network theory - gateflix.in · passive network elements are those which are capable only of...

NETWORK THEORY

For ELECTRICAL ENGINEERING

INSTRUMENTATION ENGINEERING ELECTRONICS & COMMUNICATION ENGINEERING

SYLLABUS Network graph, KCL, KVL, Node and Mesh analysis, Transient response of dc and ac networks, Sinusoidal steady‐state analysis, Resonance, Passive filters, Ideal current and voltage sources, Thevenin’s theorem, Norton’s theorem, Superposition theorem, Maximum power transfer theorem, Two‐port networks, Three phase circuits, Power and power factor in ac circuits.

ANALYSIS OF GATE PAPERS ELECTRONICS ELECTRICAL INSTRUMENTATION

Exam Year 1 Mark Ques.

2 Mark Ques. Total

1 Mark Ques.

2 Mark Ques. Total

1 Mark Ques.

2 Mark Ques. Total

2003 4 7 18 3 6 15 ‐ 1 2 2004 5 5 15 1 7 15 ‐ ‐ ‐ 2005 5 6 17 4 7 18 ‐ 2 4 2006 6 ‐ 6 2 6 14 1 4 9 2007 2 4 10 ‐ 7 14 2 4 10 2008 2 7 16 2 6 14 2 7 16 2009 3 4 11 2 6 14 ‐ 2 4 2010 2 3 8 3 4 11 2 2 6 2011 3 3 9 3 5 13 2 3 8 2012 4 4 12 5 6 17 4 6 16 2013 3 6 15 2 3 8 3 5 13

2014 Set‐1 2 4 10 2 2 6 3 4 11 2014 Set‐2 2 4 10 3 2 7 ‐ ‐ ‐ 2014 Set‐3 2 4 10 3 3 9 ‐ ‐ ‐ 2014 Set‐4 2 4 10 ‐ ‐ ‐ ‐ ‐ ‐ 2015 Set‐1 4 3 10 4 3 10 3 5 13 2015 Set‐2 3 3 9 3 3 9 ‐ ‐ ‐ 2015 Set‐3 3 2 7 ‐ ‐ ‐ ‐ ‐ ‐ 2016 Set‐1 1 2 5 4 5 14 3 3 9 2016 Set‐2 4 2 8 5 4 13 ‐ ‐ ‐ 2016 Set‐3 1 3 7 ‐ ‐ ‐ ‐ ‐ ‐ 2017 Set‐1 2 3 8 1 2 5 4 4 12 2017 Set‐1 0 4 8 2 2 6 ‐ ‐ ‐

2018 2 3 8 2 3 8 3 3 9

NETWORK THEORY

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

Topics Page No

1. NETWORK BASICS

1.1 Introduction 1 1.2 Classifications of Network Elements 2 1.3 Circuit Components 3 1.4 Kirchoff's Laws 6 1.5 Mesh & Nodal Analysis 7 1.6 Equivalent Circuits 9

Gate Questions

2. NETWORK THEOREMS

2.1 Introduction 49 2.2 Superposition Theorem 49 2.3 Thevenin's & Norton's Theorem 50 2.4 Maximum Power Transfer Theorem 51 2.5 Tellegen's Theorem 53 2.6 Reciprocity Theorem 53 2.7 Substitution Theorem 54 2.8 Millman's Theorem 54 2.9 Duality Principle 54

Gate Questions

3. TRANSIENTS

3.1 Introduction 89 3.2 Steady State & Transient Response 89 3.3 DC Transients 91

Gate Questions

4. AC ANALYSIS

4.1 Introduction 129 4.2 Sinusoidal Steady state analysis 129 4.3 Series Circuits 130 4.4 Parallel Circuits 132

Gate Questions

5. RESONANCE

5.1 Introduction 149

CONTENTS

13

58

99

136


5.2 Series Resonance 149 5.3 Parallel Resonance 152

Gate Questions

6. GRAPH THEORY

6.1 Graph of a Network 157 6.2 Incidence Matrix 159 6.3 Tie‐Set Matrix 159 6.4 F‐Cut Set Matrix 163

Gate Questions

7. COUPLED CIRCUITS

7.1 Introduction 174 7.2 Mutual Inductance 174 7.3 The Coupling Coefficient 176 7.4 Series Connection of Coupled Inductors 176 7.5 Parallel Connection of Coupled Coils 177 7.6 Ideal Transformer 177

8. TWO PORT NETWORKS

8.1 Introduction 179 8.2 Open Circuit Impedance (Z) Parameters 179 8.3 Short Circuit Admittance (Y) Parameters 180 8.4 Transmission (A B C D) Parameters 180 8.5 Inverse Transmission (A' B' C' D') Parameters 181 8.6 Hybrid (H) Parameters 181 8.7 Inverse Hybrid (G) Parameters 182 8.8 Conditions for Reciprocity and Symmetry 182 8.9 Inter Relationships of Different Parameters 182 8.10 Interconnection of Two‐Port Network 184

Gate Questions

9. NETWORK SYNTHESIS

9.1 LC, RC, RL Impedance & Admittance Functions 211 Gate Questions

10. ASSIGNMENT 217

155

164

189

212


1.1 INTRODUCTION

In terms of the atomic theory concept, an electric current in an element is the time rate of flow of free electrons in the element. The material may be classified as • Conductors, where availability of free

electrons is very large, as in the casemetals.

• Insulators, where the availability offree electrons is rare, as in case of glass,mica, plastics etc.

• Other materials, such as germaniumand silicon called semiconductors,may play a significant role inelectronics. Thermally generated electrons are available as free electronsat room temperature, and act asconductors, but at 0 Kelvin they act asinsulator.

1.1.1 CHARGE

According to basic physics, we know that there are two types of charges: Positive (corresponding to proton) and Negative (corresponding to electron). The fundamental unit of charge is the coulomb[c]. A single electron has a charge of 191.602 10 C−− × and a single proton has a charge of 191.602 10 C−+ × where one coulomb is defined as one ampere second. Charge in coulomb Q = It where, I is current in ampere and t is time in second.

1.1.2 CURRENT

The phenomenon of transferring charge from one point is a circuit to another is termed as “electric current”. The eclectic current, denoted by either ‘I’ or ‘i’ .The unit of electric current is ‘ampere’ which is denoted by ‘A’. Electric current mathematically expressed as

QI = (ampere)t

Where, I is the current Q is the charge

T is the time The current through a circuit element is the

time derivative of the electric charge i.e.dqi =dt

(c/s) or (Ampere)

Where, dq is small change in charge. dt is small change in time.

1.1.3 VOLTAGE

All opposite charges possess a certain amount of potential energy because of the separation between them. The difference is potential energy of the charges is called the potential difference. The potential difference in electrical terminology is known as voltage, and is denoted either by ‘V’ or ‘v’. Voltage is expressed in terms of energy (W) per unit charge (Q) i.e.

W dwV = Or ν = Q dq

(J/c) or (volt) Where, dw is the small change in energy dq is the small change in charge. One Volt, is the potential difference between two points when one joule of energy is used to pass one coulomb of charge from one point to the other

1.1.4 ENERGY

Energy is capacity for doing work. Energy may exist in many forms such as mechanical, chemical; electrical is called ‘Joule’. Energy is denoted by ‘W’. The energy delivered to a circuit element over the time interval (to, t) is given by

( )0

t

t

W= p x dx∫

1 NETWORK BASICS


1.1.5 POWER

Power is the rate of change of energy. It is denoted by ‘P’ or ‘p’, unit of power is ‘Watts’.

( ) Energy WPower p = =time t

or dwP = dt

Where, dw is the change in energy dt is the change in time

we can also write, dwP = dt

dw dq= × dt dt

P = v × iwattsSo, the instantaneous power p(t) delivered to a circuit element is the product of the instantaneous value of voltage v(t) and current i(t) of the element

( )P = v t × i(t)

1.2 CLASSIFICATION OF NETWORK ELEMENTS

1) ACTIVE AND PASSIVEActive network elements are thosewhich are capable of delivering powerto some external device. Specifically anactive element (energy sources likevoltage and current sources) is capableof delivering an average power greaterthan zero to some external device overan infinite time interval. For example,ideal sources are active elements.Passive network elements are thosewhich are capable only of receivingpower. A passive element is defined asone that cannot supply average powerthat is greater than zero over an infinite

time interval. For example, passive elements like inductors and capacitor capable of storing a finite amount of energy. Resistor is also a passive element.

2) BILATERAL AND UNILATERALIn Bilateral elements, the voltage–current relation is same for the currentflowing in either direction. For theentire time‘t’ element offers the sameimpedance for the different directionsof the same current flow and hence theresistor is said to bilateral.

A unilateral element has different relations between voltage and current for the two possible directions of current. Vacuum diodes, silicon diodes and metal rectifiers are examples of unilateral elements.

3) LINEAR AND NONLINEAR ELEMENTSAn element is said to be linear (device islinear if it is characterized by anequation of the form y=mx, where m isconstant) if its voltage-currentcharacteristics is at all times a straightline through the origin. For example,the current passive through a resistor isproportional to the voltage appliedthrough it, and relation is expressed asV IORV IR∝ =A linear element or network is onewhich satisfies the principle ofsuperposition, i.e. the principle ofhomogeneity and additivity. An elementwhich does not satisfy the aboveprinciple is called a non linear element.

4) LUMPED AND DISTRIBUTEDLumped elements are those elementswhich are very small in size and inwhich simultaneous actions takes place


for any given cause at the same instant of time. For example, capacitors, resistors, inductors and transformers are lumped elements. Distributed elements are those which are not electrically separable fro analytical purpose. For example, a transmission line which has distributed resistance, inductance and capacitance along its length may extend for hundreds of miles.

1.3 CIRCUIT COMPONENTS

1. THE RESISTANCE (R)The property of a material to restrictthe flow electron is called resistance.

Note: Current in a Resistor always enters from the positive terminal.

The basic ohm’s law in the electromagnetictheory form can written as J E= σ

r r

Where,

J =The current density, 2i AJS m =

E = Electric field along the length of

conductor, v VEml

=

σ = Conductivity of the conductor. i v v is s

σσl

l ⇒ = ⇒ =

v R i⇒ = → Ohm’s law in circuit theoryform.

Limitation: The Ohm’s law i.e. a linear voltage and current relation is valid only when the proportionality constant R is kept constant i.e. temperature is kept constant. v Ri→ =

v Ri→ =

dqv Rdt

∴ =

ViR

⇒ =

i Gv⇒ =• Electric power ,

p = vi2Ri.i i R(watts)= =

(v iR)=

Also, vp v.R

=

ViR

Q =

2VR

= (Watts)

• Electrical energy ,W= pdt∫

2W= i Rdt⇒ ∫ (J)2( p=i R)Q

2VW= dtR

⇒ ∫ (J)

2VpR

Q

=

The V-I characteristic of resistance:

Observation: From the characteristics, it can be observed that the resistor is a linear, passive, bilateral and time invariant in v-i plane. • Resistor is linear because the v-i

characteristic is a straight line passingthrough origin.

• Resistor is passive because the slope ofv-i characteristics is positive.

• Resistor is bilateral because even if wechange the direction of flow of currentthe R value is same.


• For all time ‘t’ the v-i characteristics hasone slope which implies that its value issame and hence it is said to be timeinvariant, which is possible only atconstant temperature

1) THE INDUCTANCE (L)A wire of certain length, when twistedin the form of a coil becomes a basicinductor.By Faraday’s law of electromagneticinduction, a time varying currentthrough coil produces a time varyingmagnetic flux, which induces voltageacross the coil.

Note: -In the presence of source V the current in an inductor enters from positive terminal.

The current voltage relation is given by div Ldt

=

t1i VdtL −∞

⇒ = ∫

Electric power, p V i= diLi (watts)dt

=

Electric energy, w pdt= ∫diw Li dt (J)dt

= ∫

Also, 2di d 1Li Lidt dt 2

=

2d 1w Li dtdt 2 ∴ = ∫

( )2w J1 Li2

=

Conclusions: • If I = constant then

( )dV L cons tan t 0dt

= =

i.e, inductor acts as short circuit to dc.• An ideal inductor never dissipates

energy, only story it is the formmagnetic field.

Note: Inductor is also a linear, passive, bilateral and time invariant.

2) THE CAPACITANCE (C)Any two conducting surfaces separatedby an insulting medium exhibit theproperty of a capacitor. A capacitorstores energy in the form of on electricfield.

q v∝ q=cv

The current-voltage relation is given bydvi cdt

=

t1v i dtc −∞

⇒ = ∫

Electric power, p V i= dvcv (watts)dt

=

Electric energy, = ∫w pdt

dvcv dtdt

= ∫

Again, 2dv d 1v ( cv )dt dt 2

=


2d 1w cv dtdt 2 ∴ = ∫

21w cv2

(J)=

Conclusions:

• If V = constant then i = ( )dc constant =0dt

i.e., capacitor acts as open circuit to dc. • An ideal capacitor doesn’t dissipate

energy but stress is in the form of electric field.

Note: Capacitor is also a linear, passive, bilateral and time invariant. The relation between V and i in L and C elements

L: diV Ldt

=

If 1 1 2 2i V ,i V→ → Then

1 2 xi i V (let)+ →

11

diV Ldt

=

22

diV Ldt

=

1 2i ix

d dV L L

dt dt= +

x 1 2V V V= + Since, linearity property is satisfied for inductor hence the v-i relation is linear for inductor.

C: dvi cdt

=

Following the similar steps it can be proved that the v-i relationship is linear for capacitor also. 3) THE SOURCES

a) Independent voltage source: Ideal: An ideal voltage source is a two –terminal element which supplies a constant voltage to a load and is independent of the load connected to it.

VL = VS for all value of is The load can be a resistor of any value. The V-I characteristics of voltage source is nonlinear hence it is a non-linear element. Also the ideal voltage source is active and unilateral. In the above circuits we have connected 2Ω and 5Ω resistors to a 10v source and we observe that in both the cases 10v would be dropped across them, irrespective of the current. In the first circuit 5A current flows through 2Ω load whereas in second circuit 2A current flows through 5Ω load.

Practical: A practical voltage source has a resistance in series with ideal voltage source. In a practical voltage source the load voltage changes with a change in load resistance. Ideal voltage source doesn’t exist in reality. Every source has internal resistance.


By KVL⇒ L S S SV V i R= −Here, load voltage depends upon load current Si .

b) Independent current source:Ideal:-An ideal current source is a twoterminal element which supplies aconstant current to a load resistor ofany value.Again ideal current source doesn’t existis reality as every current source has aninternal impedance.

Practical: A practical current source has a resistance in parallel to the ideal current source. Here the load current depends on the value of load connected across it.

KCL⇒ LL S

S

Vi IR

= −

Here load current depends upon load voltage.

Observation: When L L SV 0theni i= =

LV 0 short circuit= ⇒ So, we can conclude that the current always chooses a minimum resistance path.

c) The dependent (or) controlledsources:

• Dependent sources are the idealsources in which the value of source isdetermined by a voltage or currentexisting at some other location in thenetwork.

• Dependent sources are said to be linear,active and a bilateral with respect tocontrolled variable only. The presenceof these elements makes the networklinear, active and bilateral.

• Dependent sources are said to besources that is an active element in thepresence of at least one independentsource then only the controlled variablewill be non-zero and the magnitude ofsource will be non-zero.

1.4 KIRCHOFF’S LAWS

1) KCLIt is defined at a node


The simple node is an interconnection of only two branches, whereas principle node is an interconnection of at least 3 branches. Def: In an electric circuit, for any of its nodes at any time‘t’ the algebraic sum of branch currents leaving the node is zero. B y KCL ⇒

leaving curent 0=∑

1 2 3 4 5i i i i i 0− − + + + =

1 2 3 4 5i i i i i⇒ + = + +i.e. sum of entering current=sum of leaving currents

Also, dqidt

= ⇒

3 51 2 4dq dqdq dq dqdt dt dt dt dt

+ = + +

1 2 3 4 5q q q q q+ = + +Conservation of charges

2) KVLDef: It is defined in a loop or mesh i.e. ina closed path. In an electric circuit forany of its loop at any time ‘t’ thealgebraic sum of branch voltagesaround the loop is zero

KVL ⇒ S 1 2 3V V V V 0− − − =

S 1 2 3V V V V= + +

Since W dqV & iq dt

= = hence charges

and current in series remain same. 31 2 RR Rs WW WW

q q q q⇒ = + +

1 2 3s R R RW W W W⇒ = + +

So the KVL expresses conservation of energy.

1.5 MESH AND NODAL ANALYSIS Mesh and nodal analysis are two basic important techniques used in finding solutions for a network. Mesh or nodal analysis to a particular problem depends mainly on the number of voltage sources or current sources.

1) Nodal analysisNodal analysis is used to find nodevoltage is a circuit. Nodal analysis is acombination of KCL and ohm’s law i.e.Nodal analysis = KCL + ohm’s lawWhen we talk about the voltage at acertain point of a circuit we imply thatmeasurement is performed betweenthat point and some other point in thecircuit. In most cases that other point isreferred to as ground.Steps:i) Identify the number of principle

nodes (having more than twobranches).

ii) Assign the node voltages w. r. t. theground node, whose voltage isalways equal to zero.

iii) By using KCL at every principlenode write the equations.

Example: As shown in fig. node 1, 2, 3 are principle node because more than two branches are connected to them. And node 3 is assumed as the reference node.


Considering node 1:

By KCL at node 1: 1 2 3I I I 0− + + = By KCL

1 1 21

1 2

V 0 V VI 0R R− −

− + + = 'using ohm slaw --

---- (1) Where, 1 2V & V are voltages at node 1 & 2respectively.

Considering node 2

By KCL at node 2 : 4 5 6I I I 0By KCL+ + =

2 1 2 2

2 3 4 5

V V V 0 V 0 0R R R R− − −

+ + =+

' u sin g ohm slaw∴ ……… (2) By solving (1) and (2) we can find voltages at each node.

2) Mesh analysisMesh analysis is used to find branchcurrent in a circuit. Mesh analysis iscombination of KVL and ohm’s law.Mesh analysis = KVL + ohm’s Law

Mesh analysis is also known as Loop current method. It’s quite similar to the nodal analysis, except KCL we use KVL.

Steps: i) Identify the number of meshes.ii) Assign the mesh currents in the

clockwise direction.iii) By using KVL write the equations.

Example:

As shown in circuit, I1 and I2 are current in loop (1) and loop (2 ).

Considering Loop 1 By KVL

( )1 1 1 1 2 2V I R I I R 0− − − = (1)

Considering Loop 2 By KVL

( )2 2 1 2 2 3V I I R I R 0− − − − = (2)

By solving neq (1) & (2) we can find the currents 1I & 2I .

3) Super node AnalysisWhen there is an ideal voltage sourcebetween two principle nodes, it isdifficult to apply the technique of nodalanalysis because the current across anideal voltage source is unknown.Therefore to overcome such situationwe use super node analysis technique.

Example:


The current through an ideal voltage source can be any value; it is not possible to write the nodal equation independently; hence the super node procedure is followed here. To apply super node technique write nodal equations at node 1 2V &V simultaneously, while doing so do not consider the branch containing ideal voltage source. Super node

n1 2eq atV & V is given by

1 2V V1 1 01 1

− + + − =

1 2V V 2+ = (1)Inside the super node always KVL is written By KVL

1 2V 1 V 0− − =

1 2V V 1− = (2)

By solving neq (1) & (2) 1V 1.5V= and

21V V2

=

4) Super mesh AnalysisWhen there is an ideal current sourcebetween two loops, it is difficult toapply mesh analysis because the voltageacross an ideal current source isunknown. Therefore to overcome suchsituation we use super mesh analysistechnique.

Example:

The voltage across an ideal current source can be any value, it is not possible to write the mesh equations for the meshes (1) & (3) independently hence the super mesh procedure is followed. To apply super mesh technique, write a combined KVL neq in loop (1) & (2) and do not consider the branch containing ideal current source.

Super mesh neq in Loop (1) & (2) is given by

1 1 2 3V I R I R 0− − = ---------- (1)Inside the super mesh KCL is written.

1 2I I I= − ------------ (2)

By solving neq (1) & (2) we can find value of 1 2I & I .

1.6 EQUIVALENT CIRCUITS

Two elements are said to be in series only when the currents through the elements are the same and they are said to be in parallel only when the voltage across the elements are same. Impedances in series and admittances in parallel can be added.

1) Impedances in series

Zeq = Z1 + Z2

R

L

C

Z RZ j L

1Zj C

Ωω Ω

Ωω

==

=

R: eq 1 2R R R= +

L: eq 1 2L L L= +

C: eq 1 2

1 1 1j c j c j c

= +ω ω ω

eq 1 2

1 1 1c c c

⇒ = +

If 1 2c c c= =

Then eqcc2

=

The voltage division principle


eqV Z I=eq 1 2

V VIZ Z Z

⇒ = =+

1 1 11 2

V V Z I ZZ Z

→ = = +

2 2 21 2

V V Z I ZZ Z

→ = = +

2) Impedance in parallel

eq 1 2Y Y Y= +

eq 1 2

1 1 1Z Z Z

= +

R

L

C

Z RZ jwL

1ZjwC

ΩΩ

Ω

==

=

R: eq 1 2

1 1 1R R R

= +

L: eq 1 2

1 1 1= +L L L

C: eq 1 2C C C= + Current Division Principle

We have, eq 1 2

1 1 1Z Z Z

= +

1 2eq

1 2

Z ZZZ Z

∴ =+

eqV Z I→ = (By ohm’s law)

1 2

1 2

Z Z IZ Z

= +

21

1 1 2

ZVI IZ Z Z

→ = = +

22

2 1 2

ZVI IZ Z Z

→ = = +

3) The Star Delta Transformation

4) to Y∆ Conversion:If n / w∆ − is given then 1 2 3Z Z and Z are known

1 3A

1 2 3

Z ZZZ Z Z

=+ +

,

1 2B

1 2 3

Z ZZZ Z Z

=+ +

,

2 3C

1 2 3

Z ZZZ Z Z

=+ +

If 1 2 3Z Z Z Z= = =

Then A B CZZ Z Z3

= = =

Y to ∆Conversion: If Y n / w− is given then A B CZ , Z and Z are known.

A B1 A B

C

Z , ZZ Z ZZ

= + + ,

B C2 B C

A

Z , ZZ Z ZZ

= + + ,

A C3 A C

B

Z , ZZ Z ZZ

= + +

If A B CZ Z Z Z= = =

Then 1 2 3Z Z Z 3Z= = =


5) The source TransformationAny practical voltage source can beconverted into its equivalent practicalcurrent source and vice-versa by usingsource transformation.It is a simplification technique whicheliminates the extra nodes present inthe network. Source transformation isnot applicable to the ideal sources.

The source transformation is applicable for the dependent sources also, provided the control variable is outside the branches, where the source transformation is applied.

Here I1 is outside the branches having

current source of 1I A2

and 20Ω

resister, hence source transformation is applicable.

Example: Determine Vx using voltage division rule.

Solution: 10Ω resistor connected in parallel with 10Ω resistor So, Req 10 ||10=

10 10Req 510 10

×= = Ω

+

So, we can redraw the circuit as follows- By voltage division rule,

x10 2V

2 3 5×

=+ +

x20V10

=

xV 2V=

Example: Count the number of branches and nodes in the circuit. If xi 3A= & the 18V sourcedelivers 8𝐴𝐴 of current. What is the value of RA?

Solution: Number of branches = Number of nodes = 3 By KCL 1i 3 13 0+ − =Again by KCL

28 10 C ' 0− − + =

2C ' 18A=Voltage across 6Ω

6V 18VΩ =Voltage across AR

RAV 18V=

Parallel branches have same voltage∴ Byohm’s law

RA 2 AV i R=

A18 18R=

AR 1= Ω


Example: Determine current through the 3Ω resistor in the circuit?

Solution: Total current following through 3Ω & 6Ω combination is

( ) 12sin sin t 12sin sin ti t 2sin sin t4 3 || 6 4 2

= =+ +

( )i t 2sin sin t A= By current division rule,

( )36i t 2sin sin t

6 3= ×

+

( )34i t sin sin t A3

=

Example: Compute the voltage across each current source.

Solution: By source transformation

Nodal at 2V ⇒

2 32 1 2 V VV V V 01 3 4

−−+ + =

2 1 2 2 312V 12V 4V 3V 3V 012

− + + −=

1 2 312V 19V 3V 0− + − = ______ (1)Nodal at 1V ⇒

1 31 2 V VV V3 01 2

−−+ + =

1 2 1 32V 2V V V 6− + − = −

1 2 33V 2V V 6− − = − ________ (2)Nodal at 3V ⇒

3 1 3 2 3V V V V V 35 02 4 5− − −

+ + =

3 1 3 2 310V 10V 5V 5V 4V 140 020

− + − + −=

1 2 310V 5V 19V 140− − + = _______(3)3 2 112 19 310 5 19

− −∆ = − −

− −

1

6 2 10 19 3140 5 19

− − −∆ = −

− −

2

3 6 112 0 310 140 19

− −= − −−

∆

3

3 2 612 19 010 5 140

− −= −− −

∆

14241V 5.231 272=

∆= =∆

1V 5.23v=

33

3120V 11.47272

= = =∆∆

3V 11.47v=


Q.1 The voltage oe in the figure is

a) 2V b) 4 V3

c )4V d )8V [GATE-2001]

Q.2 If each branch of a Delta circuit has

impedance 3Z , the each branch ofthe equivalent Wye circuit has impedance.

a )Z3

b)3Z

c) 3 3Z d ) Z3

[GATE-2001]

Q.3 The voltage oe in the figure is

a)48 V b)24 V c)36 v d) 28 V

[GATE-2001]

Q.4 The dependent current source shown in the figure

a)delivers 80W b)absorbs 80W c)delivers 40 W d)absorbs 40W

[GATE-2002]

Q.5 If the 3-phase balanced source in the figure delivers 1500 W at a leading

power factor 0.844, then the values of 𝑍𝑍𝐿𝐿 (in ohm is

a) 90 32.44∠ ° b) 80 32.44∠ °c) 80 32.44∠− ° d) 90 32.44∠− °

[GATE-2001]

Q.6 The minimum number of equations required to analyze the circuit shown in the figure is

a)3 b)4 c)6 d)7

[GATE-2003]

Q.7 Twelve 1Ω resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is a) 5Ω

6b) 1Ω

c) 6Ω5

d) 3Ω2

[GATE-2003]

Q.8 The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is

a) 1 2L L M+ + b) 1 2L L M+ −

c) 1 2L L 2M+ + d) 1 2L L 2M+ − [GATE-2004]

GATE QUESTIONS(EC)


Q.9 The transfer function ( ) 0

i

V (S)H sV (S)

=

of an R-L-C circuit is given by

( )6

2 610H s .

S 20S 10+ + The Quality

factor (Q-factor of this circuit is a) 25 b) 50c) 100 d) 5000

[GATE-2004]

Q.10 For the circuit shown in the figure, the initial conditions are zero. Its

transfer function ( ) C

i

V (S)H sV (S)

= is

a) 2 6 6

1S 10 S 10+ +

b)6

2 3 610

S 10 S 10+ +

c)3

2 3 610

S 10 S 10+ +d)

6

2 6 610

S 10 S 10+ + [GATE-2004]

Q.11 Impedance Z as shown in the given figure is

a) j29Ω b) j9Ωc) j19Ω d) j39Ω

[GATE-2005]

Q.12 If 1 2 4R R R R= = = and 3R 1.1R=in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is

a)0.238 V b)0.138 V c)-0.238 V d)1 V

[GATE-2005]

Q.13 In the interconnection of ideal source shown in the figure, it is known that the 60V source is absorbing power.

Which of the following can be the value of the current source I? a)10A b)13Ac)15A d)18A

[GATE-2009]

Q.14 In the circuit shown, the power supplied by the voltage source is

a) W0 b) 5Wc)10W d)100W

[GATE-2010]

Q.15 In the circuit shown below, the current I is equal to

a)1.4 0°A∠ b) 2.0 0°A∠c) 2.8 0°A∠ d) 3.2 0°A∠

[GATE-2011]


Q.16 If A BV V 6V,− = then C DV V− is

a) -5V b) 2Vc) 3V d) 6V

[GATE-2012]

Q.17 The average power delivered to an impedance ( )4 j3− Ω by a current5cos(100πt 100)+ A is a) 44.2W b) 50Wc) 62.5W d)125

[GATE-2012]

Q.18 In the circuit shown below, the current through the inductor is

a) 2 A1 j+

b) 1 A1 j−+

c) 1 A1 j+

d) 0A

[GATE-2012]

Common Data Questions 19 & 20 Consider the following figure

Q.19 The current IS in Amps in the voltage source, and voltage Vs is Volts across the current source respectively, are a)13,-20 b)8,-10 c)-8,20 d)-13,20

[GATE-2013]

Q.20 The current in the 1Ω resistor in Amps is a)2 b)3.33 c)10 d)12

[GATE-2013]

Q.21 Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection are scaled by a factor k,k 0> , the elements of the corresponding star equivalent will be scaled by a factor of

a) 2k b) kc)1/k d) k

[GATE-2013]

Q.22 The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage

WX1V =100V is applied across WX to get an open circuit voltage YZ1Vacross YZ .Next, an ac voltage

YZ2V =100V is applied across YZ to get an open circuit voltage WX2Vacross WX. Then YZ1 WX1V / V ,

WX2 YZ2V / V are respectively.

a)125/100and80/100b)100/100and80/100c)100/100and100/100d) 80/100and80/100

[GATE-2013]


Q.23 Three capacitors 1 2C ,C and 3Cwhose values are 10μF,5μF,and 2μFrespectively, have breakdown voltages of 10V, 5V and 2Vrespectively. For the interconnection shown below, the maximum safe voltage in volts that can be applied across the combination , and the corresponding total charge in μC stored in 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

[GATE-2013]

Q.24 Consider the configuration shown in the figure which is a portion of a larger electrical network

For R =1Ω , and currents i1= 2A, i4 = -1A, i5 = -4A, which one of the following is TRUE? a) 6i 5A=b) 3i 4A= −c) Data is sufficient to conclude

that the supposed currents areimpossible

d) Data is insufficient to identify thecurrent i2, i3, and i6

[GATE-2014]

Q.25 A Y-network has resistances of 10Ω each in two of its arms, while the

third arm has a resistance of 11Ω in the equivalent — network, the lowest value (in Ω among the three resistances) is _________.

[GATE-2014]

Q.26 Consider the building block called 'Network N' shown in the figure. Let C =100µF and R = 10kΩ

Two such blocks are connected in cascade, as shown in the figure.

The transfer function 3

1

V (s)V (s)

of the

cascaded network is

a) s1 s+

b)2

2

s1 3s s+ +

c) 2

1s

s +

d) s2 s+

[GATE-2014]

Q.27 In the circuit shown in the figure, the value of node voltage V2 is

a) 22 + j 2 V b) 2 + j 22 Vc) 22 ‒ j 2 V d) 2‒ j 22 V

[GATE-2014]


Q.28 For the Y-network shown in the figure, the value of R1 (in Ω) in the equivalent ∆ -network is___________.

[GATE-2014]

Q.29 The circuit shown in the figure represents

a)Voltage controlled voltage sourceb)Voltage controlled current source c)Current controlled current source d)Current controlled voltage source

[GATE-2014]

Q.30 The magnitude of current (in mA) through the resistor R2 in the figure shown is _______.

[GATE-2014]

Q.31 The equivalent resistance in the infinite ladder network shown in the figure is Re.

The value of Re/R is ________. [GATE-2014]

Q.32 In the given circuit, the values of 1V and 2V respectively are

a) 5V, 25V b) 10V, 30V c)15V, 35V d) OV, 20V

[GATE-2015]

Q.33 In the circuit shown, the voltage Vx (in Volts) is _________.

[GATE-2015]

Q.34 An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that 1R = 3kΩ,

2R =6kΩ and 3R = 9kΩ, and that the diode is ideal.

RMS current rmsI (in mA through the diode) is _______.

[GATE-2016]

Q.35 In the given circuit, each resistor has a value equal to 1Ω .


What is the equivalent resistance across the terminals a and b? a)1/6 Ω b) 1/3 Ωc) 9/20Ω d) 8/15 Ω

[GATE-2016]

Q.36 In the circuit shown in the figure, the magnitude of the current (in amperes through R2) is _________.

[GATE-2016]

Q.37 In the figure shown, the current i (in ampere) is ________.

[GATE-2016]

Q.38 A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10Ω, 5Ω, 2Ω are provided. Consider all possible permutation of the given resistors into the positions A,B,C and identify the configuration with maximum possible overall resistances. The ratio of maximum to minimum values of the resistances (up to second decimal place) is___________ .

[GATE-2017]

Q.39 Consider the network shown below with 1R 1= Ω , 2R 2= Ω and 3R 3= Ω . The network is connected to constant voltage source of 11V.

The magnitude of the current (in amperes, accurate to two decimal places) through the source is _______.

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (a) (d) (a) (d) (a) (a) (d) (b) (d) (b) (c) (a) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (b) (a) (b) (c) (d) (c) (b) (b) (d) (a) 29.09 (b) (d) 10 29 30 31 32 33 34 35 36 37 38 39 (c) 2.8 2.618 (a) 8 0.68 (d) 5 -1 2.143 8

ANSWER KEY:


Q.1 (c) Applying KCL

o o oe 12 e e 04 4 4−

+ + = o3e 12⇒ =

o e 4V∴ =

Q.2 (a)

YZ 3Z=V Y 3Z 3Z∆⇒ =

YZZ

3∆=

Q.3 (d) Applying source conversion

o o oe 80 e e 16 012 12 6− −

+ + =

o4e 112=

o112e 28V

4= =

Q.4 (a)

Applying KVL, 1V20 5I 5 I 05

− − + =

20 10I 20 0− − =I 0⇒ =

∴Only dependent source acts. 1V 4A

5=

Power delivered 2I R 16 5 80W= = × =

Q.5 (d) p p3V l cosθ 1500=

L L

L

V V3 cosθ 15003 3Z

=

2L

LV .cosθZ

1500=

2400 0.8441500×

= 90Ω=

1θ cos (0.844)−= 32.44= As power factor is leading, load is capacitive so angle will be negative. θ 32.44°= −

Q.6 (a) As voltage at 1 node is known ∴ using nodal analysis only 3 equations required.

Q.7 (a)

abi i iV 1 1 13 6 3

= × + × + ×

abeq

V 5R Ωi 6

⇒ = =

Q.8 (d)

If current enters the dotted terminals of coil 1 then a voltage is developed across coil 2 whose higher potential is at dotted terminals.

12

L dIMdI MdI dIV Ldt dt dt dt

−= + − +

EXPLANATIONS


( )1 2dIL L 2Mdt

= + −

eqdIV Ldt

=

Q.9 (b) Characteristic equation

2 6S 20S 10= + + 6o

oωQ ,ω 10

BW= =

310 1000Q 5020 20

= = =

Q.10 (d)

( ) 2

11sCH s 1 s LC sCR 1R sL

sC

= =+ ++ +

( )( )

2 3 6

3 6

1s 10 10 100 10

s 10 10 100 10 1

− −

−

=× × ×

+ × × × +

( )6

6 2 2 6 61 10H s

10 S s 1 S 10 S 10−= =+ + + +

Q.11 (b) 1 2 3 m mX X X X 2X 2X= + + + −

5j 2 j 2 j 20j 20j= + + + − = 9j (one additive & other subtractive)

Q.12 (c) ( )a 1 2V 5 R R= =

3b

3 4

R 1.1V 10 10R R 2.1

= × = ×+

a bV V V= − V 0.238V= −

Q.13 (a)

In the given circuit, the current through the branch of 60 V source is (12-I) as shown in Fig. The source of 60V absorbs power, only if P=(12-I) 60 is +ve. i.e., I<12. The value of the current source, I can only be 10 A given in option (a), as the currents given in other options are more than 12 A.

Q.14 (a)

The current through all the branches are marked as shown in Fig 1. Apply KVL to outer loop ( ) ( )1 12 I 3 2 I 2 10+ + + =

14I 10 10+ =

1I 0= ∴ Power supplied by 10V 10 0 0W= × =

Q.15 (b) SZ 7 0°Ω(using Y Δtransformation)= ∠ −

14 0°I 2 0°A7 0°∠

= = ∠∠

Q.16 (a)

From the given circuit, A BV V 6V− =

A BAB DC

V VI 3A I2−

= = =


KCL at ‘D’ gives C D

C DV V 2 3 0, V V 5V

1−

+ + = − = −

Q.17 (b) The load consists of a resistance and a capacitance of this, only R is passive and consumes power So 2

rmP Ri=25 4 50W

2 = × =

note rms value of A cos ωt2

A =

Q.18 (c) Assume current as shown,

By applying current division rule in upper part of the circuit

L1 1I 1 0

1 j 1 j= × ∠ =

+ +

Q.19 (d)

Across AB voltage drop is 10 V S SI 13A,V= −

S10 1020V2 I 02 1

= = + + =

SI 5 10 2 13= − − + = −

Q.20 (c)

10current through 1Ω 10A1

⇒ = =

Q.21 (b) Consider a b cR x;R k;R k= = =

2a b

Stara b c

R R kRR R R 3k

+= =

+ +

StarR k∝

Q.22 (b)

W 1 2 W 1V 100 V turns ratio V 125× ×= ⇒ = × =

YZ1 2V 0.8 V 100V= × = When YZ2V 100V=

Thevenin’s circuit seen by 2-2’ will be as follows

thV 100V= An thR 0.2 || 0.8= negligible

'2 2V 100V−∴ =

W 2V ×∴

'2 2100V 1V turns ratio

1.25−×

= × = 80V=

Q.23 (d)


2 3eq 1

2 3

C CC CC C

= ++

11.5μF= Safety voltage = 7V Q CV=

eq safetyQ C V⇒ = ×11.5 7 80μC× ;

Q.24 (a) Given i1 = 2A

i4 = –1A

i5 = – 4A KCL at node A, i1 + i4 = i2 ⇒ i2 = 2 –1 = 1A .1. KCL at node B, i2 +i5 =i3

⇒ i3 = 1– 4 = -3AKCL at node C, i3 + i6 =i1

⇒ i6 = 2 – (-3) = 5A

Q.25 (29.09Ω)

X = 29.09Ω

X= (10)(10) (10)(11) (10)(11)11

+ + Ω

Y= 32Ω

y = (10)(10) (10)(11) (10)(11)10

+ + Ω

Z= 32Ω

z = (10)(10) (10)(11) (10)(11)10

+ + Ω

i.e, lowest value among threeresistances is 29.09 Ω

Q.26 (b) Two blocks are connected in cascade, Represent in s-domain,

3

1

V (s) R.R1 1 1V (s) R R R Rsc SC SC

= + + + +

( ) [ ]R.R

1 1 R. 2R SC 1 1 RSCsc sc SC

+ + +

=( )

2 2

2 2 2

S C R.R1 2R SC RSC R S C+ + +

2 6 6 3 3

2 6 4 12 6 4

S .100 100 10 10 10 10 10 10S 100 10 10 10 3S 100 10 10 1

− −

− −

× × × × × × ×=

× × × × + + × × +2

32

1

V (s) SV (s) 1 3S S

=+ +

Q.27 (d)

KVL for V1 & V2

o1 2V V 10 | 0− = …(1)

0

1 2V -V =+10|0


KCL at super node:

-4 0 1 2 2V V V| 0 0 .(2)j3 6 j6

+ + + = …−

01 2 2V V V 4| 0j3 6 j6+ + =

−From (1) & (2),

o02 2 2V 10 | 0 V V 4| 0

j3 6 j6+

+ + =−

01 2 22

V V V 10V 4| 0j3 6 j6 j3

+ + = + − ( )2V 2 j22 Volts∴ = −

Q.28 (10Ω)

R1 = ( )( ) ( )( ) ( )7.5 5 3 5 7.5 (3)7.5

+ +Ω

R1=10 Ω

Q.29 (c)

The dependent source represents a current controlled current source

Q.30 (2.8) By source transformation

By KVL, 20 –10k.I + 8 = 0

28I10k

⇒ = I 2.8mA⇒ =

Q.31 (2.618)

→For an infinite ladder network, if all resistance are having same value of R then equivalent resistance is

1 5 .R2

+ →For the given network, we can split in to R is in series with Requivalent

equR R 1.618R 2.618equRR

⇒ = + ⇒ =

Q.32 (a) By nodal analysis -5 I I 2I 0+ + + = 4I 5=

I A54

=

1 = 4I =V 5volts ( )2 1V =4 5 +V 1 20 25 voltsV= + =

Q.33 (8)

Apply KCL at point P


20xV + 0.25

10x xV V− + 0.5 5xV =

x1 0.75V 0.5 520 10

+ + =

x x5V V V58

= ⇒

=

Q.34 (0.68 to 0.72)

Q.35 (d) Let assume all resistance as R, then by using start-delta transformation

2

ab4R 8R 32R 5 8RR5 5 25 12R 1

a5

s R 1= ×⇒ = = =P Ω

ab8R

15= Ω .

Q.36 (5) Let current through 1R = L

xx

VI 0.04V5

=⇒ +

x x xV V 4VI5 25 25

⇒ = − =

Applying KVL, x x4V V5 8

26

5 50 × + ×=

125

60 25⇒= =xx

V V

Thus current through L25R5

= =

5amps

Q.37 (-1) Nodal equation at V V-8 V V-8 V+ + + =01 1 1 1

4V 16⇒ =By using KCL at node 'a'.

1 11 i 8 4 - 1

0 i 5A+ = ⇒ =−

KCL at b

- 14 i i 0 4 5 i 01+ + = ⇒ − + + =

I 1A⇒ = −

Q.38 2.143 The connection of resistors is as shown below

Given resistors are: 10Ω , 5Ω and 2Ω The maximum resistance possible is

( ) ( )T maxR 10 5 2= Ω+ Ω Ω

10 80107 7

= + Ω = Ω


The minimum resistance possible is ( ) ( )T minR 2 10 5= Ω+ Ω Ω

10 1623 3

= + Ω = Ω

( )

( )

T max

T min

R 80 / 7R 16 / 3

=

15 2.1437

= =

Q.39 8 From the symmetry of given circuit, we can conclude that point B and C are equipotential points since the resistive distance of both the points from terminals A as well as from F are equal. Similar reasoning can be applied to identify that points D and E are also equipotential. We can remove the resistance connected between two equipotential points as current through it will always be zero.

Hence, we can re-draw the simplified circuit as shown in figure, which can be further simplified as shown below,

We can find current supplied by voltage source,

eq

VIR

=

Where

eq1 1 1 3R2 2 2 2

= + +

eq

1 32 2R 1 1 32 2

×= +

+

3 / 412

= +

eq11R8

= Ω

Here

eq

VIR

=

1111/ 8

=

I 8A=


Q.1 Given two coupled inductors 1L and

2L , their mutual inductance Msatisfies

a) 2 21 2M L L= + b) 1 2L L

M2+

>

c) 1 2M L L> d) 1 2M L L≤[GATE-2001]

Q.2 Two incandescent light bulbs of 40W and 60 W rating are connected in series across the mains. Then a) the bulbs together consume

100W. b) the bulbs together consume 50 W.c) the 60 W bulb glows brighter.d) the 40 W bulb glows brighter.

[GATE-2001]

Q.3 Consider the star network shown in figure. The resistance between terminal A and B with terminal C open is 6Ω , between terminals B and C with terminal A open is 11Ω , and between terminals C and A with terminal B open is 9Ω .Then

a) A B CR 4 ,R 2 ,R 5= Ω = Ω = Ω

b) A B CR 2 ,R 4 ,R 7= Ω = Ω Ωc) A B CR 3 ,R 3 ,R 4= Ω = Ω = Ωd) A B CR 5 ,R 1 ,R 10= Ω = Ω = Ω

[GATE-2001]

Q.4 In the circuit shown in figure it is found that the input ac voltage i(V )

and current i are in phase. The

coupling coefficient is 1 2

MKL L

=

where M is the mutual inductance between the two coils. The value of K and the do polarity of the coil P-Q are

a) K=0.25 and dot at Pb) K=0.5 and dot at Pc) K=0.25 and dot at Qd) K=0.5 and dot at Q

[GATE-2002]

Q.5 A segment of a circuit is shown in figure R CV =5V,V = 4sinsin2t the voltage LV is given by

a) 3 8cos 2t− b) 32sin 2tc)16sin 2t d)16cos 2t

[GATE-2003]

Q.6 Figure shows the waveform of the current passing through an inductor of resistance 1Ω and inductance 2H. The energy absorbed by the inductor in the first four seconds is

GATE QUESTIONS(EE)


a)144J b)98J c)96J d)168J

[GATE-2003]

Q.7 In figure, the potential difference between points P and Q is

a) 12 V b)10V c)-6V d)8V

[GATE-2003]

Q.8 In figure, the value of R is

a)10Ω b)18Ωc) 24Ω d)12Ω

[GATE-2003]

Q.9 In figure, the value of the source voltage is

a)12V b)24V c)30V d)44V

[GATE-2004]

Q.10 In figure, the admittance values of the elements in Siemens are

RY 0.5 j0,= + LY 0 j1.5,= −

CY 0 j0.3= + respectively. The valueof I as a phasor when the voltage E across the elements is 10∠0°V is

a) 1.5+j0.5 b) 5-jc) 0.5+j1.8 d) 5-j12

[GATE-2004]

Q.11 In figure the value of resistance R in Ω is

a)10 b)20 c)30 d)40

[GATE-2004]

Q.12 In figure a b cR ,R and R are 20Ω,10Ω and 10Ω respectively. The resistance 1 2R ,R and 3R in Ω of anequivalent star connection are

a) 2.5, 5, 5 b) 5, 2.5, 5c) 5, 5, 2.5 d) 2.5, 5, 2.5

[GATE-2004]

Q.13 The rms value of the current in a wire which carries a d.c. current of 10A and a sinusoidal alternating current of peak value 20A is a)10A b)14.14A c)15A d)17.32A

[GATE-2004]


Q.14 In the figure given below the value of R is

a) 2.5Ω b) 5.0Ωc) 7.5Ω d)10.0Ω

[GATE-2005]

Q.15 The RMS value of the voltage ( )u t 3 4cos(3t)= + is

a) 17V b) 5Vc) 7V d) ( )3+2 2 V

[GATE-2005]

Q.16 The RL circuit of the figure is fed from a constant magnitude, variable frequency sinusoidal voltage source 𝑉𝑉𝐼𝐼𝐼𝐼 .At 100Hz, the R and L elements each have a voltage drop RMSu .If the frequency of the source is changed to 50Hz, then new voltage drop across R is

a) RMS5u8

b) RMS2u3

c) RMS8u5

d) RMS3u2

[GATE-2005]

Q.17 For the three phase circuit shown in the figure the ratio of the currents

R Y BI :I :I is given by

a)1:1: 3 b)1:1:2c)1:1:0 d) 1:1: 3/2

[GATE-2005]

Q.18 The circuit shown in the figure is energized by a sinusoidal voltage source 1V at a frequency which causes resonance with a current of I.

The phasor diagram which is applicable to this circuit is a) b)

c) d)

[GATE-2006]

Q.19 An energy meter connected to an immersion heater (resistive) operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units (kWh) in 1 hour. The heater is removed from the supply and now connected to a 400 V peak to peak square wave 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

[GATE-2006]


Q.20 The state equation for the current 1Ishown in the network shown below in terms of the voltage xV and the independent source V, is given by

a) 1x 1

dI 51.4V 3.75I Vdt 4

= − − +

b) 1x 1

dI 51.4V 3.75I Vdt 4

= − − −

c) 1x 1

dI 51.4V 3.75I Vdt 4

= − + +

d) 1x 1

dI 51.4V 3.75I Vdt 4

= − + −

[GATE-2007]

Q.21 In the figure below all phasors are with reference to the potential at point “0”. The locus of voltage phasor YXV as R is varied from zero to infinity is shown by

a) b)

c) d)

[GATE-2007]

Q.22 A 3V dc supply with an internal resistance of 2Ω supplies a passive non-linear resistance characterized by the relation 2

NL NLV I .= Thepower dissipated in the non-linear resistance is a)1.0W b)1.5Wc) 2.5W d) 3.0W

[GATE-2007]

Q.23 The Thevenin’s equivalent of a circuit operating at ω=5rad/s has

ocV =3.71 -15.9°∠ and

oZ =2.38-j0.667Ω . At this frequency, the minimal realization of the Thevenin’s impedance will have a a) Resistor and a capacitor and an

inductorb) Resistor and a capacitorc) Resistor and an inductord) Capacitor and an inductor

[GATE-2008]

Q.24 Assuming ideal elements in the circuit shown below, the voltage abVwill be

a) -3V b) 0Vc) 3V d) 5V

[GATE-2008] Q.25 In the circuit shown in the figure,

the value of the current i will be given by

a) 0.31A b) 1.25 Ac) 1.75A d) 2.5A

[GATE-2008]


Q.26 The current through the 2kΩ resistance in the circuit shown is

a) 0mA b)1mAc) 2mA d) 6mA

[GATE-2009]

Q.27 How many 200W/220 V incandescent lamps connected in series would consume the same total power as a single 100 W/220V incandescent lamp? a) Not possible b) 4c) 3 d) 2

[GATE-2009]

Q.28 The equivalent capacitance of the input loop of the circuit shown is

a)2μF b)100μF c)200μF d) 4μFs

[GATE-2009]

Q.29 For the circuit shown, find out the current flowing through the 2Ω resistance. Also identity the changes to be made to double the current through the 2Ω resistance.

a) ( )s5A;putV 20V=

b) ( )s2A;putV 8V=

c) ( )s5A;putI 10A=

d) ( )s7A;putI 12A= [GATE-2009]

Q.30 As shown in the figure, a 1Ω resistance is connected across a source that has a load line v+i=100 . The current through the resistance is

a) 25A b) 50Ac)100A d) 200A

[GATE-2010]

Q.31 If the 12Ω resistor draws a current of 1A as shown in the figure, the value of resistance R is

a) 4Ω b) 6Ωc)8Ω d) 18Ω

[GATE-2010]


a) 2 A1 j+

b) 1 A1 j−+

c) 1 A1 j+

d) 0A

[GATE-2012]

Q.33 A two –phase load draws the following phases currents: ( ) ( )1 m 1i t I sin ωt ,= −∅ ( )2 mi t I=

These currents are balanced if 1∅is equal to a) 2−∅ b) 2∅


c) ( )2π / 2−∅ d) 2π2

∅ +

[GATE-2012]

Q.34 The average power delivered to an impedance ( )4-j3 Ω by a current 5cos(100πt+100) A is a) 44.2W b) 50Wc) 62.5W d)125

[GATE-2012]

Q.35 If A BV -V =6V, then C DV -V is

a) -5V b) 2Vc) 3V d) 6V

[GATE-2012]

Q.36 Three capacitors 1 2C , C and 3Cwhose values are 10μF,5μF, and 2μF respectively, have breakdown voltages of 10V, 5V and 2Vrespectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in μC stored in the effective capacitance across the terminals are respectively,


[GATE-2013]

Q.37 Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 μH and

240μH. Their mutual inductance in μH is ___.

[GATE-2014]

Q.38 The voltage across the capacitor, as sown in the figure, is expressed as

( ) ( ) ( )t 1 1 1 2 2 2v t A sin t A sin t= ω −θ + ω −θ

The value of 1A and 2A respectively, are a) 2.0 and 1.98 b) 2.0 and 4.20c) 2.5 and 3.50 d) 5.0 and 6.40

[GATE-2014]

Q.39 The total power dissipated in the circuit, show in the figure, is 1kW.

The voltmeter, across the load, reads 200 V. The value of XL is

[GATE-2014]

Q.40 The line A to neutral voltage is 10∠15°V for a balanced three phase star-connected load with phase sequence ABC. The voltage of line B with respect to line C is given by a)10 3 105 V∠ ° b)10 3 105 V∠ °c)10 3 75 V∠− ° d) 10 3 90 V− ∠− °

[GATE-2014]

Q.41 The power delivered by the current source, in the figure, is 1V

[GATE-2014]


Q.42 The voltages developed across the 3 Ω and 2 Ω resistors show in the figure are 6 V and 2 V respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5 V voltage source?

a) 5 b) 7c) 10 d) 14

[GATE-2015]

Q.43 In the given circuit, the parameter k is positive, and the power dissipated in the 2Ω resistor is 12.5 W. The value of k is ____.

[GATE-2015]

Q.44 The current I (in ampere) in the 2Ω resistor of the given network is _____.

[GATE-2015]

Q.45 In the given network V1 = 100∠0°V, V2 = 100∠120° V, V3 = 100∠+120° V. The phasor current i (in Ampere) is

a) 173.2∠-60° b) 173.2∠120°c) 100.0∠-60° d) 100.0∠120°

[GATE-2015]

Q.46 AR and BR are the input resistances of circuits as shown below. The circuits extend infinitely in the direction shown. Which one of the following statements is TRUE?

a) A BR R= b) A BR R 0= = c) A BR R< d) B A AR R / (1 R )= +

[GATE-2016]

Q.47 In the portion of a circuit shown, if the heat generated in 5Ω resistance is 10 calories per second then heat generated by the 4Ω resistance, the calories per second, is_____

[GATE-2016]

Q.48 In the given circuit, the current supplied by the battery, in ampere, is


[GATE-2016]

Q.49 In the circuit shown below, the node voltage VA is V.

[GATE-2016]

Q.50 The voltage (v) and current (A) across a load are as follows. v(t) =100 sinω(t), i(t) = 10sin(ωt - 60°) + 2sin(3ωt) + sin(5ωt) The average power consumed by the load, in W, is_______.

[GATE-2016]

Q.51 In the circuit shown below, the voltage and current are ideal. The voltage (Vout) across the current source, in volts, is

a) 0 b) 5c) 10 d) 20

[GATE-2016]

Q.52 The equivalent resistance between the terminals A and B is…………

[GATE-2017, Set-1]

Q.53 The power supplied by the 25V source in the figure shown below is…………….W.

[GATE-2017, Set-1]

Q.54 The equivalent impedance eqZ for the infinite ladder circuit shown in the figure is

a) j12Ω b) j12− Ω

c) j13Ω d) 13Ω

[GATE-2018]

Q.55 A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t = 0 as shown in the figure. Assume ( )i 0 0= , ( )v 0 0=. Which one of the following circular loci represents the plot of ( )i t

versus ( )v t ?

a)


b)

c)

d)

[GATE-2018]

Q.56 A three-phase load is connected to a three-phase balanced supply as shown in the figure. If

0anV 100 0 V= ∠ , 0

bnV 100 120 V= ∠−

and 0cnV 100 240 V= ∠− (angles are

considered positive in the anticlockwise direction), the value of R

for zero current in the neutral wire is _________ (up to 2 decimal places).

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (d) (d) (b) (c) (b) (c) (c) (d) (c) (d) (b) (a) (d) (c) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (a) (c) (a) (a) (b) (a) (a) (a) (b) (a) (b) (a) (d) (a) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (b) (b) (b) (c) (d) (b) (a) (d) 35 (a) 17.34 (c) 3 (a) 43 44 45 46 47 48 49 50 51 52 53 54 55 56 0.5 0 (a) (d) 2 0.5 11.42 250 (d) 3 250 (a) (b) 5.77

ANSWER KEY:


Q.1 (d) 1 2M K L L=

Where K=coefficient of coupling 0 K 1< <Q

1 2M L L∴ ≤

Q.2 (d) 1PR

∝Q

Therefore resistance of 40W bulb > resistance of 60 W bulb. For series connection, current through both the bulbs will same

2P l R= (for series connection). Power consumed by 40 W bulb > power consumed by 60W bulb. Hence, the 40W bulb glows brighter.

Q.3 (b) When C is open, AB A BR R R= +

6Ω= When B is open, AC A CR R R= +

9Ω= When A is open,

BC B CR R R 11Ω= + = On solving above equations

A B CR 2Ω,R 4Ω and R 7Ω= =

Q.4 (c) Input ac voltage and current will be in phase only at resonance condition i.e.

C LX X= j12−

( ) ( )j8 j8 2k j8 j8= + + ×

12 8 8 16k= + +4 1k 0.25

16 4⇒ = − = − = −

Hence coupling will be opposite

Q.5 (b) By KCL,

P Q C Ll l l l 0+ + + =

C L2 1 l l 0+ + + = But, Cl C dv / dt= ×

1 d / dt(4sin 2t)= × (8cos 2t)=

Ll (2 1 8cos 2t)∴ = − + +3 8cos 2t= − −

Q.6 (c) For 0 t 2< < s current varies linearly with time and given as ( )i t 3t=and for 2 s t 4s< < current is constant, ( )i t 6A.= The energyabsorbed by the inductor (Resistance neglected) in the first 2 sec,

1 2

T

L L L0

diE Li dt E Edt

= = +∫2

L10

diE Li dtdt

= ∫

2

0

2 3t 3dt= × ×∫2 2

0

2t18 t dt 1802

= = ×∫418 0 36J2 = × − =

The energy absorbed by the inductor in (2 4)→ second

2

4

L2

diE Li dtdt

= ∫

4

2

2.6.0dt 0J= =∫A pure inductor does not dissipate energy but only stores it. Due to resistance, some energy is dissipated in the resistor. Therefore, total energy absorbed by the inductor is the sum of energy stored in the inducer tor and the energy dissipated in the resistor. The

EXPLANATIONS


energy dissipated by the resistance in 4 sec.

T2

R0

E i Rt= ∫2 4

2 2

0 2

(3t) 1dt 6 1dt= × + ×∫ ∫2 4

2

0 2

(9t) dt 36 1dt= +∫ ∫ 3 2 4t9 36t

0 23= × +

( )89 36 23= × + × 24 72 96j= + =

LV L(di / dt) 2 2 8sin 2t∴ = = × ×32sin 2t=

Note: KCL is based on the law of conversation of charges. ∴Dot will at Q

Q.7 (c) Given: RV 10V=By KCL

P PV 10 V2 02 8−

+ + = …..(i)

Q QV 10 V2 0

4 6−

− + = …..(ii)

Q.8 (d) By KCL,

P P PV 40 V 100 V 01 14 2− −

∴ + + =

P22V 660=

PV 30V∴ = Potential difference between node x and y = 60 V by taking KCL at node y

40 30l 5 01−

− − + =

l 5A∴ =60R 12Ω5

∴ = =

From equation(i)∴ ( )P P4 V 10 2 8 V 0− + × + =

P P4V 40 16 V 0− + + =

P5V 24 0− =

PV 4.8=

( )Q Q6 V 10 2 4 6 4V− − × × + 0=

Q10V 108 0− =

QV 10.8∴ =

P QV V 6V∴ − = −

Q.9 (c)

Method –I Using KCL,

a aV E V 1 06 6−

+ − =

a2V E 6⇒ − = …….(i) Where

aE V 26−

=

aE V 12⇒ − = …….(ii) Solving eq (i) & (ii) We get

aV 18V and E 30V= = Current through the branch a-Q I = 1 + 2 = 3A

aV 6l 6 3 18V= = × =

aE V 26−

=E 18 2

6−

⇒ = E 30V⇒ =


Q.10 (d)

( )R RI Y E 0.5 j0 10 0° 5A= = + × ∠ =

( )Y LI Y E 0. j1.5 10 0° j15A= = + × ∠ = −

( )C CI Y E 0 j0.3 10 0° j3A= = + × ∠ =

R Y CI I I I= + +

( )5 j15A j3= + +5 j12A= −

Q.11 (b)

p PV 100 V 2 010 10−

+ + =

P2V 100 20 0− + =

P80V 40V2

∴ = =

PV 40R 20Ω2 2

∴ = = =

Q.12 (a)

Given: aR 20Ω=

bR 10Ω= And cR 10Ω=

b c1

a b c

R RRR R R

=+ +

10 10 2.5Ω20 10 10

×= =

+ +c a

2a b c

R RRR R R

=+ +

10 20 5Ω20 10 10

×= =

+ +

a b3

a b c

R RRR R R

=+ +

20 10 5Ω20 10 10

×= =

+ +Remember: If all the branches of ∆-connection has same impedance Z then impedance of branch of Y-

connection be Z/3

Q.13 (d) R.M.S value of d.c Current

dc10A l= = R.M.S value of sinusoidal Current

dc20 A l

2 = =

R:M:S value of resultant, 2 2

R dc acl l l= +

Q.14 (c) The Resultant (R) when viewed from voltage

Source 100 12.58

= =

R 10 ||10 12.5+ = R 12.5 10 ||10∴ = − 12.5 5 7.5Ω= − =

Q.15 (a) R.M.S value of d.c Voltage

(rms)dcV 3V= =

R.M.S value of a.c. Voltage (rms)a.cV=

( )4 / 2 V=

R.M.S value of the voltage

( )223 4 / 2= +


9 8 17V+ =

Q.16 (c) At f 100Hz=

R LV V= as R & L are series connected, current through R & L is same, so

LIR lX lωL= =

LR X ωL⇒ = =

in2 2

L

VIR X

=+

in in2 2

V V2RR R

= =+

R rmsV u IR= =

in inV VR2R 2

= × =

in rmsV 2u⇒ = …..(i)At f=50Hz

LX f∝

So, LL L

X50 RX ' X100 2 2

= × = =

( )' in

22L

VIR X '

=+

in in2

2

V 2V5RRR

2

= = +

' ' inR R in

2V 2V I R V5R 5

= = =

From eq. (i)

( )'R rms

2V 2u5

= ×

rms rms2 2 8u u

55=

Q.17 (a)

Assuming phase-sequence to be RYB Taking RYV as the reference

RYV V 0°= ∠

YBV V 120°= ∠−

BRV V 240°= ∠−

RB BRR

1 1

V VlR R

= = −

1 1

V 240° V 60°R R

∠−= − = ∠−

YBY

1 1

V V 120°lR R

∠−= =

Using KCL R Y Bl l l 0+ + =

B1 1

V V60° 120° l 0R R

∠− + ∠− + =

B1

Vl 3 90°R

⇒ = ∠

So R Y B1 1 1

V V Vl : l : l : : 3R R R

=

1:1: 3=

Q.18 (a) As circuit is under resonance, V1& I should be in phase.

Q.19 (b) Assuming resistance of the heater = R

i) When heater connected to 230 V,50 Hz source, energy consumedby the heater =2.3 units or 2.3kWh in 1 hour Power consumedby the heat

energy 2.3kWhtime period 1hour

= =

1P 2.3kW= rms value of the input voltage

rmsV 230V= = 2rms

1VP

R= =

23 2302.3 10 23Ω

R⇒ × = =


ii) When heater connected to 400V(peack to peck) square wavesource of 150 Hz

Vrms value of the input voltage 1

T 22

rms0

1V V dtT

= ∫

( )

1T 2

T222

T02

1 200 dt 200 dtT

= + −

∫ ∫

rmsV 200V= = 2 2

3rms2

V 200P 10 kWR 23

−= = ×

1.739kW=

Q.20 (a) 1

1 2 xdiV 3 I I V 0.5 0dt

− + − − =

…..(1)

12 x

di5I 0.5 0.2Vdt

− =

12 x

di0.5 5T 0.2V 0dt

− + = ………..(2)

Eliminating I2 for eq(1) and (2) we get

1x 1

dI 51.4V 3.75I Vdt 4

= − − +

Q.21 (a)

Let capacitive reactance = XC

C C

V 0° V 0° 2VIR jX R jX∠ + ∠

= =− −

Using KVL, YXV IR V 0+ − =

YXV V IR⇒ = −

YXC

2VV V RR jX

= − − ( )C

C

V R jX 2VRR jX− −

=−

( )( )

C

C

V R jXR jX

+=

−Method -1

CYX

C

R jXV VR jX +

= − − When R=0

CYX

C

0 jXV V V0 jX +

= − = −

c

YXc

X1 jRV X1 jR

+ = −

when R →∞YXV V= −

Method -2 C

YXC

R jXV VR jX −

= − −

1 CXV 180° 2 tanR

∠ − = −

Magnitude of YXV V=

2 2 1 1 CC

2 2 1 CC

XR X tan tanR

V 180°XR X tanR

∠∠

∠

− −

−

+ = × − +

So, option (c) and (d) cannot be correct, as magnitude is 2V in these two options.

Angle of 1 CYX

XV 108° 2 tanR

− = −

When R = 01

YXV 180° 2 tan ( )−∠ = − ∞

= 0 0180 2x90 0− = When R= ∞


1YXV 180° 2 tan (0) 180°−∠ = − =

On the basic of above analysis, the locus of YXV is drawn below:

Q.22 (a)

23 2I I I 1A∴ = + ⇒ = Power delivered by source

3 1 3W= × = Power dissipated by 2Ω resistor.

2I 2 2W= × = ∴ Power dissipated in non-linear element

3 2 1W= − =

Q.23 (b) Thevenin’s Impedance

0Z 2.38 j0.667Ω= − As real part is not zero, so Z0 has resistor

[ ]0Im Z j0.667= −Case-I

0Z has capacitor [ ]0(as Im Z is negative) Case-II

0Z has both capacitor and inductor, but inductive reactance ∠ capacitive reactance at ω 5= rad/sec For minimal realization case–(i) is considered. Therefore, 0Z will have a resistor and a capacitor.

Q.24 (a) abV 2i 5 2 5 3V= − = − = −

Q.25 (b)

a5 1V 2.5V

2×

= =

Also, ( )ab a b4V 4i 4 V V 4i= ⇒ − = ………(1)

Also, a

b ab a b bV1V 4V V V V

4 2= × = − ⇒ =

1.25V= ( )4 2.5 1.25∴ −

( )( )4i from 1 i 1.25A= ⇒ =

Q.26 (a) As the ABCD bridge is balanced,

CSI 0=

Q.27 (d) For a lamp, 2P KV=For 2

200ω 200lamp,K220V 220=

Consider n lamps connected in series, Total power consumed

2n K 110 100= × × =2

2200n 110 100 n 2200

⇒ × × = ⇒ =

Q.28 (a) Assume a 1A current source at input terminals,

1I 1A∴ =

Applying KVL ( ) ( )in 1 1 CV i 1 50i jX 0− + − − =

[ ]in 2 CV i 2 j50X⇒ = −Input impedance

inC

1

V 2 j50Xi

= = −

As imaginary part is negative, input impedance has equivalent capacitive reactance CX eq.


Ceq. CX 50X=

eq.

1 50ωC ωC

=

50 1ω 100 2ω

= =×

eq.C 2μF=

Q.29 (b) The relevant circuit is shown in fig. As the voltage across 2Ω = 4V

4I 2A2

= =

In order to double the current through 2Ω resistance, SV is to be doubled (Put SV 8V= )Note that the 5A source has no effect on the answer. However it gives 3A current through the voltage source as shown in fig.

Q.30 (b) 'V i 100 and V i.1(by ohm s law)+ = =

2i 100 i 50A∴ = ⇒ =

Q.31 (b) Current through R=1A By KVL, 1.R 6 12+ =

R 6⇒ = Ω



L1 1I 1 0

1 j 1 j= × ∠ =

+ +

Q.33 (d) 1 m 1i l sin(ωt )= −φ

2 m 2i l cos(ωt )= −φ

m 2πl sin ωt2

φ = + −

As these two currents are balanced 1 2i i 0+ =

( )1 2πsin ωt sin ωt 02

φ φ ⇒ − + + − =

1 22 1

π 1 π2sin ωt cos 02 4 2 2

φ φφ φ

+ − + − + =

2 11 π π2 2 2

φ φ ⇒ − + =

1 2π2

∴φ = + φ


rmP Ri=25 4 50W

2 = × =


A =

Q.35 (a)


A BAB DC

V VI 3A I2−

= = =


C DV V 2 3 0,V V 5V

1−

+ + = − = −


Q.36 (d)

2 3eq 1

2 3

C CC CC C

= ++

11.5μF=

Safety voltage = 7V Q CV=

eq safetyQ C V⇒ = ×11.5 7 80μC× ;

Q.37 (35 μH) Two possible series connections are 1. Aiding then L equation= L1 +L2 + 2M. 2. Opposing then L equation = L1 + L2 –2M L1+L2+2M=380μ H ……(1) L2 +L2 –2M = 240 μH ….. (2) From 1 & 2, M = 35 μH

Q.38 (a) By using super position theorem, 1.

1Cϑ — When 20 sin 10t voltage source is acting, Network function

( )

11j cH j 1 (10j 1)R

j c

ωω

ω

= ⇒++

( )1

1C

1t 20sin sin(10t tan (10))101

ϑ −= −

2. ( )2C tϑ When 10 sin 5t current

source is acting

2

100 1 0.21 0.2C j

jϑ ×

= ×−−

2C2 j

1 0.2 jϑ −

=−

( )( )2C 22

2t .sin(5t θ )1 0.2

ϑ = −+

( )2C 2t 1.98sin sin(5t θ )ϑ = −

c 1 2V (t) 2sin sin(10t- )+1.98(5t- )θ θ=By comparing with given expression, 1A =2.0

2A =1.98

Q.39 (17.34 𝛀𝛀) Total power dissipated in the circuit is 1kW. P = 1kW 1000 = 12.1+12.R. 1000 = (2)2.1 + (10)2.R. R=9.96Ω

V 200Z 201 10

= ⇒ =

2 2LZ R X= +

⇒ ( )22 2LX Z R= −

2 2 2LX (20) (9.96)= − ⇒ LX 17.34Ω=

Q.40 (c)

L phV 3V=

3 10 10 3= × =If AV 100=

BC

A

thenV 10 3 90givenV 1015

= −= °

∴ 90+15¯ 7 ¯n 5

BC

nV =10 3 - =10 3 -

Q.41 (3 Watts)

KCL at node xV

x x1 V V21 1−

+ =


xV 1.5V= Power delivered by current source is = 2x 1.5 = 3 watts

Q.42 (a) 6VI 2A3

= =Ω

2VI 1A3

= =Ω

I 1 2+ = I 1A P 5 1 5W== × =

Q.43 (0.5) 2ΩP 12.5W=

2Ω12.5i 2.5

2= =

0V 2 2.5 5V= × =

02.5 KV 5+ =

0KV 5= 2.5 1K 0.55 2

= = =

Q.44 (0) The Network is balanced Wheatstone bridge. ⇒ i = 0Amp

Q.45 (a) ( ) ( )i 3 2 3V V V V

ij j

− −− = +

−100 0° 100 120° 100 120° 100 120°i

1 90° 1 90°∠ ∠ ∠ ∠

∠ ∠− − −

− = +−

i 173.2 60°∠= −

Q.46 (d) By comparing 2 networks on the

input side we can say that A

B A BA

RR 1/ /R R1 R

= ⇒ =+

Q.47 (2) Here the power information regarding the resistor is given because

E calorieP wattt sec

= = =

5P 10Ω→ =

⇒ 55

V 10 V 505

ΩΩ= ⇒ =

Ω→ 4ΩP is asked

( )2 24Ω

4Ω

V 1 4P 504 4 4 6

= = + 1 16 calorie50 24 100 sec

= × × =

Q.48 (0.5)

If we write KCL at node × then 1

1 2 2II 2I I2

= ⇒ =

Write KVL in the outer boundary of network

( ) ( )1 2 1 21 1 I 2 I 0 1 I 2I− × − × = ⇒ = +

11 1 1

I 11 I 2 1 2J I 0.5A2 2

= + ⇒ = ⇒ = =

= 0.5A Q.49 (11.42)

ALL the branch currents are expressed interval of VA now writing KCL at node A

A 11 1 1V 2I 5 15 5 10

⇒ + + + = +

AA

V 102 1V 2 65 10 10

− ⇒ + + =


A2 1 2V 6 25 10 10

⇒ + + = +

A A7 80V 8 V 11.42V

10 7 ⇒ = ⇒ = =

Q.50 (250) The instantaneous power of load is p(t) = V(∈)i(t) [(100sinωt)(10sin(ωt-60)]+ [(100sinωt)(2sin3ωt)]+[(100sinωt)(5sin5ωt)]

→ since, ( )T

avg0

P P t dt= ∫ in the above

expression only 1st term will result non zero answer Remaining 2 terms wiII be 0. → So directly consider

( ) ( )P t = 100sinωt 10 sinωt-60

( ) [ ]P t =100sinsinωt 10(sinωt-60) ( )avg rms rms v 1P V I cos θ θ= −

100 10 cos(60)2 2

=

1000 1 250watt2 2

= =

Q.51 (d)

( )outV 5 2 10= × +

20V=

Q.52 3 Consider the following circuit diagram,

After rearrangement we get,

Now

AB6R 1 0.8 35

= + + = Ω

Q.53 250 Using KCL at node, we get I 0.4I 14+ = I 10A= P 25 10 250W= × =

Q.54 (a) Given: The circuit is given below,

The given network consists only reactive element, the equivalent figure can be represented as,

The above circuit can be re-drawn as shown below,

The equivalent impedance ( )eqZ is

given as


eqeq

eq

j4 ZZ j9

j4 Z×

= ++

( ) ( )eq eqeq

eq

j9 Z j4 Z 36Z

Z j4+ −

=+

( )2eq eqZ j9 Z 36 0− + =

( ) ( )2eq eqZ j12 j3 Z 36 0− + + =

( ) ( )eq eqZ j12 Z j3 0− + + =

eqZ j12= [ eqZ j3= − is neglected

as the whole circuit is a inductive nature]

Hence, the correct option is (a).

Q.55 (b)

Given: The circuit with ( ) ( )i 0 0 v 0 0= = =

The equivalent circuit is represented as,

( ) 2

55sI s 1 s 1s

s

= =++

( )i t 5sin t=

( ) ( )t

0

1V t i t dtC

= ∫

t

0

1 5sin tdt1

= ∫

( ) ( )V t 5 1 cos t= −

The circular loci of ( )i t is given by,

Q.56 5.77

Given:

0anV 100 0 V= ∠


0bnV 100 120 V= ∠−

0cnV 100 240 V= ∠−

a b c nI I I I 0+ + = =

100 0 100 120 100 240 0R j10 j10∠ ∠− ∠−

+ + =−

100 0 10 210 10 150 0R∠

+ ∠− + ∠− =

100 0 8.66 5j 8.66 5j 17.32 10jR∠

= − + − = −

100 0 8.66 5j 8.66 5j 17.32 10jR∠

= − + − = −

Equating the real part of the equation

100 17.32R

=

100R 5.77317.32

= = Ω

Hence, the value of R is 5.773Ω


2.1 INTRODUCTION

This chapter introduces a number of theorems that have application throughout the field of electricity and electronics. Not only can they be used to solve networks such as encountered in the previous chapter, but they also provide an opportunity to determine the impact of a particular source or element on the response of the entire system. In most cases, the network to be analyzed and the mathematics required to find the solution are simplified.

2.2 SUPERPOSITION THEOREM

If the circuit has more than one independent (voltage and/or current) sources, one way to determine the value of variable (voltage across the resistance or current through a resistance) is to use nodal or mesh current methods as discussed in detailed in chapter1. Alternative method for any linear network, to determine the effect of each independent source (whether voltage or current) to the value of variable (voltage across the resistance or current through a resistance) and then the total effects simple added. This approach is known as the superposition.

Def: In a linear network with several independent sources, the response in a particular branch when all the sources are acting at a time is equal to the linear sum of individual responses calculated by considering one independent source act at a time while considering one source other sources are replaced by their internal impedance. → All the ideal voltage sources are

eliminated from the network by shorting the sources because the

internal impedance theorems that have application throughout the field of electricity and electronics. Not only can they be used to solve networks such as encountered in the previous chapter, but they also provide an opportunity to determine the impact of a particular source or element on the response of the entire system. In most cases, the network to be analyzed and the independent (voltage and/or current) sources, one way to determine the value of variable (voltage across the resistance or current through a resistance) is to use nodal or mesh current methods as discussed in detailed in chapter 1. Alternative method for any linear network, to determine the effect of each independent source (whether voltage or current) to the value of variable (voltage across the resistance or current through a resistance) and then the total effects simple added. This response in a particular branch when all the sources are acting at a time is equal to the linear sum of individual responses calculated by considering one independent source act at a time while considering one source other sources are replaced by their internal impedance of voltage source is zero.

→ All the ideal current sources are eliminated from the network by opening the sources, because the internal impedance of current source is infinity

2 NETWORK THEOREMS


→ Don’t disturb the dependent sources present in the network.

2.2.1 PROPERTIES OF SUPERPOSITION THEOREM

1) This theorem is applicable only forlinear network with R, L, C transformerand linear controlled sources as itselements.

2) The presence of dependent sourcesmakes the network an active hencesuper position is applicable to bothactive and as well as passive networks.

2.2.2 HOMOGENITY PRINCIPLE

It is the principle obeyed by all the linear networks. In a linear network if the excitation is multiplied with constant ‘k’ then response in all the other branches of the network is also multiplied with the same constant ‘k’.

In the above figures when 20V source in fig. a) is multiplied by 3 times to get 60Vsource in fig. b the current in 4Ω branch also gets multiplied by 3 times. Note: When multiple sources are present use the superposition theorem first and later homogeneity principle to get the response in a particular branch in the network.

2.3 THEVENIN’S AND NORTON’S THEOREM

A simple circuit as shown in fig. 1) is considered to illustrate the concept of equivalent circuit and it is always possible to view even a very complicated circuit in terms of much simpler equivalent source and load circuits. Subsequently the reduction of computational complexity that involves in solving the current through a branch for different values of load resistance (RL) is also discussed. In many applications, a network may contain a variable component or element while other elements in the circuit are kept constant. If the solution for current (I) or voltage (V) or power (P) in any component of network is desired, in such cases the whole circuit need to be analyzed each time with the change in component value. In order to avoid such repeated computation, it is desirable to introduce a method that will not have to be repeated for each value of variable component. Such tedious computation burden can be avoided provided the fixed part of such networks could be converted into a very simple equivalent circuit that represents either in the form of practical voltage source known as Thevenin’s voltage source (VTH = magnitude of voltage source, RTH = internal resistance of the source) or in the form of practical current source known as Norton’s current source ( IN = magnitude of current source, RTH = internal resistance of the source). In true sense, this conversion will considerably simplify the analysis while the load resistance changes.

Let us consider the circuit shown in fig. 8.1(a). Our problem is to find a current


through RL using different techniques; the following observations are made. • If we use Mesh Analysis then 3

equations need to be solved. • If Nodal Analysis is used then 2

equations need to be solved. • By superposition theorem we need to

find the current through RL considering each source at a time and replacing other sources by their internal impedance. Suppose, the value of RL is changed then the above three techniques need to be applied again right from the beginning. To avoid all the above problems the circuit contained inside the fence in fig. (1) with two terminals A & B, is replaced by the simple equivalent voltage source (as shown in fig. 2) or current source (as shown in fig. 3).

Def: Thevenin’s theorem states that any two terminal linear network having a number of voltage and current sources and resistances can be replaced by a simple equivalent circuit consisting of a single voltage source called Thevenin’s voltage (VTH) in series with a resistance called Thevenin’s resistance (RTH), where the value of the Thevenin’s voltage source is equal to the open circuited voltage across the two terminals of the network and Thevenin’s resistance is equal to the equivalent resistance measured between the terminals.

Def: Norton’s theorem states that any two terminal linear networks with current sources, voltage sources and resistances can be replaced by an equivalent circuit consisting of a current source called Norton’s current (IN) in parallel with Thevenin’s resistance (RTH). The value of the current source is the short circuit current between the two terminals of the network.

2.4 MAXIMUM POWER TRANSFER THEOREM

In an electric circuit, the load receives electric energy via the supply sources and converts that energy into a useful form. The maximum allowable power received by the load is always limited either by the heating effect (in case of resistive load) or by the other power conversion taking place in the load. The Thevenin and Norton models imply that the internal circuits within the source will necessarily dissipate some of power generated by the source. A logical question will arise in mind, how much power can be transferred to the load from the source under the most practical conditions? In other words, what is the value of load resistance that will absorbs the maximum power from the source? This is an important issue in many practical problems. To answer the above questions we make use of Maximum Power Transfer Theorem. This theorem is applicable only when the load is a variable otherwise choose the minimum internal impedance of the source, which results a maximum current through a fixed load and hence a maximum power dissipation across the load.

2.4.1 UNDER THE VARIABLE LOAD CONDITIONS

Case1: Source resistance ‘RS’ is fixed and Load resistance ‘RL’ is variable


S

s L

VIR R

=+

The maximum power transfer theorem states that, “maximum power is delivered from a source to a load when the load resistance is equal to the source resistance”. →Current in the circuit is

S

s L

VIR R

=+

→Power delivered to the load RL is 2

LP I R (W)=

∴ ( )

2S L

2s L

V RP (W)R R

=+

( )

2S L

2L L s L

V RdP ddR dR R R

= +

( ) ( ) ( ) ( )

22S s L L s L

4s L

V R R 2R R R

R R

+ − +=

+

When L

dP 0dR

=

( ) ( )( )2s L L s LR R 2R R R 0∴ + − + =

We get s LR R= Now,

L Smax atR RP P | ==

→ ( ) L S

2S

max atR R2s L

V RLP |R R

==+

( )

2S

max 2s S

V RSPR R

⇒ =+

( )2 2V VS SP W or (W)max 4RS 4RL

⇒ =

Power delivered by voltage source (VS) = Power absorbed by source resistance ‘RS’ Power absorbed by ′RL′

S LT R RP P P= +

L S

2 2T s L R RP I R I R | == +

2 2S S

S L

V V4R 4R

= +

2

TVSP2RS

=

Efficiency ( ) usefulpowerTotalpower

η =

2S

2S

V4RLV2RL

=

12

η =

% 50%η = So, the efficiency of maximum power transfer theorem is almost 50%

Case2: Source impedance Zs is fixed and load impedance ZL is variable

Current in the circuit is

( )S

S L S L

VIR R j X X

=+ + +

Power delivered to the load impedance is 2

LP I R (w)=

( ) ( )

2S L

22 2

S L S L

V RP (w)R R X X

= + + +

( ) ( )

2S L

2 2S L S L

V RPR R X X

=+ + +

Case2.1: Only load resistance RL is variable Now differentiating power w.r.t LR & equate to zero

( ) ( ) ( )

( ) ( )

2 2 2 2S L S L S L

22 2S L S L

R R X X VS VS RL 2R 2RdpdRL R R X X

+ + + − +=

+ + +


dp 0dRL

=

We get ( )22L S LR Rs X X+ + Ω

Case2.2: Only load reactance ‘XL’ is variable Now differentiate power w.r.t 𝑋𝑋𝐿𝐿& equate to zero

( ) ( ) ( )

( ) ( ) 2 2 2

S L S L S L22 2L

S L S L

R R X X 0 Vs RL(2X 2X )dPdX R R X X

+ + + − +=

+ + +

L

dP 0dX

=

We get S LX X 0+ =

Case2.3: Both Load resistance ‘RL’ & reactance ‘XL’ are varied simultaneously

Here both 2 (i) and 2 (ii) are satisfied simultaneously i.e.

( )22L S LR Rs X X= + + [from 2 (i)] and

L SX X 0+ = [from 2 (ii)] Combining above two conditions, we get

2L L SR Rs 0 & X X 0= + + =

L SR R= & L SX X= − We have, L L LZ R jX= +

S SR jX= − ( )L S L SR R & X X= = −Q

L SZ Z *=

Case3: Source impedance is fixed and load resistance RL variable

It is a special case of 2(i) with LX 0=

From 2 i) ( )22L S LR Rs X X= + +

2 2LR Rs Xs⇒ = +

L SR Z=

2.5 THE TELLEGAN’S THEOREM

Tellegan’s theorem is valid any lumped network which may be linear or non-linear, passive or active, time–varying or time invariant. It states that, the algebraic sum of all the powers delivered by some elements in the network is equal to the power observed by the remaining elements present is the network. When the current enters of the negative terminal of an element then that element will deliver the power otherwise it absorbs the power. The sources can either deliver the power or they can absorb the power whereas the passive, elements will always absorb the power, since the current enters from positive terminal in the R, L, C elements in the presence of sources.

2.6 RECIPROCITY THEOREM

A linear, passive and a bilateral network, the ratio of response to the excitation is constant even through the source is interchanged from the input terminal to the output terminals.

From fig a & fig b by reciprocity theorem 1 2

1 2

V VI I

= Or, V cons tan tI=

Proof:


40Req7

= Ω

20 7I 7 A40 2

= × =

17 / 2I 4 2A4 3

= × =+

From above figures it can be concluded that the current remains 2A even if the 20V source is moved from input terminals to the output terminals.

2.6.1 FEATURES OF RECIPROCITY THEOREM

1) This theorem is applicable only for thelinear, passive, bilateral network i.e.networks with R, L, C and transformeras its elements, so called reciprocalnetwork.

2) The presence of the dependent sourcesmakes the network an active and hencethe Reciprocity theorem is notapplicable, so called the non-reciprocalnetwork.

2.7 SUBSTITUTION THEOREM

In a linear network any passive elements can be equivalently substituted by a ideal voltage source or an ideal current source, provided the original passive element or the substituted active sources absorbs the same power then only all the other branch currents and voltages are kept constant.

2.8 MILLIMAN’S THEOREM

Milliman’s theorem states that in any network, if the voltage sources 1 2 nV ,V .V…in series with internal resistances

1 2 nR ,R .R… respectively, are in parallel, then these sources may be replaced by a single voltage sources V’ in series with ‘R’ as shown in fig.

' 1 1 2 2 n n

1 2 3 n

V G V G V GVG G G G

+ +…+=

+ + +…+

1 2 3 n

1 1 1 1 1.R ' R R R R

= + + +… +

'1 2 nG G G G= + +…+

'

1 2 n

1 1RG ' G G G

= =+ +…+

Note: In the above case if the polarities of the source V2 are reversed then V2 replaced by V2 in the expression for V’.

2.9 THE DUALITY PRINCIPLE

The network and its dual are the same w. r. t performance point of view but the elements and connections point of view they are not equal. In electrical circuit itself there are pairs of terms which can be interchanged to get new circuits such pair of dual terms is given below.` Current voltage↔ Open short↔ L C↔ R G↔ Z Y↔ series parallel↔ KCL KVL↔ Star delta↔ Thevenins Nortons↔Ri(t) Gv(t)↔

di(t) dv(t)L Cdt dt

↔


( ) ( )1 1i t dt v t dtC L

↔∫ ∫To draw dual of the network, the following steps are to be followed; 1) In each loop of a network place a node,

and place an extra node, called thereference node, outside the network.

2) Draw the lines connecting adjacentnodes passing through each elementand also to the reference node, byplacing the dual of each element in theline passing through original elements.

Example: Find the current passing through the 3Ω resistor in the circuit shown in fig by superposition theorem

Solution: By superposition theorem Consider 20V voltage source and open circuit 5A current source We get,

By KVL, 1 120-5(i )-3(i )=0

120=8i

120i8

=

1i 2.5A= Now consider 5A current source and short circuit 20V voltage source. We get,

By current division rule,

25i 5

5 3= ×

+

2i 3.125A= So, the total current I passing through 3Ω resister is

1 2I i i= + I 2.5 3.125= + I 5.625A=

Example: Determine Thevnin’s and Norton’s equivalent circuit across ‘AB’ for the given circuit

Solution: Let Vth is voltage across terminal AB

By applying nodal we get, TH THV 50 V 25 0

10 5− −

+ =

TH THV 50 2V 50 0− + − =

TH3V 100=

THV 33.3V= RTH is the resistance seen into the terminals AB. To find RTH the two voltage sources are removed & replaced with short circuit

TH10 5R (10 || 5) 3.3310 5

×= = = Ω

+So, the Thevenin’s and Norton’s equivalent circuits are

Example:


Determine the values of load resistance when the load resistance drawn maximum power. Also find the value of the maximum power.

Solution: The source delivers the maximum power when load resistance is equal to the source resistance

LR 25= Ω

The current 50I25 RL

=+

5025 25

=+

I 1A= The maximum power delivered to the load

2P I R= ( )2P 1 25= ×

P 25W= Example: Draw the dual network for the given network shown in fig

Solution: Place nodes in each loop and one reference node outside the circuit joining the nodes through each element & placing the dual of each element we get.

The dual of circuit is redrawn as shown below

Example: Verify Tellegen’s Theorem

Solution: While verifying the Tellegen’s theorem don’t disturb the original network for evaluating the voltages & currents in each & every element of the network.

By KVL, 20 5i 5 0− − = i 3A= Powers due all the elements of the network are:

20vP 20 3 60W= × = (del)

5P 15 3 45WΩ = × = (abs)

5vP 5 3 15W= × = (abs) Power delivered 60W Power absorbed∴ = =

i.e. the conservation of power. Example: Calculate the current I shown in figure using Milliman’s theorem

Solution: According to Milliman’s theorem, the two voltage sources can be replaced by a single voltage source in series with resistance shown


' 1 1 2 2

1 2

V G V GVG G

+=

+

( ) ( )1 110 202 5

1 12 5

+ =

+

'V 12.86V= '

1 2

1 1R 1.431 1G G2 5

= = = Ω+ +

Therefore, the current passing through the 3Ω resistor is

12.86I3 1.49

=+

I 2.9A=


Q.1 In the figure, the switch was closed for a long time before opening at t = 0. The voltage xV att 0+= is

a) 25V b) 50Vc) 50V− d) 0V

[GATE-2002]

Statement for linked answer questions: The circuit for Q.2 and Q.3 is given Assume that the switch S is in position 1 for a long time and thrown to position 2 at t=0

Q.2 At +t=0 the current 1i is

a) V2R− b) V

R−

c) V4R− d) zero

[GATE-2003]

Q.3 ( )1I S and ( )2I S are the Laplace

transforms of ( )1i t and ( )2i trespectively. The equations for the loop currents ( )1I S and ( )2I S for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t=0, are

a) 1

2

1 VR Ls Ls I (s)Cs sI (s)1 0Ls R

Cs

+ + − = − +

b) 1

2

1 VR Ls Ls I (s)Cs sI (s)1 0Ls R

Cs

+ + − − = − +

c) 1

2

1 VR Ls Ls I (s)Cs sI (s)1 0Ls R Ls

Cs

+ + − − = − + + +

d) 1

2

1 VR Ls Ls I (s)Cs sI (s)1 0Ls R Ls

Cs

+ + − = − + + [GATE-2003]

Q.4 For the R-L circuit shown in the figure. The input voltage

( )iV t u(t)= . The current i(t) is

a) b)

c) d)

[GATE-2004]

GATE QUESTIONS(EC)


Q.5 The circuit shown in the figure has initial current ( ) ( )- -

L Li 0 =1i 0 =1 Athrough the inductor and an initial voltage ( )-

CV 0 =-1V across the

capacitor. For input ( )v t =u(t) theLaplace transform of the current i(t)for t³0 is

a) 2S

S S 1+ +b) 2

S 2S S 1

++ +

c) 2S 2

S S 1−+ +

d) 2S 2

S S 1−+ +

[GATE-2004]

Q.6 A square pulse of 3 volts amplitude is applied to C-R circuit shown in the figure. The capacitor is initially uncharged .The output voltage V2at time t=2 sec is

a) 3V b) -3Vc) 4V d) -4V

[GATE-2005]

Q.7 A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

a) 0.5A b) 2.0Ac) 1.0A d) 0.0A

[GATE-2006]

Q.8 In the figure shown, assume that all the capacitors are initially uncharged. If ( )iV t 10u(t)= Volts,then V0 (t) is given by

a)8e 0.004t Volts−b) 0.004t8(1 e )Volts−−

c) ( )8u t Voltsd) 8 Volts

[GATE-2006]

Q.9 In the circuit shown, cV is 0 volts att=0 sec. For t>0 , the capacitor current ic(t) Where t is in seconds, is given b

a) 0.50exp( 25t)mA−b) 0.25exp( 25t)mA−c) 0.50exp( 12.5t)mA−d) 0.25exp( 6.25t)mA−

[GATE-2007]

Q.10 In the following circuit, the switch S is closed at t = 0 . The rate of change

of current di (0 )dt

+ is given by

a) 0 b) S SR IL

c) ( )S SR R IL

+d) ∞

[GATE-2008]


Q.11 The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows For ( )2nT t 2n 1 T,≤ < +

( )n 0,1,2= …S1 to P1 and S2 to P2 For ( ) ( )2n 1 T t 2n 2 T+ ≤ < +

( )n 0,1,2= … , S1 to Q1 and S2 to Q2

Assume that the capacitor has zero initial charge. Given that u(t) is a unit step function, the voltage cV (t) across the capacitor is given by

a) ( ) ( )n

n 01 tu t nT

∝

=

− −∑

b) ( ) ( )n

n 1u t 2 1 u(t nT)

∝

=

+ − −∑

c) ( ) ( )n

n 1tu t 2 1 (t nT)u(t nT)

∞

=

+ − − −∑

d) ( ) ( )t nT t nT T

n 10.5 e 0.5e

∞− − − − −

=

− + ∑ [GATE-2008]

Common Data Questions 12 & 13: The following series RLC circuit with zero initial condition is exited by a unit impulse functions δ(t)

Q.12 For t>0 , the voltage across the resistor is

a) 3 1t t

2 21 e e3

− − −

d)1 t2 3t 1 3te cos cos sin sin

2 23−

−

c) 1 t22 3te sin

23−

d) 1 t22 3te cos

23−

[GATE-2008]

Q.13 For t > 0, the output voltage Vc(t) is 1 3t t2 22 e e

3− −

−

a) 1 t22 te

3

−

b) 1 t22 3e cos t

23−

c) 1 t22 3e sin t

23

d) t22 3e sin t

23−

[GATE-2008]

Q.14 The switch in the circuit shown was on position ‘a’ for a long time and is moved to position ‘6’ at time t=0.The current i(t) for t > 0 is given by

a) ( )125t0.2e u t mA−

b) ( )1250t20e u t mA−

c) ( )1250t0.2e u t mA−

d) ( )1000t20e u t mA−

[GATE-2009]


Q.15 The time domain behavior of an RL circuit is represented by

( ) RtL

0di t

L Ri V (1 Be sint)u(t)dt

−= + For

an initial current of ( ) 0Vi 0R

= , the

steady state value of the current is given by

a) 0Vi(t)R

→ b) 02Vi(t)®R

c) 0Vi(t) (1 B)R

→ + d) 02Vi(t) (1 B)R

→ +

[GATE-2009]

Q.16 In the circuit shown, the switch S is open for a long time and is closed at t=0 . The current i(t) for t 0+≥ is

a) ( ) 1000ti t 0.5 0.125e A−= −

b) ( ) 1000ti t 1.5 0.125e A−= −

c) ( ) 1000ti t 1.5 0.125e A−= −

d) ( ) 1000ti t 0.375e A−= [GATE-2010]

Q.17 In the circuit shown below the initial charge on the capacitor is 2.5mC, with the voltage polarity as indicated. The switch is closed at time =0 .The current i(t) at a time t after the switch is closed is

a) ( ) ( )3i t 15exp 2 10 t A= − ×

b) ( ) ( )3i t 5exp 2 10 t A= − ×

c) ( ) ( )3i t 10exp 2 10 t A= − ×

d) ( ) ( )3i t 5exp 2 10 t A= − − ×

[GATE-2010]

Q.18 In the following figure, 1C and 2Care ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t=0 .The current i(t) for all t is

a) Zerob) A step functionc) An exponentially decaying functiond) An impulse function

[GATE-2012]

Q.19 For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z1 of the first section to the input impedance Z2 of the second section is a) 2 1Z Z= b) 2 1Z Z= −

c) *2 1Z Z= − d) *

2 1Z Z= − [GATE-2014]

Q.20 In the circuit shown in the figure, the value of capacitor C(in mF) needed to have critically damped response i(t) is

[GATE-2014]

Q.21 In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t > 0?


a) ( ) /I t (1 ), msec5 23 3

−= − t τe τ -

b) ( ) t /I t 5 (1- ), msec2e2 3

− τ =τ=

c) ( ) t /I t (1 ), 3 ce se2

m5 − τ =τ= −

d) ( ) t /I t (1 ), 3 me sec52

− τ =τ= −

[GATE-2014]

Q.22 In the circuit shown in the figure, the value of V0(t) (in Volts) for t →∞ is __.

[GATE-2014]

Q.23 In the circuit shown, the switch SW is thrown from position A to position B at time t = 0. The energy (in 𝜇𝜇,J) taken from the 3V source to charge the 0.1 μF capacitor from 0V to 3V is

a) 0.3 b) 0.45c) 0.9 d) 3

[GATE-2015]

Q.24 The switch has been in position 1for along time and abruptly changes to position 2 at = 0

If time t is in seconds, the capacitor voltage Vc (in volts) for t>0 is given by a)4(1- exp(-t/0.5)) b)10- 6 exp(-t/0.5) c)4(1- exp(-t/0.6)) d) 10- 6exp(-t/0.6)

[GATE-2016]

Q.25 Assume that the circuit in the figure has reached the steady state before time t = 0 when the 3Ω resistor suddenly burns out, resulting in an open circuit. The current i(t) (in ampere) at t = 0+ is _____.

[GATE-2016]

Q.26 In the circuit shown, the voltage VIN(t) is described by:

VIN(t) =

𝟎𝟎, 𝒇𝒇𝒇𝒇𝒇𝒇 𝒕𝒕 < 𝟎𝟎𝟏𝟏𝟏𝟏𝟏𝟏𝒇𝒇𝟏𝟏𝒕𝒕𝟏𝟏, 𝒇𝒇𝒇𝒇𝒇𝒇 𝒕𝒕 ≥ 𝟎𝟎

Where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value 2 Amperes is______________.

[GATE-2017]

Q.27 The switch in the circuit, shown in the figure, was open for a long time and is closed at t=0.


The current i(t) (in ampere) at t=0.5 seconds is________________

[GATE-2017]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (a) (c) (c) (b) (b) (a) (c) (a) (b) (c) (b) (d) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 (a) (a) (a) (d) (c) 10 (a) 31.25 (c) (d) 1 0.3405 8.16

ANSWER KEY:


Q.1 (c) When switch was closed circuit was in steady state,

Li (0 ) 2.5A− =

At +t=0

V IR⇒ =2.5 20 50V= × =

xV 50V∴ = − (Polarity of Vx is given reverse of V)

Q.2 (a) At t 0−= in steady state

( ) ( )1 2i t i t 0= =

( )cV 0 V− =

At t 0+=

1 1i R V i R 0− − − =

1Vi

2R−

∴ =

Q.3 (c) When switch is in position 2,

KVL in loop (1),

( ) ( ) ( )1 1 1 2V 1I s .R I s . I s I (s) sL 0s sC

+ + + − =

( ) ( )1 21 VI s R sL I s .sL

sC s ⇒ + + = = −

KVL in loop 2,

( ) ( ) ( ) ( )2 1 2 21I s I s sL I s R I s . 0

sC− + + =

( ) ( )1 21I s .sL I s R sL 0

sC ⇒ − + + + =

( )( )

1

2

1R sL sL I ssC1 I ssL R sL

sC

+ + − − + +

Vs

0

− =

Q.4 (c)

( ) V(s) 1I ss 2 s(s 2)

= =+ +

( ) 1 1 1 1I ss(s 2) 2 S s 2

= = − + +

( ) ( )2t1i t 1 e2

= − −

At ( )t 0, i t 0= =

( )t , i t 0.5= ∞ =

( )1t , i t 0.312

= =

EXPLANATIONS


Graph (c) satisfies all conditions.

Q.5 (b) KVL,

( ) ( ) ( )0

Ldi(t) 1v t Ri t i t dtdt C

∞

= + + ∫Taking L.T on both sides,

( ) ( ) ( ) ( ) ( )c 0VI(s)v s RI s LsI s LI 0sC S

++= + − + +

( ) ( )1 I(s) 1I s sI s 1S S S

⇒ = + − + −

22 I(s)1 s S 1S S

+ = + +

( ) 2s 2I s

S S 1+

=+ +

Q.6 (b) 6 3 4RC 0.1 10 10 10− −= × × =

100μs= As RC is very small, so steady state will be reached in 2 sec.

cV 3V=

2 cV V 3V= − = −

Q.7 (a)

∴ ( ) ( )L.i 0 1mv i 0 0.5A+ += ⇒ =

Q.8 (c) For the given circuit

( )( )

( )( ) ( )

0 2

i 1 2

V S Z SV S Z S Z S

=+

Where ( ) 22

2 2

RZ S1 R C S

=+

,

where 2 2R 4K ,C 1μF= =Ω

( ) 11

1 1

RZ S1 R C S

=+

,

where 1 1R 1K ,C 1μF= =Ω

2 2 1 1R C R C= 3 6 310 4 10 sec 4 10 sec− −= × × = × ( )( )

0 2

i 1 2

V S R 4V S R R 5

= =+

( ) ( ) ( ) ( )0 i 0 i4V S V S ,V t 0.8V t5

= =

For ( ) ( ) ( )i 0V t 10u t ,V t 8u(t)= =

Q.9 (a) Given

( ) ( )tT

i C f i fV 0 V t V V V e−

= ⇒ = + −

( )25t5 1 e−= −

( )

( )

cC

6 25t

25t

dvi t C.dt

4 10 5 e 25

0.5e mA

− −

−

=

= × ×− ×−

=

Q.10 (b) In the circuit shown,

( )( )

( )

S Si f

S

eq S

eq S

I .RI 0; I ;R R

R R R ;

L LT R R R

= =+

= +

= = +

( )

( )

S S

S S Sft 0

S

I .RR R I .RIdi | Ldt T L

R R+=

+= = =

+

Q.11 (c) The waveform of voltage VC(t) is shown below.


In mathematical form, ( ) ( ) ( ) ( )( ) ( )

CV t tu t 2 t T u t 2t

2 t 2T u t 2T

= − − −

+ − − …

( ) ( ) ( ) ( )n

n 1tu t 2 1 t nT u t nT

∞

=

= + − − −∑

( )CSV S=

Q.12 (b) ( ) ( ) ( )( )R CV S R.I S 1. CSV S= =

( ) ( )R 2

SV SS S 1

⇒ =+ +

( )2 2

1 1S 2 21 3S 22

+ −= + +

( )t2

R

t2

3V t e .cos t2

1 3e .Sin t23

−

−

= −

Q.13 (d)

( ) ( )C

2i

1V (S) 1S1V (S) S S 11 S S

= =+ ++ +

( ) ( ) ( )i iV t δ t V S 1= ⇒ =

( )( )

( )C C22

1V S V t31S 2 2

∴ = ⇒

+ +

= t22 3e .Sin t

23−

Q.14 (b)

When S is in position ‘a’,VC(0−)100V After S is moved to b,(for t ≥ 0)

i fV 100V;V 0;R 5K;0.8 0.2C 0.16μF

1

= = =×

= =

( )t 1250t0.0008

CV t 100.e 100e− −= =

( ) ( ) ( )c 1250tV ti t 20e u t mA

5K−= =

Q.15 (a) The steady state supply across R-L circuit is V0

( ) 0Vi R∴ ∞ =

Q.16 (a) ( )L ii 0 0.75A I− = =

( )L fi 0.5A I∞ = =3

310T 15 10 sec15

−−= × =

( ) 1000tLi t 0.5 0.25e−∴ = +

( ) ( )1000ti t 1.5 0.5 0.25e / 2−= − −1000t 0.5 0.125e−= −

Q.17 (a) ( ) ( )Q 0 2.5mC V 0− −= − ⇒

( ) 3

6

Q 0 2.5 10 50VC 50 10

− −

−− ×

= = = −×

x

R

eq

for t 0,V 50V;V 100V;τ R : C 0.5m sec

≥ = −=

= =

( ) ( )2000tCV t 100 150e u(t)−∴ = −

( ) cC

dVi t C.dt

∴ =

( )6 2000t50 10 150 2000 e u t− −= × × × ×

( )2000t15e u t A−=

Q.18 (d) When the switch in closed at t = 0


Capacitor C1 will discharge and will get charge since both C1 and C2 are ideal and there is no –resistance in the circuit charging and discharging time constant will zero. Thus current will exist like an impulse function.

Q.19 (c) Two cascaded sections

Z1=Output impedance of first section Z2=Input impedance of second section For maximum power tranfer, upto 1st section is Z1 = Z1

∗ ZL = Z2⇒ Z1

∗

Q.20 (10mF) By KVL,

( )di(t) 1V(t) Ri(t) L i t dtdt C

= + + ∫Differentiate with respect to time,

2

R.di(t) R di(ti) i(t)0 . 0dt L dt LC

= + + =

2

2

.d i(t) R di(t) i(t). 0dt L dt LC

+ + =

2

1,2

R R 4L L LCD

2

− ± − =

2

1,2R R 1D

2L 2L LC− = −

For critically damped response,

2

2

R 1 4LC F2L LC R

= ⇒ =

Given, L=4H; R=40Ω

C=( )24 4 10mF40×

⇒

Q.21 (a) Vc(t) = VR2(t) = Vfinal

t

initial finalV V e τ−

−

3 62. 10 103equ equR Cτ −= ⇒ × ×

22 ||13equR K K K= ⇒ Ω

1equC Fµ=2 sec3

mτ =

0initialV volts=

.2 105.3 3final s sV V= = = volts

VR2(t)=tτ10 10 e

3 3−

−

VR2(t) = tτ10 1 e

3− −

volt

( )tτ

R2R 2V (t) 5i t 1 e2K

mA3

− = =⇒ −

Q.22 (31.25)

For t ,→∞ i.e., at steady state, inductor will behave as a shot circuit and hence B5V= .ix By KCL at node B, BV10− + ,

x x x50 2 i8

i 0i− + = =

( )0V t 5= ( ) ( )0x250t V t8

i ⇒ =

31.25volts=

Q.23 (c) So the capacitor in initially uncharged i.e. Vc(0) = 0


→The capacitor will be charged to supply voltage 3V when the switch is in position B for ∞ time. → So we need to find capacitor voltage

t /Vc(t) Vc( ) [Vc(0)Vc( )]e− τ= ∞ + ∞ /3 3 te τ−= −

6120 0.1 10 12 secRCτ µ−= = × × = c

cdVi Cdt

=

= (0.1 610 ) [3 3 ]td e

dtτ

−−× −

= 3 0.1 te τ

τ−× = 0.3

12t

e τ−

=

→So instantaneous power of source = V(t)i(t)

( ) ( )t t0.3 0.9eP e

12 1t 3

2τ τ

− − = =

0

( )E P d t∞

→ = ∫

= 0

0.9 ( )12

te d tτ

∞−

∫t

0

0.9 0.9) e12 1

(2

ττ τ∞−

=

=

= 12 6 0.91012

−× ×

= 0.9 μJ

Q.24 (d) → At t = 0 switch in position 1 and

since the capacitor is open circuited

( )c2V = 10=4V

2+30

→At t = infinity switch is in position 2 and since the capacitor is open circuited

( )cV ∞ = (5) 2 = 10V → Time constant τ = ( )thR C= 4+2 0.1=0.6sec

→ cV (t) = ( )cV ∞ + [ cV (0) ‒ ( )cV ∞-t/τe

= [ ]10 4 10+ − -t/0.6e 10 6= − -t/0.6e

Q.25 (1) At t = 0 , the circuit is on steady state i .e. the capacitor is open circuited so the circuit will be

] [3F 2 3 3V [ V V - V ] 4V= + =Ω Ω Ω

2F 3V V 6V= =Ω at t = 0+ when 3Ω is open circuited, the capacitors will have an ideal voltage source of values 4V and 6V so the circuit will be

So the current through 2Ω resistor at

t = 0+ should be 42 2+

= 1A

Q.25 0.3405 sec

Q.26 8.16A


Q.1 In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor a) Is always zerob) Can never be greater than the

input voltagec) Can be greater than the input

voltage; however it is 90° out ofphase with the input voltage

d) Can be greater that the inputvoltage, and is in phase with theinput voltage

[GATE-2001]

Q.2 In the circuit shown if figure, what value of C will cause a unity power factor at the ac source?

a) 68.1μF b)165μFc) 0.681μF d) 6.81μF

[GATE-2003]

Q.3 A first order, low pass filter is given with R 50Ω= & C 5μF.= What is the frequency at which the gain of the voltage transfer function of the filter is 0.25? a)4.92 kHz b)0.49kHz c)2.46 kHz d)24.6 kHz

[GATE-2003]

Q.4 A series R-L-C circuit has R 50= ΩL 100 H= µ and C 1 F= µ . The lower half power frequency of the circuit is a) 30.55 kHz b) 3.055kHzc) 51.92 kHz d) 1.92 kHz

[GATE-2003]

Q.5 In the circuit of figure, the magnitudes of LV and CV are twice

that of RV .Given that f = 50 ZH , the inductance of the coil is

a) 2.14 b)5.30Hc)31.8mH d)1.32H

[GATE-2003]

Q.6 The value of Z in figure which is most appropriate to cause parallel resonance at 500Hz is

a)125.00mH b)304.20μF c) 2.0μF d) 0.05μF

[GATE-2004]

Q.7 The R-L-C series circuit shown is supplied from a variable frequency voltage source. The admittance- locus of the R-L-C network at terminals AB for increasing frequency ω is

a) b)

GATE QUESTIONS(EE)


c) d)

[GATE-2007]

Q.8 The resonant frequency for the given circuit will be

a) 1 rad/s b) 2 rad/sc) 3 rad/s d) 4 rad/s

[GATE-2008]

Q.9 Two magnetically uncoupled inductive coils have Q factors 1q and

2q at the chosen operating frequency. Their respective resistances are 1R and 2R . When connected in series, their effective Q factor at the same operating frequency is a) 1 2q q+b) 1 2(1/ q ) (1/ q )+

c) ( ) ( )1 1 2 2 1 2q R q R / R R+ +

d) ( ) ( )1 2 2 1 1 2q R q R / R R+ +[GATE-2013]

Q.10 A series RLC circuit is observed at two frequencies. At 1 1krad / sω = , we note that source voltage

1V 100 0 V= ∠ ° results in current

1I 0.03 31= ∠ ° . At cot = 2 krad/s, the source voltage 2V 100 0 V= ∠ °results in a current 2I 2 0 V= ∠ ° A. The closest values for R,L,C out of the following options are a) R 50 ;L 25mH;C 10 F;= Ω = = µ

b) R=50Ω;L=10mH;C=25μF;c) R 50 ;L 50mH;C 5 F;aµ= Ω = =d) R 50 ;L 5mH;C 50 F;µ= Ω = =

[GATE-2014]

Q.11 An inductor is connected in parallel with a capacitor as shown in the figure.

As the frequency of current i is increased, the impedance (Z) of the network varies as a)

b)

c )

d )

[GATE-2015]

Q.12 In the circuit shown below, the supply voltage is 10sin (10000 volts. The peak value of the steady state


current through the 1Ω resistor, in amperes, is ____.

[GATE-2016]

Q.13 The circuit below is excited by a sinusoidal source. The value of R, in Ω, for which the admittance of the circuit becomes a pure conductance at all frequencies, is ___.

[GATE-2016] Q.14 In the balanced 3-phase, 50Hz,

circuit shown below, the value of inductance (L) is 10mH. The value of the capacitance (C) for which all the line current ar zero, in millifarads, is

[GATE-2016]

Q.15 The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is a)11 b)12 c)13 d)14

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (a) (c) (b) (c) (d) (a) (c) (c) (b) (b) 1 14.14 3.04 15 (b)

ANSWER KEY:


Q.1 (c) In a series RLC circuit, at resonance

L sourceV jQV= and C sourceV jQV= −

also for C sourceQ 1, V V> > Hence option (c) is correct.

Q.2 (a) 1Y jωC

30 40°= +

∠jωC 0.0255 j0.0214= + −

= (ωC0.0 0255 j .0214)+ −

( )Real Y jImag(Y)= +To have a unity power factor at ac source i.e. resonancecondition, ( )Imag Y 0=

ωC 0.0214 0⇒ − =ω 2π 50= ×Q

0.0214C 68.1μF100π

∴ = =

Q.3 (c)

o

i

1V 1jωCT.F 1V 1 jωCRR

jωC

= = =++

2

1Gain1 (ωCR)

∴ =+

( )26

10.251 ω 5 10 50−

=+ × × ×

On solving 3ω 15.49 10 r / s= ×

15.49f 2.46kHz2π

⇒ = =

Q.4 (b)

0 6 6

1 1ωLC 100 10 10− −

= =× ×

510 r / s=4

6R 50ω 50 10 r / sL 100 10−

∆ = = = ××

22

lower 0ω ωω ω2 2

∆ ∆ = + −

( )25 525 5 10 5 1010

2 2

× × = + − 510 1 6.25 2.5= + −

50.193 10 r / s= ×Hence,

5lower

lowerω 0.193 10f

2π 2π×

= =

3065Hz 3.055kHz= ;

Q.5 (c) R L CV V i(V V )= + −

Since L C L RV V & V 2 V= = Therefore, the circuit is at resonance and

RV V= Quality factor

L L R

R R

V V 2V 2V V V

= = = =

As we know 0ω LQR

=

2πf L25×

⇒ =

L 31.8mH⇒ =

Q.6 (d) At resonance, the circuit should be in unity power factor ∴ Hence ‘Z’ should be capacitive Admittance of the parallel circuit

EXPLANATIONS


1 1Y 0jLω 1/ jCω

= + =

1 Cω 0Lω−

+ =

∴ ( )2 2

1 1CL ω 2 2π 500

= =× × ×

0.05μF=

Q.7 (a) Admittance of the series connected RLC

1Y1R j ωLωC

= + −

22

1R j ωLωCY1R 1ωLωC

− − = + −

By rationalization Separating, real and imaginary part of admittance.

22

RRe[Y]1R ωLωC

= + −

For any value ofω , the real part of always positive.

When 1ωL orωC

=

At 01ωLC

= (resonance)

[ ] 1Re YR

= (maximum value)

( )m 22

1ωLωCI Y

1R ωLωC

− − = + −

22

1 ωLωC

1R ωLωC

− = + −

At 01ωLC

= (resonance)

Imaginary part of zero

( )mI Y 0⇒ =

For 00<ω<ω1 ωLωC

>

Therefore [ ]mI Y 0>

For 0ω < < ∞ω 1 ωLωC

<

Therefore, Im[Y] > 0 On the basis of above analysis, the admittance locus is

Q.8 (c) Input impedance

1z jωL RjωC

= +

Rz jωL1 jωRC

= ++

1 1 jωz j0.1ω1 jω 1 jω

−∴ = + ×

+ −

21 jωj0.1ω1 jω−

= ++

2 21 ωj 0.1ω

1 ω 1 ω = + − + +

At resonance, imaginary part must be zero.

2ω0.1ω 0

1 ω− =

+

210.1

1 ω=

+

2ω 1 10+ = 2ω 9∴ =

ω 3rad / sec∴ =


Q.9 (c) 1 2

1 22 2

ωL ωLQ QR R

= = =

After series connection ( )1 2

1 2

ω L LQ

R R+

=+

1 2 21R Q R

1 2

QQ

R R+=+

Q.10 (b) Given 1 1V 100 0 V;I 0.03 31= ° = ° at

1ω 1000r / sec=

12 0 V;I 2 0V 100= ° = ° at

2ω 2000r / sec=

2

2

100i.e R 50I 2ϑ

Ω= = ⇒

01

L c01

v 100 0Z R j(X X )I 0.03 31

= ⇒ ⇒ + −

1 L CX X31 tanR

Φ − − = °⇒

L CX Xtan 31R− ⇒ ° =

11

1Lc

tan 31R

ωω

−

° =

11

1 0 Lc

.600 50ωω

⇒ − = ×

11

1 30.0L 4q

⇒ − = ω

ω c

22

1L 0c

ωω

− = ….. (2)

1 21000r / sec; 2000r / secω ω= = From 1 and 2, C = 25μF L=10mH

Q.11 (b)

L CZ Z / /Z=

ωLωC

ωLωC

1jjZ1jj

×=

+

( )ωL

2

jZ1 ( 1)ω LC

=+ −

2

ωLZ j1 ω LC = −

Q.12 (1)

W =1000, the various impedance at this frequency are

3250 f 4mH 6

jZ || Z ( j1000 4 10 )1000 250 10µ

−−

− = × × × × →

( ) ( )j4 || j4 open circuit= − = ∞ =

324 500 6

j|| ( j1000 500 10 )1000 250 10

−−

− → × × × × f mHZ Z

( j500) || ( j500)= − = ∞ = open circuit Since both LC pair parallel combination becomes open then the circuit can be redrawn as

→ 1Ω10sin sin1000tI sin sin1000t

4 1 5= =

+ +→ So peak value of 1ΩI 1A=

Q.13 (14.14) Admittance becomes pure conductance means the imaginary part of Y must be zero which imply resonance condition.


Let first get Y expression in terms of L, C then by equalizing imaginary part we will get the answer.

1 1Y 1R j L RC

ωω

= ++ −

22 22

jRR j L CR ( L) 1R

C

ω ωω

ω

+−= +

+ +

eqIm g Y 0 ⇒ =

22 22

1L C

R ( L) 1RC

ω ωω

ω

=+ +

⇒

⇒ Cross multiplying

( ) ( )2

22 21 1 1L R L R LC C C

ω ω ωω ω ω

+ = +

2 1R LC

ωω

⇒ − 1 1L 0C C

ωω ω

= − =

2 L 1R L 0C C

ωω

⇒ − − = Now by looking into above equation we can say that if

2 LRC

− then it will have no

depending on frequency for

resonance ⇒ 2 LRC

=

So

6

L 0.02R 10 2 14.14C 100 10

Ω−= = = =×

Q.14 (3.04)

LI 0 2ph= ⇒ = ∞

L Cph

L C

jωL -jX X 3 ωcZ = = jωL -jX +X -

3 ωc

ωL 1b ωc

=

3

331410 10 C−

=× ×

C=3.04mF

Q.15 (b) Given: (i) Number of nodes = 8 (ii) Number of independent loop = 5 In graph theory, number of independent loop represents number of links b n 1= − +

Hence,

b n 1 5− + =

b 8 1 5− + =

b 12=

Number of branch = 12

Hence, the correct option is (b).


Q.1 The power supplied by the dc voltage source in the circuit shown below is

a) 0W b) 1.0Wc) 2.5W d) 3.0W

[GATE-2008]

Q.2 The current I supplied by the dc voltage source in the circuit shown below is

a) 0A b) 0.5Ac)1A d) 2A

[GATE-2008]


a) 2 A1 j+

b) 1 A1 j−+

c) 1 A1 j+

d) 0A

[GATE-2012]

Q.4 The source network S is connected to the load network Las shown by dashed lines.

The power transferred from S to L would be maximum when RL is a) 0Ω b) 0.6Ωc) 0.8Ω d) 2Ω

[GATE-2009]

Q.5 In the dc circuit shown in the adjoining figure, the node voltage

2V at steady state is

a) 0V b) 1Vc) 2V d) 3V

[GATE-2010]

Q.6 A 100Ω , 1W resistor and a 800Ω , 2 W resistor are connected in series. The maximum dc voltage that can be applied continuously to the series circuit without exceeding the power limit of any of the resistors is a) 90V b) 50Vc) 45V d) 40V

[GATE-2010]

Q.7 The impedance looking into nodes 1 and 2 in the given circuit is

GATE QUESTIONS(IN)


a) 50Ω b)100Ωc) 5kΩ d)10.1kΩ

[GATE-2012]

Q.8 If A BV V 6V,− = then C DV V− is

a) 5V− b) 2Vc) 3V d) 6V

[GATE-2012]

Q.9 Assuming both the voltage sources are in phase the value of R for which maximum power is transferred from circuit A to circuit B is

a) 0.8Ω b) 1.4Ωc) 2Ω d) 2.8Ω

[GATE-2012]

Q.10 A source ( )Sv t Vcos100πt= has aninternal impedance of (4 j3)Ω+ . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in Ω should be a)3 b)4 c)5 d)7

[GATE-2012]

Q.11 The transfer function 2

1

V (S)V (S)

of the

circuit shown below is

a) 0.5s 1s 1

++

b) 3s 6s 2++

c) s 2s 1++

d) s 1s 2++

[GATE-2013]

Q.12 Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection are scaled by a factor, the elements of the corresponding star equivalent will be scaled by a factor of

a) 2k b) kc) 1/ k d) k

[GATE-2013]

Q.13 The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage

WX1V 100V= is applied across WX to get an open circuit voltage YZ1Vacross YZ .Next, an ac voltage

YZ2V 100V= is applied across YZ to get an open circuit voltage WX2Vacross WX. Then, YZ1 WX1V / V ,

WX2 YZ2V / V are respectively.

a)125 /100and80 /100b100 /100and80 /100 ) c)100 /100and100 /100d) 80 /100and80 /100

[GATE-2013]

Q.14 In the circuit shown below, if the source voltage SV 100 53.13° V= ∠


then The venin’s equivalent voltage in volts as seen by the load resistance LR is

a)100 90°∠ b) 800 0°∠c)800 90°∠ d)100 60°∠

[GATE-2013]

Q.15 Time domain expressions for the voltage v1(t) and v2 are given as v1(t) =Vm sin o(10t 130 )− &

( ) o2 mv t V cos(10t 130 )− Which one

of the following statements is TRUE? a) ( )1 2v t leadv by130°

( )1 2b)v t lagsv (t)by130°

( ) ( )1 2c)v t lagsv t by 130°−

d) ( ) ( )1 2v t lagsv t by 130°− [GATE-2014]

Q.16 The circuit shown in the figure contains a dependent current source between A and B terminals. The Thevenin’s equivalent resistance in kΩ between the terminals C and D is ___________.

[GATE-2014]

Q.17 A load resistor RL is connected to a battery of voltage E with internal resistance Ri through a resistance RS a shown in the figure. For fixed values of RL and Ri, the value of RS (≥ 0) for maximum power transfer to RL is.

a) 0 b) RL - Ri

c) RL d) RL + Ri

[GATE-2015]

Q.18 Consider the circuits shown in the figure. The magnitude of the ratio of the currents, i.e., |I1/I2|, is ________.

[GATE-2015]

Q.19 The current in amperes through the resistor R in the circuit shown in the figure is_____ A.

[GATE-2015]

Q.20 Three currents i1, i2 and i3 meet at a node as shown in the figure below. If i1 =3cos(ω t)ampere, i2 = 4sin (ω t) ampere and i3= I3 cos (ω t+𝜃𝜃 )ampere, the value of I3 in ampere is __________.

[GATE-2016]

Q.21 The current IX in the circuit given below in milliampere is ____________.

[GATE-2016]


Q.22 A circuit consisting of dependent and independent source is shown in the figure. If the voltage at Node-1 is -1 V, then the voltage at Node-2 is _____________V.

[GATE-2017]

Q.23 In the given circuit, the mesh currents 1 2 3,I I and I are

a) 1 2 31 , 2 3I A I A and I A= = =

b) 1 2 32 , 3 4I A I A and I A= = =

c) 1 2 33 , 4 5I A I A and I A= = =

d) 1 2 34 , 5 6I A I A and I A= = = [GATE-2018]

Q.24 In the figure, an RLC load is supplied by a 230 V, 50 Hz single phase source. The magnitude of the reactive power (in VAr) supplied by the source is ____.

[GATE-2018]

Q.25 The Thevenin equivalent circuit representation across terminals p-

q of the circuit shown in the figure is

a) 1 V source in series with150kΩ

b) 1 V source in parallel with100kΩ

c) 2 V source in series with150kΩ

d) 2 V source in parallel with200kΩ

[GATE-2018]

Q.26 A series R-C circuit is excited by a 1 0V∠ sinusoidal ac voltage source. The locus diagram of the phasor current ( )AI x iy= + , when C is

varied, while keeping R fixed, is

a)


b)

c)

d)

[GATE-2018]

Q.27 In the given circuit, superposition is applied. When V2 is set to 0V, the current is –6A. When V1 is set to 0 V, the current I1 is +6A. Current I3 (in A) when both sources are applied will be (up to two decimal places) _____.

[GATE-2018]


Q.1 (d) The given circuit in Fig.1 is simplified as shown in Fig.2 and Fig. 3

From Fig.3, I 1= A, power supplied by 3 V d.c source

1P V I 3 1 3W= = = × =

Q.2 (a)

The circuit is shown in Fig . Voltage across 1 1V=Ω

1I 1A∴ = Applying KCL at node ,P

1I 1 I+ = Or I 1 1,+ = I 0∴ =



L1 1I 1 0

1 j 1 j= × ∠ =

+ +

Q.4 (c) The circuit is shown in Fig.

L

7I2 R

=+

Power supplied by the 10V source,

SL L

10 7 70P2 R 2 R

×= =

+ +Power dissipated in 2Ω resistance,

( )2

2 2L

98P I 22 R

= × =+

Power transferred to the network,

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (d) (a) (c) (c) (b) (c) (a) (a) (a) (c) (d) (b) (b) (c) 15 16 17 18 19 20 21 22 23 24 25 26 27 (a) 20 (a) 1 1 5 10 2 (a) 67.36 (c) (a) 1

ANSWER KEY:

EXPLANATIONS


( )( )t L S 2 2

L L

70 98P R P P2 R 2 R

= − = −+ +

tP would be maximum , if L

dPt 0dR

=

( ) ( )2 3L L

70 298 02 R 2 R

−− − × =

+ +

LL

196 196 9870,2 R2 R 70 35

= + = =+

L98 28R 2 0.835 35

= − = = Ω

Q.5 (b) The given circuit is shown in Fig. 1 At steady state i.e., as t →∞ , capacitor behaves as open circuit. The circuit at steady state is shown in Fig .2

Apply voltage division across 1 2R 2 and R 1= =Ω Ω

Node voltage 2

11 2

R 1V 9 9R R 3

= × = ×+

3V= 1V is divided between

1 2C 10μF and C 20μF= = Apply, again voltage division across

1C and 2C

12

1 2

C 10V 3 3 1VC C 30

= × = × =+

Q.6 (c) Resistor1: 100 ,1WΩ Resistor1: 800 ,2WΩ Maximum current that resistor can withstand

11 1I A

100 10= =

Similarly 22 1I A

800 20= =

If these two resistors are connected in series

Then maximum value of 1I20

=

V 1(100 800)∴ = +

( )1 900 45Volts20

= =

Q.7 (a)

After connecting voltage source of V

( )( ) ( )1 2 b b bV V 10K i 100 I 99i i ;= ⇒ − = + +

b b10000i 100I 100 i− = + ×

b100I 10000i= +

b b20000i 100I i= ⇒−

100 II20000 200− − = =

[ ]b bV 100 I 99i i= + +

I100 I 100200

− = + 50I=

thV 50IR 50I I

= = = Ω

Q.8 (a)



A BAB DC

V VI 3A I2−

= = =


C DV V 2 3 0, V V 5V

1−

+ + = − = −

Q.9 (a) Power transferred from circuit A to circuit

7 6 10RB VIR 2 R 2

+ = = + +

( )42 70R

R 2+

=+

10 3 7I2 R 2 R−

= =+ +

7RV 3 IR 32 R

= + = ++

6 10R2 R+ = +

( ) ( ) ( ) ( )( )

2

4R 2 70 42 70R 2 R 2dP

dR R 2

+ − + +=

+

( ) ( ) ( )270 R 2 42 70R 2 R 2+ = + +

( )5 R 2 2(3 5R)⇒ + = +5R 10 6 10R⇒ + = +4 5R⇒ = R 0.8Ω⇒ =

Q.10 (c) 2 2

L thR Z 4 3 5= = + = Ω

Q.11 (d)

2

1

V ?V

=

( )( )

1 22 1

1 1 2 11 2

1R RC S 1 CV C S1V RC S C C1R C S C S

+ += =

+ + +

Substituting the values we get 2

1

V S 1V S 2

+=

+


2a b

Stara b c

R R kRR R R 3k

+= =

+ +

StarR k∝

Q.13 (b)

W 1 2 W 1V 100 V turns ratio V 125× ×= ⇒ = × =

YZ1 2V 0.8 V 100V= × = When YZ2V 100V=

Thevenin’s circuit seen by 2-2’ will be as follows

thV 100V= And thR 0.2 || 0.8= negligible

'2 2V 100V−∴ =


W 2V ×∴

'2 2100V 1V turns ratio

1.25−×

= × =

80V=

Q.14 (c) TH L1V 10V=

418c

L1V 100 53.13V tan j4

3 j4 3∠

− = = − ×

+

L1V 80 90°= ∠

THV 800 90°= ∠

Q.15 (a) Given ( )1 mv t V sin(10t 130°= − )

( )2 mv t V cos(10t 10°= − )-v2

( )1 mv t V cos(10t 130° 90°= − − )

( )1 mv t V cos(10t 220°= − )- 1v𝜙𝜙 →+ ve in A.C.W 𝜙𝜙→+ ve in C.W

( ) ( )1 2v t leadsv t by130°

Q.16 (20) When dependent source is present; to find RTh ,VTh &ISC are required. VTh (= VO.C ) :

By Nodal Analysis, x x

3 4

V 10 V 05*10 10−

− =

20xV Volts⇒ =i.e VTh =( VO.C) =20 Votls

s.cI

s.c10I 1mA

10K= ⇒

ThTh 3

S.C

V 20R 20kI 1 10−∴ = ⇒ ⇒

×Ω

Q.17 (a) When load is constant, we should see for what value of resistance current will be maximum and .

LRP max=

Q.18 (1) By reciprocity theorem, I1 = I2

So 1

2

II

1=

Q.19 (1) Using supermesh analysis Mesh 1, 3 form super mesh

1 2 32I I 2I 0− + =

1 2 321 I 2I 0⇒ − + = … (1)Writing KCL at Q

1 3I I 1− = − ….(2) Writing KCL on mesh 2

2 1 32I I I 1− + = ….(3) Solution Equipment 1,2,3

1 2 3I 0, I I 1A= = = Current through R is 3I 1A= Note: Strictly saying it is an ambiguous question as the direction of current is not mentioned, so it could be –1A as well.

Q.20 (5) ( ) ( ) ( )1 2 3By KCL i t i t i t= +

( ) ( ) ( )3 1 2 i t i t – i t=

By phasor I3 = 1In

- 2In

[ ]3 0 – [4 90 553.13 = < < − =

( )3 i t 5cos(t 53.13)= +So by comparison I3 = 5.

Q.21 (10)


By kVL, 1 =[100( Ix -10)+ 100Ix] 103 1000= 200Ix- 1000 Ix= 10mA

Q.22 2 By KCL at node 2

12 R 122

V 4V VV2I ..(i)1/ 3 0.5

+ −= +

R1V 1....(ii)= −

11 R 22

V 4V VI ....(iii)

0.5− −

=

Substituting equation (ii) and (iii) in equation (i)

2V 2V=

Q.23 (a)

Writing the KVL equations, In outer most loop through 5V

1 32 5 0− + − =I I

1 32 5.....( )⇒ + =I I iIn the loop containing current source

1 1 2 3 2 32 6( ) 1( ) 1( ) 0− − − − − − =I I I I I I

1 1 2 3 2 32 6 6 0⇒ − − + − + − =I I I I I I

1 2 38 7 2 0..............( )⇒ − + − =I I I ii By writing KCL at node X, we have

3 1 2= +I I

3 1 2........................( )⇒ − =I I iiiBy solving the above equations, or by verifying each options to satisfy the above equations, we have

1 21 , 2= =I A I A

3 3=and I A

Q.24 67.36

1230 0

162.6 162.6∠ °

=+

Ij

230 0 1 45230 45

∠ °= = ∠− °

∠ °

2230 0

230 90∠ °

=∠− °

I

1 90= ∠ °

1 2= +I I I1 45 1 90= ∠− °+ ∠ °

1 12 2

= − +j j

1 112 2

= + −

j

1 0.2932

= + j

0.765 22.51= ∠ All the given information are in the RMS form by default Q | | | | sin[ ]= −θ θsource source source V IV I

230 0.765 sin(0-22.51)= × ×= 67.36VAR

Q.25 (c) Thevenin voltage is the open circuit voltage across the defined terminal PQ


100 4 2V100 100thV = × = +

Thevenin resistance is obtained by making all the independent source value as 0V, i.e., independent ideal voltage source (4V in this case) should be replaced by short circuit.

[ ]100 100 100 1502

th

th

R kV V

= + = Ω

=

P

Q.26 (a)

We want the locus of I, so its expression is

1 0I 1∠

=+

ωR

j C

1

22

1 1tan1

− = ∠ +

ω

ω

RCR

C

In the above equation C is variable

When C = 0, I 0 90°= ∠

When 1C = , I 0°∞ = ∠R

So the locus diagram should start with 900 axis with magnitude 0, and should end with 00 axis with magnitude 1/R.

Q.27 1

2 2Case-1: 0 V , 6 AV I= = −


( )2 6

1 63

2 612 2A

6 6

V VVI

Ω Ω

Ω

= Ω Ω

−= = = −

Q P

1 1Case-2: 0V , 6AV I= =

( )

( ) ( )

6 3

11 63

1 113 3 3

2 618 3A

6 6By superposition, by referring above 2 circuits we can say

2 3 1A

Ω Ω

Ω

= Ω Ω

= = =

= + = − + − =

V VVI

I I I


3.1 INTRODUCTION

In chapters 1 and 2 (Basics and Theorems) we have considered dc resistive networks (networks with dc sources and resistances) in which voltages and currents were independent of time (constant w .r .t time). Whenever inductor (L) and capacitor (c) are included in the circuit the analysis which was done for resistive circuits is invalid. The voltage and current relationship for inductor and capacitor are given below: Inductor:

( ) ( ) ( ) ( ) ( )t

LL L L L

0

di t 1V t L i t V t dt i 0dt L

= ⇒ = +∫Capacitor:

( ) ( ) ( ) ( ) ( )t

cc c c c

0

dV t 1i t C V t i t dt V 0dt c

= ⇒ = +∫From the above equations we can observe that the voltage and current equations have differential terms which make the analysis complicated. e.g.: Let’s consider a simple example.

Resistive Network: - Network with inductor and capacitor:-

1 1 1 1 2KVL V I R I R⇒ = +

( )t

22 2

0

di (t) 1KVL v L i t dtdt c

⇒ = + ∫

Voltage and current are constant Voltage and current are time dependent. From the above example it is very much clear that it is very easy to solve the equation for purely resistive circuit as they are simple linear equations whereas the circuits containing inductors and capacitors have differential terms and voltage and currents will be a function of time. Hence, they have to be solved as differential equations.

3.2 STEADY STATE AND TRANSIENT RESPONSE

In a network containing energy storing elements (i.e. L or c) when the excitation (or input) is changed, the current and voltages change from one state to another state. The behavior of the voltage or current when it is changed from one state to another is called the transient state. e.g:- Consider a circuit which contains aswitch.

In the above circuit S is a switch and the arrow indicates that at time t=0 the switch is closed. When the switch is closed, suddenly a huge amount energy flows from source to the circuit. This state of the circuit just after the switch is closed is called transient state. The resulting voltages and currents change w.r.t. time and they are called transient response. A circuit having constant sources (i.e. dc sources or sources with same frequency, ω) connected for long time is said to be in steady state. Current and voltage do not change with time in steady state. Ideally after infinite amount of time and practically

3 TRANSIENTS


after 5 time constants circuit enter into steady state.

3.2.1 SOLUTIONS OF DIFFERENTIAL EQUATION

The complete solutions of a differential equation is given by Complete solution = complementary function + Particular integral (or) = Natural response + Forced response (or) = Transient response + steady state response

Example: Find the solution of the differential equation given below

2

2d y dy5 4y 1

dtdt+ + =

Solution: Complementary function ( )2D 5D 4 y 0+ + =

2m 5m 4 0+ + = ( )( )m 4 m 1 0+ + =

m 4, 1= − − Complementary function,

4t t1 2y c e c e− −= +

Particular integral ot

2 21 1y 1 e

D 5D 4 D 5D 4= =

+ + + +1 (D a 0)4

= = =

Complete solution 4t t1 2

1c e c e4

− −= + +

Transient Steady state Response Response

3.2.2 TERMINOLOGY IN TRANSIENTS

Consider the same circuit with a switch as considered earlier

t 0= ⇒Time of closing the switch t 0−= ⇒Time just before the switch is closed (steady state) t 0+= ⇒Time just after closing the switch (transient state) t →∞⇒ Infinite amount of time after closing the switch (steady state)

3.2.3 THE BEHAVIOR OF INDUCTOR AND CAPACITOR AT -t=0 , AT +t=0 AND AS t →∞

3.2.4 THE INDUCTOR CURRENT AND CAPACITOR VOLTAGES AT -t=0 AND AT

+t=0 INSTANTS

L: ( ) ( )t

L L1i t V t dtL −∞

= ∫

( ) ( )0 t

L L0

1 1V t dt V t dtL L

−

−−∞

= +∫ ∫


( ) ( )t

L L0

1i 0 V t dtL −

−= + ∫At t 0+= above equation becomes

( ) ( ) ( )0

L L L0

1i 0 i 0 V t dtL

+

−

+ −= + ∫

( ) ( )L Li 0 i 0+ −=

( 0+Q and 0− are nearly equal) Hence the inductor current can’t change instantaneously from 0− to 0+

C: ( )t

c c1V (t) i t dtc −∞

= ∫

( ) ( )0 t

c c0

1 1i t dt i t dtc c

−

−−∞

= +∫ ∫

( ) ( )t

c c0

1V 0 i t dtc −

−= + ∫At t 0+= above equation can be written as

( ) ( ) ( )t

c c c0

1V 0 V 0 i t dtc −

+ −= + ∫

( ) ( )c cV 0 V 0+ −=

( 0 and 0− +Q are nearly same) Hence, the capacitor voltage can’t change instantaneously from 0 to 0− +

3.2.4 THE EQUIVALENT CIRCUITS

L: Inductor with initial current I0

In the figure shown above the inductor L constrains an initial current Io. The initial current can be replaced by a current source in parallel to the inductor. C: Capacitor with initial voltage V0

In the figure shown above the capacitor C has an initial voltage𝑉𝑉0. The initial voltage can be replaced by a voltage source in series with the capacitor.

Note: Whatever we have discussed so far forms the basis to solve the transient problems. Once the student is well understood with these basics the transient problems are very easy to solve.

3.3 DC TRANSIENTS

In dc transients we would have dc sources. Whenever there is sudden application of voltages or currents on the circuit (due to opening or closing the switch) circuit would be in transient state. We will consider the following cases in dc transients 1. Source free circuits

• Source free RL circuit• Source free RC circuit• Source free RLC circuit

2. RL and RC circuits with sources.3. RLC circuits with sources, only initial

condition (t 0 )+= and final conditiont →∞will be found.

4. RLC circuits with sources, LaplaceTransform Approach of solving thetransient problems.

3.3.1 SOURCE FREE CIRCUITS

In these circuits when the switching action is performed (i.e. either switch is open or closed) the source will be disconnected from the circuit. Before switching action the memory elements (i.e. L or c) will store the maximum energy. At -t=0 the circuit will be in steady state and inductor will be short circuited, capacitor will be open circuited. After the switching action is performed at +t=0 the circuit will be in transient state and sources wouldn’t be connected to the circuit. Under such condition the energy stored by the inductor


and capacitor would be dissipated across the resistors present in the circuit.

a) Source free RL circuit:Consider a simple RL circuit as shownin figure

Our aim is to find the current across the inductor L(i (t)) and voltage across the inductor L(V (t)) after switch is closed.

Step1: Finding the initial current -

L(i (0 )) across the inductor To find the initial current across inductor consider the circuit at -t=0 . The switch will remain closed at t=0 circuit is in steady state hence inductor will be short circuited.

( )L si 0 I− = Since the current always

chooses law resistance path, We have already proved that ( ) ( )L Li 0 i 0− +=

( ) ( )L s Li 0 I i 0− +∴ = =

The energy in the inductor is given by

( ) ( )2 2L L s

1 1E 0 Li 0 LI2 2

− += =

Hence the inductor has stored the energy at t 0−=

Step2: Finding the current ( )Li t fort 0≥

Applying KVL in loop (I), we get L

LdiRi L 0dt

− − =

LL

di R i 0dt L

+ =

LRD i 0L

+ = Solving the above differential equation, we get

( )R tL

Li t ce−

=

At L st 0, i I= =

∴ 0sI ce=

sc I=

( )R tL

L si t I e−

∴ =for t 0≥

Lτ =R

sec is called the time constant

of RL circuits

( )t

L si t I e−τ∴ = for t 0≥

( ) LL

di (t)V t Ldt

=R tLe

sRLI e

L

− − =

tR

LsRI e

−= − for t 0≥ Graphically the current and voltage can be shown as


At ( )t 1

L s s st , i I e I e 0.368I− −τ= τ τ = = =

at ( ) 2L s st 2 , i 2 I e 0.135I−= τ τ = =

at ( ) 3L s st 3 , i 3 I e 0.0498I−= τ τ = =

at ( ) 4L s st 4 , i 4 I e 0.0183I−= τ τ = =

at ( ) 5L s st 5 , i 5 I e 0.0067I−= τ τ = =

After 5 time constants the current decays by 99% of the initial current and hence we say that transient exist for5τ . After 5τ the circuit enters into steady state.

b) Source free RC circuit:

Consider a simple RC circuit as shown in figure

Again we have to find ci (t) and cV (t)(t)after switch is closed.

Step1: Finding the initial voltage

c(V (0 ))− across the capacitor Capacitor will be open circuited as the circuit is in steady state

( ) ( )VSv 0 v 0c c=− +=

2c c

1E (0 ) cv (0 )2

− −=

1 2cVs2=

Here, the capacitor has stored the energy at t 0−=

Step2: Find ( )c cv t andi (t) for t 0≥

For t 0≥ we can’t short circuit or open circuit the capacity because t 0≥includes t 0+= (transient state) as well as t →∞ (steady state)

By KCL⇒ R Ci i 0+ =

c cV dVc 0R dt

+ =

c1D V 0

RC + =

( )tRc

cV t ke−

= for t 0≥ Using initial condition

( )c sV 0 V

( )c sV 0 k V= =

( )tRc

c sV t V e−

∴ =

RCsecsecττ = is called the time constant of RC circuits

( )t

c sV t V e−τ= for t 0≥

( )tc

c sdV (t) 1i t c cV e

dtτ

τ

− − = =

( )ts

cVi t eR

−τ= − for t 0≥


c) Source free RLC circuit:

Let understand this topic with the help of an example.

Here, we have to determine ( )Li t and

( )cV t for t 0≥

Step1: Finding initial conditions

Li (0 )− and cV (0 )−

Note: Number of initial condition to be found = Number of inductors and capacitors present in the network. In this problem there is one inductor and one capacitor hence two initial conditions are required The input ( )20u t− is as shown below

( )20;&t 0

20u t0;&t 0

≤− = >

At t 0−= the above circuit can be drawn as

( )L L20i 0 20mA i (0 )1k

− += = =

( )c cV 0 20v V (0 )− += =

Step2: Finding Li (t) and cV (t) for t 0≥

In the above circuit the RL and RC are two independent circuits since there is a short circuit in parallel. The circuit is a source free circuit Time constant of RL ckt,

4L

R 1K 10 secL 0.1

τ = = =

Time c constant of RC ckt, 3 9

c RC 10 10 200 10−τ = = × × ×32 10−= × sec

We know that for a source free RL circuit

( ) ( ) 4L

t t10

L Li t i 0 e 20e (mA)− −τ−= =

Similarly for source free RC ckt

( ) ( ) 3c

t t2x10

c cV t V 0 e 20e −− −τ−= =

500t20e v−=

Note: The procedure explained in the above example is valid only when RL and RC circuit can be separated otherwise we must use Laplace Transform which will be explained in RLC circuits later in this chapter.

3.3.2 RL AND RC CIRCUITS WITH SOURCES

These are the second category of circuits in which source will remain connected after the switching action. In source free circuits we have observed that the current across inductor and voltage across capacitor was exponentially decaying but here they will be rising exponentially.


• In a circuit if there are many resistors,inductors and sources (i.e. RL circuitwith sources) then current acrossinductor is given by

( ) ( ) ( ) ( )t

L L L Li t i i 0 i e τ−+ = ∞ + − ∞

Where, LR

τ = sec is the time constant

of RL circuits • In a circuit if there are many resistors

capacitors and sources (i.e. RC circuitwith sources) then voltage acrosscapacitor is given by

( ) ( ) ( ) ( )t

c c c cV t V V 0 V e τ−+ = ∞ + − ∞

Where, RCτ = sec is the time constantof RC circuit

3.3.3 PROCEDURE TO SOLVE RL AND RC CIRCUITS (WITH SOURCE OR SOURCE FREE)

1) Draw the circuit for t 0−= Find thecurrent across inductors and voltageacross capacitors at t 0−=

2) Draw the circuit for t 0≥

case1: if the source doesn’t affect the inductor current (in RL circuit) or capacitor voltage (in RC circuit) then the circuit is source free. In such case,

Inductor current, ( ) ( ) tL Li t i 0 e

−− τ= for

t 0≥

Where, L secR

τ = for RL circuit.

Capacitor voltage ( ) ( ) tc cV t V 0 e

−− τ= for

t 0≥ Where, RCsecτ = for RC circuit.

case2:If the source affects the circuit then it is a circuit with source and in such circuits, Inductors current

( ) ( ) ( ) ( )t

L L L Li t i i 0 i e τ−− = ∞ + − ∞

for t 0≥

Where L secR

τ = for RL circuits

Capacitor voltage

( ) ( ) ( ) ( )t

c c c cV t V V 0 V e−− τ = ∞ + − ∞

for t 0≥ Where, RCsecτ = for RC circuit.

3) Also if some value of current or voltageis desired at t=0+ (transient state) thendraw the circuit at t=0+. Find thedesired value by the application of KCL,KVL, Nodal or Mesh analysis

Example: For the circuit shown in figure, the switch is in position 1 for a long time and it is moved to position 2 at t 0= Determine, ( ) ( )L Li 0 , v 0 ,i (t)+ + for t 0≥

Solution: Step1) Finding initial condition (t 0 )−=

Note: - Inductor is short circuited as the circuit is in steady state at t 0−= Applying current division principle

( ) ( )L L8i 0 5 4A i 0

8 2− += × = =

+

Step 2) Circuit for t 0≥


Step3) since there is no source connected to the inductor for t 0≥ hence it is a source free circuit. The current across inductor in a source free circuit is given by

( ) ( )t

L Li t i 0 e−τ= for t 0≥

L 3 3 secReq 2 12 14

τ = = =+

(Req is obtained by finding equivalent resistance across L)

( )t

14t3314

Li t 4e 4e−

−∴ = = for t 0≥ Also we can find 𝑉𝑉𝐿𝐿(𝑡𝑡) using the relation

( ) LL

di (t)V t Ldt

=

14t3 143 4e

3− − = ×

( )14t

3LV t 56e

−= − for t 0≥

Step 4) Circuit at t 0+=To find the value of v(0 )+ across the 12Ωresistor we must go into t 0+= circuit is in transient state. Inductor will become open circuited.

Current enters in a resistor from positive terminal. Hence ( )v 0 12 4+ = − ×

( )v 0 48v+ = −

Example: Determine the values of ( )cV t and ( )ci tfor t 0≥ for the circuit shown in figure

Solution: The input ( )20u t can beshown as

( )20v, for & t 0

20u t0v, for & t 0

>= <

Step1) Finding initial condition (at t 0 )−=

Since the value of source is at t 0−= hence ( ) ( )c cV 0 0v V 0− += =

Step2) Circuit for t 0≥

Step3) Since the source is connected to the circuit for t 0≥ hence to find the value of cV (t) we need to employ the following formula

( ) ( ) ( ) ( )t

c c c cV t V V 0 V e−τ= ∞ + − ∞ for

t 0≥ We need to find the value of ( )cV ∞ and τto use above formula. Step4) Finding ( )cV ∞As t →∞ circuit is in steady state and capacitor becomes open circuited

Using voltage division principle


( )c2V 20 10v

2 2∞ = × =

+Finding time constant

Reqτ = 𝐶𝐶 ( )2 || 2 1 1= + × ( )1 1 1= + ×2 1 2sec= × =

ckt to find time constant. Voltage source is short circuited. ∴ Substituting in the formula we get

( )t2

cV t 10 [0 10]e−

= + −

( )t2

cV t 10 10e−

= − for t 0≥

( ) ( ) ( )tc 2

c cdV t

i t c i t 5edt

−= ⇒ = for t 0≥

3.3.4 RLC CIRCUITS WITH SOURCES: ONLY INITIAL CONDITION (t 0 )+= AND FINAL CONDITION (AS t →∞ )

• In these circuit we won’t be abledetermine the time constant of thecircuit. Also in these circuits you wouldbe asked to find the initial conditionsat(t 0 )+= where the circuit would be intransient state or the final conditions(as t →∞ ) ,where the circuit would bein steady state.

• These circuit would be purely resistiveas the inductors and capacitors areeither short circuited or open circuitedin transient state t 0+= and steady state(as t →∞ )

Example: Determine the current through the battery at t 0+= and as t →∞ .

Solution: Finding initial condition (at t 0−= ) We need to find ( )Li 0− and ( )cV 0− in this

case

As the source is not connected to the circuit ( ) ( )L Li 0 0A i 0− += =

( ) ( )c cV 0 0V V 0− += =

Circuit at +t=0

The current through the battery is 10i(0 ) A 10A1

+ = =

Circuit at t = ∞

The current through the battery is 10i( ) A 10A1

∞ = =

3.3.5 RLC CIRCUITS WITH SOURCES USING LAPLACE TRANSFORM APPROACH


This case is similar to the previous case as here also we will consider RLC circuit and we won’t be able to find their time constant. But here we would go one step further and determine the values of current and voltages for t 0≥ which is possible only using Laplace Transform Approach (LTA). Let’s consider a problem

Example: The switch S in the circuit shown in figure below is open for a long time. At t=0, S is closed connecting the voltage source to the circuit for t 0≥ , obtain the voltage cVacross the capacitor.

Solution:- Finding initial conditions (at t 0−= )

As the switch is not connected to the circuit hence ( ) ( )L Li 0 0A i 0− += =

( ) ( )c cV 0 0V V 0− += =

for t 0≥ ,

Now, here we can’t determine the time constant and we have to find the voltage across capacitor for t 0≥ so we convert the above network into s- domain

Nodal at cV (s)⇒

( ) ( ) ( )c c c

5V s V s V s5 022s 4 1 s

−+ + =

+

( ) ( ) ( ) ( )c c c5V s 2s 4 V s s(s 2)V s5 0

(2s 4)

− + + + +=

+

( )2c

51 2s 4 s 2s V ss

+ + + + =

( )c 2 25 5V s

(s 4s 5) s(s 4s 5)= =

+ + + +

( )c 2A Bs CV sS s 4s 5

+= +

+ +

( )c s 0 s 025A sV s | s | 1

s(s 4s 5)= == = =+ +

( )c 21 Bs CV sS s 4s 5

+= +

+ +( )2

2 2s 45 5 Bs c s5

s(s 45 5) s(s 4s 5)+ + + +

=+ + + +

( ) ( )25 s 1 B 4 c s 5= + + + +B 1 0 B 1+ = ⇒ = − 4 C 0 c 4+ = ⇒ =

( )c 21 s 4V sS s 4s 5

− −= +

+ +

21 s 4S s 4s 5

+= −

+ +

( )21 s 4S s 2 1

+= −

+ +

( ) ( )2 21 s 4 2S s 2 1 s 2 1

+= − −

+ + + +

( ) 2t 2tcV t 1 e cos cos t e sin sin t− −= − − for

t 0≥ ( ) 2t

cV t 1 e (cos cos t sin sin t)−= − + for t 0≥


Q.1 When the angular frequency 𝜔𝜔 in the figure is varied from 0 to ∞ , the locus of the current phasor 2 I is given by

a)

b)

c)

d)

[GATE-2001]

Q.2 A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q=100.If each of R, L and C is doubled from its original value, the new Q of the circuit is

a) 25 b) 50c) 100 d) 200

[GATE-2003]

Q.3 An input voltage ( )v t 10 5= cos

(t+10° ( )10 5 cos 2t 10° V+ + appliedto a series combination of resistance R 1Ω= and an inductance L=1H. . The resulting steady state current i(t) in ampere is a) ( ) 110cos t 55° 10cos(2t 10° tan 2)−+ + + +

b) ( ) 310cos t 55° 10 cos(2t 55)2

+ + +

c) ( ) 110cos t 35° 10cos(2t 10° tan 2)−− + + −

d) ( ) 310cos t 35 10 cos(2t 35°)2

− + −

[GATE-2003]

Q.4 The circuit shown in the figure, with 1 1R Ω,L H,C 3F3 4

= = = has input

voltage ( )V t sin 2t= .The resultingcurrent i(t) is

a) 5sin(2t 53.1°)+ b) 5sin(2t 53.1°)−c) 25sin(2t 53.1°)+ d) 25sin(2t 53.1°)−

[GATE-2004]

Q.5 For the circuit shown in the figure, the time constant RC=1ms . The input voltage is ( ) 3

iV t 2 sin10 .= The output voltage ( )0V t is equal to

GATE QUESTIONS(EC)


a) 3sin(10 t 45°)− b) 3sin(10 t 45°)+

c) 3sin(10 t 53°)− d) 3sin(10 t 53°)+[GATE-2004]

Q.6 Consider the following statements S1 & S2

S1: At the resonant frequency the impedance of a series R-LC circuit is zero.

S2: In a parallel G-L-C circuit, increasing the conductance G results in increase in its Q factor.

Which one of the following is correct? a) 1S is FALSE and 2S is TRUE b) Both 1S and 2S are TRUE c) 1S is TRUE and 2S is FALSE d) Both 1S and 2S are FALSE

[GATE-2004]

Q.7 The condition on R, L and C such that the step response y(t) in the figure has no oscillations, is

a) 1 LR2 C

≥ b) LRC

≥

c) LR 2C

≥ d) 1RLC

=

[GATE-2005]

Q.8 In a series RLC circuit,

R 2k ,L 1H= Ω = and 1C F400

= µ . The

resonant frequency is

a) 42 10 Hz× b) 41 10 Hz×π

c) 410 Hz d) 42 10 Hzπ× [GATE-2005]

Q.9 For the circuit shown in the figure, the instantaneous current 1i (t) is

a) 10 3 902

∠ °A b) 10 3 902

∠− °A

c) 5 60 A∠ ° d) 5 60 A∠− °[GATE-2005]

Q.10 In the AC network shown in the figure, the phasor voltage ABV (in Volts ) is

a) 0 b) 5 30°∠c)12.5 30°∠ d)17 30°∠

[GATE-2007]

Q.11 An AC source of RMS voltage 20V with internal impedance

SZ (1 2j)Ω= + feeds a load of impedance LZ =(7+4j)Ω in the figure below. The reactive power consumed by the load is

a) 8VAR b) 28VARc) 16VAR d) 32 VAR

[GATE-2009]

Q.12 For the parallel RLC circuit, which one of the following statements is NOT correct? a) The bandwidth of the circuit

decreases if R is increasedb) The bandwidth of the circuit

remains same if L is increasedc) At resonance, input impedance is

a real quantity


d) At resonance, the magnitude ofinput impedance attains itsminimum value

[GATE-2010]

Q.13 The current I in the circuit shown is

a) j1A− b) j1Ac) 0A d) 20A

[GATE-2010]

Q.14 The circuit shown below is driven by a sinusoidal input

i pV V cos(t / RC)= . The steady state output V0 is

a) ( )pV / 3 cos(t / RC)

b) ( )pV / 3 sin(t / RC)

c) ( )pV / 2 cos(t / RC)

d) ( )pV / 2 sin(t / RC)

[GATE-2011]

Q.15 Two magnetically uncoupled inductive coils have Q factors 1q and

2q at the chosen operating frequency. Their respective resistances are 1R and 2R . When connected in series, their effective Q factor at the same operating frequency is a) 1 2q q+b) 1 2(1/ q ) (1/ q )+

c) ( ) ( )1 1 2 2 1 2q R q R / R R+ +

d) ( ) ( )1 2 2 1 1 2q R q R / R R+ +[GATE-2013]

Q.16 A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is a) (18 + j 1.5) kVAb)(18 -j 1.5) kVA c) (20 + j 1.5) kVAd)(20 - j 1.5) kVA

[GATE-2014]

Q.17 A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is ___.

[GATE-2014]

Q.18 A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage Vs, the resistance R, the capacitance C, and the current i(t) is given below:

Ve = Ri(t)+ ( )t

0

1 i u duC ∫ Which one of

the following represents the current f(t)? a) b)

c) d)

[GATE-2014]


Q.19 The steady state output of the circuit shown in the figure is given by ( )y t A( ) sin( tω ω)ω )(= + φ .If the amplitude | A( ) | 5ω 0.2= ’ then thefrequency ω is

a) 13R C

b) 23R C

c) 1R C

d) 2R C

[GATE-2014]

Q.20 In the circuit shown the average value of the voltage abV (in Volts) in steady state condition is _______.

Q.21 An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is a) 1

2 LCπb) 1

2 LCπ21 CR

L−

c) 12 LCπ 21 L

R C− d) 1

2 LCπ21 CR

L −

[GATE-2015]

Q.22 The figure shows an RLC circuit with a sinusoidal current source.

At resonance, the ratio L RI / I ,i.e., the ratio of the magnitudes of the

inductor current phasor and the resistor current phasor, is ___________.

[GATE-2016]

Q.23 In the RLC circuit shown in the figure, the input voltage is given by ( ) ( ) ( )t 2cos 200t 4sin 500tiv = +

The output voltage ( )0V t is

a) ( ) ( )cos 200t 2sin 500t+

b) ( ) ( )2cos 200t 4sin 500t+

c) ( ) ( )sin 200t 2cos 500t+

d) ( ) ( )2sin 200t 4cos 500t+

[GATE-2016]

Q.24 In the circuit shown, the positive angular frequency 𝜔𝜔 (in radians per second) at which the magnitude of the phase difference between the voltage V1 and V2 equals 𝜋𝜋

4 radians,

is__________.

[GATE-2017]

Q.25 The figure shows an RLC circuit excited by the sinusoidal voltage 100cos(3t) Volts, where t is in seconds. The ratio 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑉𝑉2

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑉𝑉1

is______________.


[GATE-2017]

Q.26 In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V. The ratio𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑣𝑣𝑜𝑜𝑎𝑎𝑎𝑎𝑎𝑎𝑣𝑣𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎𝑎𝑎 𝑎𝑎ℎ𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑣𝑣𝑜𝑜𝑎𝑎𝑎𝑎𝑎𝑎𝑣𝑣𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎𝑎𝑎 𝑎𝑎ℎ𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎

is_______.

[GATE-2017]

Q.27 For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is ______.

[GATE-2018]

Q.28 For the circuit given in the figure, the voltage cV (in volts) across the capacitor is

a) ( )1.25 2 sin 5t 0.25− π

b) ( )1.25 2 sin 5t 0.125− π

c) ( )2.5 2 sin 5t 0.25− π

d) ( )2.51.25 2 sin 5t 0.125− π


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (b) (c) (a) (a) (d) (c) (b) (a) (d) (b) (d) (a) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (d) (b) 0.408 (a) (b) 5 (b) 0.316 (b) 1 2.6 0.2 0.316 (c)

ANSWER KEY:


Q.1 (a)

( ) m2 m

22

E cosωt jωCi t E 01 1 jωCRRjωC

= = ∠++

( ) m2 2 2 2 1

2

E 0ωC 90°i t1 ω C R tan ωCR−

∠ ∠∠ =

+ ∠

( ) 1m2 2 2 2

E ωCi t 90 tan ωCR1 ω C R

−= ∠ −+

At ( ) ( ) m2 2

2

Eω 0,i t 0,ω ,i tR

= = = ∞ =

Fig (a) satisfies both conditions.

Q.2 (b) ofQ

BW=

o1f

2π LC=

RBWL

=

2Characteristic equati Rs 1sL L

nC

o = + +

Or 1 LQR C

=

When R, L, C are doubled,' 1Q Q 50

2= =

Q.3 (c)

( ) ( )v t 10 2 cos(t 10°)i tR jωL 1 1j

10 5 cos(2t 10°)1 2j

+= =

+ +

++

+

( )

1

10 2 cos(t 10 )i t2 45

10 5 cos(2t 10 )5 tan 2−

+ °=

∠ °

+ °+

∠( ) ( )

1

i t 10cos t 35

10cos(2t 10 tan 2)−

∴ = − °

+ + −

Q.4 (a) ( ) ( )i t V t .Y=

1

1 1Y V(t) jωcR jωL

= + +

1

1 1Y V(t) jωcR jωL

= + +

sin 2t[3 2j 6 j]= − + sin 2t[3 4j]= +

1 45sin 2t tan3

−= ∠

( )5sin 2t 53.1°= +

Q.5 (a)

( ) ( )0 i

1jωCV t V t1R

jωC

=+

31 2 sin10 t1 jωCR

=+

33 3

1 2 sin10 t1 j 10 10−=+ × ×

( ) 30V t sin(10 t 45°)= −

Q.6 (d) S1: Impedance of series RLC circuit at resonant frequency is minimum

1Z R j ωLωC

= + −

1ωL 0ωC

− =

Z R= (Purely resistive)

2S : CQ RL

=

1 1 CG QR G L

= ⇒ =

G ↑ then Q ↓ if C and L are same

Q.7 (c) Transfer function

EXPLANATIONS


1sC

1R sLsC

=+ +

21

s LC sCR 1=

+ +

2

1Y(s) LC

R 1U(s) s sL LC

=+ +

nR2ξωL

=

n1ωLC

=

Rξ LC2L

=

R Cξ2 L

=

For no oscillations , ξ ≥ 1 R C 12 L

≥

LR 2C

≥

Q.8 (b)

06

1 1f2π LC 12π 1 10

400−

= =× ×

3 410 20 10 Hz2π π×

= =

Q.9 (a) When 5 0°∠ is acting along, ( )1i t 5 0°= − ∠ (as 10 60°∠ is kept

open) When 10 60°∠ is acting alone. ( )1i t 10 60°= ∠ (as 5 0°∠ is kept

open) ( )1i t 10 60° 5 0°= ∠ − ∠5 8.66j 5= + − ( )1i t 8.66j=

( ) 10i t 5 3 90° 3 90°2

= ∠ = ∠

Q.10 (d) ABV current impedance= ×

( ) ( )5 30° (5 3J || 5 3J )= ∠ × − + 345 30°10

= ×∟ 17 30°= ∠

Q.11 (b) The RMS current in the load is given by

S L

20IZ Z

=+

r 208 j6

=+

104 j3

=+

110 3tan5 4∠ − = −

1 32 tan

4∠ − = −

rmsI 2= , reactive power 2rms LI X=

4 4 16VAR= × = Also note that the active power consumed by the load

2rms LI R 4 7= = × =28W

Q.12 (d) This is standard concept of parallel resonant circuit

Q.13 (a)

Q.14 (a) Redrawing the circuit s –domain


( ) ( )1

1R.1 sCV s R I s I(s)1sC RsC

= + + +

( ) ( )11 SCR RV s I s I(s)

sC 1 SCR+

= ++

………………….(i) i pV V cos(t / RC)∴ =

So here, 1ωRC

=

Now,

( ) ( ) ( )i1 jωCR RV S I s I(s)

jωC 1 jωCR+

= ++

Put, 1ωRC

=

So, ( ) ( )i

1 j R RV S I(s)j 1 j

+= + +

( ) ( )i3RV S

1 j=

+…………………..(ii)

( ) ( ) ( )iV SI s 1 j

3R= × +

Now, ( )01V s R || I(s)

sC =

( )0

1R.sCV s I(s)1RsC

⇒ =+

( ) ( ) ( )i0

V SRV s . 1 j1 SCR 3R

⇒ = ++

( ) ( ) ( )i0

V SRV s . 1 j1 j 3R

⇒ = ++

( ) ( )i0

V SV s

3⇒

In time domain,

( )0 i1v t v (t)3

=

( ) p0

V tv t cos3 RC

=

Q.15 (d) When the switch in closed at t=0 Capacitor 1C will discharge and 2Cwill get charge since both 1C and 2C

are ideal and there is no –resistance in the circuit charging and discharging time constant will zero. Thus current will exist like an impulse function.

Q.16 (b)

Load 1:

I

P 10kwcos 0.8 S P jQ 10 j7.5KVA

Q P tan 7.5KVARφφ

= = = − = −= =

Load 2: S=10KVA

Cos 0.8φ = sin QS

φ =

Cos PS

φ =

P P 8kw10

0.8 → ==

Q= 6KVAR S1= P+ jQ= 8 + j6 Complex power delivered by the source is SI + SII=18-j1.5KVA

Q.17 (0.408)

( )( )2

0

1= ∫

T

rmsx x t dtT

2 Tt0 t 2Tx(t)T0 2 t T

≤ ≤=

≤ ≤

( )T 2 T2

2

0 T/2

1 2 .t .dt 0 .dtT T

= + ∫ ∫


T3 2

20

1 4 t.T T 3

=

3

rms 3

4 T 1X . 0.4083T 8 6

= ⇒ ⇒

Q.18 (a) In a series RC circuit, →Initially at t = 0, capacitor charges

with a current of sVR

and in steady

state at t = ∞, capacitor behaves like open circuit and no current flows through the circuit → So the current i(t) represents an exponential decay function

Q.19 (b)

By nodal method,

( )oV 1| 0 V

R 1jωc

−+ ( )

V2

jωc

= 0

o1 j c 1| 0j cR 2 R

V ωω + + =

V= 22 3jωRC+

Y= V 12 2 jω3RC⇒

+

Given |A(2 2 2

1 1ω) |4 4 9R c .ω

= ⇒+

2ω3RC

⇒ =

Q.20 (5)

Q.21 (b) e LRY Y Y= +

( )1Y=jωC+ =j

jωL+R

2 2 2

( )( )

R j LCR L

ωωω

−+

+Placing Imaginary part to zero we get option (B).

Q.22 0.316 At resonance,

L m

R m

I QI= =QI I

For parallel circuits Q = R CL

= 10

6

3

10 1010 10

−

−

××

= 0.316

Q.23 (b) ( )iV t 2cos 200t 4 sin 500t= + , since

there are 2 frequency term output will also have 2 frequency term→ If we take 4sin500t first i.e. W = 500 then on the output section, this parallel LC combination have LCZ = ∞ , SO it is open circuit and Vo = Vi

So w.r.t. 4sin 500t output must be 4sin500t without any change in amplitude and phase, this is satisfied by only option B.

Q.24 1rad/sec

Q.25 2.6


3rad / secω = ( )1Z 4 j3= + Ω

( )2Z 5 j12= − Ω2 2

2 2V i z i 5 12 13 i= = + =2 2

1 1V i z i 4 3 5 i= = + =

2

1

V 13 i 13 2.6V 5 i 5

= = =

Q.26 0.2

Given that, V and I have same phase. So, the circuit is in resonance. At resonance,

c RV QV=

c

R

Amplitudeof V QAmplitudeof V

=

1 L 1 5 0.2R C 5 5

= = =

Q.27 0.316 Given: Switch is closed at t 0=

At t 0−=

Source is disconnected current through the inductor before switching

( ) ( )Li 0 i 0 0− −= =

The find, loop current that is inductor current after t 0= , we can use

( ) ( ) ( ) ( ) t /L L L Li t i i 0 i e , t 0− − τ = ∞ + − ∞ ≥

( )Li 0 0− =

To find ( )Li ∞ :

(Inductor will behave as short circuit)

( )L1i 0.5A

1 1∞ = =

+

( ) ( )L1 1 1 i 0− + + ∞ =

To find τ :

eq

1R

τ = for t 0>

1 0.5sec1 1

τ = =+

( ) ( ) t /0.5Li t 0.5 0 0.5 e−= + −

( ) t /0.5Li t 0.5 1 e− = −

Value of loop current at t 0.5sec=

( ) 0.5/0.5Li 0.5 0.5 1 e− = −

( ) 1Li 0.5 0.5 1 e 0.316A− = − =


Q.28 (c)

Given:

1 2R R 100k= = Ω , ( ) ( )iV t 5sin 5t=

We have to find the voltage across the capacitor.

Network can be further simplified by taking the equivalent of two series resistance as shown. From the given input

( ) ( )iV t 5sin 5t= .

5rad / sec∴ω=

Using voltage division,

( ) ( )cc i

c

ZV t V tR Z

= +

Where iZcj C

=ω

and R 200k=

( )C

ij CV t 5sin 5t iR

j C

ω= + ω

( )C1V t 5sin 5t

1 j RC

= + ω

( )C 3 6

1V t 5sin 5t1 j 5 200x10 1 10−

= + × × × ×

( )C1V t 5sin 5t

1 j1

= +

( )C1V t 5sin 5t

2 / 4 = ∠π

( )C5V t sin 5t

42π = −

( )2.5 2 sin 5t 0.25= − π


Q.1 In the figure 1Z 10 60°,= ∠−

2Z 10= 60°,∠ 3Z 50 53.13°= ∠Thevenin impedance seen from X-Y is

a) 56.66 45°∠ b) 60 30°∠c) 70 30°∠ d) 34.4 65°∠

[GATE-2003]

Q.2 Two ac source feed a common variable resistive load as shown in figure .Under the maximum power transfer condition, the power absorbed by the load resistance RL is

a)2200W b)1250W c)1000W d)625W

[GATE-2003]

Q.3 In the given figure, the Thevenin’s equivalent pair (voltage, impedance), as seen at the terminals P-Q is given by

a) (2V,5Ω) b) (2V,7.5Ω)c) (4V,5Ω) d) (4V,7.5Ω)

[GATE-2005]

Q.4 In the figure the current source is 1 0A,R 1∠ = Ω , the impedances are

cZ j= − Ω , and LZ 2j= Ω . The Thevenin equivalent looking into the circuit across X-Y is

a) ( )2 0V, 1 2j∠ + Ω

b) ( )2 45 V, 1 2j∠ ° − Ω

c) ( )2 45 V, 1 j∠ ° + Ω

d) ( )2 45 V, 1 j∠ ° + Ω

[GATE-2006]

Statements for Linked Answer Questions Q.5 & Q.6

Q.5 For the circuit given above, the Theremin’s voltage across the terminals A and B is a)1.25V b) 0.25Vc)1V d) 0.5V

[GATE-2009]

Q.6 For the circuit given above, the Thevenin’s resistance across the terminals A and B is a) 0.5kΩ b) 0.2kΩc)1kΩ d) 0.11kΩ

[GATE-2009]


GATE QUESTIONS(EE)


a) 50Ω b)100Ωc) 5kΩ d)10.1kΩ

[GATE-2012]

Q.8 Assuming both the voltage sources are in phase the value of R for which maximum power is transferred from circuit A to circuit B is

a) 0.8Ω b) 1.4Ωc) 2Ω d) 2.8Ω

[GATE-2012]

Q.9 A source ( )sV t =Vcos100πt has an internal impedance of 4+j3Ω . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in Ω should be a)3 b)4 c)5 d)7

[GATE-2013]

Q.10 Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection are scaled by a factor k,k>0 , the elements of the corresponding star equivalent will be scaled by a factor of

a) 2k b) kc)1/k d) k

[GATE-2013]

Q.11 In the circuit shown below, if the source voltage SV 100 53.13° V= ∠ then the venin’s equivalent voltage in volts as seen by the load resistance LR is

a)100 90°∠ b) 800 0°∠c)800 90°∠ d)100 60°∠

[GATE-2013]

Q.12 Assuming an ideal transformer. The Thevenin's equivalent voltage and impedance as seen from the terminals x and y for the circuit in figure are

a)2sin (ωt), 4Ω b)1sin (ωt), 1Ω c)1sin (ωt), 2Ω d)2sin(ωt), 0.5Ω

[GATE-2014]

Q.13 A non-ideal voltage source Vs has an internal impedance of Zs If a purely resistive load is to be chosen that maximizes the power transferred to the load, its value must be a) 0b) real part of ZS

c) magnitude of ZS

d) complex conjugate of ZS

[GATE-2014]


Q.14 The Norton's equivalent source in amperes as seen into the terminals X and Y is

[GATE-2014]

Q.15 For the given circuit, the Thevenin equivalent is to be determined. The Thevenin voltage, VTh (in volt), seen from terminal AB is

[GATE-2015]

Q.16 In the circuit shown below, the maximum power transferred to the resistor R is……….W

[GATE-2017, Set-1]

Q.17 In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is

a) 1nF b) 1µFc) 10nF d) 10 µF

[GATE-2017, Set-2]

Q.18 For the network given in figure below, the Thevenin’s voltage Vab is

a) -1.5V b) -0.5Vc) 0.5V d) 1.5V

[GATE-2017, Set-2]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (d) (a) (d) (d) (b) (a) (a) (c) (b) (c) (a) (c) 2 15 16 17 18

3.36 3.025 (d) (a)

ANSWER KEY:


Q.1 (a) By Thevenin’s theorem

th x y 1 2 3Z Z Z || Z Z−= = +

1 23

1 2

Z Z Z(Z Z )

×= +

+

( )10 60 10 60 (50 53.13)10 60 10 60∠− × ∠−

+ ∠∠− + ∠−

56.66 45°= ∠

Q.2 (d) For obtaining power absorbed by

LR under maximum power transfer condition. We find thevenin’s equivalent circuit across LR

ZthI calculated by short circuiting the votage sources.

( ) ( )thZ 6 j8 || 6 j8 3 j4Ω= + + = +

th thV 110 0° V 90 0° 06 8j 6 j8− ∠ − ∠

+ =+ +

thV 100 0°= ∠

For maximum power transfer 2 2 2 2

L th thR R X 3 4 5Ω= + = + =

( )th

L

V 100l3 j4 R 8 j4

= =+ + +

11.18 26.56°A= ∠ Power absorbed by RL (max).

2 2Ll R 11.18 5 625W= = × =

Q.3 (a) To calculate thR (seen at terminals P-Q), voltage source is short- circuit

thR 10 ||10 5Ω= =

thV = Open–circuit voltage at the terminals P-Q

th4V 10 2V

10 10= × =

+Thevenin’s equivalent circuit

EXPLANATIONS


Q.4 (d) To calculate Thevenin’s impedance, current –source is open -circuited

th L CZ R Z Z= + + 1 2j j= + −

1 jΩ+ Open-circuit voltage at terminals X-Y

thI Z= ×

( )1 0 1 j= ∠ × +

2 45°= ∠ volts

Q.5 (d) Thevenin’s voltage, OCV 0.5V=

Q.6 (b) Thevenin’s resistance is calculated using the circuit shown in fig (1) and (2), where independent voltage source is short circuited

Write the loop equation: AB ABV 3V+ =

( )3 3 3ABI 10 10 10 V− −−× = −

3AB5V 10=

3AB

THV 10R 0.2k

I 5∴ = = =Ω Ω

Q.7 (a)

After connecting voltage source of V ( )( ) ( )1 2 b b bV V 10K i 100 I 99i i ;= ⇒ − = + +

b b10000i 100I 100 i− = + ×

b100I 10000i= +

b b20000i 100I i= ⇒−100 II

20000 200− − = =

[ ]b bV 100 I 99i i= + +

I100 I 100200

− = + 50I=

thV 50IR 50I I

= = = Ω

Q.8 (a) Power transferred from circuit A to

circuit 7 6 10RB VI

R 2 R 2+ = = + +

( )42 70R

R 2+

=+

10 3 7I2 R 2 R−

= =+ +

7RV 3 IR 32 R

= + = ++

6 10R2 R+ = +

( ) ( ) ( ) ( )( )

2

4R 2 70 42 70R 2 R 2dP

dR R 2

+ − + +=

+


( ) ( ) ( )270 R 2 42 70R 2 R 2+ = + +

( )5 R 2 2(3 5R)⇒ + = +5R 10 6 10R⇒ + = +4 5R⇒ =R 0.8Ω⇒ =

Q.9 (c) 2 2

L thR Z 4 3 5= = + = Ω


2a b

Stara b c

R R kRR R R 3k

+= =

+ +

StarR k∝

Q.11 (c) TH L1V 10V=

418c

L1V 100 53.13V tan j4

3 j4 3∠

− = = − ×

+

L1V 80 90°= ∠

Q.12 (a) xy ocVϑ =

2sin sin1 2

xyinxy oc t

ϑϑ ϑ ϑ ω= ⇒ = =

2

xy2R 100 41

= × ⇒

2sin sinth tϑ ω=

Q.13 (c) For minimum power transferred to the load, Its value must be L sR Z=

2 2L s sR R X= +

Q.14 (2)

( ) 55

=SC NI I

Q.15 (3.36) 1 V 2ith =

[ ]2 1 i i1 i 2i i1= + + = +

( )i 1 20i 2i1= − +

21i 2i1∴ = As the frequency of current i is increased, the impedance (Z) of the network varies as

SCI 1A= thR 4Ω=


Q.1 In the circuit shown in the following figure the input voltage Vi (t) is constant at 2V for time

1<t s and then it changes to 1V.The output voltage, v0 (t), 2 s after the change will be

a) ( )exp 2− − V

b) ( )1 exp 2− + − V

c) ( )exp 2− V

d) ( )1 exp 2− − V[GATE-2006]

Q.2 In the circuit shown in the following figure, the current through 1Ω the resistor is

a) ( )1 5cos 2t A+

b) ( )5 cos 2+ t A

c) ( )1 5cos 2t A−d) 6A

[GATE-2007]

Q.3 For the circuit shown below the steady–state current I is

a) 0Ab) ( )5 2 cos 1000t A

c) π5 2 cos 1000t A4

−

d) 5 2A [GATE-2008]

Q.4 For the circuit shown below the voltage across the capacitor is

a) ( )10 j0 V+ b) ( )100 j0 V+

c) ( )0 j100 V+ d) ( )0 j100 V− [GATE-2008]

Statement for Linked Answer Questions 5& 6: In the circuit shown below the steady –state is reached with the switch K open Subsequently the switch is closed at time t 0=

GATE QUESTIONS(IN)


Q.5 At time t 0+= , is 2dldt

a) 5A / s− b) 10 A / s3

−

c) 0A / s d) 5A / s [GATE-2008]

Q.6 At time t 0+= ,current I is

a) 5 A3

− b) 0A

c) 5 A3

d) A∞

[GATE-2008]

Q.7 In the circuit shown below, the switch, initially at position 1 for a long time, is changed to position 2 at t 0=

The current i through the inductor for t 0≥ is a) 20t1 e A−− b) 20t1 e A−+c) 20t1 2e A−+ d) 20t2 e A−−

[GATE-2011]

Q.8 The average power delivered to an impedance ( )4 j3− Ω by a current5cos(100πt 100)+ A is a) 44.2W b) 50Wc) 62.5W d)125

[GATE-2012]

Q.9 In the following figure, 1C and 2Care ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t 0= .The current i(t) for all t is

a) Zerob) A step functionc) An exponentially decaying functiond) An impulse function

[GATE-2012]

Statement for Linked Answer Questions 10 & 11: In the circuit shown, the three voltmeter reading 1 2 3V 220V,V 122V,V 136V= = =

Q.10 If RL=5Ω,the approximate power consumption in the load is a)700W b)750W c)800W d)850W

[GATE-2012]

Q.11 The power factor of the load a) 0.45 b) 0.50c) 0.55 d) 0.60

[GATE-2012]

Q.12 Two magnetically uncoupled inductive coils have Q factors 1qand 2q at the chosen operating frequency. Their respective resistances are 1R and 2R . When connected in series, their effective Q factor at the same operating frequency is a) 1 2q q+b) 1 2(1/ q ) (1/ q )+

c) ( ) ( )1 1 2 2 1 2q R q R / R R+ +

d) ( ) ( )1 2 2 1 1 2q R q R / R R+ +[GATE-2013]

Q.13 Three capacitors 1 2C ,C and 3Cwhose values are 10μF,5μF, and 2μF respectively, have breakdown


voltages of 10V , 5V and 2Vrespectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination , and the corresponding total charge in μCstored in the effective capacitance across the terminals are respectively,


[GATE-2013]

Q.14 The circuit shown in figure was at steady state for t < 0 with the switch at position ‘A’. The switch is thrown to position ‘B ’at timet = 0. The voltage V(volts) across the 10Ω resistor at time t = 0+ is _____.

[GATE-2014]

Q.15 The average real power in watts delivered to a load impedance ZL

=(4- j2 )Ω by an idea current source i(t) 4sin (ωt 20 )+ ° t A is____.

[GATE-2014]

Q.16. In the circuit shown in the figure, initially the capacitor is uncharged. The switch ‘S’ is closed at t = 0.Two milliseconds after theswitch is closed, the current through the capacitor (in mA) is___.

[GATE-2014]

Q.17 A capacitor ‘C’ is to be connected across the terminals ‘A ‘and ‘B’ as shown in the figure so that the power factor of the parallel combination becomes unity. The value of the capacitance required in μF is

[GATE-2014]

Q.18 The capacitor shown in the figure is initially charged to +10 V. The switch closes at time t = 0. Then the value of VC(t) in volts at time t = 10 ms is ___V.

[GATE-2015]

Q.19 The circuit shown in the figure is in series resonance at frequency fc Hz. The value of Vc in volts _______ V.


[GATE-2015]

Q.20 A current i(t) shown in the figure below is passed through a 1 F capacitor that had zero initial charge. The voltage across the capacitor for t > 2 s in volt is _________.

[GATE-2016]

Q.21 In the circuit shown below ( )1 2 [V V+ = 1 sin (2𝜋𝜋10000t )+1sin(2𝜋𝜋30000t)] V. The RMS value of the current through the resistor R will be minimum if the value of the capacitor C in microfarad is____.

[GATE-2016]

Q.22 In the circuit shown below, VS=101∠ 0V, R =10Ω and =L =100Ω. The current IS is in phase with VS. The magnitude of I S in milliampere is _________.

[GATE-2016]

Q.23 The voltage v(t) shown below is applied to the given circuit. v(t) = 3 V for t < and v(t)= 6V for t > 0. The

value of current i(t) at t = 1s, in ampere is ______ .

[GATE-2016] Q.24 A series R-L-C circuit is excited

with a 50 V, 50Hz sinusoidal source. The voltages across the resistance and capacitance are shown in the figure. The voltage across the inductor (VL) is _______V.

[GATE-2017]

Q.25 The current response of a series R-L circuit to a unit step voltage is given in the table. The value of L is ___________H.

t in s 0 0.25 0.5 0.75 1.0 … ∞i(t) in A 0 0.197 0.316 0.388 0.432 … 0.5

[GATE-2017]

Q.26 In the circuit diagram, shown in the figure, S1 was closed and S2 was open for a very long time. At t=0, S1 is opened and S2 is closed. The voltage across the capacitor, in volts, at t=5𝜇𝜇s is___________

[GATE-2017]


Q.27 A series R-L-C circuit is excited with an A.C voltage source. The quality factor (Q) of the circuit is given as Q=30. The amplitude of current in ampere at upper half-power frequency will be_______.

[GATE-2017]

Q.28 For the circuit, shown in the figure, the total real power delivered by the source to the loads is_________kW.

[GATE-2017]

Q.29 A coil having an impedance of (10 +j100) is connected in parallel to a variable capacitor as shown in figure. Keeping the excitation frequency unchanged, the value of the capacitor is changed so that parallel resonance occurs. The impedance across terminals p-q at resonance (in Ω) is ____.

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (a) (a) (d) (b) (a) (d) (b) (d) (b) (a) (c) (d) -30 15 16 17 18 19 20 21 22 23 24 25 26 27 28 32 1.51 187.24 3.678 100 8 0.28 100 1.632 50 1 1.52 6.36 1.86 29

1010

ANSWER KEY:


Q.1 (a) For t <1, the circuit is shown in Fig.1

C behaves as open circuit after long time .

,At t 1−∴ = the circuit is shown in Fig. 2 Current is zero,

( )0v 1 0V−∴ = and

( )cv 1 2V−∴ = As capacitor voltage cannot change instantaneously, the circuit for t 1+= is shown in Fig 3.

( ) ( )c cv 1 v 1_ 2V+ = =

( )0v 1 1V+∴ = −

At steady state as t → ∞ , the capacitor behaves as open circuit as shown in Fig. 4 Current is zero,

( ) ( )0 cv 0V andv 1V∴ ∞ = ∞ = Initial value of ( )0v t

1V,at t 1+= − = Final value of ( )0v t 0V,= at t ∞=

( )tτ

0v t 1e−

= −

where τ Time constant RC= =For R 1M ,C 1μF, τ 1sec= = =Ω

( )0v t 2sec∴ = after the change

at 2t 1 1e V−+= = −

Q.2 (a) Due to ac source,

1Ri 1/1 1A= =

Due to ac source,

2R

11i 5 0°

1 1 J0.51 J2

∠= × + +

5 0°= ∠ Ri (1 5cos cos 2t)A∴ = +

Q.3 (a) As t → ∞ voltage source become 0V

Q.4 (d) Circuit is under resonance

10 0°I 1 0°A10∠

= = ∠

CV 100JV= −

Q.5 (b) From Fig 3 : Write the Outer loop equation :

( ) 2dl (t)I t 1 1 10 5dt

× + + =

At t 0+= ,

( ) ( )2dl 50 5 I 0 5dt 3+ −= − − = − +

10 A / sec3

= −

Q.6 (a) For the circuit is shown in Fig .1

EXPLANATIONS


After a long time , at t 0 L−=behaves as short circuit and C behaves as open circuit . The relevant circuit is shown in Fig .2

At ( ) ( ) ( ), 1 L 2t 0 I 0 0,i 0 I 0− − − −= = =

( )c0andv 0 10V−= =

For ,t 0> the switch , K is closed and the relevant circuit is shown in Fig.3

( )Li t and ( )cv t cannot changeinstantaneously.

( ) ( ) ( )L 2 2i 0 I 0 I 0 0+ + −∴ = = =

( ) ( )c cv 0 v 0 10V+ −= =

At t 0+= , the state of the circuit is shown in Fig. 4

( ) ( )15I 0 I 0 A3− += = −

Q.7 (d)

For t<0, the status of the circuit is shown in Fig.1 inductor behaves as short circuit after a long time.

( ) 10i 0 1A10−∴ = =

For t > 0 , the status of the circuits shown in Fig .2 Current through the inductor cannot change instantaneously.

( ) ( )i 0 i 0 1A+ −∴ = = initial value(I.V) After a long time inductor behaves as short circuit .

( )i 2A∴ ∞ = = final value (F.V)

Time constant L 1 1τ , 20R 20 τ

= = =

( ) ( )tτi t IV F.V I.V I e for t 0

−

= + − − ≥

( ) ( )20t 20t1 2 1 1 e 2 e− −= + − − = −


rmP Ri=25 4 50W

2 = × =


A =

Q.9 (d) When the switch in closed at t=0


Capacitor 1C will discharge and

2C will get charge since both 1Cand 2C are ideal and there is no –resistance in the circuit charging and discharging time constant will zero. Thus current will exist like an impulse function.

Q.10 (b) L: R 5= Ω

RL 3 1V V cos cos 136 0.45[FromQ 7]= φ = ×

RLV 61.2V∴ =

L

2R

RLV

P 749ω 750R

= =

Q.11 (a) Phasor diagram

1 1 2V 220V V 122V= =

3V 136V=

31 2 vv v→→ →

= +

By parallelogram law of addition of vectors

2 21 2 3 2 3V V V 2V V cos= + + φ

∴by using options, cos 0.45φ =

Q.12 (c) 1 2

1 22 2

ωL ωLQ QR R

= = =

After series connection ( )1 2

1 2

ω L LQ

R R+

=+

1 2 21R Q R

1 2

QQ

R R+=+

Q.13 (d)

2 3eq 1

2 3

C CC CC C

= ++

11.5μF= Safety voltage = 7V Q CV=

eq safetyQ C V⇒ = ×11.5 7 80μC× ;

Q.14 (-30) At t < 0, switch is at position “A” and it was at steady state.

Steady state inductor behaves like short circuit In a source free circuit,

( )tτ

0L Ii .et−

=

( )t

Lτ3.i et

−=

( )1

L

2t53.ei t

−=

( ) 12 /53.v t [1 ]0 te−−=

( ) 12 /5v t 30 010 30[ ;− +− ⇒ = == −te At t V Volts

Q.15 (32) Given ZL =(4- j2)Ω Current source oi(t)4sin( t 20 )Aω + The average real power P = 2Irms .R

Watts P = 24 .4 32

2watts =

Q.16 (1.51) When switch „S‟ is closed At t=0 :


ci (0+) = 5 2.5mA2k

⇒

At t =∞ :

c ( ) 0.A∞ =i

c (t)i = (Initial Value – Final Value )/ Final Valuete τ− +

seceq eqR Cτ = 1 *4Kτ µ= 4=τ msec

3t /4*10 3c (i [2.5 0]e 10 () )t 0

−− − + = −

( ) 2502.5 −= tci t e mA

At t=2* 10-3sec ;

( )3ci 2*10 1.51mA− =

Q.17 (187.24) Power factor of the parallel combination is unity. i.e., in total impedance equate theimaginary term to zero, which gives the capacitance required.

[ ] 14 1 ||totalZ jj Cω

= + Ω

[ ] 1 14 j . 4 j 1jωC *

1 14 j 1 4 j 1ωC ωC

+ − − = + − − −

ωC

total 2

j4 1 14 j 1C C CZ

116 1C

ω ω ω

ω

− + − − = + −

2 2 2 2

total 2

4 1 1 16jC C C CZ

116 1C

ω ω ω ω

ω

− − + = + −

Equate the imaginary term to zero

2 2

16 16 1 1Cw C 17C Cω ω ω

+ = ⇒

1 187.2417 2 50

C Fµπ

⇒ = =× ×

Q.18 3.678V It is a source free network where capacitor voltage

( )

( )3

0100

100 10 10 3.673

0110

10010 10 10

tte

ct

e

Vc

V t VRC

V e

τ

−

−

−

− × × =−

= >= = =

× =

Q.19 (100V) Under series resonance condition, voltage across L, C is Q times of the supply voltage.

CV QV=

Q= 6

1 L 1 0.1 100R C 10 0.1 10−= =

×

CV QV= ( )( )100 1 100V= =

Q.20 (8)

( ) ( ) ( ) ( )1 1idt 8[ 1 2 8[ 1 2 ]1cV u t u t dt r t r t

C= = − − − = − − −∫ ∫

( )cV t 2 8V> =

Q.21 (0.28 to 0.283)

Q.22 (100) In phase means circuit is under resonance and the admittance seen by source must be real i.e. imaginary part of Yeq = 0


⇒Yeq = 1 1R jωL (1/ jωL)

++

= 2 2

R jωL jωLR (ωL)

−+

+

( ) 2 2 2

R jωL 10Real yR (ωL) 100 (100)

+→ = =

+ +

= 11010

I= VY = 101 0.1 100mA1010

= =

Q.23 (1.63)2 ( / )( ) ( ) [ (0 ) ( )] t

L L L Li t i i i e τ− −= ∞ + − ∞ At t = , 6∞ supply is v

→1.5 11.5in

LR

τ = = =

( )L ti→ = ( ) 2 tLi t e−= −

( )Li t = 2 1.632te−− =

Q.24 50

( )22R L CV V V V= + −

( )22L50 50 V 50= + −

LV 50=

Q.25 1H

( )RtLVi t 1 e

R−

= −

From given data at t = ∞ V 0.5R=

V 1= R 2= Ωt 0.25=

( )2 0.25L0.197 0.5 1 e

−−

= −

L 1H=

Q.26 1.52 At t 0−=

( )CV 0 1− =

At t = ∞

( )C2V 3 2V

2 1∞ = =

+

( ) ( ) ( ) ( )t

RCC C C CV t V 0 V e V

−+ = − ∞ + ∞

[ ]6

t2 10 1031 2 e 2

−−

× ×= − +

At t 5 sec= µ

CV 1.52V=

Q.27 6.36

01LC

ω =

3 6

110 10 4 10− −

=× × ×

5000rad / sec= 0LQRω

=

( )35000 10 1030

R

−×=

5R3

= Ω

At 0V 15I 9VR 5 / 3

ω → = = =

At 2I 9 6.36A2 2

ω → = =

Q.28 1.86 0 0

1 2I I I 5 0 5 30= + = ∠ + ∠I 9.659 14.94= ∠P VIcos= θ


( )200 9.659 cos 14.94= × ×1.866kW=

Q.29 1010 It is given that the circuit is under resonance, and we know at this resonant frequency

Img 0 Img 0eq eqY or Z = =

In a parallel circuit, from calculation point of view, it is wised to deal with

Img 0eqY =

From the given circuit we can say,

1 1= +

RL C

YZ Z

1 11

= ++

ωω

R j Lj C

2 2( )−

= ++

ω ωω

R j L j CR L

2 2 2 2( ) ( )

= + − + +

ωωω ω

R Lj CR L R L

At resonance frequency

Im 0 = eqg Y

So,

2 2( )=

+ ωeqRY

R L

2 21 ( )+= =

ωeq

eq

R LZY R

2 210 100 101010+

= =


4.1 INTRODUCTION

A sinusoidal current or voltage at a given frequency is characterized by only two parameters amplitude and a phase angle. The complex representation of the voltage or current is also characterized by these same two parameters. Phasor representation is defined only for cosinusoidal signal. All the sinusoidal signal are converted into cosinusoidal by subtracting 90° from phase A real sinusoidal current ( ) mi t I cos cos( t )= ω +ϕ is expressed as the

real part of a complex quantity by invoking Euler’s identity. ( ) j( t )

mi t Re I e ω +ϕ=

j t jmRe I e eω ϕ=

j tRe Ie ω=

We represent the current as a complex quantity by dropping the instruction Re . Thus adding an imaginary component to the current without affecting the real component and suppressing the factor j te ω

We get j

mI I e ϕ= - Exponential Form

mI I= ∠ϕ - Polar Form

mI I [cos jsin ]= ∅+ ∅ -Trigonometric Form The above abbreviated complex representation is the phasor representation. The process by which we change i(t) into I is called a phasor transformation from the time domain to the frequency domain_(Shown in fig below)

Similarly, j

mV V e ∅= -Exponential Form

mV V= ∠∅ -Polar Form

mV V cos jsin )= ∅+ ∅ -Trigonometric Form

Example: Transform the time domain voltage ( ) ( )v t 100cos 400t 30= − ° volts into the

frequency domain. Solution: We know that ( ) mv t V cos( t )= ω +∅ &

( ) ( )v t 100cos 400t 30= − ° (Given)Thus time –domain expression is already is in the form of a cosine wave with a phase angle. Thus, suppressing 𝜔𝜔=400 rad/sec.

mV V= ∠∅

i.e. ( )v t 100 30 volts= ∠− °

4.2 SINUSOIDAL STEADY STATE ANALYSIS

A circuit having constant source is said to be in steady state if the currents and voltages do not change with time. Thus circuits with currents and voltages having constant amplitude and constant frequency sinusoidal functions are also considered to be in a steady state. Inductors and capacitors are kept as it is while applying sinusoidal steady state analysis. The analysis of AC in the steady state is generally carried out in the phasor domain i.e. the KCL, KVL, ohm’s law, Nodal and mesh analysis and the source transformation are written in the Phasor domain.

4.2.1 THE PHASOR DIAGRAM The Phasor diagram is a pictorial representation of all the phase voltages and current in a network. The Phasor is a

4 AC ANALYSIS


rotating vector, which rotates in the anticlockwise direction with angular frequency ‘ω’ in the time domain. e.g. For the Phasor diagram shown below

In Fig. a), the length of the arrow represent the magnitude of the sine wave and angle ‘θ’ represents the angular position of the sine wave. In Fig. b), the magnitude of the sine wave is 3 and phase angle is 30° . A Phasor diagram can be use to represent the relation between two or more sine waves of the same frequency. e.g

In the above fig sine wave C lag behind B by45° , sine wave A lead sign wave B by 30° .

4.3 SERIES CIRCUITS The impedance diagram is useful for analyzing series ac circuits. The series circuits can divided in RL, RC and RLC series circuits. In the series ac circuits, one must draw the impedance diagram We know that

Rz R R 0= Ω = ∠ °

LL L L whereX Lz j L jX X 90 | =ω= ω Ω = Ω = ∠ °

cC c C 1whereX

C

1 1z j jX X 90 |j C C =

ω

= Ω = − = − = ∠− °ω ω

4.3.1 THE SERIES RL CIRCUIT

If we apply a sinusoidal input to a RL circuit the current in the circuit and all voltages across the elements are sinusoidal. In the analysis of a RL series circuit we can find the impedance, current, phase angle and voltage drop.

(Network is in steady state ) We know,

RV IR=The resistor voltage R(V ) and current (I) are in phase with each other (as shown in fig. b) inductor voltage L(V ) leads the source voltage (𝑉𝑉𝑆𝑆). The phase angle between current and voltage in a pure inductor is always 90° (as shown in fig(b))

L LV IZ=

LIX 90= ∠ °

Fig (b) shows 𝑉𝑉𝑅𝑅 and I are in phase and the source voltage ‘V’ is the phasor sum of 𝑉𝑉𝑅𝑅 and 𝑉𝑉𝐿𝐿

2 2R LV V V= +

The phase angle between resistor voltage and source voltage is

1 L

R

VtanV

− ∅ =


The power factor RVcosV

= ∅ =

(Lag)

4.3.2 SERIES RC CIRCUIT When a sinusoidal voltage is applied to an RC series circuit, the current in the circuit and voltages across each of the elements are sinusoidal. The series RC circuit is shown in fig.

Here the resistor voltage (VR) and current are in phase with each other (shown in fig. b) RV IR= The capacitor voltage ‘VC’ lags behind the source voltage. The phase angle between the current and the capacitor voltage is always 90°

C CV IZ=

C CV IX 90= ∠− °

Here I leads Vc by 90°, VR and I are in phase The source voltage is given by

2 2R CV V V= +

The phase angle between the resistor voltage and the source voltage is

1 C

R

VtanV

− ∅ =

The power factor is RVpowerfactor cos (lead)

V = ∅ =

4.3.3 SERIES R-L-C CIRCUITS A series R-L-C circuits is the series combination of resistor, inductor and capacitor.

Case1: When L CV >V We know, RV IR=

L LV IX 90= ∠

C CV IX 90= ∠− °

The source voltage v is given as

( )22R L CV V V V= + −

The phase angle is 1 L C

R

V VtanV

− −∅ =

And power factor is RVp.f cos (lag)

V = ∅ =

Case2: When 𝑽𝑽𝑪𝑪 > 𝑽𝑽𝑳𝑳

( )22

R C LV V V V= + − And phase angle is

1 C L

R

V VtanV

− −∅ =

And power factor is RVp.f cos (lead)

V = ∅ =


Case 3: When L CV =V

Example: A sine wave generator supplies a 50HZ, 10V rms signal to a 2 𝑘𝑘Ω resistor in series with a 0.1μF capacitor as Shown in fig. Determine the total impedance Z, current I, Phase angle θ, capacitive voltage Vc and resistive voltage VR.

Solution: To find the impedance Z, we first solve for Xc

C 6C

1 1X 3184.72 f 2 500 0.1 10−= = = ΩΠ Π× × ×

Total impedance Z (2000 j3184.7)= − Ω

( ) ( )2 2Z 2000 3184.7= +

Z 3760.6= Ω Capacitive voltage

C CV IX= 3

CV 2.66 10 3184.7−= × ×

CV 8.47V=Resistive voltage RV IR=

32.66 10 2000−= × × RV 5.32V=

Phase Angle, 1 1C

R

V 8.47tan tan 57.87V 5.32

θ − − = = = °

Example:

In the circuit shown below, Determine the total impedance, current I, phase angle and the voltage across each elements.

Solution: To find the impedance z, all first solve for Vc and XL

C 6C

1 1X2 f 2 50 10 10−= =Π Π× × ×

CX 318.5= Ω

L LX 2 f 6.28 0.5 50= Π = × ×

LX 157= Ω Total impedance in rectangular form z (10 j157 j318.5)= + − Ω z 10 j(157 318.5)= + − Ω z 10 j161.5= − ΩHere, C LX X>

( ) ( )2 2z 10 161.5∴ = +

100 26082.2= +z 161.8= Ω

Current SV 50I 0.3AZ 161.8

= = =

I 0.3A∴ = Voltage across the resistor

RV IR 0.3 10 3V= = × = Voltage across the capacitive reactance

C CV IX 0.3 313.5 95.55V= = × = Voltage across the Inductive reactance

L LV IX 0.3 157 47.1V= = × =

4.4 PARALLEL CIRCUITS

The parallel circuits can divide in RC, RL and RLC circuits. In parallel A.C. circuits, the voltage is the same across each element.

We know that RVIR

=


LL L L

V V VI 90Z X 90 X

= = = ∠− °∠ °

CC C C

V V VI 90Z X 90 X

= = = ∠ °∠− °

4.4.1 PARALLEL RL CIRCUITS

Parallel RL circuit shown below

Current I is distributed into two parallel branches as resistive current (IR) and inductor current (IL) (which is given as follow)

R LL L L

V V V VI and I 90R Z X 90 X

= = = = ∠− °∠ °

Figure shows - IR and v are in phase and source current is phasor sum of IR and IL

2 2R LI I I∴ = +

The phase angle between resistive currant and inductive current is

1 L

R

ItanI

− ∅ =

The power factor RIcos (lag)I

= ∅ =

4.4.2 PARALLEL RC CIRCUIT

Parallel RC circuit Shown below

Current I is divided into two parallel branches as resistive current (𝐼𝐼𝑅𝑅) shown below

R CC C C

V V V VI and I 90R Z X 90 X

= = = = ∠ °∠− °

As Shown in figure above,𝐼𝐼𝑅𝑅 and v are in phase and current (I) Phasor sum of IR

and IC 2 2R CI I I∴ = +

The phase angle between resistive current and inductive current is ---

1 C

R

ItanI

− ∅ =

The power factor RIcos (lead)I

= ∅ =

4.4.3 PARALLEL RLC CIRCUIT

Parallel R-L-C circuit Shown below

Current I is divided into three parallel branches as, resistive current IR Inductive current IL and capacitive current IC which is given as bellow

RVIR

=

LL

VI 90X

= ∠− °

CC

VI 90X

= ∠ °

Case 1: When C LI I> We know


RVIR

=

LL

VI 90X

= ∠− ° & CC

VI 90X

= ∠ °

Source current I is given as

( )22R C LI I I I= + −

Phase angle is given as 1 C L

R

I ItanI

− −∅ =

And power factor is

RIp.f cos (lead)I

= ∅ =

Case2: when 𝐈𝐈𝐋𝐋 > 𝐈𝐈𝐂𝐂

The source current I is given as

( )22R L CI I I I= + −

Phase angle is given as 1 L C

R

I ItanI

− −∅ =

And power factor is given as

RIp.f cos (lag)I

= ∅ =

Example: A 50Ω resistor is connected in parallel with an inductive reactance of 30Ω . A 20 V signal is applied to the circuit find the total impedance and line current in the circuit shown in fig.

Solution: Since the voltage across each element in parallel ac circuit is same--- So current in the resistive branch R(I ) is

RV 20 0I 0.4AR 50

∠ °= = =

And current in the inductive branch

LL

V 20 0I 0.66 90X 30 90

∠ °= = = ∠− °

∠ °

Total current I 0.4 j0.66= − Total current in polar form is I 0.77 58.8= ∠− ° Here current lags behind the voltage by58.8° Total impedance

V 20 0zI 0.77 58.8

∠ °= =

∠− °z 25.97 58.8= ∠ °Ω

Example: In the circuit shown below determine the values of the following 1) Total current (I)2) Total impedance (Z)3) Phase angle ()

Solution: We know that, L LX 2 f= π

( )2 50 (0.1)= π

LX 31.42= Ω From given circuit, 10Ω resistor is in series with the parallel combination of 20Ω and j31.42Ω ∴Total impedance (z) is ------


( )( )20 j31.42z 10 10

20 j31.42= + = +

+628.4 90

37.24 57.52∠ °

∠ °z 24.23 j9.06= + In polar form z 25.87 20.5= ∠ ° Here current lags behind the applied voltage by 20.5

Total current VIZ

=

2025.87 20.5

=∠ °

I 0.77 20.5= ∠− ° The phase angle between voltage & current is

20.5θ = °


Q.1 A unit step voltage is applied at t 0 to a series RL circuit with zero initial conditions. a) It is possible for the current to

be oscillatory.b) The voltage across the resistor at

t 0+= is zero.c) The energy stored in inductor in

the steady state is zero.d) The resistor current eventually

falls to zero. [GATE-2001]

Q.2 Consider the circuit shown in figure. If the frequency of the source is 50 Hz, then a value of 𝑡𝑡0 which results in a transient free response is

a) 0ms b) 1.786msc) 2.71ms d) 2.91ms

[GATE-2002]

Q.3 An 11V pulse of 10μs duration is applied to the circuit shown in figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is

a) 11V b) 5.5Vc) 6.32V d) 0.96V

[GATE-2002]

Q.4 In the circuit shown in figure, the switch S is closed at time (t=0). The

voltage across the inductance at +t=0 , is

a) 2V b) 4Vc) -6V d) 8V

[GATE -2003]

Q.5 In figure, the capacitor initially has a charge of 10 Coulomb. The current in the circuit one second after the switch S is closed will be

a)14.7A b)18.5A c)40.0A d)50.0A

[GATE-2004]

Q.6 In the figure given, for initial capacitor voltage is zero. The switch is closed at t=0. The final steady –state voltage across the capacitor is

a)20V b)10V c)5V d)0V

[GATE -2005]

Q.7 The circuit shown in the figure is in steady state, when the switch is closed at t=0. Assuming that the inductance is ideal, the current through the inductor at +t=0 equals

GATE QUESTIONS(EE)


a) 0A b) 0.5Ac) 1A d) 2A

[GATE-2005]

Statement for linked Answer Questions Q.8 & Q.9 A coil inductance 10H and resistance 40Ω is connected as shown in the figure. 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 Q.8 For the value of obtained in (a), the

time taken for 95% of the stored energy to be dissipated is close to a) 0.10sec b) 0.15secc) 0.50sec d)1.0sec

[GATE-2005]

Q.9 if, at +t=0 , the voltage across the coil is 120 V , the value of resistance R is

a) 0Ω b) 20Ωc) 40Ω d) 60Ω

[GATE-2005]

Q.10 An ideal capacitor is charged to a voltage oV and connected at t=0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let

01LC'

ω = the voltage across the

capacitor at time t>0 is given by a) oV b) o 0V cos( t)ω

c) o 0V sin(ω t) d) 0ω to 0V e cos(ω t)−

[GATE-2006]

Q.11 In the figure transformer T1 has two secondaries, all three windings having the same number of turns and with polarities as indicated. One secondary is shorted by a 10Ωresistor R and the other by a 15 mF capacitor. The switch SW is opened t=0 when the capacitor is charged to 5 V with left plate as positive at (t+0+) the voltage VP and Current I are

a) -25V,00Ab) Very large voltage, very large

currentc) 5.0 V, 0.5 Ad) -5.0V,-0.5A

[GATE-2007]

Q.12 In the circuit shown in the figure, the current source I=1A , voltage source V 5V= , 1 2 3R R R 1 ,= = = Ω

1 2 3L L L 1H,= = = 1 2C C 1F= = . The currents (in A) through R3 and the voltage source V respectively will be

a) 1, 4 b) 5, 1c) 5, 2 d) 5, 4

[GATE-2006]

Q.13 In the circuit shown in figure, switch Sw1 is initially CLOSED and 2Sw is OPEN. The inductor L carries a current of 10 A and the capacitor is charged to 10 V with polarities as indicated. 2Sw Is initially caps at t=0 and 1Sw is OPENED at t=0 The


current through C and the voltage across L at +t=0 is

a) 55A,4.5V b) 5.5A,45Vc) 45A,5.5V d) 4.5A,55V

[GATE-2007]

Q.14 The time constant for the given circuit will be

a) 1 S9

b) 1 S4

c) 4 S d) 9 S [GATE-2008]

Statement for Linked Answer Questions Q.15 & Q.16 The current i (t) sketched in the figure flows through an initially uncharged 0.3nFCapacitor.

Q.15 The capacitor charged up to 5ms, as per the current profile given in the figure, is connected across an inductor of 0.6mH. Then the value of voltage across the capacitor after 1us will approximately be a)18.8V b) 23.5Vc) -23.5V d) -30.6V

[GATE-2008] Q.16 The charge stored in the capacitor at

=5μs, will be a)8nC b)10nC c)13nC d)16nC

[GATE-2008]

Q.17 In the figure shown, all elements used are ideal. For time 1t<0,Sremained closed and 2S open. At

1t=0,S is opened and 2S is closed. If the voltage c2V across the capacitor

2C at t=0 is zero, the voltage across the capacitor combination at +t=0will be

a)1V b) 2Vc)1.5V d) 3V

[GATE-2009]

Q.18 The switch in the circuit has been closed for a long time. It is opened att 0= . At +t=0 , the current through the 1μF capacitor is

a) 0A b)1Ac)1.25A d) 5A

[GATE-2010]

Statement for Linked Answer Questions Q.19 & Q.20

The L-C circuit shown in the figure has an inductance L 1mH= and a capacitance C 10μF=


Q.19 The initial current through the inductor is zero, while the initial capacitor voltage is 100V. The switch is closed at t=0 . The current i through the circuit is: a) ( )35cos 5 10 t A×

b) ( )45sin 10 t A

c) ( )310cos 5 10 t A×

d) ( )410sin 10 t A

[GATE-2010]

Q.20 In the following figure, 1C and 2Care ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0 .The current i(t) for all t is

a) Zerob) A step functionc) An exponentially decaying

functiond) An impulse function

[GATE-2012]

Statement for Linked Answer Questions Q.21 & Q.22: In the circuit shown, the three voltmeter reading

1 2 3V =220V,V =122V,V =136V

Q.21 If RL=5Ω, the approximate power consumption in the load is a)700W b)750W c)800W d)850W

[GATE-2012]

Q.22 The power factor of the load a) 0.45 b) 0.50c) 0.55 d) 0.60

[GATE-2012]

Q.23 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

02 01V > V , the voltage Vc (t) across capacitor is

a) ( ) t /RCc 02 01v t V (1 e ) V−= − − −

b) ( ) t /RCc 02 01v t V (1 e ) V−= − − +

c) ( ) t /RCc 02 01v t V (1 e ) V−= − − −

d) ( ) ( )( )t /RCc 02 01 01v t V V 1 e V−= − + − +

[GATE-2014]

Q.24 The circuit shown in the figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by v(t) = 12sint. If the circuit is in steady state, then the rms value of the current (in Ampere) flowing in the circuit is______

[GATE-2015]

Q.25 A series RL circuit is excited at t=0 by closing a switch as shown in the figure. Assuming zero initial conditions, the

value of 2

2

d Idt

at t =0+ is


a) VL

b) VR−

c) 0 d) 2

-RVL

[GATE-2015]

Q.26 In the circuit shown, switch S2 has been closed for a long time. A time t = 0 switch Si is closed At t = 0+, the rate of change of current through the inductor, in amperes per second, is ____.

[GATE-2016]

Q.27 A resistance and a coil are connected in series and supplied from a single phase, 100V, 50Hz ac source as shown in the figure below. The rms values of plausible voltage across the resistance (VR) and coil (Vc) respectively, in volts, are

a) 65, 35 b) 50, 50c) 60, 90 d) 60, 80

[GATE-2016] Q.28 In the circuit shown below, the

initial capacitor voltage is 4V. Switch1S is closed at t = 0. The charge (in μ

C) lost by the capacitor from t =25μs to t =100μs is ______.

[GATE-2016] Q.29 The switch in the figure below was

closed for a long time. It is opened at t=0. The current in the insuctor of 2H for 0t , is

a) 42.5 te b) 45 te

c) 0.252.5 te d) 0.255 te

[GATE-2017, Set-1]

Q.30 The voltages across the circuit in the figure, and the current through it are given that the following expressions:

( ) ( )iV t 5 10cos t 60= − ω +

( ) ( )ii t 5 X cos t= + ω

Where 100 rad / secω = π . If the average power delivered to the circuit is zero then the value of X (in ohm) is _____________ (up to two decimal places)

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (b) (b) (c) (b) (a) (b) (c) (b) (c) (b) (d) (d) (d) (c) (d) 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 (c) (a) (b) (d) (d) (b) (a) (d) 10 (d) 2 (c) 6.99 (a) 10

ANSWER KEY:


Q.1 (b) At t 0+= inductor works as open circuit, hence complete source voltage drops across it and consequently, current through the resistor R is zero. Hence, voltage across the resistor at t 0+= is zero, and further with time it rises according to

( )RtL

RV t R.(1 e )u(t)−

−= −

Q.2 (b) For transient response,

( )0ωLtan ωtR

=

( )02π 50 0.01tan 2π 50 t

5× ×

× × =

10

π2π 50 t tan5

− × × =

32.14° 0.561rad= =

00.561t 1.786ms100π

= =

Q.3 (c) ( ) ( ) ( ) ( ) t /RC

C C C CV t V V V 0 e−= ∞ − ∞ −

( )

6

3 9

10 1010 10 11 1011

CV peak 10 (10 0)e

−

−

×

× × ×= − −

( )110 1 e 6.32V= − =

net10[WhereR 10 ||1 kΩ,11

= =

( )C10V 11 10V

10 1∞ = × =

+ And

∵ pulse of duration 10μs is applied.

Hence capacitor charges till 10 μs and then starts discharging, so VC will maximum at t= 10μs]

1

112 l 0

2

Eh | 0.25E == =

Q.4 (b) Before closing the switch, the circuit was not energized, therefore, current through inductor and voltage across capacitor are zero. After closing the switch, at t = 0+ inductor acts as open –circuits and capacitor acts as short circuit. Equivalent circuit at t 0+=

10l3 4 || 4

=+

2A=

( )LV 0 l (4 || 4)+ = ×

2 2 4V= × =

Q.5 (a) Method -1 Using KVL,

dq100 R q / cdt

=

dq100C R qdt

= +

0

0 t

0q

dq 1 dt100C q RC

=−∫ ∫

1/RC0100C q (100C q )e−− = −

t /RC0(100C q )dqi edt RC

−−= =

t /RC 1e 40e 14.7A− −= = Method -2 At (t=0) switch is closed

EXPLANATIONS


( ) ( )C CQV 0 V 0C

− += =

10 20V0.5

=

( )CV 100V∞ =

( ) ( ) ( ) ( ) t /RCC C C CV t V V 0 V e+ −= ∞ + − ∞

( ) t /1CV t 100 (20 100)e−= + −

( ) CC

dV (t)i t Cdt

∴ =

t0.5 80 ( e )−= ×− × −t40e−=

( ) 1C t 1i t l 40e−= =14.71A=

Q.6 (b) At ( )t 0 ,→ + the capacitors act asshort- circuit. At ( )t ,∞→ the capacitor will become open circuit.

∴ Voltage across capacitor 20 10 10V

10 10× =

+

Q.7 (c) Before closing the switch at 0- , the circuit is in steady state. So, inductor behaves as short-circuit

( )L10i 0 1A10

−= = =

After closing the switch at t 0+= Current through inductor cannot change abruptly

( ) ( )L Li 0 i 0 1A+ −∴ = = =

Q.8 (b) The circuit (in s-domain)

( ) ( )20l s

20 40 40 10s=

+ + +20 2

10s 100 S 10= =

+ +

( ) [ ]-1 -1 2i t =L l(s) =LS+10

10t2e−= Or ( ) ( )effRL

Li t i 0 e t−+=

(20 40 40) t 10t102e 2e+ +

−= = Initial stored energy in inductor

( )20 L

1W Li 02

+=

21 10 2 20Joules2

= × × =

Remaining energy in inductor 1 0W 0.05W=0.05 20= × 1Joule=

2L

1 Li 12

=

21

1 10 i 12× × =

11i 0.44725

= =

10T1i 2e−=

Let at t=T current decrease to i 10T0.4472 2e−=

T 0.15sec=

Q.9 (c) Before moving the switch, at t = 0− The circuit is in steady state and inductor behaves as short-circuit. The circuit at t = 0−


( )L120i 0 2A

20 40− = =

+After moving the switch, at t 0+=Current through inductor cannot change abruptly So, ( ) ( )L Li 0 i 0 2A+ −= =

( )L LV i 0 20 R+= × + 120 2 (20 R)= × +R 40Ω=

Q.10 (b) The relevant circuit is shown in Fig.

2 1ω0 LC=

It is a standard LC circuit. With ( ) ( )c 0 0V t V cos ω t=

( )0 0or V sin ω t 90°+

Q.11 (d)

All the three windings has same number of turns, so magnitude of induced emf’ s in all the three windings will be same i.e.,

p S TV V V= =

Polarity of the windings is decided on the basis of dot –convention. As capacitor is charged to 5 V with left plate as positive. So, 1T is positive wrt T2

T T1 T2V V V 5V= − = As 2T has negative polarity. so 1Phas negative polarity Therefore, p P1 P2V V V 5V= − = −Similarly, S1 has negative polarity So, S S1 S2V V V 5V= − = −

SR

V 5I 0.5AR 10

−= − = −

Q.12 (d) At steady state, Inductor acts as short & capacitor acts open

Current through 35R 5A1

= =

∴ Current delivered by 5V source 5 1 4A= − =

Q.13 (d) Equivalent circuit t 0+= at is,

By nodal analysis, V (V 10)10 010 10

−− + =

2V 10 100⇒ − = V 55V⇒ =(55 10)I 4.5A

10−

⇒ = =

Q.14 (c)


eq eq2C F;R 63

= = Ω

eq eqT R .C 4sec⇒ = =

Q.15 (d) Capacitor charged up to 5μs, so total charge stored in capacitor =Q=13nC Voltage across the capacitor before connecting to indictor

9

0 9Q 13 10X 43.33VC 0.3 10

−

−×

= = =×

Voltage across the capacitor at time t

( )cV t at t 1μs=

( ) [ ]c t 1μs 0 0 1 μsV t | V cosω t= ==

60 3 9

1ω t 1 100.6 10 0.3 10

−

− −= × ×

× × ×2.357rad 135°= = ( )c t 1μsV t | 43.33 cos135°= = ×30.6V≈ −

Q.16 (c)

Charged stored in the capacitor = area under i-t curve

1 2Q A A= +

( ) ( ) ( ) ( )6 3 3 61 12 10 4 10 4 2 10 5 2 102 2

− − − −= × × × + + × × − ×

96 34 10 13nC2

−× = + × =

Q.17 (a) The status of the circuit at t 0−= is shown in fig .1

The status of the circuit at t 0+= is shown if Fig. 2

1

1 2

3 C 3 1V 1VC C 3× ×

= = =+

Q.18 (b) ( ) ( )c cV 0 4V V 0 4V− += ⇒ =

( ) ( )cc

V 0i 0 1A

4

++ = =

Q.19 (d) Initial current through the indictor is zero and capacitor voltage is charged upto voltage ( )cV 0 10V.=As current through inductor and voltage across can-not change abruptly. So, after closing the switch ( ) ( )L Li 0 i 0 0+ −= = And

( ) ( )c CV 0 V 0 100V+ −= =

The circuit is s –domain

( )100

sI s1sL

sC

= +

2

100 11L s

LC

= +

( )22

C 1/ LC100L S 1/ LC

= +

Taking inverse Laplace transform


( ) [ ]1i t L I(S)−=

C 1100 sin sin tL LC

=

6

3 3 6

10 10 1100 sin t1 10 1 10 10 10

−

− − −

×= × × × × × ×

( ) ( )4i t 10sin 10 t A=

Q.20 (d) When the switch in closed at t = 0 Capacitor 1C will discharge and 2Cwill get charge since both 1C and

2C are ideal and there is no –resistance in the circuit charging and discharging time constant will zero. Thus current will exist like an impulse function.

Q.21 (b) L: R 5= Ω

RL 3 1V V cos cos 136 0.45[FromQ 7]= φ = ×

RLV 61.2V∴ = L

2R

RLV

P 749ω 750R

= =

Q.22 (a) Phasor diagram

1 1 2V 220V V 122V= =

3V 136V=

31 2 vv v→→ →

= +

By parallelogram law of addition of vectors

2 21 2 3 2 3V V V 2V V cos= + + φ

∴by using options, cos 0.45φ =

Q.23 (d)

When switching is in position 1

( )tz

c eV t (Initial — final) final value−

= +

( )tRC

c 01V t V 1 e− = −

When switch is in position 2 Initial value is

( )tRC

c 01V t V 1 e− = −

Final value is —V02

( )t2RC

c 01 02 01V t V V V 1 e− = − −

Q.24 (10) 1 1Y(S)

Z(S) (1 j )ω= =

+

1

2

1Y(S) tan (u)1 ω

−= ∠ −+

inv (t) 8 12sin t= +

11 12i(t) 8 tan (0) sin(t 4)1 0 1 0

−= ∠− + −+ +

12 1 1i(t) 8. sin t cos t1 0 2 2

= − + i(t) 8 6sin t 6cos t= + −

2 22

rms6 6I 8 102 2

= + + =

Q.25 (d) RtL

LVi i (t) 1 eR

− − = −


RtL Ldi V e

dt L

− =

2

2 2

di R Vdt L

RtLe−

= −

2

2 2t 0

di RVdt L

=

= −

Q.26 (2) At t = 0- Network is in steady state with S1 opens S2 (closed) So we can say iL (0-)

= 3 1.5A2= At t = 0+ indicator

behaves as ideal current source of 1.5A if we draw the network at t = 0+, both switch closed Writing Nodes equation at VL(0+) node

( )+L

1 1 3 3V 0 = + = + -1.51 2 1 2

LV (0 ) 2+⇒ =

( ) ( )Ldi 0 di 0L 2 2A / sec

dt dt

+ +

=⇒ ⇒=

Q.27 (c)

Q.28 (6.99) It is given ( )CV 0 4− =

R=5Ω, C+4f so 1 40000RC

=

→ ( ) ( ) t /τ 40000tc CV t V 0 e ; t 0 4e− − −= > =

→ We are asked to find the charge last by capacitor From t = 25μs to 100 μs We know in a capacitor Q = CV Or ∆Q = C(∆V)

( ) ( )c cV 25μsec V 10μsecQ C∆ = − → 40000t

cV 4e−=

→ 6c 40000 100 10V

4e 1.47t 25μsec

−− × ×= ==

→ 6c 40000 100 10V

4e 0.073t 100μsec

−− × ×= ==

→ [ ]5 1.47 0.073 6.99μsQ∆ = − =

Q.29 (a) From the given circuit, consider the following circuit diagram

After rearrangement,

For t 0≥

( )0I i 0−=

2.5A=We can write

( )RtL

0i t I e−

=

( ) 4ti t 2.5e A−=

Q.30 10 Given

(i) ( ) ( )iV t 5 10cos t 60= − ω +

(ii) ( ) ( )ii t 5 X cos t= + ω

The average power is given by,

avg 0 0 01 01 1P V I V I cos= − φ

Where

01 01 1V I cosφ = Fundamental Power


Hence,

( )avg10 XP 5 5 cos 60 0

2×

= × − =

10X 1252 2

= ×

100X 1010

= =

Hence, the value of X is 10Ω


5.1 INTRODUCTION

The resonance in the electric circuit is because of the presence of both the energy storing elements i.e. the inductor and the capacitor. At resonant frequency rω , the inductor and capacitor will exchange the energy freely as a function of time, which results in sinusoidal oscillations either across the inductor or across the capacitor. In two terminal electrical network containing at least one inductor and one capacitor, we define resonance as the condition which exists when the input impedance of the network is purely resistive. A network is in resonance when the voltage and current at the network input terminals are in phase. In many electrical circuits, resonance is a very important phenomenon. The study of resonance is very useful, particularly in the area of communications. For example, the ability of radio receiver to select a certain frequency, transmitted by a station and to eliminate frequencies from other stations is based on the principle of resonance. The resonance may be classified into two groups 1) Series Resonant circuit2) Parallel Resonant circuit

5.2 SERIES RESONANCE

As shown in figure above the resistor, inductor and capacitor are connected in series, so the circuit is series RLC circuit. A sinusoidal voltage ‘VS’ sends a current ‘I’ through the circuit. In series RLC circuit, the current lag behind or leads the applied

voltage depending upon the values of XL and XC Consider series RLC circuit shown in figure (1). The total impedance for the series RLC circuit is

R L CZ Z Z Z= + +1Z R j L

j c= + ω +

ω1Z R j Lc

ωω

= + −

It is clear from the circuit that the current SVI

Z=

The circuit is said to be in resonance if current is in phase with the applied voltage. In series RLC circuit resonance occurs when L cX X= . The frequency at which the resonance occurs is called the resonant frequency (fr). Since L cX X= the impedance in series RLC circuit is purely resistive. At resonant frequency (fr) the voltage across capacitance and inductance are equal in magnitude. At resonance,

L CX X=

rr

1LC

ω =ω

r1 (rad / sec)LC

ω =

1fr (Hz)2 LC

=π

5.2.1 IMPEDANCE OF A SERIES RESONANT CIRCUIT

The impedance of a series RLC circuit is 1Z R j Lc

ωω

= + −

22 1Z R L

Cω

ω ⇒ = + −

5 RESONANCE


As shown in above graph • At zero frequency, |Z|is infinitely large.• At resonant frequency fr, |Z|=R• At frequencies f<fr, |Z| increases.• At frequencies f>fr, |Z|, increases.

5.2.2 THE VARIATION OF XC AND XL WITH FREQUENCY

• In series RLC circuit, the current lagsbehind or leads the applied voltagedepending upon the values of XC and XL.At zero frequency, both XC and Z areinfinitely large, and XL is zero because atzero frequency the capacitor acts asopen circuit and the inductor acts as ashort circuit. As frequency increases, XC

decreases and XL increases. While XC

causes the total current to lead theapplied voltage. When c LX X> thecircuit is predominantly capacitive (asshown in the graph above).

• In series RLC circuit, resonance occurswhen L CX X= . The frequency at which the resonance occurs is called theresonant frequency (fr). Since, L CX X=the impedance in a series RLC circuit ispurely resistive and Z=R (as shown inthe graph above)

• At frequencies above the resonantfrequencyfr, XL is larger than XC causingZ to increase. XL causes the total currentto lag behind the applied voltage. WhenXL>XC the circuit is predominantlyinductive (as shown in the graph above)

5.2.3 FREQUENCY RESPONSE OF SERIES RLC CIRCUIT

The figure above shows the variation of current I with frequency for small values of R. As shown in fig. the frequency f1 is the lower cut-off frequency, the frequency f2 is the upper cut-off frequency. The bandwidth or BW is defined as the frequency difference between f2 and f1

2 1BW f f (Hz)∴ = − The unit of BW is Hertz (Hz). There is another relationship for BW which is given by

r2 1

fBW f fQ

= − =

Where, Q is known as quality factor. The upper and lower cut off frequencies is sometimes called the half power frequencies. At resonant frequency the power is

2max maxP I R=

At frequency f1, the power is 2 2

max max1

I I RP R22

= =

Similarly, at frequency f2 , the power is – 2 2

max max2

I I RP R22

= =

5.2.4 MAGNIFICATIONS

We know that, in series RLC circuit, a sinusoidal voltage ‘VS’ sends a current I through the circuit.


So, sVIR

=

At = rω ω

1) RV IR= sV RR

= R sV V⇒ =

2) L LV IZ= sr

V jω LR

= L sV QV 90°⇒ = ∠

3) C CV =IZ s

r

V 1=R jω c C sV QV 90°⇒ = ∠−

Where, rω LQR

= where, r

1Qω CR

=

5.2.5 PHASOR REPRESENTATION

At the resonant frequency fr, the voltage across capacitance and inductance are equal in magnitude, since they are 180° out of phase with each other, they cancel each other and hence zero voltage appear across the LC combination .

5.2.6 QUALITY FACTOR (Q) AND ITS EFFECT ON BANDWIDTH

The Quality factor (Q) is the ratio of the reactive power in the inductor or capacitor to the true power in the resistance in series with the coil or capacitor. The Quality factor,

max imum energy storedQ 2energy dissipated per cycle

= π×

In an inductor, the maximum energy stored is given by

2LI2

=

and energy dissipated per cycle = power × Periodic time for one cycle.

2I R T2

= ×

2I RT2

=

So,

2

2

LI2Q 2

I RT2

= π×

LQ 2RT

= π×

L 1Q 2 T1 fRf

π Q = × =

2 fLQRπ

=

LQRω

=

1 1Q L at resonanceCR C

ωω ω

Q = =

In series RLC, the quality factor L 1Q

R cRω

= =ω

L 1 1 LQR cR R Cω

= = =ω

So, Q-factor is a function of only circuit constants. As Q is high then the circuit is said to be more selective and oscillation produced are high quality in nature.

Example: In circuit shown in figure, determine the circuit constants when the circuit draws a maximum current at 10μF with a 10v, 100Hz supply. When the capacitance is changed to12μF , the current that flows through the circuit becomes 0.707 times its maximum value. Determine Q of the coil at 900 rad/sec. Also find the maximum current that flows through the circuit

Solution:


At resonant frequency, the circuit draws maximum current. So, the resonant frequency rf 100Hz=

r1f

2 LC=

π

( )2r

1Lc 2 f

=π

( )26

1 0.25H10 10 2 100−

= =× π×

We have, 1L RC

ω − =ω

61900 0.25 R

900 12 10−× − =× ×

R 132.4⇒ = ΩThe quality factor

L 900 0.25Q 1.69R 132.4ω ×

= = =

The maximum current n the circuit is 10I 0.075

132.4= =

Example: In the circuit shown in fig, a maximum current of 0.1A flows through the circuit when the capacitor is at 5μFwith a fixed frequency and a voltage of 5V.Determine the frequency at which the circuit resonates, the bandwidth, the quality factor Q and the value of resistance at resonant at resonant frequency .

At resonance, the current is maximum in the circuit. It is given by

VIR

=

V 5 R 50I 0.1

∴ = = = Ω

The resonant frequency is

r1LC

ω =

6

1

0.1 5 10−=

× ×1414.2rad / sec=

r1414.2f 225HZ

2⇒ = =

πThe quality factor

L 1414.2 0.1Q 28R 50ω ×

= = =

Since, rf QBW

=

The Bandwidth, rf 225BW 80.36Hz

Q 2.8= = =

5.3 PARALLEL RESONANCE

As a shown in fig. above the resistor, capacitor and inductor are connected in parallel, so the circuit is parallel RLC circuit. The current flowing through resistor, inductor and capacitor are

R L CI , I and I respectively. Consider parallel RLC circuit shown in fig. The total admittance fr parallel RLC circuit is:

R L CY Y Y Y= + + 1 1Y j cR j L

= + + + ωω

1 1Y j( c )R L

= + ω −ω

It is clear from the circuit that current I = YV So voltage across parallel elements is same. Parallel resonance occurs when C LX X= the frequency at which resonance occurs is called the resonant frequency. When

C LX X= , the two branch currents are equal in magnitude and 180° out of phase with each other.


The condition for resonance occurs when C LX X=

At resonance, rr

1 cL= ω

ω

r1 (rad / sec)LC

ω =

r1f (Hz)

2 LC=

π

5.3.1 ADMITTANCE OF A PARALLEL RESONANT CIRCUIT

As shown in above graph- • At zero frequency, Y is infinitely large.

• At resonant frequency rf , 1YR

=

• At frequencies rf f< , Y increases.

• At frequencies rf f> , Y increases.

5.3.2 THE VARIATION OF 𝑿𝑿𝑳𝑳 AND 𝑿𝑿𝑪𝑪WITH FREQUENCY

• At zero frequency both1ωL

and Y are

large and ωC is zero. So 1 >ωCωL

the

circuit is predominantly Inductive.

• In parallel RLC circuit, resonance occurs

when 1 = ωCωL

. The frequency at which

the resonance occurs is called the

resonant frequency (fr) since 1 CL= ω

ω

the admittance in parallel RLC circuit is

purely resistive and IYR

=

• At frequencies above resonant

frequency fr, cw is large than 1ωL

so

1ωCωL

> the circuit is predominantly

capacitive.

5.3.3 MAGNIFICATION

We know that in parallel RLC circuit, voltage is

I IV IRIYR

= = =

The Response at resonance At rω=ω

1) RVIR

=IRR

= RI I⇒ =

2) LL

VIZ

=r

IRjω L

= LI QI 90⇒ = ∠− °

3) cc

VIZ

= r

r

IR jω CRI1jω C

= = cI QI 90⇒ = ∠ °

Where r

RQL

=ω

where rQ CR= ω

5.3.4 PHASOR REPRESENTATION

At the resonant frequency, when L CX ,X=

the two branch current are equal in magnitude and 1800 of phase with each


other. Therefore, the two current cancel out each other and the total current is zero.

Example: Determine the resonant frequency of the circuit

Solution: 1 1z 10 j4 ||j j

ωω ω

= + +

214

102j 4

ω

ωω

−= +

−

21j(4 )

10 24

−ω= −

ω−ω

At resonant frequency imaginary terms equals to zero

2r

rr

14024

−ω

=ω −

ω

2r

14 0⇒ − =ω r

1 rad / sec2

⇒ω =


Q.1 The differential equation for the current i(t) in the circuit of the figure is

a) ( )2

2d i di2 2 i t sin t

dtdt+ + =

b) ( )2

2d i di2 2i t cos t

dtdt+ + =

c) ( )2

2d i di2 2 i t cos t

dtdt+ + =

d) ( )2

2d i di2 2i t sin t

dtdt+ + =

[GATE-2003]

Q.2 Consider the network graph shown in the figure. Which one of the following is NOT a ‘tree’ of this graph?

a) b)

c) d)

[GATE-2004]

Q.3 In the following graph, the number of trees (p) and the number of cut sets (Q) are

a) P 2,Q 2= = b) P 2,Q 6= =c) P 4,Q 6= = d) P 4,Q 10= =

[GATE-2008]

1 2 3 (c) (b) (c)

ANSWER KEY:

GATE QUESTIONS(EC)


Q.1 (c) Applying KVL,

( ) ( )di(t) 1sin t i t 2 L i t dtdt C

= × + + ∫

( ) ( )di(t)sin t 2i t 2 i t dtdt

= + + ∫Differentiating with respect to t

( )2

22di(t) 2d i(t)cos t t i(t)

dt dt= + +

Q.2 (b) It is forming a closed loop. So it can’t be a tree.

Q.3 (c) Different tress (P) are shown below.

Different cut sets (Q) are shown below

So P = 4, Q = 6

EXPLANATIONS


6.1 GRAPH OF A NETWORK

Graph theory deals with graphs of networks and provide information that helps in the formulation of network equations. If each element or a branch of a network is represented on a diagram by a line irrespective of the characteristics of the elements, we get a graph. Network equations can be easily written by converting the network into a graph. A network is an interconnection of active elements (voltage and current source) and passive elements (resistor, capacitor, and inductor) as shown in fig (1)

If two or more branches or elements of a network intersect at a single point, that point is known as node and it is denoted by ‘n’. As shown in above network there are four nodes represented by number (1, 2, 3, 4). Two nodes joined by line segment is known branch, which is denoted by ‘b’ and represented by letters (a, b, c…..). In above network there are six branches. While constructing graph from above network represent network element by lines, internal impedance of ideal voltage source is zero and replaced by short circuit and internal impedance of ideal current source is infinite and replaced by open circuit. The graph of network (shown in fig (1)) can be drawn as (shown in fig (2))

Fig (2) has four nodes and six branches. This graph is called as undirected graph as directions are not given. Each branch or edge of the graph carries an arrow to indicate its orientation. A graph whose branches are oriented is called a directed or oriented graph as shown in fig (3)

In a complete graph between any pair of node only one branch is connected for all the combinations. The number of edges of a complete graph with n nodes is

n(n 1)2−

From given graph, we have 4 nodes So, node(n) 4= and branches

6 GRAPH THEORY


n(n 1) 4(4 1) 62 2− −

= = =

In a connected graph all the nodes are connected by at least one branch otherwise it is said to be unconnected graph.

Sub graph is a graph with less number of branches as compared with the original graph. The rank of a graph is (n-1) where n is the number of nodes or vertices of the graph.

6.1.1 TREES, COTREES AND LOOPS

A network consists of n nodes which are interconnected in some way, by b edges or branches. If we start at any node, and return to the staring node such a closed path formed by network branch is known as loop. A tree is a connected sub graph of a network which consists of all the nodes of the original graph but containing no closed loops. The number of tree branches are (n-1) where ‘n’ is the number of nodes orvertices of a graph. The branches of a tree are called twigs; those branches that are not in a tree are called links or chords. All the links of a tree together (constitute the complement of the corresponding tree) is called the cotree. Or cotree can be also defined as ‘It is a tree formed with all the removed branches from the network graph in order to construct a tree. The cotree branches are called links or chords. The number of links or chords is

( )b n 1 b n 1− − = − +Let us consider following directed graph.

The above graph contains 4 nodes & 6 branches. A tree has (n 1)− i.e. (4 1) 3− = branches and co tree (b n 1)− + i.(6 4 1) 3− + = branches. Following figures shows possible trees and co-trees

Tree cotree (2,5,6) (1,3,4)

(2,4,5) (1,3,6)

(1,4,6) (2,3,5) Note: • Cotree may consist of a closed loop.• Also, tree + cotree = original graph

6.1.2 PROPERTIES OF TREES

1) In a tree there exists one and only onepath between any pair of nodes.


2) Every connected graph has at least onetree.

3) The number of terminal nodes or endvertices of every tree are at least two.

4) A connected sub graph of a connectedgraph is a tree if there exists all thenodes of the group.

5) Each tree has (n-1) branches, where n isthe number of nodes of the tree.

6) The rank of tree is (n-1).This is also therank of the graph to which the treebelongs.

7) The number of possible trees ofcomplete graph is given by (n 2)n − .e.g. For the complete graph if nodes n=4then number of possible trees

(4 2)4 16−= = possible trees.

6.2 INCIDENCE MATRIX

In the above oriented graph, from the arrows indicated in the graph, it is possible to tell which branches are incident at which nodes and what is the orientation relative to the nodes. The most convenient way in which this incidence information can be given is in a matrix form known as incidence matrix ‘A’ For a graph with n nodes and b branches, the complete incidence matrix ‘A’ is rectangular matrix of order n×b whose elements have following value 1) When element incident to node and its

direction is away from node then entry of that element in incidence matrix is represented by ‘+1’

2) When element incident to node and itsdirection is towards the node then

entry of that element in incidence matrix represented by ‘-1’

3) And entry of the element representedby‘0’ when branch is not connected to that node.

For the above fig, the complete incidence matrix A is given as – nodes

bran esch→↓

a b c d e f1 1 0 1 0 0 12 1 1 0 1 0 0

A3 0 1 0 0 1 14 0 0 1 1 1 0

− − = − − − −

6.2.1 PROPERTIES OF INCIDENCE MATRIX

1) The rank of incidence matrix is equal tothe rank of graph (n 1)= −

2) For any graph the incidence matrix isunique and its order is n×b

3) The determinant of the incidencematrix of a closed loop is always equalto zero

4) The sum of all the entries in a column iszero.

5) The number of non-zero elements (1’sand -1’s) in a row is called the degree ofa node.

6) If the degree of a node is two, then itindicates that two branches are incidentat the node and these are in series.

7) Column of ‘A’ with unit entries in twoidentical rows corresponds to twobranches with same end nodes andhence they are in parallel.

6.3 FUNDAMENTAL LOOP MATRIX OR TIE –SET MATRIX


For a given tree of a graph, addition of each link between any two nodes forms a loop called the Fundamental loop. In a loop there exists a closed path and circulating current, which is called the ‘link current.’ The current in any branch of a graph can be found by using link currents. The Fundamental loop formed by one link has a unique path in the tree joining the two nodes of the link. This loop is called f-loop or a tie-set Def: Fundamental circuits or Fundamental loops are the minimum number of loops or mesh equations required to solve a given network. Steps: 1. Select a tree2. By adding one link to the existing tree

will result one f-loop at a time.3. Select fundamental loop current

direction as in the link currentdirection.

Example: Consider connected graph shown above, it has 4 nodes and 6 branches. One of its trees is arbitrarily selected having twig or tree branches (4, 5, 6) and link corresponding to it is (1, 2, 3) as shown below

Let 1 2i , i , 6..i… be the branch currents with directions shown in fig above and 1 2v , v ,

6..v… are the branch voltages. Let us add a link (say link represented by1) in its proper place in selected tree (shown in fig. (3)).The loop current i1 is formed by branches 1, 5 and 6. The direction of loop current is in the direction of added link as a shown in fig (3)

From fig. (3) KVL can be written as: 1 5 6v v v 0+ − = _________ (1)

Now By adding the other link branches 2 and 3 we can form two more fundamental loops as follow-

From fig. (4) f-loop can be written as: 2 4 5v v v 0+ − = _________ (2)

From fig. (5) f loop can be written as: 3 4v v 0− = _________ (3)

Equation (1),(2) (3) can be written in matrix form as follow- loop

Branches →↓

1 2 3 4 5 6


1

21

32

43

5

6

vv

I 1 0 0 0 1 1 0v

I 0 1 0 1 1 0 0v

I 0 0 1 1 0 0 0vv

− − = −

Note: In general the order of tie-set matrix is (b n 1) b− + × i.e. (link × branches) The elements in tie-set or fundamental loop matrix is defined as follow- 1) When direction of loop current and

reference direction of loop coinciderepresent that element by ‘+1’

2) When direction of loop current andreference direction of loop are oppositerepresent that element by ‘-1’

3) When any branches is not infundamental loop represent it by ‘0’

6.3.1 PROPERTIES OF TIE-SET MATRIX

1) The rank of the tie-set matrix is(b n 1)− +

2) Since every link will result one tie set ata time, for any graph the number offundamental loops equal to link(b n 1)− +

3) Every fundamental loop consists of onlyone link in its representation.

4) Since every tree will result one tie-setmatrix at a time, for any graph thenumber of tie-set matrices are alwaysequal to the number of trees

5) For a complete graph there are n(n−2)

tie-set matrices.

6.3.2 CUTSET A cutset is a minimal set of branches of a connected graph such that the removal of these branches divides the graph into exactly two parts. Example:

Fig (1) shown above is not a complete graph since there are two branches (3 & 4) between nodes III and IV The removal of the branch 1 and 5 divides the graph into exactly two parts. So (1, 5) may be a cut-set (as shown in fig.2).

Cut set (1, 5) Other possible cut-sets are

Cut-set (1, 2)

Cut-set (5, 2)

Cut-set (2, 3, 4)


Cut-set (5, 3, 4)

Cut-set (1, 3, 4)

Not a valid Cut-set As shown in fig (3) if we remove branches 1, 2 and 5 then it divides the graph into 3 isolated parts so it is not at all a cut-set.

6.3.3 FUNDAMENTAL CUT-SET

Fundamental cut-set are the minimum number of nodal equations required to solve a given network. Steps: 1) Select a tree.2) By removing one tree branch at a time

will result one f-cut set at a time.3) Select the f-cut set direction as in the

tree branch direction.

Example:

Note: It is not a compete graph

Let’s select a tree (1, 7, and 5) as shown below

Now remove twig (say twig 1) from selected tree in such a way that tree disconnects into two parts and it also cuts all the links which go from one part of this disconnected tree to the other, together constitute a cutest, we call this a fundamental cut-set as shown in fig below.

So from above we get fundamental f-cut set (1, 2, and 4). Similarly consider twig 7 then f-cut set drawn as follows:


f-cut-set obtained after twig 7 removed is (4,7,6,3) Similarly consider twig 5 removed, we can draw cut set as:

So we get fundamental cutest (5, 6)

6.4 F–CUT SET MATRIX (QC)

f cutsetBranches

−→

↓

C

1 2 3 4 5 6 71

1 1 0 1 0 0 0Q 7

0 0 1 1 0 1 15

0 0 0 0 1 1 0

− = − − −

Where, 1) When the direction of the f-cut-set and

that of the other branches coincide thenis represented by ‘1’

2) When the direction of the f cut-set andthat of the other branches are oppositethen it is represented by ‘-1’

3) When f-cut-set doesn’t contain anybranch then that branch is representedby ‘0’

Properties: 1) The rank of the f-cut set matrix is n-12) Since every tree branch will result one f

–cut-set at a time, for any graph thenumber of f-cut-set equal to number of twigs equal to n-1

3) Every f-cut set contain only one treebranch is its representation.

4) Since every tree will result one f-cut setmatrix at a time, for any graph thenumber of f-cut-set matrices are alwaysequal to the number of trees.

5) For a compete graph there are ( )n 2n − f-cut set matrices.


Q.1 In the figure, the value of the load resistor R which maximize the power delivered to it is

a)14.14Ω b)10Ωc) 200Ω d) 28.28Ω

[GATE-2001]

Q.2 A source of angular frequency 1 rad/sec has source impedance consisting of 1Ω resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is a) 1Ω resistanceb) 1Ω resistance in parallel with 1H

inductancec) 1Ω resistance in series with 1F

capacitord) 1Ω resistance in parallel with 1 F

capacitor[GATE-2003]

Q.3 The maximum power that can be transferred to the load resistor LRfrom the voltage source in the figure is

a)1W b)10W c)0.25 W d)0.5W

[GATE-2005]

Q.4 An independent voltage source in series with an independence

S S SZ R jX= + delivers a maximum

average power to a load impedance L Z when

a) L S SZ =R +jX b) L SZ =Rc) L SZ =jX d) L S SZ =R -jX

[GATE-2007]

Q.5 For the circuit shown in the figure, the Thevenin voltage and resistance looking into X-Y are

a) 4 / 3V,2Ω b) 4V,2 / 3Ω

c) 24 V,3 3Ω

d) 4V,2Ω

[GATE-2007]

Q.6 The Thevenin equivalent impedance ZTh between the nodes P and Q in the following circuit is

a) 1 b) 11 ss

+ +

c) 12 ss

+ + d) 2

2s s 1

s 2s 1+ ++ +

[GATE-2008]

Q.7 In the circuit shown, what value of LR maximizes the power delivered

to LR ?

GATE QUESTIONS(EC)


a)2.4Ω b) 83Ω

c) 4 Ω d) 6Ω[GATE-2009]

Q.8 In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

a) 6.4-j4.8 b) 6.56-j7.87c)10+j0 d)16+j0

[GATE-2011]

Q.9 In the circuit shown below, the value of LR such that the power transferred to LR is maximum is

a) 5Ω b)10Ωc)15Ω d) 20Ω

[GATE-2011]


a) 50Ω b)100Ωc) 5kΩ d)10.1kΩ

[GATE-2012]

Q.11 Assuming both the voltage sources are in phase the value of R for which

maximum power is transferred from circuit A to circuit B is

a) 0.8Ω b) 1.4Ωc) 2Ω d) 2.8Ω

[GATE-2012] Q.12 A source ( )sV t =Vcos100πt has an

internal impedance of 4+j3Ω . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in Ω should be a)3 b)4 c)5 d)7

[GATE-2013]

Q.13 In the circuit shown below, if the source voltage SV 100 53.13° V= ∠ then the venin’s equivalent voltage in volts as seen by the load resistance RL is

a)100 90°∠ b) 800 0°∠c)800 90°∠ d)100 60°∠

[GATE-2013]

Q.14 Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance a) in series with a current sourceb) in parallel with a voltage sourcec) in series with a voltage sourced) in parallel with a current source

[GATE-2014]


Q.15 In the figure shown, the value of the current I (in Amperes) is_________.

[GATE-2014]

Q.16 In the circuit shown in the figure, the angular frequency ω (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b' is purely resistive, is ___________.

[GATE-2014]

Q.17 In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is ______.

[GATE-2016]

Q.18 In the circuit shown below, Vs is a constant voltage source and LI is a constant current load.

The value of LI that maximizes the power absorbed by the constant current load is

a) sV4R

b) sV2R

c) sVR

d) ∞

[GATE-2016]

Q.19 Consider the circuit shown in the figure.

The Thevenin equivalent resistance (in Ω) across P-Q is____________.

[GATE-2017]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (c) (c) (d) (d) (a) (c) (a) (c) (a) (a) (c) (c) (d) 15 16 17 18 19 0.5 (2r/sec) 0.8 (b) 1

ANSWER KEY:


Q.1 (a) ωL 10Ω= R for max power transfer

10 jωL,= − (Complex conjugate of eqZ )

10 10j= − 10 2 45°= ∠ 14.14Ω=

Q.2 (c) L S SZ R jX= −

LZ 1 1j∴ = −

Q.3 (c) For maximum power transfer,

L SR R 100Ω= = 2

maxV 5 5P 0.25WR 100

×∴ = = =

L(V acrossR 5V)=

Q.4 (d) For maximum power transfer,

L S S SZ Z * R JX= = − Q.5 (d)

For OCVApply nodal analysis,

x xx x

V V 2i2 V 0&V i2 1

−∴ − + + = =

x xx

V V2 2V 2i 0 22 2

⇒ − + − = ⇒ =

OC xV V 4V⇒ = =Similarly, shI =2A

OCth th

sh

VV 4V;R 2I

∴ = = = Ω

Q.6 (a) Replace 10V by short and 1A by open,

( )th PQ

1s 1 (1 )SZ Z 112 S S

+ +∴ = = =

+ +

Q.7 (c) For maximum power transfer, RL =Rth

OCV 100V=

( )xsh

100 V100I8 4

+= +

Also xV =-50V,

shI 12.5 12.5 25A∴ = + =

OCth

sh

VR 4I∴ = = Ω LR 4⇒ = Ω

Q.8 (a)

( )N sh PQ25I I 16 0°

40 J30∠= = ×

+

1 38 tan 8 36.864

∠ ∠− = − = −

hence Norton current is N SCI I 36.86°= = ∠−

( )NI 6.4 j4.8 A= −

Q.9 (c)

For Maximum power transfer to RL L thR R=

ThR = Thevenin’s resistance seen across the terminals of RLinto the rest of the NW. The relevant circuit

EXPLANATIONS


is shown in Fig 1, where the independent current source is open circuited and the voltage sources are short circuited.

Q.10 (a)

After connecting voltage source of V ( )( ) ( )1 2 b b bV V 10K i 100 I 99i i ;= ⇒ − = + +

b b10000i 100I 100 i− = + ×

b100I 10000i= + b b20000i 100I i= ⇒−100 II

20000 200− − = =

[ ]b bV 100 I 99i i= + +

I100 I 100200

− = + 50I=

thV 50IR 50I I

= = = Ω

Q.11 (a) Power transferred from circuit A to

circuit 7 6 10RB VIR 2 R 2

+ = = + +

( )42 70R

R 2+

=+

10 3 7I2 R 2 R−

= =+ +

7RV 3 IR 32 R

= + = ++

6 10R2 R+ = +

( ) ( ) ( ) ( )( )

2

4R 2 70 42 70R 2 R 2dP

dR R 2

+ − + +=

+

( ) ( ) ( )270 R 2 42 70R 2 R 2+ = + +

( )5 R 2 2(3 5R)⇒ + = +5R 10 6 10R⇒ + = + 4 5R⇒ =R 0.8Ω⇒ =

Q.12 (c) 2 2

L thR Z 4 3 5= = + = Ω Q.13 (c)

TH L1V 10V= 418c

L1V 100 53.13V tan j4

3 j4 3∠

− = = − ×

+

L1V 80 90°= ∠

THV 800 90°= ∠ Q.14 (d)

Norton's theorem

Q.15 (0.5)

Apply KCL at node

V, V 5 V15 15−

− + = 0

30 volts4

V⇒ =

V 2current I 0.50Amperes15 4

⇒ = ⇒ ⇒

Q.16 (2 r/sec) Norton's equivalent impedance

11* . 121 .11 .2

N

jZ

jj

ω

ωω= +

+= . 1

2j

j jωω ω+

+

( )2

N 2

2 ω jωZ

[2jω ω ]− +

=−


( )2 2

N 4 2

ω 2 jω . ω 2jωZ

ω 4ω

− − + ⇒ +

Equating imaginary term t zero i.e., 3ω -4ω=0

( )2ω ω 4 0 ω 2r / sec⇒ − = ⇒ =

Q.17 (0.8) To find maximum power transferred to load we need to obtain thevenin equivalent of the circuit → obtaining Voc

o2V 5 2V

3 2=

+=

oc40V 100

10 40=

+=

04V = ×100×2=160V5

→obtaining Isc

o2V 5 2V

3 2= =

+

= osc

100V 200I 20mA10 10

= = =

ocin

sc

V 160R 8kI 20

= = = Ω

so the network is

→ for MPT R = 8k 2 2

+nmcr

+n

V 160P = =4R (4×8)×103

= 0.8watt.

Q.18 (b) Under maximum power transfer condition, half of thV is dropped

across thR and remaining thV2

dropped across load.

→ So we can say under MPT sV2

will appear on the load

so LI = s

sVV2

R

−= sV

2R

Q.19 1Ω


Q.1 The graph of an electrical network has N node and B branches. The number of links L, with respect to the choice of a tree, is given by a) B-N+1 b) B+Nc) -B+1 d) N-2B-1

[GATE -2002]

Q.2 The matrix A given below is the node incidence matrix is a network. The columns correspond to branches of the network while the rows correspond to nodes. Let

[ ]T1 2 6V v v v= … denote the vector of

branch voltages while [ ]T1 2 6I i i i= …that of branch currents. The vector

[ ]T1 2 3 4E e e e e= denotes the vector ofnode voltages relative to a common ground.

1 1 1 0 0 00 1 0 1 1 0

A1 0 0 0 1 1

0 0 11 0 1

− − = − − − −

Which of the following statements is true? a) The equations 1 2 3v -v +v = 0,

3 4 5v + v -v = 0 are KVL equationsfor the network for some loops

b) The equations 1 3 6v -v -v = 0,

4 5 6v + v -v = 0 are KVL equations for the network for some loops c) E=AVd) AV=0 are KVL equations forthe network

[GATE-2007]

Q.3 The number of chords in the graph of the given circuit will be

a)3 b)4 c)5 d)6

[GATE-2008]

Q.4 The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KCL equations and the number of independent KV L equations, respectively, are a) 2 and 5 b) 5 and 2c) 3 and 4 d) 4 and 3

[GATE-2016]

Q.5 The voltage ( )v t across theterminals a and b as shown in the figure, is a sinusoidal voltage having a frequency 100radian / sω = . When the inductor current ( )i t in phase

with the voltage ( )v t , the magnitudeof the impedance Z (in Ω ) seen between the terminals a and b is ________ (up to 2 decimal places).

[GATE-2018]

GATE QUESTIONS(EE)


1 2 3 4 5 (a) (b) (a) (d) 50

ANSWER KEY:


Q.1 (a) Number of links = B - (N - 1) = B – N + 1

Q.2 (b) For the given node–to branch incidence matrix

1 1 1 0 0 00 1 0 1 1 0

A1 0 0 0 1 1

0 0 11 0 1

− − = − − − −

The graph of the network is shown in fig

Where N=4, B=6 From the graph it can be observed that i) 1 2 3 3 4 5V V V 0,V V V 0− + = + − =

are not KVL equations as set ofbranches (1,2,3) & (3,4,5) do notform closed paths.

ii) 1 3 6V V V 0− − = and

4 5 6 0V V V+ − = are KVL equations for the loops (1,3,6)and (4,5,6) From the matrix, A itcan be concluded that(i) + E A V≠(ii) A V 0= are not KVL

equations

Q.3 (a) The graph of the given circuit is shown in Fig Number of nodes =N=4

Number of branches =B=6 Number of tree branches = (N-1) = 3 Number of links =L=B-(N-1) =3

∴ Statement in options (b) is true.

Q.4 (d) No of branches = 7 Nodes = 5 No of KCL equations = No of Modal equations = n —1 = 5 —1 = 4 No of KVL equations= No of Mesh equations= b-(n —1) = 7-4 = 3 Since no information given regarding how many simple & principal node, if we assume all principal nodes then the answer for nodal is 5 —1

Q.5 50 Given:

1100j C

Z j Lj100C

ω = + ω − ω

22 2

j100 j100C C j L1100

C

− + ω ω = + ω+ω

EXPLANATIONS


Since, the power factor is unity, unity power factor represents resonance. So, imaginary part of Z is zero only real part exist.

( )2 2

22 2

100CRe Z 1100

C

ω=+ω

( )2100

100 C 1=

ω +

( )26

100

100 100 100 10 1−=

× × × +

( )Re Z 50= Ω

Hence, the magnitude of the impedance Z seen between the terminals a and b is50Ω .


7.1 INTRODUCTION

Two circuits are said to be ‘coupled’ when energy transfer takes place from one circuit to the other, when one of the circuit is energized. Whenever, a current flows through a conductor, whether as ac or dc, a magnetic field is generated around that conductor. A majority of electrical circuits in practice are conductively or electromagnetically coupled. Certain coupled elements are frequently used in network analysis & synthesis e.g. transformer.

7.2 MUTUAL INDUCTANCE

Mutual inductance is a property associated with two or more coils or inductors which are in close proximity and the presence of common magnetic flux which links the coil. A transformer is such a device whose operation is based on mutual inductance. A transformer consists of two coils of wire separated by a small distance and is commonly used to convert ac voltage to higher or lower values depending on the application. A current flowing in one coil establishes a magnetic flux about that coil and also about a second coil nearby. The time varying flux surrounding the second coil produces voltage across the terminals of the second coil; this voltage is proportional to the time rate of change of the current flowing through the first coil.

( ) di(t)v t Ldt

=

Where, ( )v t is the voltage across the coil

( )i t is the current through the coilL is the inductance of the coil.

7.2.1 COEFFICIENT OF MUTUAL INDUCTANCE

As shown in fig a simple model of two coils L1 and L2 sufficiently close together that the flux produced by a current 1i (t) flowing through 1L establishes an open circuit voltage 2v (t) across the terminals of L2

( ) 12 1

di (t)v t Mdt

=

Where, v2 is the voltage induced in coil 2L and M1 is the coefficient of proportionality (OR) coefficient of mutual inductance (OR) simple mutual inductance. Similarly from Fig (b), we can write

( ) 21 2

di (t)v t Mdt

=

1M and M2 are two mutual inductances are involved in determining the mutually induced voltages in the two coils, it can be shown from energy considerations that the two coefficients are equal and, therefore need not be represented by two different letters. Thus 1 2M M M= =

( ) 12 1

di (t)v t M voltsdt

∴ = and

( ) 21 2

di (t)v t M voltsdt

=

Mutual inductance is also measured in Henrys

7.2.2 DOT CONVENTION

Dot convention is used to establish the choice of correct sign for the mutually

7 COUPLED CIRCUITS


induced voltage in coupled circuits. Circular dot marks and /or special symbols are placed at one end of each of two coils which are mutually coupled to simplify the diagrammatic representation of the windings around its core.

Consider above circuit which shows a pair of linear time invariant, coupled inductor with self inductance L1 and L2 and mutual inductance M. Currents i1 and i2, each arbitrarily assumed entering at dotted terminals and voltage v1 & v2 are developed across the inductors. The voltage across L1is, thus composed of two parts and is given by

( ) 1 21 1

di (t) di (t)v t L Mdt dt

= ± (1)

The first term on the RHS of the above neq is the self induced voltage due to i1 and the second term represents the mutually induced voltage due to i2. Similarly,

( ) 2 12 2

di (t) di (t)v t L Mdt dt

= ± (2)

Although, the self induced voltages are designated with positive sign and mutually induced voltages can be either positive or negative depending on the direction of the winding of the coil and can be decided by the presence of the dots placed at one end of each of the two coils. The convention is as follows-

If the current enters or leaves the dotted terminals of both the coils then the mutual inductance term is positive (as shown in fig. (1) & fig. (2)) The assumed current i1 and i2 produced flux in the core that are additive. The terminal ‘a’ and ‘c’ of the two coils attain similar polarities simultaneously. The two terminals are positive & identified by two dots (show in fig. (1)). The other possible location of the dots is the other ends of the coil to get additive fluxes in the core i.e. at b and d terminal (shown in fig (2)). It is concluded that the mutually induced voltage is positive when current i1 and i2 both enter (or leave) the windings by the dotted terminals:

i.e. 12

div Mdt

=

If the current in one winding enters at the dot marked terminals and the current in the other winding leaves at the dot-marked terminal , the voltage due to self and mutual induction in any coil have opposite sign (shown in fig (3) &(4))

i.e. 12

div Mdt

= −


7.3 THE COUPLING COEFFICIENT The amount of coupling between the inductively coupled coils is expressed in terms of the coefficient of coupling, which is defined as

1 2

MkL L

=

Where, M mutual inductance between the coils L1 self inductance of the first coil L2 self inductance of the second coil

Coefficient of coupling is always less than unity and has maximum values of 1. If the value of k=1 then it called perfect coupling The coefficient k is a non negative number and is independent of the reference directions of the currents in the coil. If the two coils are a great distance apart in space, the mutual inductance is very small, and k is very small.

7.4 SERIES CONNECTION OF COUPLED INDUCTORS

Let there be two inductors connected in series with self inductances L1 and L2 and mutual inductance M. Two kinds of series connections are possible

1) Series aiding (fig a): In case of seriesaiding connection, the currents in bothinductors at any instant of time are inthe same direction relative to liketerminals (shown in fig a) For thisreason, the magnetic fluxes of selfinduction and of mutual inductionlinking with each element add together.For the series aiding circuit 1 2Φ & Φare the flux produced by coil 1 and 2respectively, and flux produced by theinductor is given by iLΦ =So the total flux

1 2Φ Φ Φ= + Where, 1 1 1 2L i MiΦ = +

2 2 2 1Φ L i Mi= +

1 1 2 2 2 1Φ L i Mi L i Mi∴ = + + + Since 1 2i i i= =

1 2L L L 2M= + +

2) Series opposition (fig b): In case ofseries opposition the current in the twoconductors at any instant of time are inopposite direction relative to liketerminals (shown in fig b)So, for the series opposition

1 2Φ Φ Φ= +Where, 1 1 1 2Φ L i Mi= −

2 2 2 1Φ L i Mi= −

1 1 2 2 2 1Φ L i Mi L i Mi∴ = − + −Since 1 2i i i= =

1 2L L L 2M= + −In general, the inductance of twoinductively coupled elements in seriesis given by

1 2L L L 2M= + ±• Positive sign is applied to the series

aiding connection• Negative sign is applied to the series

opposition connection.


7.5 PARALLEL CONNECTION OF COUPLED COILS

Consider two inductors with self inductances 1L and 2L connected parallel which are mutually coupled with mutual inductance M shown in fig.

21 2

eq1 2

L L MLL L 2M

−=

+ −

21 2

eq1 2

L L MLL L 2M

−=

+ +

7.6 IDEAL TRANSFORMER

Transformers are used to transfer energy from one circuit to another circuit through mutual induction. The transformer winding to which the supply source is connected is called the primary and voltage across it is called primary voltage, while the winding connected to load is called the secondary and voltage across it is called secondary voltage correspondingly i1 and i2 are the currents in the primary and secondary windings .

1 2 1 1

2 1 2 2

V I N LV I N L

= = =

Where, 1

2

NN

is turns ratio of transformer

Input impedance of transformer is 21

in L2

NZ ( ) ZN

=

7.6.1 EQUIVALENT CIRCUIT

1 21 1

di diV L Mdt dt

= + (1)

2 12 2

di diV L Mdt dt

= + (2)

Add and subtract 1diMdt

in equation (1) and

2diMdt

in equation (2), equation (1)

becomes 1 2 1 1

1 1di di di diV L M M Mdt dt dt dt

= + + −

11 1 1 2

di dV (L M) M (i i )dt dt

= − + + (3)

Equation (2) becomes 2 1 2 2

2 2di di di diV L M M Mdt dt dt dt

= + + −

22 1 2

di d (L M) M (i i )dt dt

= − + + (4)

Using (3) and (4) the equivalent circuit can be drawn as

Example: Calculate the effective inductance of the circuit shown in figure across terminal A and B


Solution: ab 1 2 3 12 23L L L L 2M 2M= + + − + 8 10 6 2 4 2 526H

= + + − × + ×=

Example: The equivalent inductance between the terminals 1 and 2 is 4 H with the open terminals 3 and 4, and it is 3H with shorted terminals 3 and 4. Determine the value of k

Solution: The equivalent circuit of the above network can be as shown in figure.

When terminals 3 and 4 are open-circuited

12 pL L M M= − +

p4 L= When terminals 3 and 4 are short circuited

( )12L 2 M || M (4 M)= − + −

( )2 M M3 4 M

2−

= + −

22M M3 4 M2−

= + −

2M3 M 4 M2

= − + −

M 2H=

s pM k L L=

P S

M 2 1k2L L 4 2

= = =×

1k2

=


8.1 INTRODUCTION

Any network may be represented schematically by a rectangular box. A pair of terminals at which a signal may enter or leave a network is called a port. A network having only one pair of terminals is called as one port network. (As shown in fig a)

Fig a. one port Network A network having two pairs of terminals is known as two port network. Usually one pair represents the input (1 1')− and the other represents the output (2 2’)− (as shown in fig b)

Fig b. two port Network As shown in fig b, input terminal has two variables 1v , I1 and output terminal has two variables 2 2v , I . Out of these four variables, we can select two variables as independent variables represented as 24C i.e. 6 ways and hence six sets of two port network parameters, they are 1) z-parameter2) y-parameters3) Transmission (ABCD) parameters.4) Inverse transmission (A’B’C’D’)

Parameters5) Hybrid (h) parameters6) Inverse hybrid (g) parameters

8.2 OPEN CIRCUIT IMPEDANCE (Z) PARAMETERS

The z-parameters of a two port network having voltages 1 2v , v in terms of the current 1 2I , I are as shown in figure. Here

1 2v , v are dependent variables and 1 2I , I are independent variables. By kVL at port 1 we get

1 11 1 12 2v z I z I= + ---- (1) By kVL at port 2 we get

2 21 1 22 2v z I z I= + ---- (2) Where 11 12 21 22z , z , z , z are the network functions, and are called impedance (z) parameters we may write matrix

[ ] [ ]neq v z [I]=

Thus 1 11 12 1

2 21 22 2

v z z Iv z z I

=

From eqn (1)

2

111 I 0

1

vz |I == …..Driving point impedance

Where, Z11 is driving point impedance at port 1 with port 2 open circuited. It is called the open circuit input impedance. Similarly, from eqn (2)

2

221 I 0

1

vz |I == … Transfer impedance

Where, 21z Is transfer impedance at port 1 with

port 2 open-circuited. It is also called the open circuit forward transfer impedance. From eqn (1)

1

112 I 0

2

vz |I == …Transfer impedance

8 TWO PORT NETWORKS


Where, 12z is transfer impedance at port 2 with

port 1 open circuited. It is also called the open circuit reverse transfer impedance. Similarly from eqn (2)

1

222 I 0

2

vz |I == …Driving point impedance

Where, 22z is open circuit driving point impedance at port 2 with port 1 open circuited. It is also called the open circuit output impedance.

8.3 SHORT CIRCUIT ADMITTANCE (Y) PARAMETERS

The y-parameters of a two port network expressing the port current 1 2I andI in terms of voltages 1 2v and v are as shown in the figure. Here 1 2I , I are dependent variables and 1 2v & v are independent variables By KCL at port (1)

1 11 1 12 2I Y v Y v= + -----------------------------(1) By KCL at port (2)

2 21 1 22 2I Y v y v= + ----------------------------(2) Where, 11 12 21 22Y ,Y , Y , Y are the network function and are also called the admittance (y) parameters. We may write matrix

[ ] [ ]neq I Y [v]=

Thus 1 11 12 1

2 21 22 2

I Y Y vI Y Y v

=

From eqn(1)

2

111 v 0

1

IY |v == …Driving point admittance

Where,Y11 is driving point admittance at port 1 with port 2 short-circuited. It is also called the short circuit input admittance. Similarly from eqn(2)

2

221 v 0

1

IY |v == …Transfer admittance

Where, 21Y is the transfer admittance at port 1 with port 2 short-circuited. It is also called short circuited forward transfer admittance. Similarly from eqn(1)

1

112 v 0

2

IY |v == …Transfer admittance

Where,Y12 is the transfer admittance at port 2 with port 1 short circuited .It is also called the short circuit reverse transfer admittance. Similarly from eqn (2)

1

222 v 0

2

IY |v == …Driving point admittance

Where 22y is driving point admittance at port 2 with port 1 short circuited .It is also called the short circuit output admittance.

8.4 TRANSMISSION (A B C D) PARAMETERS

In transmission parameters, the input variables 1 1v & I at port 1-1’ called the sending end are expressed in terms of output variable 2 2v & I at port 2 are known as the receiving end, Transmission Parameters are defined by –

1 2 2v Av BI= − ---------------------------------(1)

1 2 2I Cv DI= − -------------------------------- (2) The negative sign is used with I2 and not for parameter B and D. The parameters A, B, C and D are called transmission parameters. They can be expressed in matrix from as follow

1 2

1 2

v vA BI IC D

= −

The matrix A BC D

is called transmission

matrix. For given network, with port 2


open circuited, then from equation (1) and (2) we get

2

1I 0

2

v |v ==A

and 2

1I 0

2

I |v ==C

Now when port 2 is short circuited, then from eqn(1) & (2) we get ---

2

1v 0

2

v |I ==

−B and

2

1v 0

2

I |I ==

−D

8.5 INVERSE TRANSMISSION (A’ B’ C’ D’) PARAMETERS

In inverse transmission parameters, output variables v2 and I2 are expressed in terms of input variables v1 andI1. Inverse transmission parameters are defined by –

2 1 1v A'v B'I= − -------- (1)

2 1 1I C'v D'I= − ------- (2) The coefficients A’,B’,C’,D’ in the above equations are known as inverse transmission parameters. For given network, when port 1 is open

1(I 0)= then from eqn(1) & (2) we get ---

1

2I 0

1

v' |v ==A

and 1

2I 0

1

I' |v ==C

When port1 is short circuited v1 = 0 then from eqn(1) &(2) we get

1

2v 0

1

v' |I ==

−B and

1

2v 0

1

I' |I ==

−D

8.6 HYBRID (h) PARAMETERS

The hybrid matrices describe a two port, when the voltage of one port and the current of other port are taken as the independent variables. From above network, if voltage at port 1 and current at port 2 are taken as dependent variables, we can express them in terms of 1I an 2v

1 11 1 12 2v h I h v= + ---- (1)

2 21 1 22 2I h I h v= + ---- (2) Where, 11 12 21 22h ,h ,h & h are called hybrid Parameters can be expressed in matrix form as follows

1 11 12 1

2 21 22 2

v h h II h h v

= From eqn(1) & (2) when 2v 0= the port 2 is short circuited Then

2

111 v 0

1

vh |I ==

Short circuited input impedance 11

1Y

2

221 v 0

1

Ih |I ==

Short circuit forward current gain 21

11

YY

Similarly from eqn(1) & (2) Let port 1 be open i.e. I1 = 0

1

112 I 0

2

vh |v ==

Open circuit reverse voltage gain 1222

zz

1

222 I 0

2

Ih |v ==

Open circuit output admittance 22

1z

Since h-parameters represent dimensionally impedance, admittance, voltage gain and current gain, these are called hybrid parameters.


8.7 INVERSE HYBRID (g) PARAMETERS

In inverse hybrid parameters current at the input port 1I and voltage at output port 2vcan be expressed in terms of 2I and 1vThe equations are as follows

1 11 1 12 2I g v g I= + ---- (1)

2 21 1 22 2v g v g I= + ---- (2) Where, 11 12 21 22g ,g ,g ,g are inverse hybrid parameters. They can be expressed in matrix as

1 11 12 1

2 21 22 2

I g g vv g g I

= From neq (1) & (2), when I2 = 0 i.e. port (2) open circuited, we get

2

111 I 0

1

Ig |v ==

Open circuit input admittance 11

1z

2

221 I 0

1

vg |v ==

Open circuit voltage gain From neq (1) & (2) when v1 = 0 i.e. port1 is short circuited, we get

1

112 v 0

2

Ig |I ==

Short circuit reverse current gain

1

222 v 0

2

vg |I ==

Short circuit output impedance 22

1Y

8.8 CONDITIONS FOR RECIPROCITY AND SYMMETRY

8.9 INTER RELATIONSHIPS OF DIFFERENT PARAMETERS

1) Z-parameters in terms of YparametersZ-parameters can be expressed in termsof Y-parameters as shown bellow[ ] 1[ ]−=Z Y orIt can be expressed in terms of matrixform as given bellow

111 12 11 12

21 22 21 22

z z Y Yz z Y Y

−

= Since

[ ]1

1 11 12

21 22

Y Y adJ[Y]YY Y Y∆

−−

=

So, [ ] 22 12

21 11

Y YadJ Y

Y Y−

= − And 11 22 12 21Y Y Y Y Y∆ = −

So, 22 2111 21

Y Yz zY Y

−= =∆ ∆

12 1112 22

Y Yz zY Y

−= =

∆ ∆

2) Y-Parameter in terms of Z-parametersY-parameters can be expressed in termsof Z-parameters as shown below-[ ] 1Y z− = or

It can be expressed in terms of matrixform as given below-

111 12 11 12

21 22 21 22

Y Y z zY Y z z

−

=


Since [ ]1

1 11 12

21 22

z z AdJ[z]zz z z

−− = = ∆

[ ] 22 12

21 11

z zAdJ z

z z−

= − 11 22 12 21∆ = −z z z z z

So, 22 2111 21

z zY Yz z

−= =∆ ∆

12 1112 22

z zY Yz z

−= =

∆ ∆

3) A B C D Parameters in terms of zparametersThe ABCD parameters equations are –

1 2 2v Av BI= − ------ (1)

1 2 2I Cv DI= − ------ (2)And z-parameters equations are-

1 11 1 12 2v z I z I= + ------ (3)

2 21 1 22 2v z I z I= + ------ (4) eqn(4) can be expressed in terms of 1Ias follows

2 22 21

21 21

v z II2 2

= − ------(5)

Comparing above eqn(5) with eqn (2) We get,

21

1C2

= & 22

21

2D2

=

Put eqn (5) in eqn (3) We get

2 221 11 2 12 2

21 21

v zv z I z Iz z

= − +

2 22 11

1 11 2 12 221 21

v z zv z I z Iz z

= − +

11 11 22 12 211 2 2

21 21

z z z z zv v Iz z

−= −

----- (6)

Compare eqn (6) with eqn (1) We get

11 11 22 12 21

21 21 21

z z z z z zA & Bz Z Z

− ∆= = =

Where, 11 22 12 21z z z z z∆ = −

So, 11

21

zAz

=21

zBZ∆

=

21

1Cz

= 22

21

zDz

=

4) A B C D Parameters in terms of Y- ParameterThe A B C D parameter equations are

1 2 2v Av BI= − -------- (1)

1 2 2I Cv DI= − ------- (2)And y- parameter equations are –

1 11 1 12 2I Y v Y v= + ------- (3)

2 21 1 22 2I Y v Y v= + ------- (4)From eqn (4)

2 221 2

21 21

I Yv vY Y

= −

221 2 2

21 21

Y 1v v IY Y−

∴ = + ------- (5)

Comparing eqn (1) & (5) we get 22

21 21

Y 1A & BY Y− −

= =

Now put eqn (5) in eqn (3)

221 11 2 2 12 2

21 21

Y 1I Y v I Y vY Y

−= + +

11 22 112 2 12 2

21 21

Y Y Yv I Y vY Y

−= + +

11 22 1112 2 2

21 21

Y Y YY v IY Y

−= + +

11 22 12 21 112 2

21 21

Y Y Y Y Yv IY Y

−= − +

-----(6)

Compare eqn (2) & (6) 11 22 12 21

21 21

Y Y Y Y yCY Y− −∆

= =

11

21

YDY−

=

So, 22

21

YAY−

=21

1BY−

=

21

yCY−∆

= 11

21

YDY−

=


8.10 INTERCONNECTION OF TWO-PORT NETWORK

1) Series connection

Fig. Series connection of two –port networks.

As shown in fig, two –port networks ANand BN connected in series. For network AN the z-parameter eqnmatrix form is-

1a 11a 12a 1a

2a 21a 22a 2a

v z z Iv z z I

=

----(I)

Similarly, for network Nb 1b 11b 12b 1b

2b 21b 22b 2b

v z z Iv z z I

=

----- (II)

From figure we can write 1 1a 1bI I I= = -----(1)

2 2a 2bI I I= = -----(2)

1 1a 1bv v v= + ------(3) and

2 2a 2bv v v= + ------(4)

( )n1 1a 1beq v v v3 ⇒ = +

From z-parameters neq (I) & (II) & (1) (2) Put value of 1av & 1bv in above neq weget

( ) ( )1 11a 1a 12a 2a 11b 1b 12b 2bv z I z I z I z I= + + +

( ) ( )1 11a 11b 1 12a 12b 2v z z I z z I= + + + ----(5)

( )n2 2a 2beq v v v4 ⇒ = +

( )2 21a 1a 22a 2a 21b 1b 22b 2bv z I z I (z I z I )= + + +

( )2 21a 21b 1 22a 22b 2v z z I (z z )I= + + + ---(6) So we have z –parameters eqn

1 11 1 12 2v z I z I= + ------ (7)

2 21 1 22 2v z I z I= + ------ (8) Comparing neq (5), (6), (7) & (8) we get

11 11a 11bz z z= +

12 12a 12bz z z= +

21 21a 21bz z z= +

22 22a 22bz z z= +So, the overall z-parameter matrix for series –connected two- port network is simply the sum of z matrices of each individual two port networks connected in series.

2) Parallel connection

Fig. Parallel connection of two port network.

As shown in fig., two-port networks NA and BN connected in parallel. For network Na the y parameter equations are

1a 11a 1a 12a 2aI Y V Y V= +

2a 21a 1a 22a 2aI Y V Y V= + Similarly, for network Nb

1b 11b 1b 12b 2bI Y V Y V= +

2b 21b 1b 22b 2bI Y V Y V= + Assuming that the parallel connection requires that

1 1a 1bV V V= =

2 2a 2bV V V= =

1 1a 1bI I I= =And 2 2a 2bI I I= = By combing above equations

1 1a 1bI I I= +

( )11a 1a 12a 2a 11b 1b 12b 2bY V Y V (Y V Y V )= + + +

( )1 11a 11b 12a 12b 2I Y Y V1 (Y Y )V= + + + ---(1)


and 2 2a 2bI I I= + ( ) ( )21a 1a 22a 1b 21b 1b 22b 2bY V Y V Y V Y V= + + +

( )2 21a 21b 1 22a 22b 2I Y Y V (Y Y )V= + + + ---(2)The y- parameter are

1 11 1 12 2I Y V Y V= + ------- (3)

2 21 1 22 2I Y V Y V= + ------- (4) Combining eqn (1), (2),(3),(4) we get

11 11a 11bY Y Y= +

21 21a 22bY Y Y= +

12 12a 12bY Y Y= +

22 22a 22bY Y Y= + So, the overall y- parameter matrix for parallel connected two port network is simply the sum of y-matrix of each individual two-port network connected in parallel

3) Cascade connection

Fig: - Two ports networks in cascade connection.

The transmission parameter representation is useful in cascaded two port network. For the network aN , the transmission parameter equations are

1a a a 2a

1a a a 2a

V A B VI C D I

= −

---- (1)

Similarly, for the network Nb , 1b b b 2b

1b b b 2b

V A B VI C D I

= −

---- (2)

From figure we can assume following observations

1 1a 2a 1b 2 2bI I , I I and I I = − = =

1 1a 2a 1b 2 2bV V ,V V a nd V V= = = The overall transmission parameter of the combined network Na and Nb can written in the matrix form as

1a a a 2a1

1a a a 2a1

V A B VVI C D II

= = −

( )nfrom- - eq 1

a a 1b

a a 1b

A B VC D I

=

a a b b 2b

a a b b 2b

A B A B VC D C D I

= − n--from eq (2)

a a b b 2

a a b b 2

A B A B VC D C D I

= −

2

2

VA BIC D

= −

So a a b b

a a b b

A B A BA BC D C DC D

=

So, the overall transmission parameter matrix for cascaded two port networks is simply the matrix product of transmission parameter matrices of each individual two-port network in cascade

4) T and 𝛑𝛑 representationIt is possible to express the elements ofT-network in terms of z- parameters orABCD parameters as explained below-

z-parameters of the network

2

111 I 0 a c

1

VZ | Z ZI == = +

2

221 I 0 C

1

VZ | ZI == =

1

222 I 0 b C

2

VZ | Z ZI == = +

1

112 I 0 C

2

VZ | ZI == =


It is possible to express the elements of the π-network in terms of y-parameters or ABCD parameters as explained below

Y-parameters of the network

2

111 V 0 1 2

1

IY | Y YV == = +

2

221 V 0 2

1

IY | YV == = −

1

222 V 0 3 2

2

IY | Y YV == = +

1

112 V 0 2

2

IY | YV == = −

5) Lattice networksOne of the common four –terminal two-port network is the lattice or bridgenetwork. The lattice network for z-parameter is shown in figure below

a b11 22

Z ZZ Z2+

= =

b a12 21

Z ZZ Z2−

= =

Example: Determine all the two port network parameters of an ideal transformer

Solution: We know that 2 2 2 1

1 1 1 2

N n V IN n V I

= = =−

So, 2 2

1 1

V nV n

= ------- (1)

And 1 2

2 1

I nI n

=−

-----(2)

⇒we know ABCD parameter 𝑒𝑒𝑒𝑒𝑛𝑛 -----1 2 2V AV BI= − ------ (3)

1 2 2I CV DI= − ------ (4) From eqn (1)

11 2 2

2

nV V 0In

= − ------ (5)

From eqn (2) 2

1 2 21

NI 0V IN

= − ------- (6)

Comparing eqn (3) & (5) and (4) & (6) We get,

1

2

2

1

n 0nA B

C D n0n

=

Now, we know that ' ' ' ' 1A B C D [ABCD]−=

So, ' ' ' 'A B C D parameters are 2

1

1

2

n 0nA' B'

C' D' n0n

=

We have h-parameter equations- 1 11 1 12 2V h I h V= +

2 21 1 22 2I h I h V= + From eqn (1)

11 1 2

2

nV 0I Vn

= +

From eqn (2) 1

2 1 22

nI I 0Vn

= − +

So,

1

211 12

21 22 1

2

n0nh h

h h n 0n

= −

We have 1[g] [h]−=


So,

2

111 12

21 22 2

1

n0ng g

g g n 0n

− =

Note: In an ideal transformer it is impossible to express 1 2V ,V in terms of 1Iand 2I hence z-parameter doesn’t exist similarly the y-parameter doesn’t exist

Example: Find all the two port network parameter for the below network

Solution: It’s a Lattice network. Comparing the above network with standard Lattice network we get

a b c dZ 1Ω,Z 3Ω,Z 3ΩZ 1Ω= = = =Using the formula derived in Lattice network

a b11 22

Z Z 1 3Z Z 2Ω2 2+ +

= = = =

b a12 21

Z Z 3 1Z Z 1Ω2 2− −

= = = =

2 1Z

1 2

=

Y: 1 1Y Z AdjZZ

−= =

Z 3= 2 1

AdjZ1 2

− = −

2 11Y1 23

− = −

T: 1 2 2V 2I I= + ----- (1)

2 1 2V I 2I= + ----- (2)

T (ABCD) parameter is the relation between 1V and 2I in terms of 1I and 2VForm equation (2) ⇒

1 2 2I V 2I= − ----- (3) Substitution in equation (1) we get

( )1 2 2 2V 2 V 2I I= − +

2 22V 3I= − ----- (4) Equation (3) and (4) can be rewritten as

1 2 2V 2V 3I= −

1 2 2I V 2I= −

1 2

1 2

V V2 3I I1 2

= −

2 31 2

=

1’ : ’ −− =T T T 2 311 21

− = −

h:−h parameters is the relation between 1V and 2I in terms of 1I and 2V

From equation (3) we get

2 1 21 1I I V2 2

= − + ------- (5)

Substituting equation (5) in (4) we get

1 2 1 2 2 11 1 1 3V 2V 3 I V V I2 2 2 2

= − − + = +

-----

(6) From (5) & (6)

1 1

2 2

3 1V I2 2I V1 1

2 2

= − g:-

1 12 2

312 2

−


Q.1 The admittance parameter 12Y in the 2-port network in Figure is

a)-0.2mho b)0.1 mho c)-0.05 mho d)0.05 mho

[GATE-2001]

Q.2 The Z parameters 11Z and 21Z for the 2-port network in the figure are

a) 11 216 16Z ; Z

11 11−

= Ω = Ω

b) 11 216 4Z ; Z

11 11= Ω = Ω

c) 11 216 16Z ; Z

11 11−

= Ω = Ω

d) 11 214 4Z ; Z

11 11= Ω = Ω

[GATE-2001]

Q.3 The impedance parameters 11Z and

12Z of the two–port network in the figure are

a) 11 12Z 2.75Ω and Z 0.25Ω= =

b) 11 12Z 3Ω and Z 0.5Ω= =

c) 11 12Z 3Ω and Z 0.25Ω= =

d) 11 12Z 2.25Ω and Z 0.5Ω= = [GATE-2003]

Q.4 For the lattice circuit shown in the figure, a bZ j2 and Z 2 .= Ω = Ω The values of the open circuit impedance

parameters 11 12

21 22

Z ZZ

Z Z

=

are

a) 1 j 1 j1 j 1 j− +

+ + b)

1 j 1 j1 j 1 j− +

− + −

c) 1 j 1 j1 j 1 j+ +

− − d)

1 j 1 j1 j 1 j+ − +

− + + [GATE-2004]

Q.5 The ABCD parameters of an ideal n:1 transformer shown in the figure

aren 00 X

The value of X will be

a)n b) 1n

c) 2n d) 21n

[GATE-2005]

Q.6 The parameters of the circuit shown in the figure are

GATE QUESTIONS(EC)


a) 0.1 0.10.1 0.3

−

b) 10 11 0.05

−

c) 30 2020 20

d) 10 1

1 0.05 −

[GATE-2005]

Q.7 A two-port network is represented by ABCD parameters given by

1 2

1 2

V VA BI IC D

= −

If port -2 is

terminated by RL, then input impedance seen at port-1 is given by

a) L

L

A BRC DR++

b) L

L

AR CBR D

++

c) L

L

DR ABR C

++

d) L

L

B ARD CR++

[GATE-2006]

Q.8 In the two port network shown in the figure below, Z11 and Z21 are, respectively

a) e 0r and rβ b) 00 and r−βc) 00 and rβ d) e 0r and r −β

[GATE-2006]

Statement for Linked Answer Questions 9 & 10 A two Part network shown below is exited by external dc sources. The voltages and current are measured with voltmeters

1 2V ,V and ammeters 1 2A ,A (All assumed to be ideal) as indicated. Under following switch conditions, the readings obtained are:

(i) 1S open− , 2 cl dS ose−

1 1 2 2A 0A, V 4.5V,V 1.5V,A 1A= = = =

(ii) 1 cl dS ose− , 2S open−

1 1 2 2A 4A,V 6V,V 6V,A 0A= = = =

Q.9 The h-parameter matrix for this network is

a)3 31 0.67

− −

b)3 1

3 0.67− −

c)3 31 0.67

d)3 13 0.67

− −

[GATE-2008]

Q.10 The z-parameter matrix for this network is

a)1.5 1.54.5 1.5

b) 1.5 4.51.5 4.5

c)1.5 4.51.5 1.5

d) 4.5 1.51.5 4.5

[GATE-2008]

Q.11 For the two–port network shown below, the short -circuit admittance parameter matrix is

a)4 2

S2 4

− −

b) 1 0.5

S0.5 1

− −

c) 1 0.5

S0.5 1

d) 4 2

S2 4

[GATE-2010]

Common Data for questions 12 and 13 With 10V dc connected at port A in the linear nonreciprocal two –port network shown below, the following were observed: (i) 1Ω connected at port B draws a current

of 3A


(ii) 2.5Ω connected at port B draws a current of 2A

Q.12 With 10 V dc connected at port A, the current drawn by7Ω connected at port B is a)3/7A b)5/7A c)1A d)9/7A

[GATE-2012]

Q.13 For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is a) 6V b) 7Vc)8V d)9V

[GATE-2012]

Q.14 In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in S) is _____.

[GATE-2014]

Q.15 For the two-port network shown in the figure, the impedance (Z) matrix (in Ω)is

a)6 2442 9

b)9 88 24

c)9 66 24

d)42 66 60

[GATE-2014]

Q.16 Consider a two-port network with the transmission matrix:

A BC D

T

= .

If the network is reciprocal, then a) -1T = Tb) 2T =Tc) Determinant (T) = 0d) Determinant (T) = 1

[GATE-2016]

Q.17 The z-parameter matrix for the two-port network shown is

2j jj 3 2jω ωω ω

+

Where the entries are in Ω .suppow (jω) = bR + jω . Then the value of

bR (in Ω) equals to __________. [GATE-2016]

Q.18 The z-parameter matrix 11 12

21 22

z zz z

a)2 22 2

− −

b) 2 22 2

c)9 36 9

−

d) 9 36 9

[GATE-2016]

Q.19 The ABCD matrix for a two-port network is defined by:

1 2

1 2

V VA BI C D I

= −


The parameter B for the given two- port network (in ohms, correct to

two decimal places) is ________.

[GATE-2018]


Q.1 (c) 1 3 3 11 12

3 2 3 21 22

Y Y Y Y YY Y Y Y Y+ −

= − +

12 3Y Y= −

121Y 0.05mho20

= − = −

Q.2 (c) 1 11 1 12 2V Z I Z I= +

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

2

111 I 0

1

VZ |I ==

⟹ Applying KVL in LHS loop 1 1 1 1E 2I 4I 10E 0− − + =

1 111E 6I=

1

1

E 6 ΩI 11

⇒ =

2

221 I 0

1

VZ |I ==

KVL in RHS loop,

2 1 1E 4I 10E 0− + =

2 1 16E 4I 10 I 0

11⇒ − + × =

1 1bE I

11 =

2 1 111E 44I 60I 0⇒ − + =

2

1

E 16ΩI 11

= −

Q.3 (a) Using ∆ − Y conversion

12 1 2R 0.5

4 4×

= = =

21 1 1R 0.25

4 4×

= = =

32 1 2R 0.5

4 4×

= = =

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (c) (a) (d) * (d) (d) (b) (a) (c) (a) (c) (b) 1.24 15 16 17 18 19 (c) (d) 3 (a) 4.8

ANSWER KEY:

EXPLANATIONS


1 3 3

3 2 3

Z Z ZZ Z Z+

+ 11 1 3Z Z Z 2.5 0.25= + = + 2.75=

12 3Z Z 0.25= =

Q.4 (d) For lattice network, Z-parameter is given as

a b a b

a b a b

Z Z Z Z2 2

Z Z Z Z2 2

+ −

− +

a bZ =2j,Z =2Ω1 j j 1j 1 1 j+ −

− +

Q.5 1 2

1 2

V VA BI IC D

=

2 1

1 2

I V nI V 1= =

1 2 2V AV BI= −

1 2 2I CV DI= −

2

1I 0

2

VA | nV == =

2

1 1V 0

2 2

I V 1D |I V n== = =

Q.6 (d) 1 11 1 12 2V h I h V= +

2 21 1 22 2I h I h V= +

2 1

1 111 V 0 12 I 0

1 2

V Vh | h |I V= ==

2 1

2 221 V 0 22 I 0

1 2

I Ih | h |I V= == =

When 2V 0=

1 2I I= −

221

1

I 1 hI= − =

1 1V 10I= 1

1

V 10I

⇒ =

1I 0=

When 1 2V V= 112

2

V h 1V

⇒ = =

(as no. drop in 10Ω resistance) 2 2V 20I=

222

2

I hV

⇒ =

1 0.0520

= =

10 11 0.05

−

Q.7 (d) The ABCD parameter equations are given by,

1 2 2V AV BI= −

1 2 2I CV DI= − When the network is terminal by

LR (fig.1),

2 2 LV I R= −

1 2 2in

1 2 2

V AV BIZI CV DI

−= =

− 2 L 2 L

2 L 2 L

AI R BI AR BCI R DI CR D

− − += =− − +

Q.8 (b)

1

112 I

2

VZ |I

=

=0 (∵ current source will be open).


2

1 0221 I 0 0

1 1

I .rVZ | rI I=

−β= = = −β

Q.9 (a) 1 11 1 12 2V h I h V= + 2 21 1 22 2I h I h V= + From given Z-parameters,

2 1 22I I V3

= − +

1 1 1 22V 1.5I 4.5 I V3

= + − +

1 23I 3V= − +

3 3

H1 0.67

− ∴ = −

Q.10 (c) 1 11 1 12 2V Z I Z I= + 2 21 1 22 2V Z I Z I= +

Case (1) 11 12

1

VI 0 Z 4.5I

⇒ = ∴ = =

222

2

VZ 1.5I

= =

Case (2) 12 11

1

VI 0 Z 1.5I

⇒ = ∴ = =

216Z 1.54

= =

1.5 4.5

Z1.5 1.5

=

Q.11 (a)

2

111 V 0

1

I 1Y | 4V 0.25== = =

1

222 V 0

2

I 1Y | 4V 0.25== = =

2

221 V 0 12

1

IY | 2 YV == = − =

4 2

[Y] S2 4

− ∴ = −

Q.12 (c) As per the given conditions, we can draw the following two figures.

Let Vth and Rth be Thevenin voltage & resistance as seen from partB.

th thV 3R 3= + ………. (1)

th th V 2R 5= + ………. (2) Solving (1) & (2)

thR 2Ω= So, thV 3x2 3 9V= + =

( )9i

2 7 Ω=

+

Q.13 (b)

So, th

7x1V 7 / 3x23

= +21 7V3

= =

The Open circuit voltage at port B is 7V

Q.14 (1.24) If two, π −n/ws are connected in parallel, The y-parameter are added i.e., equ 1 2y y y= +


1

2 13 3y

1 23 3

− = −

2

11 2y1 12

− = −

equ

5 53 6y

5 56 3

− = −

12

11 11

21

11 11

y1y y

hy y

y y∆

−

=

where ∆y = 11 22 12 21y y y y− − The value of h22 = ∆y

= 5 5 5 53 3 6 6

− − − ∆y = 2.0833

115

3y = ∴ h 22=1.24

Q.15 (c) For the two-part network

1 1 130 10 30

1 1 130 60 3

Y mat

0

rix

+ −

− +

=

[ ] 1matrixZ Y −=

10.1333 0.03330.0333 0.0

Z5

−− −

=

9 66 24

Z =

Q.16 (d)

Q.17 (3) a c c

matrixc b c

Z +Z ZZ =

Z Z +Z

b cZ +Z =3+2jω

cZ =jω ⇒ bZ =3+jω⇒

Q.18 (a) Since the given network is symmetric and reciprocal Z11 = Z22 Z12 = Z21

2

111

1 0

3 6 23 6I

VzI

=

×=⇒ = =

+

2

221

1 0I

VzI

=

⇒ = We know

2 1 21 2V V z⇒− = −=

So 11 12

21 22

z z 2 2z z 2 2

− = −

Q.19 4.8 Given:

1 2

1 2

V VA BI C D I

= −

It can be re-written as

1 2 2V AV BI ...(i)= −

21 2I CV DI ....(ii)= −

From equation (i),

2

1

2 V 0

VB .......(iii)I

=

= −


We can write two KVL equations for the given electrical network as,

1 1 2V 7I 5I .......(iv)= +

( )2 1 2V 5I 7I ....... v= +

At 2V 0=

1 20 5I 7I= +

1 27I I5

= −

Substituting in equation (iv)

1 2 27V 7 I 5I5

= × − +

1 224V I5

= −

1

2

V 24B 4.8I 5

= − = =


Q.1 A passive two-port network is in a steady state. Compared to its input, the steady state output can never offer a) higher voltageb) lower impedancec) greater powerd) better regulation

[GATE-2001]

Q.2 A two-port network, Shown in figure is described by the following equations:

1 11 1 12 2l Y E Y E= + , 2 21 1 22 2l Y E Y E= +

The admittance parameters11 12 21Y ,Y ,Y , 22and Y for the network

shown are a) 0.5 mho, 1mho, 2 mho and 1

mho respectivelyb) 1

3mho, 1

6− mho, 1

6− mho and

13

mho respectively

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

d) 25

− mho, 37

− mho, 37

mho and

25

mho respectively

[GATE-2003]

Q.3 The h-parameters for a two–port network are defined by

1 11 12 1

2 21 22 2

E h h ll h h E

=

For the two –port network shown in figure, the value of 12h is given by

a) 0.125 b) 0.167c) 0.625 d) 0.25

[GATE-2003]

Q.4 The Z matrix of a 2- port network as

given by 0.9 0.20.2 0.6

The element Y22 of the corresponding Y matrix of the same network is given by a) 1.2 b) 0.4c) -0.4 d) 1.8

[GATE-2004]

Q.5 For the two port networks shown in the figure the Z-matrix is given by

a) 1 1 2

1 2 2

Z Z ZZ Z Z

+ +

b) 1 1

1 2 2

Z ZZ Z Z +

c) 1 2

2 1 2

Z ZZ Z Z +

d) 1 1

1 1 2

Z ZZ Z Z +

[GATE-2005]

Q.6 Two networks are connected in cascade as shown in the figure. With the usual notation the equivalent A, B, C and D constants are obtained.

GATE QUESTIONS(EE)


Given that C 0.025 45= ∠ ° , the value of Z2 is

a)10 30°Ω∠ b) 40 -45°Ω∠c)1Ω d) 0Ω

[GATE-2005]

Q.7 The parameters of the circuit shown in the figure are iR 1M ,= Ω

oR 10 ,= Ω 6A 10 V / V.= If iV 1 V,= µ then

output voltage, input impedance and output impedance respectively are

a)1V, ,10∞ Ω b)1V,0,10Ωc)1V,0,∞ d)10V, ,10∞ Ω

[GATE-2006]

Q.8 The parameter type and the matrix representation of the relevant two port parameters that describe the circuit shown are

a) 0 0

z parameters, 0 0

b) 1 0

h parameters, 0 1

c) 0 0

h parameters 0 0

d) 1 0

z parameters 0 1

[GATE-2006]

Q.9 The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d) respectively. It has an impedance matrix Z with parameters denoted by ijZ A 1Ω resistor is connected in series with the network at port 1 as shown in the figure. The impedance matrix of the modified two–port network (shown as a dashed box) is

a) 11 12

21 22

z 1 z 1z z 1+ +

+

b) 11 12

21 22

z 1 zz z 1+

+

c) 11 12

21 22

z 1 zz z+

d) 11 12

21 22

z 1 zz 1 z

+ +

[GATE-2010]

Common Data for questions Q.10 and Q.11 With 10V dc connected at port A in the linear nonreciprocal two–port network shown below, the following were observed: (i) 1Ω connected at port B draws a

current of 3A (ii) 2.5Ω connected at port B draws a

current of 2A


[GATE-2012]


Q.11 For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is a) 6V b) 7Vc)8V d)9V

[GATE-2012]

Q.12 In a linear two port network, when 10 V is applied to port 1, a current of 4 A flows through port 2 when it is short circuited. When 5 V is applied to Port 1, a current of 1.25 A flows through a 1Ω resistance connected across Port 2. When 3 V is applied to Port 1, the current (in Ampere) through a 2Ω resistance connected across Port 2 is __________

[GATE-2015]

Q.13 The z-parameters of the two port network shown in the figure are Z11=40Ω, Z12 = 60Ω, Z21=80Ω, Z22=100Ω. The average power delivered to RL = 20Ω, in watts, is____.

[GATE-2016]

Q.14 For the given 2-port network, the value of transfer impedance Z21 in ohms is…….

[GATE-2017, Set-2]

Q.15 In the two-port network shown, the

11h parameter (where 111

1

VhI

= when

2V 0= ) in ohms is ______ (up to 2 decimal places)

[GATE-2018]


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (b) (d) (d) (d) (b) (a) (c) (c) (c) (b) 0.545 35.55 3 15 0.5

ANSWER KEY:


Q.1 (c) For a passive two-port network, output power can never be greater than input power.

Q.2 (b) Using KVL,

( )1 1 1 2E 2l 2 l l= + +Again using KVL,

( )2 2 1 2E 2l 2 l l= + +

[ ] 4 2z

2 4

⇒ =

[ ] [ ] 1y z −=

( ) ( )4 212 44 4 2 2

− = −× − ×

11 12

21 22

1 1Y Y 3 6Y Y 1 1

6 3

− −

Q.3 (d)

1

112 | 0

2

Eh |E ==

Or 12h is ratio of 1E to 2E for the input open- circuited condition. Two methods are provided to solve the problem. Method -1 Assuming

1l 0=

( )2 2

2E El

2 2 4 || 4 4= =

+ +

( )2

xll 4

2 2 4= ×

+ +

2 2 2l E E12 2 4 8

= =

2 21 x

E EE 2l 28 4

= = =

1

2

E 0.25E

⇒ =

Method -2

Using ∆-Ytransformation

( ) ( )1 1 1 2E 4 0.5 l 1 l l= + + +

1 25.5l l= + … (a)

( ) ( )2 2 1 2E 2 1 l 1 l l= + + +

1 2l 4l= + … (b)Put 1l 0= in eq (a) and (b)

1 2E l= And 2 2E 4l=

1

112 l 0

2

Eh | 0.25E == =

Q.4 (d)

[ ] 0.9 0.2z

0.2 0.6

=

[ ] [ ] 1

0.6 0.20.2 0.9

y z[0.9 0.6 0.04]

−

− − = =

× −1.2 0.40.4 1.8

− = −

22y 1.8=

Q.5 (d)

EXPLANATIONS


Two method are provided to solve the problem. Method-1

1 2 1 1v v i z= =

So 2

111 i 0 1

1

vz | zi == =

(i) 2

221 i 0 1

1

vz | zi == =

(ii)When 1i 0=

1 2 1v i z=

⇒ 1

112 i 0 1

2

vz | zi == =

(iii) 2 2 1 2v i (z z )= +

⇒1

222 i 0 1 2

2

vz | z zi == = +

Method -2 (iv) ( )1 1 2 1v i i z= +

1 1 1 2z i z i= +

(v) ( )2 2 1 1 1 2v z i z i i= + +

1 1 1 2 2z i (z z )i= + +(vi)From eq (i) to(iv)

1 1

1 1 2

Z Zz matrix

Z Z Z

− = +

Q.6 (b)

1 2 2V AV Bl= +

1 2 2l CV Dl= +

( )2 2 1 2V Z l l= +

(i) 2

1I 0

2

lC |V ==

Putting 12=0 in equation (i) 2 2 1V Z l=

2

22 I 0

1

V 1 1Z |l C 0.025 45=⇒ = = =

∠ °

2 40 45 Ω= ∠Z °

Q.7 (a) 6 6

0V 10 10 1V−= × =

111

1

VZI

= →∞

222 0

2

VZ R 10I

= → =

Q.8 (c) 1 11 1 12 2I h V h I= +

2 21 1 22 2V h V h I= +

Since port -1 is open – circuit, 1I 0=Port -2 is sort – circuit, 2V 0=

2

111 I 0

1 1

I 0h | 0V V== = =

1

112 V 0

2 2

I 0h | 0I I== = =

2

221 L 0

1 1

V 0h | 0V V== = =

1

222 V 0

1 2

V 0h | 0I I== = =

So, h − parameters 11 12

21 22

h h 0 0h h 0 0

= =

Q.9 (c) The Impedance matrix mZ of the modified network is calculated from fig. given below:


[ ]1 1

2 2

V Iz

V I

=

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

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

( )S 2 1V 1 I V= × +

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

( )11 1 12 21 Z I Z I= + +

[ ]S 1m

2 2

V Iz

V I

=

11 12m

21 22

Z 1 Zz

Z Z+

=

[ ] 1 0Z

0 0

= +


Let Vth and Rth be Thevenin voltage & resistance as seen from part B.

th thV 3R 3= + ……….(1)

th th V 2R 5= + ……….(2) Solving (1) &(2)

thR 2Ω= So, thV 3x2 3 9V= + =

( )9i

2 7 Ω=

+

Q.11 (b)

So, th7x1V 7 / 3x23

= +

21 7V3

= =

The Open circuit voltage at port B is 7V.

Q.12 0.545 1 11 1 12 1I y v y v= +

2 2I 0.4 3 0.6[2I ]= × −

2 21 1 22 2I y v y v= + 2=1.2-1 2I.

21 214 10y y 0.4= → = 2I 0.545A=

1 221.25 0.4v 1.25y 0.4= + =

22y 0.6= −

Q.13 35.55 In the given terminated 2 port network the Z matrix is known and for load of 20Ω we want to find power on the load. → The get it assuming RL as load let first obtain the thevenin equivalent of 2 port → Thevenin equivalent means Vth & Rth

2th 2 I =0

V = V i.e., O.C voltage of port 2

SC 2 2I =-I /V =0 i.e., s.c current of port 2,

thin

SC

VR =I

→ Evaluation of thVThe Z matrix equation is 1 1 2V =40I +60I 2 1 2V =80I +100I


In the above two equations if 2I 0=then 1 1V =40I (1) 2 1V =80I (2) From the input side we can say ( )1 1V =20-10I⇒ 1 120-10I =40I

⇒ 2 12V =80I =80× =32V5

So th 2V =V =32 Evaluation of SCIIn the Z matrix equation if we put

2V 0= then

1 1 1V =40I +60I …. (5)

1 20=80I +100I …. (4)

1 2 1 1 210 100I =- I &V =20-10I =20+ I8 8

Using these in equation 1V & 1I in equation 3

2 2 2100 40020+ I =- I +60I

8 8⇒ 2 2 2160+100I =-400I +480I⇒ 2 2160 20I I 8A= − ⇒ = − SC 2I =-I =8A

→ inin

SC

V 32R = = =4ΩI 8

→Now the ckt is from port 2is ( )2

20Ω 20ΩP I 20=

32 20 35.55watt4 20

= = +

Q.14 3

AR 1= Ω

BR 1= Ω

CR 1/ 2= ΩAfter rearrangement consider the following circuit.

From the circuit diagram we get, 2

211

VZ 3I

= = Ω

Q.15 0.5 Given: The two port network is shown below,

To find the value of2

111

1 V 0

VhI

=

= ,the

equivalent circuit is shown below,

Applying KVL in loop (1),

( )1 1 1 2V I I I 0− − + =


1 1 2V 2I I= +

Applying KVL in loop (2),

( ) ( )1 2 1 2I I 2I I 0+ + + =

1 23I 2I 0+ =

From equation (i) and (ii),

11 1

3IV 2I2

= −

1

1

V 3 12 0.5I 2 2

= − = =

Hence,

2

111

1 V 0

Vh 0.5I

=

= =

Hence, the correct answer is 0.5


Q.1 The DC voltage gain 0

i

VV

in the

following circuit is given by

a) 2

1 2

RAvR R+

b) 1

1 2

RAvR R+

c) 2o

1 2

RAv RR R

++

d) Av

[GATE-2007]

Q.2 For the circuit shown below the input resistance

111 11 I2 0

1

VR R |I == = is

a) -3Ω b) 2Ωc) 3Ω d) 13Ω

[GATE-2008]

Common Data for questions 3 and 4 With 10V dc connected at port A in the linear nonreciprocal two –port network shown below, the following were observed: i) 1Ω connected at port B draws a

current of 3Aii) 2.5Ω connected at port B draws a

current of 2A


[GATE-2012]

Q.4 For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A . If 8 V dc is connected to port A, the open circuit voltage at port B is a) 6V b) 7Vc)8V d)9V

[GATE-2012]

Q.5 Considering the transformer to be ideal, the transmission parameter ‘A’ of the 2 -port network shown in the figure below is

a)1.3 b)1.4 c)0.5 d)2.0

[GATE-2013]

Q.6 The output voltage of the ideal transformer with the polarities and dots shown in the figure is given by

a) NV1sin ωt b) -NV1sin ωt

c) V1sin ωt d) -V1sin ωt[GATE-2015]

GATE QUESTIONS(IN)


Q.7 The connection of two 2-port networks is shown in the figure. The ABCD parameters of N1 and N2 networks are given as

The ABCD parameters of the combined 2-port network are

a) 2 50.2 1 b) 1 2

0.5 1

c) 5 20.5 1 d) 1 2

0.5 5

[GATE-2017]


Q.1 (a)

( )2

i1 2

RV VR R

=+

( )2

0 V V1 2

RV A V AR R

= =+

Q.2 (d) ( )1 1 2 1 3 3 3V I 3I 2 I 2V V andV= + + + +

( )1 3 1 32 I 2V 2I V= + ⇒ =

For ( )2 1 1 1 1 1I 0V I 2 I 4I 2I= = + + +

2

111 I 0

1

VR | 13ΩI =⇒ = =


Let Vth and Rth be Thevenin voltage & resistance as seen from part B.

th thV 3R 3= + ……….(1)

th th V 2R 5= + ……….(2) Solving (1) &(2)

thR 2Ω= So, thV 3x2 3 9V= + =

( )9i

2 7 Ω=

+

Q.4 (b)

So, th7x1V 7 / 3x23

= +

21 7V3

= =

The Open circuit voltage at port B is 7V.

Q.5 (a)

Q.6 (b) First mark the mutual voltage polarity using dot convention (when a reference current enters at dot of one coil, it generates positive polarity on dot terminal of other coil) →Using Transformer ratio

1 2 3 4 5 6 7

(a) (d) (c) (b) (a) (b) (a)

ANSWER KEY:

EXPLANATIONS


2 2

i 1

V NV N

=

2 1 iNV V NV sin sinωt1

⇒ = =

o 2 iV V NV sin sinωt→ = − = −

Q.7 (a)


9.1 RC, RL, LC IMPEDANCE & ADMITTANCE FUNCTIONS

1) For the LC impedance function thepoles and zeros are alternate and lie onthe jω axis

e.g. ( )( ) ( )( )( )

2 2

2 2

S 1 S 3Z s

s S 2 S 4

+ + +=

+ +

2) For the RL impedance function thepoles and zero are alternate and lie onlyon the negative real axis and nearest tothe origin is zero .It can be at the origin.e.g.

3) For the RC impedance function the polesand zeros are alternate and lie only onthe negative real axis and nearest to theorigin is a pole. It can be at the origin.e.g.

4) For the RLC impedance function, thepoles are complex conjugate pair andthey are symmetrical wrt the negativereal axis.

Note: In the above cases instead of impedance function if admittance function is given then they are converted into the impedance function first and then above tests are performed RL impedance Function = RC admittance function and Vice versa Admittance = impedance or admittance.

Example: The driving point impedance function of a network is

( )( )S 2 (S 4)

F(s)S 1 (S 3)+ ++ +

.Then the function is

a) an RL impedance functionb) an RC admittance functionc) LC impedance functiond) RL admittance function

Solution: If F(s) is impedance function then it is an RC function because pole is nearest to the origin. If F(s) is admittance function then it is an RL function because in that case zero would be nearest to the origin Hence, F(s) is RC impedance or RL admittance function.

9 NETWORK SYNTHESIS


Q.1 The driving–point impedance Z(s) of a network has the pole–zero locations as shown in the figure. If Z(0) = 3,then Z(s) is

a) 23(s 3)

S 2S 3+

+ +b)

22(s 3)

S 2S 2+

+ +

c) 23(s 3)

S 2S 2−

− −d)

22(s 3)

S 2S 3−

− − [GATE-2003]

Q.2 The first and the last critical frequency of an RC- driving point impedance function must respectively be a) a zero and a poleb) a zero and a zeroc) a pole and a poled) a pole and a zero

[GATE-2005]

Q.3 The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements are a pole and a zero respectively. The above property will be satisfied by a) RL network onlyb) RC network onlyc) LC network onlyd) RC as well as RL networks

[GATE-2006]

Q.4 A negative resistance Rneg is connected to a passive network N having driving point impedance Z1(s) as shown below. For Z2 (s) to be positive real.

a) ( )neg 1R Re Z j ,≤ ω ∀ω

b) ( )neg 1R Z j ,≤ ω ∀ω c) ( )neg 1R Im Z j ,≤ ω ∀ω

d) ( )neg 1R Z j ,≤ ∠ ω ∀ω [GATE-2006]

Q.5 The RC circuit shown in the figure is

a) a low–pass filterb) a high-pass filterc) a band–pass filter

d)a band-reject filter [GATE-2007]

Q.6 Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that

of Filter 2 be B2 .The Value of 1

2

BB

is

a) 4 b) 1

GATE QUESTIONS(EC)


c) 12

d) 14

[GATE-2007]

Q.7 The driving point impedance of the following networks is given by

( ) 20.2SZ s

S 0.1s 2=

+ +The component values are

a) L 5H,R 0.5 ,C 0.1F= = =Ω

b) L 0.1H,R 0.5 ,C 5F= = =Ω

c) L 0.1H,R 0.5 ,C 5F= = =Ω

d) L 0.1H,R 2 ,C 5F= = =Ω

[GATE-2008]

Q.8 If the transfer functions of the following network is

0

i

V (S) 1V (S) 2 sCR

=+

The value of the load resistance RL is a) R/4 b) R/2c) R d) 2R

[GATE-2009]

Q.9 The transfer function 2

1

V (S)V (S)

of the

circuit shown below is

a) 0.5s 1s 1

++

b) 3s 6s 2++

c) s 2s 1++

d) s 1s 2++

[GATE-2013]

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

ANSWER KEY:


Q.1 (b)

( ) ( )1 2

K(s z)Z sS P (S P )

−=

− −

= ( )

K(s 3)S 1 j (S 1 j)

++ + + −

( )( )2

K(s 3)Z ss 1 j2

+=

+ −

= ( )2K(s 3)s 1 1

+

+ +

( ) 0Z 0 | 0ω− = 3K 3 K 22

⇒ = ⇒ =

( ) 22(s 3)Z s

S 2S 2+

∴ =+ +

Q.2 (b) For stability poles and zero interlace on real axis. Since its RC, first pole should come and zero at last.

Q.3 (b)

Q.4 (a) ( ) ( )2 1 negZ s Z s R= +

⇒ ( ) ( )( ) ( )( )2 neg e 1 1Z s R R Z (s J Im Z s= + +

neg&R 0< For ( )2Z s to be +ve & real, Re

( )( ) ( )( )1 neg e 1 negZ s R R Z s R≥ − ⇒ ≥

Q.5 (c) ( )

( )( )1

01

i

R || /scV (S)1V (S) R R || /scSC

=+ +

( )0

2 2 2i

V (S) SRCV (S) S R C 3SRC 1

⇒ =+ +

Put ( )( )

0

i

V JωS Jω,

V Jω= ∴

( )2 2 2

JωRC G(Jω)1 ω R C 3JωRC

= =− +

Asω 0→ , ( )G Jω =0Asω ∞→ , ( )G Jω =0

At ( ) J1 1ω G JωRC, 33J= = =

∴ Filter is band –pass filter.

Q.6 (d) 1 2

1 21 2 2 1

B LR R 1B ;BL L 4B L= = ⇒ = =

Q.7 (a) ( ) 1Z S R SL

SC=

( )1RSL. SC

1 LRSL R. SC C=

+ +

( )2

RSLSC

S RCL R SLSC

=+ +

( ) ( )

( )

2

2

SRLZ SS RCL SL R

1S. SCS 1S RC LC

⇒ =+ +

=+ +

( )2

0.2SS 0.18 2

=+ +

1 0.2 C 5FC = ⇒ =

1 0.1 R 2ΩRC = ⇒ =

1 2 L 0.1HLC = ⇒ =

Q.8 (c)

( )( )

( )( )

L0

i L

1R ||V S SC1V S R R || SC

=+

EXPLANATIONS


( )( )

( )( )( )

L

L

L

L

1R . SC1R SC

1R R . SC1R SC

+=

+

+

( )

( )

L

L

L

L

R1 SCR

RR 1 SCR

+=

+ +

( )0 L

i L L

V (S) R 1V (S) R R SCRR 2 SCR

⇒ = =+ + +

LR = R Satisfies above equation

Q.9 (d)

2

1

V ?V

=

( )( )

1 22 1

1 1 2 11 2

1R RC S 1 CV C S1V RC S C C1R C S C S

+ += =

+ + +

Substituting the values we get 2

1

V S 1V S 2

+=

+


Q.1 The driving point input impedance seen from the source sV of the circuit shown below, in Ω , is ____.

[GATE-2016]

Q.1 (20) The Driving point impedance is nothing but the ratio of voltage to current from the defined port. In this case it is Vs

Is

Writing KCL at node x x x

s 1V VI 4V 03 6

− + − + =

Substituting these in Eq(1)

s s s ss s

V 2I V 2II 8I 03 6− −

− + − → =

⇒ s s1 1 2 2V + =I 1+ +8+3 6 3 6

Q.2 A major advantage of active filters is that they can be realized without using a) op-amps b) inductorsc) resistors d) capacitors

[GATE-2016]

⇒ ( )s sV 2+1 =I (6+4+48+2)

⇒ s

s

V 60 20ΩI 3

= =

Q.2 (b) Inductive coils are bulky in nature.

1 2 20 (b)

ANSWER KEY:

GATE QUESTIONS(EE)

EXPLANATIONS


Q.1 The v-i characteristic of an element is shown in the figure given. The element is

a) Non linear, active, non-bilateralb) Linear, active, non-bilateralc) Non-linear, passive, non-bilaterald) Non-linear, active, bilateral

Q.2 The incandescent bulbs rated respectively as P1 and P2 for operation at a specified mains voltage are connected in series across the mains as shown in the figure. Then the total power supplied by the mains to the two bulbs

a) 1 2

1 2

P PP P+

b) 2 21 2P P+

c) 1 2(P P )+ d) 1 2P P×

Q.3 Consider the circuit as shown above which a current-dependent current

source has. The value 1

2

VV is

a) 1 b) 2

c)αα

++

21 d)

2 +αα

Q.4 Consider the following circuit:

Which one of the following statements is correct? a) Passive and linearb) Active and linearc) Passive and non-lineard) Active and non-linear

Q.5 Consider the circuits A and B. For what values respectively of I and R, the circuit B is equivalent to circuit A?

a)3A, 40Ω b)4 A, 24 Ω c)1 A, 100 Ω d)2 A, 100 Ω

Q.6 For the circuit given in the figure the power delivered by the 2volt source is given by

a) 4W b) 2Wc) -2W d) -4W

Q.7 The current in the given circuit with a dependent voltage source is a) 10 A b) 12 Ac) 14 A d) 16 A

Q.8 A certain network N feeds a load resistance R as shown in figure-I. It

ASSIGNMENT QUESTIONS


consumes a power of ‘P’W. If an identical network is added as shown in figure-II, the power consumed by R will be

a) Less than Pb) Equal to Pc) Between P and 4Pd) More than 4P

Q.9 In the circuit shown in Fig., the current I through 2 resistor is

a) – 94.34 mA b) -70.34 mAc) 70.34 mA d) 94.34 mA

Q.10 For the circuit as shown above, if E = E1 and I is removed, then V = 5 volts. If E = 0 and I = 1A, then V = 5 volts. For E = E1 and I replaced by a resistor of 5Ω, what is the value of V in volts?

a) 5.0 b) 2.5c) 7.5 d) 3.5

Q.11 Consider the circuit in the below figure. What is the power delivered by the 24 V source?

a) 96 W b) 144 Wc) 192 W d) 288 W

Q.12 Consider the following circuit:

What is the value of current I in the 5 Ω resistor in the above circuit? a) 0 A b) 2 Ac) 3 A d) 4 A

Q.13 In the circuit shown above, if the current through the resistor R is zero, what is the value of I?

a) 1A b) 2Ac) 3A d) 4A

Q.14 In the network shown above, what is the current I in the direction shown?

a) 0 b) 1/3 Ac) 5/6 A d) 4

Q.15 The current I in the network in Fig. is

a) 1A b) 3 Ac) 5A d) 7A

Q.16 The current i in figure is

a) 0.5 A b) 5/6 Ac) 1.5 A d) 2.5 A

Ω


Q.17 In Fig., the voltage source

a) Delivers 200/3 Wb) Absorbs 100 Wc) Delivers 100 Wd) Absorbs 200/3 W

Q.18 The nodal voltage V1 in fig. is

a) – 13.5 V b) – 6. 75 V c) – 4.5 V d) 0 V

Q.19 What is the current through the 2Ωresistance for the circuit as shown above?

a) 5A b) 4Ac) 3A d) 2A

Q.20 What is the value of I for the above shown circuit, if V = 2 volts?

a) 2 A b) 4 Ac) 6 A d) 8 A

Q.21 In the circuit, V1 = 40V when R is 10Ω. When R is zero, the value of V2 will be

a) 40V b) 30Vc) 20V d) 10V

Q.22 The circuit shown in figure has the following source values: A= 600 t u (t + 1) V, VB = 600 (t + 1) u(t) V and IC = 6(t – 1) u(t – 1) A, where u(.) denotes the unit step function. For this circuit, the current i at t = - 0.5s will be

a) – 9 A b) – 6 A c) – 1 A d) 0 A

Q.23 Match List I (Quantities) with List II (Units) and select the correct answer using the codes given below the lists:

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

Q.24 For the circuit shown above, what is the value I?

a) 10 A b) 6 Ac) 3.7 A d) 3 A

Q.25 If the voltage V across 10Ωresistance is 10V, what is the voltage


E of the voltage source in the circuit shown above?

a) – 50V b) – 10V c) + 10V d) + 50V

Q.26 Consider the following circuit: If V1 = 5V and V2 = 3V, then what is the input impedance of the CRO in the circuit?

a) 1 MΩ b) 1.5 MΩc) 3MΩ d) 5MΩ

Q.27 Consider the following circuit: What is power delivered to resistor R in the circuit?

a) -15Vb) 0Wc) 15 Wd) Cannot be determined unless the

value of R is known

Q.28 Consider the following circuit: What is the current I in the circuit?

a) 0A b) 2Ac) 5A d) 6A

Q.29 Three parallel resistive branches are connected across a d.c. supply. What

will be the ratio of the branch currents I1:I2:I3 if the branch resistances are in the ratio R1:R2:R3: : 2 : 4 : 6? a) 3 : 2 : 6 b) 2 : 4 : 6c) 6 : 3 : 2 d) 6 : 2 : 4

Q.30 For the circuit as shown above, what is the value of I?

a) 4A b) 3Ac) 2 A d) 1 A

Q.31 In the circuit shown above, when is the power absorbed by the 4Ωresistor maximum?

a) R = 0 b) R = 2Ωc) R = 4Ω d) R = ∞

Q.32 In the circuit shown in the figure, the power consumed in the resistance R is measured when one source is acting at a time, these values are 18 W, 50W and 98 W. When all the sources are acting simultaneously, the possible maximum and minimum values of power in R will be

a) 98W and 18Wb) 166W and 18Wc) 450W and 2Wd) 166W and 2W

Q.33 According to maximum power transfer theorem, when is the


maximum power absorbed by one network from another network? a) The impedance of one of the

networks is half that of the other b) The impedance of one is the

complex conjugate of the other c) The impedance of one is equal to

that of the other d) Only the resistive parts of both

are equal

Q.34 In the circuit shown in the given figure, RL will absorb maximum power when its value is

a) 2.75 Ω b) 7.5 Ω c) 25 Ω d) 27 Ω

Q.35 In a linear circuit, the superposition

principle can be applied to calculate the

a) Voltage and power b) Voltage and current c) Current and power d) Voltage, current and power Q.36 In the circuit shown in the given

figure, power dissipated in the 5 Ω resistor is

a) Zero b) 80 W

c) 125 W d) 405 W

Q.37 If two identical 3A, 4Ω Nrton equivalent circuits are connected in parallel with like polarity, the combined Norton equivalent circuit will be

a) 3A, 8Ω b) 6A, 8Ω c) 0A, 2Ω d) 6A, 2Ω

Q.38 Which of the following theorems can be applied to any network-linear or

non-linear, active or passive, time-variant or time-invariant?

a) Thevenin theorem b) Norton theorem c) Tellegen theorem d) Superposition theorem Q.39 In the circuit shown below, if the

source voltage Vs = 100∠ 53.130 V then the Thevnin’s equivalent voltage in Volts as seen by the load resistance RL is

a) 100 ∠ 900 b) 800∠ 00

c) 800∠ 900 d) 100∠ 600 Q.40 Match List –I with List-II and select

the correct answer using the code given below the Lists:

List – I (Term) A. Norton equivalent of one port

B. Open-circuit output admittance C. Reciprocal network D. Transmission parameters

List –II (Concept) 1. Network where loop and node

equation have a symmetric coefficient matrix

2. Hybrid parameter h22 3. Parameters where V1 and I1

expressed as functions of V2 and –I2

4. Current source in parallel with the impedance

Code: A B C D a) 1 3 4 2 b) 4 2 1 3 c) 1 2 4 3 d) 4 3 1 2 Q.41 Consider the following statements: Network NA in figure (A) can be

replaced by the network NB shown in figure (B), when Ic and Rc, respectively, are

1. 5A and 2Ω 2. 10A and 1Ω


3. 15A and Ω21 4.30A and Ω

51

Which of these statements given

is/are correct? a) 1 only b) 2, 3 and 4 c) 1, 2, 3 and 4 d) 2 and 3 Q.42 Consider the following properties of

a particular network theorem: 1. The theorem is not concerned

with type of elements. 2. The theorem is only based on the

two Kirchoff’s laws. 3. The reference directions of the

branch voltages and currents are arbitrary except that they have to satisfy Kirchoff’s laws.

Which one of the following theorems has the above characteristics?

a) Thevenin’s theorem b) Norton’s theorem c) Tellegen’s theorem d) Superposition theorem Q.43 Consider the following circuit:

What should be the value of

resistance R, in the above circuit it has to absorb the maximum power from the source?

a) 8/3 ohms b) 3/8 ohms c) 4 ohms d) 8 ohms Q.44 A network with independent

sources and resistors shown above in figure (a) has a Thevenin voltage VT and Thevenin resistance RT. What

are the Norton equivalent current IN

and resistance RN in the figure (b)?

a) T T L

T LT L

T L

V R R,R +RR R

R +R

b) TN T

T

V , R =RR

c) TN L

T

V ,R RR

=

d) Noneof theabove Q.45 The terminal volt-ampere

conditions of a linear reciprocal network N are shown in the figure (a). What is the current I corresponding to the terminal conditions shown in the figure (b)?

a) – 1 A b) 9 A

c) 10 A d) 11 A Q.46 What is the value of R required for

maximum power transfer in the network shown above?

a) Ω2 b) Ω4 c) Ω8 d) Ω16

Q.47 What are the source voltage and

source resistance, respectively for the Thevenin’s equivalent circuit as seen from the terminals indicated in the circuit given above?


a) 20V,24Ω b) 20V,48Ω c) 20V,4.8 Ω d) 20V,12Ω

Q.48 What is the Thevenin resistance

seen from the terminals AB of the circuit shown above in the figure?

a) Ω2 b) Ω4

c) Ω8 d) Ω12

Q.49 In the network in fig, the mesh current I and the input impedance seen by the 50V source, respectively, are

a) 125 11A and

13 8Ω

b) 150 13A and

13 8Ω

c) 150 11A and13 8

Ω

d) 125 13A and13 8

Ω

Q.50 Match List – I with List – II and

select the correct answer using the code given below the Lists:

List – I (Theorem/Law)

A. Norton’s theorem 1.Effects of independent sources in a linear circuit are additive B. Superposition theorem 2.Law of non- accumulation of charge holds good at nodes C. Thevenin’s theorem

D. Kirchhoffs current law List – II (Property)

1. Effects on independent sources in a linear circuit are addictive

2. Law of non-accumulation of charge holds good at nodes

3. Current source with shunt resistor

4. Voltage source with series resistor

Code: A B C D

a) 2 4 1 3 b) 3 1 4 2 c) 2 1 4 3

d) 3 4 1 2

Q.51 For the network shown above I = (0.2V – 2) A, (I = the current delivered by the voltage source V). The Thevenin voltage Vth and resistance Rthfor the network N across the terminals AB are respectively

a) –10V, 5 Ω b) 10V, 5 Ω c) –10V, 0.2 Ω d) 10V, 0.2 Ω Q.52 In the circuit, S was initially open. At

time t=0, S is closed. When the current through the inductor is 6A, the rate of change of current through the resistor is 6A/s. The value of the inductor would be


a) 1H b) 2H

c) 3H d) 4H Q.53 In the circuit in fig, the switch S is

closed at t = 0. Assuming that there is no initial charge in the capacitor, the current iC (t) for t > 0 is

a) 2t

S RCV e AR

− b)

2tS RCV e A

2R−

c) 1

S RCV e A2R

− d)

1S RCV e A

R−

Q.54 In the circuit shown in the given

figure, the values of i(0+) and I (∞), will be, respectively

a) zero and 1.5A b) 1.5 A and 3A

c) 3A and zero d) 3A and 1.5A Q.55 In the circuit shown in the given

figure, C1 = C2 = 2F and the capacitor C1 has a voltage of 20V when S is open.

If the switch S is closed at t = 0, the

voltage 2cv will be a

a) Fixed voltage of 20V b) Fixed voltage of 10V c) Fixed voltage of -10V d) Sinusoidal voltage

Q.56 The circuit shown in the given figure is in the steady state with the switch S closed

The current i(t) after S is opened at t=0 is

a) A decreasing exponential b) An increasing exponential c) A damped sinusoid d) Oscillatory Q.57 A resistor R of 1 Ω and two

inductors L1 and L2 of inductances 1H and 2H, respectively, are connected in parallel. At some time, the currents through L1 and L2 are 1A and 2A, respectively. The current through R at time t = ∞ will be

a) zero b) 1A c) 2A d) 3A

Q.58 In the circuit shown, the switch is

moved from position A to B at time t = 0. The current i through the inductor satisfies the following conditions.

1.i(0) 8A= −

di2. (t 0) 3A / sdt

= =

3.i( ) 4A∞ =

The value of R is a) 0.5 ohm b) 2.0 ohm c) 4.0 ohm d) 12 ohm Q.59 For the circuit in fig, the switch was

kept closed for a long time before opening it at time t = 0. The voltage vL (0+) is


a) – 10V b) – 1V c) 0 V d) 10V Q.60 For the circuit given, what is the

expression for the voltage v?

a) i cv v+ b) cv

c) cc

dvRC vdt

− d) cc

dvRC vdt

+

Q.61 A step voltage is applied to the

circuit shown. What is the transient current response of the circuit?

a) Undamped sinusoidal b) Overdamped

c) Underdamped d) Critically damped Q.62 The circuit shown above is under

steady-state condition with the switch closed. The switch is opened at t = 0. What is the time constant of the circuit?

a) 0.1 s b) 0.2 s

c) 5 s d) 10 s Q.63 Consider the following statements

regarding the use of Laplace transforms and Fourier transforms in circuit analysis:

1. Both make the solution of circuit problems simple and easy.

2. Both are applicable for the study of circuit behavior for t – α to α.

3. Both convert differential equations to algebraic equations.

4. Both can be used for transient and steady state analysis.

Which of the above statements are correct? a) 1, 2, 3 and 4 b) 2, 3 and 4 only c) 1, 2 and 4 only d) 1, 3 and 4 only

Q.64 If the switch S in the circuit shown

above is opened at t = 0, what are the values of V(0+) and d V(0+)/dt, respectively?

a) 100V, 10,000 V/s

b) 100 V, - 10,000 V/s c) –100 V, 10,000 V/s d) –100 V, -10,000 V/s Q.65 The response of a linear, time-

invariant system to a unit step is s(t) = (1 – e-t/RC) u(t), where u(t) is the unit step. What is the impulse response of this system?

a) t /RCe− b) e-t/RC u(t) c) 1/RC e-t/RCu(t) d) (t)δ

Q.66 The relation between input x(t) and output y(t) of a continuous-time system is given by dy(t) 3y(t) x(t).

dt+ =

What is the forced response of the system when x(t) = k (a constant)?

a) k b) k/3 c) 3k d) 0 Q.67 In the circuit shown below, the

switch in open for a long time and closed at time t = 0. What is the


current through the switch after the switch is closed?

a) Zero b) 1 A c) 2 A d) 5 A

Q.68 In the circuit shown above, the switch is closed at t = 0. What is the initial value of the current through the capacitor?

a) 0.8 A b) 1.6 A c) 2.4 A d) 3.2 A

Q.69 The switch of above circuit was

open for long, and at t = 0 it is closed. What is the final steady state voltage across the capacitor and the time – constant of the circuit?

a) 0 V and 0.1 sec b) 20V and 0.2 sec c) 10 V and 0.2 sec d) 10V and 0.1 sec

Q.70 In the circuit shown, VC is 0 volts at t

= 0 sec. For t >0, the capacitor current ic(t), where t is in seconds, is given by

a) 0.50 exp (-25 t) mA b) 0.25 exp (-25 t) mA

c) 0.50 exp (-12.5 t) mA

d) 0.25 exp (-6.25 t) mA Q.71 The network shown above is

initially at rest. What is the initial current I when the switch S is closed at t = 0?

a) 0A b) 5A c) 10A d) 20A

Q.72 For series R-L-C circuit, the

characteristic 2 R 1equation is given as s s 0.

L LC+ + =

RIf is denoted by and2L

α1LC

by then under theconditionβ 2 2of , thesystem will beβ > α a) Critically damped

b) Under damped c) Undamped

d) Over damped

Q.73 In the above circuit, the switch has been in position 1 for quite a long time. At t = 0 the switch is moved to position 2. At this position what is the time constant?

a) 0.1 s b) 1 s c) 0.11 s d) 1.11 s Q.74 In the above circuit, the switch is

open for a long time. At time t = 0, the switch is closed. What are the initial and final values of voltages across the inductor?


a)0V and 0V b)0V and 80V c)80 V and 0V d)80V and 80V Q.75 The voltage applied to an R-L circuit

at t = 0 when switch is closed is 100 cos (100t + 30°). The circuit resistance is 80 Ω and inductance is 0.6 H (in which initial current is zero). What is the maximum amplitude of current flowing through the circuit?

a) 1A b) 2A c) 5A d) 10A

Q.76 The current in the network is

a) |t – 1 + e-t | u(t) b) |t2 – t + e-t | u(t) c) |t + 1 + e-t | u(t) d) |t – 1 – e-t | u(t)

Q.77 The switch shown Fig, is ideal and

has been in position 1 for t < 0. If the switch is moved to position 2 at t = 0, then v0 for t > 0 is given by

a) 0 V

b) 2 + 2(1 – e-100t) c) 2(1 – e-1000t)

d) 2 e-1000t V Q.78 Voltage and current expressions for

the above circuit are given at t ≥ 0 as

v =, 50t 50t125e V,i 5e A− −= .The value of L will be

a) 0.005 H b) 0.05 H

c) 0.5 H d) 5 H Q.79 In the circuit shown above, switch S

is closed at t = 0. The time constant of the circuit and initial value of current i(t) are

a) 30 sec, 0.5 A b) 60 sec, 1.0 A c) 90 sec, 1.0 A d) 20 sec, 0.5 A

Q.80 The circuit as shown above is in the

steady state. The switch S is closed at t=0. What are the values of v and dvdt

at t = 0?

a) 0 and 4 b) 4 and 0 c) 2 and 0 d) 0 and 2 Q.81 The value of the current i(t) in

amperes in the above circuit is

a) 0 b) 10 c) 10 e-t d) 10 (1 – e-t)


Q.82 In the circuit shown above, the switch is closed after a long time. The current is(0+) through the switch is

a) 1A b) 2/3A

c) 1/3A d) 0A

Q.83 The value of V that would result in a steady-state current of 1A through the inductor in the above circuit is

a) 30V b) 15V c) 20V d) 25V

Q.84 The circuit shown in the figure is in

steady state before the switch is closed at t = 0. The current iS(0+) through the switch is

a) 1/3 A b) 2/3 A

c) 1 A d) 0 A Q.85 Consider the following statements:

1. Voltage across a capacitor cannot change abruptly

2. Voltage across an inductor cannot change abruptly

3. Current through a capacitor cannot change abruptly

4. Current through an inductor cannot change abruptly

Which of these statements are correct?

a) 1 and 2 only b) 2 and 3 only c) 3 and 4 only d) 1 and 4 only Q.86 In the circuit shown, the initial

current I0 through the inductor is given in the figure. The initial value of the voltage across the inductor V0(0+) is

a) 12.5 V b) 5.0 V

c) 10.0 V d) 0.0 V Q.87 Initially, the circuit shown in the

given figure was relaxed. if the switch is closed at t=0, the values of

i(0+), di/dt (0+) and ( )2

2

d i 0dt

+ will

respectively be

a) 0, 10 and -100 b) 0, 10 and 100 c) 10,100 and 0 d) 100, 0 and 10

Q.88 For the above shown network, the

function 0

i

V (s)G(s)V (s)

= is 2

4ss 4s 20+ +

when R is 2 ohm. What is the value of L and C?

a) 0.3 H and 1 F b) 0.4H and 0.5F c) 0.5H and 0.1F d) 0.5H and 0.01F

Q.89 A unit step u (t-5) is applied to the

RL network. The current i is given by


a) 1- e-t b) [1-e-(t-5)] u(t-5) c) (1- e-t) u(t-5) d) 1-e-(t-5)

Q.90 The response of a network is

i(t)=Kte-α t for t 0≥ where α is real positive. The value of‘t’ at which the i(t) will become maximum, is

a) α b) 2α c) 1/α d) α2

Q.91 If I = - 10 e-2t, the voltage of the

source of the given circuit, Vs is given by

a) -10 e-2t b) -20 e-2t

c) 20 e-2t d) -30 e-2t Q.92 The steady state in the circuit,

shown in the given figure is reached with S open. S is closed at t =0. The current I at t =0+ is

a) 1 A b) 2 A c) 3 A d) 4 A

Q.93 The system function H(s) = .1

1+s

for

an input signal cost, the steady state response is

a) 1 cos t42π −

b) cost

c) cos t4π −

d) 1 cos t

2

Q.94 For the circuit shown in the given

figure, if C=20µF, v(0-)=-50V and dv(0 ) 500V / S

dt

−

= , then R is

a) 2K b) 3K c) 5K d) 10K Q.95 In the circuit shown in the given

figure, the switch is closed at t = 0. The current through the capacitor will decrease exponentially with a time constant:

a) 0.5s b) 1s c) 2s d) 10s

Q.96 A unit step current of 1A is applied

to a network whose driving point

impedance is Z(s) = 2

V(s) s 3I(s) (s 2)

+=

+;

then the steady state and initial values of the voltage developed across the source are respectively

a) 3 V,1V4

b) 1 3V, V4 4

c) 3 V,0V4

d) 31V, V4

Q.97 The current i in a series R-L circuit

with R = 10 Ω and L = 20 mH is given by i = 2sin 500t A. If v is the


voltage across the R-L combination, then i

a) Lags v by 450 b) is in-phase with c) Leads v by 450 d) lags v by 900

Q.98 The input impedance of a series RLC

circuit operating at frequency02ω = ω , ω0 being the resonant

frequency, is

a) 0LR j2

ωΩ −

b) 0LR j

2ω

Ω +

c) 0(R j 2 L)− ω Ω d) 0(R j 2 L)+ ω Ω Q.99 Which one of the following

theorems can be conveniently used to calculate the power consumed by the 10 Ω resistor in the network shown in the given figure?

a) Thevenin’s theorem b) Maximum power transfer

theorem c) Millman’s theorem d) Superposition theorem

Q.100 In the circuit shown in the given

figure, the current supplied by the sinusoidal current source I is

a) 28A b) 4A c) 20A d) not determinable from the data

given. Q.101 In the circuit, if the power dissipated

in the 6Ω resistor is zero then V is

a) o20 2 45∠ b) o3020 ∠

c) o4520∠ d) o30220 ∠ Q.102 In the circuit shown in the above

figure, switch K is closed at t=0. The circuit was initially relaxed. Which one of the following sources of v(t) will produce maximum current at t=0+ ?

a) Unit step b) Unit impulse c) Unit ramp d) Unit step plus unit ramp

Q.103 Consider the following statements: If a network has an impedance of (1-

j) as a specific frequency, the circuit would consist of series

1. R and C 2. R and L 3. R, L and C Which of these statements are

correct? a) 1 and 2 b) 1 and 3 c) 1, 2 and 3 d) 2 and 3

Q.104 In the transformer shown in the

given figure, the inductance measured across the terminal 1 and 2 was 4H with open terminals 3 and 4. It was 3H when the terminal 3 and 4 were short circuited. The coefficient of coupling would be

a) 1 b) 0.707


c) 0.5 d) indeterminate due to insufficient

data. Q.105 The circuit shown in the figure, will

act as an ideal current source with respect to terminals A and B, when frequency is

a) Zero b) 1 rad/s c) 4 rad/s d) 16 rad/s

Q.106 A series LCR circuit with

LR 10 |X | 20= Ω = Ω , and |Xc| = 20Ω is connected across an ac supply of 200Vrms. The rms voltage across the capacitor is a) o200 -90 V∠ b) o200 90 V ∠ c) o400 90 V∠ d) o400 -90 V∠

Q.107 In the circuit, if |I1| = |I2| = 10A

a)

-1 -11 2

8 8I will lead bytan , I will lag by tan6 6

b)-1 -1

1 26 6I will lead by tan ,I will lag by tan8 8

c) 1 1

1 28 8I will lag by tan , I will lead by tan6 6

− −

d) 1 1

1 26 8I will lag by tan , I will lead by tain8 6

− −

Q.108 In a two element series network, the

voltage and current respectively are given as.V (t) = 50 sin (314 t) + 50 sin (942 t) V, i (t) = 10 sin (314 t + 600) + 8 sin (942 t + 450) A, then the

power factor of the network is approximately:

a) 0.9 b) 0.6 c) 0.3 d) 0.1

Q.109 A series R–L circuit is to be

connected to an a.c. source v(t) = Vm sin ( iω +φ )volt. Which one of the following is correct? The transient current will be absent if the source is connected at a time t0 such that a) ωt0 = 0

b) ωt0 =2π

c) ωt0 = tan-1L

Rω

d) ωt0 has any arbitrary value

Q.110 A series R – L – C circuit is switched

on to a step voltage V at t = 0. What is the initial and final value of the current in the circuit, respectively?

a)V/R, V/R b)Zero, Infinity c)Zero, Zero d)Zero, V/R Q.111 A lossy capacitor is represented by

an ideal capacitor C with a high resistance R in parallel. What is the Q of the circuit at frequencyω ?

a) ωCR b) 1/(ωCR) c) ωC/R d) R/(ωC)

Q.112 Two coils are coupled in such a way

that the mutual inductance between them is 16mH. If the inductances of the coils are 20mH and 80mH respectively, the coefficient of coupling is:

a) 0.01 b) 0.4 c) 0.1 d) 0.0025

Q.113 When is a series RLC circuit over

damped?

a) 2

2

R 14L LC

= b) 2R 1

4L C<

c) 2R 1

4L C> d)

2

2

R 14C LC

=


Q.114 Consider the following circuit: For the above circuit, which one of

the following statements is correct? The voltage V0 is independent of R, if the input signal frequency ω

a) 1is

LC b) 1is

2 LC

c) is LC d) Has any value Q.115 Consider the following circuit: For

what value of ω, the circuit shown exhibits unity power factor ?

a) 1LC

b) 2 2

1

LC R C +

c) 2 2

1

LC R C −

d) 1RC

Q.116 An RLC series circuit has a

resistance R of 20 Ω and a current which lags behind the applied voltage by 45o. If the voltage across the inductor is twice the voltage across the capacitor, what is the value of inductive reactance?

a) 10 Ω b) 20Ω c) 40Ω d) 60Ω

Q.117 R and C are connected in parallel

across a sinusoidal voltage source of 240V. If the currents through the source and the capacitor are 5A and 4A, respectively; what is the value of R?

a) 24 Ω b) 48 c) 80 Ω d) 240 Ω Q.118 A parallel circuit consists of two

branches: one with a pure capacitor and the other has resistor of 5 Ω in series with a variable inductor. To this circuit on ac voltage of fixed

value and frequency is connected. The circuit will exhibit two resonances if a) The reactance of the capacitor is

less than 10Ω. b) The reactance of the capacitor is

greater than 10Ω. c) The reactance of the capacitor

equals 10Ω d) The capacitor is removed by a

short circuit.

Q.119 A series R-L-C circuit, excited by a 100V variable frequency source, has a resistance of 10Ω and an inductive reactance of 50 Ω at 100 Hz. If the resonance frequency is 500 Hz, what is the voltage across the capacitor at resonance?

a) 100V b) 500V c) 2500V d) 5000V

Q.120 In a series RLC, circuit, the locus of

the tip of the admittance phasor in the complex plane as the frequency is varied, is a) A semicircle in the upper half of

the G-B plane having the centre

at 1 ,0R

and radius R1

b) A circle in the upper half of the G-B plane having the centre at

( )1 ,0

2R

and radius ( )R2

1

c) A semicircle in the bottom half of the G-B plane having the centre

at ( ) ( )

1 1,0 and radius2R 2R

−

d) A semicircle in the upper half of the G-B plane having the centre

at 1 1,0 and radiusR R

−

Q.121 Which one of the following

statements is not correct for the circuit shown at resonant frequency?


a) The current is maximum b) The equivalent impedance is real c) The inductive and capacitive

reactance are equal in magnitude

d) The quality factor equals LC

R1

Q.122 A parallel circuit has two branches.

In one branch, R and L are in series and in the other branch, R and C are in series. The circuit will exhibit unity power factor when

a) LRC

= b) R LC=

c) CRL

= d) LRC

=

Q.123 x(t): Input voltage y(t): Output voltage Consider the circuit shown above: What is the natural response of this

system?

a) A sinusoid with constant

amplitude b) A growing sinusoid c) Zero d) A decaying sinusoid

Q.124 Consider the following statements: When a series R –L –C circuit is

under resonance 1. Current is maximum through R 2. Magnitude of the voltage across

L is equal to that across C 3. The power factor of the circuit is

unity

Which of the statements given above are correct?

a) 1, 2 and 3 b) 1 and 2 only c) 2 and 3 only d)1 and 3 only Q.125 Width of resonance curve in an R – L

– C network is determined by which one of the following?

a)R alone b)L alone c)C alone d)All R, L and C

Q.126 What is the average power for

periodic non-sinusoidal voltages and currents? a) The average power of the

fundamental component alone b) The sum of the average powers

of the harmonics excluding the fundamental

c) The sum of the average powers of the sinusoidal components including the fundamental

d) The sum of the root mean square power of the sinusoidal components including the fundamental

Q.127 A coil is tuned to resonance at 500

kHz with a resonating capacitor of 36pF. At 250 kHz, the resonance is obtained with resonating capacitor of 160pF. What is the self – capacitance of the coil?

a) 2.66pF b) 5.33pF c) 8pF d) 10.66pF

Q.128 Which one of the following relations for power is not correct?

a) P = VIcosϕ b) P = Re part of [VI*]

c) P = Re part of [V*I] d) P =VIsinϕ

Q.129 For the circuit shown in the above

figure, what is the natural frequency?


a) 1 M rad/s b) 2 M rad/sc) 3 M rad/s d) 5 M rad/s

Q.130 If the transmission parameters of the above network are A=C=1, B=2 and D=3, then the value of Zm is

a) Ω1312 b) Ω

1213

c) Ω3 d) Ω4

Q.131 The impedance matrices of two, two-port networks are given by

3 22 3

and 15 55 25

. If these two

networks are connected in series, the impedance matrix of the resulting two-port network will be

a)3 52 25

b)18 77 28

c)15 25 3

d)indeterminate

Q.132 A two port network is reciprocal, if and only if a) 2211 ZZ =b) BC AD 1− = −c) 12 21Y Y= −d) 12 21h h=

Q.133 The input voltage V1 and current I1 for a linear passive network is given by V1 = AV2 + BI2 and I1 = CV2 + DI2 Now consider the following network:

Which one of the following is the

transfer matrix A BC D

of the

network shown?

a) 1 00 10

b) 1 100 1

c) 0 1

10 0

d) 0 101 0

Q.134 For an ideal step-down (n:1) transformer, which one of the following is the ABCD parameter matrix?

a) n 11 n

b) n 00 n

c) n 00 1/ n

d) n 1/ n

1/ n 1

Q.135 What is the expression for h12 in respect of the network shown:

a) 2 1

1 2

Z ZZ Z

−+

b) 1 2

2 1

Z ZZ Z+−

c) 1 2

1 2

Z ZZ Z+−

d) 1 2

1 2

Z ZZ Z−+

Q.136 Two two-port networks are connected in parallel. The combination is to be represented as a single two-port network. The parameters of this network are obtained by addition of the individual a) z-parametersb) h – parameterc) y-parametersd)ABCD parameters


Q.137 Two 2-port networks with

transmission matrices

TA = B

1 2 2 4and T

0.1 4 0.5 3

=

are connected in cascade. Which is the transmission matrix of the combination?

a) 3 10

2.2 12.4

b) 3 6

0.2 12.4

c) 1 10

2.0 12.0

d) 3 10

12.4 2.2

Q.138 Which one of the following is the

transmission matrix for the network shown in the figure given above?

a) 1 1 yzy z

+

b) 1 yz z

y 1+

c) 1 zy 1 yz +

d) 1 1 yzz y

+

Q.139 What is the value of the parameter

h12 for the 2-port network shown in the figure given above?

a) 0.125 b) 0.167

c) 0.250 d) 0.625 Q.140 The currents I1 and I2 at the output

of 2-port network can be written as I1 = 5V1 – V2 I2 = - V1 + V2

Which one of the following gives the parameters of an equivalent πnetwork shown above?

a) y1 = 4 , y2 = 0, y3 = 1 b) y1 = 4 , y2 = 4 , y3 = 1 c) y1 = 1 , y2 = 1 , y3 = 1 d) y1 = 4 , y2 =0, y3 = 2 Q.141 Which of the following are the

conditions for a two port passive network to be a reciprocal one?

1. z12 = z21 2. y12 = y21 3. h12 = -h21 Select the correct condition from the

code given below: a) Only 1 and 2 b) Only 2 and 3 c) Only 1 and 3 d) 1, 2 and 3 Q.142 What is the open circuit impedance

Z’11(s) of the network shown in the figure given above?

a) 10 + 2s b) 410

s −

c) 410s

+

d) 10 – 2s

Q.143 What is the value of z21 for the

network shown above?

a) - 2Ω b) –1/2 Ω

c) 1Ω d) 2Ω


Q.144 The circuit shown in the figure above

a) Is reciprocal but not symmetrical b) Is not reciprocal but symmetrical c) Is both reciprocal and

symmetrical d) Is neither reciprocal nor

symmetrical

Q.145 The circuit shown in the above figure

1. is reciprocal 2. has Z11 = 2, Z22 = 2 3. has Z11 = 4, Z22 = 2 4. has Z11 = 0, Z22 = 2 Select the correct answer using the

code given below:

a) 1 and 3 b) 1 and 2 c) 1 and 4 d) 3 only

Q.146 A reciprocal two-port network is symmetrical if a) ∇ A = 1 b) A = C

c) z11 = z22 d) ∆ y = 1 Q.147 With respect to transmission

parameters, which one of the following is correct?

a) A and B are dimensionless b) B and C are dimensionless c) A and D are dimensionless d) B and D are dimensionless Q.148 A Two-port network has z11 =

13/35, z12 = z21 = 2/35, z22 = 3/35. Its y11 and y12 parameters will, respectively, be

a) 3, -2 b) 3, 2 c) 13, -2 d) 13, 2

Q.149 What are the ABCD parameters of the single element circuit given above?

a) 1 Z0 1

b) 1 1Z 0

c) 1 Z1 0

d) Z 11 1

Q.150 For determining the network

functions of a two-port network, it is required to consider that a) All initial conditions remain

same b) All initial conditions are zero c) Part of initial conditions are

equal to zero d) Initial conditions vary depending

on nature of network Q.151 If a two-port network is reciprocal

as well as symmetrical, which one of the following relationships is correct?

a) Z12 = Z21 and Z11 = Z22 b) Y12 = Y21 and Y11 = Y22 c) AD – BC = 1 and A = D d) All of the above Q.152 Match List I with List II and select

the correct answer using the code given below the lists:

List-I (Network parameter) A. Z11 B. A C. C D. Z22 List-II (Measure under open-circuit conditions)


1) 21

2

V I 0I

=

2) 12

2

V I 0V

=

3) 12

1

V I 0I

=

4) 12

2

I I 0V

=

Codes: A B C D

a) 1 4 2 3 b) 3 4 2 1 c) 1 2 4 3 d) 3 2 4 1 Q.153 If the connection of two two-ports is

such that the transmission matrix of the overall network is the product of the transmission matrices of the individual networks, what type of connection is it?

a) Series connection b) Cascade connection c) Parallel connection d) none of the above Q.154 In the case of ABCD parameters, if

all the impedances in the network are doubled, then a) A and D remain unchanged, C is

halved and B is doubled b) A, B, C and D are doubled c) A and B are doubled, C and D are

unchanged d) A and D are unchanged, C is

doubled and B is halved Q.155 Match List I with List II and select

the correct answer using the code given below the lists:

List I (Excitation)

List II (Two-port parameters)

A. I1, I2 B. V1, V2 C. I1, V2

D. V1, I2

1. y 2. z 3. g 4. h

Codes: A B C D

a) 1 2 3 4 b) 4 2 3 1 c) 1 3 2 4 d) 4 3 2 1 Q.156 With reference to the above

network the value of Z11 will be

a) – 3 b) 3

c) – 1 d) – 5 Q.157 A two-port network satisfies the

following relations: 4I1 + 8I2 = 2V1 8I1 + 16 I2 = V2 1. The network is reciprocal 2. Z11 = 4 and Z12 = 8 3. Z21 = 8 and Z22 = 16 4. Z11 = 2 and Z12 = 4

Which of these relations are correct?

a)1, 2, 3 and 4 b)2 and 3 only c)3 and 4 only d)1 and 2 only

Q.158 In the circuit shown, 2-port network

N has Z11 = 103 Ω, Z12 = 10Ω, Z21 = -106Ω and Z22 =104Ω. The current

gain 2

1

I isI


a) -50 b) +50c) +20 d) -20

Q.159 In the 2-port network shown in the figure, the value of Y12 is

a) 1 mho3

− b) 1 mho3

+

c) 3mho− d) 3mho+

Q.160 The h parameters h11 and h22 are related to z and y parameters as a) h11=z11 and h22=1/z22

b) h11= z11 and h22=y22

c) h11= 1/y11 and h22=1/z22

d) h11=1/y11 and h22=y22

Q.161 The lattice has the following impedances ZA = 3 + j4, Zn = 3-j4. Then the Z- parameters would be

a)3 j4 0

0 3 j4+

− b)

3 j4j4 3

− −

c)3 j4 3

3 3 j4−

+ d)

j4 33 j4−

+

Q.162 Which one of the following gives the correct short circuit parameter matrix Y for the network shown

a) 0.7 0.50.5 0.8

− −

b) 0.7 0.50.5 0.8

−

c) 0.8 0.50.5 0.7

− −

d) 0.7 0.50.5 0.8

−

Q.163 Assertion A: The fundamental loop of a linear directed graph contains four twigs and two links corresponding to a given tree. Reason R: In a linear directed graph, a link forms a closed loop. a) Both A and R are individually

true and R is the correctexplanation of A.

b) Both A and R are individuallytrue but R is NOT the correctexplanation of A.

c) A is true but R is false.d) A is false but R is true

Q.164 The number of edges in a compete graph of n vertices is

a) n (n-1) b) 2

)1n(n −

c) n d) n-1

Q.165 Which one of the following is a cut set of the graph shown in the figure?

a) 1, 2, 3 and 4 b) 2, 3, 4 and 6c) 1, 4, 5 and 6 d) 1, 3, 4 and 5

Q.166 The network has 10 nodes and 17 branches. The number of different node pair voltages would be a) 7 b) 9c) 10 d) 45

Q.167 The dual of a parallel R-C circuit is a a) Series R-C circuitb) Series R-L circuitc) Parallel R-C circuitd) Parallel R-C circuit


Q.168 For a network of 11 branches and 6

nodes, what is the number of independent loops?

a) 4 b) 5 c) 6 d) 11

Q.169 Which one of the following

statements is not correct? a) A tree contains all the vertices of

its graph. b) A circuit contains all the vertices

of its graph. c) The number of f-circuits is the

same as the number of chords. d) There are at least two edges in a

circuit. Q.170 Consider the following graph:

Which one of the following is not a tree of the above graph? a) b)

c) d)

Q.171 What is the total number of trees for

the graph shown above?

a) 4 b) 8 c) 12 d) 16

Q.172 For the network graph shown in the figure given above, which one of the following is not a tree?

a) b)

c) d)

Q.173 The graph of a network is shown in

figure above. Which one of the figures shown below is not a tree of the graph?

a) b)

c) d)

Q.174 A network has 4 nodes and 3

independent loops. Number of branches in the network? a) 5 b) 6 c) 7 d) 8

Q.175 Consider the following statements with regard to a complete incidence matrix: 1. The sum of the entries in any

column is zero.


2. The rank of the matrix is n – 1 where n is the number of nodes.

3. The determinant of the matrix of a closed loop is zero.

Which of the statements given above are correct?

a) 1 and 2 only b) 2 and 3 only c) 1 and 3 only d) 1, 2 and 3

Q.176 What is the number of chords of a

connected graph G of n vertices and e edges?

a) n (n – 1)/2 b) n – 1 c) e – n – 1 d) e – n + 1

Q.177 Consider a circuit which consists of

resistors and independent current sources, and one independent voltage source connected between the nodes i and j. The equations are obtained for voltage of n unknown nodes with respect to one reference node in the form

1

2

n

VV

[ ]

V

∆

=

MM

M MM

What are the elements of the∆ ? a) All conductance’s b) All resistances c) Mixed conductance’s and

constant d) Mixed conductance’s and

resistances Q.178 Consider the directed graph shown above: What is its incidence matrix?

a)1 1 0

0 1 11 0 1

− − − − −

b)1 0 11 1 00 1 1

− −

c)1 1 0

0 1 11 0 1

− − −

d)1 0 11 1 0

0 1 1

− − −

Q.179 Number of fundamental cut-sets of

any graph will be a) Same as the number of twigs b) Same as the number of tree

branches c) Same as the number of nodes d) Equal to one

Q.180 In a network with twelve circuit

elements and five nodes, what is the minimum number of mesh equations?

a) 24 b) 12 c) 10 d) 8 Q.181 The maximum number of trees of

the graph in fig, is

a)16 b)25 c)100 d)125 Q.182 Consider the spanning tree of the

connected graph: What is the number of fundamental cut-sets?

a)15 b)16 c)8 d)7 Q.183 Match List X with List Y for the tree

branches 1, 2, 3 and 8 of the graph shown in the given figure and select the correct answer using the codes given below the lists:


List X List Y Twigs 4,5,6,7 Links 1,2,3,8 Fundamental cutest 1,2,3,4 Fundamental loop 6,7,8

Codes: A B C D

a) I II III IV b) III II I IV c) I IV III II d) III IV I II Q.184 Driving point impedance

4s1)s(sZ(s) 2

2

+

+=

is not

realizable because the a) Number of zeroes is more than

the number of poles b) Poles and zeroes lie on the

imaginary axis c) Poles and zeroes do not

alternate on imaginary axis d) Poles and zeroes are not located

on the real axis

Q.185 The function s32s ++ can be realized

a) Both as a driving point impedance and as a driving point admittance

b) As an impedance, but not as an admittance

c) As an admittance, but not as an impedance

d) Neither as impedance nor as admittance

Q.186 What is the minimum number of elements required to realize a given driving point susceptance function? a) One greater than the total

number of internal poles & zeros

b) Equal to the total number of internal poles and zeros

c) One less than the total number of internal poles and zeros

d) None of the above Q.187 For the network shown in the figure

given above, what is the value of z(s)?

a) 2s 2s 2

s 1+ ++

b) 2

s 2(s 1)++

c) 2

s 1s 2s 2

++ +

d) 2(s 1)

(s 2)++

Q.188 If Y1 and Y2 are the RC driving point

admittance and impedance functions, respectively, such that

12

2

Y (s 1)(s 3)Y (s 2)

+ +=

+ then, the value of

Y2 can be

s 31.s 2++

s 22.s 1++

s 23.(s 1)(s 3)

++ +

4. s 2+

Which of the above are correct? a) 1 and 2 b) 2 and 3

c) 3 and 4 d) 1 and 4 Q.189 A reactive network has poles at ω =

0, 4000 rad/s, and infinity and zeros at ω = 2000 and 6000 rad/s. The impedance of the network is –j 700 Ohm at 1000 rad/s. What is the correct expression for the driving point impedance?

a) 2 6 2 6

2 2 6

( 4 10 )( 36 10 )j(0.1 ) Ohm( 16 10 )

ω − × ω − ×− ω

ω ω − ×

b)

2 2 6

2 6 2 6

( 16 10 )j(0.1 ) Ohm( 4 10 )( 36 10 )

ω ω − ×ω

ω − × ω × ×

c) 2 6 2 6

2 2 6

( 4 10 )( 36 10 )j(0.1 ) Ohm( 16 10 )

ω − × ω − ×ω

ω ω − ×


d) 2 2 6

2 6 2 6

( 16 10 )j(0.1 ) Ohm( 4 10 )( 36 10 )

ω ω − ×− ω

ω − × ω − ×

Q.190 Consider the following expression for the driving point impedance:

)4()9)(1(2)( 2

22

+++

=ss

sssZ

1. It represents an LC circuit2. It represents an RLC circuit3. It has poles lying on the jω axis4. It has a pole at infinite frequency

and a zero at zero frequencyWhich of the statements given above are correct? a) 2 and 4 b) 1 and 3c) 1 and 4 d) 2 and 3

Q.191 Which one of the following functions is an RC driving point impedance?

a) s(s 3)(s 4)(s 1)(s 2)+ ++ +

b) (s 3)(s 4)(s 1)(s 2)+ ++ +

c) (s 3)(s 4)s(s 1)(s 2)+ ++ +

d) (s 2)(s 4)(s 1)(s 3)+ ++ +

Q.192 Laplace transform of sin (ωt +α ) is

a) 2 2

s cos sinsα +ω α+ω

b) 2 2 cossω

α+ω

c) 2 2

s sins

α+ω

d) 2 2

s sin cossα +ω α+ω

Q.193 For an RC driving-point impedance function ZRC (s) a) ZRC (0) ≥ ZRC(∞ )b) ZRC (0) = ZRC(∞ ) onlyc) ZRC(0) ≤ ZRC (∞ )d) ZRC (0) > ZRC (∞ ) only


1 2 3 4 5 6 7 8 9 10 11 12 13 14 (b) (c) (c) (a) (b) (b) (b) (a) (a) (b) (d) (a) (d) (a) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (c) (a) (b) (a) (d) (c) (a) (c) (a) (d) (b) (b) (b) (a) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (c) (d) (a) (c) (d) (c) (b) (a) (d) (c) (c) (b) (b) (c) 43 44 45 46 47 48 49 50 51 52 53 54 55 56 (a) (b) (d) (c) (b) (a) (b) (b) (b) (b) (a) (c) (d) (a) 57 58 59 60 61 62 63 64 65 66 67 68 69 70 (a) (a) (a) (d) (d) (a) (c) (b) (c) (b) (a) (a) (d) (a) 71 72 73 74 75 76 77 78 79 80 81 82 83 84 (c) (b) (a) (c) (a) (a) (a) (c) (d) (c) (c) (c) (a) (a) 85 86 87 88 89 90 91 92 93 94 95 96 97 98 (d) (a) (a) (c) (b) (c) (b) (b) (a) (c) (a) (c) (a) (b) 99 100 101 102 103 104 105 106 107 108 109 110 111 112 (d) (c) (a) (b) (b) (c) (c) (a) (c) (b) (c) (c) (a) (b)

113 114 115 116 117 118 119 120 121 122 123 124 125 126 (c) (a) (c) (c) (c) (b) (c) (b) (d) (a) (a) (a) (d) (d)

127 128 129 130 131 132 133 134 135 136 137 138 139 140 (b) (d) (d) (a) (b) (b) (b) (c) (a) (c) (a) (c) (c) (a)

141 142 143 144 145 146 147 148 149 150 151 152 153 154 (d) (c) (b) (a) (c) (c) (c) (a) (a) (b) (d) (d) (b) (a)

155 156 157 158 159 160 161 162 163 164 165 166 167 168 (a) (d) (c) (b) (a) (c) (b) (a) (d) (b) (d) (d) (b) (c)

169 170 171 172 173 174 175 176 177 178 179 180 181 182 (d) (c) (d) (d) (d) (b) (d) (d) (a) (c) (a) (d) (d) (d)

183 184 185 186 187 188 189 190 191 192 193 (a) (c) (a) (b) (c) (c) (c) (b) (d) (d) (d)

ANSWER KEY:


Q.1 (b) • The given function similar to

y x= which is Non linear,• The element is active since it has

two different slopes. One is +veslope (for i > 0) and other is – ve(for i < 0)

• For different current directionsimpedance is different hencenon-bilateral

Q.2 (c) Let, I be the current 2

1 1P I R=2

2 2P I R=Power Supplied by the source,

21 2P I (R R )= +

2 21 2 1 2I R I R P P= + = +

Q.3 (c)

Nodal at 2V ⇒

2 1 2V V V i 0R R−

+ −α = ------ (1)

Also 2 1V ViR− = −

------ (2)

Substituting (2) in (1) we get 2

1

V 1V 2

+α=

+α

Q.4 (a)

Since the independent sources cancel each other therefore the circuit will consist of only resistances. Hence the system is Linear and passive.

Q.5 (b)

Nodal⇒ xy xyV 120 V 60

40 60+ −

+

xyV 96=

xyR 40 || 60 24Ω= =

Q.6 (b)

KCL at node A⇒ xi 2 1 0− + − =

xi 1A= − ( )2vP 2 1= × −

2w(delivered)= − Power is delivered since current is entering from –ve terminal. absP due to 2V source ( )delP 2w 2w= = − − =

Q.7 (b)

EXPLANATIONS


KVL ⇒ b b24 I 2V V 4I 0− + − − = b24 5I V 0− + = ------- (1) Also bV 3I= ------- (2) From (2) and (1) I 12A= Q.8 (a) Q.9 (a)

By applying source transformation to the dependent source we get

Nodal at V ⇒

xV 51VV 15 V 07 7 19

+−+ + =

xV 51V2V 15 07 19

+−+ = -------- (1)

xV 51VI19+

= -------- (2)

xV 2I= -------- (3) (by ohm’s law) Solving (1) (2) and (3) we get I 94.34mA= − Q.10 (b)

Here ThV 5v= (when 1E E= and I is removed)

ThV 5R 5ΩI 1

= = = (For E=0, I=1A and

V = 5 volt)

V 2.5v= Q.11 (d)

R

24I 4A6

= =

R2I 8A= By KCL⇒ 1 RI 3I 12A= = Power delivered by 24v source 24 12 288w= × = Q.12 (a)

Nodal at 1V ⇒

1 1 2V 5 V V 1 05 2− −

+ − =

1 27V 5V 20− = --------- (1) Nodal at 2V ⇒

2 1 2 2V V V V 5 02 2 4− −

+ + =

1 22V 5V 5− + = --------- (2) Solving (1) &(2) 1 15V 25 V 5= ⇒ =

1V 5I 0A5−

= =

Q.13 (d)

KVL in Loop I ⇒


1 12I 2I 16 0− − − = 1 14I 16 I 4A− = ⇒ = − KVL in Loop II⇒ x 1V 2I 0 0+ + = x 1V 2I 8v= − = KVL in Loop III ⇒ xV 2I 16 0− − + = I 4A⇒ = Q.14 (a)

By applying source transformation we get

10 10I 0A5−

= =

Q.15 (c)

By finding the Thevenin’s equivalent of the circuit across 1Ω resistor

Th A BV V V= −

25 5 20 10v2 2 2

= − = =

Th1 1R 1 1 1 1 1Ω2 2

= + = + =

10I 5A2

= =

Q.16 (a)

Applying source transformation we get

3 1i A

6 2= =

Q.17 (b)

Nodal at xV ⇒

x xx

V 20 V 0.3V 01 10−

+ − =

xV 25v= 20 25I 5A

1−

= = −

20VP 20 5 100w= ×− = − (Delivered) (Since current is entering from –ve

terminal) ( )abs delP P 100w 100w= − = − − = Q.18 (a) Nodal at V1 ⇒

1 11

V 9 V3 0 V 13.5v3 9+

+ + = ⇒ = −

Q.19 (d)

kVL ⇒ V 6v 20= − 5V 20− = − V 4volt=

By ohm’s law, 4I 2A2

= =

Q.20 (c)


V 2volt=

KVL in Loop (I) ⇒ x xV 1 2 0 V 3volt− − = ⇒ = By KCL I 2 4 6A⇒ = + = Q.21 (a)

The Value of R doesn’t affect the value of V1 Hence when R=0,

1V 40v= . Since R=0 means circuit hence 2 1V V 40v= =

Q.22 (c) At t 0.5s= −

( )AV 600 0.5 300v= − = −

BV 0= CI 0= Hence the circuit can be modified as

300I 1A

300−

= = −

Q.23 (a)

Time constant of RL Ckt = Time constant of RC(dimensionally)

L RCR= (sec)

From above equation:

1R secL

−→

RC sec→

2L LR R ohmC C= ⇒ = →

( )2

0 01 1 radω ω secLCLC

= ⇒ = →

Q.24 (d)

Nodal at Vx ⇒

x xV 10 V 8 01 5−

+ − =

x x5V 50 V 40 0− + − =

xVI 3A5

= =

Q.25 (b)

V 10v= KCL ⇒ x xI 1 5 0 I 4+ − = ⇒ = − KVL ⇒ ( )E 20 V 0− − − = E 10v= − Q.26 (b)

1 2V VI 2μA1M−

= =

2in

V 2R 1.5MΩI 2μ

= = =

Q.27 (b) It is a balanced wheat stone bridge

30 60 1 160 120 2 2

= ⇒ =

Current through R is zero Power 2I R 0w= = Q.28 (a) By source Transformation


The circuit on both sides of 10Ω is same hence I 0=

Q.29 (c) Let the resistance be 2x, 4x, 6x

Since the resistances are connected in parallel, voltage is same. Currents are in the ratio of V V V: : 6 : 3 : 22x 4x 6x

⇒

Q.30 (d)

Nodal at Vx ⇒

x xV 3 V 6 02 1−

+ − =

x xV 3 2V 12− + = xV 5v=

5 3I 1A2−

= =

Q.31 (a)

When the load (4Ω) is constant the minimum resistance proved maximum power at load.

For resistance to be minimum R = 0Ω Q.32 (c)

( )2

1 2 3P P P P= ± ±

( )2

maxP 48 50 98= + + 450W=

( )2

minP 98 50 18= − − 2W=

Q.33 (b)

When impedance of one network is the complex conjugate of the other

L SZ Z= Q.34 (c) Rth across RL is 25Ω

For maximum power across L thR R 25= = Ω Q.35 (b)

Superposition theorem can be applied to calculate voltage and current only .It can’t be used to find power.

Q.36 (a)

Applying source Transformation we

get Nodal ⇒ x xV V 204 05 14

−− + + =

Q.37 (d)

Q.38 (c)

Tellegen theorem is applicable to any network linear or nonlinear, active or passive, time variant or time-invariant

Q.39 (c) Vth is open circuit voltage across RL 2I 0= due to open circuit KVL in input loop ⇒

1 1

SS 1 L L S

3VV 3I V 0 V V3 j4

− − = ⇒ = −+

Sj4V

3 j4

= +

1th L Sj4 4 90°V 10V 10 V 10 100

3 j4 5

= = = +

∟

800 90°= ∟ Q.40 (b)

Norton equivalent: Current source in parallel with impedance

Open circuit output admittance:


1

222 I 0

2

Ih |V ==

Reciprocal network: Network where loop and equation have a symmetric coefficient matrix Transmission Parameter:

1 2 2V AV BI= − 1 2 2I CV DI= − Q.41 (b) For fig (A) Voltage across 1Ω 5v= Current Across 1Ω 5A=

1. C C 1Ω

2 10AI 5A,R 2Ω, I 53 3

= = ⇒ = × =

2. C C 1Ω 1ΩI 10A,R 1Ω, I 5AV= = ⇒ = 5V= 3. C C 1Ω

1I 15A, R Ω, I2= = ⇒

1Ω

12 15 5A,V 5V32

= × = =

4. C C 1Ω1I 30A,R Ω, I5= = ⇒

1Ω

15 30 5A,V 5V15

= × = =

Q.42 (c)

Tellegan’s theorm is applicable for any type of element. It is based on KCL & KVL. Voltage and current direction can be arbitrary.

Q.43 (a)

Th8R R 8 || 4 Ω3

= = =

Q.44 (b)

Q.45 (d) Considering 30v source alone

130I 4 2.5 4 10A12

= × = × =

(By homogeneity principle) Considering 6v source alone 2I 1A= (By reciprocity) By superposition 1 2I I I 11A= + = Q.46 (c) ThR R= Calculated across R

Th5 20R 5 || 20 4 4 8Ω

25×

= + = + =

Q.47 (b)

Th 1 2V V V= −

40 60100 100100 100

= × − ×

40 60 20v= − = − ThV 20v= ThR 60 | 40 60 | 40= +

60 402 48Ω100×

= × =

Q.48 (a)

4 124 ||12 3Ω

16×

= =

AB1 1 1R4 6 12

= + +

AB3 2 1 6 R 2Ω

12 12+ +

= = ⇒ =


Q.49 (b)

( ) inR 2 ||1 1 ||1 1= + +

2 1 1 ||13

× = +

13 Ω8

=

1in

50 400I AR 13

= =

11 150I I A5 131

3

= × =+

Q.50 (b)

Norton’s theorem → current source with shunt resistor

Q.51 (b)

When V 0, I 2= = − (short circuit current) When I 0, V 10= = (open circuit voltage)

Th ThV 10V 10vR 5 5ΩI 2

= = = = − =−

Q.52 (b) ( ) ( ) ( ) ( )

tτ

L L L Li t i i 0 i e−

= ∞ + − ∞

( ) ( )L LLi 0 0,i 18A, τ Lsec1

= ∞ = = =

( )tτ

Li t 18 18e−

= − --------- (1)

tτ6 18 18e

−= − (Given ( )Li t 6= )

tτ 12 2e

18 3−

= =

From (1) ( ) tL τdi t 118e

dt τ− − ⇒ = −

18 26 L 2HL 3

= × ⇒ =

Q.53 (a)

( ) ( ) ( ) ( )tτ

C C C CV t V V 0 V e fort 0−

= ∞ + − ∞ ≥

( ) ( ) SC C

V RCV 0 ,V , τ2 2

∞ = =

( )2t

S S RCC

V VV t e2 2

−

= −

( )2t

C S RCC

dV V 2i t c c edt 2 RC

− = = − −

2t

S RCV eR

−

=

Q.54 (c)

at t 0+=

( ) 3i 0 3A1

+ = =

at t →∞ ( )i 0∞ = (Capacitor gates open circuital) Q.55 (d) Since the given circuit is a LC tank circuit hence the voltage VC2 will be sinusoidal.

Q.56 (a)

fort 0≥ converting the circuit in s-domain

( ) 2I s2 s

=+

( ) 2t i t 2e for t 0−= ≥ Q.57 (a)


As t →∞ , the inductor gets short circuited and current through resistor R would be zero.

Q.58 (a) ( ) ( ) ( ) ( )

tτ


= ∞ + − ∞

2τ secR

=

( )tτ

Li t 4 12e−

= −

( ) tL τdi t 112e

dt τ− − =

( )Ldi 0 12R 6Rdt 2

= =

( )Ldi 03 1R 0.5 36 2 dt

Ω

= = = = Q

Q.59 (a) ( )Li 0 1A− =

( ) ( ) tτ

L Li t i 0 e−−=

L 5mτ 0.5mR 10

= = =

( ) 2000tLi t e for t 0−∴ = ≥

( ) 2000tLL

diV t L 10edt

−= = −

( )LV 0 10V+ = − Q.60 (d)

CIR V= +

CdVI Cdt

=

CC

dVV RC Vdt

= +

Q.61 (d)

ξ R C=2 L

(For series RLC ckt)

𝛏𝛏 =1

Hence, critically damped Q.62 (a)

( )Li 0 1A− = KVL ⇒ L0.5V 10i V 0− − = L0.5V 10i 0− − =

LL

di 10i 0dt

− − =

( ) 10tLi t Ce−=

Comparing with standard form

( )( )tτ

Li t Ce−

=

1τ 0.1sec10

= =

Q.63 (c) Q.64 (b)

Current across inductor at t 0−= is ( ) ( )L Li 0 0 i 0− += = As t →∞

( )Li 1A∞ =

( ) ( ) ( ) ( )tτ


= ∞ + − ∞

t0.01 L 11 e τ

R 100− = − = =

100t1 e−= −

( ) 100tLL

di (t)V t L 100edt

−= =


4 100tLdv (t)dv(t) 10 edt dt

−= =

dv(0 ) v10,000 sdt

+

= −

( )v 0 100v+= (From figure) Q.65 (c)

Step response, ( ) ( )tRCs t 1 e u(t)

−= −

given Impulse response,

( ) ( )dh t s tdt

=

( ) ( ) ( )t tRC RC 11 e δ t e u(t)

RC− − − = − + −

tRC1 e u(t)

RC−

=

Q.66 (b) Forced response = particular integral

( ) ( )D 3 y t k+ =

( ) 0tky t e k / 3D 3

= =+

Q.67 (a)

( )V 0 10− =c v

The current through the switch is zero since 0A current flows through 2Ω resistor. i(0 ) OA+ =

Q.68 (a)

t 0−= ( ) ( )L Li 0 4A i 0− += =

( ) ( )c cV 0 4v V 0− += =

t 0+=

( ) 12 4 8i 02.5 2.5

+ −= =

3.2A=

KCL at node (A) ( )c3.2 4 i 0+= +

( )ci 0 0.8A+ = −

( )ci 0 0.8A+ =

Q.69 (d)

Steady state is achieved as t→∞ ( )cV 10v∞ = τ RC (10k | |10k) 20μ= = × 5k 20μ 0.1sec= × = Q.70 (a) ( ) ( )c cV 0 0 V 0− += =

( )cV 5v∞ =

( )tτ

cV t 5 5e−

= −

( ) 6τ 20kΩ || 20kΩ 4 10−= ×

3 6 210 4 10 10 4 10 sec− −= × × × = × ( )

tτ

cV t 5 5e−

= −

( ) ( )c 6 25tc

cdV ti t 4 10 125e

dt− − = = ×

25t0.5e mA−=


Q.71 (c)

( ) 50I 0 10A5

+ = =

Q.72 (b)

2 R 1s s 0L LC

+ + =

2n n

1 1ω ω βLC LC

= ⇒ = =

nR 1 R2ξω ξ LCL 2 L

= ⇒ =

2 2αξ 1( β α )β

= < ∴ >

Hence, the system is under damped.

Q.73 (a)

3 6100 10 10 0.1sec−= = × × =τ RC

Q.74 (c)

( ) ( )L L

100i 0 10A i 010

− += = =

( )LV 0 100 20 80v+ = − =

( )LV 0v∞ =

Q.75 (a)

Transfer function,

( ) ( )( ) ( )

I s 1H sV s Z s

= =

1R sL

=+

( ) ω 1001H jω |

80 j60= =+

8161 tan

100−

= −∟

m1i(t) V

100= ×

1100 1A100

= × =

Q.76 (a)

( ) 2 2

V(s) 1 A B CI ss 1 s (s 1) s s s 1

= = = + ++ + +

( ) 2

1 1 1I ss s s 1

= − ++

( ) ( )ti t t 1 e u(t)−= − + Q.77 (a)

Let iL be the inductor current

( ) ( )L L10i 0 2A i 05

− += = =

( )L20i 2A10

∞ = =

( ) ( ) ( ) ( )tτ

L L L Li t i i 0 (i e−

= ∞ + − ∞ 2=

( ) LL

diV t L 0dt

= =

Q.78 (c) ( ) 50tV t 125e−=

( ) 50ti t 5e−=

L 1τR 50

= = RL50

⇒ =

50t

50t

V(t) 125eR 25i(t) 5e

−

−= = =

25L 0.5H50

∴ = =

Q.79 (d)


τ RC=

3 610 20sec3 6×

= × =+

( ) 5i 0 0.5A10

+ = =

Q.80 (c) ( ) ( )L Li 0 1A i 0− += =

( ) ( )V 0 2V V 0− += =

at t 0+=

( )ci 0 0+ =

( )dv 0c 0

dt

+

=

this gives ( )dv 0

0dt

+

=

Q.81 (c)

( ) 10I s1 s

=+

( ) ti t 10e−=

Q.82 (c)

( )L Li (0 ) 1A i 0− += =

( ) ( )3 36 8V 0 4 V 09 3

− += × = =

( ) ( )6 64V 0 V 03

− += =

So the current

( )s1i 0 A3

+ =

Q.83 (a) As t →∞ , circuit would be in steady state

V 30v(by kVL)= Q.84 (a) ( ) ( )

1 1L Li 0 1A i 0− += =

( ) ( )2 2L Li 0 0 i 0− += =

( ) ( )c cV 0 4v V 0− += =

By KVL ⇒

( )112 4 2i 0 A

12 3+ −= =

By KCL ⇒


( ) ( )1 si 0 i 0 1+ ++ =

( )s2 1i 0 1 A3 3

+ = − =

Q.85 (d)

Voltage across capacitor and current across inductor can’t change abruptly.

Q.86 (a) Nodal at V1⇒

1 1V 10 V 1.2 0

2.5 2.5−

+ − =

12V 10 3− =

113V 65v2

= =

KVL in Loop (I) ⇒ ( )oV 0 12.5v+ = Q.87 (a)

( )i 0 OA+ =

( )LV 0 10v+ =

( )Ldi 0

L 10A / sdt

+

⇒ =

KVL⇒

( ) ( )010 10 0

++ − − −

dii

dt

6

1 010 10− =× ∫idt

( ) ( ) ( )2

2 5

di 0 d i 0 i 010 0

dt dt 10

+ + +

−− − − =

(on differentiating)

( )2

22

d i 0100A / s

dt

+

= −

Q.88 (c) The transfer function of series RLC circuit is

Vo(S) R1Vi(S) R sL

SC

=+ +

22

RSSRC LR 1S LC SRC 1 s SL LC

= =+ + + +

Comparing with given transfer function

R 4 L 0.5HL= ⇒ =

1 20 C 0.1FLC

= ⇒ =

Q.89 (b)

( )5seI s

s(s 1)

−

=+

5s 1 1es s 1

− = − +

( ) ( )( t 5)u t 5 e u t 5− −= − − −

( )( t 5)[1 e ]u t 5− −= − −


Q.90 (c) ( ) ti t kte−α=

To find the condition for maximum value di 0dt

=

( )αt αtke kte α 0− −+ − =

1t α=

To check whether maximum or minimum at 1t α=

2

21t α

d idt

=

= negative, hence 1t α= is

condition for maximum value.

Q.91 (b)

( ) 2tI t 10e−= −

( ) 10I ss 2−

=+

( ) ( )1 1

1010S s 2I s I s 1s 2 s

− − += = +

( ) ( ) ( )2 1I s I s I s= +

10(s 1)s 2

− +=

+

( ) ( )210(s 1) 10V s

s 2 (s 1) s 2− + −

= =+ + +

By KVL ⇒

( ) ( )s 210 20V s V s

s 2 s 2−

= − =+ +

( ) 2tsV t 20e−= −

Q.92 (b)

( ) ( )c cV 0 4v V 0− += =

t 0+= Nodal at Vx⇒

xx

V2 V 4 02

− + + − =

x3V 62

=

xV 4v=

( ) xVI 0 2A2

+ = =

Q.93 (a)

Given ( ) 1H ss 1

=+

( ) 1H j1 j

ω =+ ω

Input signal ( )r t cos cos t=

1

1 1H( j ) 451 j 2ω=

ω = = − °+

∟

( ) 1 cos2

c t t4π −

=

Q.94 (c)

( ) ( )c

dv 0i 0 C

dt

++ =

Also, ( ) ( )v 0 50 v 0− += =

( ) ( )c

dv 0i 0 C

dt

++∴ =

-2 A=10 S


( )c

50R 5ki 0+

= = Ω (From figure)

Q.95 (a) τ RC=

1(1| |1).1 0.5sec2

= = =

Q.96 (c) ( ) ( ) ( )v s z s & s=

2

s 3s(s 2)

++

Initial value, ( ) ( )sv 0 lt v s 0+→∞= =

Final value

( ) ( ) ( )s3v v 0 lt sv s4

+→∞∞ = = =

steady state value. Q.97 (a) RV 20 90°= −∟ RV 20 0°= ∟

The angle between V and I is

L1 1

R

V 20θ tan tanV 20

− − = = 45°=

For RL circuit the power factor is lagging in nature. Hence i lags v by45°.

Q.98 (b)

( ) 1z jω R jωLjωc

= + +

( ) 1z jω R j ω.Lωc

= + −

( )0ω 2ω

z jω=

00

1R j 2ω L2ω C

= + −

00

1ω Lω C

=

Q

0ω LR j Ω2

= +

Q.99 (d)

Two different frequencies are operating on the circuit i.e. (ω 100,ω 200)= = simultaneously.

Since LZ =jωL, c1Z

jωC= is frequency

dependent hence super position theorem can be used more conveniently.

Q.100 (c) 2 2

R LI I R 144 256= + = +

400 20A= = Q.101 (a)

I 0= (Given) Hence V1=V2

Nodal at V1 ⇒ 1

1VV 20 0° 0j1

− + =∟

1 220 0° 20V 45° V

1 j 2= = =

−∟

∟

Nodal at V2 ⇒

2 22

V V v 0 2V v v5 5

−= + = ⇒ = ⇒

20 2 45°= ∟ Q.102 (b) For unit impulse input

( ) ( ) 1i 0 i 0L

+ −= +

For all other inputs ( ) ( )i 0 i 0 0+ −= = Q.103 (b) For RC circuit, cz R jx= − For series RLC circuit,


L cz R j(x x )= + − For parallel RLC circuit,

C Lz R j(x x )= + − Hence, it can be RC or RLC circuit.

Q.104 (c) The T-equivalent circuit is given by When 3 & 4 are open circuited

eq p pL L M M L= − + =

pL 4H=

When 3 & 4 are short circuited ( )eq p sL L M || M L M= − + −

( )ps

p

L M M3 L M

L−

= + −

s pM 2 k L L 2= ⇒ =

1k 0.52

⇒ = =

Q.105 (c) For the circuit to behave as an ideal current source,

ABZ = ∞s 1

16 ss 1

16 s

×= ∞

+

2

ss 16

= ∞+

2 2s 16 ω 16 ω 4rad / sec= − ⇒ − = − ⇒ =

Q.106 (a) Since, L cx x 20Ω= = the circuit is at resonance.

cV Qv 90°= −∟

cL xx 20Q 2R R 10

= = = =

cV 200 90°= −∟

Q.107 (c)

2 22 R CI I I= +

2R10 I 64= +

RI 6A=

( )221 R L CI I I I= + −

LI 16A=

1 L C1

R

I Iθ tanI

− −=

1 8tan6

− =

1 C2

R

Iθ tanI

− =

1 8tan6

− =

2I will lead by 1 8tan6

−

, I1will lag

by 1 8tan6

−

Q.108 (b) Power Factor = 0.6

Q.109 (c) For transient free response

1o

ωLωt tanR

− =

Q.110 (c)


( )i 0 0− =

( )i 0 0+ =

( )i 0∞ =

Q.111 (a) For parallel circuit Q ωCR= Q.112 (b) 1 2M k L L=

3

61 2

M 16 10 16k 0.440L L 20 80 10

−

−

×= = = =

× × Q.113 (c) Q.114 (a)

At resonant frequency L & C will act as short circuit

L1z R || jωL R

jωc

= + +

L

1R jωLjωc

z R1R jωLjωc

+

= ++ +

at resonance , j- term =0

01ω rad / secLC

=

Q.115 (c)

Unity power factor exhibited by the circuit at resonance.

1 1Y 1jωL Rjωc

= ++

22 2

jRj ωc1ωL R

ω c

+−= +

+

j-term = 0 at resonance

0 2 2

1ωLC R c

∴ =−

Q.116 (c) Z 20 j20= + (From given condition)

L CZ R j(x x )= + − (For serial RLC circuit)

L CV 2V= (Given)

L C L CZ 2Z x 2x⇒ = ⇒ = L Cx x 20− = (From (1)) C C2x x 20− =

Cx 20Ω= Lx 40Ω= Q.117 (c)

2 2

R CI I I= +

2 2R CI I I= −

2 25 4 3= − =

R

V 240R 80ΩI 3

= = =

Q.118 (b)

c L

1 1Yjx R jx

= +− +

L2 2

c L

R jxjx R x

−= +

+

At resonance, j-term =0

L2 2

c L

x1x R x

=+

2 2L c Lx x x R 0− + =


Damping factor, ξ 1< − for two

resonance

cc

x 1 x 2R2R

− < − ⇒ > cx 10Ω⇒ >

Q.119 (c)

Voltage across capacitor At resonance; cV QV 90°= ∠−

ωL 50 at 100Hz= 1L Hz

4π⇒ =

o

12π 500ω L 4πQ 25R 10

× ×= = =

cV 2500 90°= ∠−

Q.120 (b) Q.121 (d)

For series RLC, at resonant frequency 1. Current is maximum 2. Equivalent impedance is R

(which is purely real) 3. L cx x=

4. 1 LQR C

=

Q.122 (a)

1 1Y 1R jωL Rjωc

= ++ +

2 2 22

2 2

jRR jωL ωc1R ω L R

ω c

+−= = +

+ +

At ω=ω0 j-term =0

o 02 2 2

202 20

1ω L ω C

1R ω L Rω c

=+ +

2o

o

L 1R ω L 0C ω C

− − =

2 L LR RC C

⇒ = ⇒ =

Q.123 (a)

For LC circuit the voltage across inductor or capacitor is sinusoidal with constant amplitude.

Q.124 (a)

For series R-L-C circuit under resonance

1. Current is maximum 2. VL =VC

3. Power factor =cosϕ=1 Q.125 (d)

BW of series or parallel R-L-C circuit is given by

ofBWQ

=

o1fLC

=

1 LQR C

= for series RLC

CQ RL

= for parallel RLC

Hence Bandwidth depends on R, L and C.

Q.126 (d) Q.127 (b)

Let inductance be L and Self capacitance be C pF.

3

12

1 500 102π L(C 36) 10−

= ×+ ×

3

12

1 250 102π L(C 160) 10−

= ×+ ×


Solving above two equations we get C = 5.33 pF Q.128 (d)

Power in option a, b, and c are active powers where as P = VI sinϕ is reactive power.

Q.129 (d)

6 12

1 1ωLC 400 10 100 10− −

= =× × ×

5Mrad / sec= Q.130 (a) ⇒ ABCD parameter equations are ___ 1 2 2 1V AV BI V= − ⇒ 2 21V 2I= − ______ (1) 1 2 2 1I CV DI I= − ⇒ 2 21V 3I= − _____ (2)

And 2

1

1 I 0

VZinI

=

=

And 2 2V 10I= − ( )( )

2 2in

2 2

10I 2I 10 2 12z10I 3I 10 3 13

− − − −∴ = = =

− − − −

in12z13

=

Q.131 (b)

18 77 28

⇒ If two networks are connected in series the impedance matrix of the resulting two port network will be the addition of the two impedance matrix.

Q.132 (b) BC AD 1− = − ⇒ Conditions for reciprocity ----- Z parameter 21 12Z Z→ = Y parameter 21 12Y Y→ = ABCD parameter BC AD→ − 1= − h parameter 12 21h h→ = −

Q.133 (b)

1 100 1

⇒Given 1 2 2

1 2 2

V AV BI(1)

I CV DI= +

= +

And from given n/ω

1 2 2

1 2

V V 10I(2)

I I= +

=

After comparing (1) & (2) we get

A B 1 10C D 0 1

=

Q.134 (c)

n 00 1/ n

We have 1 2 2V AV BI= − (1) 1 2 2I CV DI= − (2)

1 2

2 1

V IV I 1

∴ = − =n

So, 1 2V nV= (3)

And 1 21I In

= − (4)

After comparing eqn (1), (2), (3) & (4)

We get A B n 0C D 0 1/ n

=

Q.135 (a)

2 112

1 2

Z ZhZ Z

−=

+

⇒ In symmetrical lattice network

( )11 1 2 221Z Z Z Z2

= + =

( )12 21 2 11Z Z Z Z2

= = −

( )

( )

2 112

1222

1 2

1 Z ZZ 2h 1Z Z Z2

−= =

+

2 112

1 2

Z ZhZ Z

−∴ =

+


Q.136 (c) Y parameter

⇒ Combination of two port networks – i) Series connection → z parameter ii) Parallel connection → Y

parameter iii) series–parallel connection → h

parameter iv) Parallel–series connection → g

parameter v) cascade connection → ABCD

parameter Q.137 (a)

3 10

2.2 12.4

⇒Transmission matrix of the combination – A BT T .T=

1 2 2 4

T0.1 4 0.5 3

=

3 10

T2.2 12.4

=

Q.138 (c)

1 ZY 1 yz +

1 2 2V AV BI⇒ = − 1 2 2I CV DI= − Making O/P port open –circuited i.e. 2I 0= 1 2V V A 1= ⇒ = 1 2I yV c y= ⇒ =

Now making output port short circuited

i.e. 2V 0=

2 1

1/ yI .I1 zy

− =+

2 11I I

1 yz− =

+

D 1 yz⇒ = + 1 2V zI= − B z⇒ =

So, 1 z

Ty 1 yz

= +

Q.139 (c)

1

112

2 I 0

VhV

=

=

3 28I I

8 8=

+

23

II2

⇒ = 24 2 3 2

II I I I2

⇒ = − = −

24

II2

⇒ =

1 3 2V 4I 2I= = ----- (1) 2 2 4V 4I 8I= + 2 2 2V 4I 4I= + 2 2V 8I= ------ (2)

1 2

2 2

V 2I 1 0.250V 8I 4

= = =

Q.140 (a) 1 2 3Y 4 ,Y 0 ,Y 1 ,= = =J J J ⇒ We have Y parameter eqn 1 11 1 12 2I Y V Y V= + 2 21 1 22 2I Y V Y V= + Comparing above eqn –


11Y 5= 12Y 1= − 21Y 1= − 22Y 1= For π network, 11 1 3Y Y Y= + 22 2 3Y Y Y= + 12 21 3Y Y Y= = − So,

1Y 4= J 2Y 0= J 3Y 1= J Q.141 (d)

1, 2, and 3 Condition for reciprocity –

z parameter 12 21z z⇒ = y parameter 12 21y y⇒ = h parameter 12 21h h⇒ = − Q.142 (c)

( )'11

4Z s 10s

= +

⇒ Open ckt impedance

( )'11

1Z s 10 1 s4

= +

( )'11

4Z s 10s

= +

Q.143 (b) 21Z 1/ 2Ω= − ⇒ This is symmetric lattice network.

( )11 221 3Z Z 2 1 Ω2 2

= = + =

( )12 211 1Z Z 1 2 Ω2 2

−= = − =

Q.144 (a) The circuit is reciprocal but not symmetrical. ⇒ From given circuit, 1 1 1 1 2V 3I RI R(I I )= − + − +

1 1 2V (2R 3)I RI∴ = − + ------ (1) 2 2 1 2V RI R(I I )= + + 2 1 2V RI 2RI∴ = + ------ (2) We have, z parameter eqn - 1 11 1 12 2V Z I Z I= + 2 21 1 22 2V Z I Z I= +

Comparing with eqn (1) & (2) We get

11Z (2R 3)= − 12Z R= 21Z R= 22Z 2R=

Condition for reciprocity 12 21Z Z⇒ =

Condition for symmetry 11 22Z Z⇒ = Therefore, the circuit is reciprocal but not symmetrical.

Q.145 (c) 1 & 4

⇒ from given circuit ____ By KVL in mesh (1)

1 1 1 1 2V 2I I 1(I I )= + + + 1 2V I= ----- (1) By KVL in mesh (2)

2 2 2 1V 1I 1(I I )= + + 2 1 2V I 2I= + ----- (2) We have z parameter 1 11 1 12 2V Z I Z I= + 2 21 1 22 2V Z I Z I= +

Comparing with eqn (1) & (2) We get

11 12Z 0, Z 1= = 21 22Z 1.Z 2= =

Condition for reciprocity 12 21Z Z⇒ = 1=

⇒ and 11 22Z 0, Z 2= = Q.146 (c) 11 22Z =Z

⇒ Condition of symmetry for various parameters are given below-


z parameter 11 22Z Z⇒ = y parameter 11 22y y⇒ = h parameter h 1⇒∆ = ABCD parameter A D⇒ = Q.147 (c)

A and D are dimensionless. ⇒ Equation for transmission parameter are 1 2 2V AV BI= −

1 2 2I CV DI= − It is clear that A and D are dimensionless. Q.148 (a) 3,-2

22 2211

11 22 12 21

Z ZYz Z Z Z Z

⇒ = =∆ −

3 / 3513 3 2 2.35 35 35 35

=−

11Y 3=

And 12 1212

11 22 12 21

Z ZYz Z Z Z Z

− −= =

∆ −

2 / 3513 3 2 235 35 35 35

=−

12Y 2= − Q.149 (a)

1 z0 1

1 2I I= − ----- (1) 1 2 1V V I Z− = 2I= −z 1 2 2V V I⇒ = − z We have ABCD parameter___ 1 2 2V AV BI= − 1 2 2I CV DI= −

Comparing with equation (1)7(2) We get

A B 1 zC D 0 1

=

Q.150 (b) All initial condition are zero. Q.151 (d) All of the above

⇒ Condition for reciprocity & symmetry.

Z parameter 12 21 11 22Z Z & Z Z= = Y parameter 12 21 11 22Y Y & y y= = ABCD parameter AD-BC 1 & A D= = Q.152 (d) A B C D ⇒ 3 2 4 1

Network Parameter Measure under open circuit conditions

a) 11Z2

1

1 I 0

VI

=

b) A2

1

2 I 0

VV

=

c) C2

1

2 I 0

IV

=

d) 22Z1

2

2 I 0

VI

=

Q.153 (b) Cascade connection

a a b b

a a b b

A B A BA BC D C DC D

⇒ =

is valid for cascade connection i.e.

Q.154 (a)

A and D remain unchanged. C is halved and B is doubled

⇒ Relation between ABCD & z – parameter of the two port network

11

21

ZAZ

=

2211Z12

21

ZB z

Z= −


21

1CZ

=,

22

21

ZDZ

=

Q.155 (a)

Q.156 (d) -5 ⇒ We have z parameter eqn 1 11 1 12 2V Z I Z I= + 2 21 1 22 2V Z I Z I= +

Open circuit output terminal

2

111

1 I 0

VZI

=

=

With I2 = 0 , draw circuit Apply KVL in loop (1) & (2)

1 1 1 2V 5I 5I 5I= + +

1 1 2V 10I 5I= + (1) And ( ) 2 1 z10 5 I 5I 4v+ + =

( )2 1 1 215I 5I 4 5 I I⇒ + = +

z 1 2v I I∴ = + 2 1I 3I⇒ = − Net value ofI2 in eqn (1) 1 1 1V 10I 15I= − 1 1V 5I= −

2

1

1 I 0

VI

=

⇒ 11Z 5⇒ = −

Q.157 (c) 3 & 4 only

⇒ Given equation can be rewritten as

1 1 22V 4I 8I= + 1 1 2V 2I 4I= + ______ (1)

And 2 1 2V 8I 16I= + ______ (2) Comparing these equations with standard z parameter equation we get –

11 12Z 2Ω, Z 4Ω= = 21 22Z 8Ω, Z 16Ω= = 12 21Z Z∴ ≠ Given network is not a reciprocal network. Q.158 (b) ⇒ From given figure – 3

2 2V I 10 10= − × × For two port network we have 2 21 1 22 2V Z I Z I= +

Putting values of V2, Z21&Z22 in above equation we get,

3 6 42 1 2I 10 10 10 I 10 I− × × = − +

( )4 4 62 1I 10 10 10 I⇒ − + = −

62

41

I 10 100 50I 2 10 2

⇒ = = =×

Q.159 (a)

1 mho3

−

⇒ Short circuit input port. Applying KCL at node V

2V VV V 01 1 1

−+ + =

23V V⇒ = _______ (1)

Also 1V I1= −

1v I⇒ = − Putting value in equation (1) ( )1 23 I V− =

1

112

2 V 0

I 1Y mhoV 3

=

⇒ = = −

Q.160 (c)


11 2211 22

1 1h & hy z

= =

⇒ We have

2 2

1 111 11

1 1v 0 v 0

V Ih & yI V

= =

= =

1 1

2 222 22

2 2I 0 I 0

I vh & zv I

= =

= =

Q.161 (b)

4

4

3 jj 3

− −

11 22 A B1z z (Z Z )2

= = +

4 41 (3 j 3 j )2

= + + − ( )1 62

=

11 22z z 3= =

12 21 B A1z z (Z Z )2

= = −

4 41 (3 j 3 j )2

= − − − 41 ( 2 j )2

= −

12 21 4z z j= = −

11 12 4

21 22 4

z z 3 jz z j 3

− ∴ = −

Q.162 (a)

0.7 0.50.5 0.8

− −

11 A By y y 0.2 0.5 0.7= + = + = 12 21 By y y 0.5= = − = − 22 B Cy y y 0.5 0.3 0.8= + = + =

Q.163 (d) A is false but R is true

⇒ (i) In a linear directed graph, a link forms a closed loop. (ii) The fundamental loop of a linear directed graph contains only one link and a number of twigs.

Q.164 (b)

n(n 1)2−

⇒ Number of edges 2C

n(n 1)n2−

= =

Q.165 (d) (1, 3, 4, 5) is cut-set of graph.

Q.166 (d) Node pair voltage

2 2C C

10 9n 10 452 1×

= = =×

Q.167 (b) Series R-L circuit ⇒ In dual circuit, the elements are replaced as R G↔ L C↔ V I↔ Series Parallel↔

Node mesh↔ Q.168 (c) Loop branch – node 1⇒ = + b n 1= − + 11 6 1= − + 6= Q.169 (d)

There are at least two edges in a circuit.

Q.170 (c)


⇒A tree is a connected sub graph of a connected graph containing all the nodes of the graph but containing no loops.

Q.171 (d) ⇒ Total number of trees n 2n −= 4 24 −= = 16 Q.172 (d)

⇒ A tree never contains a loop. Q.173 (d)

⇒ A tree is a connected sub graph of a connected graph containing all he nodes of the graph but containing no loops.

Q.174 (b) 6 L b n 1⇒ = − + 3 b 4 1= − + b 6=

Q.175 (d) 1, 2 and 3. Q.176 (d) e-n+1 Q.177 (a) All conductance. Q.178 (c)

1 1 0

0 1 11 0 1

− − −

Q.179 (a)

Same as number of twigs. ⇒ Number of fundamental cut-sets of any graph will be same as number of twigs.

Q.180 (d)

L b n 1⇒ = − + ( )Branch b 12,node(n) 5= = Minimum number of mesh eqn L 12 5 1= − + L = 8 Q.181 (d)

⇒The maximum number of tree is (n 2)n − Where, n is node =5

(n 2) (5 2) 3n 5 5 125− −∴ = = = Q.182 (d) ⇒ No. of node =8

No. of fundamental cut-sets = (n −1)

(8 1) 7= − = Q.183 (a) ⇒ Twigs 4,5,6,7→ Links 1,2,3,8→ Fundamental cut-set 1,2,3,4→ Fundamental loop 6,7,8→ Q.184 (c)

For LC circuit, the poles and zeroes must be alternate and lie on imaginary axis.

Q.185 (a)

The function 3s 2s

+ + can be realized

as driving point impedance or as driving point admittance function. If it is impedance function then

1R 2Ω,L 1H,C F3

= = =

If it is admittance function then


1G 2 ,L H,C 1F3

= = =J

Q.186 (b) The number of elements required

to realize a given driving point susceptance function is equal to total number of poles and zeros.

Q.187 (c)

Admittance, ( ) 1Y s s 1s 1

= + ++

2 21 (s 1) s 2s 2s 1 s 1

+ + + += =

+ +

( ) 2

1 s 1zy(s) s 2s 2

s += =

+ +

Q.188 (c) For RC driving point impedance function a pole must be near to the imaginary axis and for RC driving point admittance function a zero must be near to the imaginary axis.

1. ( )2s 3Y Z ss 2+

= =+

( )( )( )

2

1 3

s 1 s 3Y Y(s)

s 2+ +

⇒ = =+

is not possible. 2. ( ) ( )2 1

s 2 s 3Y Z s Y Y ss 1 s 2+ +

= = ⇒ = =+ +

is not possible.

3. ( ) ( )( )2s 2Y Z s

s 1 s 3+

= =+ +

( )11Y Y s

s 2⇒ = =

+is possible.

4. ( ) ( )2 1Y Z s s 2 Y Y s= = + ⇒ =

( )s 1 (s 3)(s 2)+ +

=+

is possible.

Q.189 (c) For

( ) ( )( )( )

2 6 2 6

2 2 6

ω 4 10 ω 36 10z jω j(0.1ω)

ω ω 16 10

− × − ×=

− ×

( )z j1000 j700= −

Q.190 (b) Since poles and zeroes of z(s) are purely imaginary and are alternate hence, the function is an LC circuit.

Q.191 (d) For RC driving point impedance function poles and zeroes are alternate and must lie on negative real axis, nearest to imaginary axis is pole. This is true for

( ) ( )( )s 2 (s 1)

z ss 1 (s 3)+ +

=+ +

Q.192 (d) ( )L sin ωt α+

L[sin ωtcosα cos ωt sinα]= + ]

2 2 2 2

ωcos α s sinαs ω s ω

= ++ +

2 2

s sinα ωcos αs ω++

Q.193 (d) For RC driving point impedance

( ) ( )RC RCz 0 z> ∞

e.g. ( )RC1z s R

SC= +

( ) ( )RC RCz 0 z R= ∞ ∞ =


network theory - gateflix.in · passive network elements are those which are capable only of...

Documents