GATE EE 2015 Solved Paper

8/21/2019 GATE EE 2015 Solved Paper

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GATEELECTRICAL ENGINEERING

Solved Paper ( 2014-1996 )

RK Kanodia

Ashish Murolia

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NODIA & COMPANY


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GATE Electrical EngineeringSolved Paper (2014 - 1996 )RK Kanodia & Ashish Murolia

Copyright By NODIA & COMPANY

Information contained in this book has been obtained by author, from sources believes to be reliable. However,neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein,and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out ofuse of this information. This book is published with the understanding that NODIA & COMPANY and its author

are supplying information but are not attempting to render engineering or other professional services.

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SYLLABUS

GENERAL ABILITY

Verbal Ability : English grammar, sentence completion, verbal analogies, word groups,instructions, critical reasoning and verbal deduction.

Numerical Ability :Numerical computation, numerical estimation, numerical reasoning anddata interpretation.

ENGINEERING MATHEMATICS

Linear Algebra:Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus:Mean value theorems, Theorems of integral calculus, Evaluation of definite andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series.Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gaussand Greens theorems.

Differential equations: First order equation (linear and nonlinear), Higher order lineardifferential equations with constant coefficients, Method of variation of parameters, Cauchys

and Eulers equations, Initial and boundary value problems, Partial Differential Equations andvariable separable method.

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

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

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

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

ELECTRICAL ENGINEERING

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


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Signals and Systems:Representation of continuous and discrete-time signals; shifting andscaling operations; linear, time-invariant and causal systems; Fourier series representation ofcontinuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.

Electrical Machines: Single phase transformer equivalent circuit, phasor diagram, tests,

regulation and efficiency; three phase transformers connections, parallel operation; auto-transformer; energy conversion principles; DC machines types, windings, generatorcharacteristics, armature reaction and commutation, starting and speed control of motors;three phase induction motors principles, types, performance characteristics, starting andspeed control; single phase induction motors; synchronous machines performance, regulationand parallel operation of generators, motor starting, characteristics and applications; servo andstepper motors.

Power Systems: Basic power generation concepts; transmission line models and performance;cable performance, insulation; corona and radio interference; distribution systems; per-unitquantities; bus impedance and admittance matrices; load flow; voltage control; power factorcorrection; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuitbreakers; system stability concepts, swing curves and equal area criterion; HVDC transmissionand FACTS concepts.

Control Systems:Principles of feedback; transfer function; block diagrams; steady-state errors;Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; statespace model; state transition matrix, controllability and observability.

Electrical and Electronic Measurements:Bridges and potentiometers; PMMC, moving iron,dynamometer and induction type instruments; measurement of voltage, current, power, energyand power factor; instrument transformers; digital voltmeters and multimeters; phase, timeand frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.

Analog and Digital Electronics:Characteristics of diodes, BJT, FET; amplifiers biasing,equivalent circuit and frequency response; oscillators and feedback amplifiers; operationalamplifiers characteristics and applications; simple active filters; VCOs and timers;combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators;sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture,

programming and interfacing.

Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs,GTOs, MOSFETs and IGBTs static characteristics and principles of operation; triggeringcircuits; phase control rectifiers; bridge converters fully controlled and half controlled;principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

***********


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PREFACE

This book doesnt make promise but provides complete satisfaction to the readers. Themarket scenario is confusing and readers dont find the optimum quality books. This bookprovides 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 theconcepts of the problems are addressed in the relevant chapters. The aim of the book isto avoid the unnecessary elaboration and highlights only those concepts and techniqueswhich are absolutely necessary. Again time is a critical factor both from the point of viewof preparation duration and time taken for solving each problem in the examination. Sothe problems solving methods is the books are those which take the least distance to thesolution.

But however to make a comment that this book is absolute for GATE preparation will bean inappropriate one. The theory for the preparation of the examination should be followedfrom the standard books. But for a wide collection of problems, for a variety of problemsand the efficient way of solving them, what one needs to go needs to go through is therein there in the book. Each unit (e.g. Networks) is subdivided into average seven number ofchapters on an average each of which contains 40 problems which are selected so as to avoidunnecessary redundancy and highly needed completeness.

I shall appreciate and greatly acknowledge the comments and suggestion from the users ofthis book.

R. K. KanodiaAshish Murolia


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CONTENTS

SP 1 Electric Circuit and Fields SP 1 - 83

SP 2 Signals and Systems 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 Electrical & Electronics Measurement SP 344 - 389

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

***********


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Chapter 1 Electric Circuits and Fields Page 1

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CHAPTER 1ELECTRIC CIRCUITS AND FIELDS

YEAR 2014 EE01 ONE MARK

EE SP 1.1 The three circuit elements shown in the figure are part of an electric circuit. Thetotal power absorbed by the three circuit elements in watts is _____.

EE SP 1.2 C0 is the capacitance of a parallel plate capacitor with air as dielectric (asin figure (a)). If, half of the entire gap as shown in figure (b) is filled with adielectric of permittivity re , the expression for the modified capacitance is

(A) C2

1 r0 e+^ h (B) C r0 e+^ h(C) C

2 r0 e (D) C 1 r0 e+^ h

EE SP 1.3 A combination of F1m capacitor with an initial voltage Vv 0 2c =-^ h in serieswith a 100 W resistor is connected to a mA20 ideal dc current source by operatingboth switches at st 0= as shown. Which of the following graphs shown in theoptions approximates the voltage vsacross the current source over the next fewseconds ?


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Page 2 Electric Circuits and Fields Chapter 1


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nodiaEE SP 1.4 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 TWO MARKS.

EE SP 1.5 In the figure, the value of resistor Ris / ohmsI25 2+^ h , where Iis the currentin amperes. The current Iis _____.

EE SP 1.6 The following four vector fields are given in Cartesian co-ordinate system. Thevector field which does not satisfy the property of magnetic flux density is(A) y z xa a ax y z

2 2 2+ +

(B) z x ya a ax y z2 2 2+ +

(C) x y za a ax y z2 2 2+ +

(D) y z x z x y a a ax y z2 2 2 2 2 2+ +

YEAR 2014 EE02 ONE MARK

EE SP 1.7 Two identical coupled inductors are connected in series. The measured inductancesfor the two possible series connections are H380 m and H240 m . Their mutualinductance in Hm is _____.


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EE SP 1.8 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 V V>o o2 1 , the voltagev tc hacross the capacitor is

(A) v t V e V 1 /c ot RC

o22

1=- - --^ ^h h

(B) v t V e V1 /c ot RC

o22

1= - +-^ ^h h

(C) v t V V e V 1 /c o ot RC

o2 12

1=- + - --^ ^ ^h h h

(D) v t V V e V 1/

c o o

t RC

o2 1

2

1= - - +-

^ ^ ^h h hEE SP 1.9 A parallel plate capacitor consisting two dielectric materials is shown in the

figure. The middle dielectric slab is placed symmetrically with respect to theplates.

If the potential difference between one of the plates and the nearest surface ofdielectric interface is 2 Volts, then the ratio :1 2e e is(A) 1 : 4 (B) 2 : 3

(C) 3 : 2 (D) 4 : 1

YEAR 2014 EE02 TWO MARKS

EE SP 1.10 The voltage across the capacitor, as shown in the figure, is expressed as

v tc h sin sinA t A t 1 1 1 2 2 2w q w q= - + -^ ^h h

The values of A 1and A 2respectively, are(A) 2.0 and 1.98 (B) 2.0 and 4.20

(C) 2.5 and 3.50 (D) 5.0 and 6.40


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EE SP 1.11 The magnitude of magnetic flux density B^ hat a point having normal distancedmeters from an infinitely extended wire carrying current of IA is d

I

20

p

m (inSI units). An infinitely extended wire is laid along the x-axis and is carryingcurrent of 4 A in the ve+ xdirection. Another infinitely extended wire is

laid along the yaxis and is carrying 2 A current in the ve+ ydirection. 0m ispermeability of free space. Assume i , j , kto be unit vectors along x, yand zaxes respectively.

Assuming right handed coordinate system, magnetic field intensity, H atcoordinate , ,2 1 0^ hwill be(A) /weber mk

23 2p

(B) /A mi34p

(C) /A mk23p

(D) /A m0

YEAR 2014 EE03 ONE MARK.

EE SP 1.12 Let f x y y z z x v 2 2 2:d = + +^ h , where f and v are scalar and vector fieldsrespectively. If y z xv i j k = + + , then fv : d is(A) x y y z z x 2 2 2+ + (B) xy yz zx 2 2 2+ +

(C) x y z+ + (D) 0

EE SP 1.13 The line A to neutral voltage is V10 15c for a balanced three phase starconnected load with phase sequence ABC. The voltage of line Bwith respect toline Cis given by

(A) V10 3 105c

(B) V10 105c

(C) V10 3 75c- (D) V10 3 90c-

EE SP 1.14 The driving point impedance Z s^ hfor the circuit shown below is

(A)s s

s s

23 1

3

4 2

++ + (B)

s

s s

22 4

2

4 2

++ +

(C)s s

s

11

4 2

2

+ ++ (D)

s s

s

11

4 2

3

+ ++


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EE SP 1.15 A non-ideal voltage sourceVshas an internal impedance of Zs. If a purely resistiveload is to be chosen that maximizes the power transferred to the load, its valuemust be(A) 0 (B) real part of Zs

(C) magnitude of Zs (D) complex conjugate of Zs

YEAR 2014 EE03 TWO MARKS

EE SP 1.16 The power delivered by the current source, in the figure, is ____.

EE SP 1.17 A perfectly conducting metal plate is placed in x-yplane in a right handedcoordinate system. A charge of 32 20pe+ coulombs is placed at coordinate

, ,0 0 2^ h. 0e is the permittivity of free space. Assume i , j , kto be unit vectorsalong x, yand zaxes respectively. At the coordinate , ,2 2 0^ h, the electricfield vector E (Newtons/Coulomb) will be

(A) k2 2 (B) k2-

(C) k2 (D) k2 2-

EE SP 1.18 A series RLC circuit is observed at two frequencies. At /krad s11w = , we notethat source voltage VV 100 01 c= results in a current . AI 0 03 311 c= . At

/krad s22w = , the source voltage VV 100 02 c= results in a current AI 2 02 c=

. The closest values for R, L , Cout of the following options are(A) R 50 W= ; mHL 25= ; FC 10 m= ;

(B) R 50 W= ; mHL 10= ; FC 25 m= ;

(C) R 50 W= ; mHL 50= ; FC 5 m= ;

(D) R 50 W= ; mHL 5= ; FC 50 m= ;


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EE SP 1.19 The Nortons equivalent source in amperes as seen into terminals Xand Yis____.

YEAR 2013 ONE MARK

EE SP 1.20 Consider a delta connection of resistors and its equivalent star connection as shownbelow. If all elements of the delta connection are scaled by a factor k, k 0> , theelements of the corresponding star equivalent will be scaled by a factor of

(A) k2 (B) k

(C) /k1 (D) k

EE SP 1.21 The flux density at a point in space is given by 4 2 8 /Wb mB xa kya a x y z2= + +v v v v .

The value of constant kmust be equal to(A) 2- (B) .0 5-

(C) .0 5+ (D) 2+

EE SP 1.22 A single-phase load is supplied by a single-phase voltage source. If the currentflowing from the load to the source is 10 150 Ac+ - and if the voltage at the loadterminal is 100 60 Vc+ , then the(A) load absorbs real power and delivers reactive power

(B) load absorbs real power and absorbs reactive power

(C) load delivers real power and delivers reactive power

(D) load delivers real power and absorbs reactive power

EE SP 1.23 A source cosv t V t 100s p=^ h has an internal impedance of j4 3 W+^ h . If a purelyresistive load connected to this source has to extract the maximum power out ofthe source, its value in Wshould be(A) 3 (B) 4

(C) 5 (D) 7


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EE SP 1.24 The transfer functionV s

V s

1

2 ^ hh of the circuit shown below is

(A) .ss

10 5 1

++ (B)

ss

23 6

++

(C)ss

12

++ (D)

ss

21

++

YEAR 2013 TWO MARKS

EE SP 1.25 A dielectric slab with mm mm500 500# cross-section is 0.4 mlong. The slab

is subjected to a uniform electric field of 6 8 /kV mmE a ax y= +v v v . The relativepermittivity of the dielectric material is equal to 2. The value of constant 0e is8.85 10 /F m12#

- . The energy stored in the dielectric in Joules is(A) .8 85 10 11#

- (B) .8 85 10 5# -

(C) .88 5 (D) 885

EE SP 1.26 Three capacitors C1, C2 and C3 whose values are 10 Fm , 5 Fm , and 2 Fm respectively, have breakdown voltages of 10 V, 5 Vand 2 Vrespectively. For theinterconnection shown below, the maximum safe voltage in Volts that can beapplied across the combination, and the corresponding total charge in Cm storedin the effective capacitance across the terminals are respectively,

(A) 2.8 and 36 (B) 7 and 119

(C) 2.8 and 32 (D) 7 and 80

EE SP 1.27 In the circuit shown below, if the source voltage 100 53.13 VVS c+= then theThevenins equivalent voltage in Volts as seen by the load resistance RL is


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(A) 100 90c+ (B) 800 0c+

(C) 800 90c+ (D) 100 60c+

YEAR 2012 ONE MARK

EE SP 1.28 The impedance looking into nodes 1 and 2 in the given circuit is

(A) 05 W (B) 100 W

(C) 5 kW (D) 10.1kW

EE SP 1.29 In the circuit shown below, the current through the inductor is

(A) Aj1

2+

(B) Aj11

+-

(C) Aj1

1+

(D) 0 A

EE SP 1.30 A system with transfer function ( )( )( )( )

( )( )G s

s s s

s s

1 3 49 22

=+ + +

+ +

is excited by ( )sin tw . The steady-state output of the system is zero at(A) 1 /rad sw=

(B) /rad s2w=

(C) /rad s3w=

(D) /rad s4w=

EE SP 1.31 The average power delivered to an impedance (4 3)j W- by a current5 (100 100)cos Atp + is(A) 44.2W (B) 50 W

(C) 62.5 W (D) 125 W


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EE SP 1.32 In the following figure, C1and C2are ideal capacitors. C1has been charged to 12V before the ideal switch Sis closed at .t 0= The current ( )i tfor all tis

(A) zero (B) a step function

(C) an exponentially decaying function (D) an impulse function

YEAR 2012 TWO MARKS

EE SP 1.33 If 6 VV VA B- = then V VC D- is

(A) 5 V- (B) V2

(C) V3 (D) V6

EE SP 1.34 Assuming both the voltage sources are in phase, the value of R for whichmaximum power is transferred from circuit A to circuit Bis

(A) 0.8 W (B) 1.4 W

(C) 2 W (D) 2.8 W

Common Data for Questions 35 and 36 :

With 10 Vdc connected at port A in the linear nonreciprocal two-port networkshown below, the following were observed :(i) 1 Wconnected at port Bdraws a current of 3 A

(ii) 2.5 Wconnected at port Bdraws a current of 2 A


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EE SP 1.35 With 10 Vdc connected at port A , the current drawn by 7 Wconnected at portBis(A) 3/7 A (B) 5/7 A

(C) 1 A (D) 9/7 A

EE SP 1.36 For the same network, with 6 Vdc connected at port A , 1 Wconnected at portBdraws 7/3 .A If 8 Vdc is connected to port A , the open circuit voltage at portBis(A) 6 V (B) 7 V

(C) 8 V (D) 9 V

Statement for Linked Answer Questions 37 and 38 :

In the circuit shown, the three voltmeter readings are 220 ,VV1 = 122 ,VV2= 136 VV3= .

EE SP 1.37 The power factor of the load is(A) 0.45 (B) 0.50

(C) 0.55 (D) 0.60

EE SP 1.38 If 5RL W= , the approximate power consumption in the load is(A) 700 W (B) 750 W

(C) 800 W (D) 850 W

YEAR 2011 ONE MARK

EE SP 1.39 The r.m.s value of the current ( )i tin the circuit shown below is

(A) A21 (B) A

21

(C) 1 A (D) A2


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EE SP 1.40 The voltage applied to a circuit is ( )cos t100 2 100p volts and the circuit draws

a current of ( / )sin t10 2 100 4p p+ amperes. Taking the voltage as the referencephasor, the phasor representation of the current in amperes is

(A) /10 2 4p- (B) /10 4p-

(C) /10 4p+ (D) /10 2 4p+

EE SP 1.41 In the circuit given below, the value of Rrequired for the transfer of maximumpower to the load having a resistance of 3 Wis

(A) zero (B) 3 W

(C) 6 W (D) infinity

YEAR 2011 TWO MARKS

EE SP 1.42 A lossy capacitor Cx, rated for operation at 5 kV, 50 Hz is represented by anequivalent circuit with an ideal capacitor Cpin parallel with a resistor Rp. The

valueC

pis found to be 0.102

F and value of 1.25MR

p W=

. Then the power lossand tan dof the lossy capacitor operating at the rated voltage, respectively, are(A) 10 W and 0.0002 (B) 10 W and 0.0025

(C) 20 W and 0.025 (D) 20 W and 0.04

EE SP 1.43 A capacitor is made with a polymeric dielectric having an re of 2.26 and adielectric breakdown strength of 50 kV/cm. The permittivity of free space is 8.85pF/m. If the rectangular plates of the capacitor have a width of 20 cm and alength of 40 cm, then the maximum electric charge in the capacitor is(A) 2 C (B) 4 C

(C) 8 C (D) 10 C

Common Data For Q. 44 and 45

The input voltage given to a converter is vi 100 (100 )sin Vt2 p=The current drawn by the converter is

10 (100 /3) 5 (300 /4)sin sini t t2 2i p p p p= - + + 2 (500 /6)sin At2 p p+ -

EE SP 1.44 The input power factor of the converter is(A) 0.31 (B) 0.44

(C) 0.5 (D) 0.71

EE SP 1.45 The active power drawn by the converter is(A) 181 W (B) 500 W

(C) 707 W (D) 887 W


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Common Data For Q. 46 and 47

An RLC circuit with relevant data is given below.

EE SP 1.46 The power dissipated in the resistor Ris(A) 0.5 W

(B) 1 W(C) W2

(D) 2 W

EE SP 1.47 The current ICin the figure above is(A) 2 Aj- (B) Aj

2

1-

(C) Aj2

1+ (D) 2Aj+

YEAR 2010 ONE MARK

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 Fm capacitor is

(A) 0 A (B) 1 A

(C) 1.25 A (D) 5 A

EE SP 1.49 As shown in the figure, a 1 Wresistance is connected across a source that has aload line v i 100+ = . The current through the resistance is

(A) 25 A (B) 50 A

(C) 100 A (C) 200 A


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YEAR 2010 TWO MARKS

EE SP 1.50 If the 12 Wresistor draws a current of 1 A as shown in the figure, the value ofresistance Ris

(A) 4 W (B) 6 W

(C) 8 W (D) 18 W

EE SP 1.51 The two-port network P shown in the figure has ports 1 and 2, denoted byterminals (a,b) and (c,d) respectively. It has an impedance matrix Z withparameters denoted by Zi j. A 1 Wresistor is connected in series with the networkat port 1 as shown in the figure. The impedance matrix of the modified two-portnetwork (shown as a dashed box ) is

(A) ZZ

ZZ

1 11

11

21

12

22+ +

+e o (B) ZZ ZZ1 111 21 1222+ +e o(C)

Z

Z

Z

Z

111

21

12

22

+e o (D) ZZ

Z

Z

1

111

21

12

22

+

+e o

YEAR 2009 ONE MARK

EE SP 1.52 The current through the 2 kWresistance in the circuit shown is

(A) 0 mA (B) 1 mA

(C) 2 mA (D) 6 mA

EE SP 1.53 How many 200 W/220 V incandescent lamps connected in series would consumethe same total power as a single 100 W/220 V incandescent lamp ?(A) not possible (B) 4

(C) 3 (D) 2


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YEAR 2009 TWO MARKS

EE SP 1.54 In the figure shown, all elements used are ideal. For time ,t S0< 1remained closedand S2open. At ,t S0 1= is opened and S2is closed. If the voltageVc2across the

capacitor C2at t 0= is zero, the voltage across the capacitor combination att 0= +will be

(A) 1 V (B) 2 V

(C) 1.5 V (D) 3 V

EE SP 1.55 The equivalent capacitance of the input loop of the circuit shown is

(A) 2 Fm (B) 100 Fm

(C) 200 Fm (D) 4 Fm

EE SP 1.56 For the circuit shown, find out the current flowing through the 2 Wresistance.Also identify the changes to be made to double the current through the 2 Wresistance.

(A) (5 ; 30 )VA Put VS= (B) (2 ; 8 )VA Put VS=

(C) (5 ; 10 )IA Put AS= (D) (7 ; 12 )IA Put AS=

Statement for Linked Answer Question 57 and 58 :


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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) 1kW (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 ONE MARK

EE SP 1.59 The number of chords in the graph of the given circuit will be

(A) 3 (B) 4

(C) 5 (D) 6

EE SP 1.60 The Thevenins equivalent of a circuit operation at 5w= rads/s, has3.71 15.9V Voc += -

% and 2.38 0.667Z j0 W= - . At this frequency, the minimalrealization of the Thevenins impedance will have a(A) resistor and a capacitor and an inductor

(B) resistor and a capacitor

(C) resistor and an inductor

(D) capacitor and an inductor

YEAR 2008 TWO MARKS

EE SP 1.61 The time constant for the given circuit will be

(A) 1/9 s (B) 1/4 s

(C) 4 s (D) 9 s

EE SP 1.62

The resonant frequency for the given circuit will be


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(A) 1 rad/s (B) 2 rad/s

(C) 3 rad/s (D) 4 rad/s

EE SP 1.63

Assuming ideal elements in the circuit shown below, the voltageVabwill be

(A) 3 V- (B) 0 V

(C) 3 V (D) 5 V

Statement for Linked Answer Question 64 and 65

The current ( )i tsketched in the figure flows through a initially uncharged 0.3nF capacitor.

EE SP 1.64 The charge stored in the capacitor at 5t sm= , will be(A) 8 nC

(B) 10 nC

(C) 13 nC

(D) 16 nC

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 thecapacitor after 1 sm will approximately be

(A) 18.8 V(B) 23.5 V

(C) 23.5 V-

(D) 30.6 V-


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EE SP 1.66 In the circuit shown in the figure, the value of the current iwill be given by

(A) 0.31 A (B) 1.25 A

(C) 1.75 A (D) 2.5 A

EE SP 1.67 Two point charges 10Q C1 m= and 20Q2= mC are placed at coordinates (1,1,0)and ( , , )1 1 0- - respectively. The total electric flux passing through a plane

0z

2=

will be(A) 7.5 Cm

(B) 13.5 Cm

(C) 15.0 Cm

(D) 22.5 Cm

EE SP 1.68 A capacitor consists of two metal plates each 500 500# mm2and spaced 6 mmapart. The space between the metal plates is filled with a glass plate of 4 mmthickness and a layer of paper of 2 mm thickness. The relative primitivities ofthe glass and paper are 8 and 2 respectively. Neglecting the fringing effect, the

capacitance will be (Given that .8 85 100 12#e = - F/m )(A) 983.3 pF

(B) 1475 pF

(C) 637.7 pF

(D) 9956.25 pF

EE SP 1.69 A coil of 300 turns is wound on a non-magnetic core having a mean circumferenceof 300 mm and a cross-sectional area of 300 mm2. The inductance of the coilcorresponding to a magnetizing current of 3 A will be

(Given that 4 100 7#m p= - H/m)(A) 37.68 Hm

(B) 113.04 Hm

(C) 3.768 Hm

(D) 1.1304 Hm

YEAR 2007 ONE MARK

EE SP 1.70 Divergence of the vector field

( , , ) ( ) ( ) ( )cos cos sinV x y z x xy y i y xy j z x y k 2 2 2=- + + + + +t t tis(A) cosz z2 2

(B) sin cosxy z z 2 2+

(C) sin cosx xy z -

(D) None of these


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YEAR 2007 TWO MARKS

EE SP 1.71 The state equation for the current I1in the network shown below in terms of thevoltageVXand the independent source V, is given by

(A) 1.4 3.75dt

dIV I V

45

X1

1=- - +

(B) 1.4 3.75dt

dII V

45VX1 1= - -

(C) 1.4 3.75dtdI V I V45X1 1=- + +

(D) 1.4 3.75dt

dII V

45VX1 1=- + -

EE SP 1.72 The R-L-C series circuit shown in figure is supplied from a variable frequencyvoltage source. The admittance - locus of the R-L-C network at terminals AB forincreasing frequency wis

EE SP 1.73 In the circuit shown in figure. Switch SW1is initially closed and SW2 is open.The inductor L carries a current of 10 A and the capacitor charged to 10 V withpolarities as indicated. SW2is closed at t 0= and SW1is opened at t 0= . The


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current through Cand the voltage across L at ( )t 0= + is

(A) 55 A, 4.5 V (B) 5.5 A, 45 V

(C) 45 A, 5.5 A (D) 4.5 A, 55 V

EE SP 1.74 A 3 V DC supply with an internal resistance of 2 Wsupplies a passive non-linearresistance characterized by the relation V INL NL

2= . The power dissipated in the

non linear resistance is(A) 1.0 W (B) 1.5 W

(C) 2.5 W (D) 3.0 W

EE SP 1.75 In the figure given below all phasors are with reference to the potential at point'' ''O. The locus of voltage phasorVYXas Ris varied from zero to infinity is shownby

EE SP 1.76 The matrix A given below in the node incidence matrix of a network. The columnscorrespond to branches of the network while the rows correspond to nodes.Let [ ..... ]V VV V T1 2 6= denote the vector of branch voltages while [ ..... ]I i i i

T1 2 6=

that of branch currents. The vector [ ]E e e e e T1 2 3 4= denotes the vector of nodevoltages relative to a common ground.


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1

0

1

0

1

1

0

0

1

0

0

1

0

1

0

1

0

1

1

0

0

0

1

1

-

-

-

-

- -

R

T

SSSSS

V

X

WWWWW

Which of the following statement is true ?(A) The equations ,V V V V V V 0 01 2 3 3 4 5- + = + - = are KVL equations for the

network for some loops

(B) The equations ,V V V V V V 0 01 3 6 4 5 6- - = + - = are KVL equations for thenetwork for some loops

(C) E AV=

(D) AV 0= are KVI equations for the network

EE SP 1.77 A solid sphere made of insulating material has a radius Rand has a total charge

Qdistributed uniformly in its volume. What is the magnitude of the electric fieldintensity, E, at a distance ( )r r R0< < inside the sphere ?

(A)R

Qr

41

03pe (B)

R

Qr

43

03pe

(C)r

Q

41

02pe (D)

r

QR

41

03pe


An inductor designed with 400 turns coil wound on an iron core of 16 cm2crosssectional area and with a cut of an air gap length of 1 mm. The coil is connectedto a 230 V, 50 Hz ac supply. Neglect coil resistance, core loss, iron reluctance and

leakage inductance, ( 4 10 )H/M07

#m p=-

EE SP 1.78 The current in the inductor is(A) 18.08 A (B) 9.04 A

(C) 4.56 A (D) 2.28 A

EE SP 1.79 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 ONE MARK

EE SP 1.80 In the figure the current source is 1 0 , 1RA+ W= , the impedances are Z jC W=- and 2Z jL W= . The Thevenin equivalent looking into the circuit across X-Y is


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(A) 0 , (1 2 )j2 V+ W+ (B) 2 45 , (1 2 )jV+ W-%

(C) 2 45 , (1 )jV+ W+% (D) 45 , (1 )j2 V+ W+%

YEAR 2006 TWO MARKS

EE SP 1.81 The parameters of the circuit shown in the figure are 1R Mi W= 10 , 10R A V/V0

6W= = If 1v Vi m= , the output voltage, input impedance andoutput impedance respectively are

(A) 1 , ,10V 3 W (B) 1 ,0,10V W

(C) 1 , 0,V 3 (D) 10 , , 10V 3 W

EE SP 1.82 In the circuit shown in the figure, the current source 1I A= , the voltage source5 , 1 , 1 , 1V R R R L L L C C V H F1 2 3 1 2 3 1 2W= = = = = = = = =

The currents (in A) through R3and through the voltage source Vrespectivelywill 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 portparameters that describe the circuit shown are

(A) z parameters,0

0

0

0= G (B) h parameters,1

0

0

1= G(C) h parameters,

0

0

0

0= G (D) z parameters,1

0

0

1= G


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EE SP 1.84 The circuit shown in the figure is energized by a sinusoidal voltage source V1ata frequency which causes resonance with a current of I.

The phasor diagram which is applicable to this circuit is

EE SP 1.85 An ideal capacitor is charged to a voltage V0and connected at t 0= across anideal inductor L. (The circuit now consists of a capacitor and inductor alone). Ifwe let

LC0

1w = , the voltage across the capacitor at time t 0> is given by(A) V0 (B) ( )cosV t0 0w

(C) ( )sin tV0 0w (D) ( )cosV e tt

0 0

0

w

w-

EE SP 1.86 An energy meter connected to an immersion heater (resistive) operating on anAC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. Theheater is removed from the supply and now connected to a 400 V peak squarewave source of 150 Hz. The power in kW dissipated by the heater will be(A) 3.478 (B) 1.739

(C) 1.540 (D) 0.870

EE SP 1.87 Which of the following statement holds for the divergence of electric and magneticflux 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


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YEAR 2005 ONE MARK

EE SP 1.88 In the figure given below the value of Ris

(A) 2.5 W (B) 5.0 W

(C) 7.5 W (D) 10.0 W

EE SP 1.89 The RMS value of the voltage ( )u t ( )cos t3 4 3= + is(A) 17 V

(B) 5 V

(C) 7 V

(D) (3 2 )2 V+

EE SP 1.90 For the two port network shown in the figure the Z-matrix is given by

(A)Z

Z Z

Z Z

Z

1

1 2

1 2

2+

+= G (B) ZZ Z

Z

Z

1

1 2

1

2+= G

(C)Z

Z

Z

Z Z

1

2

2

1 2+= G (D) Z

Z

Z

Z Z

1

1

1

1 2+= G

EE SP 1.91 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 (B) 10 V

(C) 5 V (D) 0 V

EE SP 1.92 If Ev is the electric intensity, ( )E4 4# v is equal to

(A) Ev (B) Ev

(C) null vector (D) Zero


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YEAR 2005 TWO MARKS


A coil of inductance 10 H and resistance 40 W is connected as shown in thefigure. After the switch S has been in contact with point 1 for a very long time,it is moved to point 2 at, t 0= .

EE SP 1.93 If, at t = 0+, the voltage across the coil is 120 V, the value of resistance Ris

(A) 0 W (B) 20 W

(C) 40 W (D) 60 W

EE SP 1.94 For the value as obtained in (a), the time taken for 95% of the stored energy tobe dissipated is close to(A) 0.10 sec (B) 0.15 sec

(C) 0.50 sec (D) 1.0 sec

EE SP 1.95 The RL circuit of the figure is fed from a constant magnitude, variable frequencysinusoidal voltage source Vi n. At 100 Hz, the Rand L elements each have avoltage drop RM Sm .If the frequency of the source is changed to 50 Hz, then newvoltage drop across Ris

(A) u85

RMS (B) u32

RMS

(C) u58

RMS (D) u23

RMS

EE SP 1.96 For the three-phase circuit shown in the figure the ratio of the currents : :I I IR Y Bis given by


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(A) : :1 1 3 (B) : :1 1 2

(C) : :1 1 0 (D) : : /1 1 3 2

EE SP 1.97

The circuit shown in the figure is in steady state, when the switch is closed att 0= .Assuming that the inductance is ideal, the current through the inductor att 0= +equals

(A) 0 A(B) 0.5 A

(C) 1 A

(D) 2 A

EE SP 1.98 The charge distribution in a metal-dielectric-semiconductor specimen is shown inthe figure. The negative charge density decreases linearly in the semiconductoras shown. The electric field distribution is as shown in


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EE SP 1.99 In the given figure, the Thevenins equivalent pair (voltage, impedance), as seenat the terminals P-Q, is given by

(A) (2 5 )V, W

(B) (2 , 7.5 )V W

(C) (4 , 5 )V W

(D) (4 , 7.5 )V W

YEAR 2004 ONE MARK

EE SP 1.100 The value of Z in figure which is most appropriate to cause parallel resonanceat 500 Hz is

(A) 125.00 mH (B) 304.20 Fm

(C) 2.0 Fm (D) 0.05 Fm

EE SP 1.101 A parallel plate capacitor is shown in figure. It is made two square metal platesof 400 mm side. The 14 mm space between the plates is filled with two layers ofdielectrics of 4re = , 6 mm thick and 2re = , 8 mm thick. Neglecting fringing offields at the edge the capacitance is

(A) 1298 pF (B) 944 pF

(C) 354 pF (D) 257 pF

EE SP 1.102 The inductance of a long solenoid of length 1000 mm wound uniformly with 3000turns on a cylindrical paper tube of 60 mm diameter is(A) 3.2 mH (B) 3.2 mH

(C) 32.0 mH (D) 3.2 H


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YEAR 2004 TWO MARKS

EE SP 1.103 In figure, the value of the source voltage is

(A) 12 V (B) 24 V

(C) 30 V (D) 44 V

EE SP 1.104 In figure, Ra, Rband Rcare 20 W, 20 Wand 10 Wrespectively. The resistances R1, R2and R3in Wof an equivalent star-connection are

(A) 2.5, 5, 5 (B) 5, 2.5, 5

(C) 5, 5, 2.5 (D) 2.5, 5, 2.5

EE SP 1.105 In figure, the admittance values of the elements in Siemens are0.5 0, 0 1.5, 0 0.3Y j Y j Y j R L C= + = - = + respectively. The value of Ias a phasor

when the voltage Eacross the elements is 10 0 V+ %

(A) 1.5 0.5j+ (B) j5 18-

(C) .5 .j0 1 8+ (D) 5 j12-

EE SP 1.106 In figure, the value of resistance Rin Wis

(A) 10 (B) 20

(C) 30 (D) 40


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EE SP 1.107 In figure, the capacitor initially has a charge of 10 Coulomb. The current in thecircuit 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

EE SP 1.108 The rms value of the current in a wire which carries a d.c. current of 10 A and asinusoidal alternating current of peak value 20 A is(A) 10 A (B) 14.14 A

(C) 15 A (D) 17.32 A

EE SP 1.109 The Z-matrix of a 2-port network as given by.

.

.

.

0 9

0 2

0 2

0 6= GThe element Y22of 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 ONE MARK

EE SP 1.110 Figure Shows the waveform of the current passing through an inductor ofresistance 1 Wand inductance 2 H. The energy absorbed by the inductor in thefirst four seconds is

(A) 144 J (B) 98 J

(C) 132 J (D) 168 J

EE SP 1.111 A segment of a circuit is shown in figure 5 , 4 2sinv V v t R c= = .The voltage vL is


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(A) 3 8 cost2- (B) 2sint32

(C) sint16 2 (D) 16 2cost

EE SP 1.112

In the figure, 10 60 , 10 60 , 50 53.13 .Z Z Z

1 2 3+ + += - = =% % %

Thevenin impedanceseen form X-Y is

(A) .56 66 45+ % (B) 60 30+ %

(C) 70 30+ % (D) .34 4 65+ %

EE SP 1.113 Two conductors are carrying forward and return current of +I and I- as shown

in figure. The magnetic field intensity Hat point P is

(A)dIYp

(B)dI Xp

(C)d

I Y2p

(D)d

I X2p

EE SP 1.114 Two infinite strips of width w m in x-direction as shown in figure, are carryingforward and return currents of +I and I- in the z- direction. The strips areseparated by distance of x m. The inductance per unit length of the configuration

is measured to be L H/m. If the distance of separation between the strips insnow reduced to x/2 m, the inductance per unit length of the configuration is

(A) L2 H/m (B) /L 4H/m

(C) /L 2 H/m (D) L4 H/m


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YEAR 2003 TWO MARKS

EE SP 1.115 In the circuit of figure, the magnitudes of VL and VCare twice that of VR. Giventhat 50f Hz= , the inductance of the coil is

(A) 2.14 mH (B) 5.30 H

(C) 31.8 mH (D) 1.32 H

EE SP 1.116 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 resistanceRL is

(A) 2200 W

(B) 1250 W

(C) 1000 W

(D) 625 W


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EE SP 1.118 In figure, the value of Ris

(A) 10 W

(B) 18 W

(C) 24 W

(D) 12 W

EE SP 1.119 In the circuit shown in figure, the switch S is closed at time (t = 0). The voltageacross the inductance at t 0= +, is

(A) 2 V

(B) 4 V

(C) 6- V

(D) 8 V

EE SP 1.120 The h-parameters for a two-port network are defined by

E

I

h

h

h

h

I

E12

1121

1222

12

== = =G G GFor the two-port network shown in figure, the value of h12is given by

(A) 0.125

(B) 0.167

(C) 0.625

(D) 0.25


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EE SP 1.121 A point charge of +I nC is placed in a space with permittivity of .8 85 10 12#-

F/m as shown in figure. The potential difference VPQbetween two points P andQ at distance of 40 mm and 20 mm respectively fr0m the point charge is

(A) 0.22 kV (B) 225- V

(C) .2 24- kV (D) 15 V

EE SP 1.122 A parallel plate capacitor has an electrode area of 100 mm2, with spacing of0.1 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

EE SP 1.123 A composite parallel plate capacitor is made up of two different dielectric materialwith different thickness (t1and t2) as shown in figure. The two different dielectricmaterials are separated by a conducting foil F. The voltage of the conductingfoil is

(A) 52 V (B) 60 V

(C) 67 V (D) 33 V

***********


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SOLUTION

SOL 1.1 Correct answer is 330 W.

We redraw the given electric circuit as

From the given circuit, we have the following observations for the connectedbatteries:

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 80 8 15 2# # #= - +^ ^h h6 @ 1000 640 30= - -

W330=

SOL 1.2 Correct option is (A).

We redraw the given parallel plate capacitors as

(a) (b)

For the capacitor shown in figure (a), we have

C0 dAoe

= ...(1)

where, A =Area of the parallel plate capacitor

d=distance between the plates

Again, the capacitor shown in figure (b) can be considered as the two capacitors

C1and C2connected in parallel. So, we have


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C1 d

Ao2 e=

and C2 d

Ao r2 e e=

Therefore, we get the net capacitance as Cnet C C1 2= +

/ /

d

A

d

A

dA

dA

dA2 2

2 2 21o o r o o r o r

e e e e e e ee= + = + = +^ h ...(2)

Thus, from equations (1) and (2), we get

CnetC2

1o re= +^ hSOL 1.3 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 VV 0 2c =-^ h )

So, from the circuit, we get

Vs V0 t 0= = -

Again, we consider the circuit for t 0= +. By s-domain analysis, we have the

equivalent voltage across capacitor as

Vs^ hSc

Is

S

V 0c= ++^ ^h h

So, we draw the equivalent circuit for given initial voltage across capacitor as


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Therefore, we redraw the complete circuit for t 0= +as

Applying KVL, we get

ScI

I52

100-

+ +b l Vs=or s s s s

2 20 101001 100 20 10

3

6

3

# ##

# #- + +- -

Vs

Thus, we obtain

Vs s s s2 20 10 2

2

3#=- + +

s

2 1024

#=

So, V ts h t2 104#=i.e. the plot of source voltage is a straight line passing through origin, as shown

below.

SOL 1.4 Correct option is (B).

Undesirable property of electrical insulating material is its high relativepermittivity re^ h.

SOL 1.5 Correct answer is 10 A.

We redraw the given circuit as

Applying KVL in the circuit, we have

300 I R=

300 I I252

= +b l


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300 I I252

2

= +

I I50 6002 + - 0=

I I I60 10 6002 + - - 0=

I I I60 10 60+ - +^ ^h h 0= I I10 60- +^ ^h h 0=So, AI 10= , A60-

Now, for the two obtained values of current, we get

R 25210 30= + =b l

R 25260 5= - =-b l (Resistance can not be

negative)

Thus, the current through the circuit is I A10=

SOL 1.6 Correct option is (C).

According to Maxwell fourth equation

B:D 0=

i.e. the divergence of magnetic field is zero. Now, we check this property for each

of the given vectors.

Option (A): B y z xa a ax y z2 2 2= + +

So, B:D xy yz zx 02 2 2

2

2

2

2

2

2= + + =

Option (B): B z x ya a ax y z2 2 2= + +

So, B:Dxz

yx

zy 02 2 2

2

2

2

2

2

2= + + =

Option (C): B x y za a ax y z2 2 2= + +

So, B:Dxx

yy

zz

2 2 2

2

2

2

2

2

2= + +

x y z2 2 2 0!= + +

Option (D):

B y z x z x y a a ax y z2 2 2 2 2 2

= + +

So, B:Dxy z

yx z

zx y 02 2 2 2 2 2

2

2

2

2

2

2= + + =

Thus, the vector given in option (C) does not satisfy the property of magnetic

flux density.

SOL 1.7 Correct answer is 35.

There are two possible connections(1) Series adding connection


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Equivalent inductance

L eq L L M21 2= + +

380 L L M2= + +

orL M+

190=

...(1)(2) Series opposing connection

Equivalent inductance

L eq L L M21 2= + -

240 L L M2= + -or L M- 120= ...(2)

Substituting eq (2) from eq (1), we get

L2 70=

or L mH35=

SOL 1.8 Correct option is (D).

For t 0< , the switch was at position 1 and circuit is in steady state. In steady

state capacitor behaves as open circuit. So

Therefore v 0C-^ h Vo1=-

Step 2 : For t 0> , the switch moves to position 2. Again in steady state capacitor

acts as an open circuit.

vC 3^ h Vo2=Time constant t R CTh=RT his the Thevenin resistance across the capacitor terminal after t 0= . So

t RC2=

Now, at any time t 0> , capacitor voltage is given by

v tC h v v v e 0 /C C C t3 3= + - t-^ ^ ^h h h6 @


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V V V e /o o ot RC

2 1 22= + - - -6 @

or, V V V e /o o ot RC

2 1 22= + - - -6 @

or, v tC h V e V e 1o RCt o RCt2 2 1 2= - -- -^ h v tC h V e V e V e V e 1 1 1

/o RC

t

o RC

t

o RC

t

ot RC

2 2 1 2 1 2 12

= - - - - + -- - - -

^ ^ ^h h hor, v tC h V V e V 1o o RCt o2 1 2 1= - - +-^ ^h hSOL 1.9 Correct option is (C).

As shown in Figure above this arrangement is equivalent to series combination

of three capacitors C1, C2and C3. We have

C1 /dA

dA

441 1e e= =

C2 /dA

dA

2 22 2e e

= =

C3 /dA

dA

441 1e e= =

Now given that voltage across first capacitor is C1is 2 V, so we write

VC1/

j C j C j C

j CV

1 1 11

2s

1 2 3

1#

w w w

w=

+ +=

C C C C C C

C C 102 3 1 3 1 2

2 3#+ +

2= V 10 Vs=^ hSubstituting values of C1, C2and C3, we get

8 16 88 10

1 2 12

1 2

1 2#

e e e e ee e

+ + 2=

16 16

81 2 1

21 2

e e ee e

+

51=

2 22 1

2

e ee+

51=

2 22 1e e+ 5 2e=

2 1e 3 2e=

2

1

e

e

2

3=

:1 2e e :3 2=


Since there are two sources with different frequencies ( 101w = , 52w = ), we use


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superposition to obtain v tC h.First consider source sinv t t20 10=^ h . Open circuit current source and redrawingcircuit in domain with /secrad101w = .

VC1

j

j

1101

101

20#=+

(Using voltage division)

j1 101 20#= +

VC1 tan10120 101= - -

A 1 . AV10120 1 99C1= = =

Now, consider current source sini t t10 5=^ h . Short circuit voltage source andredrawing circuit in frequency domain with /secrad52w = .

IC2

j1

51

1 10 0c=+

j

j

1 55

10 0# c= +

VC2 I j51

C2 #=

VC2 j1 510 0c

=+

VC2 tan26

10 5= - -

A 2 . AV26

10 1 96C2= = =



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H By wire on x-axisd

Ik

2p=

k14

21

# p=

H by wire on y-axis dI k2p= -^ h k

22

21

# p= -^ h

H on (2, 1, 0) k k24

21

p p= -

H /A mk23p

=


Given the relation

fv:d ^ h x y y z z x2 2 2= + + ...(1)where f, vare scalar and vector field respectively. From the properties of vector

field, we have

f v:d ^ h v f f v : :d d= -^ ^h hor v f: d h f v f v : :d d= +^ ^h h ...(2)Again, we have

v y z xi j k= + +

So, v:d

x

y

y

z

z

x

2

2

2

2

2

2= + +

0=

Therefore, we get

f v:d h 0=Substituting it in equation (2), we get

fv : d h fv:d= ^ hThus, fv : d h x y y z z x2 2 2= + + [from equation (1)]


We sketch the phasor diagram for three phase star connected load as

Since, line A to neutral voltage is

A V10 15c=


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So, for balance 3-phase, we have

B 10 105c= -

and C 10 225c= -

Therefore, the voltage of lineB

with respect to lineC

is vBC v vB C= -

10 105 10 225c c= - - -

10 3 75c= -


We have the impedance circuit,

The capacitor and inductor can be replaced with its equivalent reactance (in

frequency domain) as shown below.

So, we have the equivalent circuit as

Solving the above circuit, we get the equivalent driving point impedance as

Z s^ h ||ss s

s1 1= + +b l

||ss s

s1 12= + +b l

s

s ss

s ss

1 1

1 1

2

2

#= +

+ +

+

ss ss

2132

= +++

or Z s^ hs s

s s

23 1

3

4 2

=+

+ +


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According to Gausss law, the total electric flux coming out of the concentric

spherical surface is equal to the charge enclosed by the surface, i.e.

Q D ds:=

#Since, the electric potential is defined as v

r

Q

4 0pe=

So, Q v r4 0# pe=

r4 0pe= (since, v 1= volt)

Hence, D ds:# r4 0pe=

SOL 1.16 Correct answer is 2.

To obtain the Norton equivalent along X-Yterminal, we redraw the given circuit

as

In the network, 5 W resistor is short-circuit. So, the equivalent circuit is

Now, writing the KVL equation in two loops, we have

I I I5 5 5 sc1 1- - -^ h 0= ...(1) . .I I I5 2 5 2 5sc sc 1- - - +^ h 0= ...(2)Solving equation (1),

I I10 5sc1 - 5= I2 1 I 1sc= + ...(3)

Solving equation (2),

.I I7 5 5sc 1- .2 5=

or I I3 2sc 1- 1=


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or I I3 1sc sc - - 1= [using equation (3)]

or I2sc 2=

Thus, Isc A1=

i.e. the Norton equivalent source as seen into terminals Xand Yis 1 A.

SOL 1.17 Correct answer is 3 W.

To obtain the required unknown, we redraw the given circuit as

Applying KCL at node Vx, we have

V V1

11

x x- + 2=

or V2 1x- 2=

or Vx 23=

Therefore, the power delivered by current source is obtained as

=Voltage across current source #current through

current source

W23 2 3#= =


We have the given system of a charge and a perfectly conducting metal plate as

shown below.

Due to presence of conducting plane, we can assume an image charge for the

system as shown below


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nodiaSo, we can observe from the figure that electric field vector along x-yplane

is cancelled, i.e. zero. Therefore, the electric field vector acts along negative z

-direction and given as

E"

E E k2 21 2= + - td ^n h ...(1)

where E1and E2are the field intensity due to the charge Qand Q- at point

( , , )2 2 0 . So, we obtain

E1( ( ) ( ) ( ) )r

Q

41

41

2 2 2

32 20

20 2 2 2 2

0

pe pepe

= =+ +

2=

Similarly, we get, E 22=

Substituting these values in equation (1), we obtain

E"

k22

22= + - te ^o h

k2=- t


Consider the circuit shown below with non-ideal voltage source Vs, an internal

impedance Z R jX s s s= + , and purely resistive load RL .

Current through the circuit is given by

IR R jX

V

s L s

s=+ +

or IR R X

V

s L s

s

2=

+ +

^ ^h hTherefore, the power transferred to load is P I RL

2=

R R X

V R

s L s

s L2 2

2

=+ +^^ h h

For maximum power, we have


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dRdP

L

0=

or[ ]

R R X

R R X R R R 1 2

s L s

s L s L s L

2 2 2

2 2

+ +

+ + - +

^^

^ ^

h h

h h 0=

or R R R R X R R R 2 2 2s L s L s s L L2 2 2 2+ + + - - 0=

or R Xs s2 2+ RL

2=

or RL R Xs s2 2= +

i.e. to maximise the power transferred to pure resistive load (RL ), its value must

be equal to the magnitude of Zsof internal circuit.


Given the current and voltage at different frequencies as

At krad11w = /s, V1 100 0= .I 0 03 311 c=

At krad22w = /s, V2 100 0= I 2 02 c=

From the given data, it is clear that voltage and current are in phase at 22w = .

So, it satisfies the resonance condition. Therefore, we have

RIV

2100 50 W= = =

Also, for the resonance circuit, we have

LwC1w

=

or L10 23 # # C2 101 3# #=

or L4 106# C1= ...(1)

Again, at 1w = krad/s, we have

cos31cZR=

or cos31cR L

C

R

12 2ww

=+ -b l

or cos31c LC

2500 1010

150

33

= + -b l ...(2)Solving equations (1) and (2), we get

L mH10= and FC 25. m

SOL 1.21 Correct option is (B).In the equivalent star connection, the resistance can be given as

RC R R RR R

a b c

b a=+ +

; RB R R RR R

a b c

a c=+ +

and RA R R RR R

a b c

b c=+ +

So, if the delta connection components Ra, Rband Rcare scaled by a factor kthen

RAl kR kR kR k R k R

a b c

b c=

+ +^ ^h h

kkR R R

R R

a b c

b c2

=+ +

k RA=

Hence, it is also scaled by a factor k


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Given, flux density

Bv x a ky a a 4 2 8x y z= + +v v v

Since, magnetic flux density is always divergence less.

i.e., B$dv 0=

So, for given vector flux density, we have

B$dv k4 2 0 0= + + =

or, k 2=-

SOL 1.23 Correct option is (B).Consider the voltage source and load shown in figure

We obtain the power delivered by load as

Pdelivered I V*L L= 10 150 10 60c c= +^ ^h h 100 210c=

cos sinj1000 210 1000 210c c

= + . j866 025 500=- -As both the reactive and average power (real power) are negative so, power is

absorbed by load. i.e., load absorbs real as well as reactive power.


For the purely resistive load, maximum average power is transferred when

RL R XTh T h 2 2= +

where R jXTh Th + is the equivalent thevinin (input) impedance of the circuit.

i.e., RL 4 32 2= + 5 W=

SOL 1.25 Correct option is (D).For the given capacitance, C F100m= in the circuit, we have the reactance.

XC sc1=

s 100 101

6# #

= - s104=

So,V s

V s

1

2 ^ hh 10 10

s s

s

10 10 10

44

44

4=+ +

+

ss

21=

++

SOL 1.26

Correct option is (B).Energy density stored in a dielectric medium is obtained as

wE /J mE21 2 2e=

The electric field inside the dielectric will be same to given field in free space


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only if the field is tangential to the interface

So, wE 2 10 /mm21 6 80

2 2 2 6 2e #= +^ hTherefore, the total stored energy is

WE W dvEv

=# 100 10 / .mm mm500 500 0 40 6 2 2# #e # #= ^ ^h h .100 10 0 4 25 100

6 4# # # # #e=

.8 85 10 1012 13# #= - 88.5J=


Consider that the voltage across the three capacitors C1, C2and C3areV1,V2and

V3respectively. So, we can write

VV

3

2 CC

2

3= ....(1)

Since, Voltage is inversely proportional to capacitance

Now, given that

C1 F10 m= ; 10VVmax

1 =

^ h C2 F5 m= ; 5 VV max2 =^ h C3 F2 m= ; 2VV max3 =^ hSo, from Eq (1) we have

VV

3

2 52=

for V max3^ h 2=We obtain, V2 0.8 volt5

2 2 5


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For evaluating the equivalent thevenin voltage seen by the load RL , we open the

circuit across it (also if it consist dependent source).

The equivalent circuit is shown below

As the circuit open across RL so

I2 0=

or, 40j I2 0=

i.e., the dependent source in loop 1is short circuited. Therefore,

VL1 jj V

4 34 s

=+

^ h VT h V10 L1= 100 53.13j

j

4 340

c=+

.100 53.13

5 53 1340 90

c

cc=

800 90c=


Applying nodal analysis at top node.

1 1

j

V V

10

101 1c c+ +

+ 1 0c=

( 1 1) 1 1j jV 01 c+ + + j1=

V1 j1 11=

+-

Current I11

j j

jV

10

11

1 11 c=

+=

-+

+

( )A

j j

j

j1 11=

+ =

+


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( )G s( )( )( )

( )( )s s s

s s

1 3 49 22

=+ + +

+ +

( )G jw ( )( )( )

( )( )j j j

j

1 3 4

9 22

w w w

w w

= + + +

- + +

The steady state output will be zero if

( )G jw 0=

92w- + 0=

w 3 /rad s=


We put a test source between terminal 1, 2 to obtain equivalent impedance

ZT h IV

test

test=

Applying KCL at top right node

9 1 99k kV V I100test test b+ + - Itest=

99k

V VI

10 100t est t est

b+ - Itest= ...(i)

But Ib 10kk kV V

9 1test test =-+

=-

Substituting Ibinto equation (i), we have

10 10k kV V V

10099test test test + + Itest=

V V10 10100

100test test

3#

+ Itest=

V1002 test Itest=

ZT h 50IV

test

test W= =


In phasor form Z j4 3= -

Z 5 .36 86cW= -

I 5 A100c=

Average power delivered :

P .avg cosZI21 2 q=

25 5 36.86cos21

# c#= 50 W=


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Alternate Method:

Z (4 3)j W= -

I 5 (100 100)cos Atp= +

Pavg Re I Z212= $ .

( ) ( )Re j21 5 4 32# #= -" ,

100 50 W21

#= =


The s-domain equivalent circuit is shown as below.

( )I s( )/ ( )

C s C s

v s

C C

v

1 10

1 10c c

1 2 1 2

=+

=+

(12 )VC CC C

1 2

1 2=+b l (0) 12 VvC =

12Ceq=

Taking inverse Laplace transform for the current in time domain,

( )i t ( )C t12 eqd= (Impulse)


In the given circuit, V VA B- 6 V=

So current in the branch, IAB 3 A2

6

= =We can see, that the circuit is a one port circuit looking from terminal BDas

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 Bwill be equal to the

incoming current from Dto Cas shown

i.e. IDC 3 AIAB= =


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The total current in the resistor 1Wwill be

I1 2 IDC= + (Writing KCL at node D)

2 3 5 A= + =

So, VCD ( )I1 1#= - 5 V=-


We obtain Thevenin equivalent of circuit B.

Thevenin Impedance :

ZTh R=

Thevenin Voltage :

VTh 3 V0c=

Now, circuit becomes as

Current in the circuit, I1 R210 3=

+-

Power transfer from circuit toA B

P ( ) 3I R I12 2

1= +


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10 3 10 3R R

R23

2

2

=+-

++-: :D D ( ) ( )RR R249 2 212= + + +

( )

( )R

R R

2

49 21 22= +

+ +

( )42 70

R

R

2 2=

++

dRdP

( )( ) ( ) ( )

RR R R

22 70 42 70 2 2 04

2=

++ - + + =

(2 )[(2 )70 (42 70 )2]R R R+ + - + 0=

R R140 70 84 140+ - - 0=

56 R70=

R 0.8 W=


When 10 V is connected at port A the network is

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let

Thevenin voltage is V , VTh10 with 10 Vapplied at port A and Thevenin resistanceis RT h.

IL R RV,10 V

T h L

Th= +

For 1RL W= , 3 AIL=

3R

V

1,10V

T h

T h=

+ ...(i)

For 2.5RL W= , 2 AIL=

2.R

V

2 5,10 V

T h

Th=+

...(ii)

Dividing above two

23 .

RR

12 5

Th

T h=+

+

R3 3Th+ R2 5Th= +

RTh 2 W=

Substituting RThinto equation (i)

V ,10VT h 3(2 1) 9 V= + =


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Note that it is a non reciprocal two port network. Thevenin voltage seen at port

Bdepends on the voltage connected at port A . Therefore we took subscript

V ,10 VTh . This is Thevenin voltage only when 10 Vsource 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

For 7RL W= , IL 1 ARV

2 2 79,10 V

L

Th

= + = + =


Now, when 6 Vconnected at port A let Thevenin voltage seen at port BisV ,6 VT h

. Here 1RL W= and AI 37

L=

V , VTh6 R 37 1

37

Th# #= + 2 7 V37

37

#= + =

This is a linear network, so VThat port Bcan be written as

VTh V1a b= +

where V1is the input applied at port A .

We have 10 VV1 = , 9 VV ,10VT h =

9 10a b= +

...(i)When 6 VV1 = , 9 VV , VTh6 =

7 6a b= + ...(ii)

Solving (i) and (ii)

a .0 5= , 4b=

Thus, with any voltage V1applied at port A , Thevenin voltage or open circuit

voltage at port Bwill be

So, V ,Th V1 .V0 5 41= +

For V1 8 V=

V ,8 VT h . V0 5 8 4 8 oc#= + = = (open circuit voltage)


By taking ,V V1 2and V3all are phasor voltages.

V1 V V2 3= +

Magnitude of ,V V1 2and V3are given as


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V1 220 V= , 122 VV2= , 136 VV3=

Since voltage across Ris in same phase with V1and the voltage V3has a phase

difference of qwith voltage V1, we write in polar form

V1 V V02 3c q= +

V1 cos sinV V jV 2 3 3q q= + +

V1 ( )cos sinV V jV 2 3 3q q= + +

V1 ( ) ( )cos sinV V V2 32

22q q= + +

220 ( ) ( )cos sin122 136 1362 2q q= + +

Solving, power factor

cosq .0 45=


Voltage across load resistance

VRL cosV3 q= .136 0 45#= 61.2 V=

Power absorbed in RL PL ( . )

750 WRV

561 2

L

RL2 2

-= =


The frequency domains equivalent circuit at 1 / secradw= .

Since the capacitor and inductive reactances are equal in magnitude, the net

impedance of that branch will become zero.

Equivalent circuit

Current ( )i t1

(1 )sin sin At tW

= =

rms value of current irms A2

1=


Voltage in time domain

( )v t ( )cos t100 2 100p=

Current in time domain


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( )i t ( / )sin t10 2 100 4p p= +

Applying the following trigonometric identity

( )sin f ( )cos 90cf= -

So, ( )i t 10 (100 /4 /2)cos t2 p p p= + -

10 (100 /4)cos t2 p p= -

In phasor form, I /2

10 2 4p= -


Power transferred to the load

P I RL2= 1

R R R

0th L

L

2

=+b l

For maximum power transfer Rth, should be minimum.

Rth RR

66 0=+

=

R 0=

Note: Since load resistance is constant so we choose a minimum value of Rth


Power loss.

( )0 W

RV

1 25 10

5 102

p

rated2

6

3 2

#

#= = =

For an parallel combination of resistance and capacitor

tand. .C R

12 50 1 25 0 102

1p p # # #w p

= =

.401 0 025= =

SOL 1.44

Correct option is (C).Charge Q CV=

dAVr0

e e= ( )A

dV

r0e e= C dAr0e e=

Q Qmax=

We have 8.85 10 /F mc014e #=

- , 2.26re = , 20 40 mA c2

#=


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dV 50 10 kV/cm3#=

Maximum electrical charge on the capacitor

whend

V 50 /kV cmd

V

max

= =

b lThus, Q 8.85 10 2.26 20 40 50 1014 3# # # # # #= - 8 Cm=


vi 100 (100 )sin Vt2 p=

Fundamental component of current

ii1 10 (100 /3)sin t2 Ap p= -

Input power factor

pf cosII1( )

rms

rms1f=

I( )1rmsis rms values of fundamental component of current and Irmsis the rms value

of converter current.

pf /3 0.44cos10 5 2

102 2 2

p=+ +

=


Only the fundamental component of current contributes to the mean ac input

power. The power due to the harmonic components of current is zero.

So, Pin cosV Irms rms 1 1f= 100 10 /3cosp#=

500 W=


Power delivered by the source will be equal to power dissipated by the resistor.

P /cosV I 4s s p=

1 /cos W2 4 1# p= =


IC RL / /I I 2 4 2 4s p p= - = - -

/ / / /cos sin cos sinj j2 4 4 4 4p p p p= + - -^ ^h h$ . /sinj2 2 4p= j2=


For t 0< , the switch was closed for a long time so equivalent circuit is


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Voltage across capacitor at t 0=

(0)vc 44 15

#= = V

Now switch is opened, so equivalent circuit is

For capacitor at 0t= +

(0 )vc+ (0) 4vc= = V

current in 4 Wresistor at t 0= +,(0 )

i v

4

1c1 = =+

A

so current in capacitor at t 0= +, ( )i i0 1c 1= =+ A


Thevenin equivalent across 1Xresistor can be obtain as following

Open circuit voltage vth 100= V ( 0)i=

Short circuit current isc 100= A ( 0vth= )

So, Rth iv

sc

th= 1100100 W= =

Equivalent circuit is

i 501 1100=+

= A


The circuit is

Current in RWresistor is

i 2 1 1= - = A

Voltage across 12 Wresistor is

VA 1 12#= 12= V

So, iR

V 6A= - 61

12 6 W= - =


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V Z I Z I 1 11 1 12 2= + V Z I Z I 1 11 1 12 2= +l l l l l

V Z I Z I 2 21 1 22 2= + V Z I Z I 2 21 1 22 2= +l l l l l

Here, ,I I I I 1 1 2 2= =l l

When 1R W= is connected

V1l 1V I1 1 #= + l V I1 1= +

V1l Z I Z I I 11 1 12 2 1= + +

V1l ( 1)Z I Z I 11 1 12 2= + +

So, Z11l 1Z11= +

Z12l Z12=Similarly for output port

V2l Z I Z I 21 1 22 2= +l l l l Z I Z I 21 1 22 2= +l l

So, Z Z21 21=l , Z Z22 22=l

Z-matrix is ZZ

Z

Z

Z

111

21

12

22=

+> H


In the bridge

R R1 4 1R R2 3= =So it is a balanced bridge

I 0= mA


Resistance of the bulb rated 200 W/220 V is

R1 PV

1

2

= ( )

242200220 2

W= =

Resistance of 100 W/220 V lamp is

RT PV

2

2

=

( )

484100

220 2

W= =

To connect in series

RT n R1#=

484 242n#=

n 2=


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For 0t< , S1is closed and S2is opened so the capacitor C1will charged upto 3

volt.

(0)VC1 3= Volt

Now when switch positions are changed, by applying charge conservation

(0 )C Veq C1+ (0 ) (0 )C V C V C C1 21 2= +

+ +

(2 1) 3#+ 1 3 2 (0 )VC2# #= + +

9 3 2 (0 )VC2= + +

(0 )VC2+ 3= Volt


Applying KVL in the input loop

(1 1) 10 ( 49 )v ij C

i i1

1 13

1 1# w- + - + 0=

v1 2 10 50i j C i13 1 1# w= +

Input impedance, Z1 iv

1

1= 2 10( / )j C5013

#w

= +

Equivalent capacitance, CeqC50

= 2F

F50

100nn= =


Voltage across 2Xresistor, VS 2= V

Current, I2WV2S=

24 2= = A

To make the current double we have to take

VS 8= V


To obtain equivalent Thevenin circuit, put a test source between terminals AB

Applying KCL at super node


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V V V2

52 1

P P S- + + IS=

5 2V V VP P S- + + 2IS=

2 2V VP S+ 2 5Is= +

V VP S+ 2.5IS= + ...(1)

V VP S- 3VS=

& VP 4VS=

So, 4V VS S+ 2.5IS= +

5VS 2.5IS= +

VS 0.2 0.5IS= + ...(2)

For Thevenin equivalent circuit

VS I R VS th th = + ...(3)

By comparing (2) and (3),

Thevenin resistance Rth 0.2 kW=

SOL 1.59 Correct option is (D).From above Vth 0.5= V


No. of chords is given as

l 1b n= - +

b" no. of branches

n"no. of nodes

l"no. of chords

6b= , n 4=

l 6 4 1= - + 3=


Impedance Zo 2.38 0.667j W= -

Constant term in impedance indicates that there is a resistance in the circuit.

Assume that only a resistance and capacitor are in the circuit, phase difference

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