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
Circuit / Network Analysis 1.3
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.
1.4 Circuit / Network Analysis
(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
For sufficiently small signals, we may express
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
For sufficiently small signals, we may express
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
Circuit / Network Analysis 1.5
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
1.6 Circuit / Network Analysis
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)
Circuit / Network Analysis 1.7
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
1.8 Circuit / Network Analysis
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
Circuit / Network Analysis 1.9
(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
1.10 Circuit / Network Analysis
(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
Circuit / Network Analysis 1.11
(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
1.12 Circuit / Network Analysis
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
Circuit / Network Analysis 1.13
(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
Correct choice - (a)
1.14 Circuit / Network Analysis
5. The short circuit current
323121
3SC ZZZZZZ
EZI
10 200100 200 100 1000 200 1000
= 6.2 mA
Correct choice - (a)
6. The Thevenin’s voltage
V 121031036
36VE 33ThOC
Correct choice - (c)
7. The Thevenin’s Resistance
k 26363R Th
Correct choice - (c)
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
Correct choice - (c)
10. The open circuit voltage
3OC SC ThE I R 10 10 10 1000
= 100 V
Correct choice - (a)
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
Correct choice - (b)
14. hr is defined as constI2
1r
2VVh
hf is defined as constV1
2f
2IIh
Correct choice - (c)
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
Correct choice - (a)
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
Correct choice - (c)
Circuit / Network Analysis 1.15
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
Correct choice - (a)
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
Correct choice - (c)
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
Correct choice - (d)
20. From Q. No. 18 and Q. No. (19) assuming x = r1 andy = r2, we find that x > y. So, r1 > r2
Correct choice - (a)
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
Correct choice - (c)
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
Correct choice - (a)
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)
1.16 Circuit / Network Analysis
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
Correct choice - (b)
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
Correct choice - (a)
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 .
Correct choice - (b)
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
Correct choice - (b)
Power consumed P1 = i2 R1
P2 = i2 R2
Circuit / Network Analysis 1.17
Since they are in series
11
2 2
RP 4P 9R
Correct choice - (d)
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
Correct choice - (c)
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
Correct choice - (a)
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
Correct choice - (c)
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
Correct choice - (c)
34. Z11 is defined as, Z11 = CA0I1
1 ZZIV
2
Z22 is defined as, Z22 = CB0I2
2 ZZIV
1
Correct choice - (c)
35. Under steady state, i.e., at resonance, XL = XC and oppositeto each other. So, the entire voltage is across R only.
Correct choice - (a)
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
1.18 Circuit / Network Analysis
or A5V I80
5
80I
VZ Ai
Correct choice - (a)
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
Correct choice - (c)
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
Correct choice - (a)
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
Correct choice - (c)
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
Correct choice - (b)
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
Correct choice - (b)
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
Correct choice - (d)
46. Let the potential at A be V.
Applying KCL, we have
V V 2I I4 2
Circuit / Network Analysis 1.19
38
IV
Correct choice - (d)
47. Correct answer - (c)
48. Equivalent Z = 50 Ω
Equivalent V = 100 V
Correct choice - (c)
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
Correct choice - (b)
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
1.20 Circuit / Network Analysis
or 1116
IEZ
1
221
Correct choice - (c)
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
Correct choice - (b)
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
Correct choice - (a)