOL 7.65
3.3
99 # 3.3
mA
mA =
0.33 + 3.3
[0.33 + 3.3] # 100
..(1)
Similarly,
v+ - vin
v -0
+ +
=0
R
1
c jwC m
nodia
v+ - vin + v+ (jwCR) = 0
v+ (1 + jwCR) = Vin
By equation (1) and (2) we have
2vin
= vin + vout
1 + jwCR
2
- 1E = vout
vin ;
1 + jwCR
(1 - jwCR)
vout = vin
1 + jwCR
..(2)
q =p
www.nodia.co.in
www.gatehelp.com
Page 450
Analog Electronics
Chapter 7
So,
= 16 (2 - ID' ) 2
ID' = 4 (4 + I'D2 - 4ID' )
4I'D2 - 17 + 16 = 0
I'D2 = 2.84 mA
SOL 7.67
nodia
Applying node equation at negative terminal of op-amp,
0 - vin + 0 - vx = 0
1
10
vx - 0 + vx - vout + vx - 0 = 0
At node x
10
10
1
SOL 7.68
...(1)
vx + vx - vout + 10vx = 0
12 vx = vout
vx = vout
12
vin + vx = 0
1
10
vin =- vout
120
vout =- 120
vin
www.nodia.co.in
www.gatehelp.com
Chapter 7
Analog Electronics
Page 451
Vout = + 4 Volt
In the negative half cycle diode D1 conducts and D2 will be off so the circuit is,
Applying KVL
Vin - 10I + 4 - 10I = 0
Vin + 4 = I
20
Vin =- 10 V (Maximum value in negative half cycle)
So,
I = - 10 + 4 =- 3 mA
20
10
Vin - Vout = I
10
- 10 - Vout =- 3
10
10
nodia
Vout =- (10 - 3)
Vout =- 7 volt
***********
www.nodia.co.in
www.gatehelp.com
Page 452
Digital Electronics
Chapter 8
CHAPTER 8
DIGITAL ELECTRONICS
ONE MARK
(D) 625
nodia
YEAR 2014 EE01
EE SP 8.2
TWO MARKS
The interfacing circuit makes use of 3 Line to 8 Line decoder having 3 enables
lines E1 , E 2 , E 3 . The address of the device is
(A) 50 H
(B) 5000 H
(C) A0 H
(D) A000 H
EE SP 8.3
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
nodia
YEAR 2014 EE02
EE SP 8.4
ONE MARK
EE SP 8.5
EE SP 8.6
Page 453
TWO MARKS
The SOP (sum of products) form of a Boolean function is S ^0, 1, 3, 7, 11h, where
inputs are A, B , C , D (A is MSB, and D is LSB). The equivalent minimized
expression of the function is
(A) ^B + C h^A + C h^A + B h^C + D h
(B) ^B + C h^A + C h^A + C h^C + D h
(C) ^B + C h^A + C h^A + C h^C + D h
(D) ^B + C h^A + B h^A + B h^C + D h
A JK flip flop can be implemented by T flip-flops. Identify the correct
implementation.
www.nodia.co.in
www.gatehelp.com
Page 454
Digital Electronics
Chapter 8
nodia
EE SP 8.7
ONE MAKR
A state diagram of a logic gate which exhibits a delay in the output is shown in
the figure, where X is the dont care condition, and Q is the output representing
the state.
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 455
nodia
EE SP 8.10
TWO MARKS
Two monoshot multivibrators, one positive edge triggered ^M1h and another
negative edge triggered ^M2h, are connected as shown in figure
www.nodia.co.in
www.gatehelp.com
Page 456
EE SP 8.11
Digital Electronics
Chapter 8
nodia
A 3-bit gray counter is used to control the output of the multiplexer as shown
in the figure. The initial state of the counter is 000 2 . The output is pulled high.
The output of the circuit follows the sequence
(A) I 0 , 1, 1, I1 , I 3 , 1, 1, I2
(B) I 0 , 1, I1 , 1, I2 , 1, I 3 , 1
(C) 1, I 0 , 1, I1 , I2 , 1, I 3 , 1
(D) I 0 , I1 , I2 , I 3 , I 0 , I1 , I2 , I 3
YEAR 2013
EE SP 8.12
ONE MARK
A bulb in a staircase has two switches, one switch being at the ground floor
and the other one at the first floor. The bulb can be turned ON and also can
be turned OFF by any one of the switches irrespective of the state of the other
switch. The logic of switching of the bulb resembles
(A) and AND gate
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 457
(B) an OR gate
(C) an XOR gate
(D) a NAND gate
YEAR 2013
EE SP 8.13
nodia
(A) XY
(C) XY
EE SP 8.14
TWO MARKS
(B) XY
(D) XY
The clock frequency applied to the digital circuit shown in the figure below is
1 kHz. If the initial state of the output of the flip-flop is 0, then the frequency of
the output waveform Q in kHz is
(A) 0.25
(B) 0.5
(C) 1
(D) 2
YEAR 2012
EE SP 8.15
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 458
Digital Electronics
Chapter 8
(D) XY Z , XYZ, XY Z , XY Z
EE SP 8.16
EE SP 8.17
(D) 10
nodia
In this circuit, the race around
(A) does not occur
EE SP 8.18
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 459
YEAR 2011
EE SP 8.19
ONE MARK
(A) 1
(B) 0
(C) X
(D) X
YEAR 2011
EE SP 8.20
TWO MARKS
nodia
A two bit counter circuit is shown below
It the state QA QB of the counter at the clock time tn is 10 then the state QA QB
of the counter at tn + 3 (after three clock cycles) will be
(A) 00
(B) 01
(C) 10
EE SP 8.21
(D) 11
POP D
(A) DAD H
PUSH D
POP
DAD
INX
(B)
INX
INX
PUSH
POP H
(C) DAD D
PUSH H
XTHL
INX D
(D) INX D
INX D
XTHL
H
D
H
H
H
H
www.nodia.co.in
www.gatehelp.com
Page 460
Digital Electronics
Chapter 8
YEAR 2010
EE SP 8.22
TWO MARKS
The TTL circuit shown in the figure is fed with the waveform X (also shown).
All gates have equal propagation delay of 10 ns. The output Y of the circuit is
nodia
Statement For Linked Answer Questions: 23 and 24
The following Karnaugh map represents a function F .
EE SP 8.23
EE SP 8.24
(B) F = X Y + YZ
(D) F = X Y + Y Z
www.nodia.co.in
www.gatehelp.com
Chapter 8
EE SP 8.25
Digital Electronics
Page 461
nodia
When a CALL Addr instruction is executed, the CPU carries out the following
sequential operations internally :
Note: (R)
means content of register R
((R)) means content of memory location pointed to by R.
PC
means Program Counter
SP
means Stack Pointer
(A) (SP) incremented
(B)
(PC)!Addr
(PC)!Addr
((SP))!(PC)
(C) (PC)!Addr
(SP) incremented
((SP))!(PC)
((SP))!(PC)
(SP) incremented
(D)
((SP))!(PC)
(SP) incremented
(PC)!Addr
YEAR 2009
EE SP 8.26
ONE MARK
The following circuit has a source voltage VS as shown in the graph. The current
through the circuit is also shown.
www.nodia.co.in
www.gatehelp.com
Page 462
EE SP 8.27
Digital Electronics
Chapter 8
The increasing order of speed of data access for the following device is
(I) Cache Memory
(II) CD-ROM
(III) Dynamic RAM
(IV) Processor Registers
EE SP 8.28
nodia
The complete set of only those Logic Gates designated as Universal Gates is
(A) NOT, OR and AND Gates
(B) XNOR, NOR and NAND Gates
(C) NOR and NAND Gates
(D) XOR, NOR and NAND Gates
YEAR 2009
EE SP 8.29
TWO MARKS
EE SP 8.30
TWO MARKS
If the voltage Vi is made + 2.5 V, the voltage waveform at point P will become
www.nodia.co.in
www.gatehelp.com
Chapter 8
EE SP 8.31
Digital Electronics
Page 463
nodia
Content
26FE
00
26FF
01
2700
02
2701
03
2702
04
The content of stack (SP), program counter (PC) and (H,L) are 2700 H, 2100 H
and 0000 H respectively. When the following sequence of instruction are executed.
2100 H: DAD SP
www.nodia.co.in
www.gatehelp.com
Page 464
Digital Electronics
Chapter 8
2101 H: PCHL
the content of (SP) and (PC) at the end of execution will be
(A) PC = 2102 H, SP = 2700 H
(B) PC = 2700 H, SP = 2700 H
(C) PC = 2800 H, SP = 26FE H
(D) PC = 2A02 H, SP = 2702 H
YEAR 2007
EE SP 8.33
ONE MARK
A, B, C and D are input, and Y is the output bit in the XOR gate circuit of the
figure below. Which of the following statements about the sum S of A, B, C, D
and Y is correct ?
nodia
(A) S is always with zero or odd
(B) S is always either zero or even
(C) S = 1 only if the sum of A, B, C and D is even
(D) S = 1 only if the sum of A, B, C and D is odd
YEAR 2007
EE SP 8.34
TWO MARKS
The switch S in the circuit of the figure is initially closed, it is opened at time
t = 0 . You may neglect the zener diode forward voltage drops. What is the
behavior of vout for t > 0 ?
to + 5 V at t = 12.98 ms
to + 5 V at t = 2.57 ms
to - 5 V at t = 12.98 ms
to - 5 V at t = 2.57 ms
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 465
nodia
EE SP 8.36
EE SP 8.37
TWO MARKS
A student has made a 3-bit binary down counter and connected to the R-2R
ladder type DAC, [Gain = (- 1 kW/2R)] as shown in figure to generate a staircase
waveform. The output achieved is different as shown in figure. What could be the
possible cause of this error ?
www.nodia.co.in
www.gatehelp.com
Page 466
Digital Electronics
Chapter 8
nodia
EE SP 8.39
EE SP 8.40
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 467
YEAR 2005
ONE MARK
EE SP 8.41
The 8085 assembly language instruction that stores the content of H and L
register into the memory locations 2050H and 2051H , respectively is
(A) SPHL 2050H
(B) SPHL 2051H
(C) SHLD 2050H
(D) STAX 2050H
EE SP 8.42
nodia
(A) JK flip-flop
(C) T flip-flop
YEAR 2005
EE SP 8.43
EE SP 8.44
TWO MARKS
Select the circuit which will produce the given output Q for the input signals X1
and X2 given in the figure
If X1 and X2 are the inputs to the circuit shown in the figure, the output Q is
www.nodia.co.in
www.gatehelp.com
Page 468
Digital Electronics
(A) X1 + X2
(C) X1 : X2
EE SP 8.45
Chapter 8
(B) X1 : X2
(D) X1 : X2
nodia
(A) at 1
(B) at 0
(D) unstable
YEAR 2004
EE SP 8.46
The digital circuit using two inverters shown in figure will act as
YEAR 2004
EE SP 8.48
ONE MARK
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 469
(A) A $ D + B $ C $ D
(B) AD + B $ C $ D
(C) (A + D) (B $ C + D )
(D) A $ D + BC $ D
EE SP 8.49
EE SP 8.50
EE SP 8.51
nodia
The digital circuit shown in figure generates a modified clock pulse at the output.
Choose the correct output waveform from the options given below.
www.nodia.co.in
www.gatehelp.com
Page 470
EE SP 8.52
Digital Electronics
Chapter 8
nodia
In the Schmitt trigger circuit shown in figure, if VCE (sat) = 0.1 V , the output logic
low level (VOL) is
(A) 1.25 V
(C) 2.50 V
(B) 1.35 V
(D) 5.00 V
YEAR 2003
EE SP 8.53
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 471
Counter contains
(A) the number of instructions in the current program that have already been
executed
(B) the total number of instructions in the program being executed.
(C) the memory address of the instruction that is being currently executed
(D) the memory address of the instruction that is to be executed next
YEAR 2003
EE SP 8.55
TWO MARKS
nodia
(A) RA = 3.62 kW, RB = 3.62 kW
(B) RA = 3.62 kW, RB = 7.25 kW
(C) RA = 7.25 kW, RB = 3.62 kW
(D) RA = 7.25 kW, RB = 7.25 kW
EE SP 8.56
EE SP 8.57
(D) XY + YZ + X Z
The shift register shown in figure is initially loaded with the bit pattern 1010.
Subsequently the shift register is clocked, and with each clock pulse the pattern
gets shifted by one bit position to the right. With each shift, the bit at the serial
input is pushed to the left most position (msb). After how many clock pulses will
the content of the shift register become 1010 again ?
(A) 3
(C) 11
(B) 7
(D) 15
www.nodia.co.in
www.gatehelp.com
Page 472
EE SP 8.58
Digital Electronics
(A) J = X, K = Y
(C) J = Y, K = X
EE SP 8.59
EE SP 8.60
Chapter 8
(B) J = X, K = Y
(D) J = Y , K = X
nodia
A memory system has a total of 8 memory chips each with 12 address lines and
4 data lines, The total size of the memory system is
(A) 16 kbytes
(B) 32 kbytes
(C) 48 kbytes
(D) 64 kbytes
The following program is written for an 8085 microprocessor to add two bytes
located at memory addresses 1FFE and 1FFF
LXI H, 1FFE
MOV B, M
INR L
MOV A, M
ADD B
INR L
MOV M, A
XOR A
On completion of the execution of the program, the result of addition is found
(A) in the register A
(B) at the memory address 1000
(C) at the memory address 1F00
(D) at the memory address 2000
************
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 473
SOLUTIONS
SOL 8.1
SOL 8.2
nodia
Correct option is (B).
Given the interfacing circuit,
As the output port is at 2 ^010h. Hence, the input to the interfacing circuit is
I2 I1 I 0 = 010
or
A15 A14 A13 = 010
Now, for E1 to be enable, we have
or
A12 A11 = 10
Therefore, we have the value at address lines as
A 15 A 14 A 13 A 12 A 11
0 1 0 1 0
By default at starting the other address lines A10 A 9 .....A 0 should be zero. Thus,
we have the overall port address as
A 15 A 14 A 13 A 12 A 11 A 10 A 9 A 8 A 7 A 6 A 5 A 4 A 3 A 2 A 1 A 0
0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0
which is equivalent to 5000 H
SOL 8.3
www.nodia.co.in
www.gatehelp.com
Page 474
Digital Electronics
Chapter 8
nodia
SOL 8.4
SOL 8.5
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 475
Here we enter 0s in those cells correspond to the max terms present in the given
POS function.
Fmin = ^B + C h^A + C h^A + B h^C + D h
So,
SOL 8.6
nodia
Present State
Next State
Flip-flop Input
Qn
Qn + 1
www.nodia.co.in
www.gatehelp.com
Page 476
Digital Electronics
Chapter 8
T = ^J + q n h^K + q n h
Thus the logic diagram showing the conversion of T flip-flop to J -K flip-flop is
show below.
SOL 8.7
nodia
2005H
carry
2006H
2007H
So
RAR
DCR B
JNZ 2005
B
03H
02H
01H
00H
SOL 8.8
C
1
1
0
0
A
01H
80H
COH
60H
For the state diagram, we form the truth table (irrespective of the current state)
as
A
Thus, from the truth table, it is clear that the logic gate is NAND gate
www.nodia.co.in
www.gatehelp.com
Chapter 8
SOL 8.9
Digital Electronics
Page 477
nodia
www.nodia.co.in
www.gatehelp.com
Page 478
Digital Electronics
Chapter 8
and
Q2 = 1
So, the output of AND gate is 1; i.e. M1 is triggered. Hence, output Q1 is high
for duration T1 . Since, M2 is negative edge triggered, So M2 will be off for the
duration T1 and hence, output vo will be as
After T1 time, the output Q1 becomes low, and the M2 gets ON (due to negative
edge triggering). Hence, the output vo again goes high for duration T2 . Thus, the
complete output waveform is
SOL 8.11
nodia
Correct option is (A).
We have the given logic circuit
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
100
SOL 8.12
Page 479
I2
Y(Bulb)
up(1)
up(1)
OFF(0)
Down(0)
Down(0)
OFF(0)
up(1)
Down(0)
ON(1)
Down(0)
up(1)
ON(1)
nodia
When the switches A and B are both up or both down, output will be zero (i.e.
Bulb will be OFF). Any of the switch changes its position leads to the ON state
of bulb. Hence, from the truth table, we get
Y = A5B
i.e., the XOR gate.
SOL 8.13
www.nodia.co.in
www.gatehelp.com
Page 480
Digital Electronics
Chapter 8
nodia
SOL 8.15
F = XY + XY
1prime
44 2
44 3
implicants
SOL 8.16
Y = 1, when A > B
A = a1 a 0, B = b1 b 0
a1
a0
b1
b0
Total combination = 6
www.nodia.co.in
www.gatehelp.com
Chapter 8
SOL 8.17
Digital Electronics
Page 481
nodia
SOL 8.18
SOL 8.19
Qn + 1 = A Qn + AQn
Qn + 1 = Qn
Qn + 1 = Qn
X 0 = Q , X1 = Q
(toggle of previous state)
www.nodia.co.in
www.gatehelp.com
Page 482
Digital Electronics
Chapter 8
Y = X 5 X = X X + XX
= XX + X X = X + X = 1
SOL 8.20
SOL 8.21
SOL 8.22
nodia
Clock pulse
QA
QB
QA (n + 1)
QB (n + 1)
Initially(tn )
tn + 1
tn + 2
tn + 3
Output Y is written as Y = X 5 B
Since each gate has a propagation delay of 10 ns.
www.nodia.co.in
www.gatehelp.com
Chapter 8
SOL 8.23
Digital Electronics
Page 483
nodia
F = X Y + YZ
SOL 8.24
SOL 8.25
SOL 8.26
SOL 8.27
www.nodia.co.in
www.gatehelp.com
Page 484
Digital Electronics
Chapter 8
SOL 8.29
nodia
2s complement of (- B) = 0 0 0 1 0 0 0 0
A + (- B) = A - B = 0 0 0 1 0 0 0 0 = 10 H
SOL 8.30
SOL 8.31
Correct option is ()
F = X Z + Y Z + XYZ
In POS form
F = (Y + Z) (X + Z) (X + Y + Z )
Since all outputs are active low so each input in above expression is complemented
F = (Y + Z ) (X + Z ) (X + Y + Z)
SOL 8.32
SP = 2700 H
PC = 2100 H
HL = 0000 H
Executing given instruction set in following steps,
DAD SP & Add register pair (SP) to HL register
HL = HL + SP
HL = 0000 H + 2700 H
HL = 2700 H
PCHL & Load program counter with HL contents
PC = HL = 2700 H
So after execution contents are,
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 485
PC = 2700 H, HL = 2700 H
SOL 8.33
SOL 8.34
nodia
-t
RC
-t
RC
vc (t) = 20 (1 - e )
Voltage at positive terminal of op-amp
v+ - vout v+ - 0
+
=0
100
10
v+ = 10 vout
11
Due to zener diodes, - 5 # vout # + 5
So,
v+ = 10 (5) V
11
Transistor form - 5 V to + 5 V occurs when capacitor charges upto v+ .
So
20 (1 - e - t/RC ) = 10 # 5
11
1 - e - t/RC = 5
22
17 = e - t/RC
22
t = RC ln ` 22 j = 1 # 103 # .01 # 10 - 6 # 0.257 = 2.57 msec
17
Voltage waveforms in the circuit is shown below
www.nodia.co.in
www.gatehelp.com
Page 486
SOL 8.35
Digital Electronics
Chapter 8
nodia
a
So
SOL 8.36
So,
SOL 8.37
Correct option is ()
SOL 8.38
SOL 8.39
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 487
Then DCR H will be executed and it goes to Loop again, since L is of 8 bit so
no more decrement possible and it terminates.
SOL 8.40
SOL 8.41
SOL 8.42
nodia
Correct option is (C).
Let the present state is Q(t), so input to D-flip flop is given by,
D = Q (t) 5 X
Next state can be obtained as,
Q (t + 1) = D
= Q (t) 5 X
and
= Q (t) X + Q (t) X
= Q (t), if X = 1
Q (t + 1) = Q (t), if X = 0
SOL 8.44
SOL 8.45
www.nodia.co.in
www.gatehelp.com
Page 488
Digital Electronics
Chapter 8
SOL 8.47
nodia
SOL 8.48
Y = (A $ BC + D) (A $ D + B $ C )
= (A $ BCD) + (ABC $ B $ C ) + (AD) + B C D
= A BCD + AD + B C D
= AD (BC + 1) + B C D
= AD + B C D
SOL 8.49
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 489
nodia
So total no. of instruction cycles are
n = 1+6+5+1
= 13
SOL 8.50
SOL 8.51
www.nodia.co.in
www.gatehelp.com
Page 490
Digital Electronics
Chapter 8
SOL 8.52
nodia
Correct option is (B).
In the given circuit Vi = 0 V
So, transistor Q1 is in cut-off region and Q2 is in saturation.
5 - IC RC - VCE(sat) - 1.25 = 0
5 - IC RC - 0.1 - 1.25 = 0
5 - IC RC = 1.35
V0 = 1.35
SOL 8.53
V0 = 5 - IC R
Cin
Sum
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 491
Inputs are,
SOL 8.54
SOL 8.55
nodia
Frequency
...(2)
...(1)
www.nodia.co.in
www.gatehelp.com
Page 492
SOL 8.57
Digital Electronics
Chapter 8
X = X1 5 X 0 , Y = X 2
Serial Input Z = X 5 Y = [X1 5 X0] 5 X2
Truth table for the circuit can be obtain as.
nodia
Clock pulse
Serial Input
Shift register
Initially
1010
1101
0110
0011
0001
1000
0100
1010
Qn
Qn+1
www.nodia.co.in
www.gatehelp.com
Chapter 8
Digital Electronics
Page 493
SOL 8.59
nodia
Correct option is (A).
Total size of the memory system is given by.
= (212 # 4) # 8 bits
SOL 8.60
www.nodia.co.in
www.gatehelp.com
Page 494
Power Electronics
Chapter 9
CHAPTER 9
POWER ELECTRONICS
ONE MARK
The figure shows the circuit of a rectifier fed from a 230 V (rms), 50 Hz sinusoidal
voltage source. If we want to replace the current source with a resistor so that
the rms value of the current supplied by the voltage source remains unchanged,
the value of the resistance (in ohms) is _____.
(Assume diodes to be ideal.)
nodia
EE SP 9.2
Figure shows four electronic switches (i), (ii), (iii) and (iv). Which of the switches
can block voltages of either polarity (applied between terminals a and b) when
the active device is in the OFF state ?
TWO MARKS
The figure shows the circuit diagram of a rectifier. The load consists of a resistance
10 W and an inductance 0.05 H connected in series. Assuming ideal thyristor and
ideal diode, the thyristor firing angle (in degree) needed to obtain an average
load voltage of 70 V is ____.
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.4
Power Electronics
Page 495
Figure (i) shows the circuit diagram of a chopper. The switch S in the circuit
in figure (i) is switched such that the voltage vD across the diode has the wave
shape as shown in figure (ii). The capacitance C is large so that the voltage
across it is constant. If switch S and the diode are ideal, the peak to peak ripple
(in A) in the inductor current is ______
nodia
EE SP 9.5
The figure shows one period of the output voltage of an inverter a should be
chosen such that 60c < a < 90c. If rms value of the fundamental component is
50 V, then a in degree is _____.
ONE MARK
www.nodia.co.in
www.gatehelp.com
Page 496
Power Electronics
Chapter 9
EE SP 9.8
TWO MARKS
A fully controlled converter bridge feeds a highly inductive load with ripple free
load current. The input supply (vs ) to the bridge is a sinusoidal source. Triggering
angle of the bridge converter is a = 30c. The input power factor of the bridge is
nodia
EE SP 9.9
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of
50 ms is applied to the SCR. The maximum value of R in W to ensure successful
firing of the SCR is ______.
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 497
If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in
ms), for which output is LOW, is _____.
EE SP 9.11
nodia
The converter is operated at a firing angle of 30c. Assuming the load current ^I 0h
to be virtually constant at 1 p.u. and transformer to be an ideal one, the input
phase current waveform is
www.nodia.co.in
www.gatehelp.com
Page 498
Power Electronics
Chapter 9
n
EE SP 9.12
EE SP 9.13
nodia
If the load current is sinusoidal and is zero at 0, p , 2p ....., the node voltage VAO
has the waveform
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 499
nodia
YEAR 2013
EE SP 9.14
TWO MARKS
Thyristor T in the figure below is initially off and is triggered with a single
pulse of width 10 ms . It is given that L = b 100 l mH and C = b 100 l mF . Assuming
p
p
latching and holding currents of the thyristor are both zero and the initial charge
on C is zero, T conducts for
www.nodia.co.in
www.gatehelp.com
Page 500
Power Electronics
(A) 10 ms
(C) 100 ms
EE SP 9.15
Chapter 9
(B) 50 ms
(D) 200 ms
The separately excited dc motor in the figure below has a rated armature current
of 20 A and a rated armature voltage of 150 V. An ideal chopper switching
at 5 kHz is used to control the armature voltage. If La = 0.1 mH , Ra = 1 W ,
neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the
rated torque at the rated speed and the rated field current is
nodia
(A) 0.4
(B) 0.5
(C) 0.6
(D) 0.7
In the figure shown below, the chopper feeds a resistive load from a battery
source. MOSFET Q is switched at 250 kHz, with duty ratio of 0.4. All elements
of the circuit are assumed to be ideal
EE SP 9.16
EE SP 9.17
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.18
EE SP 9.19
Power Electronics
Page 501
In the interval when V0 < 0 and i 0 > 0 the pair of devices which conducts the
load current is
(B) Q 3, Q 4
(A) Q1, Q2
(C) D1, D2
(D) D 3, D 4
nodia
Appropriate transition i.e., Zero Voltage Switching ^ZVS h/Zero Current Switching
^ZCS h of the IGBTs during turn-on/turn-off is
(A) ZVS during turn off
(B) ZVS during turn-on
(C) ZCS during turn off
(D) ZCS during turn-on
YEAR 2012
ONE MARK
EE SP 9.20
EE SP 9.21
EE SP 9.22
TWO MARKS
In the circuit shown, an ideal switch S is operated at 100 kHz with a duty ratio
of 50%. Given that Dic is 1.6 A peak-to-peak and I 0 is 5 A dc, the peak current
in S , is
www.nodia.co.in
www.gatehelp.com
Page 502
Power Electronics
(A) 6.6 A
(C) 5.8 A
Chapter 9
(B) 5.0 A
(D) 4.2 A
EE SP 9.23
EE SP 9.24
nodia
The rms value of load phase voltage is
(A) 106.1 V
(C) 212.2 V
(B) 141.4 V
(D) 282.8 V
EE SP 9.25
ONE MARK
A three phase current source inverter used for the speed control of an induction
motor is to be realized using MOSFET switches as shown below. Switches S1 to
S6 are identical switches.
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.26
Power Electronics
Page 503
nodia
Circuit turn-off time of an SCR is defined as the time
(A) taken by the SCR turn to be off
(B) required for the SCR current to become zero
(C) for which the SCR is reverse biased by the commutation circuit
(D) for which the SCR is reverse biased to reduce its current below the holding
current
YEAR 2011
EE SP 9.27
TWO MARKS
If the maximum value of load current is 10 A, then the maximum current through
the main (M) and auxiliary (A) thyristors will be
(A) iM max = 12 A and iA max = 10 A
(B) iM max = 12 A and iA max = 2 A
(C) iM max = 10 A and iA max = 12 A
(D) iM max = 10 A and iA max = 8 A
www.nodia.co.in
www.gatehelp.com
Page 504
Power Electronics
Chapter 9
EE SP 9.29
nodia
The kVA rating of the input transformer is
(A) 53.2 kVA
(B) 46.0 kVA
(C) 22.6 kVA
YEAR 2010
EE SP 9.30
ONE MARK
The power electronic converter shown in the figure has a single-pole doublethrow switch. The pole P of the switch is connected alternately to throws A and
B. The converter shown is a
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.32
Power Electronics
Page 505
nodia
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
(A) 150 V
(B) 210 V
(C) 300 V
(D) 100p V
YEAR 2009
EE SP 9.33
ONE MARK
EE SP 9.34
TWO MARKS
The circuit shows an ideal diode connected to a pure inductor and is connected
to a purely sinusoidal 50 Hz voltage source. Under ideal conditions the current
waveform through the inductor will look like.
www.nodia.co.in
www.gatehelp.com
Page 506
EE SP 9.35
Power Electronics
nodia
Chapter 9
In the chopper circuit shown, the main thyristor (TM) is operated at a duty ratio
of 0.8 which is much larger the commutation interval. If the maximum allowable
reapplied dv/dt on TM is 50 V/ ms , what should be the theoretical minimum value
of C1 ? Assume current ripple through L 0 to be negligible.
(A) 0.2 mF
(B) 0.02 mF
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
(C) 2 mF
EE SP 9.37
Page 507
(D) 20 mF
nodia
(A) P-I, Q-II, R-III, S-IV
(C) P-IV, Q-III, R-I, S-II
YEAR 2008
EE SP 9.38
ONE MARK
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 ?
www.nodia.co.in
www.gatehelp.com
Page 508
Power Electronics
YEAR 2008
EE SP 9.40
Chapter 9
TWO MARKS
The truth table of monoshot shown in the figure is given in the table below :
Two monoshots, one positive edge triggered and other negative edge triggered,
are connected shown in the figure, The pulse widths of the two monoshot outputs
Q1 and Q2 are TON and TON respectively.
1
nodia
The frequency and the duty cycle of the signal at Q1 will respectively be
1
(A) f =
, D= 1
5
TON + TON
TON
1
(B) f =
, D=
TON + TON
TON + TON
TON
(C) f = 1 , D =
TON
TON + TON
TON
(D) f = 1 , D =
TON
TON + TON
1
EE SP 9.41
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
(A) 0.65
(B) 0.78
(C) 0.85
(D) 0.866
EE SP 9.42
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 509
If the firing pulses are suddenly removed, the steady state voltage (V0) waveform
of the converter will become
EE SP 9.43
nodia
A single phase source inverter is feeding a purely inductive load as shown in the
figure. The inverter is operated at 50 Hz in 180c square wave mode. Assume that
the load current does not have any dc component. The peak value of the inductor
current i0 will be
(A) 6.37 A
(C) 20 A
(B) 10 A
(D) 40 A
EE SP 9.44
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
EE SP 9.45
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.
www.nodia.co.in
www.gatehelp.com
Page 510
Power Electronics
Chapter 9
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 I0 = 10 A will be
(A) 44c
(B) 51c
(C) 129c
(D) 136c
EE SP 9.46
nodia
In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5.
A large capacitor is connected across the load. The inductor current is assumed
to be continuous.
The average voltage across the load and the average current through the diode
will respectively be
(A) 10 V, 2 A
(B) 10 V, 8 A
(C) 40 V 2 A
(D) 40 V, 8 A
YEAR 2007
ONE MARK
EE SP 9.47
EE SP 9.48
EE SP 9.49
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 511
(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.
EE SP 9.50
nodia
(C) False, because it can be operated both as Current Source Inverter (CSI) or
a VSI
(D) False, because MOSFETs can be operated as excellent constant current
sources in the saturation region.
YEAR 2007
EE SP 9.51
TWO MARKS
2
V rms
- V 12
# 100
V1
where V1 is the rms value of the fundamental component of the output voltage.
The THD of output ac voltage waveform is
(A) 65.65%
(B) 48.42%
(C) 31.83%
(D) 30.49%
THD =
EE SP 9.52
EE SP 9.53
EE SP 9.54
In the circuit of adjacent figure the diode connects the ac source to a pure
inductance L.
www.nodia.co.in
www.gatehelp.com
Page 512
Power Electronics
Chapter 9
(B) 180c
(D) 360c
nodia
(A) 0 ms < t # 25 ms
(C) 50 ms < t # 75 ms
(B) 25 ms < t # 50 ms
(D) 75 ms < t # 100 ms
EE SP 9.56
(B) 470 kW
(D) 4.7 W
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.57
Power Electronics
EE SP 9.58
Page 513
ONE MARK
nodia
EE SP 9.59
www.nodia.co.in
www.gatehelp.com
Page 514
Power Electronics
Chapter 9
YEAR 2006
EE SP 9.60
EE SP 9.61
EE SP 9.62
nodia
(A) 0.0%
(B) 19.6%
(C) 31.7%
(D) 53.9%
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) IDF = 0.867; IPF = 0.828
(B) IDF = 0.867; IPF = 0.552
(C) IDF = 0.5; IPF = 0.478
(D) IDF = 0.5; IPF = 0.318
A voltage commutation circuit is shown in figure. If the turn-off time of the
SCR is 50 msec and a safety margin of 2 is considered, then what will be the
approximate minimum value of capacitor required for proper commutation ?
(A) 2.88 mF
(C) 0.91 mF
EE SP 9.63
TWO MARKS
(B) 1.44 mF
(D) 0.72 mF
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.64
EE SP 9.65
Power Electronics
Page 515
(A) 23.8 A
(B) 15 A
(C) 11.9 A
(D) 3.54 A
nodia
EE SP 9.66
The minimum time in msec for which the SCR M should be ON is.
(A) 280
(B) 140
(C) 70
(D) 0
EE SP 9.67
www.nodia.co.in
www.gatehelp.com
Page 516
Power Electronics
Chapter 9
YEAR 2005
ONE MARK
EE SP 9.68
EE SP 9.69
EE SP 9.70
2
3
(D)
400
3
nodia
EE SP 9.71
TWO MARKS
The figure shows the voltage across a power semiconductor device and the
current through the device during a switching transitions. If the transition a
turn ON transition or a turn OFF transition ? What is the energy lost during
the transition?
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 517
(A) Only P
nodia
(B) P and Q
(C) P and R
(D) R and S
EE SP 9.73
The given figure shows a step-down chopper switched at 1 kHz with a duty ratio
D = 0.5 . The peak-peak ripple in the load current is close to
(A) 10 A
(C) 0.125 A
EE SP 9.74
(A) 45c
(C) 90c
EE SP 9.75
(B) 0.5 A
(D) 0.25 A
(B) 135c
(D) 83.6c
www.nodia.co.in
www.gatehelp.com
Page 518
Power Electronics
(C) 4 kg-m2
Chapter 9
(D) 4 Nm2
YEAR 2004
EE SP 9.76
ONE MARK
nodia
(D) both the base-emitter and base-collector junctions are forward biased
EE SP 9.77
The circuit in figure shows a full-wave rectifier. The input voltage is 230 V (rms)
single-phase ac. The peak reverse voltage across the diodes D 1 and D 2 is
(A) 100 2 V
(B) 100 V
(C) 50 2 V
(D) 50 V
EE SP 9.78
(A) 10000 W
(B) 1600 W
(C) 1200 W
(D) 800 W
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.79
Power Electronics
Page 519
(A) 400 Hz
(B) 800 Hz
(C) 1200 Hz
(D) 2400 Hz
nodia
YEAR 2004
EE SP 9.80
(A) 33.8 W
(C) 7.5 W
EE SP 9.81
(B) 15.0 W
(D) 3.8 W
The triac circuit shown in figure controls the ac output power to the resistive
load. The peak power dissipation in the load is
(A) 3968 W
(C) 7935 W
EE SP 9.82
TWO MARKS
(B) 5290 W
(D) 10580 W
Figure shows a chopper operating from a 100 V dc input. The duty ratio of the
main switch S is 0.8. The load is sufficiently inductive so that the load current is
ripple free. The average current through the diode D under steady state is
www.nodia.co.in
www.gatehelp.com
Page 520
EE SP 9.83
Power Electronics
Chapter 9
(A) 1.6 A
(B) 6.4 A
(B) 8.0 A
(D) 10.0 A
Figure shows a chopper. The device S 1 is the main switching device. S 2 is the
auxiliary commutation device. S 1 is rated for 400 V, 60 A. S 2 is rated for 400 V,
30 A. The load current is 20 A. The main device operates with a duty ratio of
0.5. The peak current through S 1 is
nodia
(A) 10 A
(C) 30 A
(B) 20 A
(D) 40 A
EE SP 9.84
EE SP 9.85
A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load.
Figure shows the reversing torque and the speed during the transient. The
moment of inertia of the drive is
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 521
YEAR 2003
EE SP 9.86
ONE MARK
Figure shows a thyristor with the standard terminations of anode (A), cathode
(K), gate (G) and the different junctions named J1, J2 and J3. When the thyristor
is turned on and conducting
nodia
(A) J1 and J2 are forward biased and J3 is reverse biased
(B) J1 and J3 are forward biased and J2 is reverse biased
(C) J1 is forward biased and J2 and J3 are reverse biased
(D) J1, J2 and J3 are all forward biased
EE SP 9.87
EE SP 9.88
The speed/torque regimes in a dc motor and the control methods suitable for the
www.nodia.co.in
www.gatehelp.com
Page 522
Power Electronics
Chapter 9
EE SP 9.89
List-II
P.
Field Control
1.
Q.
Armature Control
2.
3.
4.
Codes:
(A) P-1, Q-3
nodia
(C) c1 - 3 m
2
(D)
3 -1
YEAR 2003
EE SP 9.90
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 9
EE SP 9.91
EE SP 9.92
Power Electronics
Page 523
(A) 0.48 A
(B) 1.2 A
(C) 2.4 A
(D) 1 A
nodia
An inverter has a periodic output voltage with the output wave form as shown
in figure
When the conduction angle a = 120c, the rms fundamental component of the
output voltage is
(A) 0.78 V
(B) 1.10 V
(C) 0.90 V
(D) 1.27 V
EE SP 9.93
With reference to the output wave form given in above figure , the output of the
converter will be free from 5 th harmonic when
(A) a = 72c
(B) a = 36c
(C) a = 150c
(D) a = 120c
EE SP 9.94
***********
www.nodia.co.in
www.gatehelp.com
Page 524
Power Electronics
Chapter 9
SOLUTIONS
SOL 9.1
nodia
As the diodes are ideal, so if we are maintaining same current through resistor
then Vab also does not change, i.e.
Vab = 230 V
and
Iab = 10 A
Hence, the required value of resistance is
Rab = Vab = 230 = 23 W
10
Iab
SOL 9.2
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 525
nodia
SOL 9.3
or
or
...(i)
1 + cos a = 1.3526
cos a = 0.3526
www.nodia.co.in
www.gatehelp.com
Page 526
Power Electronics
a = cos-1 ^0.3526h
= 69.35c
Thus,
SOL 9.4
Chapter 9
nodia
Since, the circuit is a buck regular. So, we have
Vo = Vs # a
= 100 # 0.05
0.1
(a = duty cycle)
= 50 V
Therefore, peak to peak inductor ripple current is obtained as
TI = Vo # Dt
L
50 # 0.05 ^m sech
=
1 mH
= 2.5 A
SOL 9.5
Since, the output voltage is an odd function, so we compute only sine terms for
the given period
Vmax
= 2
T
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 527
180 + a
360 - a
360
360 - a
= 100 6^1 - cos ah - ^cos a + cos ah + ^1 - cos ah + ^- cos a - cos ah-^- 1 + cos ah@
p
= 100 64 - 8 cos a@
p
...(i)
= 400 61 - 2 cos a@
p
Since, the rms value is
Vrms = 50 V
Vmax = 50
2
Substituting it in equation (i), we get
400 ^1 - cos ah = 50
p 2
50 p 2 E
or
cos a = 1 ;1 - #
400
2
nodia
a = 77.15c
Thus,
SOL 9.6
www.nodia.co.in
www.gatehelp.com
Page 528
Power Electronics
Chapter 9
nodia
-3
= tan-1 b 2p # 50 # 16 # 10 l - 45c
5
Voltage across the device is zero is vo = vi
If input and output voltage are identical then Vo rms = 230 V
230
Io rms = Vo rms =
2
2
Z
^5 h + ^2p # 50 # 16 # 10-3h
= 230
5 2
rms value of current through SCR
230
Ilrms = Io rms =
= 23 A
2
5 2# 2
SOL 9.9
-3
# 50 # 10
-6
ii = 0.023 = 23 mA
Rs =
100
= 5882.35 W
17 # 10-3
www.nodia.co.in
www.gatehelp.com
Chapter 9
SOL 9.10
Power Electronics
Page 529
nodia
Also, we have the upper and lower trigger level voltages as
UTL = 1.7 V
LTL = 0.9 V
For the capacitor, the voltage across it, is given as
Vc (0) = 1.7 V
Vc (3) = 0 V
and
Vc (t ) = 0.9
So, we obtain the time t after which output triggers as
0.9 = 0 + [1.7 - 0] e-t/RC
3
-6
= 0.64 ms
SOL 9.11
R = Io # 2R = 2Io
3
3R
I
o
current through
B and Y =
3
Phase an conduct from 90 to 150
current through
www.nodia.co.in
www.gatehelp.com
Page 530
Power Electronics
where B conducts,
Chapter 9
Io = 2I
3
R and Y = Io
3
current through
B = 2Io
3
The current through D -phase winding
current through
nodia
current through primary, I p = kIs
SOL 9.12
...(i)
Vs = 0
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 531
Vs = 100 sin wt = 0
So,
wt = np
or
t = p = 10 ms
100p
Therefore, the current is obtained as
1 t V dt = IdI
L t s
0
10 ms
10 ms
1
or
Vs dt =
I =1
100 sin wt + d ^wt h
3
L 2.5 ms
31.83 # 10 2.5 ms
1
cos wt t = 10 ms
=
-3 # 100 9w Ct = 2.5 ms
31.83 # 10
1
= c 1 + 1m # 100 #
p
100
31.83 # 10-3
2
= 17.07 A
or
SOL 9.13
nodia
Correct option is (D).
It is given that load current is sinusoidal, so continuous conduction.
SOL 9.14
L = 100 mH
p
C = 100 mF
p
When the circuit is triggered by 10 ms pulse, the thyristor is short circuited and
so, we consider
IC = Im sin wt
Therefore, voltage stored across capacitor is
VC = 1 IC dt
C
= Vm ^1 - cos wt h
where w is angular frequency obtained as
w = 1 = 100 1 -6 = p # 10 4
^ p h # 10
LC
So,
T = 1 = 2p = 200 ms
w
f
As
IC = Im sin wt oscillates between - ve and - ve half
cycle so, circuit is conducting for only half of cycle and thyristor is open after
half cycle.
i.e.,
the conduction period = T = 100 ms
2
SOL 9.15
www.nodia.co.in
www.gatehelp.com
Page 532
Power Electronics
Chapter 9
^T " Torqueh
I = 6Ia^rotatedh@^0.5h = 10 A
N = Nrated ,
I f = I f rated " rated field current
nodia
At the rated conditions,
E = V - Ia^ratedh Ra
or,
SOL 9.16
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 533
0=
&
DTs ^- io h + ^1 - D h Ts ^Is - io h
2
(D is duty ratio)
....(1)
nodia
Hence, from Eq. (1)
Is =
SOL 9.18
i0 = 1 = 5 A
0.6 3
1-D
In this case the + ve terminal of V0 will be at higher voltage. i.e. V0 > 0 and so
i 0 > 0 (i.e., it will be + ve ). Now, when the Q1 , Q2 goes to OFF condition we
consider the second case.
Case II : When Q 3 , Q 4 ON and Q , Q2 OFF :
In this condition, - ve terminal of applied voltage V0 will be at higher potential
i.e., V0 < 0 and since, inductor opposes the change in current so, although the
polarity of voltage V0 is inversed, current remains same in inductor i.e. I 0 > 0 .
This is the condition when conduction have been asked.
In this condition ^V0 > 0, I 0 > 0h since, IGBTs cant conduct reverse
currents therefore current will flow through D 3, D 4 until ID becomes negative.
Thus, D 3 and D 4 conducts.
SOL 9.19
SOL 9.20
www.nodia.co.in
www.gatehelp.com
Page 534
Power Electronics
Chapter 9
nodia
We note that, for continuous load current, the flywheel diode conducts from p
to p + a in a cycle. Thus, fraction of cycle that freewheel diode conducts is a/p.
Thus fraction of cycle that freewheel diode conducts is a/p.
SOL 9.21
SOL 9.22
SOL 9.23
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
VL =
Page 535
2V = 2
300
3 dc
3 #
Vdc = 300 V
= 141.4 V
SOL 9.24
P = 3 # VL
R
= 3#
SOL 9.25
SOL 9.26
SOL 9.27
(141.4) 2
- 3 kW
20
nodia
0.1 # 10-6
1 # 103
= 12 A
Maximum current through auxiliary thyristor
IA (max) = I 0 = 10 A
SOL 9.28
www.nodia.co.in
www.gatehelp.com
Page 536
Power Electronics
Chapter 9
3 VL IL =
3 # 400 # 6 # 14
p
= 7.5 kVA
SOL 9.30
nodia
Vout = DVin
where, D = Duty cycle and D < 1
SOL 9.31
Vdc =
0
2 2 Vdc
cos a
p
1
2 2 Vdc
cos 0c
p
= 300p
2 2
300 =
Vdc
At a
= 60c, Vdc = ?
2
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 537
SOL 9.33
SOL 9.34
nodia
for 0 < wt +p,
at 100pt = p/2 ,
iL = 0 , C = 0
iL =- 100 cos pt
iL (peak) = 1 Amp
SOL 9.35
www.nodia.co.in
www.gatehelp.com
Page 538
SOL 9.36
Power Electronics
Chapter 9
SOL 9.37
SOL 9.38
nodia
Correct option is (C).
Characteristics are as
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 539
R + jXL = 50 + 50j
tan f = wL = 50 = 1
50
R
f = 45c
so, firing angle a must be higher the 45c, Thus for 0 < a < 45c, V0 is
uncontrollable.
SOL 9.39
SOL 9.40
nodia
1
TON1 + TON 2
TON 2
D =
TON1 + TON 2
f =
In this case
and,
SOL 9.41
SOL 9.42
www.nodia.co.in
www.gatehelp.com
Page 540
Power Electronics
Chapter 9
nodia
Here
So
= 20 A
SOL 9.44
2
10 = 8.16 A
3#
where
1 -1
DF2
THD =
1 2
b 0.955 l - 1 = 31%
V - 2Ia + 150 = 0
Ia = V + 150
2
` I1 = 10 A, So
V =- 130 V
2Vm cos a =- 130
p
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
2#
Page 541
SOL 9.46
SOL 9.47
nodia
Correct option is (A).
Firing angle
Overlap angle
so,
20 =
`
`
a = 25c
m = 10c
I 0 = Vm [cos a - cos (a + m)]
wLs
Ls = 0.0045 H
V0 = 2Vm cos a - wLsI 0
p
p
-3
= 2 # 230 2 cos 25c - 2 # 3.14 # 50 # 4.5 # 10 # 20
3.14
3.14
= 187.73 - 9 = 178.74c
Displacement factor = V0 I 0 = 178.25 # 20 = 0.78
230 # 20
Vs Is
SOL 9.48
www.nodia.co.in
www.gatehelp.com
Page 542
Power Electronics
Chapter 9
SOL 9.49
SOL 9.50
SOL 9.51
nodia
THD =
Vrms - V 12
# 100
V1
2
Vrms = b
Here
150 V = 0.91V
s
180 l s
V1 = Vrms(fundamental)
= 0.4Vs sin 75c
p# 2
= 0.8696Vs
(0.91Vs) 2 - (0.87Vs) 2
(0.87Vs) 2
= 31.9%
THD =
SOL 9.52
e = V0 = Km wm
440 = Km # 1500 # 2p
60
Km = 2.8
cos a = 0.37
at this firing angle
Vt = 3 2 # 440 # (0.37)
p
= 219.85 V
Ia = 1500 = 34.090
440
Isr = Ia 2/3 = 27.83
p.f. =
Vt Is
= 0.354
3 Vs Isr
www.nodia.co.in
www.gatehelp.com
Chapter 9
SOL 9.53
Power Electronics
Page 543
nodia
= 1.059 W/A
SOL 9.54
SOL 9.55
SOL 9.56
SOL 9.57
I g max = 150 mA
7
= 46.67 W
R = Vm =
Ig max
150 mA
t = T,
ia = iL = 0.25
www.nodia.co.in
www.gatehelp.com
Page 544
Power Electronics
Chapter 9
So,
SOL 9.59
SOL 9.60
nodia
so,
SOL 9.61
a = 60c, IL = 10 A
www.nodia.co.in
www.gatehelp.com
Chapter 9
SOL 9.62
Power Electronics
and,
so,
4 # 10 sin 60c
Is(fundamental)
p
distortion factor =
= # 2
Is
10 # 2/3
= 0.955
input power factor = 0.955 # 0.5 = 0.478
SOL 9.63
Page 545
T = RC ln 2
T
100
=
= 2.88 mF
C =
50 # 0.693
R # 0.693
nodia
Correct option is (A).
Let we have
so
R solar = 0.5 W , I 0 = 20 A
Vs = 350 - 20 # 0.5 = 340 V
340 = 3 # 440 #
p
2 cos a
cos a = 55c
So each thyristor will reverse biased for 180c - 55c = 125c.
SOL 9.64
So
www.nodia.co.in
www.gatehelp.com
Page 546
Power Electronics
I 0 (avg) =
Chapter 9
1 [2 2
# 230 cos 38c - 200 (p - 2 # 0.66)]
2p # 2
= 11.9 A
SOL 9.65
In this given circuit minimum gate pulse width time= Time required by ia rise
up to iL
i2 = 100 3 = 20 mA
5 # 10
i1 = 100 [1 - e- 40t]
20
nodia
`
SOL 9.66
SOL 9.67
SOL 9.69
V0 =
2 Vrms = 400 2 V
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 547
SOL 9.70
SOL 9.71
SOL 9.72
SOL 9.73
nodia
Duty ratio a = 0.5
1
= 10- 3 sec
1 # 10- 3
Ta = L = 200 mH = 40 msec
5
R
T =
here
(1 - e- aT/Ts) (1 - e- (1 - a) T/Ta)
Ripple = Vs =
G
R
1 - e- T/Ts
100
(TI) max = Vs =
4fL 4 # 103 # 200 # 10- 3
= 0.125 A
SOL 9.74
so,
If whether
Then
so
SOL 9.76
= 7 Nm
= 2 rad/sec2
= Ia
= Tst - TL = 8 Nm
I = 8 = 4 kgm2
2
TL
a
T
T
www.nodia.co.in
www.gatehelp.com
Page 548
SOL 9.77
Power Electronics
Chapter 9
so,
SOL 9.78
SOL 9.79
nodia
Correct option is (C).
SOL 9.80
SOL 9.81
R = 0.15 W
I = 15 A
1 # p/w I 2 Rdt
=
(2p/w) 0
= w # 102 # 0.15 # p/w = 7.5 W
2p
Vdc =
sin p/2 2
= 2 # 230 2 = p 1 'a2p - p k + b
= 317.8 V
2 l1G
4
4 #p
2
(317.8) 2
losses = V dc =
= 10100 W
100
R
SOL 9.82
SOL 9.83
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
I = I 0 + VS C/L = 20 + 200
SOL 9.84
Option ( ) is correct.
SOL 9.85
Page 549
2 # 10- 6 = 40 A
200 # 10- 6
nodia
so
a =;
and
500 - (- 1500)
2p
2
E # 60 = 418.67 rad/sec
0.5
T = 40 Nm
T = Ia
I = T # 40 = 0.096 kgm2
a
418.67
SOL 9.86
SOL 9.87
SOL 9.89
a a = 30c
www.nodia.co.in
www.gatehelp.com
Page 550
Power Electronics
SOL 9.90
Option ( ) is correct.
SOL 9.91
Chapter 9
nodia
V -t are applied to L is = (60 - 12) Ton = 48Ton
DI = 48Ton
L
-3
= 48 # 0.2 #-10
= 0.48 A
3
20 # 10
SOL 9.92
4VS
b np l^sin nd h^sin nwt h^sin np/2h
` RMS value of fundamental component
Vrms(fundamental) = 4VS sin d # 1
2p
a = 120c, 2d = 120c & d = 60c
Output voltage V0 =
n = 1, 3, 5
www.nodia.co.in
www.gatehelp.com
Chapter 9
Power Electronics
Page 551
nodia
www.nodia.co.in
www.gatehelp.com
Page 552
Engineering Mathematics
Chapter 10
CHAPTER 10
ENGINEERING MATHEMATICS
ONE MARK
nodia
(D) The systems would have no solution for any values of b1 and b2
EE SP 10.2
EE SP 10.3
Let f ^x h = xe-x . The maximum value of the function in the interval ^0, 3h is
(A) e-1
(B) e
-1
(C) 1 - e
(D) 1 + e-1
The solution for the differential equation
d 2 x =- 9x
dt2
with initial conditions x ^0 h = 1 and dx
= 1, is
dt t = 0
(B) sin 3t + 13 cos 3t + 23
(A) t2 + t + 1
(C)
EE SP 10.4
EE SP 10.5
1
3
sin 3t + cos 3t
(D) cos 3t + t
Let S be the set of points in the complex plane corresponding to the unit circle.
(That is, S = "z : z = 1, ). Consider the function f ^z h = zz * where z* denotes
the complex conjugate of z . The f ^z h maps S to which one of the following in
the complex plane
(A) unit circle
(B) horizontal axis line segment from origin to (1, 0)
(C) the point (1, 0)
(D) the entire horizontal axis
x ^ t h is nonzero only for Tx < t < Txl, and similarly, y ^ t h is nonzero only for
Ty < t < Tyl. Let z ^ t h be convolution of x ^ t h and y ^ t h. Which one of the following
statements is TRUE ?
(A) z ^ t h can be nonzero over an unbounded interval.
(B) z ^ t h is nonzero for t < Tx + Ty
(C) z ^ t h is zero outside of Tx + Ty < t < Txl+ Tyl
(D) z ^ t h is nonzero for t > Txl+ Tyl
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 553
TWO MARKS
EE SP 10.7
A fair coin is tossed n times. The probability that the difference between the
number of heads and tails is ^n - 3h is
(A) 2-n
(B) 0
n
-n
(C) Cn - 3 2
(D) 2-n + 3
EE SP 10.8
The line integral of function F = yzi , in the counterclockwise direction, along the
circle x2 + y2 = 1 at z = 1 is
(A) - 2p
(B) - p
(C) p
(D) 2p
EE SP 10.9
EE SP 10.10
nodia
1
be the z -transform of a causal signal x 6n@.
1 - z-3
Then, the values of x 62@ and x 63@ are
(A) 0 and 0
(B) 0 and 1
(C) 1 and 0
(D) 1 and 1
Let X ^z h =
ONE MARK
EE SP 10.11
Which one of the following statements is true for all real symmetric matrices ?
(A) All the eigenvalues are real
(B) All the eigenvalues are positive
(C) All the eigenvalues are distinct
(D) Sum of all the eigenvalues is zero
EE SP 10.12
Consider a dice with the property that the probability of a free with n dots
showing up proportional to n . The probability of the face with three dots showing
up is____.
EE SP 10.13
www.nodia.co.in
www.gatehelp.com
Page 554
EE SP 10.14
EE SP 10.15
Engineering Mathematics
Chapter 10
EE SP 10.16
TWO MARKS
nodia
To evaluate the double integral
substitution
# e#
0
^y/2h + 1
y/2
2x - y
dx dy , we make the
2 l o
(A)
(C)
EE SP 10.17
EE SP 10.18
# c # 2udu mdv
0
# c # udu mdv
0
(B)
(D)
# c # 2udu mdv
0
# c # udu mdv
0
The minimum value of the function f ^x h = x3 - 3x2 - 24x + 100 in the interval
6- 3, 3@ is
(A) 20
(B) 28
(C) 16
(D) 32
YEAR 2014 EE03
ONE MARK
EE SP 10.19
EE SP 10.20
www.nodia.co.in
www.gatehelp.com
Chapter 10
EE SP 10.21
Engineering Mathematics
Page 555
EE SP 10.22
EE SP 10.23
TWO MARKS
2
Integration of the complex function f ^z h = 2 z , in the counterclockwise
z -1
direction, around z - 1 = 1, is
(A) - pi
(B) 0
(C) pi
(D) 2pi
The mean thickness and variance of silicon steel laminations are 0.2 mm and
0.02 respectively. The varnish insulation is applied on both the sides of the
laminations. The mean thickness of one side insulation and its variance are 0.1
mm and 0.01 respectively. If the transformer core is made using 100 such varnish
coated laminations, the mean thickness and variance of the core respectively are
(A) 30 mm and 0.22
(B) 30 mm and 2.44
nodia
(C) 40 mm and 2.44
EE SP 10.24
EE SP 10.25
EE SP 10.26
ONE MARK
EE SP 10.27
# Fv : dlv evaluated
x1
0
(B) only one solution > H = > H
x2
0
(D) multiple solutions
www.nodia.co.in
www.gatehelp.com
Page 556
EE SP 10.28
Engineering Mathematics
Chapter 10
The curl of the gradient of the scalar field defined by V = 2x2 y + 3y2 z + 4z2 x is
(A) 4xyavx + 6yzavy + 8zxavz
(B) 4avx + 6avy + 8avz
(C) ^4xy + 4z2h avx + ^2x2 + 6yz h avy + ^3y2 + 8zx h avz
(D) 0
EE SP 10.29
EE SP 10.30
EE SP 10.31
nodia
EE SP 10.32
# zz
, is
2
2
(D) 35
(A) - 4p
(C) 2 + p
EE SP 10.33
TWO MARKS
(B) 0
(D) 2 + 2i
1
A Matrix has eigenvalues - 1 and - 2 . The corresponding eigenvectors are > H
-1
1
and > H respectively. The matrix is
-2
1 1
(A) >
- 1 - 2H
1 2
(B) >
- 2 - 4H
-1 0
(C) >
0 - 2H
0 1
(D) >
- 2 - 3H
YEAR 2012
EE SP 10.34
-1
ONE MARK
www.nodia.co.in
www.gatehelp.com
Chapter 10
EE SP 10.35
EE SP 10.36
Engineering Mathematics
Page 557
(B) e p/2
(D) 1
Given f (z) =
EE SP 10.37
With initial condition x (1) = 0.5 , the solution of the differential equation
t dx + x = t , is
dt
(A) x = t - 1
(B) x = t 2 - 1
2
2
nodia
2
(C) x = t
2
(D) x = t
2
YEAR 2012
EE SP 10.38
EE SP 10.39
TWO MARKS
-5 -3
1 0
Given that A = >
, the value of A3 is
and I = >
H
2 0
0 1H
(A) 15A + 12I
(B) 19A + 30I
(C) 17A + 15I
(D) 17A + 21I
(D) 46
EE SP 10.40
A fair coin is tossed till a head appears for the first time. The probability that
the number of required tosses is odd, is
(A) 1/3
(B) 1/2
(C) 2/3
(D) 3/4
EE SP 10.41
The direction of vector A is radially outward from the origin, with A = krn .
where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is
(A) - 2
(B) 2
(C) 1
(D) 0
EE SP 10.42
=0
t = 0-
www.nodia.co.in
www.gatehelp.com
Page 558
Engineering Mathematics
Chapter 10
YEAR 2011
EE SP 10.43
EE SP 10.44
ONE MARK
(B) (+ 1, - 1, + 1)
(D) (- 1, + j, - j)
A point Z has been plotted in the complex plane, as shown in figure below.
nodia
The plot of the complex number Y = 1 is
Z
EE SP 10.45
With K as a constant, the possible solution for the first order differential equation
dy
= e-3x is
dx
(B) - 13 e3x + K
(A) - 13 e-3x + K
(C) - 13 e-3x + K
(D) - 3e-x + K
YEAR 2011
EE SP 10.46
TWO MARKS
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 559
Assuming the initial values are x1 = 0.0 and x2 = 1.0 , the jacobian matrix is
10
(A) >
0
0
(C) >
10
EE SP 10.47
- 0.8
- 0.6H
- 0.8
- 0.6H
10 0
(B) >
0 10H
10 0
(D) >
10 - 10H
EE SP 10.48
EE SP 10.49
EE SP 10.50
nodia
2 1
The matrix [A] = >
is decomposed into a product of a lower triangular
4 - 1H
matrix [L] and an upper triangular matrix [U]. The properly decomposed [L] and
[U] matrices respectively are
1 1
2 0
1 1
1 0
and >
(B) >
and > H
(A) >
H
H
H
0 -2
4 -1
0 1
4 -1
1 0
2 1
2 0
1 1.5
(C) > H and >
(D) >
and >
H
H
4 1
0 -1
4 -3
0 1H
The two vectors [1,1,1] and [1, a, a2] where a = _- 12 + j
3
2
i, are
(A) Orthonormal
(B) Orthogonal
(C) Parallel
(D) Collinear
YEAR 2010
ONE MARK
1
EE SP 10.51
EE SP 10.52
www.nodia.co.in
www.gatehelp.com
Page 560
Engineering Mathematics
Chapter 10
t)
(D) 3 (ti + tj + k
YEAR 2010
EE SP 10.53
EE SP 10.54
EE SP 10.55
EE SP 10.56
EE SP 10.57
TWO MARKS
A box contains 4 white balls and 3 red balls. In succession, two balls are
randomly and removed form the box. Given that the first removed ball is white,
the probability that the second removed ball is red is
(A) 1/3
(B) 3/7
(C) 1/2
(D) 4/7
J1 1 0N
O
K
An eigenvector of P = K0 2 2O is
K0 0 3O
(A) 8- 1 1 1BT
P
L
T
(C) 81 - 1 2B
(B) 81 2 1BT
(D) 82 1 - 1BT
nodia
2
For the differential equation d x2 + 6 dx + 8x = 0 with initial conditions x (0) = 1
dt
dt
and dx
= 0 , the solution is
dt t = 0
(B) x (t) = 2e- 2t - e- 4t
(A) x (t) = 2e- 6t - e- 2t
(C) x (t) =- e- 6t + 2e- 4t
(D) x (t) = e- 2t + 2e- 4t
(B) a discontinuity
(D) a maximum
YEAR 2009
EE SP 10.58
ONE MARK
EE SP 10.59
TWO MARKS
f (x, y) is a continuous function defined over (x, y) ! [0, 1] # [0, 1]. Given the two
constraints, x > y2 and y > x2 , the volume under f (x, y) is
(A)
y=1
x= y
#y = 0 #x = y
f (x, y) dxdy
(B)
y=1
x=1
#y = x #x = y
2
f (x, y) dxdy
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
(C)
y=1
x=1
#y = 0 #x = 0
f (x, y) dxdy
(D)
Page 561
y= x
#y = 0
x= y
#x = 0
f (x, y) dxdy
EE SP 10.60
Assume for simplicity that N people, all born in April (a month of 30 days), are
collected in a room. Consider the event of at least two people in the room being
born on the same date of the month, even if in different years, e.g. 1980 and 1985.
What is the smallest N so that the probability of this event exceeds 0.5 ?
(B) 7
(A) 20
(C) 15
(D) 16
EE SP 10.61
nodia
EE SP 10.62
EE SP 10.63
Let x2 - 117 = 0 . The iterative steps for the solution using Newton-Raphons
method is given by
(A) xk + 1 = 1 bxk + 117 l
(B) xk + 1 = xk - 117
2
xk
xk
(C) xk + 1 = xk - xk
(D) xk + 1 = xk - 1 bxk + 117 l
2
xk
117
YEAR 2008
ONE MARKS
EE SP 10.64
EE SP 10.65
EE SP 10.66
www.nodia.co.in
www.gatehelp.com
Page 562
Engineering Mathematics
Chapter 10
TWO MARKS
EE SP 10.67
Consider function f (x) = (x2 - 4) 2 where x is a real number. Then the function
has
(A) only one minimum
(B) only tow minima
(C) three minima
(D) three maxima
EE SP 10.68
EE SP 10.69
EE SP 10.70
nodia
A is m # n full rank matrix with m > n and I is identity matrix. Let matrix
A' = (AT A) - 1 AT , Then, which one of the following statement is FALSE ?
(B) (AA') 2
(A) AA'A = A
(D) AA'A = A'
(C) A'A = I
A differential equation dx/dt = e - 2t u (t), has to be solved using trapezoidal rule
of integration with a step size h = 0.01 s. Function u (t) indicates a unit step
function. If x (0 -) = 0 , then value of x at t = 0.01 s will be given by
(A) 0.00099
(B) 0.00495
(C) 0.0099
EE SP 10.71
(D) 0.0198
Let P be a 2 # 2 real orthogonal matrix and x is a real vector [x1, x2] T with
length x = (x12 + x22) 1/2 . Then, which one of the following statements is correct ?
(A) Px # x where at least one vector satisfies Px < x
(B) Px # x for all vector x
(C) Px $ x where at least one vector satisfies Px > x
(D) No relationship can be established between x and Px
YEAR 2007
EE SP 10.72
ONE MARK
YEAR 2007
EE SP 10.73
TWO MARKS
1-x
The differential equation dx
is discretised using Eulers numerical
dt = t
integration method with a time step 3 T > 0 . What is the maximum permissible
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 563
The value of
C
(A) 2pi
where C
# (1 dz
+ z2)
is the contour z -
(D) pi tan - 1 z
(C) tan z
EE SP 10.76
= 1 is
(B) p
-1
EE SP 10.75
i
2
The integral 1
2p
(A) sin t cos t
(C) 12 cos t
2p
nodia
A loaded dice has following probability distribution of occurrences
Dice Value
Probability
1/4
1/8
1/8
1/8
1/8
1/4
If three identical dice as the above are thrown, the probability of occurrence of
values 1, 5 and 6 on the three dice is
(A) same as that of occurrence of 3, 4, 5
(B) same as that of occurrence of 1, 2, 5
(C) 1/128
(D) 5/8
EE SP 10.77
Let x and y be two vectors in a 3 dimensional space and < x, y > denote their
dot product. Then the determinant
< x, x > < x, y >
det =< y, x > < y, y >G
(A) is zero when x and y are linearly independent
(B) is positive when x and y are linearly independent
(C) is non-zero for all non-zero x and y
(D) is zero only when either x or y is zero
EE SP 10.78
The linear operation L (x) is defined by the cross product L (x) = b # x , where
T
T
b = 80 1 0B and x = 8x1 x2 x3 B are three dimensional vectors. The 3 # 3 matrix
M of this operations satisfies
R V
Sx1 W
L (x) = M Sx2 W
SSx WW
3
T X
Then the eigenvalues of M are
(A) 0, + 1, - 1
(B) 1, - 1, 1
(C) i, - i, 1
(D) i, - i, 0
www.nodia.co.in
www.gatehelp.com
Page 564
Engineering Mathematics
Chapter 10
EE SP 10.80
(B) A2 + 2A + 2I = 0
(D) exp (A) = 0
A9 equals
(A) 511A + 510I
(B) 309A + 104I
(C) 154A + 155I
(D) exp (9A)
nodia
YEAR 2006
EE SP 10.81
EE SP 10.82
TWO MARKS
The expression V =
R
#0
(A)
#0
(C)
#0 2prH (1 - r/R) dh
pR2 (1 - h/H) 2 dr
(B)
#0
(D)
#0
pR2 (1 - h/H) 2 dh
2
2prH`1 - r j dr
R
EE SP 10.83
Two fair dice are rolled and the sum r of the numbers turned up is considered
(A) Pr (r > 6) = 16
(B) Pr (r/3 is an integer) =
5
6
EE SP 10.84
www.nodia.co.in
www.gatehelp.com
Chapter 10
EE SP 10.85
Engineering Mathematics
Page 565
R V R V R V
R V R V R V
S 6 W S- 3 W S 3 W
S 4 W S 1 W S5 W
(C) S 7 W S 2 W S 9 W
(D) S 3 W S31W S3 W
SS- 1WW SS- 2 WW SS- 4 WW
SS11WW SS 3 WW SS4 WW
T X T X T X
T X T X T X
The following vector is linearly dependent upon the solution to the previous
problem
V
R V
R
S8 W
S-2 W
(B) S- 17 W
(A) S9 W
SS3 WW
SS 30 WW
TR V X
RT VX
4
S W
S13 W
(C) S4 W
(D) S 2 W
SS5 WW
SS- 3 WW
T X
T X
nodia
YEAR 2005
EE SP 10.86
ONE MARK
EE SP 10.87
EE SP 10.88
If S =
#1
3 -3
(A) - 1
3
(C) 1
2
EE SP 10.89
(B) 1
4
(D) 1
(D) x (t) = x0 e - 1
YEAR 2005
EE SP 10.90
TWO MARKS
V
R
S3 - 2 2 W
For the matrix p = S0 - 2 1 W, one of the eigen values is equal to - 2
SS0 0 1 WW
T is anXeigen vector ?
Which of the following
www.nodia.co.in
www.gatehelp.com
Page 566
EE SP 10.91
EE SP 10.92
Engineering Mathematics
R V
R V
S3 W
S- 3 W
W
S
(A) - 2
(B) S 2 W
SS 1 WW
SS- 1WW
TR V X
RT VX
S1 W
S2 W
(C) S- 2 W
(D) S5 W
SS 3 WW
SS0 WW
T X
T X
V
R
S1 0 - 1W
If R = S2 1 - 1W, then top row of R - 1 is
SS2 3 2 WW
X
T
(A) 85 6 4B
(B) 85 - 3 1B
(C) 82 0 - 1B
(D) 82 - 1 1/2B
nodia
A fair coin is tossed three times in succession. If the first toss produces a head,
then the probability of getting exactly two heads in three tosses is
(A) 1/6
(B) 1/2
(C) 3/6
EE SP 10.93
EE SP 10.94
EE SP 10.95
Chapter 10
(D) 3/4
For the equation x'' (t) + 3x' (t) + 2x (t) = 5 ,the solution x (t) approaches which of
the following values as t " 3 ?
(A) 0
(B) 5
2
(C) 5
(D) 10
***********
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 567
SOLUTION
SOL 10.1
...(i)
...(ii)
{2(ii)-(i)}
Let b1 = b2 = 1,
9x + 4z = 1
x = 1 - 4z
9
z = 1 - 9x
4
Hence these will be infinitely many solutions for any given b1 and b2 .
SOL 10.2
nodia
Correct option is (A).
Given the function,
f ^x h = xe-x
To determine the maximum value, we equate the first derivative of function to
zero, i.e.
f l^x h = e-x + x ^- e-x h = e-x - xe-x = 0
or
e-x ^1 - x h = 0
So,
x = 1, 3
Again, we check the point of maxima by substituting the obtained values in the
second derivative as
f ll^x h =- e-x + xe-x - e-x
-1
At x = 1
f ll^x h =- 2e-1 + ^1 h e^ h
=- e-1 < 0
Hence, f ^x h is maximum at x = 1, and given as
f ^1 h = ^1 h e-1 = e-1
SOL 10.3
at x = 1
s2 X ^s h - s ^1 h - ^1 h =- 9X ^s h
X ^s h^s2 + 9h = s + 1
X ^s h = s2 + 1 = 2 s + 2 1
s +9 s +9 s +9
X ^s h = 2 s + 1 d 2 3 n
s +9 3 s +9
www.nodia.co.in
www.gatehelp.com
Page 568
Engineering Mathematics
Chapter 10
nodia
...(1)
x ^0 h = 1 = A
dx
1
and
dt t = 0 = 1 = i3B & B = 3i
Substituting these values in equation (1), we get
x = cos 3t + 1 sin 3t
3
SOL 10.4
zz * = z 2
Thus, for the given set of points, i.e. z = 1; the value of function is
f ^z h = 1
i.e. f ^z h maps S to the point (1, 0) on the plane.
SOL 10.5
g ^x h = x - 6x @
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
or
So, we have the graph
Page 569
g ^x h = "x ,
As this is a periodic function, the constant term will be is average value, i.e.
T
Average value ^a 0h = 1 f ^x h dx
T 0
1
a 0 = 1 xdx
T 0
#
#
nodia
2 1
2 1
= :1 # x D = x = 0.5
2 0 2 0
SOL 10.7
(Given)
x-y = n-3
and
x+y = n
Now, we consider the two cases.
Case I: When x > y ; we have
x-y = n-3
x+y = n
2x = 2n - 3
So,
x = 2n - 3 = n - 3
2
2
As x is an integer and n is also an integer value. So, the above relation is
contradictory, and therefore
n-3 ! x
2
Case II: When x < y ; we have
-x + y = n - 3
x+y = n
y = n-3
2
Again, y is an integer and n is also an integer value. So, the above relation is
contradictory.
Thus, the required probability is zero.
SOL 10.8
Fv = yzit
x2 + y2 = 1
www.nodia.co.in
www.gatehelp.com
Page 570
Engineering Mathematics
Chapter 10
"
##
"
##
## zdxdy
nodia
# F $ dl
"
"
## 1 dxdy
=- ## dxdy =- p ^1 h
=-
(area of circle)
=- p
SOL 10.9
1
1
1 2
-3 = 1 + 3 + c 3 m + ...
1-z
z
z
1
1
= 1 + 3 + 6 + .....
z
z
Now, the z -transform is defined as
X ^z h =
...(i)
/ x6n@z
-n
n=0
...(ii)
x 62@ = 0 , x 63@ = 1
SOL 10.10
Correct answer is 3.
Given the system matrix,
R 0
1 - 1VW
S
A = S- 6 - 11 6W
SS- 6 - 11 5WW
X
T
The Eigen values for the system matrix is given by the roots of the equation
or
or
A - lI = 0
V
R
V
R 0
1 - 1W Sl 0 0W
S
S- 6 - 11 6W - S0 l 0W = 0
SS- 6 - 11 5WW SS0 0 lWW
X T
T
X
-1
-l
1
=0
= - 6 - 11 - l 6
- 6 - 11 5 - l
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
or
or
r
Page 571
[- l ^- 55 - 5l + 11l + l2 + 66h
+ 1 ^- 36 + 30 - 6lh - 1 ^66 - 66 - 6lh = 0
^- l3 - 6l2 - 11lh - 6 - 6l + 6l = 0
l3 + 6l2 + 11l + 6 = 0
So,
l =- 1, - 2 , - 3
Thus, the ratio of maximum eigenvalue to minimum eigenvalue is
l max = - 3 = 3
-1
l min
SOL 10.11
SOL 10.12
nodia
Correct answer is 0.14 .
On the first dot probability \ 1
On the second dot probability \ 2
.
.
.
SOL 10.13
SOL 10.14
f ^x h = 10 ^x - 1h2/3
Since f ^x h is a real valued function ^x - 1h can not be negative, because in that
case fractional power of negative number is imaginary. So at x = 1, the value of
f ^x h will be minimum.
Correct option is (B).
We can write
or,
So,
1i = 6ei (2np)@
1i = e-i (2np)
1i = e-2np
Therefore all values of 1i are non-negative and real number.
or
SOL 10.15
www.nodia.co.in
www.gatehelp.com
Page 572
Engineering Mathematics
Chapter 10
dy
dy
= b1l
dx
dt x
dy
dy
=
x
dt
dx
or
or
dt
1
:t = ln x & dx = x D
2
d 2y
1 d y dt - 1 dy
=
x c dt2 dx m x2 dt
dx2
2
d 2y
1 d y - dy
=
c
dt m
dx2
x2 dt2
d 2y
d 2 y dy
or
x2 2 = 2 dt
dx
dt
Substituting eq (i) and (ii) in given differential equation, we get
d 2 y dy dy
+
-y = 0
dt dt
dt2
...(i)
Now,
...(ii)
nodia
d 2y
-y = 0
dt2
Characteristic equation
m2 - 1 = 0
m =! 1
Roots are distinct real, so two possible solution
y = et and y = e-t
SOL 10.16
Since
x = et
So
# f#
8
At x =
y
2
at x =
y
+1
2
y
+1
2
y/2
2x - y
dx p dy = ?
2 l
y
2x - y
= u & x = x- = u
2
2
b
dx = du
2 y/2 - y
u = #
=0
2
y
2 a + 1k - y
=1
u = 2
2
# ; # uduEdy
0
y
2
dy
dv =
& dy = 2dv
2
v =
at y = 0 &
at y = 8 &
v =0
v =4
=
Reduced form =
# ; # uduE2dv
0
# ; # 2uduEdv
0
www.nodia.co.in
www.gatehelp.com
Chapter 10
SOL 10.17
Engineering Mathematics
Page 573
f ^x h = *0.1
0
0.2
f ^x h = *0.1
0
for x # 1
for 1 < x # 4
otherwise
x = ^- 1, 1h
x = ^- 4, - 1h , ^1, 4h
otherwise
On solving inequalities
Probability ^P h =
=
# f ^x hdx
3
-3
0.5
0.2dx +
#
1
0.1dx
nodia
= 0.2 # 0.5 + 0.1 # 3
SOL 10.18
x2 - 2x - 8 = 0
x2 - 4x + 2x - 8 = 0
x ^x - 4h + 2 ^x - 4h = 0
^x - 4h^x + 2h = 0
So,
x = 4 and x =- 2
We consider the critical point that actually fall in the given interval. So we only
take x =- 2 . Now, we find the function value at the critical point x =- 2 and
end points x =- 3 , x = 3 .
f ^- 2h = ^- 2h3 - 3 ^- 2h2 - 24 ^- 2h + 100
f ^- 2h = 128
f ^- 3h = ^- 3h3 - 3 ^- 3h2 - 24 ^- 3h + 100
f ^- 3h = 172
f ^3h = ^3h3 - 3 ^3h2 - 24 ^3 h + 100
f ^3h = 28
We have
f ^- 2h = 128 , f ^- 3h = 172 , f ^3 h = 28
From list of above values we can see that absolute minimum is 28 which occurs
at x = 3 .
SOL 10.19
q
sH
p2 + q2 pr + qs
B =>
H
pr + qs r2 + s2
www.nodia.co.in
www.gatehelp.com
Page 574
Engineering Mathematics
Chapter 10
To determine the rank of matrix A, we obtain its equivalent matrix using the
operation, a2i ! a2i - a21 a1i as
a11
p
q
A = >0 s - r q H
p
If
s- r q = 0
p
ps - rq = 0
then rank of matrix A is 1, otherwise the rank is 2.
Now, we have the matrix
or
p2 + q2 pr + qs
B =>
H
pr + qs r2 + s1
nodia
To determine the rank of matrix B , we obtain its equivalent matrix using the
operation, a2i ! a2i - a21 a1i as
a11
V
R 2
2
pr + qs
W
Sp + q
2
W
S
pr
qs
+
^
h
B =
^r2 + s2h - 2
S 0
2 W
p +q W
S
X
T
^pr + qs h2
2
2
2
If
= ^ps - rq h = 0
^r + s h - 2
p + q2
or
ps - rq = 0
then rank of matrix B is 1, otherwise the rank is 2.
= 6sin p2 t @0p = 3
=- 1
or
|x | = 1
i.e. the magnitude of displacement is 1.
Now, we have to determine the distance covered by the particle. For calculating
distance, we have to consider speed and speed cant be negative, so the distance
is given by
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
x0 =
t=3
t=0
v dt =
Page 575
t=3
t=0
p cos p t dt
2
2
t=1
= 1 - ^- 1 - 1h = 3
Thus, the difference between the distance covered and magnitude of displacement
is
Dx = x 0 - | x | = 3 - 1 = 2
SOL 10.21
nodia
#
f ^x h dx = 1
-3
Since, the minimum and maximum lifetimes of bulbs are 1 and 2 years respectively.
So, we have
1#x#2
f ^x h dx = 1
or
or
k :x D = 1
3 1
kb 8 - 1 l = 1
3
k = 3 = 0.43
7
#
1
or
Thus,
SOL 10.22
kx2 dx = 1
3 2
Here, z - 1 = 1 is a circle with radius 1 and centre at point (1,0). So, we sketch
the contour C as
www.nodia.co.in
www.gatehelp.com
Page 576
Engineering Mathematics
Chapter 10
z2
z2 - 1
nodia
# f ^z hdz
C
SOL 10.23
SOL 10.24
f ^x h = ex - 1
So, its first derivative is obtained as
f l^x h = ex
By Newton-Raphson method, we have
f ^xi h
xi + 1 = xi f l^xi h
Since, we have the initial value,
x0 = 1
So, we obtain the 1st iteration as
x
f ^x 0h
1
= x0 - e x1 = x 0 ex
f l^x 0h
1
= 1 - e -1 1 = e-1 = .3678
e
Therefore, the 2nd iteration is
x
f ^x1h
1
= x1 - e x 2 = x1 ex
f l^x1h
= x1 - 1 + e-x = 0.3678 - 1 + e-0.3678
0
= .06005
www.nodia.co.in
www.gatehelp.com
Chapter 10
SOL 10.25
Engineering Mathematics
Page 577
SOL 10.26
# Fv : dlv = 0
nodia
A =0
We have the determinant
Hence, it will have multiple solutions
SOL 10.27
i = eip/2
- i = e-ip/2
- i = !^e-ip/2h = ! e-ip/4
= !=cos d p n - i sin d p nG
4
4
1/2
www.nodia.co.in
www.gatehelp.com
Page 578
Engineering Mathematics
Chapter 10
field is always zero. So, there is no need to solve the curl and gradient.
SOL 10.29
SOL 10.30
P ^x > 1h =
#e
3-x
0<x<3
-x 3
dx = :e D = e-1 = 0.368
-1 1
nodia
so, from N - R method we obtain
f ^xn h
xn + 1 = xn f l^xn h
x 0 = 1.2
f ^x 0h = ^1.23h + 2 ^1.2h - 1 = 3.128
Also,
f l^x h = 3x2 + 2
So,
f l^x 0h = 3 ^1.2h2 + 2 = 6.32
Hence, 1 st iterative value is
f ^x 0h
x1 = x 0 f l^x 0h
= 1.2 - 3.128
6.32
Here
= 0.705
SOL 10.31
y = f ^b h = f ^2 h = 5 ^2 h2 + 10 ^2 h = 40
Therefore,
f l^c h = 40 - 15 = 25
2-1
SOL 10.32
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 579
- 4 dz
+4
It has two poles given as
# zz
z = ! 2i
Now, the contour is defined by circle z - i = 2 which is shown in the figure
below
nodia
SOL 10.33
6A@6X@ = l 6X@
where 6A@ is the matrix as l is the scalar which gives eigen values. Now, we
consider the matrix
a b
^2 # 2 matrixh
6A@ = >c dH
1
For eigen value - 1 as eigen vector is > H, so, we have
-1
a b 1
1
>c d H>- 1H =- 1 >- 1H
or,
a - b =- 1
c-d = 1
Similarly, for
a
>c
1
eigen value - 2 with eigen vector > H, we obtain
-2
1
b 1
=- 2 > H
-2
dH>- 2H
or,
a - 2b =- 2
....(1)
....(2)
....(3)
www.nodia.co.in
www.gatehelp.com
Page 580
Engineering Mathematics
c - 2d = 4
Solving Eqs. (1) and (3), we obtain
Chapter 10
....(4)
a = 0, b = 1
and solving Eqs. (2) and (4), we obtain
c =- 2, d = 3
Thus, the required matrix is
0 1
a b
>c dH = >- 2 - 3H
SOL 10.34
nodia
P &[max (x, y)] < 1 0
2
Since X and Y are independent.
P &[max (x, y)] < 1 0
2
PbX < 1 l
2
Similarly for Y :
P bY < 1 l
2
So
P &[max (x, y)] < 1 0
2
= P b X < 1 l P bY < 1 l
2
2
= shaded area = 3
4
=3
4
=3#3= 9
4
4 16
Alternate Method:
From the given data since random variables X and Y lies in the interval [- 1, 1]
as from the figure X , Y lies in the region of the square ABCD .
Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside
the region of square AEFG .
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
So,
SOL 10.35
Page 581
x=
x = ei 2
So,
SOL 10.36
p i
nodia
f (z) =
1
2 pj
# f (z) dz
C
1 - 2
z+1 z+3
C & z+1 = 1
Only pole z =- 1 inside the circle, so residue at z =- 1 is.
-z + 1
f (z) =
(z + 1) (z + 3)
(z + 1) (- z + 1) 2
= =1
= lim
2
z "- 1 (z + 1) (z + 3)
1
So
f (z) dz = 1
2 pj C
SOL 10.37
Integrating factor,
Solution has the form
Pdt
1
# dt
t
# ^Q # IF hdt + C
x # t = # (1) (t) dt + C
x # IF =
xt = t + C
2
Taking the initial condition
C =0
2
xt = t & x = t
2
2
So,
SOL 10.38
x (1) = 0.5
0.5 = 1 + C
2
www.nodia.co.in
www.gatehelp.com
Page 582
Engineering Mathematics
Chapter 10
-5 - l -3
=0
2
-l
5l + l2 + 6 = 0
l2 + 5l + 6 = 0
Since characteristic equation satisfies its own matrix, so
A2 + 5A + 6 = 0 & A2 =- 5A - 6I
Multiplying with A
A3 + 5A2 + 6A = 0
A3 + 5 (- 5A - 6I) + 6A = 0
A3 = 19A + 30I
SOL 10.39
nodia
Correct option is (B).
&
x = 4, x = 2
d 2 f (x)
= 6x - 18
dx 2
d 2 f (x)
For x = 2,
= 12 - 18 =- 6 < 0
dx2
So at x = 2, f (x) will be maximum
f (x)
SOL 10.40
max
3
5
P = 1 + b 1 l + b 1 l + ..... = 2 1 = 2
3
2
2
2
1- 4
Probability
SOL 10.41
SOL 10.42
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
dy
2
;s Y (s) - sy (0) - dt
&
Page 583
t=0
2
6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1
Y (s) [s2 + 2s + 1] = 1 - 2s - 4
Y (s) = 2- 2s - 3
s + 2s + 1
We know If,
y (t)
dy (t)
dt
then,
Y (s)
sY (s) - y (0)
(- 2s - 3) s
+2
(s2 + 2s + 1)
2
2
= - 2s - 32 s + 2s + 4s + 2
(s + 2s + 1)
1
1
= 1 +
sY (s) - y (0) = s + 2 2 = s + 1 2 +
s + 1 (s + 1) 2
(s + 1)
(s + 1)
(s + 1) 2
Taking inverse Laplace transform
dy (t)
= e-t u (t) + te-t u (t)
dt
dy
At t = 0+ ,
= e0 + 0 = 1
dt t = 0
So,
sY (s) - y (0) =
nodia
+
SOL 10.43
x3+x2+x+1 = 0
x 2 (x + 1) + (x - 1) = 0
(x + 1) (x 2 + 1) = 0
x + 1 = 0 & x =- 1
or
x2 + 1 = 0 & x =- j, j
x =- 1, - j, j
and
SOL 10.44
www.nodia.co.in
www.gatehelp.com
Page 584
Engineering Mathematics
Chapter 10
nodia
dy = e-3x dx
by integrating, we get
SOL 10.46
SOL 10.47
SOL 10.48
Variance
sp =
-a
x 2 p (x) dx
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
-a
Page 585
x2 : 1 dx
2a
3
2
3 a
= 1 :x D = 2a = a
2a 3 -a
6
3
It means square value is equal to its variance
2
2
= sp = a
p rms
3
p rms = a
3
SOL 10.49
nodia
A = 6L@6U @
only option (D) satisfies the above relation.
SOL 10.50
X1 = [1, 1, 1]
X2 = [1, a, a 2]
Dot product of the vectors
Where
R1V
S W
X 1 $ X 2 = X1 X 2T = 81 1 1BS a W = 1 + a + a2
SSa 2WW
T X
a =- 1 + j 3 = 1 - 2p/3
2
2
so,
X1, X2 are orthogonal
1 + a + a2 = 0
Note: We can see that X1, X2 are not orthonormal as their magnitude is ! 1
SOL 10.51
#0
xex dx
= 6x # e
= 6xe
dx @0 - # 1 : d
@0 - #0
x 1
dx
(x) # ex dx D dx
(1) ex dx
= (e1 - 0) - 6e
@0
x 1
www.nodia.co.in
www.gatehelp.com
Page 586
Engineering Mathematics
Chapter 10
= e1 - [e1 - e0] = 1
SOL 10.52
SOL 10.53
SOL 10.54
nodia
Correct option is (B).
Let eigen vector
X = 8x1 x2 x 3BT
Eigen vector corresponding to l1 = 1
8A
R0
S
S0
SS0
T
I BX = 0
Rx V R0V
S 1W S W
Sx2W = S0W
SSx WW SS0WW
3
T X T X
x2 = 0
x2 + 2x 3 = 0 & x 3 = 0 (not given in the option)
Eigen vector corresponding to l2 = 2
8A - l2 I B X = 0
R- 1 1 0V Rx V R0V
S
W S 1W S W
S 0 0 2W Sx2W = S0W
SS 0 0 1WW SSx WW SS0WW
3
T
X T X T X
- x1 + x 2 = 0
2x 3 = 0 & x 3 = 0 (not given in options.)
Eigen vector corresponding to l3 = 3
- l1
1 0VW
1 2W
0 2WW
X
8A - l3 I B X = 0
R- 2 1 0V Rx V R0V
S
W S 1W S W
S 0 - 1 2W Sx2W = S0W
SS 0 0 0WW SSx WW SS0WW
3
T
X T X T X
- 2x1 + x2 = 0
- x 2 + 2x 3 = 0
Put x1 = 1, x2 = 2 and x 3 = 1
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
So Eigen vector
SOL 10.55
Page 587
Rx V R1V
S 1W S W
X = Sx2W = S2W = 81 2 1BT
SSx WW SS1WW
3
T X T X
nodia
X (s) =
(s + 6)
(s2 + 6s + 8)
By partial fraction
2 - 1
s+2 s+4
Taking inverse Laplace transform
X (s) =
SOL 10.56
.....(1)
x1 + 2x2 + x 3 + 4x 4 = 2
.....(2)
3x1 + 6x2 + 3x 3 + 12x 4 = 6
or
3 (x1 + 2x2 + x 3 + 4x 4) = 3 # 2
Equation (2) is same as equation(1) except a constant multiplying factor of 3. So
infinite (multiple) no. of non-trivial solution exists.
SOL 10.57
www.nodia.co.in
www.gatehelp.com
Page 588
SOL 10.58
Engineering Mathematics
Chapter 10
Trace A = a + d =- 2
Determinent
Eigenvalue
ad - bc =- 35
A - lI = 0
a-l b
=0
c d-l
(a - l) (d - l) - bc = 0
l - (a + d) l + (ad - bc) = 0
2
l2 - (- 2) l + (- 35) = 0
l2 + 2l - 35 = 0
nodia
(l - 5) (l + 7) = 0
l1, l2 = 5, - 7
SOL 10.59
Limit of y : y = 0 to y = 1
Limit of x : x = y2 to x2 = y & x =
So volume under f (x, y)
V =
y=1
x= y
#y = 0 #x = y
f (x, y) dx dy
SOL 10.60
SOL 10.61
Option ( ) is correct.
Assume a Cubic polynomial with real Coefficients
P (x) = a 0 x3 + a1 x3 + a2 x + a 3
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 589
nodia
So
So
SOL 10.63
f (x) = x2 - 117
f' (x) = 2x
f (xk ) = x k2 - 117
f' (xk ) = 2xk = 2 # 117
2
xk + 1 = xk - x k - 117 = xk - 1 :xk + 117 D
2xk
2
xk
# F $ dl
Line
#0
(x2 + xy) dx +
y - 2 =- x
=
#2
(y2 + xy) dy
dy =- dx
So
SOL 10.64
# F $ dl
#0
[x2 + x (2 - x)] dx +
#0
2xdx +
#2
#2
0 2
y + (2 - y) y dy
2 2
y2 0
2y dy = 2 :x D + 2 ; E = 4 - 4 = 0
2 0
2 2
#0
X3 (1) dx
4 1
= :X D = 1
4 0
4
SOL 10.65
www.nodia.co.in
www.gatehelp.com
Page 590
Engineering Mathematics
Chapter 10
a (l) = lI - P = l3 + l2 + 2l + 1 = 0
Matrix P satisfies above equation
P 3 + P 2 + 2P + I = 0
I =- (P3 + P2 + 2P)
Multiply both sides by P- 1
P- 1 =- (P2 + P + 2I)
SOL 10.66
SOL 10.67
nodia
Correct option is (B).
Given function
f (x) = (x2 - 4) 2
f' (x) = 2 (x2 - 4) 2x
To obtain minima and maxima
f' (x) = 0
4x (x - 4) = 0
x = 0, x2 - 4 = 0 & x = ! 2
x = 0, + 2, - 2
2
So,
SOL 10.68
(Maxima)
(Minima)
(Minima)
f (x 0)
, Given x 0 =- 1
f' (x 0)
f (x 0) = ex - 1 = e- 1 - 1 =- 0.63212
f' (x 0) = ex = e- 1 = 0.36787
0
x1 =- 1 -
So,
(- 0.63212)
(0.36787)
=- 1 + 1.71832 = 0.71832
SOL 10.69
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 591
option (A)
AA- 1 A = A
A =A
(AA') 2 = I
(AA- 1 I) 2 = I
option (B)
(true)
(I) 2 = I
A'A = I
option (C)
(true)
A- 1 IA = I
I =I
AA'A = A'
AA- 1 IA = A =
Y A'
option (D)
SOL 10.70
(true)
nodia
(false)
dx = e- 2t u (t)
dt
x = # e- 2t u (t) dt =
# e- 2t dt
# f (t) dt ,
t = .01 s
t0
t 0 + nh
= .0099
SOL 10.71
SOL 10.72
x 12 + x 22
= X
www.nodia.co.in
www.gatehelp.com
Page 592
Engineering Mathematics
R V
Sx1W
Sx2W
=S W
ShW
SxnW
T X
So rank of V is n .
SOL 10.73
Correct option is ( ).
SOL 10.74
Chapter 10
R V
Sx1W
Sx2W
ShW
S W
SxnW
T X
dz = #
dz
2
+
(
z
i
) (z - i)
1
z
+
C
C
Contour
z- i = 1
2
P(0, 1) lies inside the circle z - i = 1 and P (0, 1) does not lie.
2
So by Cauchys integral formula
1
(z - i)
# dz 2 = 2pi lim
z"i
+
(
z
i
)
(z - i)
1
z
+
C
= 2pi lim 1 = 2pi # 1 = p
2i
z"i z + i
Given
SOL 10.75
SOL 10.76
nodia
Correct option is ( ).
SOL 10.77
SOL 10.78
Option ( ) is correct.
SOL 10.79
(- 3 - l) (- l) + 2 = 0
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 593
(l + 1) (l + 2) = 0
According to Cayley-Hamiliton theorem
(A + I) (A + 2I) = 0
SOL 10.80
(A + I) (A + 2I) = 0
A 2 + 3A + 2 I = 0
A2 =- (3A + 2I)
A 4 = (3A + 2I) 2 = (9A2 + 12A + 4I)
or
nodia
=- 255A - 254I
A =- 255A2 - 254A
=- 255 (- 3A - 2I) - 254A
= 511A + 510I
9
SOL 10.81
V =
#0
2
pR2 b1 - h l dh
H
Correct option is ( ).
SOL 10.83
SOL 10.84
Correct option is ( ).
SOL 10.85
Correct option is ( ).
SOL 10.86
www.nodia.co.in
www.gatehelp.com
Page 594
Engineering Mathematics
Chapter 10
r (P) = r (r)
r (P) " rank of matrix P
r (r) " rank of augmented matrix [P]
r = 8P : qB
SOL 10.87
SOL 10.88
nodia
Correct option is (C).
S =
# 3 x- 3 dx
-2 3
= :x D = 1
-2 1
2
SOL 10.89
A.E.
Thus solution is
From x (0) = x 0 we get
Thus
xo(t) =- 3x (t)
xo(t) + 3x (t) = 0
D+3 = 0
x (t) = C1 e- 3t
C1 = x 0
x (t) = x 0 e- 3t
SOL 10.90
SOL 10.91
SOL 10.92
www.nodia.co.in
www.gatehelp.com
Chapter 10
Engineering Mathematics
Page 595
If the toss produces head, then for exactly two head in three tosses three tosses
there must produce one head in next two tosses. The probability of one head in
two tosses will be 1/2.
SOL 10.93
xe- x (2 - x) = 0
x = 0, 2
or
nodia
at x = 0
f'' (0) = 1 (+ ve)
at x = 2
f'' (2) =- 2e- 2 (- ve)
Now f'' (0) = 1 and f'' (2) =- 2e- 2 < 0 . Thus x = 2 is point of maxima
SOL 10.94
4 u = cti 2 + tj 2 m u
2x
2y
= ti2u + tj2u
2x
2y
= xti + 2 ytj
3
At (1, 3) magnitude is
4u =
SOL 10.95
x2 + b 2 y l = 1 + 4 =
3
2
t"3
s"0
= lim s
s"0
5
=5
2
s (s + 3s + 2)
2
***********
www.nodia.co.in
www.gatehelp.com
Page 596
General Aptitudes
Chapter 11
CHAPTER 11
GENERAL APTITUDES
EE SP 11.2
ONE MARK
Which of the following options is the closest in meaning to the phrase underlined
in the sentence below ?
It is fascinating to see life forms cope with varied environmental conditions.
(A) adopt
(B) adapt to
(C) adept in
(D) accept with
nodia
Choose the most appropriate word from the options given below to complete the
following sentence.
He could not understand the judges awarding her the first prize, because he
thought that her performance was quite __________.
(A) superb
(B) medium
(C) mediocre
EE SP 11.3
EE SP 11.4
EE SP 11.5
(D) exhilarating
In a press meet on the recent scam, the minister said, The buck stops here.
What did the minister convey by the statement ?
(A) He wants all the money
(B) He will return the money
(C) He will assume final responsibility (D) He will resist all enquiries
If ^z + 1/z h2 = 98 , compute ^z2 + 1/z2h.
The roots of ax2 + bx + c = 0 are real and positive. a , b and c are real. Then
ax2 + b x + c = 0 has
(A) no roots
(B) 2 real roots
(C) 3 real roots
(D) 4 real roots
YEAR 2014 EE01
EE SP 11.6
TWO MARKS
The Palghat Gap (or Palakkad Gap), a region about 30 km wide in the southern
part of the Western Ghats in India, is lower than the hilly terrain to its north
and south. The exact reasons for the formation of this gap are not clear. It
results in the neighbouring regions of Tamil Nadu getting more rainfall from
the South West mansoon and the neighbouring regions of Kerala having higher
summer temperatures.
What can be inferred from this passage ?
(A) The Palghat gap is caused by high rainfall and high temperatures in
southern Tamil Nadu nad Kerala
(B) The regions in Tamil Nadu and Kerala that are near the Palghat Gap are
low-lying
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 597
(C) The low terrian of the Palghat Gap has a significant impact on weather
patterns in neighbouring parts of Tamil Nadu and Kerala.
(D) Higher summer temperature result in higher rainfall near the Palghat Gap
area.
EE SP 11.7
Geneticists say that they are very close to confirming the genetic roots of
psychiatric illnesses such as depression and schizophrenia, and consequently, that
doctors will be able to eradicate these diseases through early identification and
gene therapy.
On which of the following assumptions does the statement above rely ?
(A) Strategies are now available for eliminating psychiatric illnesses
(B) Certain psychiatric illnesses have a genetic basis
(C) All human diseases can be traced back to genes and how they are
expressed
nodia
(D) In the future, genetics will become the only relevant field for identifying
psychiatric illnesses
EE SP 11.8
EE SP 11.9
Own vehicle
Men
Women
Car
40
34
Scooter
30
20
Both
60
46
20
50
EE SP 11.11
ONE MARK
Choose the most appropriate phrase from the options given below to complete
the following sentence.
India is a post-colonial country because
(A) it was a former British colony
(B) Indian Information Technology professionals have colonized the world
(C) India does not follow any colonial practices
(D) India has helped other countries gain freedom
www.nodia.co.in
www.gatehelp.com
Page 598
General Aptitudes
Chapter 11
EE SP 11.12
EE SP 11.13
Column 2
1.
eradicate
P.
misrepresent
2.
distort
Q.
soak completely
3.
saturate
R.
use
4.
utilize
S.
destroy utterly
nodia
(A) 1 : S, 2 : P, 3 : Q, 4 : R
(B) 1 : P, 2 : Q, 3 : R, 4 : S
(C) 1 : Q, 2 : R, 3 : S, 4 : P
(D) 1 : S, 2 : P, 3 : R, 4 : Q
EE SP 11.14
EE SP 11.15
The value of
(A) 3.464
(C) 4.000
12 + 12 + 12 + ... is
(B) 3.932
(D) 4.444
TWO MARKS
EE SP 11.16
The old city of Koenigsberg, which had a German majority population before
World War 2, is now called Kaliningrad. After the events of the war, Kaliningrad
is now a Russian territory and has a predominantly Russian population. It is
bordered by the Baltic Sea on the north and the countries of Poland to the south
and west and Lithuania to the east respectively. Which of the statement below
can be inferred from this passage ?
(A) Kaliningrad was historically Russian in its ethnic make up
(B) Kaliningrad is a part of Russia despite it not being contiguous with the
rest of Russia
(C) Koenigsberg was renamed Kaliningrad, as that was its original Russain
name
(D) Poland and Lithuania are on the route from Kaliningrad to the rest of
Russia
EE SP 11.17
The number of people diagnosed with dengue fever (contracted from the bite of
a mosquito) in north India is twice the number diagnosed last year. Municipal
authorities have concluded that measures to control the mosquito population
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 599
EE SP 11.19
EE SP 11.20
nodia
At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand of
a clock make an angle closest to 60c ?
(B) 6 : 27 a.m.
(A) 6 : 22 a.m.
(C) 6 : 38 a.m.
(D) 6 : 45 a.m.
YEAR 2014 EE03
EE SP 11.21
ONE MARK
While trying to collect an envelope from under the table , Mr. X fell down and
II
III
Which one of the above underlined parts of the sentence is NOT appropriate ?
(A) I
(B) II
(C) III
(D) IV
www.nodia.co.in
www.gatehelp.com
Page 600
EE SP 11.22
General Aptitudes
Chapter 11
If she ____ how to calibrate the instrument, she _____ done the experiment.
(A) knows, will have
(B) knew, had
(C) had known, could have
(D) should have known, would have
EE SP 11.23
EE SP 11.24
EE SP 11.25
nodia
Which number does not belong in the series below ?
2, 5, 10, 17, 26, 37, 50, 64
(A) 17
(B) 37
(C) 64
(D) 26
Marks
Not Attempted
21
17
15
27
23
18
What is the average of the marks obtained by the class in the examination ?
(A) 1.34
(B) 1.74
(C) 3.02
(D) 3.91
YEAR 2014 EE03
EE SP 11.26
TWO MARKS
A dance programme is scheduled for 10.00 a.m. Some students are participating
in the programme and they need to come an hour earlier than the start of the
event. These students should be accompanied by a parent. Other students and
parents should come in time for the programme. The instruction you think that
is appropriate for this is
(A) Students should come at 9.00 a.m. and parents should come at 10.00 a.m.
(B) Participating students should come at 9.00 a.m. accompanied by a parent,
and other parents and students should come by 10.00 a.m.
(C) Students who are not participating should come by 10.00 a.m. and they
should not bring their parents. Participating students should come at 9.00
a.m.
(D) Participating students should come before 9.00 a.m. Parents who
accompany them should come at 9.00 a.m. All others should come at 10.00
a.m.
www.nodia.co.in
www.gatehelp.com
Chapter 11
EE SP 11.27
General Aptitudes
Page 601
EE SP 11.28
nodia
EE SP 11.29
The ratio of male to female students in a college for five years is plotted in the
following line graph. If the number of female students in 2011 and 2012 is equal,
what is the ratio of male students in 2012 to male students in 2011 ?
(A) 1 : 1
(C) 1.5 : 1
EE SP 11.30
(B) 2 : 1
(D) 2.5 : 1
www.nodia.co.in
www.gatehelp.com
Page 602
General Aptitudes
Chapter 11
YEAR 2013
EE SP 11.31
ONE MARK
EE SP 11.32
nodia
(D) Because you need not pay towards the telephone bills when you give me a
ring
EE SP 11.33
(D) 49
EE SP 11.34
EE SP 11.35
SOL 11.1
TWO MARKS
EE SP 11.36
EE SP 11.37
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
(A)
(C)
EE SP 11.38
9 ^9n + 1h
+1
10
9 ^9n - 1h
+n
8
Page 603
(B)
(D)
9 ^9n - 1h
+1
8
9 ^9n - 1h
+ n2
8
EE SP 11.39
EE SP 11.40
nodia
The set of values of p for which the roots of the equation 3x2 + 2x + p ^p - 1h = 0
are of opposite sign is
(A) ^- 3, 0h
(B) ^0, 1h
(C) ^1, 3h
(D) ^0, 3h
What is the chance that a leap year, selected at random, will contain 53 Sundays?
(A) 2/7
(B) 3/7
(C) 1/7
(D) 5/7
2012
ONE MARK
EE SP 11.41
If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321
(B) 4.33
(A) 2.23
(D) 27.64
(C) 11.37
EE SP 11.42
Choose the most appropriate alternate from the options given below to complete
the following sentence :
If the tired soldier wanted to lie down, he..................the mattress out on the
balcony.
(A) should take
(B) shall take
(C) should have taken
(D) will have taken
EE SP 11.43
Choose the most appropriate word from the options given below to complete the
following sentence :
Give the seriousness of the situation that he had to face, his........was impressive.
(A) beggary
(B) nomenclature
(C) jealousy
(D) nonchalance
EE SP 11.44
Which one of the following options is the closest in meaning to the word given
below ?
Latitude
(A) Eligibility
(B) Freedom
(C) Coercion
(D) Meticulousness
www.nodia.co.in
www.gatehelp.com
Page 604
EE SP 11.45
General Aptitudes
Chapter 11
One of the parts (A, B, C, D) in the sentence given below contains an ERROR.
Which one of the following is INCORRECT ?
I requested that he should be given the driving test today instead of tomorrow.
(A) requested that
(B) should be given
(C) the driving test
2012
EE SP 11.46
TWO MARKS
One of the legacies of the Roman legions was discipline. In the legious, military
law prevailed and discipline was brutal. Discipline on the battlefield kept units
obedient, intact and fighting, even when the odds and conditions were against
them.
Which one of the following statements best sums up the meaning of the above
passage ?
(A) Through regimentation was the main reason for the efficiency of the
Roman legions even in adverse circumstances.
nodia
(B) The legions were treated inhumanly as if the men were animals
(C) Disciplines was the armies inheritance from their seniors
(D) The harsh discipline to which the legions were subjected to led to the odds
and conditions being against them.
EE SP 11.47
Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs.
10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10
notes that Raju has is
(A) 5
(B) 6
(C) 9
(D) 10
EE SP 11.48
There are eight bags of rice looking alike, seven of which have equal weight and
one is slightly heavier. The weighing balance is of unlimited capacity. Using this
balance, the minimum number of weighings required to identify the heavier bag
is
(A) 2
(B) 3
(C) 4
(D) 8
EE SP 11.49
The data given in the following table summarizes the monthly budget of an
average household.
Category
Amount (Rs.)
Food
4000
Clothing
1200
Rent
2000
Savings
1500
Other Expenses
1800
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
(C) 81%
EE SP 11.50
Page 605
(D) 86%
EE SP 11.51
There are two candidates P and Q in an election. During the campaign, 40%
of voter promised to vote for P , and rest for Q . However, on the day of election
15% of the voters went back on their promise to vote for P and instead voted for
Q . 25% of the voter went back on their promise to vote for Q and instead voted
for P . Suppose, P lost by 2 votes, then what was the total number of voters ?
(A) 100
(B) 110
nodia
(C) 90
EE SP 11.52
ONE MARK
(D) 95
The question below consists of a pair of related words followed by four pairs of
words. Select the pair that best expresses the relations in the original pair :
Gladiator : Arena
(A) dancer : stage
(B) commuter : train
(C) teacher : classroom
(D) lawyer : courtroom
EE SP 11.53
Choose the most appropriate word from the options given below to complete the
following sentence :
Under ethical guidelines recently adopted by the Indian Medical Association,
human genes are to be manipulated only to correct diseases for which...................
treatments are unsatisfactory.
(A) similar
(B) most
(C) uncommon
(D) available
EE SP 11.54
Choose the word from the from the options given below that is most opposite in
meaning to the given word :
Frequency
(A) periodicity
(B) rarity
(C) gradualness
(D) persistency
EE SP 11.55
Choose the most appropriate word from the options given below to complete the
following sentence :
It was her view that the countrys had been ............. by foreign techno-crafts, so
that to invite them to come back would be counter-productive.
(A) identified
(B) ascertained
(C) exacerbated
(D) analysed
www.nodia.co.in
www.gatehelp.com
Page 606
General Aptitudes
Chapter 11
2011
EE SP 11.56
TWO MARKS
The fuel consumed by a motor cycle during a journey while travelling at various
speed is indicated in the graph below.
nodia
The distance covered during four laps of the journey are listed in the table below
Lap
Distance (km)
15
15
75
45
40
75
S
10
10
From the given data, we can conclude that the fuel consumed per kilometre was
least during the lap
(A) P
(B) Q
(C) R
(D) S
EE SP 11.57
The horse has played a little known but very important role in the field of
medicine. Horses were injected with toxins of disease until their blood build up
immunities. Then a serum was made from their blood. Serums to fight with
diphteria and tetanus were developed this way.
It can be inferred from the passage, that horses were
(A) given immunity to diseases
(B) generally quite immune to diseases
(C) given medicines to fight toxins
(D) given diphtheria and tetanus serums
EE SP 11.58
EE SP 11.59
Given that f (y) = y /y, and q is any non-zero real number, the value of
f (q) - f (- q) is
(A) 0
(B) - 1
(C) 1
(D) 2
EE SP 11.60
Three friends R, S and T shared toffee from a bowl. R took 1/3 rd of the toffees,
but returned four to the bowl. S took 1/4 th of what was left but returned three
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 607
toffees to the bowl. T took half of the remainder but returned two back into the
bowl. If the bowl had 17 toffees left, how many toffees were originally there in
the bowl ?
(A) 38
(B) 31
(C) 48
(D) 41
2010
EE SP 11.61
EE SP 11.62
Which of the following options is the closest in meaning to the word below ?
Circuitous
(A) Cyclic
(B) Indirect
(C) Confusing
(D) Crooked
nodia
The question below consist of a pair of related words followed by four pairs of
words. Select the pair that best expresses the relation in the original pair.
Unemployed : Worker
(A) Fallow : Land
(B) Unaware : Sleeper
(C) Wit : Jester
EE SP 11.63
ONE MARK
Choose the most appropriate word from the options given below to complete the
following sentence :
If we manage to ........ our natural resources, we would leave a better planet for
our children.
(A) unhold
(B) restrain
(C) cherish
(D) conserve
EE SP 11.64
Choose the most appropriate word from the options given below to complete the
following sentence :
His rather casual remarks on politics..................his lack of seriousness about the
subject.
(A) masked
(B) belied
(C) betrayed
(D) suppressed
EE SP 11.65
25 persons are in a room 15 of them play hockey, 17 of them play football and 10
of them play hockey and football. Then the number of persons playing neither
hockey nor football is
(A) 2
(B) 17
(C) 13
(D) 3
2010
EE SP 11.66
TWO MARKS
Modern warfare has changed from large scale clashes of armies to suppression
of civilian populations. Chemical agents that do their work silently appear
to be suited to such warfare ; and regretfully, their exist people in military
establishments who think that chemical agents are useful fools for their cause.
Which of the following statements best sums up the meaning of the above passage
www.nodia.co.in
www.gatehelp.com
Page 608
General Aptitudes
Chapter 11
?
(A) Modern warfare has resulted in civil strife.
(B) Chemical agents are useful in modern warfare.
(C) Use of chemical agents in ware fare would be undesirable.
(D) People in military establishments like to use chemical agents in war.
EE SP 11.67
EE SP 11.68
(D) 1531
5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a
wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has
2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build
the wall ?
(A) 20 days
(B) 18 days
nodia
(C) 16 days
EE SP 11.69
EE SP 11.70
(D) 15 days
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 609
SOLUTIONS
SOL 11.1
SOL 11.2
SOL 11.3
SOL 11.4
nodia
Correct option is (C).
He will assume final responsibility.
Correct answer is 96.
Given that
1 2
bz + z l = 98
...(i)
1
2
bz + 2 l = ?
z
^a + b h2 = a2 + b2 + 2a $ b
Since,
Applying this formula in equation (i), we get
1 2
bz + z l = 98
or
or
So,
SOL 11.5
z2 + 2.z. 1 + 12 = 98
z z
z2 + 12 = 98 - 2
z
z2 + 12 = 96
z
SOL 11.6
www.nodia.co.in
www.gatehelp.com
Page 610
General Aptitudes
Chapter 11
SOL 11.7
SOL 11.8
(ii)
nodia
SOL 11.9
...(ii)
Correct answer is 6.
We draw a tetrahedron structure ABCD as
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 611
nodia
SOL 11.11
SOL 11.12
SOL 11.13
SOL 11.14
www.nodia.co.in
www.gatehelp.com
Page 612
General Aptitudes
Chapter 11
= 100
SOL 11.15
12 + 12 + ...
x ^x - 4h + 3 ^x - 4h = 0
^x + 3h^x - 4h = 0
nodia
x =- 3 , 4
x =4
SOL 11.16
SOL 11.17
SOL 11.18
- x3 + x2 - x = -^- 8h + ^4h - ^- 2h = 8 + 4 + 2 = 14
For x = 4
- x3 + x2 - x
-^64h + ^16h - 4 = 52
www.nodia.co.in
www.gatehelp.com
Chapter 11
SOL 11.19
SOL 11.20
General Aptitudes
Page 613
nodia
Correct option is (A).
As shown in Figure above, at 6:00 a.m. initial angle between minute and hour
hand is 180c. As we know that hour hand completes 30c in every hour (60
minutes), so angle moved by hour hand in 1 minute is 30
60 . Similarly, minute hand
complete 360c in every hour, so angle moved by minute hand in 1 minute is 360
60 .
Let us assume that after x minutes the angle is 60c. Let us assume that after x
minutes the angle is 60c.
60c = 180c + b 360c l x - b 30c l x
60c
60c
angle moved
by hour hand
angle moved
by min hand
www.nodia.co.in
www.gatehelp.com
Page 614
General Aptitudes
Chapter 11
SOL 11.21
SOL 11.22
SOL 11.23
SOL 11.24
SOL 11.25
nodia
SOL 11.26
SOL 11.27
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 615
1.07x
60
- 50x
x
50
# 100%
nodia
= 10.83%
= 11%
SOL 11.29
SOL 11.30
^7526h8 - ^Y h8 = ^4364h8
or
^Y h8 = ^7526h8 - ^4364h8
Octal subtraction is done in same way as decimal subtraction. The only difference
is that while obtaining carry we get 8 instead of 10.
^7526h8
-^4364h8
^3142h8
SOL 11.31
SOL 11.32
SOL 11.33
www.nodia.co.in
www.gatehelp.com
Page 616
General Aptitudes
Chapter 11
TH + TTU + TW = 41
3
TTU + TW + TTH = 43
3
and
....(1)
....(2)
also, as the temperature on Thursday was 15% higher than that of Monday
....(3)
i.e.
TTH = 1.15 TM
solving eq (1), (2) and (3), we obtain
TTH = 46cC
SOL 11.34
SOL 11.35
nodia
Correct option is (D).
They were requested not to quarrel with others.
Quarrel has a similar meaning to fall out
SOL 11.36
Time Duration
8 km
1
4
hr
6 km
1
4
hr
16 km
1
4
hr
SOL 11.37
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
SOL 11.38
SOL 11.39
Page 617
3x2 + 2x + P ^P - 1h = 0
It will have the roots with opposite sign if
P ^P - 1h < 0
So it can be possible only when
nodia
0 <P<1
i.e., P is in the range ^0, 1h
SOL 11.40
SOL 11.41
Again
SOL 11.42
SOL 11.43
SOL 11.44
www.nodia.co.in
www.gatehelp.com
Page 618
General Aptitudes
SOL 11.45
SOL 11.46
SOL 11.47
Chapter 11
x + y = 14
20x + 10y = 230
Solving the above two equations we get
x = 9, y = 5
So, the no. of notes of Rs. 10 is 5.
SOL 11.48
nodia
then C2 is heavier.
Case 2 :
A1 A 2 A 3 > B1 B 2 B 3
it means one of the A1 A2 A 3 will be heavier So we will perform next weighting as:
2 nd weighting " A1 is kept on one side of the balance and A2 on the other.
it means A 3 will be heavier
if
A1 = A 2
then A1 will be heavier
A1 > A 2
then A2 will be heavier
A1 < A 2
Case 3 :
A1 A 2 A 3 < B 1 B 2 B 3
This time one of the B1 B2 B 3 will be heavier, So again as the above case weighting
will be done.
2 nd weighting " B1 is kept one side and B2 on the other
if
B 3 will be heavier
B1 = B 2
B1 will be heavier
B1 > B 2
B1 < B 2
B2 will be heavier
So, as described above, in all the three cases weighting is done only two times to
give out the result so minimum no. of weighting required = 2.
SOL 11.49
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 619
nodia
So, the area of shaded region is given by
Area of 4PQRS
TGSH )
= 60 # 60 - 2 b 1 # 45 # 45 l
2
= 1575
So, the required probability = 1575 = 7
3600 16
SOL 11.51
SOL 11.52
www.nodia.co.in
www.gatehelp.com
Page 620
General Aptitudes
Chapter 11
performs on a stage.
SOL 11.53
SOL 11.54
SOL 11.55
SOL 11.56
SOL 11.57
SOL 11.58
nodia
SOL 11.59
Now
or
SOL 11.60
Bowl Status
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 621
= x -4
3
= 2x + 4
3
= 1 :2x + 4D - 3
4 3
= x +1-3 = x -2
6
6
= 2x + 4 - x + 2
3
6
= x +6
2
= 1 a x + 6k - 2
2 2
= x +1
4
= x +6-x -1
2
4
= x +5
4
x + 5 = 17
4
x = 17 - 5 = 12
4
Now,
nodia
or
x = 12 # 4 = 48
SOL 11.61
: Not direct
: lacking clarity of meaning
: set at an angle; not straight
SOL 11.62
SOL 11.63
SOL 11.64
SOL 11.65
SOL 11.66
SOL 11.67
www.nodia.co.in
www.gatehelp.com
Page 622
General Aptitudes
Chapter 11
731 8
672 8
1623
SOL 11.69
nodia
www.nodia.co.in
www.gatehelp.com
Chapter 11
General Aptitudes
Page 623
nodia
www.nodia.co.in