ELECTRONICS COMMUNICATION ENGINEERINGNetwork solution methods: nodal and mesh analysis; Network theorems: superposition, Thevenin and ... Chapter - 1 Circuit / Network Analysis 1.1

Prof. Dr. S.L. KakaniM.Sc. (Physics), Ph.D.

Former Executive Director,Institute of Technology and Management,

Bhilwara - 311001 (Rajasthan).

Dr. K.C. BhandariAssociate Professor – Physics (Retd.),

M.L. Sukhadia University,Udaipur (Rajasthan), India.

GATEELECTRONICS

&COMMUNICATION ENGINEERING

[GATE, UGC-CSIR: NET/SET, B.Tech, M.Sc. (Elect.), B.Sc. (Pass/Hons.) (Elect.)All Other National and State Level Enterance Examinations]

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The book has been thoroughly revised and updated as per GATE 2018 and other entranceexaminations. Good number of new problems and multiple choice questions are added.GATE Electronics and Communication Engineering 2018 to 2013 papers and GATE(Physics Electronics part) 2001 to 2017 are solved and added as appendices.Five more chapters are added. Chapter on Digital Electronics is rewritten.

We hope this comprehensive, updated version will be very handy and user-friendlyfor the aspirants, who are wishing to maximize the score in GATE, UGC-CSIR: NET/SET and other entrance examinations.

We wish all success to the aspirants in their sincere effort.

Authors will be happy to receive suggestions and feedback for further improvementof the book.

S.L. Kakani

K.C. Bhandari

PREFACE

Electronics and Communication Engineering

Engineering MathematicsLinear Algebra:Vector space, basis, linear dependence and independence, matrix algebra, eigen values and eigen vectors,rank, solution of linear equations – existence and uniqueness.

Calculus:Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partialderivatives, maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series.

Differential Equations:First order equations (linear and non-linear), higher order linear differential equations, Cauchy’s and Euler’sequations, methods of solution using variation of parameters, complementary function and particular integral,partial differential equations, variable separable method, initial and boundary value problems.

Vector Analysis:Vectors in plane and space, vector operations, gradient, divergence and curl, Gauss’s, Green’s and Stoke’stheorems.

Complex Analysis:Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula; Taylor’s and Laurent’s series,Residue theorem.

Numerical Methods:Solution of non-linear equations, single and multi-step methods for differential equations, convergencecriteria.

Probability and Statistics:Mean, median, mode and standard deviation; combinational probability, probability distribution functions –Binomial, Poisson, Exponential and normal; Joint and conditional probability; Correlation and regressionanalysis.

Electronics and Communication EngineeringNetworks, Signals and Systems:Network solution methods: nodal and mesh analysis; Network theorems: superposition, Thevenin andNorton’s, maximum power transfer; Wye Delta transformation; Steady state sinusoidal analysis using phasors;Time domain analysis of simple linear circuits; Solution of network equations using Laplace transform;Frequency domain analysis of RLC circuits; Linear two port network parameters: driving point and transferfunctions; State equations for networks.

GATE SYLLABUS

Continuous-time signals: Fourier series and Fourier transform representations, sampling theorem andapplications; Discrete-time signals: discrete-time Fourier transform (DTFT), DFT, FFT, Z-transform,interpolation of discrete-time signals; LTI systems: definition and properties, causality, stability, impulseresponse, convolution, poles and zeros, parallel and cascade structure, frequency response, group delay,phase delay, digital filter design techniques.

Electronic Devices:Energy bands in intrinsic and extrinsic silicon; Carrier transport: diffusion current, drift current, mobility andresistivity; Generation and recombination of carriers; Poisson and continuity equations; P-N junction,Zener diode, BJT, MOS capacitor, MOSFET, LED, photo diode and solar cell; Integrated circuit fabricationprocess: oxidation, diffusion, ion implantation, photolithography and twin-tub CMOS process.

Analog Circuits:Small signal equivalent circuits of diodes, BJTs and MOSFETs; Simple diode circuits: clipping, clampingand rectifiers; Single-stage BJT and MOSFET amplifiers: biasing, bias stability, mid-frequency small signalanalysis and frequency response; BJT and MOSFET amplifiers: multi-stage, differential, feedback, powerand operational; Simple op-amp circuits; Active filters; Sinusoidal oscillators: criterion for oscillation, single-transistor and op-amp configurations; Function generators, wave-shaping circuits and 555 timers; Voltagereference circuits; Power supplies: ripple removal and regulation.

Digital Circuits:Number systems; Combinatorial circuits: Boolean algebra, minimization of functions using Boolean identitiesand Karnaugh map, logic gates and their static CMOS implementations, arithmetic circuits, code converters,multiplexers, decoders and PLAs; Sequential circuits: latches and flip/flops, counters, shift, registers andfinite state machines; Data converters: sample and hold circuits, ADCs and DACs; Semiconductor memories:ROM, SRAM, DRAM; 8-bit microprocessor (8085): architecture, programming, memory and I/Ointerfacing.

Control Systems:Basic control system components; Feedback principle; Transfer function; Block diagram representation;Signal flow graph; Transient and steady-state analysis of LTI systems; Frequency response; Routh-Hurwitzand Nyquist stability criteria; Bode and root-locus plots; Lag, lead and lag-lead compensation; Statevariable model and solution of state equation of LTI systems.

Communications:Random processes: autocorrelation and power spectral density, properties of white noise, filtering ofrandom signals through LTI systems; Analog communications: amplitude modulation and demodulation,

angle modulation and demodulation, spectra of AM and FM, superheterodyne receivers, circuits for analogcommunications; Information theory: entropy, mutual information and channel capacity theorem; Digitalcommunications: PCM, DPCM, digital modulation schemes, amplitude, phase and frequency shift keying(ASK, PSK, FSK), QAM, MAP and ML decoding, matched filter receiver, calculation of bandwidth,SNR and BER for digital modulation; Fundamentals of error correction, Hamming codes; Timing andfrequency synchronization, inter-symbol interference and its mitigation; Basics of TDMA, FDMA and CDMA.

Electromagnetics:Electrostatics; Maxwell’s equations: differential and integral forms and their interpretation, boundary conditions,wave equation, Poynting vector; Plane waves and properties: reflection and refraction, polarization, phaseand group velocity, propagation through various media, skin depth; Transmission lines: equations, characteristicimpedance, impedance matching, impedance transformation, S-parameters, Smith chart; Waveguides: modes,boundary conditions, cut-off frequencies, dispersion relations; Antennas: antenna types, radiation pattern,gain and directivity, return loss, antenna arrays; Basics of radar; Light propagation in optical fibers.

General Aptitude1. Verbal Ability

Word problems (Synonyms and Antonyms) Sentence completion Word Analogy

2. Critical Reasoning Arithmetical Reasoning Sitting / Learning Arrangement Logical Analytical Reasoning Verbal Deductions

3. Numerical Ability Number System Percentage Ratio and Proportion Average Simple Interest Compound Interest Time and Work Time, Speed and Distance Logarithm Progression Data Interpretation

Chapter - 1 Circuit / Network Analysis 1.1 – 1.20

Chapter - 2 Semiconductors and Diodes 2.1 – 2.57

Chapter - 3 Rectifiers, Filters, Power Supply and Voltage Regulation 3.1 – 3.23

Chapter - 4 Bipolar Junction Transistor (BJT) and Field Effect Transistor (FET) 4.1 – 4.27

Chapter - 5 Small Signal Voltage Amplifiers 5.1 – 5.30

Chapter - 6 Power Amplifiers 6.1 – 6.16

Chapter - 7 Feedback Amplifiers 7.1 – 7.14

Chapter - 8 Operational Amplifiers 8.1 – 8.25

Chapter - 9 Oscillators 9.1 – 9.15

Chapter - 10 Modulation and Demodulation 10.1 – 10.11

Chapter - 11 Digital Electronics 11.1 – 11.39

Chapter - 12 Cathode Ray Oscilloscope (CRO) 12.1 – 12.5

Chapter - 13 Wave Shaping Circuits 13.1 – 13.7

Chapter - 14 Television and Radar 14.1 – 14.5

Chapter - 15 VLSI Technology 15.1 – 15.3

Chapter - 16 OPTO–electronics (Photonics) 16.1 – 16.39

Chapter - 17 Communication Systems 17.1 – 17.67

Chapter - 18 Control System 18.1 –18.43

Chapter - 19 Signals and Systems 19.1 – 19.63

Chapter - 20 Electromagnetic Field Theory 20.1 – 20.86

Chapter - 21 Engineering Mathematics 21.1 – 21.137

Chapter - 22 General Aptitude 22.1 – 22.137

Appendix – A : Matching Questions A.1 – A.10

Appendix – B : GATE : Electronics and Communication Engineering B.1 – B.154 Papers with Solutions (2018 – 2013)

Appendix – C : GATE : Physics (Electronics) C.1 – C.39 Papers with Solutions (2017 - 2001)

CONTENTS

Circuit / Network Analysis 1.1

Chapter

1Study Aids :1.1 Passive Network

A passive network (circuit) comprises of circuit ele-ments like resistors, inductors, etc.

1.2 Active NetworkAn active network (circuit) comprises of energy sourcesas well as other circuit elements like resistors, capacitors,diode, transistors, etc. Following terms are related to boththe above networks:(a) Node : It is an equipotential surface at which two or

more elements are joined.(b) Junction : It is a point in network where two or

more elements are joined.(c) Branch : Any group of elements in series having two

terminals is called a branch.(d) Loop : Any closed path in a network is called loop.(e) Mesh : The space which a loop encloses is called as

mesh.

1.3 Linear NetworkIn a linear network, the current in all branches isproportional to the driving voltagae.

1.4 Voltage SourceReal voltage source is represented as ideal voltage sourcein series with a resistance ‘r’ as shown. The ideal voltagesource has zero resistance.

Fig. 1

1.5 Current SourceReal current source is represented as ideal current sourcein parallel with a resistance ‘r’ as shown. The ideal currentsource has infinite resistance.

Fig. 2

1.6 Kirchoff’s Laws(1) First Law (Current Law) called KCL

It states that total current entering a junction or node isexactly equal to current leaving the junction or node. Inother words, the algebraic sum of all currents enteringand leaving the node must be equal to zero.

I 0 or I(outgoing) + I(entering) = 0

In a way, it is conservation of charge.

(2) Second Law (Voltage Law) called (KVL)

It states that in any closed loop network, the total voltagearound the loop is equal to the sum of all voltage dropswithin the loop.

In a way, it is conservation of energy. In other words,we can state that the algebraic sum of the product ofcurrent and resistance in any closed loop of circuit is equalto the algebraic sum of electromotive forces (emf) presentin the loop.

IR E

1.7 Network Theorems (AC and DC)(1) Superposition theorem :

It states that current in any branch of a bilateral linearcircuit equals the algebraic sum of currents produced byeach source acting separately through the circuit. Here,the voltage source is treated as short circuit and currentsource is treated as open circuit.

The superposition does not apply to power since itis not a linear function. Power in a circuit is square ofcurrent as voltage.

While applying the theorem, following points mustbe kept in mind:

(i) It is applicable to linear circuits only.

(ii) The effect of one source is taken at a time bysuppressing other sources.

(iii) Current source is treated as open circuit whilevoltage source is treated as short circuit.

(iv) Dependent sources will remain unchanged.

Example : Find the current I in the circuit (Fig. 2).Let I1 be the current due to (– 4 V) voltage source

with current source open circuit.I2 be the current due to (2 A) current source with

voltage source short circuit.

CIRCUIT/NETWORK ANALYSIS

1.2 Circuit / Network Analysis

2

(I2 – 2)

Vx

+

–

Fig. 6

Substituting for Vx in equation (ii), we have5 I2 + 5 (– 2 I2 + 4) = 4

or – 5 I2 = 4 – 20 or I2 = A5

16

Substituting for I1 and I2, we obtain

I = I1 + I2 = A 205

1654

(2) Thevenin’s theorem :

Any two terminal network comprising of linear impedancesand generators can be replaced by an equivalent circuitconsisting of a voltage source VTh in series with aresistance (impedance ( ThZ )), RTh as shown in Fig. 7.

2

2

+

–E

Z1

Z3

Z2

ZL ZL

2

2

Rth

Z0VTh

Fig. 7

The value of VTh is the open circuit voltage between theterminals of the network and ZTh ThR is the impedancemeasured between the terminals of the network.(3) Norton’s theorem :Any two terminal network consisting of linear impedancesand generators can be replaced by an equivalent circuitconsisting of a current source ISC in parallel with animpedance Z0 as shown in Fig. 8.

2

2

+

–E

Z1

Z3

Z2

ZL ZL

2

2

Z0ISC

Fig. 8

So, total current I = I1 + I2

– +– 4V 3

1 2 Vx

I

5 Vx

+

–

2 A

Fig. 3

Applying KVL in the above circuit (Fig. 3),

2 I1 – (– 4) + 3 I1 + 5 Vx = 0

or 5 I1 + 5 Vx = – 4 … (i)

To find Vx, we have (Fig. 4)

2

I1

Vx

+

–

Fig. 4

– Vx – 2 I1 = 0 or Vx = – 2 I1

Substituting for Vx in equation (i), we have

5 I1 + 5 (– 2 I1) = – 4 or (–) 5 I1 = – (4)

I1 = 4/5 A3

2 5 Vx

+

–

1

2 A

Fig. 5

To find I2, short circuit voltage source (– 4 V) andapply KVL in the outer circuit which does not containcurrent source (Fig. 5).

Applying KVL in outer loop,

2 (I2 – 2) + 3 I2 + 5 Vx = 0

or 5 I2 + 5 Vx = 4 … (ii)

To find Vx, we have (Fig. 6)

– Vx – 2 (I2 – 2) = 0

or Vx = – 2 I2 + 4


1.8 Applications(1) Thevenin’s theorem :

(i) Remove the load resistor ZL.

(ii) Find RTh (ZTh) by shorting all the voltage sourcespresent in network (Fig. 7) as viewed back intothe terminals (2, 2)

1 3Th 2

1 3

Z ZR ZZ Z

.

(iii) With load terminals open, the open circuit voltageis calculated which is called Thevenin’s voltage(VTh) or

VTh = 30C

1 3

ZE E

Z Z

.

(2) Norton’s theorem :

(i) Remove the load resistor (ZL).

(ii) The short current (ISC) flowing between theterminals (2, 2) (Fig. 8) is found. This current isalso called Norton’s current and given by

3

SC1 2 1 3 2 3

E ZIZ Z Z Z Z Z

(iii) The Norton’s resistance RTh or Z0 of the circuitas viewed back into the open terminals (2, 2) iscalculated by shorting all the voltage source presentin network and given by

1 3Th 0 2

1 3

Z ZR Z ZZ Z

(iv) Maximum power transfer theorem : It statesthat power delivered by an active network to aload across its terminals is maximum when theload resistance (impedance) is equal to internalresistance (impedance) of DC source.

In general, ZL = ZS

or L L S SR jX R jX

So RL = RS and XL = (–)XS

(v) Reciprocity theorem : It states that if an emfapplied in one mesh of a network of linearimpedence produces certain current in the secondmesh, then the same emf acting in the second meshwill give an identical current in the first mesh. Thebasis of the theorem is the symmetry of Z and Yparameters. This theorem is not applicable to thecircuit in which dependent source is present.

1.9 Two-port Network

V1V2

I1 I1

2

2

1

1

Fig. 9

For a two-port network as shown, we may have any twoas independent parameters and rest two will be dependentparameters.The following cases are of interest.(1) Z-parameters or open circuit impedance parameters:Here, I1 and I2 are independent variables. Therefore,

1 1 1 2V f I , I

2 2 1 2V f I , I

For sufficiently small signals, we may express

V1 V2

I1I2Z11

Z12I2 Z21I1

Z122

Fig. 10

1 11 1 12 2V Z I Z I

2 21 1 22 2V Z I Z I

The above equations form the circuit as shown. (Fig. 10)

Here, Z11 = ZI = constI1

1

ZIV

is called open circuit

impedance.

Z12 = Zr = constI1

1

1IV

is called open circuit forward transfer

impedance.

Z21 = Zf = 2

1 I const2

VI

is called open circuit reverse transfer

impedance.

Z22 = Z0 =2

2 I const1

VI

is called open circuit output

impedance.


(2) Y-parameters or short circuit admittance parameters

Here, V1 and V2 are independent variables and I1 and I2are dependent variables.

V1 V2Y12y11

Y11V2

Y21V1

I1I2

Fig. 11

Therefore, 1 1 1 2I f V ,V

2 2 1 2I f V , V


I1 = Y11 V1 + Y12V2

I2 = Y21 V1 + Y22V2

The above equations form the circuit as shown in Fig. 11.

Here, Y11= Yi = 1

1 V const2

IV

is called shortt circuit input

admittance.

Y12 = Y r = 1

2 V const1

IV

is called reverse transfer

admittance.

Y21 = Yf = 2

1 V const2

IV

is called short circuit forward

transfer admittance.

Y22= Yo= 2

2 V const1

IV

is called short circuit output

admittance.

(3) h-parameters

Here, I1 and V2 are taken as independent variables and V1and I2 are dependent variables.

therefore, 1 1 1 2V f I ,V

2 2 1 2I f I ,V


V1 = h11 I1 + h12 V2

I1 = h21 I1 + h22 V2

The above equations form the circuit as shown inFig. 12.

Fig. 12

Here, h11= hI=1

1 V const2

VI

is called short circuit input

impedance.

h12= hr=1

1 I const1

VI

is called open circuit reverse

transfer voltage ratio (I1 = constant).

h21= hf =1

1 V const2

II

is called short circuit forward

transfer current ratio.

h22= ho=1

2

2 I const

IV

is called open circuit output

admittance.

Ex. 1. A four terminal network is inserted between sourceS and resistance R as shown in Fig. 13. The resistanceseen by the source S remain the same with or without thefour terminal network. What is the value of R?

5 5

20 RS

Fig. 13

Sol. The image impedance of network should beequal to R.

Image impedance SCOC ZZ

where ZOC – Open circuit impedance

ZSC – Short circuit impedance

ZOC = 20 + 5 = 25

SC20 5

Z 5 9 20 5

Image impedance 25 9 15


Ex. 2. Current I in the Fig. 14 shown is

10 V

+

1 2

3 4

I

I1

I2

Fig. 14

Sol. The total resistance seen by the battery

Reg = 1225

68

43

Current in the circuit drawn from battery

A5

2412/25

10R10I

egT

Current in upper branch A5

1843

524

43II T1

Current in lower branch A56

41

524

41II T2

Current A5

1256

518III 21

Ex. 3. For the circuit shown in Fig. 15, the voltage VAB is

50 V10 V

5 10

A

B

+

Fig. 15

Sol. For finding Thevenin’s equivalent voltage, removethe resistor across A and B.

Current in circuit A4510

5010I

Hence, VOC= 50 – (10 4) = 10 V

Resistance

3

10510510R Th

B

A

I1

5 V = 10 VOC

103R = th

Fig. 16

Thevenin’s equivalent network is as shown in Fig. 16.

Current network A56

53

1010I1

Voltage across AB = V 6565

Ex. 4. Find the current through load RL in the circuit usingNorton’s theorem.

15 14

4 R = 20 L 70 V

Fig. 17Sol.

20 32619

2

2

Fig. 18

Current through the network by short circuiting load

A

41441415

70I

Current through short circuiting branch

184

562701870

1444IISC

A326280

Internal impedance of network

19326

15415414R Th

Current through load IL using Norton’s theorem

A326280ISC

2016326

19/326II CSL

A39.0706326

326280


Ex. 5. Calculate Thevenin’s equivalent of the given network.Also calculate current into load.

100 1000

200 1 k10 V+

–RL

Fig. 19

Sol. Thevenin’s voltage

VTh = 10 V320

100200200

Internal Impedance of network

3

3200200100200100100

Current through load

mA 2.3

10003

32003/20

RRVR

LTh

ThL

MULTIPLE CHOICE QUESTIONS

1. The Thevenin’s resistance of the circuit (Fig. 20) acrossterminals (1, 2) is

100 1000

200 1 k10 V+

–RL

1

2Fig. 20

(a) 3200

3 (b) 3200

(c) 320012

(d) 3200

11

2. The Thevenin’s voltage of the circuit (Fig. 20) acrossterminals (1, 2) is

(a) 100 V11 (b)

20 V3

(c) 10 V6 (d) 10 V

3. The Thevenin’s equivalent circuit of Fig. 16 acrossterminals (1, 2) is:

Fig. 21

4. The load current in Fig. 20 is(a) 3.2 mA (b) 2.1 mA(c) 9.0 mA (d) 3.1 mA

5. The short circuit current (Fig. 22) as per Norton’s theoremis

+36 V

6 k

3 k 1 k

1

2

Fig. 22(a) 6.2 mA (b) 3.1 mA(c) 31 mA (d) 62 mA

6. The Thevenin’s voltage for the circuit (Fig. 22) acrossterminals (1, 2) is(a) 36 V (b) 18 V(c) 12 V (d) 6 V

7. The Thevenin’s resistance for the circuit (Fig. 22) whenseen across terminals (1, 2) is

(a) 9 k (b) 3k

(c) 2k (d) 1 kΩ

8. The current through load 1 kΩ (Fig. 22) is(a) 3.6 A (b) 4.8 A(c) 5.4 A (d) 4 A

9. The Thevenin’s equivalent circuit for (Fig. 22) circuit whenseen across terminals (1, 2) is

20/3 V

(3200/3)

1 k+

2

1

10 V

(3200/3)

1 k+

2

1

(b)

100

20/3 V1 k

+

2

1

(c)

20/3 V

1 k

1 k+

2

1

(d)


9 k

1 k+

2

1

(a)

2 k

1 k+

2

1

(c)

6 k

3/4 k+

2

1

(b)

6 k

1 k+

2

1

(d) Fig. 23

10. In the Norton’s circuit (Fig. 24), short circuit currentISC = 10 mA and Thevenin’s resistance ThR =10 kΩ . Theopen circuit voltage EOC is

RLRThISC

Fig . 24

(a) 100 V (b) 50 V(c) 10 V (d) 1 V

11. The open circuit impedance and short circuit impedanceof a two-port network is defined as

(a)2

1

1 I const

VI

and

2

1

1 V const

VI

respectively

(b)2

1

1 I const

VI

and

1

1

1 V const

VI

respectively

(c)2

1

1 V const

VI

and

2

1

1 I const

VI

respectively

(d)1

1

1 V const

VI

and

2

1

1 V const

VI

respectively

12. The units of open circuit impedance and open circuitadmittance are(a) ohm, ohm respectively (b) ohm, mho respectively(c) mho, mho respectively (d) No unit for both

13. The units of hybrid parameters hi, hr, hf and h0 are(a) ohm, ohm, no unit, mho respectively(b) ohm, no unit, no unit, mho respectively(c) No unit, ohm, no unit, ohm respectively(d) ohm, no unit, no unit, ohm respectively

14. The hybrid parameters hr and hf of a two-port networkare defined as

(a) 2

1

2 I const

VV

and

2

1

2 V const

II

respectively

(b) 2

2

1 I const

VV

and

2

1

2 V const

II

respectively

(c) 2

1

2 V const

VV

and

21 V const

I2I

respectively

(d) 2

2

1 I const

VV

and

2

2

1 V const

II

respectively

15. In a two-port network, correct statement is(a) hi, hf are short circuit parameters and hr, h0 are open

circuit parameters(b) hi, hf are open circuit parameters and hr, h0 are short

circuit parameters(c) hi, hr are open circuit parameters and hf, h0 are short

circuit parameters(d) hi, h0 are short circuit parameters and hr, hf are open

circuit parameters16. Assuming all resistances equal to r, the equivalent resistance

between A and B is

r1

r2

r9

r13

r10

r11r12

r3

r4

r5

r6r7

r8

A B

Fig. 25

(a) 2 r (b) 4 r/3(c) 2 r/3 (d) r

17. Eight resistors, each of value 5 , are connected as shownin Fig. 25. The equivalent resistance between A and B is

(a) 83 (b)

163

(c) 157 (d)

192

18. The equivalent resistance between A and B [Fig. 26 (a)] is

Fig. 26 (a)

(a) 1 (b) 2

(c)5 12

(d)

52

19. The equivalent resistance between points C and D[Fig. (26 (b)] is


Fig. 26 (b)

(a) 1 (b) 2

(c) 5 12

(d)

5 12

20. The equivalent resistance of circuit [Fig. 26 (a)] is r1 andthat of circuit [Fig. 26 (b)] is r2. The relation betweenthem is(a) r1 > r2 (b) r1 < r2(c) r1 = r2 (d) r1 = r2 + r

21. The current through 3 resistor (Fig. 27) is

+–

+–10 V

2

3

B

A

4 V

Fig. 27

(a)14 A3 (b) 3 A

(c) 2 A (d) 12 A5

22. The emf of the battery E shown in Fig. 28 is

Fig. 28(a) 12 V (b) 16 V(c) 18 V (d) 20 V

23. The potential difference between terminals A and B of thecircuit shown (Fig. 29) is 16 V. The current passingthrough 2 resistor is

2

4 1

6

A B

9 V 3 V

r

Fig. 29

(a) 2.5 A (b) 3.5 A(c) 4.0 A (d) zero

24. On closing the switch S, the current flowing throughswitch S is

2 4

2

20 V 5 V

S

Fig. 30

(a) 4.5 A (b) 16.0 V(c) 3.0 A (d) zero

25. The reading of voltmeter is V1 when only switch S1 isclosed. The reading of voltmeter is V2 when only switchS2 is closed and the reading of voltmeter is V3 when boththe switches S1 and S2 are closed. Then we have

Fig. 31

(a) V3 > V2 > V1

(b) V2 > V1 > V3

(c) V3 > V1 > V2

(d) V1 > V2 > V3

26. Maximum power that can be developed in resistor R in thecircuit shown (Fig. 32) is

Fig. 32

(a) 50 W (b) 75 W(c) 25 W (d) 100 W

27. Two bulbs consume same power when operated on200 V and 300 V respectively. When these bulbs areconnected in series and operated on by a DC source of500 V.(i) The ratio of potential difference across them is


(a) 3/2 (b) 4/9(c) 1/2 (d) 1/1(ii) The ratio of power consumed by them is(a) 1 : 1 (b) 1 : 2(c) 3 : 2 (d) 4 : 9

28. For maximum power transfer from source shown in Fig.33, the value of resistor Rx in the given circuit is

7.1 k

19.4 k

5.2 k

10.4 k

10 kV

Rx

Fig. 33

(a) 33.4 k (b) 17.6 k

(c) 10 k (d) 5.0 k

29. Match List I with List II and select the correct answerusing codes given belowList I List II

(Network Theorem) (Most Distinguished Property)(A) Reciprocity (1) Impedance matching(B) Thevenin (2) Bilateral(C) Superposition (3) Voltage source(D) Maximum Power (4) Linear

A B C D(a) (1) (2) (3) (4)(b) (1) (3) (2) (4)(c) (2) (3) (4) (1)(d) (2) (4) (3) (1)

30. Manifestation of law of conservation of charge is relatedto(a) Kirchoff’s current law (b) Reciprocity theorem(c) Thevenin’s theorem (d) Norton’s theorem

31. For the two-port network shown in Fig. 34, hybridparameter h21 is

I1

1 2I2

R R

R

1 2

V1V2

Fig. 34

(a) 12

(b) 12

(c) 32

(d) 32

32. For the two-port network shown in Fig. 35, Z-parametersZ11 and Z21 are

I1

1 2I2

R R

1 2

V1V2

4

10 V

2

Fig. 35

(a) 611

and 6

11 respectively

(b) 6

11 and

411

respectively

(c) 6

11 and

1116

respectively

(d) 411

and 4

11 respectively

33. Load 2 is connected across the secondary of idealtransformer (Fig. 36). The number of turns in secondaryis 40. Resistance of the source VS in primary is 8 . Formaximum power transfer in the load, the number of turnsin primary is

VS

8

NPNS

2

Fig. 36

(a) 20 (b) 40(c) 80 (d) 160

34. For the two-port network as shown in Fig. 37, Z11 andZ22 are

I1

1 2I2

1 2

V1V2

ZA ZB

ZC

Fig. 37

(a) ZA + ZC, ZB respectively(b) ZA, ZB + ZC respectively(c) ZA + ZC, ZB + zC respectively(d) ZB, ZA + ZC respectively

35. A DC voltage source is applied across a series R-L-Ccircuit. Under steady state condition, the entire voltagedrops across


(a) R only (b) C only(c) L only (d) R-L combination

36. The voltage across the terminals A and B in the circuitshown in Fig. 38 is

+1 V 2

2 1 A

B

Fig. 38

(a) 0.5 V (b) 3.0 V(c) 3.5 V (d) 4.0 V

37. The maximum power transfer to RL in the circuit shownin Fig. 39 should be

+

50 V

20 0.5 IRL

Fig. 39

(a) 16 (b) 403

(c) 60 (d) 20

38. A voltage source V and current source I are connected toa resistance network as shown in Fig. 40. The load Rdissipates a power of 4 watt where ‘V’ alone is active andit dissipates a power of 9 watt when I alone is active.When both sources are active, load R will dissipate

V Resistance Network

I

R

Fig. 40

(a) 1 W (b) 5 W(c) 13 W (d) 25 W

39. The Thevenin’s theorem when applied across A and B(Fig. 41) gives VTh and RTh as

4

5 V

+

–5 V

6

+

A

BFig. 41

(a) 5 V and 10 respectively

(b) 1 V and 2.4 respectively

(c) 7 V and 2.4 respectively

(d) 1 V and 10 respectively

40. The characteristic impedance R0 of the given network(Fig. 42) is

R1

R2

R0

R1

R2

Fig. 42

(a) 1 2R R2

(b) 1 2

1 2

2R RR R

(c) 1 2

1 2

R RR R (d) 1 2R R

41. “In any network of linear impedances, the current flowingat any point is equal to the algebraic sum of the currentscaused to flow at the point by each of the sources of emftaken separately with all other emf reduced to zero.” Theabove statement represents(a) Millmans Theorem (b) Reciprocity Theorem(c) Thevenins Theorem (d) Superposition Theorem

42. The value of R for maximum power transfer to it is

R

5 4

20 3 A

Fig. 43

(a) 2 (b) 4

(c) 8 (d) 16 43. The Y-parameters Y11, Y12, Y21, Y22 of the two-port

network shown in Fig. 44 areI1

1 2I2

R R

1 2

V1V22

2 2

Fig. 44

(a) 0.5 mho, 1 mho, 2 mho, 1 mho respectively

(b) 1/3 mho, 16

mho, 16

mho, 13 mho respectively


(c) 0.5 mho, 0.5 mho, 1.5 mho, 2 mho respectively

(d) 25 mho,

37 mho, 2 mho,

25 mho respectively

44. The voltage transfer ratio for the circuit shown in Fig. 45is

2 2

2

2 V1

A V2

2

I3 I1

2

1 A

Fig. 45

(a) 1

13 (b) 2

13

(c) 3

13 (d) 4

1345. The equivalent resistance of network between X and Y as

shown in Fig. 46 is

30

15

15

30

30

15

15

A B

X

Y

Fig. 46

(a) 150 (b) 45

(c) 55 (d) 30

46. The circuit shown in Fig. 47 has an equivalent load of

–

4

+

AI

2 I

Fig. 47

(a) 1 Ω2 (b) 83

(c) 4 (d) 2

47. For the circuit shown in Fig. 48, the current through R is1.0 A when VA = 0 and VB = 15 V. If both VA and VB areincreased by 15 V, the current flowing through R will be

3 3

6

R

6 VA

+

–VB

+

–

Fig. 48

(a) 1.0 A (b) 0.5 A(c) 3.0A (d) 0.33 A

48. The two networks shown in Fig. 49 below are equivalentas seen from the terminals A and B. The value of V (involts) and Z (in ohms) will be

20 20 A100 V+

30

V+

30 A A

BB

Fig. 49

V Z(a) 100 12(b) 60 12(c) 100 50(d) 60 30

49. For the circuit shown in Fig. 50, the Thevenin’s voltageand Thevenin’s equivalent resistance at terminals (A, B) is

5

5 0.5 I1

+

– B

A

1 A I1

I1 +10 V

C

Fig. 50

(a) 5 and 2Ω (b) 7.5 V and 2.5Ω

(c) 4 V and 2Ω (d) 3 V and 2.5Ω

50. In the bridge circuit shown in Fig. 51, R1 = R2 = R4 = Rand R3 = 1.1 R, then the reading of the voltmeter acrossA and B is

V

R4

R3R2

R1

A B+ –10 V+

–

Fig. 51

(a) 0.238 V (b) 0.138 V(c) – 0.238 V (d) 1.0 V


51. In the circuit (Fig. 52), Z1 = 10 –60o, Z2 = 10 60o andZ3 = 50 53.13o. The Thevenin’s impedance as seen fromterminal X – Y is

Z1

Z2

Z3

100 0 o+

–

X

Y

Fig. 52

(a) 56.5745o (b) 6030o

(c) 7030o (d) 34.465o

52. The value of R (in ohms) required for maximum powertransfer in the network shown (Fig. 53) is

+25 V –

5 4

20 3 A

R

Fig. 53

(a) 2 (b) 4(c) 8 (d) 16

53. A two-port Network is shown in Fig. 54. The value ofhybrid parameter h21 for the network is

R RI1 I2

++

V1 V2R

Fig. 54

(a) 12

(b) 21

(c) 32

(d) 23

54. The current i4 in the circuit (Fig. 55) shown isi1 = 5A i2 = 5A

i3 = 4Ai4

i0 = 7A

Fig. 55

(a) 12 A (b) – 12 A(c) 4 A (d) None of the above

55. The voltage e0 across resistor in the circuit (Fig. 56) shownis

4 2

2 4 +

12 V –e0

Fig. 56(a) 2 V (b) 4/3 V(c) 4 V (d) 8 V

56. The admittance parameter Y12 is the two-port network(Fig. 57) shown is

RI1 I2

++

E1 E25 10

20

Fig. 57(a) – 0.2 mho (b) 0.1 mho(c) – 0.05 mho (d) 0.05 mho

57. The Z-parameters Z11 and Z21 for the two-port network(Fig. 58) shown is

RI1 I2+

E1 E25 10

20

Fig. 58

(a) 11 126 16Z = – Ω, Z = Ω

11 11

(b) 11 126 4Z = Ω, Z = Ω

11 11

(c) 11 126 16Z = Ω, Z = – Ω

11 11

(d) 11 124 4Z = – Ω, Z =– Ω

11 1158. In the circuit (Fig. 59) shown, the current through diode

(assuming it ideal and forward biased) is

+

–10 V

4

4

1 2A

D

Fig. 59


(a) 0 A (b) 4 A(c) 1 A (d) 2 A

59. The equivalent impadance of the circuit (Fig. 60) as seenacross terminals (a), (b) is (Assume bridge to be balanced)

4

4 2

+ j 3

– j 4

2

Fig. 60

(a) 16 Ω3 (b)

8 Ω3

(c) 12 (d) 3

60. The value of resistor R connected across terminal (A, B)in the circuit (Fig. 61) which will absorb maximum poweris

4 k

4 k6 k

3 k

RA B

Fig. 61

(a) 4.0 k (b) 4.11 k

(c) 8.0 k (d) 9 k

1. (a) 2. (b) 3. (a) 4. (a) 5. (a) 6. (c) 7. (c) 8. (d) 9. (c) 10. (a)

11. (a) 12. (b) 13. (b) 14. (c) 15. (a) 16. (c) 17. (a) 18. (c) 19. (d) 20. (a)

21. (c) 22. (a) 23. (b) 24. (a) 25. (b) 26. (a) 27. (b) 28. (c) 29. (c) 30. (a)

31. (a) 32. (c) 33. (c) 34. (c) 35. (a) 36. (c) 37. (a) 38. (d) 39. (c) 40. (a)

41. (d) 42. (c) 43. (b) 44. (b) 45. (d) 46. (d) 47. (c) 48. (c) 49. (b) 50. (c)

51. (a) 52. (c) 53. (a) 54. (b) 55. (c) 56. (c) 57. (c) 58. (c) 59. (b) 60. (a)

ANSWERS

SOLUTIONS

1. Keeping terminal (1, 2) as open and short circuiting thevoltage source, the Thevenin’s resistance (RTh) is

100 200 32001000100 200 3

Correct choice - (a)2. The Thevenin’s voltage across terminals (1, 2)

OC Th10 20E V 200 V

100 200 3

Correct choice - (b)

3. The Thevenin’s equivalent circuit is shown in Fig. 62.

Th20V V3

VThRL

RTh

Fig. 62

3

3200R Th

Correct choice - (a)

4. The load current

ThL

Th L

20V 3i3200R R 1000

3

= 3.2 mA



5. The short circuit current

323121

3SC ZZZZZZ

EZI

10 200100 200 100 1000 200 1000

= 6.2 mA


6. The Thevenin’s voltage

V 121031036

36VE 33ThOC

Correct choice - (c)

7. The Thevenin’s Resistance

k 26363R Th


8. Load current

3Th

L 3Th L

V 12R 10 4 AR R 2 1 10

Correct choice - (d)

9. The Thevenin’s equivalent circuit


10. The open circuit voltage

3OC SC ThE I R 10 10 10 1000

= 100 V


11. The open circuit impedance is defined as

constI1

1i

2IVZ

and the short circuit input impedence is defined as

constV1

1i

2IVZ

Corect choice- (a)

12. Correct choice - (b)

Since unit of impedance is ohm and unit of admittance ismho.

13.constV1

1i

2IVh

unit is ohm

constI2

1r

2VVh

so it has no unit

2

2f

1 V cons t

Ih =

I

so it has no unit

constI2

2o

2VIh

SI unit 1

ohm = mho


14. hr is defined as constI2

1r

2VVh

hf is defined as constV1

2f

2IIh


15. (a) hi and hr are short circuit parameters of a two-port network.

constV1

2f

constV1

1i

22IIh and

IVh

i.e., dV2 = 0

(b) hr and ho are open circuit parameters since

constI2

2o

constI2

1r

12VIh and

VVh

or dI1 = 0


16. The circuit can be redrawn as4 R

R

4 RA B

Fig. 63

where R is equivalent of 2 R and 2 R in parallel.

Its equivalent resistance across A and B is

1 1 1 1R ' 4R R 4R

or 2R ' R3



17. The circuit can be redrawn as shown in Fig. 64.

5

A B

10

(40/3)

Fig. 64

Its equivalent resistance across A and B is

1 3 1 1R ' 40 10 5

or 8R '3


18. The given circuit [Fig. 26 (a)] can be redrawn as1

1 x

B

A

Fig. 65

Let ABx 1R x 1x 1

or xx 1

x 1

or x2 – x – 1 = 0

So, 5 1x2


19. The given circuit [Fig. 26 (b)] can be redrawn as shownin Fig. 66.

1

1 Y

D

C

Fig. 66

Let

CD

1 1 yR y

1 2 y

or y2 + y - 1 = 0

So 5 1y2


20. From Q. No. 18 and Q. No. (19) assuming x = r1 andy = r2, we find that x > y. So, r1 > r2


21. From the circuit, we have

A BV V 4V

A CV V 10V

B CV V 6V

So, potential difference across 3 is 6 V and current

flowing through it is 6 = 2A.3


22. Net resistance of the given circuit is 4 .

Current drawn from battery = Ei A4

This current is distributed in the circuit shown in Fig. 67

2 1

6

2

i 2 i3

i3

i3

i3

i+

2 1 –

2i3

Fig. 67

From the Fig. 67, we have i 13

or i = 3 A

So, 34E

E = 12 V


23. Let the current flowing through points A to B be i1 asshown in Fig. 68 and current flowing from 9 V battery be

i2, then current flowing through 2 Ω is i1 + i2.

2

4 1 A B

9 V 3 V+ + 1 i1i1

i1 + i2

Fig. 68

We know that VA – VB = 16 V ........ (i)


Applying KVL, we have

1 1 2 14i 2 i i 3 4i 16

Also applying KVL to close circuit,

9 – i2 – 2(i1 + i2) = 0 ....... (ii)

Solving equations (i) and (ii),

i1 = 1.5 A, i2 = 2A

So, current through 2 resistor = 2 + 1.5

= 3.5 A


24. Let V be the potential of junction J and current through

2 be i1 and through 4 be i2. So current through

branch JE is i1 + i2 = i3

Current 120 Vi

2

Current 25 Vi

4

Current 3V 0i

2

Subtracting these values, we have

20 V 5 V V 02 4 2

or V = 9V

and 39i 4.52


25. We know that potential difference PD (V) Resistor (R)in which current is flowing

When switch S1 is closed, 13V E4

When switch S2 is closed, 26V E7

When both switch is closed, 32V E3

So, we have 2 1 3V V V .


26. The given Circuit can be redrawn as shown in Fig. 69.

+E

R

r

Fig. 69

Maximum power that can be transferred to R is only whenr = R. Here,

1 2

1 2

r r 1×1 1r = = = Ω

r +r 1+1 2

10 10E +r 1 1E = = = 10 V1 1+1r

For maximum power transfer,

1R r r2

maxE 10I 10A1 1R r 2 2

2 2max max

1P I R [10] 50 W2

Correct choice- (a)

27. (i) Power in device, 2VP

R

2VRP

Resistance of first bulb (R1) 2

1

200P

Resistance of second bulb (R2) 2

2

300P

21 2

2 1

R P200 4R 300 P 9

Since P1 = P2

When the two bulbs are connected in series, the potentialdrop across them is proportional to their resistance.

1 1

2

V R 4V 2 9


Power consumed P1 = i2 R1

P2 = i2 R2


Since they are in series

11

2 2

RP 4P 9R


28. For maximum power, transfer the value of Rx be equal tointernal resistance of source, i.e., resistance as seen by it

x10.4 × 29.4 5.2 × 7.1R = + = 10 kΩ10.4 + 29.4 2.5 + 7.1


29. Correct choice - (c)

30. Correct choice - (a)

31. For h-parameters, we have

1 11 1 12 2V h I h V

2 21 1 22 2I h I h V

h21 is defined as h21 = 0dV1

2

2II

So, keeping V2 = 0, i.e., short circuiting terminals (2, 2),we have

1 2 2R(I I ) RI 0

or 1 2RI + 2RI = 0

or 211

2 h21)(

II


32. For Z-parameters, we have

V1 = Z11I1 + Z12I2

V2 = Z21I1 + Z22I2

Z11 is defined when I2 = 0. So, Z11 = 1

1

IV

So, from network, we have

V1 = 6I1 – 10 V1

or 11 V1 = 6 I1

or 1

1

V 6I 11

Z21 is defined when I2 = 0, and so Z21 = 2

2

IV

So, from network, we have

2 1 1V 4 I 10 V

1 16

4I 10 I11

Z21 = 1116)(

IV

1

2


33. For maximum power transfer;

Zi = ZLwhere ZL is impedance as seen in the primary and ZL isimpedance connected in secondary.

2

PL L

S

NZ Z

N

2P

LN

Z 240

240N

Z8Z2

pLi

NP = 80


34. Z11 is defined as, Z11 = CA0I1

1 ZZIV

2

Z22 is defined as, Z22 = CB0I2

2 ZZIV

1


35. Under steady state, i.e., at resonance, XL = XC and oppositeto each other. So, the entire voltage is across R only.


36. With respect to node A, applying Kirchoff’s current law,

A AV 1 V3 0

2 2

or 2VA – 1 – 6 = 0

or A7V 3.5V2

Correct choice - (c)37. For maximum power transfer to load RL, its value should

be equal to internal resistance of voltage source. So

RL = Ri, then Zi can be determined as

Applying KCL of node A, we have

A A AV V V I20 40 2 40


or A5V I80

5

80I

VZ Ai


38. (i) When source V along is active, power dissipated is 2

1 1P I R .

214 I R

or 12I

R

(ii) When source ‘I’ alone is active then power dissipated is

22 2P I R

or 9 = RI22

or 23I

R

When both sources are active, current drawn is

I = I1 + I2

and power dissipated is

P = I2 R

= (I1 + I2)2R

2 2

2 3 5.R RR R R

= 25 WCorrect choice - (d)

39. Shorting the voltage sources gives

Th6 × 4R = = 2.4 Ω6 + 4

Also, Th5V 10 6 10 3 7 V

10


40. It can be easily seen that the circuit reduces to (Fig. 70)

P Q

R2 R1

R1 R2

Fig. 70

We have R1 + R2 in series and the combination of the twogives

0 1 2 1 2

1 1 1R R R R R

So, 1 20

R RR2


41. Correct choice - (d)

42. For maximum power transfer R = Rin of voltage source,

when voltage source is reduced to zero

in5 × 20R = + 4 = 8Ω2 + 20


43. Short circuit terminals (2, 2), i.e., V2 = 0.

61)(

VIY ,mho

31

VIY

0V1

221

0V1

111

22

31

VIY ,mho

61)(

VIY

0V2

222

0V2

112

11


44. Assume that current 1 A is flowing

V2 = 1 4 = 4 V, Current 21

V 4I = = = 2 Ar 2

Current I2 = I + I1 = 1 + 2 = 3 A

Potential at A = VA = V2 + I2 r = 4 + 3 2 = 10 VV

current A3

V 10I 5Ar 2

1 A 4 2 3V = V + I × r = 10 + I + I × 2 = 10 + 8 2 = 26 V

132

264

VV

1

2


45. It is a balanced bridge. So, no current flows through branchAB i.e., it is open.

Equivalent resistance between X and Y

602

601

601

r1

r1

R1

BAABeq

Req = 30


46. Let the potential at A be V.

Applying KCL, we have

V V 2I I4 2


38

IV


47. Correct answer - (c)

48. Equivalent Z = 50 Ω

Equivalent V = 100 V


49. Applying KVL to close circuit ABC, we have

10 = –5I1 + 5(1 – I1) = 5 – 10I1

or 11I 2

AB 11 1 15V 5(1 I ) 5 1 ( ) 5 1 7.52 2 2


50. The circuit can be drawn as shown in Fig. 71.

BAR

R 1.1 R

R

I1 I2

10 V

+

–

Fig. 71

110 5I2R R

210 4.76I

(2.1)R R

AB 1 2V ( ) RI I R

2 14.76 5R I I R ( )0.238

R R

51. Here Z1 = 10 –60o = 10[cos(– 60o) + j sin (–60o)] Z2 = 10 60o = 10[cos 60o + j sin 60o] Z3 = 50–53.13o = 50[cos 53.13o + j sin 53.13o] Z1 = Z2 = 18 ZTh = Z3 + Z1 || Z2 = 10 + 30 + j 40o = 40 + j 40

22 4040 45o= 56.6 45o

Correct choice - (a)52. For maximum power transfer.

ZL =ZS or RL = RS

ZS = 5 || 20 + 4 = 444205205

= 8 ohmCorrect choice - (d)

53. For h-parameters, we have V1 = h11 I1 + h12 V2

I2 = h21I1 + h22 V2

So, 0V1

22

2IIh

For V2 = 0, output must be short circuited and we have R(I1 + I2) + RI2 = 0.

or RI1 + 2 RI2 = 0

21

II

1

2

Correct choice - (a)54. Applying Kirchoff’s current law at node, we have

I0 + I1 + I4 = 0or 5 + 7 + I4 = 0or I4 = (–) 12 A

Correct choice (b)55. Zin = 4 + [4 || (2 + 2)] = 4 + 2 = 6

Current I = 12/6 = 2 A, current resistor = 1 AVoltage across resistor = 1 4 = 4 VCorrect choice - (c)

57. For Z-parameters, we have V1 = Z11I1 + Z12I2

V2 = Z21I1 + Z22I2

When I2 = 0, Z11 = 1

1

IV

we have E1 = 6 I1 – 16 E1

or 11 E1 = 6 I1

116

IEZ

1

111

Also, E2 = 4 I1 – 10 E1.Substituting for E1,

E2 = 4 I1 – 10

1I116

or E2 = 1 11

44 I – 60 I –16= I11 11


or 1116

IEZ

1

221


58. Applying Kirchoff’s current law at A,

04

V4

10V AA

or VA = 5V

Applying Kirchoff’s current law at B,

021

VB or VB = 2

Current through diode is 5 – 2

3

= 1 AA

59. If the bridge is balanced, i.e., 4

3

2

1

ZZ

ZZ

, then no current

flows through terminals CD since they are equipotential.

Equivalent impedance Zeq = (2 + 2) || (4 + 4)

= 4 || 8 = 38


60. For maximum power to be absorbed by resistor R,

we have Zeq of the circuit as

Zeq = 6 || 3 + 4 || 4 = 816

918

4444

3636

= 4


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ELECTRONICS COMMUNICATION ENGINEERINGNetwork solution methods: nodal and mesh analysis; Network theorems: superposition, Thevenin and ... Chapter - 1 Circuit / Network Analysis 1.1