Network Theory

NETWORK THEORY

For ELECTRICAL ENGINEERING INSTRUMENTATION ENGINEERING ELECTRONICS & COMMUNICATION ENGINEERING

NETWORK THEORY 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 1 Mark 2 Mark 1 Mark 2 Mark 1 Mark 2 Mark Exam Year Ques. Ques. Total Ques. Ques. Total Ques. 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

1. NETWORK BASICS

1.1 Introduction 1.2 Classifications of Network Elements 1.3 Circuit Components 1 1.4 Kirchoff's Laws 2 1.5 Mesh & Nodal Analysis 3 1.6 Equivalent Circuits 6 Gate Questions 137 9 2. NETWORK THEOREMS

2.1 Introduction 2.2 Superposition Theorem 2.3 Thevenin's & Norton's Theorem 49 2.4 Maximum Power Transfer Theorem 49 2.5 Tellegen's Theorem 50 2.6 Reciprocity Theorem 51 2.7 Substitution Theorem 53 2.8 Millman's Theorem 53 2.9 Duality Principle 54 Gate Questions 54 54 3. TRANSIENTS 58

3.1 Introduction 3.2 Steady State & Transient Response 3.3 DC Transients 89 Gate Questions 89 91 4. AC ANALYSIS 99

4.1 Introduction 4.2 Sinusoidal Steady state analysis 4.3 Series Circuits 129 4.4 Parallel Circuits 129 Gate Questions 13013 132 5. RESONANCE 6

5.1 Introduction

149

6.1 Graph of a Network 6.2 Incidence Matrix 6.3 Tie-Set Matrix 157 6.4 F-Cut Set Matrix 159 Gate Questions 159 163 7. COUPLED CIRCUITS 164

7.1 Introduction 7.2 Mutual Inductance 7.3 The Coupling Coefficient 174 7.4 Series Connection of Coupled Inductors 174 7.5 Parallel Connection of Coupled Coils 176 7.6 Ideal Transformer 176 177 8. TWO PORT NETWORKS 177

8.1 Introduction 9 8.2 Open Circuit Impedance (Z) Parameters 9 8.3 Short Circuit Admittance (Y) Parameters 17 8.4 Transmission (A B C D) Parameters 170 8.5 Inverse Transmission (A' B' C' D') Parameters 180 8.6 Hybrid (H) Parameters 18 8.7 Inverse Hybrid (G) Parameters 181 8.8 Conditions for Reciprocity and Symmetry 181 8.9 Inter Relationships of Different Parameters 182 8.10 Interconnection of Two-Port Network 182 Gate Questions 1182 184 9. NETWORK SYNTHESIS 89

9.1 LC, RC, RL Impedance & Admittance Functions Gate Questions 211 10. ASSIGNMENT 212

217

1.1 INTRODUCTION Q I = (ampere) t In terms of the atomic theory concept, an Where, I is the current electric current in an element is the time Q is the charge rate of flow of free electrons in the element. T is the time The material may be classified as The current through a circuit element is the • Conductors, where availability of free time derivative of the electric charge i.e. electrons is very large, as in the case dq i = (c/s) or (Ampere) metals. dt • Insulators, where the availability of Where, dq is small change in charge. free electrons is rare, as in case of glass, dt is small change in time. mica, plastics etc. • Other materials, such as germanium 1.1.3 VOLTAGE and silicon called semiconductors, may play a significant role in All opposite charges possess a certain electronics. Thermally generated amount of potential energy because of the electrons are available as free electrons separation between them. The difference is at room temperature, and act as potential energy of the charges is called the conductors, but at 0 Kelvin they act as potential difference. The potential insulator. difference in electrical terminology is known as voltage, and is denoted either by 1.1.1 CHARGE ‘V’ or ‘v’. Voltage is expressed in terms of energy (W) per unit charge (Q) i.e. According to basic physics, we know that W dw there are two types of charges: Positive V = Or ν = Q dq (corresponding to proton) and Negative

(corresponding to electron). The (J/c) or (volt) fundamental unit of charge is the Where, dw is the small change in energy coulomb[c]. A single electron has a charge dq is the small change in charge. − One Volt, is the potential difference of −×1.602 10 19 C and a single proton −19 between two points when one joule of has a charge of +×1.602 10 C where one energy is used to pass one coulomb of coulomb is defined as one ampere second. charge from one point to the other Charge in coulomb Q = It where, I is current in ampere and t is time in second. 1.1.4 ENERGY

1.1.2 CURRENT Energy is capacity for doing work. Energy may exist in many forms such as The phenomenon of transferring charge mechanical, chemical; electrical is called from one point is a circuit to another is ‘Joule’. Energy is denoted by ‘W’. The termed as “electric current”. The eclectic energy delivered to a circuit element over current, denoted by either ‘I’ or ‘i’ .The the time interval (to, t) is given by unit of electric current is ‘ampere’ which t is denoted by ‘A’. Electric current W=pxdx∫ () mathematically expressed as t0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1.1.5 POWER time interval. For example, passive elements like inductors and capacitor Power is the rate of change of energy. It is capable of storing a finite amount of denoted by ‘P’ or ‘p’, unit of power is energy. Resistor is also a passive ‘Watts’. element. Energy W Power() p = = time t 2) BILATERAL AND UNILATERAL dw In Bilateral elements, the voltage– or P = dt current relation is same for the current Where, dw is the change in energy flowing in either direction. For the dt is the change in time entire time‘t’ element offers the same dw impedance for the different directions we can also write, P = of the same current flow and hence the dt resistor is said to bilateral. dw dq = × dt dt P = v × iwatts So, 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 has different P = v() t × i(t) A unilateral element relations between voltage and current for the two possible directions of 1.2 CLASSIFICATION OF NETWORK current. Vacuum diodes, silicon diodes ELEMENTS and metal rectifiers are examples of unilateral elements.

3) LINEAR AND NONLINEAR ELEMENTS An element is said to be linear (device is linear if it is characterized by an equation of the form y=mx, where m is constant) if its voltage-current characteristics is at all times a straight line through the origin. For example, 1) ACTIVE AND PASSIVE the current passive through a resistor is Active network elements are those proportional to the voltage applied which are capable of delivering power through it, and relation is expressed as to some external device. Specifically an V∝= IORV IR active element (energy sources like A linear element or network is one voltage and current sources) is capable which satisfies the principle of of delivering an average power greater superposition, i.e. the principle of than zero to some external device over homogeneity and additivity. An element an infinite time interval. For example, which does not satisfy the above ideal sources are active elements. principle is called a non linear element. Passive network elements are those which are capable only of receiving 4) LUMPED AND DISTRIBUTED power. A passive element is defined as Lumped elements are those elements one that cannot supply average power which are very small in size and in that is greater than zero over an infinite which simultaneous actions takes place

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission for any given cause at the same instant dq ∴=vR of time. For example, capacitors, dt resistors, inductors and transformers V ⇒=i are lumped elements. R Distributed elements are those which ⇒=i Gv are not electrically separable fro • Electric power , analytical purpose. For example, a p = vi transmission line which has distributed 2 resistance, inductance and capacitance =Ri.i = i R(watts) along its length may extend for (v= iR) hundreds of miles. v Also, p= v. R 1.3 CIRCUIT COMPONENTS  V  Q i = R 1. THE RESISTANCE (R) V2 The property of a material to restrict = (Watts) the flow electron is called resistance. R • Electrical energy , Note: Current in a Resistor always enters W= pdt from the positive terminal. ∫ ⇒ W=∫ i2 Rdt (J) (Q p=i2 R) V2 ⇒ W= dt (J) ∫ R  V2  Q p = The basic ohm’s law in the electromagnetic R rr  theory form can written as JE= σ The V-I characteristic of resistance: Where, iA J =The current density, J =  S m2 E = Electric field along the length of vV conductor, E =  l m σ = Conductivity of the conductor. iv l ⇒=σ ⇒=v i Observation: From the characteristics, it ssl σ can be observed that the resistor is a linear, ⇒=v Ri →Ohm’s law in circuit theory passive, bilateral and time invariant in v-i form. plane. • Resistor is linear because the v-i Limitation: characteristic is a straight line passing The Ohm’s law i.e. a linear voltage and through origin. current relation is valid only when the • Resistor is passive because the slope of proportionality constant R is kept constant v-i characteristics is positive. i.e. temperature is kept constant. →=v Ri • Resistor is bilateral because even if we →=v Ri change the direction of flow of current the R value is same.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission • For all time ‘t’ the v-i characteristics has 1 2 one slope which implies that its value is w= Li () J 2 same and hence it is said to be time invariant, which is possible only at : constant temperature Conclusions • If I = constant then d 1) THE INDUCTANCE (L) V= L() constant = 0 A wire of certain length, when twisted dt in the form of a coil becomes a basic i.e, inductor acts as short circuit to dc. inductor. • An ideal inductor never dissipates By Faraday’s law of electromagnetic energy, only story it is the form induction, a time varying current magnetic field. through coil produces a time varying Note: magnetic flux, which induces voltage Inductor is also a linear, passive, bilateral across the coil. and time invariant.

2) THE CAPACITANCE (C) Any two conducting surfaces separated by an insulting medium exhibit the property of a capacitor. A capacitor stores energy in the form of on electric field.

Note: -In the presence of source V the current in an inductor enters from positive terminal. The current voltage relation is given by di vL= dt 1 t ⇒=i Vdt qv∝ L ∫ −∞ q=cv The current-voltage relation is given by dv ic= dt t Electric power, p= Vi 1 ⇒=v i dt di c ∫ = Li (watts) −∞ dt Electric power, p= Vi Electric energy, w= pdt dv ∫ = cv (watts) dt di w= Li dt (J) Electric energy, w= pdt ∫ dt ∫ dv di d  1 2  = cv dt Also, Li=  Li  ∫ dt dt dt  2  dv d 1 2 d12 Again, v= ( cv ) ∴=w Li dt dt dt 2 ∫ dt 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission d12 a) Independent voltage source: ∴=w cv dt ∫ dt 2 Ideal: An ideal voltage source is a two – 1 w= cv2 (J) terminal element which supplies a 2 constant voltage to a load and is Conclusions: independent of the load connected to it. d • If V = constant then i = c() constant =0 dt 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 VL = VS for all value of is and time invariant. The load can be a resistor of any value. The relation between V and i in L and C The V-I characteristics of voltage source elements is nonlinear hence it is a non-linear di element. Also the ideal voltage source is L: VL= active and unilateral. dt In the above circuits we have connected If i→→ V ,i V Then 1 12 2 2 and 5 resistors to a 10v source and i12+→ i V x (let) we observe that in both the cases 10v di wouldΩ beΩ dropped across them, VL= 1 1 dt irrespective of the current. In the first circuit 5A current flows through 2 di2 VL2 = load whereas in second circuit 2A dt current flows through 5 load. Ω ddii 12 VLx = + L dt dt Practical: Ω Vx= VV 12 + A practical voltage source has a Since, linearity property is satisfied for resistance in series with ideal voltage inductor hence the v-i relation is linear for source. In a practical voltage source the inductor. load voltage changes with a change in dv load resistance. Ideal voltage source C: ic= doesn’t exist in reality. Every source has dt Following the similar steps it can be proved internal resistance. that the v-i relationship is linear for capacitor also.

3) THE SOURCES

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission By KVL⇒ VL= V S − iR SS Observation: Here, load voltage depends upon load When VL= 0theni LS = i

current iS .

b) Independent current source: Ideal:- An ideal current source is a two terminal element which supplies a constant current to a load resistor of any value. V= 0 ⇒ short circuit Again ideal current source doesn’t exist L So, we can conclude that the current always is reality as every current source has an chooses a minimum resistance path. internal impedance. c) The dependent (or) controlled sources:

Practical: A practical current source has a resistance in parallel to the ideal current source. Here • Dependent sources are the ideal the load current depends on the value of sources in which the value of source is load connected across it. determined by a voltage or current existing at some other location in the network. • Dependent sources are said to be linear, active and a bilateral with respect to controlled variable only. The presence of these elements makes the network linear, active and bilateral. • Dependent sources are said to be sources that is an active element in the presence of at least one independent source then only the controlled variable will be non-zero and the magnitude of source will be non-zero.

1.4 KIRCHOFF’S LAWS VL KCL⇒iILS= − RS 1) KCL Here load current depends upon load It is defined at a node voltage.

The simple node is an interconnection W WWRRWR ⇒=s 12 + +3 of only two branches, whereas principle qq q q node is an interconnection of at least 3 ⇒=WWsR + W R + W R branches. 123 So the KVL expresses conservation of Def: In an electric circuit, for any of its nodes at any time‘t’ the algebraic sum of energy. branch currents leaving the node is zero. 1.5 MESH AND NODAL ANALYSIS B y KCL 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 analysis leaving curent= 0 ∑ Nodal analysis is used to find node −−iiiii012345 + + + = voltage is a circuit. Nodal analysis is a combination of KCL and ohm’s law i.e. ⇒+ii12 =++ iii 345 i.e. sum of entering current=sum of Nodal analysis = KCL + ohm’s law leaving currents When we talk about the voltage at a dq certain point of a circuit we imply that Also, i = ⇒ dt measurement is performed between that point and some other point in the dq12 dqdq35 dq 4 dq +=++ circuit. In most cases that other point is dt dt dt dt dt referred to as ground. qq+=++ qqq 12 345 Steps: Conservation of charges i) Identify the number of principle nodes (having more than two 2) KVL branches). Def: It is defined in a loop or mesh i.e. in ii) Assign the node voltages w. r. t. the a closed path. In an electric circuit for ground node, whose voltage is any of its loop at any time ‘t’ the always equal to zero. algebraic sum of branch voltages iii) By using KCL at every principle around the loop is zero node 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.

KVL ⇒

VVVVS123−−− = 0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 Considering node 1: clockwise direction. iii) By using KVL write the equations.

Example:

By KCL at node 1:

−+III123 + = 0{By KCL} V0VV−− As shown in circuit, I and I are current in −+I01 + 12 ={using ohm' slaw} -- 1 2 1 loop (1) and loop (2 ). RR12 ---- (1) Where, V &V are voltages at node 1 & 2 Considering Loop 1 12 By KVL respectively. V1− IR 11 −−() I 1 I 2 R 2 = 0 (1) Considering node 2 Considering Loop 2 By KVL

−−V2() I 2 − I 1 R 2 − IR 23 = 0 (2) By solving eqn (1) & (2) we can find the

currents I1 & I2 .

3) Super node Analysis By KCL at node 2 : When there is an ideal voltage source between two principle nodes, it is I456++= I I 0{By KCL} difficult to apply the technique of nodal VVV0−−− V0 21++ 2 2 =0 analysis because the current across an R2 R 3 RR 45+ ideal voltage source is unknown. {∴ u sin g ohm' slaw} ……… (2) Therefore to overcome such situation By solving (1) and (2) we can find voltages we use super node analysis technique. at each node. Example: 2) Mesh analysis Mesh analysis is used to find branch current in a circuit. Mesh analysis is combination of KVL and ohm’s law. Mesh analysis = KVL + ohm’s Law

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The current through an ideal voltage Super mesh eqn in Loop (1) & (2) is given source can be any value; it is not possible to by write the nodal equation independently; V−− IR IR = 0------(1) hence the super node procedure is 11 2 3 followed here. Inside the super mesh KCL is written. To apply super node technique write nodal II=12 − I------(2) n equations at node V12 &V simultaneously, By solving eq (1) & (2) we can find value while doing so do not consider the branch of I12 &I . containing ideal voltage source. Super node n is given by eq atV12 & V 1.6 EQUIVALENT CIRCUITS VV −+112 + −= 10 11 Two elements are said to be in series only when the currents through the elements VV12+= 2 (1) are the same and they are said to be in Inside the super node always KVL is parallel only when the voltage across the written elements are same. By KVL Impedances in series and admittances in −− = V12 1V 0 parallel can be added. VV12−= 1 (2) n 1) Impedances in series By solving eq (1) & (2) V1 = 1.5V and 1 VV= Zeq = Z1 + Z2 2 2 ZR= Ω 4) Super mesh Analysis R When there is an ideal current source ZL = jLωΩ between two loops, it is difficult to 1 Z = Ω apply mesh analysis because the voltage C jCω across an ideal current source is R: R= RR + unknown. Therefore to overcome such eq 1 2 situation we use super mesh analysis L: Leq= LL 1 + 2 technique. 1 11 C: = + jcω jc ωω jc Example: eq 1 2 1 11 ⇒=+ ceq cc 1 2

If ccc12= = c Then c = eq 2 The voltage across an ideal current source The voltage division principle 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 eqn in loop (1) & (2) and do not consider the branch containing ideal current source.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VV ZZ12 V= ZIeq ⇒=I = = I Z ZZ+  eq 1 2 ZZ12+ V V Z →== V ZI Z  →=II = 2 11 1 + 1  ZZ12 Z1 ZZ 12+ V V Z →== V ZI Z  →=II = 2 22 2 + 2  ZZ12 Z2 ZZ 12+

2) Impedance in parallel 3) The Star Delta Transformation

Yeq= YY 1 + 2 1 11 = + Zeq ZZ 1 2 4) ∆ to Y Conversion:

If ∆−n/w is given then Z12 Z and Z 3 are known ZRR = Ω ZZ13 ZL = jwLΩ Z = , A ZZZ++ 1 123 ZC = Ω ZZ jwC Z = 12 , B ZZZ++ 1 11 123 R: = + ZZ R RR Z = 23 eq 1 2 C ZZZ++ 1 11 123 L: =+ If ZZZZ= = = L LL 123 eq 1 2 Z Then = = = C: = + ZZZABC Ceq CC 1 2 3 Current Division Principle Y to ∆ Conversion:

If Y− n/w is given then ZAB , Z and Z C are known.

ZAB ,Z ZZ1AB=++ Z , ZC Z ,Z ZZZ=++BC, 2 BC Z 1 11 A We have, = + Z ,Z =++AC Zeq ZZ 1 2 ZZZ3AC ZB ZZ12 ∴=Zeq If ZZZZABC= = = ZZ12+ Then Z123= Z = Z = 3Z →=V ZIeq (By ohm’s law)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 5) The source Transformation So, we can redraw the circuit as follows- By Any practical voltage source can be voltage division rule, converted into its equivalent practical current source and vice-versa by using source transformation. It is a simplification technique which eliminates the extra nodes present in the network. Source transformation is not applicable to the ideal sources. 10× 2 V = x 235++ 20 V = x 10

Vx = 2V

The source transformation is applicable Example: Count the number of branches and nodes in for the dependent sources also, = provided the control variable is outside the circuit. If ix 3A & the 18V source the branches, where the source delivers 8 of current. transformation is applied. What is the value of RA? 𝐴𝐴

Here I1 is outside the branches having I Solution: current source of 1 A and 2 Number of branches = Number of nodes = 3 resister, hence source transformation20Ω is applicable. By KCL i1 +− 3 13 = 0 Again by KCL

Example: −−8 10 + C'2 = 0 Determine Vx using voltage division rule. C'2 = 18A Voltage across 6

V6Ω = 18V Ω Voltage across R A

VRA = 18V {}∴Parallel branches have same voltage By Solution: 10Ω resistor connected in parallel with ohm’s law 10Ω resistor VRA= iR 2 A So, Req= 10 ||10 18= 18R A 10× 10 Req= = 5 Ω R1A = Ω 10+ 10

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Example: 12V− 12V ++− 4V 3V 3V 2 1223= 0 Determine current through the 3Ω 12 resistor in the circuit? −12V1 + 19V 23 −= 3V 0 ______(1)

Nodal at V1 ⇒ VV− VV− 30++=12 13 12

2V1− 2V 213 +− V V =− 6 3V− 2V −=− V 6 ______(2) Solution: 1 23 ⇒ Total current following through 3 & 6 Nodal at V3 combination is VVVV−−− V35 31++= 32 3 0 Ω Ω 245 12sin sin t 12sin sin t 10V− 10V +−+− 5V 5V 4V 140 i() t= = 2sin sin t 3 132 3 = 0 4++ 3|| 6 4 2 20 i() t= 2sin sin t A −10V12 −+ 5V 19V 3 = 140 ______(3) By current division rule, 3−− 21 6 ∆= −12 19 − 3 i3 () t= 2sin sin t × 63+ −−10 5 19 4 i3 () t= sin sin t A −6 −− 21 3 ∆=1 0 19 − 3 Example: −−140 5 19 Compute the voltage across each current source. 3−− 61 ∆2 =−−12 0 3 −10 140 19 3−− 26

∆3 = −12 19 0 −−10 5 140 ∆ 1424 V=1 = = 5.23 1 ∆ 272 By source transformation Solution: V1 = 5.23v ∆ 3120 V=3 = = 11.47 3 ∆ 272

V3 = 11.47v

Nodal at V2 ⇒ VVV− VV− 21++ 2 23 =0 134

Q.1 The voltage eo in the figure is power factor 0.844, then the values of (in ohm is

𝑍𝑍𝐿𝐿

4 a) 2V b) V 3 a)90∠ 32.44° b) 80∠ 32.44° c )4V d )8V c) 80∠− 32.44° d) 90∠− 32.44° [GATE-2001] [GATE-2001] Q.6 The minimum number of equations If each branch of a Delta circuit has Q.2 required to analyze the circuit impedance 3Z , the each branch of shown in the figure is the equivalent Wye circuit has impedance. Z a ) b)3Z 3 c) 3 3Z d ) Z 3 a)3 b)4 [GATE-2001] c)6 d)7 [GATE-2003] Q.3 The voltage eo in the figure is Q.7 Twelve 1Ω resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is 5 a) Ω b) 1Ω a)48 V b)24 V 6 c)36 v d) 28 V 6 3 c) Ω d) Ω [GATE-2001] 5 2 [GATE-2003] Q.4 The dependent current source shown in the figure Q.8 The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is

a)delivers 80W b)absorbs 80W c)delivers 40 W d)absorbs 40W [GATE-2002] a) LLM12++ b) LLM12+−

Q.5 If the 3-phase balanced source in the c) L12++ L 2M d) L12+− L 2M figure delivers 1500 W at a leading [GATE-2004]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V (S) Q.9 The transfer function Hs() = 0 Vi (S) of an R-L-C circuit is given by 106 Hs() . The Quality S26++ 20S 10 factor (Q-factor of this circuit is a) 25 b) 50 a)0.238 V b)0.138 V c) 100 d) 5000 c)-0.238 V d)1 V [GATE-2004] [GATE-2005]

Q.10 For the circuit shown in the figure, Q.13 In the interconnection of ideal the initial conditions are zero. Its source shown in the figure, it is known that the 60V source is VC (S) transfer function Hs() = is absorbing power. Vi (S)

Which of the following can be the 106 value of the current source I? a) 1 b) 23 6 S26++ 10 S 10 6S++ 10 S 10 a)10A b)13A c)15A d)18A 103 106 c) d) [GATE-2009] S23++ 10 S 10 6S26++ 10 S 10 6 [GATE-2004] Q.14 In the circuit shown, the power supplied by the voltage source is Q.11 Impedance Z as shown in the given figure is

a)0W b)5W c)10W d)100W [GATE-2010] a) j29Ω b) j9Ω Q.15 In the circuit shown below, the c) j1 9 Ω d) j39Ω current I is equal to [GATE-2005]

Q.12 If RRRR124= = = and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and a)1.4∠ 0°A b) 2.0∠ 0°A b is c) 2.8∠ 0°A d) 3.2∠ 0°A [GATE-2011]

Q.20 The curr Amps is a)2 ent in the b)3.33 1Ω resistor in c)10 d)12 [GATE-2013] a) -5V b) 2V c) 3V d) 6V Q.21 Consider a delta connection of [GATE-2012] resistors and its equivalent star connection as shown. If all elements Q.17 The average power delivered to an of the delta connection are scaled by impedance ()4− j3 Ω by a current a factor k,k> 0, the elements of the 5cos(100πt+ 100) A is corresponding star equivalent will a) 44.2W b)50W be scaled by a factor of c) 62.5W d)125 [GATE-2012]

Q.18 In the circuit shown below, the current through the inductor is 2 a) k b) k c)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

2 −1 VWX1 =100V is applied across WX to a) A b) A 1j+ 1j+ get an open circuit voltage VYZ1 1 across YZ .Next, an ac voltage c) A d) 0A 1j+ VYZ2 =100V is applied across YZ to

[GATE-2012] get an open circuit voltage VWX2

across WX. Then VYZ1 /V WX1 , Common Data Questions 19 & 20 are respectively. Consider the following figure VWX2 /V YZ2

Q.19 The current IS in Amps in the voltage source, and voltage Vs is a)125/100and80/100 Volts across the current source b)100/100and80/100 respectively, are c)100/100and100/100 a)13,-20 b)8,-10 d)80/100and80/100 c)-8,20 d)-13,20 [GATE-2013]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.23 Three capacitors C12 ,C and C3 third arm has a resista in whose values are 10μF,5μF, and 2μF the equivalent — network, the lowest value (in amongnce ofthe 11Ω three respectively, have breakdown resistances) is ______. voltages of 10V, 5V and 2V Ω [GATE-2014] respectively. For the interconnection shown below, the Q.26 Consider the building block called maximum safe voltage in volts that 'Network N' shown in the figure. can be applied across the Let C =100µF a combination , and the corresponding total charge in μC nd R = 10kΩ stored in the effective capacitance across the terminals are respectively,

Two such blocks are connected in cascade, as shown in the figure. 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 V (s) The transfer function 3 of the V1 (s) cascaded network is s2 a) s b) 1s+ 1++ 3s s 2 2 s s c)  d) 1+ s 2s+ [GATE-2014]

For R = Ω , and currents i1= 2A, i4 = 1 Q.27 In the circuit shown in the figure, -1A, i5 = -4A, which one of the the value of node voltage V2 is following is TRUE?

a) i6 = 5A

b) i3 = − 4A c) Data is sufficient to conclude that the supposed currents are impossible d) Data is insufficient to identify the current i2, i3, and i6 [GATE-2014] a) 22 + j 2 V b) 2 + j 22 V c) d) A Y-network has resista Q.25 [GATE-2014] each in two of its arms, while the 22 ‒ j 2 V 2‒ j 22 V nces of 10Ω

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.28 For the Y-network shown in the figure, the value of R1 ) in the equiva -network is______. (in Ω lent ∆

a) 5V, 25V b) 10V, 30V c)15V, 35V d) OV, 20V [GATE-2015]

Q.33 In the circuit shown, the voltage Vx [GATE-2014] (in Volts) is ______.

Q.29 The circuit shown in the figure represents

[GATE-2015]

Q.34 An AC voltage source V = 10 sin(t) a)Voltage controlled voltage source volts is applied to the following b)Voltage controlled current source network. Assume that R c)Current controlled current source 1 R =6kΩ and R and that the d)Current controlled voltage source 2 3 = 3kΩ, [GATE-2014] diode is ideal. = 9kΩ, Q.30 The magnitude of current (in mA) through the resistor R2 in the figure shown is ______.

[GATE-2014] RMS current Irms (in mA through the diode) is ______. Q.31 The equivalent resistance in the [GATE-2016] infinite ladder network shown in the figure is Re. Q.35 In the given circuit, each resistor has a value equal to 1 Ω .

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

Q.32 In the given circuit, the values of V1

and V2 respectively are

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission What is the equivalent resistance across the terminals a and b? a) b) c) d) 1/6 Ω [GA1/3TE Ω-2016] 9/20Ω 8/15 Ω Q.36 In the circuit shown in the figure, the magnitude of the current (in amperes through R2) is ______.

The magnitude of the current (in amperes, accurate to two decimal places) through the source is ______. [GATE-2016] [GATE-2018] 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.

are provided. Consider all possible permutationThree resistors of ofthe value given 10Ω, resistors 5Ω, 2Ω 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 R11 = Ω , R22 = Ω and R33 = Ω . The network is connected to constant voltage source of 11V.

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

Q.1 (c) 4002 × 0.844 Applying KCL = = 90Ω 1500 e− 12 e e o++= oo0 ⇒=3e 12 θ= cos−1 (0.844) = 32.44 4 44 o As ∴= eo 4V power factor is leading, load is Q.2 (a) capacitiveθ= − 32.44° so angle will be ZV = 3ZY ⇒= 3Z∆ 3ZY negative.

Z∆ Q.6 (a) ZY = 3 As voltage at 1 node is known Q.3 (d) ∴ using nodal analysis only 3 Applying source conversion Q.7 (a)equations required.

e−− 80 e e 16 o++ oo =0 12 12 6

4eo = 112 112 eo = = 28V 4 iii V= ×+ 111 ×+ × ab 363 Q.4 (a) Vab 5 V ⇒==Req Ω Applying KVL, 20−− 5I 5 I +1 = 0 i6 5 20− 10I −= 20 0 Q.8 (d) ⇒=I0

V 1 = 4A ∴Only5 dependent source acts.

=I2 R = 16 ×= 5 80W Power delivered Q.5 (d) of 3Vpp l cosθ= 1500 If current enters the dotted  VVLL terminals coil 1 then a voltage is 3 cosθ= 1500 3 3ZL developed across coil 2 whose 2 higher−MdI potentialL dI MdI is at dI dotted V .cosθ 1 L terminals.VL= +−+2 ZL = dt dt dt dt 1500

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission dI , =()L +− L 2M 12 dt source dI In the- given circuit the current VL= through the branch of 60 V only if eq dt is (12 -I) as shown in Fig. The source Q.9 (b) of 60V absorbs power, P=(12 I) 60 is +ve. i.e., I<12. The =++S26 20S 10 value of the currentn source, I can only be 10 A given in option (a), as Characteristicω equation Q,=o ω = 106 the currents given i other options BW o Q.14 (a)are more than 12 A. 103 1000 Q= = = 50 20 20

Q.10 (d) 1 1 Hs() =sC = 1 2 ++ R++ sL s LC sCR 1 sC 1 = s23() 10× 10−− ×× 100 10 6 +s() 10 ××× 1036 100 10− + 1 1 106 The current through all the Hs() = = 10−62 S++ s 1 S 2 + 10 6 S + 10 6 branches are marked as shown in Fig 1. 2I()()11++ 3 2I += 2 10 Apply KVL to outer loop Q.11 (b) 4I1 += 10 10 X=+++ X123 X X 2X m − 2X m I01 = =+++ − 5j 2j 2j 20j 20j 10V= 10 ×= 0 0W ∴ Power supplied by = 9j (one additive & other Q.15 (b) Q.12 (c)subtractive) ZS =∠− 7 0°Ω(using Y Δtransformation) = = Va 5R() 12 R 14∠ 0° I= = 2 ∠ 0°A R3 1.1 7∠ 0° Vb = ×= 10 × 10 R34+ R 2.1 Q.16 (a) VV=ab − V V= − 0.238V Q.13 (a)

VAB−= V 6V From the given circuit, VV− I= AB= 3A = I AB2 DC

Q.17 (b) Rabc= x;R = k;R = k 2 RRab+ k ConsiderRStar = = The load consists of a resistance and Rabc++ R R 3k a capacitance2 of this, only R is So P= Ri RkStar ∝ passive rmand consumes power 2 5 = ×=4 50W Q.22 (b) 2 A note rms value of A cos ωt = 2

Q.18 (c)

Assume current as shown, VW1××= 100 ⇒= V 2 turns ratio × V W1 = 125

VYZ1=×= 0.8 V 2 100V

VYZ2 = 100V When

in By apply11ing current division rule I= ×∠ 10 = L upper1j++ part of the 1j circuit -

Q.19 (d) Thevenin’s circuit seen by 2 2’ will Vth = 100V An Rth = 0.2 || 0.8 be as follows

∴=' V22− 100V negligible ∴VW2× 100V× 1 =×=V' turns ratio = 80V 22− 1.25

Across AB v Q.23 (d)

ISS= − 13A, V oltage drop is 10 V 10 10 =20V2 =++= I 0 S 21

IS =−− 5 10 + 2 =− 13

Q.20 (c)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission CC23 (10)(10)++ (10)(11) (10)(11) CCeq= + 1 z = CC23+ 10 =11.5μF Ω = 7V i.e, lowest value among three Q= CV Safety voltage Q.26 (b)resistances is 29.09 Ω ⇒=QCeq × V safety 11.5× 7; 80μC - Two blocks are connected in Q.24 (a) cascade, Represent in s domain, Given i = 2A

1 i4 = – 1A

V (s) R.R 3 = V (s) 1 11   1 RR++ + R + R sc SC  SC  R.R 11 R . 2R() SC+ 1+[] 1 + RSC i5 = – 4A sc sc SC + i4 = i2 S22 C R.R = 2 = 2 – 22 2 1 1+ 2R() SC ++ RSC R S C KCL at node A, i 2 +i5 =i 2−− 6 6 33 ⇒ i – 41 = -1A . S .100× 100 × 10 × 10 ××× 10 10 10 × 10 3 = 1. KCL at node B, i 2 64−− 12 64 3 + i =i S× 100 × 10 × 10 × 10 ++ 3S 100 × 10 × 10 + 1 ⇒ i = 1 3A 2 = 2 – (- 3 6 1 V (s) S KCL at node C, i 3 = 6 V (s) 1++ 3S S2 Q.25 ⇒(29.09Ω i ) 3) = 5A 1 Q.27 (d)

KVL for V 2

1 (10)(10)++ (10)(11) (10)(11) & V X= X = 29.09Ω 11 Ω (10)(10)++ (10)(11) (10)(11) Y=y = 32Ω 10 o V12−= V 10 | 0 …(1) Ω 0 V12 -V =+10|0 Z= 32Ω

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission KCL 28 ⇒=I ⇒= VVV I 2.8mA -4| 00 +1 ++=… 22 0 .(2) 10k at super−j3 node: 6 j6 VVV Q.31 (2.618) 1++= 224| 00 −j3 6 j6

o V+ 10 | 0 V V 0 From2++=(1) 224|& 0 (2), −j3 6 j6

VVV1 22 0 10 V2 ++ = 4| 0 + − j3 6 j6 j3

∴=−V2 () 2 j22 Volts →For an infinite ladder network, if all resistance are having same value 15+ Q.28 (10Ω) of R then.R equivalent resistance is 2

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

R ⇒=+⇒=R R 1.618Requ 2.618 ()()()()()7.5 5++ 3 5 7.5 (3) equ R = Ω 7.5 1 R Q.32 (a)

1 Q.29 (c)R =10 Ω -5+++ I I 2I = 0 By4I nodal= 5 analysis 5 I= A 4

V1 = 4I = 5volts

V21=4() 5 +V = += 20 V 1 25 volts The dependent source represents a Q.30 (current2.8) controlled current source Q.33 (8)

By source transformation

By KVL, – Apply KCL at point P

20 10k.I + 8 = 0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Vx VVxx− 0.25 Vxx V 4V x + Vx = 5 ⇒=I − = 20 10 5 25 25 1 0.75 + 0.5 Applying KVL, V++= 0.5 5 x  4V V 20 10 60 = xx×+×58 5 255 = ⇒ = Vxx 5 VV 12V  8  60 = x ⇒ V = 25 5 x 25 Q.34 (0.68 to 0.72) R = = L 5 Q.35 (d) Thus current through

- Q.37 (-1)5amps Let assume all resistance as R, then by using start delta transformation V-8 V V-8 V ++ +=0 Nodal1111 equation at V ⇒=4V 16

84− 1+- i = 0 ⇒= i 5A By using1 KCL1 at node 1 'a'.

4 - +i += i0 ⇒−++= 45i0 KCL1 at1 b ⇒=−I 1A

Q.38 2.143

2 4R 8R 32R 5 8R The connection of resistors is as ⇒=R ab P = × =as R = 1Ω 5 5 25 12R 15 shown below 8 R = Ω ab 15 . Q.36 (5) 10Ω , 5Ω and R = L 2Ω 1 Given resistors are: Vx ⇒+LetI current 0.04Vx through= RT() max = 10 Ω+() 5 Ω 2 Ω 5 The maximum resistance possible is 10 80 =10 + Ω= Ω 77

RT() min = 2 Ω+() 10 Ω 5 Ω The minimum resistance possible is 10 16 =2 + Ω= Ω 33

R T() max 80 / 7 = RT() min 16 / 3 15 = = 2.143 7

Q.39 8 We can find current supplied by B C voltageV source, I = From the symmetry of given circuit, R we can conclude that point and eq are equipotentialA points since theF resistive distance of both the points from terminals as well as from Where11 13 =++ are equal. D E R eq  2 2 22 Similar reasoning can be applied to identify that points and are also 13 equipotential. × We can remove the resistance R1= + 22 eq 13 connected between two + equipotential points as current 22 through it will always be zero. 3/4 =1 + 2

11 R = Ω eq 8

Here

V I = R eq

11 = 11/ 8

- I= 8A

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

and current i are in phase. The Q.1 Given two coupled inductors L1 and M coupling coefficient is K = L2 , their mutual inductance M LL12 satisfies where M is the mutual inductance LL12+ between the two coils. The value of a) M= LL22 + b) M > 12 2 K and the do polarity of the coil P-Q are c) M> LL12 d) M≤ LL12 [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. a) K=0.25 and dot at P b) the bulbs together consume 50 W. b) K=0.5 and dot at P c) the 60 W bulb glows brighter. c) K=0.25 and dot at Q d) the 40 W bulb glows brighter. d) K=0.5 and dot at Q [GATE-2001] [GATE-2002]

Q.3 Consider the star network shown in Q.5 A segment of a circuit is shown in figure. The resistance between figure VRC =5V,V = 4sinsin2t the terminal A and B with terminal C voltage VL is given by 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) R=Ω=Ω=Ω 4 ,R 2 ,R 5 a)3− 8cos 2t b)32sin 2t ABC c)16sin 2t d)16cos 2t b) =Ω=ΩΩ RABC 2 ,R 4 ,R 7 [GATE-2003] c) RABC=Ω=Ω=Ω 3 ,R 3 ,R 4 Q.6 Figure shows the waveform of the d) RA=Ω=Ω=Ω 5 ,R BC 1 ,R 10 current passing through an inductor [GATE-2001] of resistance 1Ω and inductance 2H. The energy absorbed by the Q.4 In the circuit shown in figure it is inductor in the first four seconds is found that the input ac voltage (Vi )

YR = 0.5 + j0, YL = 0 − j1.5,

YC = 0 + j0.3respectively. The value of I as a phasor when the voltage E a)144J b)98J is c)96J d)168J [GATE-2003] across the elements is 10∠0°V

Q.7 In figure, the potential difference between points P and Q is a) 1.5+j0.5 b) 5-j c) 0.5+j1.8 d) 5-j12 [GATE-2004]

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

a) 12 V b)10V c)-6V d)8V [GATE-2003] a)10 b)20 c)30 d)40 Q.8 In figure, the value of R is [GATE-2004]

Q.12 In figure Rab , R and R c are 20Ω, 10Ω and 10Ω respectively. The

resistance R12 ,R and R3 equivalent star connection are in Ω of an

a)10Ω b)18Ω c) 24Ω d)12Ω [GATE-2003]

Q.9 In figure, the value of the source voltage is a) 2.5, 5, 5 b) 5, 2.5, 5 c) 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)12V b)24V a)10A b)14.14A c)30V d)44V c)15A d)17.32A [GATE-2004] [GATE-2004]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.14 In the figure given below the value a)1:1: 3 b)1:1:2 of R is c)1:1:0 d) 1:1: 3/2 [GATE-2005]

Q.18 The circuit shown in the figure is energized by a sinusoidal voltage a) 2.5Ω b)5.0Ω source V1 at a frequency which c) 7.5Ω d)10.0Ω causes resonance with a current of I. [GATE-2005]

Q.15 The RMS value of the voltage u() t= 3 + 4cos(3t) is a) 17V b) 5V c) 7V d) ()3+2 2 V [GATE-2005] The phasor diagram which is applicable to this circuit is The RL circuit of the figure is fed Q.16 a) b) from a constant magnitude, variable frequency sinusoidal voltage source .At 100Hz, the R and L elements each have a voltage drop u .If the 𝐼𝐼𝐼𝐼 RMS frequency𝑉𝑉 of the source is changed to 50Hz, then new voltage drop across R is c) d)

5 2 a) uRMS b)uRMS 8 3 [GATE-2006] 8 3 c) uRMS d) uRMS Q.19 An energy meter connected to an 5 2 immersion heater (resistive) [GATE-2005] operating on an AC 230 V, 50 Hz, AC single phase source reads 2.3 units Q.17 For the three phase circuit shown in (kWh) in 1 hour. The heater is the figure the ratio of the currents removed from the supply and now I :I :I is given by RYB 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]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.20 The state equation for the current I1 Q.22 A 3V dc supply with an internal shown in the network shown below resistance of 2Ω supplies a passive in terms of the voltage V and the non-linear resistance characterized x 2 independent source V, is given by by the relation VINL= NL. The power dissipated in the non-linear resistance is a)1.0W b)1.5W c) 2.5W d) 3.0W [GATE-2007]

Q.23 The Thevenin’s equivalent of a circuit operating at ω=5rad/s has dI 5 a) 1 =−−+1.4V 3.75I V V =3.71∠ -15.9° and dtx1 4 oc

dI 5 Zo =2.38-j0.667Ω . b) 1 =−−−1.4V 3.75I V dtx1 4 At this frequency, the minimal realization of the Thevenin’s dI1 5 c) =−++1.4Vx1 3.75I V impedance will have a dt 4 a) Resistor and a capacitor and an dI1 5 inductor d) =−+−1.4Vx1 3.75I V dt 4 b) Resistor and a capacitor [GATE-2007] c) Resistor and an inductor d) Capacitor and an inductor Q.21 In the figure below all phasors are [GATE-2008] with reference to the potential at point “0”. The locus of voltage Q.24 Assuming ideal elements in the phasor V as R is varied from zero YX circuit shown below, the voltage Vab to infinity is shown by will be

a) -3V b) 0V c)3V d) 5V a) b) [GATE-2008] Q.25 In the circuit shown in the figure, the value of the current i will be given by

c) d)

a) 0.31A b) 1.25 A [GATE-2007] c) 1.75A d) 2.5A [GATE-2008]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.26 Q.30 As shown in the figure, a resistance in the circuit shown is resistance is connected across a The current through the 2kΩ source that has a load line v+i=1001Ω. The current through the resistance is

a) 0mA b)1mA c) 2mA d) 6mA [GATE-2009] a) 25A b)50A c)100A d) 200A Q.27 How many 200W/220 V [GATE-2010] incandescent lamps connected in series would consume the same Q.31 If the 12Ω resistor draws a current total power as a single 100 W/220V of 1A as shown in the figure, the incandescent lamp? value of resistance R is a) Not possible b) 4 c) 3 d) 2 [GATE-2009]

Q.28 The equivalent capacitance of the input loop of the circuit shown is a) 4Ω b) 6Ω c)8Ω d) 18Ω [GATE-2010]

Q.32 In the circuit shown below, the current through the inductor is

a)2μF b)100μF [GATE-2009] c)200μF d) 4μFs Q.29 For the circuit shown, find out the

resistance. Also identity the changes tocurrent be made flowing to double through the thecurrent 2Ω 2 −1 . a) A b) A 1j+ 1j+ through the 2Ω resistance 1 c) A d) 0A 1j+ [GATE-2012]

Q.33 A two –phase load draws the a)()5A;putVs = 20V following phases currents:

b)()2A;putVs = 8V i1m()() t= I sin ωt −∅ 1 , it2m() = I These currents are balanced if ∅ c)()5A;putIs = 10A 1 d) ()7A;putI= 12A is equal to s a) −∅ b) ∅ [GATE-2009] 2 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission π . Their mutual inductance in c) ()π/2−∅2 d)+ ∅2 2 240μH [GATE-2012] [GATE-2014] μH is ___. Q.34 The average power delivered to an Q.38 The voltage across the capacitor, as sown in the figure, is expressed as impedance ()4-j3 Ω by a current v()()() t= A sin ω t −θ + A sin ω t −θ 5cos(100πt+100) A is t 1 11 2 2 2 a) 44.2W b)50W c) 62.5W d)125 [GATE-2012]

Q.35 If VAB -V =6V, then VCD -V is The value of A1 and A2 respectively, are a) 2.0 and 1.98 b) 2.0 and 4.20 c) 2.5 and 3.50 d) 5.0 and 6.40 [GATE-2014]

Q.39 The total power dissipated in the a) -5V b) 2V circuit, show in the figure, is 1kW. c)3V d) 6V [GATE-2012]

Q.36 Three capacitors C12 , C and C3 whose values are 10μF,5μF, and 2μF respectively, have breakdown voltages of 10V, 5V and 2V The voltmeter, across the load, respectively. For the reads 200 V. The value of XL is interconnection shown below, the [GATE-2014] maximum safe voltage in Volts that can be applied across the Q.40 The line A to neutral voltage is combination, and the corresponding total charge in μC stored in the star-connected load with phase 10∠15°V for a balanced three phase effective capacitance across the sequence ABC. The voltage of line B terminals are respectively, 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]

a)2.8 and 36 b)7 and 119 Q.41 The power delivered by the current c)2.8 and 32 d)7 and 80 source, in the figure, is 1V [GATE-2013]

Q.37 Two identical coupled inductors are connected in series. The measured inductances for the two possible series c [GATE-2014] onnections are 380 μH and

figure are 6 V and 2 V respectively, withΩ and the 2 polarity Ω resistors as marked. show What in the is the power (in Watt) delivered by the 5 V voltage source? - - a) 173.2∠ 60° b) [GATE 173.2∠120°-2015] c) 100.0∠ 60° d) 100.0∠120°

Q.46 R A and R B are the input resistances of circuits as shown below. The circuits extend infinitely in the a) 5 b) 7 direction shown. Which one of the c) 10 d) 14 following statements is TRUE? [GATE-2015]

Q.43 In the given circuit, the parameter k is positive, and the power dissipated 12.5 W. The

in the 2Ω resistor is value of k is ____.

a) RRAB= b) RR0AB= =

c) RRAB< d) RBA= R / (1 + R A ) [GATE-2016] [GATE-2015] Q.47 In the portion of a circuit shown, if Q.44 resist is 10 calories per second then heat The current I (in ampere) in the 2Ω the heat generated in 5Ω resistance or of the given network is _____. generated by the 4Ω resistance, the calories per second, is_____

[GATE-2015] [GATE-2016]

Q.45 In the given network V1 In the given circuit, the current V2 3 Q.48 V. The phasor current i (in= 100∠0°V,Ampere) supplied by the battery, in ampere, is = 100∠120° V, V = 100∠+120° is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission [GATE-2016] Q.53 The power supplied by the 25V source in the figure shown below Q.49 In the circuit shown below, the node is…………….W. voltage VA is V.

[GATE-2017, Set-1]

Q.54 The equivalent impedance Z for [GATE-2016] eq the infinite ladder circuit shown in Q.50 The voltage (v) and current (A) the figure is across a load are as follows. v(t) -

The=100 averagesinω(t), i(tpower) = 10sin(ωt consumed 60°) by + 2sin(3ωt) + sin(5ωt) [GATE-2016] a) j1 2 Ω b) −Ωj1 2 the load, in W, is______. Q.51 In the circuit shown below, the c) j1 3Ω d) 13Ω voltage and current are ideal. The voltage (Vout) across the current [GATE-2018] source, in volts, is 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. Assumei0() = 0, v0() = 0 . Which one of the following circular loci represents the plot of it() a) 0 b) 5 c) 10 d) 20 versus vt() ? [GATE-2016]

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

[GATE-2017, Set-1]

[GATE-2018]

Q.56 A three-phase load is connected to a three-phase balanced supply as shown in the figure. If 0 0 Van = 100 ∠ 0 V , Vbn = 100 ∠− 120 V 0 and Vcn = 100 ∠− 240 V (angles are considered positive in the anticlockwise direction), the value of R for zero current in the neutral wire places). [GATE-2018] is ______(up to 2 decimal

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

Q.1 (d) llllPQCL+++= 0 M= K LL 12 21l++CL + l = 0 Where K=coefficient of coupling But, lC = C × dv / dt Q 0K1<< =1 × d / dt(4sin2t) ∴≤ M LL12 = (8cos 2t)

∴lL =−++ (2 1 8cos2t) Q.2 (d) =−−3 8cos 2t 1 Q P ∝ R Q.6 (c) Therefore resistance of 40W bulb > For 0t2< absorbed by the inductor (Resistance power consumed by 60W bulb. neglected) in the first 2 sec, Hence, the 40W bulb glows brighter. T di EL= Li dt = E LL + E ∫ dt 12 Q.3 (b) 0 When C is open, R= RR + 2 di AB A B = EL1 ∫ Li dt = 6Ω 0 dt When B is open, RAC= RR A + C 2 = 9Ω =∫2 ×× 3t 3dt When A is open, 0 R=+= RR11Ω On solving 2 t2 2 BC B C =18∫ t dt = 18 × above equations 0 2 0 = = R2ABΩ,R 4Ω and R C 7Ω 4 =×−=18 0 36J 2 Q.4 (c) The energy absorbed by the Input ac voltage and current will be inductor in (2→ 4) second in phase only at resonance condition 4 i.e. di 4 E= Li dt =2.6.0dt = 0J L2 ∫  ∫ XXCL= −j1 2 2 dt 2 A pure inductor does not dissipate =j8 ++ j8 2k()() j8 × j8 energy but only stores it. Due to 12=++ 8 8 16k resistance, some energy is 41 dissipated in the resistor. Therefore, ⇒=−=−=−k 0.25 16 4 total energy absorbed by the Hence coupling will be opposite inductor is the sum of energy stored in the inducer tor and the energy Q.5 (b) dissipated in the resistor. The By KCL,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission energy dissipated by the resistance ∴=VP 30V in 4 sec. Potential difference between node x T and y = 2 ER ∫ i Rt = 60 V 0 by taking KCL at node y 24 22 40− 30 =∫∫(3t) ×+ 1dt 6 × 1dt −−l5 + = 0 02 1 24 ∴= t3 24 l 5A =(9t)2 dt + 36 1dt =×+9 36t 60 ∫∫3 02 ∴=R = 12Ω 02 5 =×98 +× 36 2 =+=24 72 96j ∴From equation(i) ()3 4() VPP− 10 +×+ 2 8 V = 0 ∴VL = L(di / dt) =×× 2 2 8sin 2t = 32sin 2t 4VPP−++ 40 16 V = 0

Note: KCL is based on the law of 5VP −= 24 0 conversation of charges. V= 4.8 Dot will at Q P 6() VQQ− 10 −××+ 2 4 6 4V = 0 Q.7 (c)∴ 10VQ −= 108 0 Given: V= 10V R ∴=V 10.8 By KCL Q ∴− =− V− 10 V VPQ V 6V PP++20 = …..(i) 28 V− 10 V Q.9 (c) QQ−+20 = …..(ii) 46

Method –I Using KCL, V− EV aa+ −=10 66 ⇒ −= …….(i) Q.8 (d) 2Va E 6 By KCL, Where V−− 40 V 100 V EV− ∴PP + += P0 a = 2 1 14 2 6 ⇒− = …….(ii) 22VP = 660 E Va 12 Solving eq (i) & (ii) We get

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

~~Va = 6l =×= 6 3 18V~~

~~EV− a E− 18 = 2 ⇒=2 ⇒=E 30V 6 6~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission RR = bc Q.10 (d) R1 RRRabc++ 10× 10 = = 2.5Ω 20++ 10 10 RRca R 2 = RRRabc++ 10× 20 = = 5Ω IRR= Y E =() 0.5 + j0 ×∠= 10 0° 5A 20++ 10 10 RRab IYL= Y E =() 0. + j1.5 ×∠=− 10 0° j15A R = 3 RRR++ I= Y E =() 0 + j0.3 ×∠= 10 0° j3A abc CC 20× 10 II=++ I I = = 5Ω RYC 20++ 10 10 =++5() j15A j3 Remember: If all the branches of - =5 − j1 2 A connection has same impedance Z then impedance of branch of ∆Y- Q.11 (b) connection be Z/3

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

~~=10A = ldc R.M.S value of sinusoidal Current~~

Vp − 100 V  +P += 20 20 =Al = dc 10 10 2 2VP − 100 += 20 0 R:M:S value of resultant, 80 22 ∴==V 40V lR= ll dc + ac P 2 V 40 ∴=RP = = 20Ω Q.14 (c) 22 The Resultant (R) when viewed from voltage Q.12 (a) 100 Source = =12.5 8 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) =Vdc = 3V (rms) R.M.S value of a.c. Voltage = Va.c = ()4/ 2 V~~

~~Given: Ra = 20Ω R.M.S value of the voltage R= 10Ω 2 b =32 + () 4/ 2 And Rc = 10Ω~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 9+= 8 17V Assuming phase-sequence to be RYB Taking VRY as the reference~~

Q.16 (c) VRY = V ∠ 0° At f= 100Hz VYB = V ∠− 120° VVRL= VBR = V ∠− 240° as R & L are series connected, VVRB BR current through R & L is same, so lR = = − RR11 IR= lXL = lωL V∠− 240° V ⇒=RXL =ωL =− =∠−60° RR11 Vin I = V V∠− 120° RX22+ l =YB = L Y RR VV 11 =in = in Using KCL 22+ 2R RR lll0RYB++= VR= u rms = IR VV ∠−60° + ∠− 120° + lB = 0 VVin in RR = ×=R 11 2R 2 V ⇒=lB 3 ∠ 90° ⇒=Vin 2u rms …..(i) R1 At f=50Hz VV V So l :l :l= : : 3 ∝ RYB XfL RR11 R 1 50X R So, X'=×== X L =1:1: 3 LL100 2 2 V I' = in Q.18 (a) 2 2 As circuit is under resonance, V & I R+ () X' 1 L should be in phase. V 2V =in = in 2 5R Q.19 (b) 2 R R +  Assuming resistance of the heater = 2 R ''2Vin 2 i) When heater connected to 230 V, VIR= R = R = V in 5R 5 50 Hz source, energy consumed From eq. (i) by the heater =2.3 units or 2.3 kWh in 1 hour Power consumed ' 2 VR= ×() 2u rms by the heat 5 energy 2.3kWh 22 8 = = uurms= rms time period 1hour 5 5 P1 = 2.3kW rms value of the input voltage Q.17 (a) V2 =V = 230V =P = rms rms 1 R 2302 ⇒×=2.3 103 = 23Ω R

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ii) When heater connected to 400V V∠ 0° +∠ V 0° 2V I = = (peack to peck) square wave R−− jX R jX source of 150 Hz CC Using KVL,

~~VYX + IR −= V 0~~

⇒=−VYX V IR 2V VVYX = −  R R− jXC V() R−− jX 2VR V() R+ jX = C = C R− jXC ()R− jXC Method -1 Vrms value of the input voltage 1 R+ jXC T VV= −  1 2 YX R− jX = 2 C Vrms ∫ V dt T 0 When R=0 1 0+ jX T C 2 VVYX =−= V 2 T 0− jX 1 2 C =2002 dt +−() 200 dt T ∫∫ X 0 T 1j+ c 2 R V =  =V = 200V YX X rms 1j− c 2 2 R Vrms 200 −3 P2 = = × 10 kW when R →∞ R 23 = − =1.739kW VVYX Method -2

Q.20 (a) R− jXC VVYX = −  di1 R− jXC V− 3 I12 +− I V x − 0.5 = 0 …..(1) dt −1 XC di =V∠ 180° − 2 tan  5I−= 0.51 0.2V R 2xdt Magnitude of VVYX = di1 0.5−+ 5T2x 0.2V = 0 ………..(2)  dt 22−− 11XC R+ XC∠ tan tan  Eliminating I2 for eq(1) and (2) we R =V∠ 180° × get −X R22+ X∠ tan− 1 C dI 5 C  1 =−−+1.4V 3.75I V R dtx1 4 So, option (c) and (d) cannot be Q.21 (a) correct, as magnitude is 2V in these two options.

~~−1 XC Angle of VYX = 108° − 2 tan  R When R = 0 −1 ∠=VYX 180° − 2 tan ( ∞ ) = 18000−= 2x90 0 Let capacitive reactance = XC When R= ∞~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −1 51× ∠=VYX 180° − 2 tan (0) = 180° Va = = 2.5V On the basic of above analysis, the 2 Also, locus of VYX is drawn below: 4Vab=⇒−= 4i 4() V a V b 4i ………(1) Also, 1 V V= 4V ×= V − V ⇒ V =a b ab42 a b b =1.25V ∴−4() 2.5 1.25 Q.22 (a) =4i() from() 1 ⇒= i 1.25A

Q.26 (a) As the ABCD bridge is balanced, = I0CS ∴=32II +2 ⇒= I1A Q.27 (d) Power delivered by source For a lamp, = 2 =×=3 1 3W P KV For 200ωlamp, K = 200 Power dissipated by 2Ω resistor. 220V 2 2 220 =I ×= 2 2W Consider n lamps connected in Power dissipated in non-linear series, element Total power consumed ∴=−= 3 2 1W =××n K 1102 = 100 200 ⇒× ×2 = ⇒= Q.23 (b) n2 110 100 n 2 Thevenin’s Impedance 200 Z= 2.38 − j0.667Ω 0 Q.28 (a) As real part is not zero, so Z0 has Assume a 1A current source at input resistor terminals, Im[] Z0 = − j0.667 ∴=I1 1A Case-I

~~Z0 has capacitor (as Im[] Z0 is negative) Case-II~~

Z0 has both capacitor and inductor, but inductive reactance ∠ capacitive reactance at ω5= rad/sec Applying KVL For minimal realization case–(i) is V− i() 1 +− 50i() − jX = 0 considered. in 1 1 C ⇒=Vin i 2[] 2 − j50X C Therefore, Z0 will have a resistor and a capacitor. Input impedance Vin = =2 − j50X C Q.24 (a) i1 Vab = 2i −=−=− 5 2 5 3V As imaginary part is negative, input impedance has equivalent capacitive

~~Q.25 (b) reactance XC eq.~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission XCeq.= 50X C 11 IL = ×∠ 10 = 1 50 1j++ 1j = ωCeq. ωC Q.33 (d) 50 1 = = i1m= l sin(ωt −φ 1 ) ω× 100 2ω i2m= l cos(ωt −φ 2 ) C2eq. = μF π =lm2 sin  +−ωt φ Q.29 (b) 2 The relevant circuit is shown in fig. As these two currents are balanced As the voltage across = 4V ii12+= 0 4 = = π I 2A 2Ω ⇒sin ()ωt −φφ12 + sin +− ωt = 0 2 2 In order to double the current φφ12+ π 1π through Ω resistance, is to be 2sin ωt− + cos −+φφ21 = 0 2 VS 2 4 22 doubled (Put VS = 8V )Note that the 1 ππ ⇒ −+φφ21 = 5A source has no effect on the 22 2 answer. However it gives 3A current π through the voltage source as shown ∴φ12 = +φ in fig. 2

Q.34 (b) The load consists of a resistance and a capacitance of this, only R is passive and consumes power 2 So P= rm Ri 2 5 Q.30 (b) = ×=4 50W 2 V+= i 100 and V = i.1(by ohm' s law) A ∴2i = 100 ⇒= i 50A note rms value of A cos ωt = 2 Q.31 (b) Current through R=1A Q.35 (a) By KVL, 1.R+= 6 12 ⇒=R6Ω

~~Q.32 (c) Assume current as shown,~~

~~From the given circuit,~~

~~VAB−= V 6V VV− I= AB= 3A = I AB2 DC KCL at ‘D’ gives VV− CD++=2 3 0, V − V =− 5V By applying current division rule 1 CD in upper part of the circuit~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.36 (d) Vc (t)= 2sin sin(10t-θθ 12 )+1.98(5t- ) By comparing with given~~

~~expression, A1 =2.0~~

~~A2 =1.98~~

~~Q.39 (17.34 ) CC23 Total power dissipated in the circuit CCeq= + 1 =11.5μF CC23+ is 1kW. 𝛀𝛀 Safety voltage = 7V P = 1kW Q= CV 1000 = 12.1+12.R. 1000 = (2)2.1 + (10)2.R. ⇒=QCeq × V safety~~

11.5× 7; 80μC V 200 Z=⇒= 20 R=9.96Ω1 10 Q.37 (35 μH) 22 Two possible series connections are Z= RX + L 1. Aiding then L equation 222 XL =() ZR − = L1 +L2 + 2M. X22= (20) − (9.96) 2 X= 17.34Ω 2. Opposing then L equation ⇒L L = L1 + L2 –2M L1+L2 Q.40 (c) ⇒ L2 +L2 – +2M=380μ H ……(1) 2M = 240 μH ….. (2) Q.38 (a)From 1 & 2, M = 35 μH By using super position theorem, 1. ϑ — When 20 sin 10t voltage C1 source is acting, V= 3V Network function L ph 1 =3 ×= 10 10 3 jcω 1 If V= 100 Hj()ω = ⇒ A 1 (10 j+ 1) R + thenVBC = 10 3 − 90 jcω givenV= 1015° 1 A ϑ t= 20sin sin(10t − tan−1 (10)) n90+15¯n 75 ¯ C1 () 101 VBC =103 - =103- ∴ 2. ϑ ()t When 10 sin 5t current Q.41 (3 Watts) C2 source is acting 100× 1 ϑ = ×−0.2 j C2 1− 0.2 j −2j ϑ = C2 1− 0.2j 2 ϑ ()t= .sin(5t − θ) C22 2 KCL at node V 1+ () 0.2 x 1V− V xx+= ϑ ()t= 1.98sin sin(5t − θ) 2 C22 11

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V= 1.5V E calorie x P= = = watt Power delivered by current source t sec~~

~~is = 2x 1.5 = 3 watts →=P5Ω 10~~

V5Ω Q.42 (a) =⇒=10 V5Ω 50 5Ω 6V I= = 2A ⇒ P4Ω is asked 3Ω 2 2 ()V 14 2V → =4Ω = I= = 1A P4Ω  50 3Ω 4 44+ 6 I12+= 1 16 calorie =× ×=50 2 I= 1A 4 100 sec P = 5 ×= 1 5W Q.48 (0.5) Q.43 (0.5)

~~P2Ω = 12.5W 12.5 i= = 2.5 2Ω 2~~

V0 =×= 2 2.5 5V 2.5+= KV 5 0 If we write KCL at node × then = KV0 5 I = ⇒=1 2.5 1 I122 2I I K= = = 0.5 2 52 Write KVL in the outer boundary of network Q.44 (0) 1−()() 1 × I − 2 × I = 0 ⇒= 1 I + 2I The Network is balanced 1 2 12 I1 1 1= I1 + 2 ⇒= 1 2J 11 ⇒ I = = 0.5A 22 Q.45 (a)Wheatstone bridge. ⇒ i = 0Amp = 0.5A ()VV−−() V V Q.49 (11.42) −=i i3 + 23 −jj ALL the branch currents are expressed interval of VA now 100∠∠ 0°− 100 120° 100∠ −− 120° 100∠ 120° −=i + writing KCL at node A 1∠∠− 90° 1 90° i= 173.2∠ − 60°

~~Q.46 (d) By comparing 2 networks on the input side we can say that~~

~~R A RB= 1/ /R AB ⇒= R 1R+ A~~

~~Q.47 (2) 11 1 ⇒VA1 ++ + 2I =+ 5 1 Here the power information 5 5 10 regarding the resistor is given 21 VA − 10 because ⇒V2A  ++ = 6 5 10 10~~

Q.50 (250) Now The instantaneous power of load is 6 RAB =++ 1 0.8 =Ω 3 -60)]+ 5 p(t) = V(∈)i(t) sin ) Q.53 250 [(100sinωt)(10sin(ωt Using KCL at node, we get [(100sinωt)(2sin3ωt)]+[(100T ωt I+= 0.4I 14 = in the above = (5sin5ωt)]Pavg ∫ P() t dt I 10A 0 P=×= 25 10 250W expression→ since, only 1st term will result non zero answer Remaining 2 terms Q.54 (a) Given: The circuit is given below, P()() t = 100sinωt  10 sinωt-60  wiII be 0. → So directly consider P() t =100sinsinωt[] 10(sinωt-60)

~~Pavg= V rms I rms cos()θθ v − 1 100 10 = cos(60) 22 1000 1 The given network consists only = = 250watt 22 reactive element, the equivalent figure can be represented as, Q.51 (d)~~

~~The above circuit can-drawn be re~~

~~Vout =×+() 5 2 10 as shown below, = 20V~~

~~Q.52 3 Consider the following circuit diagram,~~

~~The equivalent impedance()Zeq is given as After rearrangement we get,~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission j4× Zeq i() t= 5sin t Zeq = + j9 j4+ Zeq 1 t Vt() = itdt() +− ∫ ()()j9Zeq j4Z eq 36 C 0 Zeq = Zeq + j4 1 t = 5sin tdt 2 ∫ Zeq−() j9 Z eq += 36 0 1 0

~~2 = − Zeq−()() j12 + j3 Z eq += 36 0 V()() t 5 1 cos t~~

~~()()Zeq− j1 2 + Z eq += j3 0~~

~~Zeq = j1 2 [ Zeq = − j3 is neglected as the whole circuit is a inductive nature]~~

~~Hence, the correct option is (a).~~

~~Q.55 (b)~~

~~Given: The circuit with i00v00()()= = =~~

~~The circular loci of itis given by, The equivalent circuit is () represented as,~~

~~Q.56 5.77 5 Given: s 5 Is() = = 2 1 s1+ 0 s + V= 100 ∠ 0 V s an~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 0 Vbn = 100 ∠− 120 V~~

~~0 Vcn = 100 ∠− 240 V~~

~~IIIIabc++== n 0~~

~~100∠ 0 100 ∠− 120 100 ∠− 240 ++ =0 R j1 0− j1 0~~

~~100∠ 0 +10 ∠− 210 + 10 ∠− 150 = 0 R~~

~~100∠ 0 =8.66 −+ 5j 8.66 −= 5j 17.32 − 10j R~~

~~Equating the real part of the equation~~

~~100 =17.32 R~~

~~100 R= = 5.773 Ω 17.32~~

~~Hence, the value of R is 5.773Ω~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 NETWORK THEOREMS~~

2.1 INTRODUCTION internal impedance theorems that have application throughout the field of This chapter introduces a number of electricity and electronics. Not only can theorems that have application throughout they be used to solve networks such as the field of electricity and electronics. Not encountered in the previous chapter, only can they be used to solve networks but they also provide an opportunity to such as encountered in the previous determine the impact of a particular chapter, but they also provide an source or element on the response of opportunity to determine the impact of a the entire system. In most cases, the particular source or element on the network to be analyzed and the response of the entire system. In most independent (voltage and/or cases, the network to be analyzed and the current) sources, one way to determine mathematics required to find the solution the value of variable (voltage across the are simplified. resistance or current through a resistance) is to use nodal or mesh 2.2 SUPERPOSITION THEOREM current methods as discussed in detailed in chapter 1. Alternative If the circuit has more than one method for any linear network, to independent (voltage and/or current) determine the effect of each sources, one way to determine the value of independent source (whether variable (voltage across the resistance or voltage or current) to the value of current through a resistance) is to use variable (voltage across the resistance nodal or mesh current methods as or current through a resistance) and discussed in detailed in chapter1. then the total effects simple added. This Alternative method for any linear network, response in a particular branch when to determine the effect of each independent all the sources are acting at a time is source (whether voltage or current) to the equal to the linear sum of individual value of variable (voltage across the responses calculated by considering resistance or current through a resistance) one independent source act at a time and then the total effects simple added. while considering one source other This approach is known as the sources are replaced by their internal superposition. impedance of voltage source is zero.

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

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 Don’t disturb the dependent sources involves in solving the current through a present in the network. branch for different values of load → resistance (RL) is also discussed. In many 2.2.1 PROPERTIES OF SUPERPOSITION applications, a network may contain a THEOREM variable component or element while other elements in the circuit are kept constant. If 1) This theorem is applicable only for the solution for current (I) or voltage (V) or linear network with R, L, C transformer power (P) in any component of network is and linear controlled sources as its desired, in such cases the whole circuit elements. need to be analyzed each time with the 2) The presence of dependent sources change in component value. In order to makes the network an active hence avoid such repeated computation, it is super position is applicable to both desirable to introduce a method that will active and as well as passive networks. not have to be repeated for each value of variable component. Such tedious 2.2.2 HOMOGENITY PRINCIPLE computation burden can be avoided provided the fixed part of such networks It is the principle obeyed by all the linear could be converted into a very simple networks. In a linear network if the equivalent circuit that represents either in excitation is multiplied with constant ‘k’ the form of practical voltage source known then response in all the other branches of as Thevenin’s voltage source (VTH = the network is also multiplied with the magnitude of voltage source, RTH = internal same constant ‘k’. 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. In the above figures when 20V source in fig. a) is multiplied by 3 times to get 60V source 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. Let us consider the circuit shown in fig. 8.1(a). Our problem is to find a current 2.3 THEVENIN’S AND NORTON’S THEOREM

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission through RL using different techniques; the following observations are made. Def: Norton’s theorem states that any two • If we use Mesh Analysis then 3 terminal linear networks with current equations need to be solved. sources, voltage sources and resistances • If Nodal Analysis is used then 2 can be replaced by an equivalent circuit equations need to be solved. consisting of a current source called • By superposition theorem we need to Norton’s current (IN) in parallel with find the current through RL considering Thevenin’s resistance (RTH). The value of each source at a time and replacing the current source is the short circuit other sources by their internal current between the two terminals of the impedance. network. Suppose, the value of RL is changed then the above three techniques need to be 2.4 MAXIMUM POWER TRANSFER applied again right from the beginning. THEOREM To avoid all the above problems the circuit contained inside the fence in fig. In an electric circuit, the load receives (1) with two terminals A & B, is electric energy via the supply sources and replaced by the simple equivalent converts that energy into a useful form. The voltage source (as shown in fig. 2) or maximum allowable power received by the current source (as shown in fig. 3). 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. Def: Thevenin’s theorem states that any This theorem is applicable only when the two terminal linear network having a load is a variable otherwise choose the number of voltage and current sources and minimum internal impedance of the source, resistances can be replaced by a simple which results a maximum current through equivalent circuit consisting of a single a fixed load and hence a maximum power voltage source called Thevenin’s voltage dissipation across the load. (VTH) in series with a resistance called Thevenin’s resistance (RTH), where the value of the Thevenin’s voltage source is 2.4.1 UNDER THE VARIABLE LOAD equal to the open circuited voltage across CONDITIONS the two terminals of the network and

~~Thevenin’s resistance is equal to the Case1: Source resistance ‘RS’ is fixed and Load equivalent resistance measured between resistance ‘RL’ is variable the terminals.~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VV22 =SS + 4RSL 4R VS2 P = T 2RS usefulpower VS Efficiency ()η= I = Totalpower RRsL+ 2 The maximum power transfer theorem VS states that, “maximum power is delivered = 4RL 2 from a source to a load when the load VS resistance is equal to the source 2RL resistance”. 1 η= Current in the circuit is 2 VS %η= 50% →I = RRsL+ So, the efficiency of maximum power Power delivered to the load RL is transfer theorem is almost 50% 2 P= I RL (W) → Case2: VR2 ∴ = SL Source impedance Zs is fixed and load P2 (W) ()RRsL+ impedance ZL is variable dP d VR2 = SL dR dR 2 LL()RRsL+ 2 +−2 + VS{}() RR sL() 2RRR LsL() = 4 ()RRsL+ Current in the circuit is dP V When = 0 I = S dR L RSL++ R jX() SL + X 2 Power delivered to the load impedance is ∴+()RRsL −() 2RRR LsL() + = 0 2 P= I RL (w) We get RRsL= 2 Now, P= P| = VRSL max atRLS R = P2 (w) 2 V RL +22 ++ P|= S ()()RSLR XX SL max2 atRRLS=  + ()RRsL 2 → 2 VRSL V RS P = ⇒=S 22 Pmax 2 ()()RRSL+ ++ XX SL ()RRsS+ VV22 Case2.1: SS ⇒=P() W or (W) Only load resistance RL is variable max 4RS 4RL Now differentiating power w.r.t R L & Power delivered by voltage source (VS) = Power absorbed by source resistance ‘RS’ equate to zero +22 ++22 − + Power absorbed by R dp {}()()RSL R X SL X VS VS RL() 2R S 2R L L = PP= + P dRL 222 TRSL R ()()RR+ ++ XX ′ ′ {}SL SL P= IR22 + IR | T s LRLS= R

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission dp RZ= = 0 LS dRL 2 2 2.5 THE TELLEGAN’S THEOREM We get RL Rs++() X SL X Ω Case2.2: Tellegan’s theorem is valid any lumped Only load reactance ‘XL’ is variable network which may be linear or non-linear, Now differentiate power w.r.t & equate passive or active, time–varying or time to zero invariant. It states that, the algebraic sum 22𝐿𝐿 dP ()()R+ R ++ X X() 0 −𝑋𝑋 Vs2 RL(2X + 2X ) = SL SL S Lof all the powers delivered by some 2 dXL +22 ++ elements in the network is equal to the {}()()RRSL XX SL power observed by the remaining elements dP = 0 present is the network. When the current dXL enters of the negative terminal of an element then that element will deliver the We get XXSL+= 0 power otherwise it absorbs the power. The Case2.3: sources can either deliver the power or they can absorb the power whereas the Both Load resistance ‘RL’ & reactance passive, elements will always absorb the ‘XL’ are varied simultaneously power, since the current enters from Here both 2 (i) and 2 (ii) are satisfied positive terminal in the R, L, C elements in simultaneously the presence of sources. i.e. 2.6 RECIPROCITY THEOREM 2 2 RL= Rs ++() X SL X [from 2 (i)] and A linear, passive and a bilateral network, XX0LS+= [from 2 (ii)] the ratio of response to the excitation is Combining above two conditions, we get constant even through the source is 2 RL= Rs0 + & XX0 LS += interchanged from the input terminal to the output terminals. RRLS= & XXLS= −

~~We have, ZLL= R + jX L~~

~~=RSS − jX ()Q RLS= R &X L = − X S~~

ZLS= Z*

~~Case3: Source impedance is fixed and load resistance RL variable~~

~~From fig a & fig b by reciprocity theorem VV V 12= Or, = cons tan t II12 I Proof:~~

~~It is a special case of 2(i) with X0L = 2 2 From 2 i) RL= Rs ++() X SL X~~

~~22 ⇒=RL Rs + Xs~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Milliman’s theorem states that in any~~

~~network, if the voltage sources V12 , V… .V n in series with internal resistances~~

R12 , R… .R n respectively, are in parallel, then these sources may be replaced by a single voltage sources V’ in series with ‘R’ 40 Req = Ω as shown in fig. 7 20 7 I= ×= 7A 40 2 7/2 I= ×= 4 2A 1 43+ From above figures it can be concluded that VG+ VG +…+ V G V' = 11 2 2 n n the current remains 2A even if the 20V GGG+ + +…+ G source is moved from input terminals to 123 n 1111 1 the output terminals. = + + +…. + R' R123 R R R n ' 2.6.1 FEATURES OF RECIPROCITY GGG=12 + +…+ G n THEOREM 11 R' = = + +…+ 1) This theorem is applicable only for the G' G12 G G n linear, passive, bilateral network i.e. Note: In the above case if the polarities of networks with R, L, C and transformer the source V2 are reversed then V2 replaced as its elements, so called reciprocal by V2 in the expression for V’. network. 2) The presence of the dependent sources 2.9 THE DUALITY PRINCIPLE makes the network an active and hence the Reciprocity theorem is not The network and its dual are the same w. r. applicable, so called the non-reciprocal t performance point of view but the network. elements and connections point of view they are not equal. In electrical circuit itself 2.7 SUBSTITUTION THEOREM there are pairs of terms which can be interchanged to get new circuits such pair In a linear network any passive elements of dual terms is given below.` can be equivalently substituted by a ideal Current↔ voltage voltage source or an ideal current source, Open↔ short provided the original passive element or LC↔ the substituted active sources absorbs the RG↔ same power then only all the other branch ZY↔ currents and voltages are kept constant. series↔ parallel KCL↔ KVL Star↔ delta Thevenins↔ Nortons Ri(t)↔ Gv(t) di(t) dv(t) LC↔ dt dt 2.8 MILLIMAN’S THEOREM

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 11 i= 3.125A itdt() ↔ vtdt() 2 CL∫∫ So, the total current I passing through 3 To draw dual of the network, the following resister is steps are to be followed; Ω Ii=12 + i 1) In each loop of a network place a node, I= 2.5 + 3.125 and place an extra node, called the I= 5.625A reference node, outside the network. 2) Draw the lines connecting adjacent Example: nodes passing through each element Determine Thevnin’s and Norton’s and also to the reference node, by equivalent circuit across ‘AB’ for the given placing the dual of each element in the circuit line passing through original elements.

~~Example: Find the current passing through the 3 resistor in the circuit shown in fig by superposition theorem Ω Solution: Let Vth is voltage across terminal AB~~

~~Solution: By superposition theorem Consider 20V voltage source and open By applying nodal we get, V−− 50 V 25 circuit 5A current source We get, TH+= TH 0 10 5~~

~~VTH−+ 50 2V TH −= 50 0~~

~~3VTH = 100~~

~~VTH = 33.3V RTH is the resistance seen into the terminals By KVL, AB. To find RTH the two voltage sources are 20-5(i11 )-3(i )=0 removed & replaced with short circuit~~

~~20=8i1 20 i = 1 8~~

~~i1 = 2.5A Now consider 5A current source and short 10× 5 R= (10 || 5) = = 3.33 Ω circuit 20V voltage source. TH 10+ 5 We get, So, the Thevenin’s and Norton’s equivalent circuits are~~

~~By current division rule, 5 i52 = × Example: 53+~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Determine the values of load resistance when the load resistance drawn maximum power. Also find the value of the maximum power.

~~Example: Verify Tellegen’s Theorem~~

Solution: The source delivers the maximum power when load resistance is equal to the source resistance RL = 25 Ω Solution: While verifying the Tellegen’s theorem 50 50 The current I = = don’t disturb the original network for + + 25 RL 25 25 evaluating the voltages & currents in each I= 1A & every element of the network. The maximum power delivered to the load P= IR2 2 P=() 1 × 25 P= 25W

~~Example: By KVL, 20− 5i −= 5 0 Draw the dual network for the given i= 3A network shown in fig Powers due all the elements of the network are:~~

~~P20v = 20 ×= 3 60W (del)~~

P5Ω = 15 ×= 3 45W (abs) P=×= 5 3 15W (abs) 5v Solution: ∴==Power delivered 60W Power absorbed Place nodes in each loop and one reference i.e. the conservation of power. node outside the circuit joining the nodes through each element & placing the dual of Example: each element we get. Calculate the current I shown in figure using Milliman’s theorem

~~Solution: According to Milliman’s theorem, the two~~

~~The dual of circuit is redrawn as shown voltage sources can be replaced by a single below voltage source in series with resistance shown~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VG+ VG V' = 11 2 2 GG12+ 11  ()10+ () 20  = 25  11 + 25 V' = 12.86V 11 R' = = = 1.43 Ω + 11 GG12 + 25 Therefore, the current passing through the 3 resistor is 12.86 I = Ω 3+ 1.49 I= 2.9A

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)~~

Q.1 In the figure, the switch was closed 1 for a long time before opening at t = R++ Ls − Ls V Cs I1 (s) + a) =  0. The voltage = is s Vx att 0 1 I2 (s) −+Ls R 0 Cs 1 R++ Ls − Ls V I (s) − b) Cs 1 =  s 1 I2 (s) −+Ls R 0 Cs 1 R++ Ls − Ls V I (s) − c) Cs 1 =  s a) 25V b) 50V 1 I2 (s) −Ls R ++ Ls + 0 c) −50V d) 0V Cs [GATE-2002] 1 R++ Ls − Ls V I (s) Statement for linked answer questions: d) Cs 1 =  s The circuit for Q.2 and Q.3 is given Assume 1 I2 (s) −Ls R ++ Ls 0 that the switch S is in position 1 for a long Cs time and thrown to position 2 at t=0 [GATE-2003]

~~+ Q.2 At t=0 the current i1 is Q.4 For the R-L circuit shown in the figure. The input voltage~~

~~Vi () t= u(t) . The current i(t) is~~

~~a) b)~~

~~a) −V b) −V 2R R c) −V d) zero 4R [GATE-2003] c) d) Q.3 IS1 () and IS2 () are the Laplace~~

~~transforms of it1 () and it2 () respectively. The equations for the~~

~~loop currents IS1 () and IS2 () for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t=0, are [GATE-2004]~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.5 The circuit shown in the figure has Q.8 In the figure shown, assume that all -- the capacitors are initially initial current iLL()() 0 =1i 0 =1 A uncharged. If = Volts, through the inductor and an initial Vi () t 10u(t) - then V0 (t) is given by voltage VC () 0 =-1V across the capacitor. For input v() t =u(t) the Laplace transform of the current i(t) for t³0 is

a)8e− 0.004t Volts b)8(1− e−0.004t )Volts c)8u() t Volts S S2+ a) b) d) 8 Volts 2 2 S++ S1 S++ S1 [GATE-2006] S2− S2− c) d) Q.9 In the circuit shown, V is 0 volts at S2 ++ S1 S2 ++ S1 c [GATE-2004] t=0 sec. For t>0 , the capacitor current ic(t)Where t is in seconds, Q.6 A square pulse of 3 volts amplitude is given b 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) 0.50exp(− 25t)mA b) 0.25exp(− 25t)mA c) 0.50exp(− 12.5t)mA a)3V b) -3V d) 0.25exp(− 6.25t)mA c) 4V d) -4V [GATE-2005] [GATE-2007] Q.7 A 2 mH inductor with some initial Q.10 In the following circuit, the switch S current can be represented as is closed at t = 0 . The rate of change di shown below, where s is the Laplace of current (0+ ) is given by Transform variable. The value of dt initial current is

~~RI a) 0 b) SS L ()RRI+ a) 0.5A b) 2.0A c) SS d) c) 1.0A d) 0.0A L [GATE-2006] [GATE∞ -2008]~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.11 The circuit shown in the figure is 31 1 −−tt used to charge the capacitor C a) ee22−  alternately from two current 3  sources as indicated. The switches d) S1 and S2 are mechanically coupled 1 − t 3t 1  3t and connected as follows 2 e cos cos− sin sin  For 2nT≤< t() 2n + 1 T, 223  1 − t  ()n= 0,1, 2 … 22 3t c) e sin  S1 to P1 and S2 to P2 3 2 1 For()()2n+≤<+ 1 T t 2n 2 T − t  22 3t d) e cos ()n= 0,1, 2 … , S1 to Q1 and S2 to Q2 3 2 [GATE-2008]

Q.13 For t > 0, the output voltage Vc(t) is 13 2 −−tt ee22−  3  1 2 − t a) te 2 Assume that the capacitor has zero 3 initial charge. Given that u(t) is a 1 23− t  b) e2 cos t unit step function, the voltage Vc (t)  3 2 across the capacitor is given by 1 ∝ t  n 23 a) ()()−−1 tu t nT c) e2 sin t ∑ 3 2 n0=  ∝ n 23− t  b) +− − 2 ut2()()∑ 1u(tnT) d) e sin t n1= 3 2 ∞ n [GATE-2008] c) tu()() t+−− 2∑ 1 (t nT)u(t − nT) n1= ∞ Q.14 The switch in the circuit shown was d) −+−−()()t nT −− t nT − T on position ‘a’ for a long time and is ∑ 0.5 e 0.5e n1= moved to position ‘6’ at time t=0.The [GATE-2008] current i(t) for t > 0 is given by

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

~~a) 0.2e−125t u() t mA b) 20e−1250t u() t mA c) 0.2e−1250t u() t mA d) 20e−1000t u() t mA Q.12 For t>0 , the voltage across the resistor is [GATE-2009]~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.15 The time domain behavior of an RL Q.18 In the following figure, C1 and C2 circuit is represented by are ideal capacitors. C1 has been di() t −Rt charged to 12 V before the ideal L Ri= V (1 + BeL sint)u(t) For dt 0 switch S is closed at t=0 .The current V i(t) for all t is an initial current of i0() = 0 , the R steady state value of the current is given by V 2V a) i(t) → 0 b) i(t)® 0 R R a) Zero V 2V c)i(t )→+0 (1 B) d) i(t )→+0 (1 B) b) A step function R R c) An exponentially decaying function [GATE-2009] d) An impulse function Q.16 In the circuit shown, the switch S is [GATE-2012] open for a long time and is closed + Q.19 For maximum power transfer at . The current i(t) for ≥ is t=0 t0 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) i() t= 0.5 − 0.125e−1000t A a) Z21= Z b) ZZ21= − −1000t b) i() t= 1.5 − 0.125e A * * c) ZZ2= − 1 d) ZZ2= − 1 c) i() t= 1.5 − 0.125e−1000t A [GATE-2014] d) i() t= 0.375e−1000t A Q.20 In the circuit shown in the figure, [GATE-2010] the value of capacitor C(in mF) Q.17 In the circuit shown below the initial needed to have critically damped charge on the capacitor is 2.5mC, response i(t) is 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 [GATE-2014]

Q.21 In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the a) i() t= 15exp() −× 2 103 t A current I(t) (in mA) for t > 0? b) i() t= 5exp() −× 2 103 t A c) i() t= 10exp() −× 2 103 t A d) i() t=− 5exp() −× 2 103 t A [GATE-2010]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 52 If time t is in seconds, the capacitor a) I() t = (1 − e−t/τ ),τ- msec 33 voltage Vc (in volts) for t>0 is given by 5 −τt/ 2 b) It() (1 -e ),=τ= m sec a)4(1- exp(-t/0.5)) 23 b)10- 6 exp(-t/0.5) 5 c) I() t= (1 − e−τt/ ), =τ 3 mse c c)4(1- exp(-t/0.6)) 2 d) 10- 6exp(-t/0.6) 5 d) I() t= (1 − e−τt/ ),=τ 3 m sec [GATE-2016] 2 [GATE-2014] Q.25 Assume that the circuit in the figure has reached the steady state before Q.22 In the circuit shown in the figure, the value of V0(t) (in Volts) for suddenly burns out, resulting in an t →∞is __. opentime tcircuit. = 0 whenThe thecurrent 3Ω resistori(t) (in ampere) at t = 0+ is _____.

~~[GATE-2014]~~

Q.23 In the circuit shown, the switch SW [GATE-2016] is thrown from position A to position B at time t = 0. The energy Q.26 In the circuit shown, the voltage (in ,J) taken from the 3V source to VIN(t) is described by: charge the 0.1 F capacitor from 0V VIN(t) = to 3V𝜇𝜇 is , < μ , 𝟎𝟎 𝒇𝒇𝒇𝒇𝒇𝒇 𝒕𝒕 𝟎𝟎 � Where𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟏𝟏t𝟏𝟏 is 𝒇𝒇𝒇𝒇𝒇𝒇in seconds.𝒕𝒕 ≥𝟎𝟎 The time (in seconds) at which the current I in the circuit will reach the value 2 Amperes is______.

~~a) 0.3 b) 0.45 c) 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 [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.~~

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

~~The current i(t) (in ampere) at t=0.5 seconds is______[GATE-2017]~~

~~ANSWER KEY:~~

~~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~~

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

~~Q.1 (c) −V ∴=i When switch was closed circuit was 1 2R in steady state, Q.3 (c) When switch is in position 2,~~

− iL (0 )= 2.5A At t=0+ KVL in loop (1), V1 Is.R1() ++ Is. 1() + Is 12() − I(s)sL0 = s sC 1V ⇒I() s R ++ sL = I() s .sL =− 12sC s KVL in loop 2, 1 Is21()()()()− IssLIsRIs. ++ 2 2 = 0 sC ⇒=V IR 1 ⇒−I()() s .sL + I s R + sL + = 0 =2.5 ×= 20 50V 12sC ∴=− Vx 50V 1 (Polarity of V is given reverse of V) R++ sL − sL x sC Is1 ()  1I() s Q.2 (a) −sL R ++ sL 2 − sC At t0= in steady state V − =  s 0

Q.4 (c) V(s) 1 it12()()= it = 0 Is() = = s2++ s(s2) V0− = V c () 1 11 1 + Is() = = − At t0= s(s2)2Ss2++ 1 it() =() 1 −− e 2t 2 At t= 0,i() t = 0 t=∞= ,i() t 0.5 − −− = 1 iR11 V iR 0 t= ,i() t = 0.31 2

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VS() ZS() 02= VSi() ZS 12()()+ Z S~~

~~R 2 Where ZS2 () = , 1+ RCS22~~

~~where R22= 4KΩ ,C = 1μF~~

Graph (c) satisfies all conditions. R1 ZS1 () = , 1+ RCS11 Q.5 (b) where R= 1KΩ , C = 1μF KVL, 11 ∞ RC= RC Ldi(t) 1 2 2 11 vt()()=++ Rit it()dt 36−− 3 dt C ∫ =10 ×× 4 10 sec =× 4 10 sec 0 VS() R 4 Taking L.T on both sides, 0 =2 = VS() R+ R 5 V + i 12 + I(s) c0() v()()() s= RI s + LsI s − LI() 0 ++ 4 sC S V0() S= V i0()()( S ,V t = 0.8V i t) 1 I(s) 1 5 ⇒ =Is()() + sIs −+ 1 − For = = S SS Vi0()()() t 10u t ,V t 8u(t) 2 I(s) +=1 s2 ++ S1 SS Q.9 (a) s2+ Given Is() = −t 2 T S++ S1 Vi=⇒ 0 V C()() t =+− V f V if Ve =51 − e−25t Q.6 (b) () =×−−63 ×= 4 dv RC 0.1 10 10 10 i() t= C. c =100μs C dt As RC is very small, so steady =4 × 10−−6() 5 ×− e 25t ×− 25 state will be reached in 2 sec. = −25t Vc = 3V 0.5e mA V=−=− V 3V 2c Q.10 (b) In the circuit shown, Q.7 (a) ISS .R Iif= 0; I = ; ()RR+ S

~~Req=() RR; + S T =LL = Req ()RR+ S~~

~~ISS .R + di If ()RRS ISS .R | + = = = dtt0= TL L ()RR+ S ∴ L.i()() 0++=⇒= 1mv i 0 0.5A Q.11 (c) Q.8 (c) The waveform of voltage VC(t) is For the given circuit shown below.~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission When S is in position ‘a’,VC(0 )100V After S is moved to b,(for t −0)~~

Vif= 100V;V = 0;R = 5K; ≥ 0.8× 0.2 C= = 0.16μF 1 −t 0.0008 −1250t VC () t= 100.e = 100e Vt() In mathematical form, i() t=c = 20e−1250t u() t mA 5K VC ()()()() t= tut −− 2t Tut − 2t +−2()() t 2T u t − 2T … Q.15 (a) ∞ n The steady state supply across R-L =tu()()()() t +−− 2 1 t nT u t − nT ∑ circuit is V0 n1= V ∴i() ∞= 0 = SVC () S R

Q.12 (b) Q.16 (a) = = − VRC()()() S R.I S 1.() CSV S iLi() 0= 0.75A = I S i∞= 0.5A = I ⇒=VSR () Lf() S2 ++ S1 () 10−3 T=×= 15 10−3 sec +−11 ()S 22 15 = −1000t 2 2 ∴=+i() t 0.5 0.25e 1 3 L S ++ −1000t 2 2 i() t=() 1.5 −− 0.5 0.25e / 2 −t 3 −1000t =2 − = 0.5 − 0.125e VR () t e .cos t 2 Q.17 (a) 13−t  −− e2 .Sin t Q()() 0=−⇒ 2.5mC V 0 3 2  − Q0() −×2.5 10−3 = = = −50V Q.13 (d) C 50× 10−6 1 ≥=− V (S) 1 for t 0, Vx 50V; C =S = 2 V= 100V; Vi (S) 1S++1 S++ S1 R ()S () τ= Req : C = 0.5m sec Vtii()()()=⇒=δt V S 1 ∴=−V t 100 150e−2000t u(t) 1 C () () ∴=VS() ⇒ Vt() CC2 dV 2 3 ∴=i t C. c ++1 C () ()S 2  dt 2 −−6 2000t =×××50 10 150 2000 × e u() t = −2000t −t  15e u() t A 232 = e .Sin t 3 2 Q.18 (d) When the switch in closed at t = 0 Q.14 (b)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission CapacitorC1 will discharge and will 44× C= 2 ⇒10mF get charge since both C1 and C2 are ()40 ideal and there is no –resistance in the circuit charging and discharging time constant will Q.21 (a) zero. Vc(t) = VR2(t) = Vfinal −t Thus current will exist like an − τ Vinitial Ve final impulse function. 2 τ =RC. ⇒× 1036 × 10− equ equ 3 Q.19 (c) 2 Two cascaded sections R=2 KK ||1 ⇒ KΩ equ 3

~~CFequ =1µ 2 τ = msec 3~~

Vinitial = 0 volts 2 10 VVfinal= s. s =5. = volts Z1=Output impedance of first 33 section 10 10 −t VR2(t)= − e τ Z2=Input impedance of second 33 section 10 − t VR2(t) = 1e− τ volt For maximum power tranfer, upto 3  st 1 section is − t VR2 (t) 5 τ Z1 = Z ⇒−it() = = 1e mA 1 R2 2K 3   ZL = Z∗2 Z1 ∗ Q.22 (31.25) Q.20 (10mF⇒) By KVL, di(t) 1 V(t)=++ Ri(t) L i() t dt dt C ∫ Differentiate with respect to time, R.di(t) R di(ti) i(t) 0.0= + += dt2 L dt LC For t→∞ , i.e., at steady state, .d2 i(t) R di(t) i(t) inductor will behave as a shot circuit + += 2 .0 dt L dt LC and hence = 5VB .ix 2 By KCL at node B, −+ , −RR4 10 VB ±−  50 L L LC − += = D = 2iixx 0i x 1,2 2 8 2 250 −RR 1 Vt0 () = 5 i ()()t⇒= Vt0x D1,2 = − 8 2L 2L LC = 31.25volts For critically damped response, 2 R 1 4L Q.23 (c) = ⇒=CF 2L LC R 2 So the capacitor in initially Given, L=4H; uncharged i.e. Vc(0) = 0

~~R=40Ω~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The capacitor will be charged to Q.25 (1) supply voltage 3V when the switch At t = 0 , the circuit is on steady state is→ in i .e. the capacitor is open circuited → So we need to find capacitor so the circuit will be voltageposition B for ∞ time. V3F=+= [ V 2ΩΩ V 3][ - V 3 Ω ] 4V −τt/ Vc(t)= Vc( ∞+ ) [Vc(0)Vc( ∞ )]e + V2F= V 3Ω = 6V at t = 0 =33 − e−t/τ −6 open circuited, the capacitors will τµ=RC =120 ×× 0.1 10 = 12 sec have an ideal voltage whensource 3Ω of is dV iC= c values 4V and 6V so the circuit will c dt be d −t = (0.1×−10−6 ) [3 3e τ ] dt 3× 0.1 −t 0.3 −t = e τ = = e τ τ 12 So instantaneous power of source = V(t)i(t) → 0.3−−tt 0.9 ()()t3= ePττ= e 12 12 at ∞ So the current through 2Ω resistor →=E∫ Pdt () t = 0+ should be 4 = 1A 0 22+ ∞ 0.9 −t = ∫ eτ dt() 0 12 ∞ − t 0.9τ  0.9 = (ττ)e=  120  12  0.9 = 12××10−6 12

Q.24 (d)= 0.9 μJ → At t = 0 switch in position 1 and since the capacitor is open circuited 2 Vc ()0 = 10=4V 2+3 →At t = infinity switch is in position 2 and since the capacitor is open Q.25 0.3405 sec circuited Q.26 8.16A Vc ()∞ = (5) 2 = 10V → Time constant

~~τ = Rth C=() 4+2 0.1=0.6sec~~

~~→ Vc (t) = Vc ()∞ + [ Vc (0) Vc ()∞ e-t/τ ‒ =10+−[] 4 10 e-t/0.6 =10 − 6 e-t/0.6~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

Q.1 In a series RLC circuit at resonance, that of VR .Given that f = 50 HZ , the the magnitude of the voltage inductance of the coil is developed across the capacitor a) Is always zero b) Can never be greater than the input voltage c) Can be greater than the input voltage; however it is 90° out of phase with the input voltage a) 2.14 b)5.30H d) Can be greater that the input c)31.8mH d)1.32H voltage, and is in phase with the [GATE-2003] input voltage [GATE-2001] Q.6 The value of Z in figure which is most appropriate to cause parallel Q.2 In the circuit shown if figure, what resonance at 500Hz is value of C will cause a unity power factor at the ac source?

a)125.00mH b)304.20 μF c) 2.0 μF d) 0.05 μF a) 68.1μF b)165μF [GATE-2004] c) 0.681μF d) 6.81μF Q.7 The R-L-C series circuit shown is [GATE-2003] supplied from a variable frequency voltage source. The admittance- Q.3 A first order, low pass filter is given locus of the R-L-C network at with R= 50Ω & C5= μF.What is the terminals AB for increasing frequency at which the gain of the voltage transfer function of the filter is 0.25? frequency ω is 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 Ω a) b) L= 100 µ H and C= 1F µ . The lower half power frequency of the circuit is a) 30.55 kHz b) 3.055kHz c) 51.92 kHz d) 1.92 kHz [GATE-2003] Q.5 In the circuit of figure, the

~~magnitudes of VL and VC are twice~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission c) d) b) R=50Ω;L=10mH;C=25μF; c) R=Ω= 50 ;L 50mH;C = 5µ F;a d) R=Ω= 50 ;L 5mH;C = 50µ F; [GATE-2014]

~~Q.11 An inductor is connected in parallel [GATE-2007] with a capacitor as shown in the figure. Q.8 The resonant frequency for the given circuit will be~~

~~As the frequency of current i is increased, the impedance (Z) of the network varies as a) 1 rad/s b) 2 rad/s a) c) 3 rad/s d) 4 rad/s [GATE-2008]~~

~~Q.9 Two magnetically uncoupled~~

~~inductive coils have Q factors q1 and~~

~~q2 at the chosen operating frequency. Their respective b)~~

~~resistances are R1 and R 2 . When connected in series, their effective Q factor at the same operating frequency is a) qq+ 12 c ) b) (1 / q12 )+ (1 / q )~~

~~c) ()()qR11++ qR 2 2 / R 1 R 2~~

~~d) ()()qR12++ qR 21 / R 1 R 2 [GATE-2013] d ) Q.10 A series RLC circuit is observed at~~

~~two frequencies. Atω=1 1krad / s , we note that source voltage~~

~~V1 = 100 ∠° 0 V results in current~~

I1 = 0.03 ∠° 31 . At cot = 2 krad/s, the source voltage V= 100 ∠° 0 V 2 [GATE-2015] results in a current I2 = 2 ∠° 0V A. The closest values for R,L,C out of Q.12 In the circuit shown below, the the following options are supply voltage is 10sin (10000 volts. a) R=Ω= 50 ;L 25mH;C =µ 10 F; The peak value of the steady state

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission r, in amperes, is ____. current through the 1Ω resisto~~

~~[GATE-2016]~~

~~Q.13 The circuit below is excited by a sinusoidal source. The value of R, in~~

~~circuit becomes a pure conductance atΩ, all for frequencies, which the is admittance ___. of the~~

~~[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]~~

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

~~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)~~

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

Q.1 (c) Q.4 (b) In a series RLC circuit, at resonance 11 ω = = = and = − 0 −− VL jQV source VC jQV source LC 100×× 1066 10 also for >> 5 Q 1, VC V source =10 r / s Hence option (c) is correct. R 50 ∆=ω = = 50 × 104 r / s L 100× 10−6 Q.2 (a) ∆∆2 1 2 ωω Yj=ωC + ωωlower=+− 0  22 30∠ 40°  =+−jωC 0.0255 j0.0214 2 = 0.0+−(ωC 0255j .0214) 2 5×× 1055 5 10 =+−()105  =Real() Y + jImag(Y) 22  To have a unity power factor at ac 5  source i.e. resonance =10 1 +− 6.25 2.5 condition, Imag() Y= 0 =0.193 × 105 r / s ⇒−ωC 0.0214 = 0 Hence, Q ω= 2π × 50 ω 0.193× 105 f =lower = 0.0214 lower ∴=C = 68.1μF 2π 2π 100π = 3065Hz; 3.055kHz

~~Q.3 (c) Q.5 (c)~~

~~V=+− VR i(V LC V )~~

~~Since VLCL= V &V = 2V R Therefore, the circuit is at resonance and~~

VVR = Quality factor 1 VV2V V jωC 1 =LL = = R = 2 T.F =o = = VV V V1 1j+ ωCR RR i R + As we know jωC ωL 1 Q = 0 ∴=Gain 2 R 1(+ ωCR) 2πf× L ⇒=2 1 0.25 = 5 − 2 ⇒=L 31.8mH 1+()ω ×× 5 106 × 50 On solving Q.6 (d) ω= 15.49 × 103 r / s At resonance, the circuit should be 15.49 in unity power factor ⇒=f = 2.46kHz Hence ‘Z’ should be capacitive 2π Admittance of the parallel circuit ∴

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 11 ⇒= Y0=+= IYm () 0 jLω 1/ jCω For 0<ω<ω0 −1 += 1 Cω0 > ωL Lω ωC 11 ∴ = = Therefore > C 22 IYm [] 0 L×ω 22××()π 500 For ω0 <ω <∞ = 0.05μF 1 < ωL Q.7 (a) ωC Admittance of the series connected Therefore, Im[Y] > 0 RLC On the basis of above analysis, the admittance locus is 1 Y = 1 Rj+−ωL ωC 1 Rj−−ωL = ωC Y 2 2 1 R1+−ωL ωC Q.8 (c) {By rationalization} Input impedance Separating, real and imaginary part 1 zj=ωL + R of admittance. jωC R = R Re[Y] 2 zj=ωL + 2 1 1j+ ωRC R +−ωL ωC For any value of ω, the real part of always positive. 1 When ωL= or ωC 1 1j− ω 1 ∴=z j0.1ω + × At ω0 = (resonance) LC 1j+−ω 1 jω 1j− ω 1 = + Re[] Y = (maximum value) j0.1ω 2 R 1j+ ω 1 1 ω −− =+− ωL 22j 0.1ω = ωC 1++ω 1ω IYm () 2 2 1 At resonance, imaginary part must R +−ωL ωC be zero. ω 1 0.1ω0−= − ωL 1+ ω2 ωC = 1 2 0.1 = 2 1 2 R +−ωL 1+ ω ωC ω2 += 1 10 1 2 At ω0 = (resonance) ∴=ω9 LC ∴=ω 3rad / sec Imaginary part of zero

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.9 (c) Z= ZLC / /Z ωL ωL QQ=12 = = 1 12 jωL × RR22 j Z = ωC After series connection 1 j + ωL()12+ L ωL Q = jωC + RR12 j Z = ωL Q1R+ Q R 2 Q = 1 22 ()1+− ( 1)ω LC RR+ 12 ωL Zj= − 2 Q.10 (b) 1 ω LC Given V= 100 0 °= V;I 0.03 31 °at 11 Q.12 (1) ω1 = 1000r / sec

~~V= 100 0°=° V;I12 2 0 at~~

ω2 = 2000r / sec ϑ 100 i.e R=2 = ⇒ 50Ω I22 v 100 00 ZR=1 ⇒ ⇒+j(XX) − I 0.03 310 Lc 1 this frequency are − −1 XXLC W =1000, the various −impedancej at −3 Φ =31 °⇒ tan  → Zµ || Z= (j1000 ×× 4 10 ) R 250 f 4mH 1000×× 250 10−6 XX− =()() −j4 || j4 =∞= open circuit ⇒tan 31 °=LC R − j −  → ××3 ZZ24f|| 500 mH −6 (j1000 500 10 ) 1 1000×× 250 10 ω − 1L =( − j500) || (j500) =∞= open circuit ω1c tan 31°= Since both LC pair parallel R combination becomes open then the 1 circuit can be redrawn as ⇒−= ω1L 0.600 × 50 ω1c 10sin sin1000t I1Ω = = sin sin1000t 1 415++ ⇒−=ω1 30.0L4q → I1Ω = 1A ω1c 1 → So peak value of ω2L0−=….. (2) ω2c

~~ωω12=1000r / sec; = 2000r / sec~~

~~Q.11 From(b) 1 and 2, C = 25μF L=10mH~~

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

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let first get Y expression in terms of L, C then by equalizing imaginary part we will get the answer. 11 Y = + R+ jLω 1 R − =⇒=∞ ωC IL 0 2ph j jωL  -j  R + R− jLω    = + ωC XXLC 3 ωc  22 2 Z=ph = R+ (ω L) 2 1 X +X jωL -j R +  LC - ωC 3 ωc ⇒= ωL 1 Im g Yeq 0 = b ωc 1 3 ωL 314 = ⇒ = ωC −3 22 2 10×× 10 C R+ (ω L) 2 1 R +  ωC Q.15 (b)C=3.04mF 2 22112 1 Given: ()ωωLR+=+ L R() ω L ⇒ Cross multiplyingωωCC ω C (i) Number of nodes = 8 (ii) Number of independent loop = 5 In graph theory, number of 2 1 ⇒−RLω independent loop represents ωC number of links =−+bn1 11   = ωL0 −=  ωωCC   Hence,

~~2 L1   ⇒−RL0ω − = − += CC ω  bn15 Now by looking into above equation b815−+= we can say that if~~

~~2 L = R − then it will have no b 12 C depending on frequency for Number of branch = 12 L R 2 = Hence, the correct option is (b). C resonanceSo ⇒ L 0.02 R= = = 10 2 = 14.14Ω C 100× 10−6~~

~~Q.14 (3.04)~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(IN)~~

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

~~The power transferred from S to L would be maximum when RL is a) 0W b) 1.0W a) 0Ω b) 0.6Ω c) 2.5W d) 3.0W c) 0.8Ω d) 2Ω [GATE-2008] [GATE-2009]~~

~~Q.2 The current I supplied by the dc Q.5 In the dc circuit shown in the voltage source in the circuit adjoining figure, the node voltage~~

~~shown below is V2 at steady state is~~

a) 0V b) 1V a) 0A b) 0.5A c) 2V d) 3V c)1A d) 2A [GATE-2010] [GATE-2008] Q.6 A 100Ω , 1W resistor and a 800Ω , 2 Q.3 In the circuit shown below, the W resistor are connected in series. current through the inductor is 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)50V c) 45V d) 40V [GATE-2010]

2 −1 a) A b) A Q.7 The impedance looking into nodes 1j+ 1j+ 1 and 2 in the given circuit is 1 c) A d) 0A 1j+ [GATE-2012] Q.4 The source network S is connected to the load network Las shown by dashed lines.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a)50Ω b)100Ω 0.5s+ 1 3s+ 6 a) b) c)5kΩ d)10.1kΩ s1+ s2+ [GATE-2012] s2+ s1+ c) d) s1+ s2+ Q.8 If −= then − is VAB V 6V, VVCD [GATE-2013]

Q.12 Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection a) −5V b) 2V are scaled by a factor, the elements c)3V d) 6V of the corresponding star equivalent [GATE-2012] will be scaled by a factor of

~~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) k2 b) k c) 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 a) 0.8Ω b) 1.4Ω [GATE-2012] V= 100V is applied across WX c) 2Ω d) 2.8Ω WX1 to get an open circuit voltage V Q.10 A source v() t= Vcos100πt has an YZ1 S across YZ .Next, an ac voltage internal impedance of (4+ j3)Ω . If V= 100V is applied across YZ to a purely resistive load connected YZ2 get an open circuit voltage to this source has to extract the VWX2 maximum power out of the source, across WX. Then, VYZ1 /V WX1 , its value in Ω should be VWX2 /V YZ2 are respectively. a)3 b)4 c)5 d)7 [GATE-2012]

~~V (S) Q.11 The transfer function 2 of the V1 (S) a)125 /100and80 /100 circuit shown below is b100 /100and80 /100 ) c)100 /100and100 /100 d)80 /100and80 /100 [GATE-2013]~~

~~Q.14 In the circuit shown below, if the~~

~~source voltage VS = 100 ∠ 53.13° V~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission then The venin’s equivalent voltage in volts as seen by the load~~

~~resistance R L is~~

~~a) 0 b) RL - Ri c) RL d) RL + Ri [GATE-2015]~~

Q.18 Consider the circuits shown in a)100∠ 90° b) 800∠ 0° the figure. The magnitude of the c)800∠ 90° d)100∠ 60° ratio of the currents, i.e., |I1/I2|, is [GATE-2013] ______. Q.15 Time domain expressions for the voltage v1(t) and v2 are given as o v1(t) =Vm sin (10t −130 ) & o [GATE-2015] v2m() t V cos(10t −130 ) Which one of the following statements is The current in amperes through the TRUE? Q.19 resistor R in the circuit shown in a) v12() t leadv by130° the figure is_____ A.

~~b)v12() t lagsv (t)by130°~~

~~c)v12()() t lagsv t by− 130°~~

d) v12()() t lagsv t by− 130° [GATE-2014] [GATE-2015] Q.16 The circuit shown in the figure contains a dependent current Q.20 Three currents i1, i2 and i3 meet at source between A and B terminals. a node as shown in the figure The Thevenin’s equivalent below. If i1 =3cos( ω t)ampere, i2 = 4sin ( ωt) ampere and i3= I3 cos ( terminals C and D is ______. ω t+ )ampere, the value of I3 in resistance in kΩ between the ampere is ______. 𝜃𝜃

~~[GATE-2016]~~

[GATE-2014] Q.21 The current IX in the circuit given below in milliampere is 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. [GATE-2016]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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-2018]~~

~~Q.25 The Thevenin equivalent circuit representation across terminals p- q of the circuit shown in the figure is~~

~~[GATE-2017]~~

~~Q.23 In the given circuit, the mesh~~

~~currents I12, I and I 3 are~~

~~a) 1 V source in series with 150kΩ~~

~~b) 1 V source in parallel with 100kΩ~~

~~c) 2 V source in series with 150kΩ~~

~~a) I12=1, A I = 2 A and I 3 = 3 A d) 2 V source in parallel with~~

~~b) I12=2, A I = 3 A and I 3 = 4 A 200kΩ~~

c) I12=3, A I = 4 A and I 3 = 5 A [GATE-2018] Q.26 A series R-C circuit is excited by a d) I=4, A I = 5 A and I = 6 A 12 3 1∠ 0Vsinusoidal ac voltage source. [GATE-2018] The locus diagram of the phasor Q.24 In the figure, an RLC load is current I=() x + iy A , when C is supplied by a 230 V, 50 Hz single varied, while keeping R fixed, is phase source. The magnitude of the reactive power (in VAr) a) supplied by the source is ____.

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

~~[GATE-2018]~~

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

~~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~~

~~EXPLANATIONS~~

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

By applying current division rule in upper part of the circuit From Fig.3, I1= A, power supplied 11 I= ×∠ 10 = by 3 V d.c source L 1j++ 1j =P = V1 I =×= 3 1 3W Q.4 (c) Q.2 (a) The circuit is shown in Fig.

~~The circuit is shown in Fig . Voltage across Ω = 7 1 1V I = ∴=I1 1A 2R+ L Applying KCL at node ,P Power supplied by the 10V source,~~

~~I1+= I1 Or I+= 1 1, 10× 7 70 PS = = ∴=I0 2R++LL 2R Power dissipated in 2Ω resistance, Q.3 (c) 98 =2 ×= Assume current as shown, PI22 2 ()2R+ L Power transferred to the network,~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 70 98 11 PRtL() =−= P S2 P − = = 2R+ 2 IA1 L ()2R+ L 100 10 dPt would be maximum , if = 21 Pt 0 Similarly IA2 = = dR L 800 20 70− 2 If these two resistors are − −×98 = 0 23 connected in series ()2R++LL() 2R 196 196 98 =70, 2 + R L = = 2+ RL 70 35 98 28 R= −= 2 = 0.8Ω L 35 35 Q.5 (b) The given circuit is shown in Fig. 1 1 Then maximum value of I = At steady state i.e., as t →∞ , 20 capacitor behaves as open circuit. ∴=V 1(100 + 800) The circuit at steady state is 1 shown in Fig .2 =()900 = 45Volts 20

~~Q.7 (a)~~

~~Apply voltage division across~~

R12= 2ΩΩ and R = 1 Node voltage After connecting voltage source of R 1 V V9=×=×2 9 1 RR+ 3 12 V= V ⇒() 10K()() −= i 100 I + 99i + i ; = 3V 1 2 b bb −10000i = 100I +× 100 i V1 is divided between b b =100I + 10000i C12= 10μF and C = 20μF b Apply, again voltage division −20000ibb= 100I ⇒ i across −−100  I  and =I =  C1 C2 20000  200  C1 10 V=× 3 =×= 3 1V V= 100[] I ++ 99ibb i 2 C+ C 30 12  −I =100 I + 100= 50I 200 Q.6 (c)  Resistor1: 100Ω ,1W V 50I Rth = = = 50Ω Resistor1: 800Ω ,2W II Maximum current that resistor can withstand Q.8 (a)

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission From the given circuit,~~

VAB−= V 6V V 2 = ? VV− AB V1 IAB= = 3A = I DC 2 + 1 VR CS ()RC S+ 1 C KCL at ‘D’ gives 21= = 12 VV− V1 RC S CC+ CD++= − =− 1R ++1 () 121 2 3 0, VCD V 5V CS 1 1 CS2 Substituting the values we get Q.9 (a) V S1+ 2 = Power transferred from circuit A V S2+ to circuit 1 7  6+ 10R  B= VI =    Q.12 (b) R2++  R2  Consider Rabc= x;R = k;R = k 42+ 70R 2 = RRab+ k ()R2+ RStar = = Rabc++ R R 3k 10− 3 7 I = = Rk∝ 2R++ 2R Star 7R V=+=+ 3 IR 3 Q.13 (b) 2R+ 6+ 10R =  2R+ 2 dP ()()()()R+ 2 70 −+ 42 70R 2 R + 2 = 4 dR ()R2+

~~VW1××= 100 ⇒= V 2 turns ratio × V W1 = 125~~

~~VYZ1=×= 0.8 V 2 100V~~

~~When VYZ2 = 100V~~

2 70()()() R+=+ 2 42 70R 2 R + 2 ⇒5() R += 2 2(3 + 5R) ⇒5R +=+ 10 6 10R ⇒=4 5R ⇒=R 0.8Ω Thevenin’s circuit seen by 2-2’ will Q.10 (c) be as follows 22 Vth = 100V And Rth = 0.2 || 0.8 RL= Z th = 43 += 5Ω negligible

~~∴=V' 100V Q.11 (d) 22−~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ∴VW2× 10 Is.c = ⇒ 1mA 100V× 1 10K =×=' V22− turns ratio V 20 1.25 ∴=Th ⇒ ⇒Ω RTh −3 20k = 80V IS.C 1× 10

Q.14 (c) Q.17 (a) VTH= 10V L1 When load is constant, we should 4 see for what value of resistance ∠ −1 Vc 100 53.13 8 current will be maximum and . VL1 == −× tan j4 3+ j4 3 P= max RL = ∠ VL1 80 90° Q.18 (1) VTH = 800 ∠ 90° By reciprocity theorem, I1 = I2 I So 1 =1 Q.15 (a) I2 Given v1m() t= V sin(10t − 130° )

~~v2m() t= V cos(10t − 10° )-v2 Q.19 (1) Using supermesh analysis v t= V cos(10t −− 130° 90° ) 1m() Mesh 1, 3 form super mesh = − )- v1m() t V cos(10t 220° v1 2I12−+ I 2I 3 = 0~~

ve in A.C.W ⇒2112 −+ I 2I 3 = 0 … (1) ve in C.W Writing KCL at Q 𝜙𝜙 →+ v12()() t leadsv t by130° II−=− 1 ….(2) 𝜙𝜙→+ 13 Writing KCL on mesh 2 Q.16 (20) 2I213−+ I I = 1 ….(3) When dependent source is Solution Equipment 1,2,3 present; to find RTh ,VTh &ISC are I= 0, I = I = 1A required. 1 23 VTh (= VO.C ) : Current through R is I3 = 1A Note: Strictly saying it is an ambiguous question as the direction of current is not mentioned, so it could be –1A as well. By Nodal Analysis, V− 10 V xx−=0 Q.20 (5) 5*1034 10 By KCL it1()()() = i 23 t + i t ⇒=Vx 20 Volts i()()() t = i t –it i.e VTh =( VO.C) =20 Votls 312 I n n s.c By phasor I3 = I1 - I 2

~~=[]3 < 0 –[4 <− 90 = 553.13~~

~~i3 () t = 5cos(t + 53.13) So by comparison I3 = 5.~~

~~Q.21 (10)~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission II31= + 2~~

~~⇒−=I31 I2...... ( iii ) By solving the above equations, or by verifying each options to satisfy the above equations, we have By kVL, 1 =[100( Ix - x] = = 103 I121, AI 2 A~~

1000= 200Ix- 1000 10)+ 100I and I3 = 3 A Ix= 10mA Q.24 67.36 Q.22 2 230∠° 0 I = By KCL at node 2 1 162.6+ j 162.6 V V2+− 4V R1 V =2 + 1 230∠° 0 2I2 ..(i) = =1 ∠− 45 ° 1/ 3 0.5 230∠° 45 = − VR1 1....(ii) 230∠° 0 I2 = V1−− 4V R2 V 230∠− 90 ° I= 1 ....(iii) 2 0.5 =∠°1 90 Substituting equation (ii) and (iii) III=12 + in equation (i) =1 ∠− 45 °+ 1 ∠ 90 ° = V2 2V 11 =−+jj 22 11 Q.23 (a) =+−j 1 22 1 = + j0.293 2 =0.765 ∠ 22.51 All the given information are in the RMS form by default

~~Qsource= |VI | source | | source sin[θθ V − I ] =××230 0.765 sin(0-22.51) = 67.36VAR~~

~~Q.25 (c) Thevenin voltage is the open Writing the KVL equations, circuit voltage across the defined In outer most loop through 5V terminal PQ −25II13 +− = 0~~

~~⇒2II13 += 5.....( i ) In the loop containing current source~~

~~−−2IIIIII112323 6( −− ) 1( −− ) 1( ) = 0~~

~~⇒−266I1 − I 1 + IIII 2323 − + − = 0~~

~~⇒−8I123 + 7 I − 2 I = 0...... ( ii ) By writing KCL at node X, we have~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission We want the locus of I, so its expression is~~

~~10∠ I = 1 R + jCω~~

~~11−1  = ∠ tan  2 ωRC 2 1 R +  ωC~~

100 In the above equation C is variable Vth = ×=4 2V 100+ 100 When C = 0, I= 0 ∠ 90° Thevenin resistance is obtained by making all the independent source 1 When C = ∞=∠ , I 0° value as 0V, i.e., independent ideal R voltage source (4V in this case) should be replaced by short circuit.

~~So the locus diagram should start Rkth =+=Ω100[] 100P 100 150 with 900 axis with magnitude 0, VV= 2 th and should end with 00 axis with magnitude 1/R.~~

~~Q.26 (a) Q.27 1~~

~~Case-1: VI22= 0 V , = − 6 A~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VV26ΩΩ=()QP26 ΩΩ V −12 I 1 =6Ω = = −2A 3 66~~

~~Case-2: VI11= 0 V , = 6 A~~

~~VV63ΩΩ=()26 ΩΩ V 18 I 11 =6Ω = = 3A 3 66 By superposition, by referring above 2 circuits we can say 1 11 I3= II 33 + =−()()2 +− 3 = 1A~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3 TRANSIENTS~~

3.1 INTRODUCTION Voltage and current are constant Voltage and current are time dependent. In chapters 1 and 2 (Basics and Theorems) From the above example it is very much we have considered dc resistive networks clear that it is very easy to solve the (networks with dc sources and resistances) equation for purely resistive circuit as they in which voltages and currents were are simple linear equations whereas the independent of time (constant w .r .t time). circuits containing inductors and Whenever inductor (L) and capacitor (c) capacitors have differential terms and are included in the circuit the analysis voltage and currents will be a function of which was done for resistive circuits is time. Hence, they have to be solved as invalid. differential equations. The voltage and current relationship for inductor and capacitor are given below: 3.2 STEADY STATE AND TRANSIENT Inductor: RESPONSE di() t 1 t In a network containing energy storing Vt() = LL ⇒= it() Vtdti0()() + LLdt L ∫ LLelements (i.e. L or c) when the excitation 0 (or input) is changed, the current and Capacitor: voltages change from one state to another dV() t 1 t it() = Cc ⇒= Vt() itdtV0()() + state. The behavior of the voltage or c ccc∫ current when it is changed from one state dt c 0 From the above equations we can observe to another is called the transient state. that the voltage and current equations have e.g:- Consider a circuit which contains a differential terms which make the analysis switch. complicated. e.g.: Let’s consider a simple example.

~~Resistive Network: - Network with inductor and capacitor:-~~

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 KVL⇒= V I R + I R sources or sources with same frequency, ) 1 11 12 connected for long time is said to be in t di (t) 1 steady state. Current and voltage do notω KVL⇒= v L2 + i() t dt 22∫ change with time in steady state. Ideally dt c 0 after infinite amount of time and practically

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission after 5 time constants circuit enter into steady state.~~

3.2.1 SOLUTIONS OF DIFFERENTIAL t0= ⇒ Time of closing the switch EQUATION t0=− ⇒ Time just before the switch is The complete solutions of a differential closed (steady state) + equation is given by t0= ⇒ Time just after closing the switch Complete solution = complementary (transient state) function + Particular integral (or) = Natural t →∞⇒Infinite amount of time after response + Forced response (or) = closing the switch (steady state) Transient response + steady state response

Example: Find the solution of the differential equation given below d2 y dy +5 += 4y 1 dt2 dt Solution: Complementary function ()D2 ++ 5D 4 y = 0 m2 + 5m += 4 0 ()()m4m1+ += 0 m=−− 4, 1 Complementary function, 3.2.3 THE BEHAVIOR OF INDUCTOR −−4t t - + y= ce12 + ce AND CAPACITOR AT t=0 , AT t=0 AND Particular integral AS t →∞ 11 y1e= = ot D22++ 5D 4 D ++ 5D 4 1 =(D = a = 0) 4 1 Complete solution =ce−−4t ++ ce t 124

Transient Steady state Response Response 3.2.4 THE INDUCTOR CURRENT AND 3.2.2 TERMINOLOGY IN TRANSIENTS CAPACITOR VOLTAGES AT t=0- AND AT Consider the same circuit with a switch as t=0+ INSTANTS considered earlier t 1 = L: iLL() t∫ V() t dt L −∞ − 110t = + ∫∫VLL() t dt V() t dt LL−∞ 0−

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 t In the figure shown above the capacitor C =− + iLL() 0∫ V() t dt has an initial voltage 0. The initial voltage L − 0 can be replaced by a voltage source in At t0= + above equation becomes series with the capacitor.𝑉𝑉 + 1 0 +−= + i0LL()() i0∫ Vtdt L() Note: Whatever we have discussed so far L 0− forms the basis to solve the transient +− problems. Once the student is well i0LL()()= i0 understood with these basics the transient + − (Q 0 and 0 are nearly equal) problems are very easy to solve. Hence the inductor current can’t change instantaneously from 0 to 0 3.3 DC TRANSIENTS 1 t − + C: Vcc (t)= i() t dt In dc transients we would have dc sources. c ∫ −∞ Whenever there is sudden application of − 110t voltages or currents on the circuit (due to = + ∫∫icc() t dt i() t dt opening or closing the switch) circuit cc− −∞ 0 would be in transient state. We will t − 1 consider the following cases in dc =V 0 + i() t dt cc() ∫ transients c 0− 1. Source free circuits At = + above equation can be written as t0 • Source free RL circuit 1 t +−= + • Source free RC circuit V0cc()() V0∫ i c() tdt − • Source free RLC circuit c 0 2. RL and RC circuits with sources. +−= V0cc()() V0 3. RLC circuits with sources, only initial (Q 0−+ and 0 are nearly same) condition (t= 0+ ) and final condition Hence, the capacitor voltage can’t change t →∞will be found. instantaneously from 0−+ to 0 4. RLC circuits with sources, Laplace Transform Approach of solving the 3.2.4 THE EQUIVALENT CIRCUITS transient problems.

~~L: Inductor with initial current I0 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 In the figure shown above the inductor L the memory elements (i.e. L or c) will store constrains an initial current Io. The initial the maximum energy. At t=0- the circuit current can be replaced by a current source will be in steady state and inductor will be in parallel to the inductor. short circuited, capacitor will be open C: Capacitor with initial voltage V0 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

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission and capacitor would be dissipated across the resistors present in the circuit.~~

~~a) Source free RL circuit: Consider a simple RL circuit as shown in figure~~

~~Applying KVL in loop (I), we get di −−Ri LL = 0 L dt di R L +=i0 dt L L Our aim is to find the current across the R D+= i0L inductor (iL (t)) and voltage across the L~~

inductor (VL (t)) after switch is closed. Solving the above differential equation, we get R Step1: Finding the initial current − t L - iL () t= ce (iL (0 )) across the inductor To find the initial current across At t= 0,iLs = I - 0 inductor consider the circuit at t=0 . Is = ce The switch will remain closed at t=0 cI= ∴ s circuit is in steady state hence inductor R − t will be short circuited. L ∴=iLs() t Ie for t0≥ L τ = sec is called the time constant R of RL circuits −t τ ∴=iLs() t Ie for t0≥

diL (t) VtL () = L − dt i0Ls() = ISince the current always R − t e L −R chooses law resistance path, = LIs e  We have already proved that L −R −+ t = L i0LL()() i0 = −RIs e for t0≥ ∴==i0−+ I i0 Graphically the current and voltage can L()() sL be shown as The energy in the inductor is given by

~~−+112 2 ELL() 0= Li() 0 = LI s 22 Hence the inductor has stored the energy at t0= −~~

~~Step2: Finding the current~~

~~iL () t fort≥ 0~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −+ v0= V = v0 cc( ) S ( ) 1 E (0−− )= cv2 (0 ) cc2 1 2 = cV 2 s Here, the capacitor has stored the energy at t0= − −t τ −1 At t=τ ,iLs() τ= I e = I s e = 0.368I s Step2: Find vcc() t andi (t) for t0≥ −2 at t= 2 τ ,iLs() 2 τ= I e = 0.135I s For t0≥ we can’t short circuit or open −3 circuit the capacity because t0≥ at t=τ 3 ,iLs() 3 τ= I e = 0.0498I s includes t0= + (transient state) as well at t= 4 τ ,i() 4 τ= I e−4 = 0.0183I Ls s as t →∞(steady state) −5 at t=τ 5 ,iLs() 5 τ= I e = 0.0067I s 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. By KCL ⇒

~~ii0RC+= b) Source free RC circuit: V dV cc+=c0 R dt Consider a simple RC circuit as shown in figure 1 D+= V0c RC −t Rc Vc () t= ke for t0≥ Using initial condition~~

~~Vcs() 0V~~

V0cs() = k = V −t Rc ∴=Vcs() t Ve Again we have to find ic (t) and Vc (t)(t) after switch is closed. ττ = RCsecsec is called the time constant of RC circuits −t Step1: Finding the initial voltage V() t= Ve τ for t0≥ − cs (Vc (0 )) across the capacitor −t dVc (t) τ −1 Capacitor will be open circuited as the ics() t= c = cV e  dt τ circuit is in steady state V −t it() = − s eτ for t0≥ c R

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −+20 iLL() 0= = 20mA = i (0 ) 1k −+ Vcc() 0= 20v = V (0 )~~

~~Step2: Finding iL (t)and Vc (t) for t0≥~~

~~c) Source free RLC circuit:~~

Let understand this topic with the help In the above circuit the RL and RC are of an example. two independent circuits since there is a short circuit in parallel. The circuit is a source free circuit Time constant of RL ckt,

R 1K 4 τ=L = =10 sec Here, we have to determine itL () and L 0.1 Time c constant of RC ckt, Vtc () for t0≥ 39− τ=c RC = 10 × 10 × 200 × 10 Step1: Finding initial conditions =2 × 10−3 sec − − iL (0 ) and Vc (0 ) We know that for a source free RL circuit −t −t Number of initial condition to be − τ 4 Note: i() t= i 0 eL = 20e10 (mA) found = Number of inductors and LL() capacitors present in the network. Similarly for source free RC ckt −t −t In this problem there is one inductor and − τ −3 V() t= V 0 ec = 20e 2x10 one capacitor hence two initial conditions cc() are required = 20e−500t v The input 20u()− t is as shown below 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. 20;&t≤ 0 20u()−= t   0;&t> 0 3.3.2 RL AND RC CIRCUITS WITH SOURCES At t0= − the above circuit can be drawn as 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.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission • In a circuit if there are many resistors, for t0≥ inductors and sources (i.e. RL circuit L Where τ= sec for RL circuits with sources) then current across R inductor is given by Capacitor voltage − + t −t it()()= i ∞+ i0 − i() ∞ eτ = ∞+− − ∞ τ LL L() L Vtcc()() V V0 c() V c() e L Where, τ= sec is the time constant for t0≥ R Where, τ=RCsec for RC circuit. of RL circuits • In a circuit if there are many resistors 3) Also if some value of current or voltage capacitors and sources (i.e. RC circuit is desired at t=0+ (transient state) then with sources) then voltage across draw the circuit at t=0+. Find the capacitor is given by desired value by the application of KCL, −t KVL, Nodal or Mesh analysis = ∞++ − ∞ τ Vtcc()() V V0 c() V c() e Where, τ=RC sec is the time constant Example: of RC circuit For the circuit shown in figure, the switch is in position 1 for a long time and it is 3.3.3 PROCEDURE TO SOLVE RL AND RC moved to position 2 at t0= Determine, CIRCUITS (WITH SOURCE OR SOURCE ++ i0,v0,i(t)LL()() for t0≥ FREE)

1) Draw the circuit for t0= − Find the current across inductors and voltage across capacitors at t0= − 2) Draw the circuit for t0≥ Solution: − case1: if the source doesn’t affect the Step1) Finding initial condition (t= 0 ) inductor current (in RL circuit) or capacitor voltage (in RC circuit) then the circuit is source free. In such case, −t − τ Inductor current, itLL() = i0e() for t0≥ Note: - L Inductor is short circuited as the circuit is Where, τ= sec for RL circuit. R in steady state at t0= − Applying current −t − τ division principle Capacitor voltage Vtcc() = V0() e for −+8 i0LL() = ×= 54Ai0 = () t0≥ 82+ Where, τ=RCsec for RC circuit. Step 2) Circuit for t0≥ case2:If the source affects the circuit then it is a circuit with source and in such circuits, Inductors current −t = ∞+− − ∞ τ itLL()() i i0 L() i L() e

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Step3) since there is no source connected Solution: The input 20u() t can be to the inductor for t0≥ hence it is a source shown as free circuit. The current across inductor in a source free circuit is given by −t τ iLL()() t= i 0e for t0≥ L 33 τ= = = sec Req 2+ 12 14 20v,for & t> 0 (Req is obtained by finding equivalent 20u() t =  resistance across L)  0v,for & t< 0 −t − 3 −14t Step1) Finding initial condition (at t= 0 ) 14 3 ∴=iL () t 4e = 4e for t0≥ Also we can find ( ) using the relation di (t) Vt() = LL 𝐿𝐿 L dt 𝑉𝑉 𝑡𝑡 −14t −14 =3 × 4e 3  3 − −14t Since the value of source is at t0= hence V() t= − 56e 3 for t0≥ −+ L V0cc()()= 0vV0 = Step2) Circuit for t0≥ Step 4) Circuit at t0= + To find the value of v(0+ ) across the 12Ω resistor we must go into t0= + circuit is in transient state. Inductor will become open circuited. Step3) Since the source is connected to the circuit for t0≥ hence to find the value

of Vc (t) we need to employ the following formula −t τ Vtcc()()()()= V ∞+ V0 cc − V ∞ e for t0≥ Current enters in a resistor from positive We need to find the value of Vc ()∞ and τ terminal. Hence to use above formula. v() 0+ =−× 12 4 Step4) Finding Vc ()∞ + v() 0= − 48v As t →∞ circuit is in steady state and capacitor becomes open circuited Example:

~~Determine the values of Vtc () and itc () for t0≥ for the circuit shown in figure~~

~~Using voltage division principle~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 V()∞= × 20 = 10v c 22+ Finding time constant τ=Req =()2 || 2 +× 1 1 𝐶𝐶 =+×()11 1 Solution: = ×= − 2 1 2sec Finding initial condition (at t0= ) − − We need to find i0L () and V0c () in this case

ckt to find time constant. Voltage source is short circuited. Substituting in the formula we get −t As the source is not connected to the circuit 2 Vc () t= 10 +− [0 10]e −+ ∴ i0LL()()= 0Ai0 = −t 2 −+ Vc () t= 10 − 10e for t0≥ V0cc()()= 0VV0 = dV() t −t + it() = cc ⇒= it() 5e2 for t0≥ Circuit at t=0 ccdt

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

• In these circuit we won’t be able determine the time constant of the The current through the battery is 10 circuit. Also in these circuits you would i(0+ )= A = 10A be asked to find the initial conditions 1 at(t= 0+ ) where the circuit would be in Circuit at t = ∞ transient state or the final conditions (as t →∞) ,where the circuit would be in steady state. • These circuit would be purely resistive as the inductors and capacitors are either short circuited or open circuited in transient state t0= + and steady state The current through the battery is (as t →∞) 10 i(∞= ) A = 10A Example: 1 Determine the current through the battery + 3.3.5 RLC CIRCUITS WITH SOURCES at t0= and as t →∞. USING LAPLACE TRANSFORM APPROACH

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 t0≥ which is possible only using Laplace Transform Approach (LTA). Nodal at V (s) ⇒ Let’s consider a problem c 5 − Vsc () Vs() Vs() Example: 5 ++=cc0 The switch S in the circuit shown in figure 2s+ 4 1 2 s below is open for a long time. At t=0, S is 5 closed connecting the voltage source to the Vsc() −+()()() 2s4Vs + cc + s(s2)Vs + circuit for t0≥ , obtain the voltage V 5 = 0 c (2s+ 4) across the capacitor. 5 1+ 2s ++ 4 s2 + 2s V() s = c s 55 Vsc () = = (s22++ 4s 5) s(s ++ 4s 5) A Bs+ C Vsc () = + S s2 ++ 4s 5 Solution:- − 5 Finding initial conditions (at t0= ) A= sVc() s | s0= = s | s0 = = 1 s(s2 ++ 4s 5) 1 Bs+ C Vsc () = + S s2 ++ 4s 5 5 s2 + 45 ++ 5() Bs + c s = s(s22++ 45 5) s(s ++ 4s 5) 5= s2 ()() 1 + B ++ 4 cs + 5 B1+= 0 ⇒ B =− 1 As the switch is not connected to the circuit 4C0+ =⇒= c 4 hence 1−− s4 −+ = = Vsc () = + i0LL()() 0Ai0 S s2 ++ 4s 5 −+ 1 s4+ V0cc()()= 0VV0 = = − S s2 ++ 4s 5 for t0≥ , 1 s4+ = − 2 S ()s2++ 1 1 s4+ 2 =−− 22 S ()s2++ 1() s2 ++ 1 −−2t 2t Vc () t=−− 1 e coscost e sinsint for Now, here we can’t determine the time t0≥ constant and we have to find the voltage −2t Vc () t=−+ 1 e (coscost sinsint) for t0≥ across capacitor for t0≥ so we convert the above network into s- domain

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)~~

Q.1 When the angular frequency in a) 25 b) 50 the figure is varied from 0 to ∞ , the c) 100 d) 200 𝜔𝜔 locus of the current phasor I2 is [GATE-2003] given by Q.3 An input voltage v() t= 10 5 cos (t+10° ++10 5 cos() 2t 10° V applied to a series combination of resistance R1= Ω and an inductance L=1H. . The resulting steady state current a) i(t) in ampere is a)10cos() t+ 55° + 10cos(2t ++ 10° tan−1 2) 3 b)10cos() t++ 55° 10 cos(2t + 55) 2 c)10cos() t− 35° + 10cos(2t +− 10° tan−1 2) b) 3 d)10cos() t−+ 35 10 cos(2t − 35°) 2 [GATE-2003]

~~Q.4 The circuit shown in the figure, with 11 R =Ω, L = H,C = 3F has input 34 c) voltage V() t= sin 2t .The resulting current i(t) is~~

~~d) 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 3 input voltage is Vi () t= 2 sin10 . [GATE-2001] The output voltage Vt0 () is equal to 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

Q.6 Consider the following statements 10 3 10 3 a) ∠90°A b) ∠−90°A S1 & S2 2 2 S1: At the resonant frequency the c)5∠ 60° A d) 5∠ − 60° A impedance of a series R-LC [GATE-2005] circuit is zero. S2: In a parallel G-L-C circuit, Q.10 In the AC network shown in the increasing the conductance G figure, the phasor voltage V (in results in increase in its Q factor. AB Volts ) is Which one of the following is correct?

~~a) S1 is FALSE and S2 is TRUE~~

~~b) Both S1 and S2 are TRUE~~

~~c) S1 is TRUE and S2 is FALSE~~

d) Both S1 and S2 are FALSE [GATE-2004] a) 0 b) 5∠ 30° Q.7 The condition on R, L and C such c)12.5∠ 30° d)17∠ 30° that the step response y(t) in the [GATE-2007] figure has no oscillations, is Q.11 An AC source of RMS voltage 20V with internal impedance

~~ZS = (1 + 2 j)Ω feeds a load of~~

impedance ZL =(7+4j)Ω in the figure below. The reactive power 1L L a) R ≥ b) R ≥ consumed by the load is 2C C L 1 c) R2≥ d) R = C LC [GATE-2005] a) 8VAR b) 28VAR Q.8 In a series RLC circuit, c) 16VAR d) 32 VAR 1 [GATE-2009] R=Ω= 2k , L 1H and CF= µ . The 400 resonant frequency is Q.12 For the parallel RLC circuit, which 1 one of the following statements is a) 2× 104 Hz b) ×104 Hz π NOT correct? a) The bandwidth of the circuit c) 4 d) π× 4 10 Hz 2 10 Hz decreases if R is increased

[GATE-2005] b) The bandwidth of the circuit remains same if L is increased For the circuit shown in the figure, Q.9 c) At resonance, input impedance is the instantaneous current is i1 (t) a real quantity

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission d) At resonance, the magnitude of Q.16 A 230 V rms source supplies power input impedance attains its to two loads connected in parallel. minimum value The first load draws 10 kW at 0.8 [GATE-2010] leading power factor and the second one draws 10 kVA at 0.8 lagging Q.13 The current I in the circuit shown is power factor. The complex power delivered by the source is a) (18 + j 1.5) kVA b)(18 -j 1.5) kVA c) (20 + j 1.5) kVA d)(20 - j 1.5) kVA [GATE-2014] a) −j1A b) j1A c) 0A d) 20A Q.17 A periodic variable x is shown in the [GATE-2010] figure as a function of time. The root-mean-square (rms) value of x is Q.14 The circuit shown below is driven ___. by a sinusoidal input

~~Vip= V cos(t / RC) . The steady state~~

~~output V0 is~~

~~[GATE-2014]~~

Q.18 A series RC circuit is connected to a a)()Vp / 3 cos(t / RC) DC voltage source at time t = 0. The b) ()Vp / 3 sin(t / RC) relation between the source voltage c) Vs, the resistance R, the capacitance ()Vp / 2 cos(t / RC) d) C, and the current i(t) is given ()Vp / 2 sin(t / RC) below: [GATE-2011] 1 t Ve = Ri(t)+ i() u du Which one of C ∫ Q.15 Two magnetically uncoupled 0 the following represents the current inductive coils have Q factors q1 and f(t)? q at the chosen operating 2 a) b) frequency. Their respective

~~resistances are R1 and R 2 . When connected in series, their effective Q factor at the same operating frequency is~~

~~a) qq12+ c) d)~~

~~b) (1 / q12 )+ (1 / q )~~

~~c) ()()qR11++ qR 2 2 / R 1 R 2~~

~~d) ()()qR12++ qR 21 / R 1 R 2 [GATE-2013] [GATE-2014]~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.19 The steady state output of the inductor current phasor and the circuit shown in the figure is given resistor current phasor, is ______. by yt() = A()ω sin(tωω + φ ))( .If the [GATE-2016] amplitude | A(ω ) |= 0.2 5 then the ’ Q.23 In the RLC circuit shown in the frequency ω is figure, the input voltage is given by

~~vi ()()()t= 2cos 200t + 4sin 500t~~

~~The output voltage V0 () t is a) cos ()() 200t + 2sin 500t 1 2 b) + a) b) 2cos()() 200t 4sin 500t 3R C 3R C c) sin()() 200t+ 2cos 500t 1 2 c) d) d) 2sin()() 200t+ 4cos 500t R C R C [GATE-2014]~~

~~Q.20 In the circuit shown the average~~

~~value of the voltage Vab (in Volts) in steady state condition is ______.~~

~~[GATE-2016]~~

Q.24 In the circuit shown, the positive angular frequency (in radians per Q.21 An LC tank circuit consists of an second) at which the magnitude of ideal capacitor C connected in the phase difference𝜔𝜔 between the parallel with a coil of inductance L having an internal resistance R. The voltage V1 and V2 equals radians, resonant frequency of the tank is______. 𝜋𝜋 4 circuit is 1 1 C a) b) 1− R2 2π LC 2π LC L

~~1 L 1 2 C c) 1− d) 1− R 2π LC RC2 2π LC L [GATE-2015]~~

~~Q.22 The figure shows an RLC circuit with [GATE-2017] a sinusoidal current source. 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 is______. 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑉𝑉1

~~At resonance, the ratio ILR /I ,i.e., the ratio of the magnitudes of the~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) 1.25 2 sin() 5t−π 0.25~~

~~b) 1.25 2 sin() 5t−π 0.125~~

~~c) 2.5 2 sin() 5t−π 0.25 [GATE-2017] d) 2.51.25 2 sin() 5t−π 0.125 Q.26 In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V. The ratio~~

𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is______. 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

~~[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 Vc (in volts) across the capacitor is~~

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

~~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)~~

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

Q.1 (a) Q.4 (a) Em cosωt jωC i()() t= V t .Y it2m() = = E ∠ 0 1 1j+ ωCR 2 R 2 + 11 jωC Y= V(t) ++ jωc Rj1 ωL E0m∠∠ωC 90° ∠=it2 () 222− 1 11 1+∠ω C R tan ωCR 2 Y= V(t) ++ jωc Rj1 ωL EmωC −1 i2 () t= ∠− 90 tan ωCR =sin2t[3 −+ 2j 6j] + 222 1 ωCR =sin 2t[3 + 4 j] Em At ω= 0,i()() t ==∞= 0,ω ,i t −1 4 22R =5sin 2t ∠ tan 2 3 Fig (a) satisfies both conditions. =5sin() 2t + 53.1° Q.2 (b) f Q.5 (a) Q = o BW 1 1 jωC V0i() t= Vt() fo = 1 2π LC R + R jωC BW = 1 L = 2 sin103 t 1j+ ωCR 2 Rs 1 Characteristic equation =++s 1 LLC = 2 sin103 t +×33 × − 1L 1 j 10 10 Or Q = 3 RC V0 () t= sin(10 t − 45°) When R, L, C are doubled, 1 Q.6 (d) Q' = Q = 50 2 S1: Impedance of series RLC circuit at resonant frequency is minimum Q.3 (c) 1 ZRj=+−ωL vt() 10 2 cos(t+ 10°) ωC it() = = Rj++ωL 1 1j 1 ωL−= 0 10 5 cos(2t+ 10°) + ωC 1+ 2j ZR= (Purely resistive) 10 2 cos(t+° 10 ) C it() = S:2 QR= 2∠° 45 L 10 5 cos(2t+° 10 ) 1 1C + GQ= ⇒= 5∠ tan−1 2 R GL G then Q if C and L are same ∴ i()() t = 10cos t −° 35 +10cos(2t +− 10 tan−1 2) Q.7 (c)↑ ↓ Transfer function

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 =∠×−5 30°()() (5 3J || 5 + 3J ) 1 = sC = 34 1 s2 LC++ sCR 1 =5∟ 30° × =17 ∠ 30° R++ sL 10 sC 1 Y(s) = LC Q.11 (b) R1 U(s) ss2 ++ The RMS current in the load is given L LC by R r 20 20 10 2ξωn = = = = L I ZZSL+ 8+ j6 4+ j3 1 ωn = 10−1  3 LC =∠ − tan  54 R ξ= LC −1 3 2L =2∠ − tan  4 RC ξ = = , reactive power = 2 2L I2rms IXrms L For no oscillations , 1 =×=4 4 16VAR Also note that the active power RC ≥ 1 ξ ≥ consumed by the load 2L =I2 R = 47 × =28W L rms L R2≥ C Q.12 (d) This is standard concept of parallel Q.8 (b) resonant circuit 11 f0 = = 2π LC 1 Q.13 (a) 2π 1×× 10−6 400 1034× 20 10 = = Hz 2ππ

~~Q.9 (a) When 5∠ 0° is acting along,~~

~~i1 () t=−∠ 5 0° (as 10∠ 60° is kept open) When 10∠ 60° is acting alone.~~

~~i1 () t= 10 ∠ 60° (as 5∠ 0° is kept open) Q.14 (a) i1 () t= 10 ∠ 60° −∠ 5 0° Redrawing the circuit s –domain =+−5 8.66j 5~~

~~i1 () t= 8.66j 10 i() t= 5 3 ∠= 90° 3 ∠ 90° 2 Q.10 (d)~~

~~VAB = current × impedance~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 are ideal and there is no –resistance R. 1 in the circuit charging and V() s=++ R I() ssC I(s) 1  1 discharging time constant will zero. sC R + sC Thus current will exist like an 1+ SCR R impulse function. V1 () s= I() s + I(s) sC 1+ SCR Q.16 (b) ………………….(i)

∴=Vip V cos(t / RC) 1 So here, ω = RC Now, 1j+ ωCR R Vi () S= I() s + I(s) jωC() 1+ jωCR Load 1: 1 P= 10kw  Put, ω =  φ = =−=− RC cos 0.8 SI P jQ 10 j7.5KVA  ()1+ jR R Q= P tanφ = 7.5KVAR So, Vi () S= + I(s) Load S=10KVA 2: j 1j+ Q 3R Cos φ=0.8 sin φ= VS() = …………………..(ii) S i + ()1j P Cos φ= VS() Is() =i ×+() 1 j S 3R P 0.8 = →=P 8kw 1 10 Now, V0 () s=  R || I(s) sC Q= 6KVAR 1 S1= P+ jQ= 8 + j6 R. Complex power delivered by the ⇒=V() ssC I(s) 0 1 source is SI + SII=18-j1.5KVA R + sC Q.17 (0.408) R VS() T i 1 2 ⇒=Vs0 () .() 1j + = 1+ SCR 3R xrms ∫ () x() t dt T 0 R VS() ⇒=Vs() .i () 1j + 0 1+ j 3R VS() ⇒ Vs() i 0 3 In time domain, 1 v() t= v (t) 0i3 V  2 T p t  t0≤≤ t v0 () t= cos T 2 3 RC x(t) =   0T  2tT≤≤ Q.15 (d) T When the switch in closed at t=0 2 2 T 12 2 Capacitor C will discharge and C =.t .dt + () 0 .dt 1 2 TT∫∫ 0 T/2 will get charge since both C1 and C2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission T Q.21 (b) 14 t3 2 = . 2  YY=e + Y LR TT 3 0 1 3 Y=jωC+ =j 4T 1 ()jωL+R Xrms =3 . ⇒⇒ 0.408 3T 8 6 ()R− jLω ωC + ()RL2+ω 22 Q.18 (a) Placing Imaginary part to zero we In a series RC circuit, get option (B). Initially at t = 0, capacitor charges V with a current of s and in steady → R state open circuit and no current flows throughat t the = ∞, circuit capacitor So behaves the current like i(t) represents an exponential decay Q.22 0.316 function → At resonance, I QI L =m =Q IIRm C For parallel circuits Q = R = 10 L 10× 10−6 = 0.316 Q.19 (b) 10× 10−3

~~Q.23 (b)~~

Vi () t= 2cos 200t + 4 sin 500t , since there are 2 frequency term output will also have 2 frequency term →If By nodal method, we take 4sin500t first i.e. W = 500 V− 1|0o V V + = 0 then on the output section, this R 1 2 parallel LC combination have Z = ∞ , ()jωc ()jωc LC SO it is open circuit and Vo = Vi 1 jω c 1|0o V ++jcω = R 2R 2 V= 2+ 3jωRC V1 Y= ⇒ So w.r.t. 4sin 500t output must be 2 2j+ ω3RC 4sin500t without any change in 11 amplitude and phase, this is Given |A( ω) |= ⇒ 22 2 satisfied by only option B. 4 4+ 9R c .ω 2 ⇒=ω Q.24 1rad/sec 3RC Q.25 2.6 Q.20 (5)

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission = ∞+− − ∞ −τt/ ≥ iLL()() t i i L() 0 i L() e ,t 0~~

~~− i0L () = 0~~

~~ω=3rad / sec~~

~~Z1 =+Ω() 4 j3~~

~~Z2 =−Ω() 5 j1 2~~

~~22 V22= i z = i 5 += 12 13 i 22 To find ∞ : V11= i z = i 4 += 3 5i iL () V 13 i 13 2 = = = 2.6 (Inductor will behave as short V 5i 5 1 circuit)~~

~~Q.26 0.2 1 i()∞= = 0.5A L 11+~~

~~Given that, V and I have same phase. So, the circuit is in resonance. At resonance,~~

VcR= QV Amplitudeof V c = Q −+1()() 1 + 1iL ∞= 0 Amplitudeof VR 1 L 15 = = = 0.2 To find τ : RC 55 1 τ= for t0> Q.27 0.316 R eq Given: Switch is closed at t0= 1 At t0= − τ= =0.5sec 11+ Source is disconnected current i t= 0.5 +− 0 0.5 e−t/0.5 through the inductor before L ()() switching = − −t/0.5 iL () t 0.5 1 e −− iL ()() 0= i0 = 0 Value of loop current at t= 0.5sec The find, loop current that is i 0.5= 0.5 1 − e−0.5/0.5 inductor current after t0= , we can L ()  use = −=−1 iL () 0.5 0.5 1 e 0.316A

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 VC () t= 5sin 5t  1+ω j RC Q.28 (c) ~~

~~1 Given: = VC () t 5sin 5t 36− 1+× j 5 × 200x10 ×× 1 10 R12= R = 100k Ω , Vi ()() t= 5sin 5t 1 VC () t= 5sin 5t  1+ j1~~

~~1 = VC () t 5sin 5t  2∠π /4~~

~~5 π VC () t= sin 5t − 2 4 We have to find the voltage across the capacitor. =2.5 2 sin() 5t −π 0.25~~

~~Network can be further simplified by taking the equivalent of two series resistance as shown. From the given input~~

~~Vi ()() t= 5sin 5t .~~

~~∴ω=5rad / sec~~

~~Using voltage division,~~

~~Zc Vtci() =  Vt() RZ+ c~~

~~i Where Zc = and R= 200k jCω~~

~~i jCω V() t= 5sin 5t  C i R + jCω~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

Q.4 In the figure the current source is Q.1 In the figure Z1 = 10 ∠− 60°, 1∠=Ω 0A,R 1 , the impedances are Z2 = 10∠60°, Z3 = 50 ∠ 53.13° Zjc =−Ω , and ZL = 2j Ω . The Thevenin impedance seen from X-Y Thevenin equivalent looking into is the circuit across X-Y is

a) 56.66∠ 45° b) 60∠ 30° c) 70∠ 30° d) 34.4∠ 65° a) 2∠ 0V,() 1 +Ω 2j [GATE-2003] b) 2∠° 45 V,() 1 − 2j Ω Two ac source feed a common Q.2 c) 2∠° 45 V,() 1 +Ω j variable resistive load as shown in figure .Under the maximum power d) 2∠° 45 V,() 1 +Ω j transfer condition, the power [GATE-2006] absorbed by the load resistance RL is Statements for Linked Answer Questions Q.5 & Q.6

a)2200W b)1250W c)1000W d)625W Q.5 For the circuit given above, the [GATE-2003] Theremin’s voltage across the terminals A and B is Q.3 In the given figure, the Thevenin’s a) b) equivalent pair (voltage, 1.25V 0.25V c) d) impedance), as seen at the terminals 1V 0.5V P-Q is given by [GATE-2009] Q.6 For the circuit given above, the Thevenin’s resistance across the terminals A and B is a) 0.5kΩ b) 0.2kΩ c)1kΩ d) 0.11kΩ [GATE-2009] a) (2V,5Ω) b) (2V,7.5Ω) The impedance looking into nodes 1 c) (4V,5Ω) d) (4V,7.5Ω) Q.7 and 2 in the given circuit is [GATE-2005]

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) k2 b) k c)1/k d) k [GATE-2013]~~

~~Q.11 In the circuit shown below, if the~~

~~source voltage VS = 100 ∠ 53.13° V a)50Ω b)100Ω then the venin’s equivalent voltage c)5kΩ d)10.1kΩ in volts as seen by the load~~

~~[GATE-2012] resistance R L is~~

~~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)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 a) 0.8Ω b)[GATE 1.4Ω -2012] terminals x and y for the circuit in c) 2Ω d) 2.8Ω figure are~~

Q.9 A source Vs () 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] b) d)2sin Q.10 Consider a delta connection of a)2sin (ωt), 4Ω 1sin (ωt), 1Ω resistors and its equivalent star [GATE-2014] c)1sin (ωt), 2Ω (ωt), 0.5Ω connection as shown. If all elements of the delta connection are scaled by Q.13 A non-ideal voltage source Vs has an a factor k,k>0, the elements of the internal impedance of Zs If a purely resistive load is to be chosen that corresponding star equivalent will maximizes the power transferred to be scaled by a factor of the load, its value must be a) 0 b) real part of ZS c) magnitude of ZS d) complex conjugate of ZS [GATE-2014]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.14 The Norton's equivalent source in amperes as seen into the terminals X Q.17 In the circuit shown below, the and Y is value of capacitor C required for maximum power to be transferred to the load is

~~[GATE-2014]~~

Q.15 For the given circuit, the Thevenin a) 1nF b) 1µF equivalent is to be determined. The c) 10nF d) 10 µF Thevenin voltage, VTh (in volt), seen [GATE-2017, Set-2] from terminal AB is Q.18 For the network given in figure below, the Thevenin’s voltage Vab is

~~[GATE-2015] a) -1.5V b) -0.5V Q.16 In the circuit shown below, the c) 0.5V d) 1.5V maximum power transferred to the [GATE-2017, Set-2] resistor R is……….W~~

~~[GATE-2017, Set-1]~~

~~ANSWER KEY:~~

~~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)~~

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

~~Q.1 (a) By Thevenin’s theorem~~

~~For maximum power transfer 2 2 22 RL= R th + X th = 34 += 5Ω~~

Vth 100 Zth= Z x− y = Z 1 || Z 2 + Z 3 l = = ()3++ j4 RL 8 + j4 ZZ12× = + Z3 =11.18 ∠ 26.56°A (Z12+ Z ) Power absorbed by RL (max). 10∠− 60 × 10 ∠− 60 +∠(50 53.13) =l22 R = 11.18 ×= 5 625W ()10∠− 60 + 10 ∠− 60 L =56.66 ∠ 45° Q.3 (a) To calculate R (seen at terminals Q.2 (d) th For obtaining power absorbed by P-Q), voltage source is short- circuit

~~R L under maximum power transfer condition. We find thevenin’s~~

~~equivalent circuit across R L~~

~~Rth = 10 ||10 = 5Ω~~

~~Vth = Open–circuit voltage at the terminals P-Q~~

~~ZthI calculated by short circuiting the votage sources.~~

~~Zth =+()() 6 j8 || 6 +=+ j8 3 j4Ω~~

~~4 V= ×= 10 2V th 10+ 10 Thevenin’s equivalent circuit~~

~~V− 110 ∠ 0° V −∠ 90 0° th+= th 0 6++ 8 j 6 j8~~

~~Vth = 100 ∠ 0°~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.4 (d) To calculate Thevenin’s impedance, current –source is open -circuited~~

~~After connecting voltage source of V~~

~~V1= V 2 ⇒() 10K()() −= i b 100 I + 99i bb + i ;~~

Zth=++ RZ L Z C −10000ib = 100I +× 100 i b =+− 1 2j j =100I + 10000ib 1j+ Ω −20000ibb= 100I ⇒ i Open-circuit voltage at terminals X- −−100 I  Y =I = 20000  200  =IZ × th V= 100[] I ++ 99i i =∠×10() 1 + j bb  −I =2 ∠ 45°volts =100 I + 100= 50I 200 Q.5 (d) V 50I Rth = = = 50Ω Thevenin’s voltage, VOC = 0.5V II

Q.6 (b) Q.8 (a) Thevenin’s resistance is calculated Power transferred from circuit A to using the circuit shown in fig (1) 7  6+ 10R  circuit B= VI =    and (2), where independent voltage R2++  R2  source is short circuited 42+ 70R = ()R2+ 10− 3 7 I = = 2R++ 2R 7R V=+=+ 3 IR 3 2R+ 6+ 10R =  2R+ 2 dP ()()()()R+ 2 70 −+ 42 70R 2 R + 2 = Write the loop equation: 4 dR ()R2+ VAB+= 3V AB −−33 3 ()I−× 10 10 = 10 − VAB 3 5VAB = 10 V 103 ∴=RAB =ΩΩ = 0.2k TH I5

~~Q.7 (a)~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 70()()() R+=+ 2 42 70R 2 R + 2 Q.14 (2) ⇒5() R += 2 2(3 + 5R) ⇒5R +=+ 10 6 10R ⇒=4 5R ⇒=R 0.8Ω

~~Q.9 (c) 22 RL= Z th = 43 += 5Ω~~

~~Q.10 (b)~~

~~Consider Rabc= x;R = k;R = k 2 RRab+ k RStar = = Rabc++ R R 3k~~

~~RkStar ∝~~

~~Q.11 (c)~~

~~VTH= 10V L1 4 V 100∠ 53.13 −1 V==c −× tan8 j4 L1 3+ j4 3~~

~~VL1 = 80 ∠ 90° 5 II() = SC N 5 Q.12 (a)~~

~~ϑxy= V oc Q.15 (3.36)~~

~~ϑ ϑ Vth = 2i1 in =⇒==xy ϑϑ2sin sinωt 12 xy oc 2 = 1[] i ++= i1 i 2i + i1 2 2 i() 1=−+ 20i 2i1 Rxy =×⇒ 100 4 1 ∴=21i 2i1~~

~~ϑωth = 2sin sin t As the frequency of current i is increased, the impedance (Z) of the Q.13 (c) network varies as For minimum power transferred to~~

~~the load, Its value must be RZLs=~~

~~22 RL= RX ss +~~

~~ISC = 1A~~

~~R4th = Ω~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(IN)~~

~~Q.1 In the circuit shown in the following figure the input voltage Vi (t) is constant at 2V for time ts< 1 and then it changes to 1V.The output voltage, v0 (t), 2 s after the change will be~~

~~a) 0A b) 5 2 cos() 1000t A π c) 5 2 cos 1000t− A 4 d) 5 2A [GATE-2008]~~

~~For the circuit shown below the a) −−exp() 2 V Q.4 voltage across the capacitor is 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 a) + b) + resistor is ()10 j0 V ()100 j0 V c) ()0+ j100 V d) ()0− j100 V through 1Ω the [GATE-2008]~~

Statement for Linked Answer Questions 5& 6: In the circuit shown below the steady – state is reached with the switch K open a) ()1+ 5cos 2t A Subsequently the switch is closed at time t0= b)()5+ cos 2tA c) ()1− 5cos 2t A d) 6A [GATE-2007]

~~Q.3 For the circuit shown below the steady–state current I is~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission dl a) Zero Q.5 At time t0= + , is 2 dt b) A step function 10 c) An exponentially decaying function a) −5A / s b) − A/s d) An impulse function 3 [GATE-2012] c) 0A / s d) 5A / s [GATE-2008] Statement for Linked Answer Q.6 At time t0= + ,current I is Questions 10 & 11: 5 In the circuit shown, the three voltmeter a) − A b) 0A reading V= 220V,V = 122V,V = 136V 3 1 23 5 c) A d) ∞ A 3 [GATE-2008]

Q.7 In the circuit shown below, the switch, initially at position 1 for a long time, is changed to position 2 Q.10 at t0= consumption in the load is a)700WIf RL=5Ω,the approximateb)750W power c)800W d)850W [GATE-2012]

The current i through the inductor Q.11 The power factor of the load for t0≥ is a) 0.45 b) 0.50 a) 1e− −20t A b) 1e+ −20t A c) 0.55 d) 0.60 c) 1+ 2e−20t A d) 2e− −20t A [GATE-2012] [GATE-2011] Q.12 Two magnetically uncoupled

~~Q.8 The average power delivered to an inductive coils have Q factors q1~~

~~impedance ()4− j3 Ω by a current and q2 at the chosen operating 5cos(100πt+ 100) A is frequency. Their respective~~

~~a) 44.2W b)50W resistances are R1 and R 2 . When c) 62.5W d)125 connected in series, their effective [GATE-2012] Q factor at the same operating frequency is~~

~~Q.9 In the following figure, C1 and C2 a) qq12+~~

are ideal capacitors. C1 has been b) (1 / q12 )+ (1 / q ) charged to 12 V before the ideal c) ()()qR++ qR / R R switch S is closed at t0= .The 11 2 2 1 2 current i(t) for all t is d) ()()qR12++ qR 21 / R 1 R 2 [GATE-2013]

~~Q.13 Three capacitors C12 ,C and C3 whose values are 10μF,5μF, and 2μF respectively, have breakdown~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission voltages of 10V , 5V and 2V respectively. 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 [GATE-2014] stored in the effective capacitance Q.17 A capacitor ‘C’ is to be connected across the terminals are across the terminals ‘A ‘and ‘B’ as respectively, shown in the figure so that the power factor of the parallel combination becomes unity. The value of the capacitance required

~~a)2.8 and 36 b)7 and 119 c)2.8 and 32 d)7 and 80 in μF is [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 [GATE-2014]~~

~~10Ω resistor at time t = 0+ is _____. Q.18 The capacitor shown in the figure is initially charged to switch closes at time t = 0. Then the value of VC(t) in volts+10 at V. time The t~~

[GATE-2014] = 10 ms is ___V. Q.15 The average real power in watts delivered to a load impedance ZL =(4- source i(t) 4sin ( ωt+ 20° ) t A is j2 )Ω by an [ ideaGATE current-2014] ____. Q.16. In the circuit shown in the figure, [GATE-2015] initially the capacitor is uncharged. The switch ‘S’ is closed at t = Q.19 The circuit shown in the figure is 0.Two milliseconds after the in series resonance at frequency fc switch is closed, the current Hz. The through the capacitor (in mA) is

~~___. value of Vc in volts ______V.~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission value of current i(t) at t = 1s, in ampere~~

~~is ______.~~

~~[GATE-2015]~~

Q.20 A current i(t) shown in the figure below is passed through a 1 F [GATE-2016] A series R-L-C circuit is excited capacitor that had zero initial Q.24 with a 50 V, 50Hz sinusoidal charge. The voltage across the source. The voltages across the capacitor for t > 2 s in volt is resistance and capacitance are

~~shown in the figure. The voltage ______. across the inductor (VL ) is ______V.~~

~~[GATE-2016]~~

Q.21 In the circuit shown below +=1 sin (2 ()VV12[ [GATE-2017] (2 30000t)] V. The RMS value of the current through𝜋𝜋 10000tthe resistor )+1sin R Q.25 The current response of a series R- will𝜋𝜋 be minimum if the value of the L circuit to a unit step voltage is capacitor C in microfarad given in the table. The value of L is

is____. t in s ______H.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] [GATE-2016] Q.26 In the circuit diagram, shown in Q.22 In the circuit shown below, VS=101 the figure, S1 was closed and S2 ∠ was open for a very long time. At current IS is in phase with VS. The t=0, S1 is opened and S2 is closed. magnitude0V, R =10Ω of I andS in milliampere =L =100Ω. The is The voltage across the capacitor, in volts, at t=5

~~______. 𝜇𝜇s is______~~

~~[GATE-2016]~~

~~Q.23 The voltage v(t) shown below is applied to the given circuit. v(t) = 3 [GATE-2017] V for t < and v(t)= 6V for t > 0. The~~

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

variable capacitor as shown in figure.+j100) isKeeping connected the in parallelexcitation to a 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]~~

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

~~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~~

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

~~Q.1 (a) where τ= Time constant = RC For t <1, the circuit is shown in For R= 1MΩ , C = 1μF, τ = 1sec Fig.1 ∴=v0 () t 2sec after the change −2 at t= 1+ = − 1e V~~

~~Q.2 (a) Due to ac source, i= 1 / 1 = 1A R1 C behaves as open circuit after Due to ac source, long time . 1~~

∴=At t 1−, the circuit is shown in 1 iR = 5∠ 0° × 2 11 Fig. 2 Current is zero, ++J0.5 1 J2 ∴=v0 () 1− 0V and =5 ∠ 0° ∴=vc () 1− 2V ∴=+iR (1 5coscos2t)A As capacitor voltage cannot change instantaneously, the circuit for Q.3 (a) = is shown in Fig 3. t1+ As t voltage source become

~~v1cc()()+ = v1_2V = 0V → ∞ ∴=−v() 1 1V 0 + Q.4 (d) Circuit is under resonance 10∠ 0° I= = 1 ∠ 0°A 10~~

VC = − 100JV At steady state as t , the capacitor behaves as open circuit Q.5 (b) as shown in Fig. 4 Current→ ∞is zero, From Fig 3 : Write the Outer loop equation : ∴v0c()() ∞= 0V andv ∞= 1V Initial dl (t) ×+2 + = value of vt0 () I() t 1 1 10 5 dt =−=1V, at t 1+ At t0= + , Final value of v() t= 0V, at t = ∞ 0 dl2 5 t ()()0+−=−− 5 I0 =−+ 5 − dt 3 = − τ v0 () t 1e 10 = − A / sec 3

~~Q.6 (a) For the circuit is shown in Fig .1~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission For t<0, the status of the circuit is shown in Fig.1 inductor behaves as short circuit after a long time. 10 ∴==i() 0− 1A 10 After a long time , at t= 0L− behaves as short circuit and C behaves as open circuit . The relevant circuit is shown in Fig .2

For t > 0 , the status of the circuits shown in Fig .2 Current through the inductor cannot change instantaneously. ∴= = initial value At t= 0−−,1 I()()() 0 = 0,i L 0 − = I 2 0 − i0()()+− i0 1A (I.V) After a long time inductor =0andv() 0− = 10V c behaves as short circuit . For t0> the switch , K is closed , ∴i() ∞= 2A = final value (F.V) and the relevant circuit is L 11 shown in Fig.3 Time constant τ= = , = 20 R 20 τ

t − itL () and vtc () cannot change  i()() t=+−− IV F.V I.V I eτ for t ≥ 0 instantaneously.   ∴=i0L22()()()++− I0 == I0 0 =+−1() 211e() −−−20t =− 2e 20t v0cc()()+−= v0 = 10V At = , the state of the circuit is t0+ Q.8 (b) shown in Fig. 4 The load consists of a resistance and a capacitance of this, only R is passive and consumes power 2 So P= rm Ri 2 5 = ×=4 50W 2 A 5 = I()() 0−+= I0 = − A note rms value of A cos ωt 1 3 2

~~Q.7 (d) Q.9 (d) When the switch in closed at t=0~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Capacitor C1 will discharge and~~

~~C2 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 CC =23 + impulse function. CCeq 1 CC23+ =11.5μF Q.10 (b) Safety voltage = 7V :R= 5Ω L Q= CV VRL= V 3 coscos φ= 136 × 0.45[FromQ 1 7] ⇒=QCeq × V safety ∴=V 61.2V RL 11.5× 7; 80μC 2 VR P=L 749ω = 750 RL R Q.14 (-30) At t < 0, switch is at position “A” Q.11 (a) and it was at steady state. Phasor diagram

~~V1= 220V 12 V = 122V~~

~~V3 = 136V →→→~~

vv12= + v3 Steady state inductor behaves like short circuit In a source free circuit, −t τ () = Ii0L .et −t τ By parallelogram law of addition of L () = 3.iet −12t vectors 5 itL () = 3.e 22 = ++ φ − V1 V 2 V 3 2V 23 V cos v() t [1 −= 3.e 12t /5 ]0 ∴by using options, cosφ= 0.45 vt() =10 −30e−+12t /5 ⇒= At t0 V = −30[; Volts Q.12 (c) Q.15 (32) ωL12 ωL QQ= = = Given ZL =(4- 12RR 22 Current source i(t)4sin(ω+ t 20o )A After series connection j2)Ω 2 The average real power P = Irms .R ωL()12+ L Q = 2 RR+ 4 12 Watts P = .4= 32watts Q1R+ Q R 2 Q = 1 22 RR+ 12 Q.16 (1.51)

~~Q.13 (d) At t=0 : When switch „S‟ is closed~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 4 1 1 16 −j −+ ω22C ωω CCC 22 ω Ztotal = 2 1 16+− 1 ωC 5 Equate the imaginary term to zero ic (0+) = ⇒ 2.5mA 2k 16 16 1 1 += ⇒C ωωCCw22 C 17ω 1 At t =∞ : ⇒=CF =187.24µ 17×× 2π 50

~~Q.18 3.678V It is a source free network where c ic (∞= ) 0.A apacitor voltage~~

−t i (t) = (Initial Value – Final Value ) t c Vt= Ve > 0 −t/τ c () 0 e + Final Value −100t 1 =10e τ =RC = τ = RCeq eq sec 100 − ××−3 = −3 100 10 10 3.67V τµ=1K *4 Vec ()10×= 10 10 τ= 4 msec −3 Q.19 (100V) =  − −−t / 4*10 3 + c (i [2.5 0]e 10 () )t0 Under series resonance condition, i() t= 2.5 e−250t mA voltage across L, C is Q times of the c supply voltage. At t=2* 10-3sec ; VC = QV −3 = ic () 2*10 1.51mA 1 L 1 0.1 Q= = =100 R C 10 0.1× 10−6 Q.17 (187.24) VC = QV =()()100 1 = 100V Power factor of the parallel combination is unity. Q.20 (8) i.e., in total impedance equate the 11 imaginary term to zero, which V=idt = 8[ ut()()()() −− 1 ut − 2 dt = 8[ rt −− 1 rt − 2 ] c C ∫∫1 gives the capacitance required. 1 Zj=[]4 + 1Ω || total jCω 1 1 []4+ j. 4−− j1 jωC ωC = * 11 4+− j1 4 −− j1 V() t>= 2 8V ωC ωC c −j4 1  1   +4 −− j1 Q.21 (0.28 to 0.283) ωωCC  ω C  Ztotal = 2 1 Q.22 (100) 16+− 1 ωC In phase means circuit is under resonance and the admittance seen by source must be real i.e. imaginary part of Yeq = 0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 11 L= 1H eq = + Rj+ ωL (1/ jωL) Q.26 1.52 ⇒YRj− ωL = + jωL At t0= − R(22+ ωL) Rj+ ωL 10 →=Real() y = R(22++ωL) 100 (100) 2 = 1 1010 101 I= VY = =0.1 = 100mA 1010 − V0C () = 1 Q.23 (1.63)2 At t = ∞ it()= i ()[(0) ∞+ i−− − i ()] ∞ e(t /)τ LL L L 2 At t =∞,6supply is v VC ()∞= 3 = 2V 21+

~~t − =+ −∞RC +∞ VtCCC()  V0() V() e V C() t − 2 ××10 10−6 =−+[]1 2e3 2 At t= 5 µ sec~~

~~L 1.5 VC = 1.52V → τ = = =1 Rin 1.5 − Q.27 6.36 →()ti = it() =2 − et L L 1 − ω= = −=t 0 iL () t 2e 1.632 LC 1 = Q.24 50 −−36 10× 10 ×× 4 10 2 2 V= VR +−() VV LC = 5000rad / sec~~

ω0L =2 +−2 Q = 50 50() VL 50 R −3 VL = 50 5000() 10× 10 30 = R Q.25 1H 5 Rt R = Ω − V L 3 it() = 1 − e R  V 15 At ω→0 I = = = 9V From given data at t = ∞ R 5/3 V I9 = 0.5 At ω→2 = =6.36A R 22 V1= R2= Ω Q.28 1.86 t= 0.25 00 I= I12 + I =∠ 5 0 +∠ 5 30 −2() 0.25 − I= 9.659 ∠ 14.94 0.197= 0.5 1 − e L  = θ  P VIcos

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission =××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Yeq 0 or Img  Z eq  0~~

~~In a parallel circuit, from calculation point of view, it is wised to deal with~~

~~= ImgYeq 0~~

~~From the given circuit we can say,~~

~~11 Y = + ZZRL C~~

~~11 = + R+ jLω 1  jCω~~

~~R− jLω = + jCω RL22+ ()ω~~

~~RLω = +−ω 22jC 22 RL++()ωω RL ()~~

~~At resonance frequency~~

~~= ImgYeq 0~~

~~So,~~

~~R Y = eq RL22+ ()ω~~

~~1RL22+ ()ω Zeq = = YReq~~

~~1022+ 100 = =1010 10~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 4 AC ANALYSIS~~

4.1 INTRODUCTION Similarly, = j∅ -Exponential Form A sinusoidal current or voltage at a given V Vem frequency is characterized by only two VV=m ∠∅ -Polar Form parameters amplitude and a phase angle. V= Vm cos ∅+ jsin ∅ ) -Trigonometric The complex representation of the voltage Form or current is also characterized by these same two parameters. Phasor Example: representation is defined only for Transform the time domain voltage cosinusoidal signal. All the sinusoidal signal = −° volts into the are converted into cosinusoidal by v()() t 100cos 400t 30 subtracting 90° from phase frequency domain. A real sinusoidal current Solution: We know that = ω +∅ & it() = Im coscos( ω t +ϕ ) is expressed as the v() t Vm cos( t ) real part of a complex quantity by invoking v()() t= 100cos 400t −° 30 (Given) Euler’s identity. Thus time –domain expression is already is j(ω t +ϕ ) i() t= Re{} Im e in the form of a cosine wave with a phase jtωϕ j angle. Thus, suppressing =400 rad/sec. = Re{} Im e e VV=m ∠∅ 𝜔𝜔 = Re{} Ie jtω i.e. v() t= 100 ∠− 30 ° volts We represent the current as a complex quantity by dropping the instruction Re {}. 4.2 SINUSOIDAL STEADY STATE Thus adding an imaginary component to ANALYSIS the current without affecting the real component and suppressing the factor e jtω A circuit having constant source is said to We get be in steady state if the currents and voltages do not change with time. Thus = jϕ - Exponential Form I Iem circuits with currents and voltages having II=m ∠ϕ - Polar Form constant amplitude and constant frequency

I= I[cosm ∅+ jsin ∅ ] -Trigonometric Form sinusoidal functions are also considered to The above abbreviated complex be in a steady state. representation is the phasor representation. Inductors and capacitors are kept as it is The process by which we change i(t) into I while applying sinusoidal steady state is called a phasor transformation from the analysis. The analysis of AC in the steady time domain to the frequency state is generally carried out in the phasor domain_(Shown in fig below) 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~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission rotating vector, which rotates in the zL= j ω L Ω= jX L Ω= X L ∠ 90 ° | whereX=ω L anticlockwise direction with angular L 11 frequency ‘ ’ in the time domain. zC= Ω=− j =− jX cC = X ∠− 90 ° | 1 = jCωω C whereXc e.g. For the Phasor diagram shown below ωC 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.

~~In Fig. a), the length of the arrow represent (Network is in steady state ) the magnitude of the sine wave and angle We know, ‘ ’ represents the angular position of the VR = IR~~

sine wave. In Fig. b), the magnitude of the The resistor voltage (VR ) and current (I) sineθ wave is 3 and phase angle is30° . are in phase with each other (as shown in A Phasor diagram can be use to represent fig. b) inductor voltage (V) leads the the relation between two or more sine L source voltage ( ). The phase angle waves of the same frequency. e.g between current and voltage in a pure 𝑆𝑆 inductor is always 𝑉𝑉90° (as shown in fig(b))

~~VLL= IZ~~

~~=IXL ∠° 90~~

In the above fig sine wave C lag behind B by 45° , sine wave A lead sign wave B by30° . Fig (b) shows and I are in phase and the 4.3 SERIES CIRCUITS source voltage ‘V’ is the phasor sum of The impedance diagram is useful for and 𝑉𝑉𝑅𝑅 𝑅𝑅 analyzing series ac circuits. The series 22 𝑉𝑉 V= VV + circuits can divided in RL, RC and RLC 𝑉𝑉𝐿𝐿 RL series circuits. The phase angle between resistor voltage In the series ac circuits, one must draw the and source voltage is impedance diagram V ∅=tan−1 L We know that  VR zR = R Ω= R0 ∠ °

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V The power factor =cos ∅=R (Lag) V~~

~~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 Case1: When VLC >V are sinusoidal. The series RC circuit is We know, VR = IR shown in fig. VLL= IX ∠ 90~~

~~VCC= IX ∠− 90 °~~

~~Here the resistor voltage (VR) and current are in phase with each other (shown in fig.~~

b) VR = IR The capacitor voltage ‘VC’ lags behind the The source voltage v is given as 2 source voltage. The phase angle between V= V2 +−() VV the current and the capacitor voltage is R LC The phase angle is always 90° = −1 VVLC− VCC IZ ∅=tan  VR VCC= IX ∠− 90 °  And power factor is V p.f= cos ∅=R (lag) V Case2: When >

~~𝑽𝑽𝑪𝑪 𝑽𝑽𝑳𝑳~~

Here I leads Vc by 90°, VR and I are in phase The source voltage is given by 22 V= VVRC + The phase angle between the resistor voltage and the source voltage is V ∅= −1 C tan  2 2 VR V= VR +−() VV CL The power factor is And phase angle is VR VV− powerfactor= cos ∅= (lead) ∅= −1 CL V tan  VR  4.3.3 SERIES R-L-C CIRCUITS And power factor is A series R-L-C circuits is the series V p.f= cos ∅=R (lead) combination of resistor, inductor and V capacitor.

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission In the circuit shown below, Determine the~~

~~Case 3: When VLC =V total impedance, current I, phase angle and the voltage across each elements.~~

Example: Solution: A sine wave generator supplies a 50HZ, To find the impedance z, all first solve 10V rms signal to a 2 resistor in series for Vc and XL with a 0.1μF capacitor as Shown in fig. 11 X = = Determine the total impedance𝑘𝑘Ω Z, current I, C −6 2fΠ C 2Π× 50 × 10 × 10 Phase angle , capacitive voltage Vc and X= 318.5 Ω resistive voltage VR. C θ XLL=Π= 2 f 6.28 × 0.5 × 50

XL = 157 Ω Total impedance in rectangular form z=+− (10 j157 j318.5) Ω z=+− 10 j(157 318.5) Ω z=−Ω 10 j161.5 Here, XX> Solution: To find the impedance Z, we first CL 22 solve for Xc ∴=z()() 10 + 161.5 11 XC = = = 3184.7 Ω =100 + 26082.2 2fΠ 2Π× 500 × 0.1 × 10−6 C z= 161.8 Ω Total impedance V 50 Z=−Ω (2000 j3184.7) Current I=S = = 0.3A Z 161.8 22 Z=()() 2000 + 3184.7 ∴= I 0.3A Voltage across the resistor Z= 3760.6 Ω VR = IR = 0.3 ×= 10 3V Capacitive voltage Voltage across the capacitive reactance VCC= IX VCC==×= IX 0.3 313.5 95.55V −3 VC = 2.66 ×× 10 3184.7 Voltage across the Inductive reactance = =×= VC = 8.47V VLL IX 0.3 157 47.1V

Resistive voltage VR = IR 4.4 PARALLEL CIRCUITS =2.66 ×× 10−3 2000 VR = 5.32V The parallel circuits can divide in RC, RL Phase Angle, and RLC circuits. In parallel A.C. circuits, the voltage is the same across each −−11VC 8.47 θ =tan = tan = 57.87 ° element. VR  5.32 V We know that I = Example: R R

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VVV Current I is divided into two parallel = = = ∠− ° IL 90 branches as resistive current ( ) shown ZLL X∠° 90 X L below VVV 𝑅𝑅 IC = = = ∠° 90 V VVV𝐼𝐼 ZCC X∠− 90 ° X C IRC= and I = = = ∠° 90 R ZCC X∠− 90 ° X C 4.4.1 PARALLEL RL CIRCUITS

~~Parallel RL circuit shown below~~

As Shown in figure above, and v are in phase and current (I) Phasor sum of I and I 𝐼𝐼𝑅𝑅 R 22 Current I is distributed into two parallel ∴=I IIRC +C branches as resistive current (IR) and The phase angle between resistive current inductor current (IL) (which is given as and inductive current is --- follow) I V VVV ∅=tan−1 C I= and I = = = ∠− 90 °  RL IR R ZLL X∠° 90 X L I The power factor =cos ∅=R (lead) I

~~4.4.3 PARALLEL RLC CIRCUIT~~

~~Parallel R-L-C circuit Shown below Figure shows - IR and v are in phase and source current is phasor sum of IR and IL 22 ∴=I IIRL + The phase angle between resistive currant and inductive current is~~

−1 I Current I is divided into three parallel ∅=tan L R IR branches as, resistive current I Inductive current IL and capacitive current IC which is I The power factor =cos ∅=R (lag) given as bellow I V I = R R 4.4.2 PARALLEL RC CIRCUIT V IL = ∠− 90 ° Parallel RC circuit Shown below XL V IC = ∠° 90 XC

~~Case 1: When IICL> We know~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V I = R R V V IL = ∠− 90 ° & IC = ∠° 90 XL XC~~

~~Solution: Since the voltage across each element in parallel ac circuit is same---~~

So current in the resistive branch (IR ) is V 20∠° 0 I= = = 0.4A R R 50 Source current I is given as And current in the inductive branch 2 2 V 20∠° 0 I= IR +−() II CL I= = = 0.66 ∠− 90 ° L X 30∠° 90 Phase angle � is given as L Total current − II− ∅=tan 1 CL I= 0.4 − j0.66 IR Total current in polar form is And power factor is I= 0.77 ∠− 58.8 ° I Here current lags behind the voltage by p.f= cos ∅=R (lead) I 58.8° Total impedance Case2: when > V 20∠° 0 z = = 𝐈𝐈𝐋𝐋 𝐈𝐈𝐂𝐂 I 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) The source current I is given as 3) Phase angle (�) 2 2 I= IR +−() II LC Phase angle is given as

− II− ∅=tan 1 LC IR And power factor is given as Solution: We know that, I XLL= 2f π p.f= cos ∅=R (lag) I =2 π() 50 (0.1) = Ω Example: XL 31.42 A 50Ω resistor is connected in parallel From given circuit, 10 resistor is in series with an inductive reactance of30Ω . A 20 V with the parallel combination of 20Ω and signal is applied to the circuit find the total j31.42Ω Ω impedance and line current in the circuit Total impedance (z) is ------shown in fig. ∴

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ()()20 j31.42 z=+=+ 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 V Total current I = Z 20 = 25.87∠° 20.5 I= 0.77 ∠− 20.5 ° The phase angle between voltage & current is θ=20.5 °

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

Q.1 A unit step voltage is applied at t 0 voltage across the inductance at to a series RL circuit with zero initial t=0+ , is conditions. a) It is possible for the current to be oscillatory. b) The voltage across the resistor at t0= + is zero. c) The energy stored in inductor in the steady state is zero. a) 2V b) 4V d) The resistor current eventually c) -6V d) 8V falls to zero. [GATE -2003] [GATE-2001] Q.5 In figure, the capacitor initially has a Q.2 Consider the circuit shown in figure. charge of 10 Coulomb. The current If the frequency of the source is 50 in the circuit one second after the Hz, then a value of 0 which results switch S is closed will be in a transient free response is 𝑡𝑡

a)14.7A b)18.5A c)40.0A d)50.0A a) 0ms b) 1.786ms [GATE-2004] c) 2.71ms d) 2.91ms [GATE-2002] Q.6 In the figure given, for initial capacitor voltage is zero. The switch Q.3 An 11V pulse of 10μs duration is is closed at t=0. The final steady – applied to the circuit shown in state voltage across the capacitor is figure. Assuming that the capacitor is completely discharged prior to applying the pulse, the peak value of the capacitor voltage is

~~a)20V b)10V c)5V d)0V [GATE -2005]~~

a) 11V b) 5.5V Q.7 The circuit shown in the figure is in c) 6.32V d) 0.96V steady state, when the switch is [GATE-2002] closed at t=0. Assuming that the inductance is ideal, the current Q.4 In the circuit shown in figure, the through the inductor at t=0+ equals switch S is closed at time (t=0). The

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 a) 0A b) 0.5A secondary is shorted by a 10Ω c) 1A d) 2A resistor R and the other by a 15 mF [GATE-2005] capacitor. The switch SW is opened t=0 when the capacitor is charged to Statement for linked Answer Questions 5 V with left plate as positive at Q.8 & Q.9 (t+0+) the voltage VP and Current A coil inductance 10H and resistance 40 is I are connected as shown in the figure. After the switch S has been in contact withΩ 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 a) -25V,00A time taken for 95% of the stored b) Very large voltage, very large energy to be dissipated is close to current a) 0.10sec b) 0.15sec c) 5.0 V, 0.5 A c) 0.50sec d)1.0sec d) -5.0V,-0.5A [GATE-2005] [GATE-2007] if, at + , the voltage across the coil Q.9 t=0 Q.12 In the circuit shown in the figure, is 120 V , the value of resistance R is the current source I=1A , voltage

~~source V= 5V , R123= R = R = 1,Ω~~

~~L123= L = L = 1H, C12= C = 1F . The currents (in A) through R3 and the voltage source V respectively will be a) 0Ω b) 20Ω c) 40Ω d) 60Ω [GATE-2005]~~

Q.10 An ideal capacitor is charged to a voltage V and connected at t=0 o a) 1, 4 b) 5, 1 across an ideal inductor L. (The c) 5, 2 d) 5, 4 circuit now consists of a capacitor [GATE-2006] and inductor alone). If we let 1 Q.13 In the circuit shown in figure, switch ω=0 the voltage across the LC ' Sw1 is initially CLOSED and Sw 2 is capacitor at time t>0 is given by OPEN. The inductor L carries a a) Vo b) Vo0 cos(ω t) current of 10 A and the capacitor is

~~−ωt0 charged to 10 V with polarities as c) Vo0 sin(ω t) d) Vo0 e cos(ω t) indicated.Sw Is initially caps at t=0 [GATE-2006] 2 and Sw1 is OPENED at t=0 The~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission current through C and the voltage [GATE-2008] across L at t=0+ is Q.16 The charge stored in the capacitor at

~~a)8nC b)10nC c)13nC=5μs, will be d)16nC [GATE-2008]~~

Q.17 In the figure shown, all elements a) 55A,4.5V b) 5.5A,45V used are ideal. For time t<0,S1 c) 45A,5.5V d) 4.5A,55V remained closed and S2 open. At [GATE-2007] t=0,S1 is opened and S2 is closed. If the voltage across the capacitor Q.14 The time constant for the given Vc2

~~circuit will be C2 at t=0 is zero, the voltage across the capacitor combination at t=0+ will be~~

~~1 1 a) S b) S 9 4 a)1V b) 2V c) 4 S d) 9 S c)1.5V d) 3V [GATE-2008] [GATE-2009]~~

Statement for Linked Answer Q.18 The switch in the circuit has been Questions Q.15 & Q.16 closed for a long time. It is opened at The current i (t) sketched in the t0= . At t=0+ , the current through figure flows through an initially the 1μF capacitor is uncharged 0.3nFCapacitor.

~~a) 0A b)1A c)1.25A d) 5A [GATE-2010]~~

Statement for Linked Answer Questions Q.19 & Q.20 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 The L-C circuit shown in the figure a)18.8V b) 23.5V has an inductance = and a c) -23.5V d) -30.6V L 1mH capacitance C= 10μF

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.19 The initial current through the Q.22 The power factor of the load inductor is zero, while the initial a) 0.45 b) 0.50 capacitor voltage is 100V. The c) 0.55 d) 0.60 switch is closed at t=0 . The current i [GATE-2012] through the circuit is: a)5cos() 5× 103 t A Q.23 The switch SW shown in the circuit is kept at position '1' for a long b)5sin() 104 t A duration. At t = 0+, the switch is moved to position '2' Assuming c)10cos() 5× 103 t A V02 >V 01 , the voltage Vc (t) across 4 d)10sin() 10 t A capacitor is [GATE-2010]

Q.20 In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0 .The current i(t) for all t is −t/RC a) vc() t=−−− V 02 (1 e ) V 01 −t/RC b) vc() t=−−+ V 02 (1 e ) V 01 −t/RC c) vc() t=−−− V 02 (1 e ) V 01 −t/RC d) vc()() t=−+ V 02 V 01() 1e − + V 01 a) Zero [GATE-2014] b) A step function c) An exponentially decaying Q.24 The circuit shown in the figure has function two sources connected in series. The d) An impulse function instantaneous voltage of the AC [GATE-2012] source (in Volt) is given by v(t) = 12sint. If the circuit is in steady Statement for Linked Answer state, then the rms value of the Questions Q.21 & Q.22: current (in Ampere) flowing in the In the circuit shown, the three circuit is______voltmeter reading

~~V123 =220V,V =122V,V =136V~~

~~[GATE-2015]~~

Q.25 A series RL circuit is excited at t=0 Q.21 the approximate power by closing a switch as shown in the consumption in the load is figure. a)700WIf RL=5Ω, b)750W Assuming zero initial conditions, the dI2 c)800W d)850W value of at t =0+ is [GATE-2012] dt2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission − a) V b) V L R [GATE-2016] -RV c) 0 d) Q.29 The switch in the figure below was 2 L closed for a long time. It is opened at [GATE-2015] t=0. The current in the insuctor of 2H for t  0 , is 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 ____. a) 2.5e4t b) 5e4t c) 2.5e0.25t d) 5e0.25t

~~[GATE-2017, Set-1]~~

The voltages across the circuit in the [GATE-2016] Q.30 figure, and the current through it are given that the following Q.27 A resistance and a coil are connected in series and supplied expressions: from a single phase, 100V, 50Hz ac Vi ()() t= 5 − 10cos ω+ t 60 source as shown in the figure below. i()() t=+ω 5 X cos t The rms values of plausible voltage i across the resistance (VR) and coil Where ω=100 π rad / sec . If the (Vc) respectively, in volts, are average power delivered to the circuit is zero then the value of X (in ohm) is ______(up to two decimal places)

~~a) 65, 35 b) 50, 50 c) 60, 90 d) 60, 80 [GATE-2016] Q.28 In the circuit shown below, the initial capacitor voltage is 4V. Switch~~

~~S1 [GATE-2018] C) lost by the capacitor from t = is closed at t = 0.______. The charge (in μ~~

~~25μs to t =100μs is~~

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

~~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~~

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

Q.1 (b) s + At t0= inductor works as open and then starts discharging, so VC circuit, hence complete source willHenc e capacitor chargess] till 10 μ voltage drops across it and E1 h12= | l= 0 = 0.25 consequently, current through the maximum1 at t= 10μ E2 resistor R is zero. Hence, voltage + across the resistor at t0= is zero, Q.4 (b) and further with time it rises Before closing the switch, the circuit according to was not energized, therefore, −Rt − current through inductor and = − L VR () t R.(1 e )u(t ) voltage across capacitor are zero. After closing the switch, at t = inductor acts as open –circuits and+ capacitor acts as short circuit. 0 Equivalent circuit at t0= +

~~Q.2 (b) For transient response, ωL = 10 tan ()ωt0 l = = 2A R 3+ 4 || 4 ×× 2π 50 0.01 + tan() 2π×× 50 t0 = V 0= l × (4 || 4) 5 L ()~~

−1 π =×=2 2 4V 2π××= 50 t0 tan  5 =32.14° = 0.561rad Q.5 (a) 0.561 Method -1 t0 = = 1.786ms Using KVL, 100π dq 100= R q / c Q.3 (c) dt −t/RC dq VtCC()()()()= V ∞− V C ∞− V0e C 100C= R + q dt 10× 10−6 0t 10 ×1039 ×× 11 10− dq 1 11 = dt VC () peak=−− 10 (10 0)e ∫∫− q0 100C q RC 0 1 =10 1 −= e 6.32V −1/RC () 100C−= q (100C − q0 )e

~~10 dq (100C− q0 ) −t/RC [WhereRnet = 10||1 = kΩ, ie= = 11 dt RC 10 e−−t/RC= 40e 1 = 14.7A VC ()∞= 11 × = 10V And 10+ 1 Method -2 s is applied.~~

~~∵ pulse of duration 10μ At (t=0) switch is closed~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −+Q Q.8 (b) V0CC()()= V0 = C The circuit (in s- 10 = 20V domain) 0.5~~

VC ()∞= 100V = ∞++− − ∞ t/RC VtCC()() V{} V0 C() V C() e −t/1 VC () t=+− 100 (20 100)e 20 dVC (t) ls() = ∴=itC () C ++ + dt ()20 40 40 10s −t 20 2 =0.5 ×− 80 × ( − e ) = = − 10s++ 100 S 10 = 40e t 2 i() t l= 40e−1 i() t =L-1[] l(s) =L -1 C t1= S+10 =14.71A R − eff −10t + L = 2e Or it() = iL () 0 e t Q.6 (b) (20++ 40 40) At →+ the capacitors act as t − ()t 0, =2e10 = 2e 10t short- circuit. At ()t,→ ∞ the Initial stored energy in inductor capacitor will become open circuit. 1 2 + W0L= Li() 0 2 1 =××=10 22 20Joules 2 Remaining energy in inductor

W10= 0.05W =0.05 × 20 20 = ×=10 10V 1Joule ∴10 Voltage+ 10 across capacitor 1 Li2 = 1 2 L Q.7 (c) 1 2 - , the ××=10 i1 1 2 circuit is in steady state. So, inductor 1 behavesBefore closing as short the-circuit switch at 0 i1 = = 0.4472 5 −10T i1 = 2e Let at t=T current decrease to i 0.4472= 2e−10T T= 0.15sec

− 10 Q.9 (c) iL =() 0 = = 1A Before moving the switch, at t = 10 The circuit is in steady state and− After closing the switch at = + t0 inductor behaves as short-circuit.0 Current through inductor cannot The circuit at t = change abruptly − +− ∴===iLL()() 0 i 0 1A 0

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

VVVpST= = − 120 Polarity of the windings is decided iL () 0= = 2A 20+ 40 on the basis of dot –convention. After moving the switch, at t0= + As capacitor is charged to 5 V with Current through inductor cannot left plate as positive. change abruptly So, T1 is positive wrt T2 +− V=−= V V 5V So, i0LL()()= i0 = 2A T T1 T2

~~As T2 has negative polarity. so P1 has negative polarity~~

Therefore, Vp=−=− V P1 V P2 5V Similarly, S has negative polarity So, V=−=− V V 5V S S11 S2 VS −5 V= i 0+ ×+ {20 R} I=−=− 0.5A LL() R R 10 120=×+ 2 (20 R) R= 40Ω Q.12 (d) At steady state, Inductor acts as Q.10 (b) short & capacitor acts open The relevant circuit is shown in Fig.

5 Current through R= = 5A 3 1 2 1 ω = 0 LC =5 −= 1 4A ∴ Current delivered by 5V source It is a standard LC circuit. Q.13 (d) With V() t= V cos()ωt c 00 Equivalent circuit t0= + at is, or V00 sin ()ω t+ 90°

~~Q.11 (d)~~

~~By nodal analysis, V (V− 10) −+10 = 0 10 10 ⇒2V −= 10 100 ⇒=V 55V (55− 10) ⇒=I = 4.5A 10~~

~~Q.14 (c)~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 The status of the circuit at t0= + is C= F; R = 6Ω eq3 eq shown if Fig. 2~~

~~⇒=T Req .C eq = 4sec~~

~~Q.15 (d) Capacitor charged up to 5 s, so total~~

μ Voltage across the capacitor before 3C× 31× connectingcharge stored to indictorin capacitor =Q=13nC V=1 = = 1V − CC+ 3 Q 13× 10 9 12 X0 = = = 43.33V C 0.3×10−9 Voltage across the capacitor at time Q.18 (b) −+ t Vcc()() 0=⇒= 4V V 0 4V

Vc () t at t= 1μs + + V0c () V() t |= [] V cos ωt ic () 0= = 1A c t1= μs 0 0 1=μs 4 1 −6 ω0 t= ×× 1 10 −−39 Q.19 (d) 0.6× 10 ×× 0.3 10 Initial current through the indictor =2.357rad = 135° is zero and capacitor voltage is Vc() t | t1= μs = 43.33 × cos135° charged upto voltage Vc () 0= 10V. ≈− 30.6V As current through inductor and voltage across can-not change

~~Q.16 (c) abruptly. So, after closing the switch +− i0LL()()= i0 = 0And +− VcC()() 0= V 0 = 100V The circuit is s –domain~~

~~Charged stored in the capacitor = area under i-t curve~~

QA=12 + A 11 =()()2 × 10−−63 ×× 4 10 +()() 4 +× 2 10 − 3 ×−× 5 2 10 − 6 100 22 63× Is() = s =+×=4 10−9 13nC 1 2 sL + sC Q.17 (a)  − 100 1 The status of the circuit at t0= is =  1 L s2 + LC shown in fig .1  C 1/ LC =  100 2 L 2 + S() 1/ LC Taking inverse Laplace transform

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission i() t= L−1 [] I(S) C1 = 100 sin sin t L LC 10× 10−6  1 =××100 sin t −3  1× 10 1× 10−−36 ×× 10 10 i() t= 10sin() 104 t A

~~Q.20 (d) When the switch in closed at t =~~

~~Capacitor C1 will discharge and C2 0 will get charge since both C1 and~~

C are ideal and there is no – −t 2 V() t= (Initial — final)z + final value resistance in the circuit charging Whence switching is in position 1 − t and discharging time constant will RC Vtc() = V 01 1e − zero.  Thus current will exist like an When switch is in position 2 impulse function. Initial value is − t Vt() = V 1e − RC Q.21 (b) c 01  Final value is —V :RL = 5Ω − t V= V coscos φ= 136 × 0.45[FromQ 7] Vt() = V V −−02 V  1e2RC RL 3 1 c 01 02 01  2 VR ∴=V 61.2V P=L 749ω = 750 RL RL R Q.24 (10) 11 Y(S) = = Q.22 (a) Z(S) (1+ jω ) Phasor diagram 1 Y(S)= ∠− tan−1 (u) V= 220V V = 122V 2 1 12 1+ ω V3 = 136V vin (t)= 8 + 12sin t →→→

~~= + 1−1 12 vv12v3 i(t)= 8 ∠− tan (0) + sin(t − 4) 10++ 10 12 1 1 i(t)= 8. sin t − cos t 10+  2 2 i(t)=+− 8 6sint 6cost~~

22 2 66 Irms =++= 8 10 By parallelogram law of addition of 22 vectors 22 V1= V 2 ++ V 3 2V 23 V cos φ Q.25 (d) − ∴by using options, V Rt cosφ= 0.45 L i−=− iL (t) 1 e R  Q.23 (d)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −Rt di V  Vc −40000 ×× 100 10−6 L = e L =4e = 1.47 dt L  t= 25μ sec di2 R −Rt → = − Ve L 22 Vc −40000 ×× 100 10−6 dt L → =4e = 0.073 di2 RV t= 100μ sec = − 22 ∆=Q 5[] 1.47 − 0.073 = 6.99μs dtt0= L → Q.26 (2) Q.29 (a) - Network is in steady state From the given circuit, consider the with S opens S2 following circuit diagram At t = 0 - say iL 1 3 (closed) So we can = = 1.5A + indicator 2 (0 ) behaves as idealAt t = current0 source of

+, both switch closed Writing Nodes1.5A if equation we draw at the VL network+ at t = After rearrangement, 0 + 11 33 VL () 0 = + = +(0 -1.5) node 12 12 + ⇒=VL (0 ) 2 ++ di() 0 diL () 0 ⇒⇒L= 2= 2A / sec dt dt For t0≥ − I0 = i0() = 2.5A We can write Rt − L i() t= Ie0 −4t Q.27 (c) i() t= 2.5e A

Q.28 (6.99) Q.30 10 − Given It is given V0C () = 4 1 R=5Ω, C+4f so = 40000 V()() t= 5 − 10cos ω+ t 60 RC i −−t/τ − 40000t VcC() t= V() 0 e ;t >= 0 4e (i) ii ()() t=+ω 5 X cos t → last by capacitor T(ihei) average power is given by, → We are asked to find the charge We know in a capacitor Q = CV P=−φ V I V I cos From t = 25μs to 100 μs avg 0 0 01 01 1

~~∆=QCVcc()() 25μsec − V 10μsec Where Or ∆Q = C(∆V) −40000t Vc = 4e V01 I 01 cos φ 1 = Fundamental Power →~~

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

~~10× X P=×−() 5 5 cos60 = 0 avg 2~~

~~10X 1 25 = × 22~~

~~100 X= = 10 10~~

~~Hence, the value of X is 10Ω~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 5 RESONANCE~~

5.1 INTRODUCTION voltage depending upon the values of XL and XC Consider series RLC circuit shown The resonance in the electric circuit is in figure (1). The total impedance for the because of the presence of both the energy series RLC circuit is

storing elements i.e. the inductor and the ZZ=++RLC Z Z capacitor. At resonant frequency ωr , the 1 Z= R +ω jL + inductor and capacitor will exchange the jcω energy freely as a function of time, which 1 results in sinusoidal oscillations either ZRj=+−ω L across the inductor or across the capacitor. ωc In two terminal electrical network It is clear from the circuit that the current V containing at least one inductor and one I = S capacitor, we define resonance as the Z condition which exists when the input The circuit is said to be in resonance if impedance of the network is purely current is in phase with the applied voltage. resistive. A network is in resonance when In series RLC circuit resonance occurs

the voltage and current at the network when XXLc= . The frequency at which the input terminals are in phase. resonance occurs is called the resonant In many electrical circuits, resonance is a frequency (f ). very important phenomenon. The study of Since XX= the impedance in series RLC resonance is very useful, particularly in the Lcr circuit is purely resistive. At resonant area of communications. For example, the frequency (f ) the voltage across ability of radio receiver to select a certain frequency, transmitted by a station and to capacitance and inductance are equal in magnitude. r eliminate frequencies from other stations is At resonance, based on the principle of resonance. = The resonance may be classified into two XXLC groups 1 ω=rL 1) Series Resonant circuit ωrC 2) Parallel Resonant circuit 1 ω=r (rad / sec) LC 5.2 SERIES RESONANCE 1 fr= (Hz) 2π LC

~~5.2.1 IMPEDANCE OF A SERIES RESONANT CIRCUIT~~

The impedance of a series RLC circuit is As shown in figure above the resistor, 1 ZRj=+−ω L inductor and capacitor are connected in ωc series, so the circuit is series RLC circuit. A 2 sinusoidal voltage ‘VS’ sends a current ‘I’ 2 1 ⇒=ZR +−ω L through the circuit. In series RLC circuit, ωC the current lag behind or leads the applied

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission • At frequencies above the resonant frequencyf , XL is larger than XC causing Z to increase. XL causes the total current to lag behindr the applied voltage. When XL>XC the circuit is predominantly inductive (as shown in the graph above)

~~5.2.3 FREQUENCY RESPONSE OF SERIES RLC CIRCUIT As shown in above graph • At zero frequency, |Z|is infinitely large. • At resonant frequency fr, |Z|=R • At frequencies ffr, |Z|, increases.~~

5.2.2 THE VARIATION OF XC AND XL WITH FREQUENCY 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

∴=−BW f21 f (Hz) The unit of BW is Hertz (Hz). There is another relationship for BW which is given by f BW=−= f f r • In series RLC circuit, the current lags 21Q behind or leads the applied voltage Where, Q is known as quality factor. depending upon the values of XC and XL. The upper and lower cut off frequencies is At zero frequency, both XC and Z are sometimes called the half power infinitely large, and XL is zero because at frequencies. zero frequency the capacitor acts as At resonant frequency the power is open circuit and the inductor acts as a P= IR2 short circuit. As frequency increases, XC max max decreases and XL increases. While XC At frequency f1, the power is 2 causes the total current to lead the I IR2 =max = max applied voltage. When > he PR1  XXcLt 2 2 circuit is predominantly capacitive (as Similarly, at frequency f , the power is – shown in the graph above). 2 2 • In series RLC circuit, resonance occurs Imax IR max 2 PR2 = = 2 2 when XXLC= . The frequency at which the resonance occurs is called the 5.2.4 MAGNIFICATIONS resonant frequency (fr). Since, XXLC= the impedance in a series RLC circuit is We know that, in series RLC circuit, a purely resistive and Z=R (as shown in sinusoidal voltage ‘VS’ sends a current I the graph above) through the circuit.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 Vs I RT So, I = = R 2 = 2 At ωωr LI Vs 2 1) VR = IR = R ⇒=VVRs So, Q2= π× R I2 RT V 2) V= IZ = s jωL⇒=V QV ∠ 90° 2 LLR r Ls L Q2= π× Vs 1 RT 3) VCC =IZ = ⇒VCs = QV ∠− 90° Rjωc L1 r Q2=×=π Q T ωL 1 1 f Where, Q = r where, Q = R R ωr CR f 2π fL Q = 5.2.5 PHASOR REPRESENTATION R ωL Q = R 11 Q=Q ω L = at resonance ωωCR C In series RLC, the quality factor ωL1 Q = = At the resonant frequency fr, the voltage Rω cR across capacitance and inductance are ωL 1 1L equal in magnitude, since they are out Q = = = 180° Rω cR R C of phase with each other, they cancel each So, Q-factor is a function of only circuit other and hence zero voltage appear across constants. As Q is high then the circuit is the LC combination . said to be more selective and oscillation produced are high quality in nature. 5.2.6 QUALITY FACTOR (Q) AND ITS EFFECT ON BANDWIDTH Example: In circuit shown in figure, determine the The Quality factor (Q) is the ratio of the circuit constants when the circuit draws a reactive power in the inductor or capacitor maximum current at 10μF with a 10v, to the true power in the resistance in series with the coil or capacitor. 100Hz supply. When the capacitance is The Quality factor, changed to12μF , the current that flows max imum energy stored through the circuit becomes 0.707 times its Q2= π× maximum value. Determine Q of the coil at energy dissipated per cycle 900 rad/sec. Also find the maximum In an inductor, the maximum energy stored current that flows through the circuit is given by LI2 = 2 and energy dissipated per cycle = power × Periodic time for one cycle. 2 I =RT ×  2 Solution:

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission At resonant frequency, the circuit draws 1 = maximum current. So, the resonant 0.1×× 5 10−6 frequency = fr 100Hz =1414.2rad / sec 1 1414.2 fr = ⇒=f= 225HZ 2π LC r 2π 1 The quality factor L = 2 ω×L 1414.2 0.1 c2f()π r Q= = = 28 1 R 50 = = 2 0.25H f 10× 10−6 () 2 π× 100 Since, r = Q BW 1 We have, ω−LR = The Bandwidth, ωC f 225 =r = = 1 BW 80.36Hz 900×− 0.25 = R Q 2.8 900×× 12 10−6 ⇒=R 132.4 Ω 5.3 PARALLEL RESONANCE The quality factor ω×L 900 0.25 Q= = = 1.69 R 132.4 The maximum current n the circuit is 10 I= = 0.075 132.4

Example: As a shown in fig. above the resistor, In the circuit shown in fig, a maximum capacitor and inductor are connected in current of 0.1A flows through the circuit parallel, so the circuit is parallel RLC when the capacitor is at 5μFwith a fixed circuit. The current flowing through frequency and a voltage of 5V.Determine resistor, inductor and capacitor are

the frequency at which the circuit IRL , I and I C respectively. resonates, the bandwidth, the quality factor Consider parallel RLC circuit shown in fig. Q and the value of resistance at resonant at The total admittance f parallel RLC circuit resonant frequency . is: r YY=++RLC Y Y 11 Y= + ++ jc ω R jLω 11 Y= + j( ω− c ) RLω At resonance, the current is maximum in It is clear from the circuit that current the circuit. It is given by I = YV V So voltage across parallel elements is same. I = R Parallel resonance occurs when XXCL= V5 the frequency at which resonance occurs is ∴===Ω R 50 I 0.1 called the resonant frequency. When The resonant frequency is XXCL= , the two branch currents are 1 equal in magnitude and 180° out of phase ω=r LC with each other.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The condition for resonance occurs when • In parallel RLC circuit, resonance occurs XX= 1 CL when = ωC . The frequency at which 1 ωL At resonance, = ω rc the resonance occurs is called the ωrL 1 1 resonant frequency (f ) since = ωC ω=r (rad / sec) ωL LC the admittance in parallelr RLC circuit is 1 f= (Hz) I r π purely resistive and Y = 2 LC R • At frequencies above resonant 5.3.1 ADMITTANCE OF A PARALLEL 1 RESONANT CIRCUIT frequency fr, w is large than so c ωL 1 ωC > the circuit is predominantly ωL capacitive.

~~5.3.3 MAGNIFICATION~~

We know that in parallel RLC circuit, voltage is As shown in above graph- II • At zero frequency, Y is infinitely large. V= = = IR Y I 1 R • At resonant frequency fr , Y = R The Response at resonance

~~• At frequencies ff< r , Y increases. At ω=ωr • At frequencies > , increases. V IR ffr Y 1) I = = ⇒=II R R R R~~

~~5.3.2 THE VARIATION OF AND V IR 2) IL = = ⇒IL = QI ∠− 90 ° WITH FREQUENCY ZL jωLr 𝑿𝑿𝑳𝑳 𝑿𝑿𝑪𝑪 3) V IR ⇒I = QI ∠° 90 Ic = = = jωr CRI c Zc 1~~

~~jωCr R Where Q = where Q= ωr CR ωrL 5.3.4 PHASOR REPRESENTATION~~

1 • At zero frequency both and Y are ωL 1 At the resonant frequency, when XLC= ,X large and ωCis zero. So >ωC the ωL the two branch current are equal in circuit is predominantly Inductive. magnitude and 1800 of phase with each

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission other. Therefore, the two current cancel out each other and the total current is zero.~~

~~Example: Determine the resonant frequency of the circuit~~

Solution: 11   z=++ 10 j4ω  ||  jjωω   1 − 4 2 =10 + ω 2 j4ω− ω 1 − j(42 ) =10 − ω 2 4ω− ω At resonant frequency imaginary terms equals to zero 1 4 − ω2 r = 0 2 4ω−r ωr 1 1 ⇒− =⇒ω = 402 r rad / sec ωr 2

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)~~

~~Q.1 The differential equation for the a) b) current i(t) in the circuit of the figure is~~

~~c) d)~~

d2 i di a) 2++= 2 i() t sin t dt2 dt d2 i di b) ++2 2i() t = cos t [GATE-2004] dt2 dt d2 i di c) 2++= 2 i() t cos t dt2 dt d2 i di d) ++2 2i() t = sin t dt2 dt Q.3 In the following graph, the number [GATE-2003] of trees (p) and the number of cut sets (Q) are Q.2 Consider the network graph shown in the figure. Which one of the following is NOT a ‘tree’ of this graph?

~~a) P= 2,Q = 2 b) P= 2,Q = 6 c) P= 4,Q = 6 d) P= 4,Q = 10 [GATE-2008]~~

~~ANSWER KEY:~~

~~1 2 3 (c) (b) (c)~~

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

~~Q.1 (c) Applying KVL, di(t) 1 sint= it() ×+ 2 L + itdt() dt C ∫ di(t) sint=++ 2it() 2 itdt() dt ∫ Differentiating with respect to t 2di(t) 2d2 i(t) cos t() t=++ i(t) dt dt2~~

~~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~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 6 GRAPH THEORY~~

~~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. Fig (2) has four nodes and six branches. A network is an interconnection of active This graph is called as undirected graph as elements (voltage and current source) and directions are not given. passive elements (resistor, capacitor, and Each branch or edge of the graph carries an inductor) as shown in fig (1) 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 If two or more branches or elements of a node only one branch is connected for all network intersect at a single point, that the combinations. The number of edges of a point is known as node and it is denoted by complete graph with n nodes is ‘n’. As shown in above network there are n(n− 1) four nodes represented by number (1, 2, 3, 2 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 From given graph, we have 4 nodes graph of network (shown in fig (1)) can be So, node(n)= 4 and branches drawn as (shown in fig (2))

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission n(n−− 1) 4(4 1) = = = 6 22 In a connected graph all the nodes are connected by at least one branch otherwise it is said to be unconnected 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. Sub graph is a graph with less number of (6−+ 4 1) = 3 branches. branches as compared with the original Following figures shows possible trees and graph. co-trees 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 Tree cotree return to the staring node such a closed (2,5,6) (1,3,4) 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 or vertices of a graph. The branches of a tree (2,4,5) (1,3,6) 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 (1,4,6) (2,3,5) number of links or chords is Note: • Cotree may consist of a closed loop. b−() n1 − =−+ bn1 • Also, tree + cotree = original graph Let us consider following directed graph. 6.1.2 PROPERTIES OF TREES

~~1) In a tree there exists one and only one path between any pair of nodes.~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2) Every connected graph has at least one entry of that element in incidence tree. matrix represented by ‘-1’ 3) The number of terminal nodes or end 3) And entry of the element represented vertices of every tree are at least two. by‘0’ when branch is not connected to 4) A connected sub graph of a connected that node. graph is a tree if there exists all the For the above fig, the complete incidence nodes of the group. matrix A is given as – 5) Each tree has (n-1) branches, where n is nodes → bran esch the number of nodes of the tree. ↓ 6) The rank of tree is (n-1).This is also the abcdef rank of the graph to which the tree belongs. 1101001  7) The number of possible trees of 2−− 1 10 10 0 A =  complete graph is given by n(n− 2) . 30 1 0 01− 1  e.g. For the complete graph if nodes n=4 400− 1−− 1 10 then number of possible trees − =4(4 2) = 16possible trees. 6.2.1 PROPERTIES OF INCIDENCE MATRIX 6.2 INCIDENCE MATRIX 1) The rank of incidence matrix is equal to the rank of graph =(n − 1) 2) For any graph the incidence matrix is unique and its order is n×b 3) The determinant of the incidence matrix of a closed loop is always equal to zero 4) The sum of all the entries in a column is zero. 5) The number of non-zero elements (1’s In the above oriented graph, from the and -1’s) in a row is called the degree of arrows indicated in the graph, it is possible a node. to tell which branches are incident at which 6) If the degree of a node is two, then it nodes and what is the orientation relative indicates that two branches are incident to the nodes. The most convenient way in at the node and these are in series. which this incidence information can be 7) Column of ‘A’ with unit entries in two given is in a matrix form known as identical rows corresponds to two incidence matrix ‘A’ For a graph with n branches with same end nodes and nodes and b branches, the complete hence they are in parallel. incidence matrix ‘A’ is rectangular matrix of order n×b whose elements have following 6.3 FUNDAMENTAL LOOP MATRIX OR value TIE –SET MATRIX 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 its direction is towards the node then

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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- From fig. (3) KVL can be written as: loop or a tie-set vvv+−= 0 ______(1) Def: Fundamental circuits or Fundamental 156 loops are the minimum number of loops or Now By adding the other link branches 2 mesh equations required to solve a given and 3 we can form two more fundamental network. loops as follow- Steps: 1. Select a tree 2. 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 current direction.

Example: Consider connected graph shown above, it From fig. (4) f-loop can be written as: has 4 nodes and 6 branches. One of its trees +−= ______(2) is arbitrarily selected having twig or tree vvv0245 branches (4, 5, 6) and link corresponding to it is (1, 2, 3) as shown below

~~From fig. (5) f loop can be written as:~~

~~vv34−= 0 ______(3) Let i12 ,i ,…..i6 be the branch currents with Equation (1),(2) (3) can be written in matrix form as follow- directions shown in fig above and v12 ,v ,~~

…..v are the branch voltages. Let us add a 6 loop link (say link represented by1) in its proper Branches → ↓ place in selected tree (shown in fig. 123456 (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)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission v1  v2 I1  1 0 00 1− 1   0  v3   I 0 1` 0 1−= 1 0 0 2  v  − 4   I3  0 0 11 0 0  0 v5  Fig (1) shown above is not a complete v 6 graph since there are two branches (3 & 4) Note: In general the order of tie-set matrix between nodes III and IV is (b−+× n 1) b i.e. (link × branches) The removal of the branch 1 and 5 divides The elements in tie-set or fundamental loop the graph into exactly two parts. So (1, 5) matrix is defined as follow- may be a cut-set (as shown in fig.2). 1) When direction of loop current and reference direction of loop coincide represent that element by ‘+1’ 2) When direction of loop current and reference direction of loop are opposite represent that element by ‘-1’ 3) When any branches is not in fundamental loop represent it by ‘0’ Cut set (1, 5) Other possible cut-sets are 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 at a time, for any graph the number of fundamental loops equal to link Cut-set (1, 2) (b−+ n 1) 3) Every fundamental loop consists of only one link in its representation. 4) Since every tree will result one tie-set matrix at a time, for any graph the number of tie-set matrices are always equal to the number of trees 5) For a complete graph there are n(n 2) Cut-set (5, 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: Cut-set (2, 3, 4)~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let’s select a tree (1, 7, and 5) as shown below~~

~~Cut-set (5, 3, 4)~~

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 Cut-set (1, 3, 4) constitute a cutest, we call this a fundamental cut-set as shown in fig below.

~~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: So from above we get fundamental f-cut set 1) Select a tree. (1, 2, and 4). 2) By removing one tree branch at a time Similarly consider twig 7 then f-cut set will result one f-cut set at a time. drawn as follows: 3) Select the f-cut set direction as in the tree branch direction.

~~Example:~~

~~Note: It is not a compete graph~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Properties: 1) The rank of the f-cut set matrix is n-1 2) Since every tree branch will result one f –cut-set at a time, for any graph the number of f-cut-set equal to number of twigs equal to n-1 3) Every f-cut set contain only one tree branch is its representation. f-cut-set obtained after twig 7 removed is 4) Since every tree will result one f-cut set (4,7,6,3) matrix at a time, for any graph the Similarly consider twig 5 removed, we can number of f-cut-set matrices are always draw cut set as: equal to the number of trees. 5) For a compete graph there are n()n2− f- cut set matrices.

~~So we get fundamental cutest (5, 6)~~

~~6.4 F–CUT SET MATRIX (QC)~~

f− cutset Branches → ↓ 1 2 3 456 7 1  1101000− Q7=  C 00−− 1 10 − 11 5  0000110 Where, 1) When the direction of the f-cut-set and that of the other branches coincide then is represented by ‘1’ 2) When the direction of the f cut-set and that of the other branches are opposite then it is represented by ‘-1’ 3) When f-cut-set doesn’t contain any branch then that branch is represented by ‘0’

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)~~

~~Q.1 In the figure, the value of the load average power to a load impedance~~

~~resistor R which maximize the ZL when power delivered to it is a) ZLS =R +jX S b) ZLS =R~~

~~c) ZLS =jX d) ZLSS =R -jX [GATE-2007]~~

~~Q.5 For the circuit shown in the figure, the Thevenin voltage and resistance looking into X-Y are a)14.14Ω b)10Ω c) 200Ω d) 28.28Ω [GATE-2001]~~

Q.2 A source of angular frequency 1 rad/sec has source impedance a) 4 / 3V, 2Ω b) 4V,2 / 3Ω consisting of 1Ω resistance in series 2 with 1 H inductance. The load that c) 4 V, d) 4V,2Ω Ω will obtain the maximum power 33 transfer is [GATE-2007] a) resistance The Thevenin equivalent impedance b) resistance in parallel with 1H Q.6 ZTh between the nodes P and Q in inductance1Ω the following circuit is c) 1Ω resistance in series with 1F capacitor d) 1Ω esistance in parallel with 1 F capacitor 1Ω r [GATE-2003]

~~The maximum power that can be 1 Q.3 a) 1 b) 1s++ transferred to the load resistor R L s from the voltage source in the figure 1 s2 ++ s1 c) 2s++ d) is s s2 ++ 2s 1 [GATE-2008]~~

~~Q.7 In the circuit shown, what value of~~

~~R L maximizes the power delivered to R ? a)1W b)10W L c)0.25 W d)0.5W [GATE-2005]~~

~~Q.4 An independent voltage source in series with an independence~~

~~ZSS= R + jX S delivers a maximum~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8 maximum power is transferred from b) Ω 3 circuit A to circuit B is a)2.4Ω [GATE-2009] c) 4 Ω d) 6Ω Q.8 In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

a) 0.8Ω b)[GATE 1.4Ω -2012] c) 2Ω d) 2.8Ω Q.12 A source Vs () t =Vcos100πt has an internal impedance of 4+j3Ω . If a purely resistive load connected to a) 6.4-j4.8 b) 6.56-j7.87 this source has to extract the c)10+j0 d)16+j0 maximum power out of the source, [GATE-2011] its value in Ω should be a)3 b)4 Q.9 In the circuit shown below, the c)5 d)7 [GATE-2013] value of R L such that the power transferred to R is maximum is L Q.13 In the circuit shown below, if the

~~source voltage VS = 100 ∠ 53.13° V then the venin’s equivalent voltage in volts as seen by the load resistance RL is~~

~~a)5Ω b)10Ω c)15Ω d) 20Ω [GATE-2011]~~

~~Q.10 The impedance looking into nodes 1 a) ∠ b) ∠ and 2 in the given circuit is 100 90° 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)50Ω b)100Ω a) in series with a current source c)5kΩ d)10.1kΩ b) in parallel with a voltage source [GATE-2012] c) in series with a voltage source d) in parallel with a current source Q.11 Assuming both the voltage sources [GATE-2014] are in phase the value of R for which

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.15 In the figure shown, the value of the Q.18 In the circuit shown below, Vs is a~~

~~current I (in Amperes) is______. constant voltage source and IL is a constant current load.~~

The value of I that maximizes the [GATE-2014] L power absorbed by the constant Q.16 In the circuit shown in the figure, current load is V V the angular frequency ω (in rad/s), a) s b) s at which the Norton equivalent 4R 2R impedance as seen from terminals V c) s d) ∞ b-b' is purely resistive, is ______. R [GATE-2016]

~~Q.19 Consider the circuit shown in the figure.~~

~~[GATE-2014]~~

~~Q.17 In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is ______.~~

~~The Thevenin equivalent resistance -Q is______.~~

~~(in Ω) across P [GATE-2016] [GATE-2017]~~

~~ANSWER KEY :~~

~~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~~

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

Q.1 (a) s++ 1 (11 ) () S ωL= 10Ω ∴=ZZth PQ = = 1 2S++1 R for max power transfer S =10 − jωL, Q.7 (c) (Complex conjugate of Zeq ) For maximum power transfer, RL = R =10 − 10j =10 2 ∠ 45° =14.14Ω th

~~Q.2 (c)~~

~~ZLS= R − jX S~~

~~∴=−ZL 1 1j~~

Q.3 (c) VOC = 100V For maximum power transfer, 100 ()100+ Vx R= R = 100Ω Ish = + LS 84 V2 55× Also V =-50V, ∴= P== 0.25W x max R 100 ∴=Ish 12.5 + 12.5 = 25A (V acrossRL = 5V) VOC ∴=R4th =Ω ⇒=R4L Ω Ish Q.4 (d) For maximum power transfer, Q.8 (a) ZLS= Z * = R S − JX S 25 I= I = 16∠ 0° × N sh PQ + Q.5 (d) ()40 J30 −1 3 =−=−8∠∠ tan 8 36.86 4 hence Norton current is

~~IN= I SC =∠− 36.86°~~

~~For VOC IN =() 6.4 − j4.8 A Apply nodal analysis, V V− 2i Q.9 (c) ∴xx −+2 V + = 0&V = i 21xx VV ⇒xx −+2 2V − 2i =⇒ 0 = 2 22x~~

~~⇒==VOC V x 4V~~

Similarly, Ish =2A V ∴=V 4V;R =OC = 2Ω For Maximum power transfer to RL th th I sh RRL= th = Thevenin’s resistance seen Q.6 (a) RTh Replace 10V by short and 1A by across the terminals of RLinto the open, rest of the NW. The relevant circuit

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission is shown in Fig 1, where the ⇒5() R += 2 2(3 + 5R) independent current source is open ⇒5R +=+ 10 6 10R ⇒=4 5R circuited and the voltage sources are ⇒=R 0.8Ω short circuited. Q.12 (c) Q.10 (a) = =22 +=Ω RL Z th 43 5 Q.13 (c)

VTH= 10V L1 4 V 100∠ 53.13 −1 V==c −× tan8 j4 L1 3+ j4 3 After connecting voltage source of V V= 80 ∠ 90° V= V ⇒() 10K()() −= i 100 I + 99i + i ; L1 1 2 b bb = ∠ VTH 800 90° −10000ib = 100I +× 100 i b =100I + 10000i −20000i= 100I ⇒ i Q.14 (d) b bb Norton's theorem −−100  I  =I = 20000  200 

~~V= 100[] I ++ 99ibb i  −I =100 I + 100= 50I 200 V 50I R= = = 50Ω Q.15 (0.5) th II~~

Q.11 (a) Power transferred from circuit A to 7  6+ 10R  circuit B= VI =    R2++  R2  Apply KCL at node 42+ 70R = V5− V ()R2+ V, −+1 = 0 5 15 10− 3 7 30 I = = ⇒V = volts 2R++ 2R 4 7R 6+ 10R V2 V=+=+ 3 IR 3 =  ⇒current I = ⇒⇒ 0.50Amperes 2R+ 2R+ 15 4 2 dP ()()()()R+ 2 70 −+ 42 70R 2 R + 2 = Q.16 (2 r/sec) 4 dR ()R2+ Norton's equivalent impedance 1 1*jω . 1 jω.1 Z =2 + = + N 1 ω + ωω 1.+ jω j .1 2 jj 2 ()2 −+ω2 jω = 2 ZN 2 70()()() R+=+ 2 42 70R 2 R + 2 [2jω− ω]

Equating imaginary term t zero i.e., condition, half of Vth is dropped ω3 -4 ω=0 V across R and remaining th ⇒ωω()2 − 4 =⇒= 0 ω 2r/sec th 2 dropped across load. V Q.17 (0.8) → So we can say under MPT s To find maximum power 2 transferred to load we need to will appear on the load V obtain thevenin equivalent of the V − s s V circuit → obtaining Voc so I = 2 = s L R 2R

~~Q.19 1Ω~~

~~2 V= 5= 2V o 32+ 40 V= = 100 oc 10+ 40 4 V = ×100×2=160V 0 5 →obtaining Isc~~

~~2 V= 5 = 2V o 32+ 100V 200 = I=o = = 20mA sc 10 10~~

~~Voc 160 Rin = = = 8kΩ Isc 20 so the network is~~

~~→for MPT R = 8k 2 2 V+n 160 P=mcr = = 0.8watt. 4R+n (4×8)×103~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

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)3 b)4 a) B-N+1 b) B+N c)5 d)6 c) -B+1 d) N-2B-1 [GATE-2008] [GATE -2002] Q.4 The graph associated with an Q.2 The matrix A given below is the electrical network has 7 branches node incidence matrix is a network. and 5 nodes. The number of The columns correspond to independent KCL equations and the branches of the network while the number of independent KV L rows correspond to nodes. Let equations, respectively, are T a) 2 and 5 b) 5 and 2 V=[] vv12 … v 6 denote the vector of T c) 3 and 4 d) 4 and 3 branch voltages while I=[] ii12 … i 6 [GATE-2016] that of branch currents. The vector T Q.5 The voltage vt() across the E= [] eeee1234 denotes the vector of node voltages relative to a common terminals a and b as shown in the ground. figure, is a sinusoidal voltage having a frequency ω= . When 111000 100radian / s  the inductor current in phase 0−− 1 01 1 0 it() A =  −1 0 00 −− 1 1 with the voltage vt() , the magnitude  0 0− 11 0 1 of the impedance Z (in Ω ) seen Which of the following statements is between the terminals a and b is true? ______(up to 2 decimal places).

~~a) The equations v12 -v +v 3 = 0,~~

~~v3 + v 45 -v = 0 are KVL equations for the network for some loops~~

~~b) The equations v136 -v -v = 0,~~

~~v4 + v 56 -v = 0 are KVL equations for the network for some loops c) E=AV [GATE-2018] d) AV=0 are KVL equations for the network [GATE-2007]~~

~~Q.3 The number of chords in the graph of the given circuit will be~~

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

~~ANSWER KEY:~~

~~1 2 3 4 5 (a) (b) (a) (d) 50~~

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

~~Q.1 (a) Number of branches =B=6 Number of links = B - (N - 1) Number of tree branches = (N-1) = 3 = B – N + 1 Number of links =L=B-(N-1) =3~~

Q.2 (b) For the given node–to branch incidence matrix 111000  0−− 1 01 1 0 A =  −1 0 00 −− 1 1 ∴ Statement in options (b) is true. 0 0− 11 0 1 Q.4 (d) The graph of the network is shown No of branches = 7 in fig 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

regarding how many simple & principalSince nonode, information if we assume given all principal nodes then the answer for Where N=4, B=6 nodal is 5 —1 From the graph it can be observed that Q.5 50 Given: i) V123−+= V V 0, V 345 +−= V V 0 are not KVL equations as set of branches (1,2,3) & (3,4,5) do not form closed paths.

~~ii) VVV136−−= 0 and~~

VVV456+−= 0 are KVL equations for the loops (1,3,6) 1 and (4,5,6) From the matrix, A it 100 jCω can be concluded that Z= +ω jL (i) + E≠ A V j 100 −  (ii) A V= 0 are not KVL ωC equations j100 j −+100 Q.3 (a) ωωCC = +ωjL The graph of the given circuit is 1 shown in Fig 1002 + ω22C Number of nodes =N=4

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission unity power factor represents resonance.Since, the power factor is unity, Z is zero only real part exist. So, imaginary part of 100 22 Re() Z = ω C 1 1002 + ω22C

~~100 = 2 ()100ω+ C 1~~

~~100 = 2 ()100××× 100 100 10−6 + 1~~

~~Re() Z= 50 Ω~~

~~Hence, the magnitude of the impedance Z seen between the terminals a and b is50Ω .~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 7 COUPLED CIRCUITS~~

7.1 INTRODUCTION 7.2.1 COEFFICIENT OF MUTUAL INDUCTANCE 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 As shown in fig a simple model of two coils electromagnetically coupled. Certain L1 and L2 sufficiently close together that the coupled elements are frequently used in flux produced by a current i (t) flowing network analysis & synthesis e.g. 1 through establishes an open circuit transformer. L1 voltage v2 (t) across the terminals of L2

7.2 MUTUAL INDUCTANCE di1 (t) vt21() = M Mutual inductance is a property associated dt with two or more coils or inductors which Where, v2 is the voltage induced in coil L2 are in close proximity and the presence of and M1 is the coefficient of proportionality common magnetic flux which links the coil. (OR) coefficient of mutual inductance (OR) A transformer is such a device whose simple mutual inductance. operation is based on mutual inductance. A Similarly from Fig (b), we can write transformer consists of two coils of wire di (t) vt() = M 2 separated by a small distance and is 12dt commonly used to convert ac voltage to M and M2 are two mutual inductances are higher or lower values depending on the 1 involved in determining the mutually application. induced voltages in the two coils, it can be A current flowing in one coil establishes a shown from energy considerations that the magnetic flux about that coil and also about two coefficients are equal and, therefore a second coil nearby. The time varying flux need not be represented by two different surrounding the second coil produces letters. voltage across the terminals of the second Thus = = coil; this voltage is proportional to the time MMM12 rate of change of the current flowing di (t) ∴=v() t M1 volts and through the first coil. 21dt di(t) di (t) vt() = L v() t= M2 volts dt 12dt Where, Mutual inductance is also measured in vt() is the voltage across the coil Henrys it() is the current through the coil 7.2.2 DOT CONVENTION L is the inductance of the coil. Dot convention is used to establish the choice of correct sign for the mutually

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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. 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 Consider above circuit which shows a pair terminal ‘a’ and ‘c’ of the two coils attain of linear time invariant, coupled inductor similar polarities simultaneously. The two with self inductance L1 and L2 and mutual terminals are positive & identified by two inductance M. Currents i1 and i2, each dots (show in fig. (1)). arbitrarily assumed entering at dotted The other possible location of the dots is terminals and voltage v1 & v2 are developed the other ends of the coil to get additive across the inductors. The voltage across fluxes in the core i.e. at b and d terminal L1is, thus composed of two parts and is (shown in fig (2)). It is concluded that the given by mutually induced voltage is positive when di (t) di (t) vt() = L12 ± M (1) current i1 and i2 both enter (or leave) the 11dt dt windings by the dotted terminals: n The first term on the RHS of the above eq di1 i.e. vM2 = is the self induced voltage due to i1 and the dt second term represents the mutually induced voltage due to i2. Similarly, di (t) di (t) vt() = L21 ± M (2) 22dt dt 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 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)) di i.e. vM= − 1 2 dt

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 7.3 THE COUPLING COEFFICIENT Let there be two inductors connected in The amount of coupling between the series with self inductances L1 and L2 inductively coupled coils is expressed in and mutual inductance M. Two kinds of terms of the coefficient of coupling, which series connections are possible is defined as M 1) Series aiding (fig a): In case of series k = LL aiding connection, the currents in both 12 inductors at any instant of time are in Where, the same direction relative to like M mutual inductance between the coils terminals (shown in fig a) For this L1 self inductance of the first coil reason, the magnetic fluxes of self L2 self inductance of the second coil induction and of mutual induction linking with each element add together. Coefficient of coupling is always less than For the series aiding circuit Φ & Φ unity and has maximum values of 1. If the 12 value of k=1 then it called perfect coupling are the flux produced by coil 1 and 2 The coefficient k is a non negative number respectively, and flux produced by the and is independent of the reference inductor is given by Φ=Li directions of the currents in the coil. If the So the total flux

~~two coils are a great distance apart in ΦΦ=12 + Φ space, the mutual inductance is very small, Where, Φ=L i + Mi and k is very small. 1 11 2 Φ2= L 22 i + Mi 1~~

~~7.4 SERIES CONNECTION OF COUPLED ∴=Φ L11 i + Mi 2 + L 2 i 2 + Mi 1 INDUCTORS Since iii12= =~~

~~L=++ L12 L 2M~~

2) Series opposition (fig b): In case of series opposition the current in the two conductors at any instant of time are in opposite direction relative to like terminals (shown in fig b) So, for the series opposition

~~ΦΦ=12 + Φ~~

~~Where, Φ1= L 11 i − Mi 2~~

~~Φ2= L 22 i − Mi 1~~

~~∴=Φ L11 i − Mi 2 + L 2 i 2 − Mi 1~~

~~Since iii12= =~~

~~L=+− L12 L 2M In general, the inductance of two inductively coupled elements in series is given by~~

~~L=+± L12 L 2M • Positive sign is applied to the series aiding connection • Negative sign is applied to the series opposition connection.~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission N Where, 1 is turns ratio of transformer 7.5 PARALLEL CONNECTION OF N2 COUPLED COILS Input impedance of transformer is

~~N1 2 Consider two inductors with self Zin= ( )Z L N2 inductances L1 and L2 connected parallel which are mutually coupled with mutual inductance M shown in fig. 7.6.1 EQUIVALENT CIRCUIT~~

di di VL=12 + M (1) 11dt dt di di VL=21 + M (2) 22dt dt di Add and subtract M 1 in equation (1) and dt di M 2 in equation (2), equation (1) dt LL− M2 LL− M2 becomes L = 12 L = 12 eq L+− L 2M eq L++ L 2M di di di di 12 12 VL=++−1 M 211 M M 11dt dt dt dt 7.6 IDEAL TRANSFORMER di d V=−++ (L M)1 M (i i ) (3) 1 1dt dt 12 Transformers are used to transfer energy Equation (2) becomes from one circuit to another circuit through di2122 di di di mutual induction. The transformer winding VL22= ++− M M M dt dt dt dt to which the supply source is connected is di2 d called the primary and voltage across it is =−++ (L2 M) M (i 12 i ) (4) called primary voltage, while the winding dt dt connected to load is called the secondary Using (3) and (4) the equivalent circuit can and voltage across it is called secondary be drawn as voltage correspondingly i1 and i2 are the currents in the primary and secondary windings .

~~Example: Calculate the effective inductance of the circuit shown in figure across terminal A VI N L and B 12= = 1 = 1 VIN21 2 L 2~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission L12 =−() 2M||M(4M) +− Solution: ()2− MM 3= +− 4M Lab=++− L 1 L 2 L 3 2M 12 + 2M 23 2 =+81062425 +−×+× 2M− M2 3= +− 4M = 26H 2 M2 Example: 3M= − +− 4M 2 The equivalent inductance between the terminals 1 and 2 is 4 H with the open M= 2H terminals 3 and 4, and it is 3H with shorted M= k LLsp terminals 3 and 4. Determine the value of k M 21 k = = = LLPS 4× 2 2 1 k = 2

~~Solution: The equivalent circuit of the above network can be as shown in figure.~~

~~When terminals 3 and 4 are open-circuited~~

~~L12= L p −+ MM~~

~~4L= p When terminals 3 and 4 are short circuited~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8 TWO PORT NETWORKS~~

8.1 INTRODUCTION 8.2 OPEN CIRCUIT IMPEDANCE (Z) PARAMETERS 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) The z-parameters of a two port network having voltages in terms of the v12 ,v

~~current I12 ,I are as shown in figure. Here~~

v12 ,v are dependent variables and I12 ,I are independent variables. Fig a. one port Network By kVL at port 1 we get A network having two pairs of terminals is v= zI + zI ---- (1) known as two port network. Usually one 1 11 1 12 2 pair represents the input (1− 1') and the By kVL at port 2 we get = + ---- (2) other represents the output (2− 2’) (as v2 zI 21 1 zI 22 2 shown in fig b) Where z11 ,z 12 ,z 21 ,z 22 are the network functions, and are called impedance (z) parameters we may write matrix eqn [][] v= z [I]

~~v1   zz 11 12  I 1  Thus  =    Fig b. two port Network v2   zz 21 22  I 2  As shown in fig b, input terminal has two From eqn (1)~~

variables v1 , I1 and output terminal has v1 z|11= I= 0 …..Driving point impedance I 2 two variables v22 ,I . Out of these four 1 variables, we can select two variables as Where, Z11 is driving point impedance at port 1 with port 2 open circuited. It is independent variables represented as 4C2 called the open circuit input impedance. i.e. 6 ways and hence six sets of two port Similarly, from eqn (2) network parameters, they are v 1) z-parameter z|= 2 … Transfer impedance 21 I2 = 0 2) y-parameters I1 3) Transmission (ABCD) parameters. Where, 4) Inverse transmission (A’B’C’D’) z21 Is transfer impedance at port 1 with Parameters port 2 open-circuited. It is also called the 5) Hybrid (h) parameters open circuit forward transfer 6) Inverse hybrid (g) parameters impedance. From eqn (1) v z|= 1 …Transfer impedance 12 I1 = 0 I2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Where, I2 Y|21= v= 0 …Transfer admittance z is transfer impedance at port 2 with 2 12 v1 port 1 open circuited. Where, Y21 is the transfer admittance at It is also called the open circuit reverse port 1 with port 2 short-circuited. It is also transfer impedance. called short circuited forward transfer Similarly from eqn (2) admittance. v2 Similarly from eqn(1) z|22= I= 0 …Driving point impedance I 1 2 I1 Y|12= v= 0 …Transfer admittance Where, is open circuit driving point 1 z22 v2 impedance at port 2 with port 1 open Where,Y12 is the transfer admittance at circuited. It is also called the open circuit port 2 with port 1 short circuited .It is also output impedance. called the short circuit reverse transfer admittance. 8.3 SHORT CIRCUIT ADMITTANCE (Y) Similarly from eqn (2) PARAMETERS I Y|= 2 …Driving point admittance 22 v1 = 0 v2

~~Where y22 is driving point admittance at port 2 with port 1 short circuited .It is also called the short circuit output admittance.~~

~~The y-parameters of a two port network 8.4 TRANSMISSION (A B C D) PARAMETERS~~

~~expressing the port current I12 andI in~~

~~terms of voltages v12 and v are as shown in~~

~~the figure. Here I12 ,I are dependent~~

variables and v12 & v are independent variables By KCL at port (1) In transmission parameters, the input I= Yv + Yv------(1) 1 11 1 12 2 variables v11 & I at port 1-1’ called the By KCL at port (2) sending end are expressed in terms of I= Yv + y v------(2) 2 21 1 22 2 output variable v22 & I at port 2 are Where, Y11 ,Y 12 ,Y 21 ,Y 22 are the network known as the receiving end, Transmission function and are also called the admittance Parameters are defined by –

(y) parameters. We may write matrix vAvBI1= 22 − ------(1) n = eq[][] I Y [v] I1= Cv 22 − DI ------(2) The negative sign is used with I and not I1   YY 11 12  v 1  2 Thus  =    for parameter B and D. The parameters A, I2   YY 21 22  v 2  n B, C and D are called transmission From eq (1) parameters. They can be expressed in I Y|= 1 …Driving point admittance matrix from as follow 11 v2 = 0 v1 vv12AB   Where,Y is driving point admittance at =    11 IICD− port 1 with port 2 short-circuited. It is also 12   called the short circuit input admittance. AB n The matrix  is called transmission Similarly from eq (2) CD matrix. For given network, with port 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission open circuited, then from equation (1) and The hybrid matrices describe a two port, (2) we get when the voltage of one port and the

v1 current of other port are taken as the A = | = I02 independent variables. From above v2 network, if voltage at port 1 and current at I1 and C = |I0= port 2 are taken as dependent variables, we v 2 2 can express them in terms of I an v Now when port 2 is short circuited, then 1 2 = + ---- (1) from eqn(1) & (2) we get --- v1 hI 11 1 hv 12 2 I= hI + hv ---- (2) v1 I1 2 21 1 22 2 B = |v0= and D = |v0= 2 2 Where, are called hybrid −I2 −I2 h11 ,h 12 ,h 21 & h 22 Parameters can be expressed in matrix 8.5 INVERSE TRANSMISSION (A’ B’ C’ D’) form as follows

~~PARAMETERS v1   hh 11 12  I 1   =    I2   hh 21 22  v 2  n From eq (1) & (2) when v02 = the port 2 is short circuited Then~~

v1 h|11= v= 0 In inverse transmission parameters, output 2 I1 variables v2 and I2 are expressed in terms 1 of input variables v1 andI1. Inverse Short circuited input impedance  Y11 transmission parameters are defined by – = − ------(1) I2 v2 A'v 11 B'I h|21= v= 0 I 2 I= C'v − D'I ------(2) 1 2 11 Short circuit forward current gain The coefficients A’, B’, C’, D’ in the above Y equations are known as inverse 21 Y11 transmission parameters. Similarly from eqn(1) & (2) Let port 1 be For given network, when port 1 is open n open i.e. I1 = 0 (I1 = 0) then from eq (1) & (2) we get --- v1 h|= = v2 12 I1 0 A '|= I0= v2 v 1 1 z I Open circuit reverse voltage gain 12 and C'|= 2 z22 I01 = v1 I2 h|22= I= 0 When port1 is short circuited v1 = 0 then v 1 from eqn(1) &(2) we get 2 1 v2 I2 Open circuit output admittance  B'|= v0= and D'|= v0= z22 −I 1 −I 1 1 1 Since h-parameters represent dimensionally 8.6 HYBRID (h) PARAMETERS impedance, admittance, voltage gain and current gain, these are called hybrid parameters.

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8.7 INVERSE HYBRID (g) PARAMETERS 8.8 CONDITIONS FOR RECIPROCITY AND SYMMETRY~~

~~In inverse hybrid parameters current at the~~

~~input port I1 and voltage at output port v2~~

~~can be expressed in terms of I2 and v1 The equations are as follows = + ---- (1) I1 gv 11 1 gI 12 2 8.9 INTER RELATIONSHIPS OF v2= gv 21 1 + gI 22 2 ---- (2) DIFFERENT PARAMETERS~~

Where, g11 ,g 12 ,g 21 ,g 22 are inverse hybrid 1) Z-parameters in terms of Y parameters. They can be expressed in parameters matrix as Z-parameters can be expressed in terms I1   gg 11 12  v 1  of Y-parameters as shown bellow  =    −1 v2   gg 21 22  I 2  []ZY= [] or n From eq (1) & (2), when I2 = 0 i.e. port It can be expressed in terms of matrix (2) open circuited, we get form as given bellow −1 I1 g|= = zz11 12  YY 11 12  11 I2 0 = v1    zz21 22  YY 21 22  1 Since Open circuit input admittance  −1 z11 −1 YY adJ[Y] []Y 11 12 = v2  g|= YY21 22 ∆Y 21 I2 = 0 v1 YY22− 12 Open circuit voltage gain So, adJ[] Y =  n −YY21 11 From eq (1) & (2) when v1 = 0 i.e. port1 And ∆= − is short circuited, we get Y YY11 22 YY 12 21 YY− I1 So, zz=22 = 21 g|12= v= 0 11 21 1 ∆∆YY I2 −YY Short circuit reverse current gain zz=12 = 11 12∆∆ 22 v2 YY g|22= v= 0 I 1 2 2) Y-Parameter in terms of Z-  Short circuit output impedance 1 parameters Y22 Y-parameters can be expressed in terms of Z-parameters as shown below- −1 []Yz= or It can be expressed in terms of matrix form as given below- −1 YY11 12  zz 11 12   =  YY21 22  zz 21 22 

zz22− 12 AdJ[] z =  4) A B C D Parameters in terms of Y- −zz21 11 Parameter The A B C D parameter equations are ∆=z zz11 22 − zz 12 21 vAvBI1= 22 − ------(1) zz22− 21 So, YY11= 21 = ∆∆zz I1= Cv 22 − DI ------(2) −zz12 11 And y- parameter equations are – YY12= 22 = ∆∆zz I1= Yv 11 1 + Yv 12 2 ------(3)

I2= Yv 21 1 + Yv 22 2 ------(4) 3) A B C D Parameters in terms of z From eqn (4) parameters IY The ABCD parameters equations are – vv=2 − 22 12YY vAvBI= − ------(1) 21 21 1 22 −Y 1 I= Cv − DI ------(2) ∴=v22 vI + ------(5) 1 22 1YY 22 And z-parameters equations are- 21 21 Comparing eqn (1) & (5) we get v1= zI 11 1 + zI 12 2 ------(3) −Y22 −1 = + ------(4) A= & B = v2 zI 21 1 zI 22 2 YY n 21 21 eq (4) can be expressed in terms of n n I1 Now put eq (5) in eq (3) as follows −Y 1 =22 ++ v2 zI 22 2 I1 Y 11 v 2 I 2 Yv12 2 I1 = − ------(5) YY21 21 2221 21 − Comparing above eqn(5) with eqn (2) YY11 22 Y 11 =v2 ++ I 2 Yv 12 2 We get, YY21 21 1 222  C = & D = −YY11 22 Y 11 = ++Yv12 2 I 2 221 221 YY21 21 Put eqn (5) in eqn (3)  We get YY11 22− YY 12 21 Y 11 =−+vI22-----(6) vz YY21 21 =2 −+ 22 v1 z 11 I 2 zI 12 2 Compare eqn (2) & (6) zz21 21 YY− YY −∆y v zz C =11 22 12 21 = v = z2−+ 22 11 I zI YY 1 11zz 2 12 2 21 21 21 21 −Y D = 11 z11 zz 11 22− zz 12 21 vv12= −  I 2----- (6) Y21 zz21 21 −Y −1 n n So, A = 22 B = Compare eq (6) with eq (1) Y Y We get 21 21 −∆y −Y z zz− zz ∆z C = D = 11 A=11 & B = 11 22 12 21 = Y21 Y21 z21 ZZ 21 21

~~Where, ∆=z zz11 22 − zz 12 21 z ∆z So, A = 11 B = z21 Z21~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8.10 INTERCONNECTION OF TWO- v1= zI 11 1 + zI 12 2 ------(7) PORT NETWORK v2= zI 21 1 + zI 22 2 ------(8) n 1) Series connection Comparing eq (5), (6), (7) & (8) we get zz11= 11a + z 11b

~~zz12= 12a + z 12b~~

~~zz21= 21a + z 21b~~

zz22= 22a + z 22b 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. Fig. Series connection of two –port networks. 2) Parallel connection

~~As shown in fig, two –port networks NA~~

~~and NB connected in series.~~

~~For network NA the z-parameter eqnmatrix form is-~~

v1a   zz 11a 12a  I 1a   =   ----(I) v2a   zz 21a 22a  I 2a  Similarly, for network Nb Fig. Parallel connection of two port v   zz  I  network. 1b= 11b 12b 1b ----- (II)      As shown in fig., two-port networks N v2b   zz 21b 22b  I 2b  A From figure we can write and NB connected in parallel. For network N the y parameter II1= 1a = I 1b -----(1) a equations are II= = I-----(2) 2 2a 2b I= YV + YV vv= + v------(3) 1a 11a 1a 12a 2a 1 1a 1b I= YV + YV and 2a 21a 1a 22a 2a Similarly, for network N vv= + v------(4) b 2 2a 2b I= YV + YV n 1b 11b 1b 12b 2b eq()3 ⇒= v1 v 1a + v 1b I2b= YV 21b 1b + YV 22b 2b n From z-parameters eq (I) & (II) & (1) Assuming that the parallel connection (2) requires that n Put value of v1a & v1b in above eq we VV1= 1a = V 1b

~~get VV2= 2a = V 2b v=+ zI zI ++ zI zI 1()() 11a 1a 12a 2a 11b 1b 12b 2b II1= 1a = I 1b~~

v1=+()() z 11a zIz 11b 1 ++ 12a zI 12b 2 ----(5) And II2= 2a = I 2b n By combing above equations eq()4 ⇒= v2 v 2a + v 2b II1= 1a + I 1b v2=+() z 21a I 1a z 22a I 2a ++ (z 21b I 1b z 22b I 2b ) =+()Y11a V 1a Y 12a V 2a ++ (Y 11b V 1b Y 12b V 2b ) v2=+() z 21a z 21b I 1 ++ (z 22a z 22b )I 2 ---(6) I=+() Y Y V1 ++ (Y Y )V ---(1) So we have z –parameters eqn 1 11a 11b 12a 12b 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission and II2= 2a + I 2b V1 V1a   AB a a  V 2a  =  =    =+++()()YV21a 1a YV 22a 1b YV 21b 1b YV 22b 2b I1 I1a   CD a a − I 2a  n I2=+() Y 21a Y 21b V 1 ++ (Y 22a Y 22b )V 2 ---(2) --from eq ()1 The y- parameter are ABV   = a a 1b I1= YV 11 1 + YV 12 2 ------(3)    CDa a  I 1b  I2= YV 21 1 + YV 22 2 ------(4) ABABa a  b b  V 2b  Combining eqn (1), (2),(3),(4) we get =     CDCDa a  b b − I 2b  YY11= 11a + Y 11b --from eqn (2) YY21= 21a + Y 22b = + ABABaa  bb V2 YY12 12a Y 12b =    CDCDaabb  −I2 YY22= 22a + Y 22b So, the overall y- parameter matrix for ABV2 =  parallel connected two port network is CD−I2 simply the sum of y-matrix of each AB ABABaa  bb  individual two-port network connected So =    in parallel CD CDCDaabb   So, the overall transmission parameter 3) Cascade connection matrix for cascaded two port networks is simply the matrix product of transmission parameter matrices of each individual two-port network in cascade

Fig: - Two ports networks in cascade 4) T and representation connection. It is possible to express the elements of The transmission parameter T-network𝛑𝛑 in terms of z- parameters or representation is useful in cascaded ABCD parameters as explained below- two port network. For the network Na , the transmission parameter equations are

V1a   AB a a  V 2a   =   ---- (1) I1a   CD a a − I 2a  Similarly, for the network Nb , z-parameters of the network V1b   AB b b  V 2b   =   ---- (2) V1 I CD− I Z11= | I= 0 = ZZ a + c 1b   b b  2b  I 2 From figure we can assume following 1 V observations Z=2 |Z = 21 I2 = 0 C I1 I1= I 1a , −= I 2a I 1b and I 2 = I 2b V V= V ,V = V a nd V = V Z=2 | = ZZ + 1 1a 2a 1b 2 2b 22 I1 = 0 b C The overall transmission parameter of I2 V the combined network Na and Nb can Z=1 |Z = 12 I1 = 0 C written in the matrix form as I2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission It is possible to express the elements of Vn So, 22= ------(1) -network in terms of y-parameters Vn11 or ABCD parameters as explained below In the π And 12= -----(2) −In21 we know ABCD parameter -----

VAVBI1= 22 − ------(3) 𝑛𝑛 ⇒ 𝑒𝑒𝑒𝑒 I1= CV 22 − DI ------(4) n Y-parameters of the network From eq (1) n I1 V=1 V − 0I ------(5) Y11= | V= 0 = YY 1 + 2 1 22 2 n V1 2 From eqn (2) I2 Y|Y== = − 21 V2 0 2 N2 V1 I12= 0V − I 2------(6) I N1 2 n Y22= | V= 0 = YY 3 + 2 Comparing eq (3) & (5) and (4) & (6) V 1 2 We get, I Y=1 |Y = − n 12 V1 = 0 2 1 V2 0 AB n2 =  CD n2 5) Lattice networks 0 One of the common four –terminal two- n1 port network is the lattice or bridge Now, we know that network. The lattice network for z- A'''' B C D= [ABCD]− 1 parameter is shown in figure below So, ABCD''''parameters are

n2 0 A' B' n1 =  C' D' n1 0 n2 We have h-parameter equations- ZZab+ ZZ11= 22 = = + 2 V1 hI 11 1 hV 12 2 ZZ− I= hI + hV ZZ= = ba 2 21 1 22 2 12 21 2 From eqn (1) n V= 0I + 1 V Example: 11 2 n2 Determine all the two port network From eqn (2) parameters of an ideal transformer n1 I2=−+ I 12 0V n2

~~n1 0 hh11 12 n2 So, =  hh21 22 n1 − 0 n2 Solution: We know that −1 NnV I We have [g]= [h] 2221= = = NnV111− I 2~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission n T (ABCD) parameter is the relation − 2 0 between V and I in terms of I and V gg11 12 n1 1 2 1 2 So, =  gg n Form equation (2) 21 22 2 0  I12= V − 2I 2 ----- (3) n1 Substitution in equation⇒ (1) we get Note: In an ideal transformer it is V2V2II1=() 2 −+ 22 impossible to express V,V12 in terms of I1 =2V22 − 3I ----- (4) and I2 hence z-parameter doesn’t exist similarly the y-parameter doesn’t exist Equation (3) and (4) can be rewritten as V1= 2V 22 − 3I

Example: I12= V − 2I 2 Find all the two port network parameter VV1223   for the below network =    II1212 −  23 =  12 T’:−= TT ’ −1 1 23− =  1 −12 Solution: It’s a Lattice network. Comparing the above h −h: parameters is the relation between and in terms of and network with standard Lattice network we V1 I2 I1 V2 get From equation (3) we get

~~Z1a=Ω, Z b = 3Ω, Z cd = 3ΩZ = 1Ω 11 I2=−+ IV 12 ------(5) Using the formula derived in Lattice 22 network Substituting equation (5) in (4) we get~~

ZZab+ 13+ 11 1 3 ZZ= = = = 2Ω V1= 2V 2 −− 3 I 1 + V 2 = V 21 + I ----- 11 22 22 22 2 2 ZZ− 31− (6) ZZ= =ba = = 1Ω 12 21 22 From (5) & (6) 3 1 21 VI  Z =  11= 22 12    IV22−11  22 −1 1 Y: Y= Z = AdjZ g:- Z 11− Z3= 22 1 3 21− 22 AdjZ =  −12 1 21− Y =  3 −12

~~T: V1= 2I 22 + I ----- (1)~~

~~V21= I + 2I 2----- (2)~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)

~~Q.1 The admittance parameter Y12 in Q.4 For the lattice circuit shown in the the 2-port network in Figure is figure, Zab=Ω=Ω j2 and Z 2 . The values of the open circuit impedance~~

~~ZZ11 12 parameters Z =  are ZZ21 22~~

~~a)-0.2mho b)0.1 mho c)-0.05 mho d)0.05 mho [GATE-2001]~~

Q.2 The Z parameters Z and Z for 11 21 1j1−+ j 1j−+ 1j the 2-port network in the figure are a)  b)  1j1j++ −+1 j1j − 1j1j++ 1j+ −+ 1j c)  d)  1j1j−− −+1j 1j + [GATE-2004] −6 16 a) Z11=Ω=Ω ;Z 21 Q.5 The ABCD parameters of an ideal 11 11 n:1 transformer shown in the figure 64 b) Z11=Ω=Ω ;Z 21 n0 11 11 are The value of X will be 6− 16 0X c) Z=Ω=Ω ;Z 1111 21 11 44 d) Z=Ω=Ω ;Z 1111 21 11 [GATE-2001]

~~Q.3 The impedance parameters Z11 and Z of the two–port network in the 1 12 a)n b) figure are n 1 c) n2 d) n2 [GATE-2005]~~

~~Q.6 The parameters of the circuit shown in the figure are a) Z11= 2.75Ω and Z 12 = 0.25Ω~~

~~b) Z311=Ω and Z 12 = 0.5Ω~~

~~c) Z311=Ω and Z 12 = 0.25Ω~~

~~d) Z11= 2.25Ω and Z 12 = 0.5Ω [GATE-2003]~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 0.1 0.1 10− 1 A112= 4A,V = 6V,V = 6V,A 2 = 0A a) b) −0.1 0.3 1 0.05 30 20 10 1 c)  d) 20 20 −1 0.05 [GATE-2005]

Q.7 A two-port network is represented by ABCD parameters given by Q.9 The h-parameter matrix for this VVAB   12= If port -2 is network is  −  II12CD   −33 −−31 a) b) terminated by RL, then input   −1 0.67 3 0.67 impedance seen at port-1 is given by A+ BR AR+ C 33 31 a) L b) L c)  d)  1 0.67 −−3 0.67 C+ DR L BRL + D DR+ A B+ AR [GATE-2008] c) L d)L BR+ C D+ CR L L Q.10 The z-parameter matrix for this [GATE-2006] network is 1.5 1.5 1.5 4.5 Q.8 In the two port network shown in a)  b) the figure below, Z11 and Z21 are, 4.5 1.5 1.5 4.5 respectively 1.5 4.5 4.5 1.5 c)  d) 1.5 1.5 1.5 4.5 [GATE-2008]

~~Q.11 For the two–port network shown below, the short -circuit admittance parameter matrix is a) re0 and β r b) 0 and−β r0~~

~~c) 0 and β r0 d) re0 and−β r [GATE-2006]~~

Statement for Linked Answer Questions 42− 1− 0.5 9 & 10 a) S b) S A two Part network shown below is exited −24 −0.5 1 by external dc sources. The voltages and 1 0.5 42 current are measured with voltmeters c) S d) S 0.5 1 24 V ,V and ammeters A ,A (All assumed to 12 12 [GATE-2010] be ideal) as indicated. Under following switch conditions, the readings obtained Common Data for questions 12 and 13 are: With 10V dc connected at port A in the (i) S1 − open , S2 − cl ose d linear nonreciprocal two –port network shown below, the following were observed: A11= 0A, V = 4.5V,V 2 = 1.5V,A 2 = 1A (i) a current of 3A (ii) S− cl ose d , S− open 1 2 1Ω connected at port B draws

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission (ii) ected at port B draws a Q.16 Consider a two-port network with current of 2A the transmission matrix: 2.5Ω conn AB T = . CD If the network is reciprocal, then a) T-1 = T Q.12 With 10 V dc connected at port A, b) T2 =T c) Determinant (T) = 0 at port B is d) Determinant (T) = 1 the current drawn by7Ω connected a)3/7A b)5/7A [GATE-2016] c)1A d)9/7A [GATE-2012] Q.17 The z-parameter matrix for the two- port network shown is Q.13 For the same network, with 6 V dc 2jωω j connected  at port B draws 7/3 A. If 8 V dc is jωω 3+ 2j connected to at port port A, A, the 1Ω open circuit voltage at port B is a) 6V b) 7V c)8V d)9V [GATE-2012] Where the entries are in Ω .suppow

~~Q.14 In the h-parameter model of the 2- (j ) = Rb + jω . Then the value of port network given in the figure R ______. ωb shown, the value of h22 (in S) is _____. [GATE-2016] (in Ω) equals to~~

~~zz11 12 Q.18 The z-parameter matrix  zz21 22~~

~~[GATE-2014]~~

Q.15 For the two-port network shown in the figure, the impedance (Z) matrix 22− 22 a)  b)  (in Ω)is −22 22 93− 93 c)  d)  69 69 [GATE-2016] 6 24 98 a)  b)  The ABCD matrix for a two-port 42 9 8 24 Q.19 network is defined by: 96 42 6 c)  d)  VV12AB   6 24 6 60 =    I12 CD − I  [GATE-2014]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The parameter B for the given two- port network (in ohms, correct to two decimal places) is ______.

~~[GATE-2018]~~

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

~~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~~

~~EXPLANATIONS~~

~~Q.1 (c) E21−+ 4I 10E 1 = 0 YY+− Y YY 13 3= 11 12 6  ⇒E21 − 4I +× 10 I 1 = 0 −+Y3 YY 23YY21 22 11~~

~~YY12= − 3 b EI11= 11~~

~~⇒11E211 −+= 44I 60I 0 E 16 2 = − Ω I1 11~~

~~Q.3 (a) 1 Using Y conversion Y =−=− 0.05mho 12 20 ∆ − Q.2 (c)~~

~~V1= ZI 11 1 + ZI 12 2~~

V2= ZI 21 1 + ZI 22 2 V 21× 2 Z|= 1 R= = = 0.5 11 I2 = 0 1 I1 44 Applying KVL in LHS loop 11× 1 R2 = = = 0.25 E111−−+ 2I 4I 10E 1 = 0 44 ⟹ × 11E= 6I 21 2 11 R3 = = = 0.5 E 6 44 ⇒=1 Ω I1 11 V Z|= 2 21 I2 = 0 I1 KVL in RHS loop,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission II12= − I 2 =−= ZZ13+ Z 3 1h21  I1 Z ZZ+ 3 23 V V= 10I ⇒=1 10 Z11=+= Z 1 Z 3 2.5 + 0.25 = 2.75 11 I1 Z12= Z 3 = 0.25 I01 =

V1 Q.4 (d) When VV12= ⇒==h112 For lattice network, Z-parameter V2 is given as (as no. drop in resistance) V= 20I ZZab+− ZZ ab 22  10Ω 22 I2  ⇒=h22 ZZab−+ ZZ ab V2 22 1 = = 0.05 Zab =2j,Z =2Ω 20 1+− j j1 10 1   j11−+ j −1 0.05

Q.7 (d) VV12AB  Q.5 =   The ABCD parameter equations are II12CD  given by, IVn 21= = VAVBI1= 22 − IV1 12 I1= CV 22 − DI VAVBI1= 22 − When the network is terminal by

~~I1= CV 22 − DI R L (fig.1), V V= − IR A=1 |n = 2 2L I02 = V2 V1 AV 22− BI Zin = = IV111 I1 CV 22− DI D|=V0= = = I2 Vn −−AI R BI AR + B 22 =2L 2 = L −−CI2L R DI 2 CR L + D Q.6 (d)~~

~~V1= hI 11 1 + hV 12 2~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V −βI .r Z|=2 =10 = −β r 21 I2 = 0 0 II11~~

~~Q.9 (a)~~

~~V1= hI 11 1 + hV 12 2~~

~~I2= hI 21 1 + hV 22 2 From given Z-parameters, 2 IIV21=−+ 2 Let Vth and Rth be Thevenin voltage 3 & resistance as seen from partB. 2 V1= 1.5I 1 + 4.5 −+ I 12 V 3~~

~~=−+3I12 3V −33 ∴=H  −1 0.67 Vth= 3R th + 3 ………. (1)~~

~~Q.10 (c) Vth= 2R th + 5 ………. (2) V1= ZI 11 1 + ZI 12 2 Solving (1) & (2)~~

~~V2= ZI 21 1 + ZI 22 2 R2th = Ω~~

~~V1 So, Vth = 3x2 += 3 9V Case (1) ⇒=∴I1 0 Z 12 = = 4.5 I1~~

~~V2 Z22 = = 1.5 I2~~

~~V1 Case (2) ⇒=∴=I2 0 Z 11 = 1.5 I1 9 i = 6 ()27+ Ω Z21 = = 1.5 4 1.5 4.5 Q.13 (b) Z =  1.5 1.5~~

Q.11 (a) I 1 Y|=1 = = 4 11 V2 = 0 V1 0.25 7x1 21 I2 1 So, V= 7 / 3x2 + = = 7V Y|22= V= 0 = = 4 th 1 3 3 V2 0.25 The Open circuit voltage at port B is I Y=2 | =−= 2Y 7V 21 V2 = 0 12 V1 42− Q.14 (1.24) ∴=[Y] S If two, n/ws are connected in −24 parallel,

~~The y-parameterπ − are added Q.12 (c) i.e., y= yy + As per the given conditions, we can equ 1 2 draw the following two figures.~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 21− 1 −1 V 36× 33 2 1 = =  ⇒=z11 = =2 y1 y2 + −12 −1 I1 I =0 36 33 2 1 2 V 55− ⇒=z 2 We know 36 21 y = I1 I =0 equ −55 2 63 V2= Vz 1⇒− 21 = −2

~~1 −y12 zz 22− So 11 12 = yy11 11  h =  zz21 22 −22 y21 ∆y yy11 11~~

where y = yy11 22−− y 12 y 21 The value of h22 = y ∆ 55    −  5  − 5   =    − ∆    33     6  6   y = 2.0833 y = 5 h 22=1.24 ∆11 3 ∴ Q.15 (c) For the two-part network 11 1 +− 30 10 30 Y matrix =  −1 11 Q.19 4.8 + 30 60 30 Given: −1 ZYmatrix = [] VV12AB   −1 = 0.1333− 0.0333    I12 CD − I  Z =  −0.0333 0.05 96 Z =  6 24

~~Q.16 (d)~~

~~Q.17 (3)~~

Zac +Z Z c It can be re-written as Z=matrix  Zc Z bc +Z V1= AV 22 − BI ...(i) Zbc +Z =3+2jω Z =jω c I= CV − DI ....(ii) 122 Zb =3+jω From equation (i), Q.18⇒ (a) ⇒ Since the given network is V1 symmetric and reciprocal Z11 = Z22 B= − ...... (iii) I2 = Z12 = Z21 V02

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission We can write two KVL equations for the given electrical network as,~~

~~V1= 7I 12 + 5I ...... (iv)~~

~~V212= 5I + 7I ...... () v~~

~~At V02 =~~

~~0= 5I12 + 7I~~

~~7 II= − 125~~

~~Substituting in equation (iv)~~

~~7 V1= 7 ×− I 22 + 5I 5~~

~~24 VI= − 125~~

~~V 24 B=−==1 4.8 I52~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

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 voltage b) lower impedance c) greater power a) 0.125 b) 0.167 d) better regulation c) 0.625 d) 0.25 [GATE-2001] [GATE-2003]

Q.2 A two-port network, Shown in figure Q.4 The Z matrix of a 2- port network as is described by the following 0.9 0.2 given by  equations: 0.2 0.6 l1= YE 11 1 + YE 12 2 , l2= YE 21 1 + YE 22 2 The element Y22 of the corresponding Y matrix of the same network is given by a) 1.2 b) 0.4 c) -0.4 d) 1.8 [GATE-2004]

The admittance parameters Q.5 For the two port networks shown in the figure the Z-matrix is given by Y11 ,Y 12 ,Y 21 , and Y22 for the network shown are a) 0.5 mho, 1mho, 2 mho and 1 mho respectively 1 1 1 b) mho, − mho, − mho and 3 6 6 1 mho respectively Z1 ZZ 12+ 3 a)  ZZ+ Z c) 0.5 mho, 0.5mho, 1.5 mho and 2 12 2 mho respectively ZZ b) 11 2 3 3  d) − mho, − mho, mho and Z1+ ZZ 22 5 7 7 ZZ 2 c) 12 mho respectively + 5 ZZ21 Z 2 [GATE-2003] ZZ11 d)  ZZ11+ Z 2 Q.3 The h-parameters for a two–port [GATE-2005] network are defined by E1   hh 11 12  l 1  Two networks are connected in  =    Q.6 l2   hh 21 22  E 2  cascade as shown in the figure. With For the two –port network shown in the usual notation the equivalent A, B, C and D constants are obtained. figure, the value of h12 is given by

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Given that C= 0.025 ∠° 45 , the value Q.9 The two-port network P shown in of Z2 is the figure has ports 1 and 2, denoted by terminals (a, b) and (c, d) respectively. It has an impedance matrix Z with parameters denoted

by Zij A 1Ω resistor is connected in series with the network at port 1 as a)10∠ 30°Ω b) 40∠ -45°Ω shown in the figure. The impedance c)1Ω d) 0Ω matrix of the modified two–port [GATE-2005] network (shown as a dashed box) is

~~Q.7 The parameters of the circuit shown~~

~~in the figure are Ri = 1M Ω ,~~

~~Ro = 10 Ω , 6 A= 10 V / V. If Vi = 1 µ V, then output voltage, input impedance and z11++ 1z 12 1 output impedance respectively are a)  z21 z1 22 +~~

z1z11+ 12 b)  z21 z1 22 + z+ 1z c) 11 12 a) b)  1V,∞Ω ,10 1V, 0,10Ω zz21 22 c)1V, 0, ∞ d)10V,∞Ω ,10 z11+ 1z 12 [GATE-2006] d)  z21+ 1z 22 Q.8 The parameter type and the matrix [GATE-2010] representation of the relevant two port parameters that describe the Common Data for questions Q.10 circuit shown are and Q.11 With 10V dc connected at port A in the linear nonreciprocal two–port network shown below, the following were observed: (i) a 00 current of 3A a) z parameters,  (ii) 1Ω connected at portrt B B drawsdraws a 00 current of 2A 10 2.5Ω connected at po b) h parameters,  01 00 c) h parameters  00  Q.10 With 10 V dc connected at port A, 10 d) z parameters  01 at port B is the current drawn by7Ω connected [GATE-2006] a)3/7A b)5/7A c)1A d)9/7A [GATE-2012]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.11 For the same network, with 6 V dc V02 = ) in ohms is ______(up to 2 decimal places) at port B draws 7/3 A. If 8 V dc is connected to at port port A, A, the 1Ω open connected circuit voltage at port B is a) 6V b) 7V c)8V d)9V [GATE-2012] [GATE-2018] 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

~~across Port 2. When 3 V is applied to Portthrough 1, athe 1Ω current resistance (in connectedAmpere) resistance connected across Port 2 is ______through a 2Ω [GATE-2015]~~

~~Q.13 The z-parameters of the two port network shown in the figure are Z11 12 21 Z22 . The average power delivered=40Ω, to Z RL= 60Ω, Z =80Ω, is____.=100Ω = 20Ω, in watts,~~

~~[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~~

~~V1 h11 parameter (where h11 = when I1~~

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

~~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~~

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

~~Q.1 (c) EE22 E1x= 2l = 2 = For a passive two-port network, 84 output power can never be greater E than input power. ⇒=1 0.25 E2 Q.2 (b) Method -2 Using KVL,~~

~~E1=++ 2l 1 2() l 12 l Again using KVL,~~

~~E2=++ 2l 2 2() l 12 l~~

42 −1 ⇒=[]z [][]yz= -Ytransformation 24 1 42− Using ∆ =  ()()44×−× 22−24 11− YY11 12 36  YY21 22 −11 E1=+()() 4 0.5 l 1 ++ 1 l 12 l 63 =5.5l12 + l … (a)

Q.3 (d) E2=+++()() 2 1l 2 1l 12 l E1 h|= =l12 + 4l … (b) 12 |1 = 0 E2 Put l01 = in eq (a) and (b) Or h12 is ratio of E1 to E2 for the El12= And E22= 4l input open- circuited condition. Two E methods are provided to solve the h=1 | = 0.25 12 l1 = 0 problem. E2 Method -1 Assuming Q.4 (d)

l01 = 0.9 0.2 []z =  0.2 0.6 0.6− 0.2  −1 −0.2 0.9 [][]yz= = [0.9×− 0.6 0.04] 1.2− 0.4 = EE − l =22 = 0.4 1.8 2 2++() 2 4 || 4 4 y22 = 1.8 l l4=2 × x ()22++ 4 Q.5 (d) l1  EE 2= 22 = 2 24 8

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission l (i) C|= 1 I02 = V2 Putting 12=0 in equation (i)~~

~~V2= Zl 21~~

V2 11 ⇒=Z|2 I0= == Two method are provided to solve 2 l1 C 0.025∠ 45° the problem. Z°=40 ∠ 45 Ω Method-1 2 v= v = iz 1 2 11 Q.7 (a) v − So z=1 |z = V= 1066 ×= 10 1V 11 i2 = 0 1 0 i1 V1 Z11 = →∞ v2 z= |z = I1 (i) 21 i2 = 0 1 i1 V2 Z22=→= R 0 10 (ii)When i01 = I2 v= iz 1 21 Q.8 (c) v1 ⇒ z= |z = I1= hV 11 1 + hI 12 2 12 i1 = 0 1 i2 V2= hV 21 1 + hI 22 2 (iii) v2= i 21 (z + z 2 ) v ⇒ z=2 | = zz + 22 i1 = 0 1 2 i2 Method -2

~~(iv) v1=() i 1 + iz 21~~

~~=zi11 + zi 12 Since port -1 is open – circuit, I01 =~~

(v) v2=++ zi 21 z 1() i 1 i 2 Port -2 is sort – circuit, V02 = = ++ I 0 z11 i (z 1 z 2 )i 2 h|=1 = = 0 11 I2 = 0 (vi)From eq (i) to(iv) VV11 ZZ I 0 −=11 h|=1 = = 0 z matrix  12 V1 = 0 ZZ11+ Z 2 II22 V 0 h|=2 = = 0 21 L2 = 0 Q.6 (b) VV11 V 0 h|=2 = = 0 22 V1 = 0 II12 So, h parameters

~~hh11 12 00 =− =  hh21 22 00 = + V1 AV 22 Bl Q.9 (c) = + l1 CV 22 Dl The Impedance matrix Zm of the~~

~~V2= Zl 21() + l 2 modified network is calculated from fig. given below:~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VI   9 11= i =  []z  + VI22   ()27Ω~~

~~V1= ZI 11 1 + ZI 12 2 Q.11 (b) V2= ZI 21 1 + ZI 22 2~~

~~VS=×+() 1I 21 V~~

~~=++IZIZI1 11 1 12 2~~

~~=++()1 Z11 I 1 ZI 12 2~~

VIS1    = []zm  7x1 VI22   So, V= 7 / 3x2 + th 3 Z+ 1Z 10 = 11 12 = + 21 zm []Z  = = 7V ZZ21 22 00 3 The Open circuit voltage at port B is 7V.

~~Q.12 0.545~~

~~I1= yv 11 1 + yv 12 1 I= 0.4 ×− 3 0.6[2I ] Q.10 (c) 22 = + As per the given conditions, we can I2 yv 21 1 yv 22 2 =1.2-1 2I. 2~~

~~draw the following two figures. 4= 10y21 →= y 21 0.4 I2 = 0.545A~~

~~1.25=+= 0.4v1 1.25y 22 0.4~~

~~y22 = − 0.6~~

Q.13 35.55 In the given terminated 2 port Let Vth and Rth be Thevenin voltage network the Z matrix is known and & resistance as seen from part B. power on the load. for load of 20Ω we wantL as load to find let first obtain the thevenin equivalent of→ 2The port get it assuming R th & Rth = + ……….(1) → Thevenin equivalent means V Vth 3R th 3 Vth =V 2 i.e., O.C voltage of port 2 I2 =0 Vth= 2R th + 5 ……….(2) ISC =-I 2 /V 2 =0 i.e., s.c current of port Solving (1) &(2) 2, R2th = Ω V R= th So, V= 3x2 += 3 9V in th ISC

~~Vth The Z matrix equation is → Evaluation of V112 =40I +60I~~

~~V21 =80I +100I 2~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission In the above two equations if I02 = then~~

~~V11 =40I (1)~~

~~V21 =80I (2) From the input side we can say~~

~~()V11 =20-10I~~

R1A = Ω 20-10I11 =40I = Ω 2 R1B ⇒ V21 =80I =80× =32V = Ω 5 RC 1/2 After rearrangement consider the ⇒So Vth =V 2 =32 following circuit. ISC In the Z matrix equation if we put Evaluation of V02 = then

~~V111 =40I +60I …. (5)~~

~~0=80I12 +100I …. (4) From the circuit diagram we get,~~

~~V2 Z321 = = Ω I1~~

~~Q.15 0.5 Given: The two port network is shown below, 10 100 I =- I &V =20-10I =20+ I 188 21 1 2~~

~~Using these in equation V1 & I1 in equation 3 100 400 20+ I =- I +60I 882 22~~

~~160+100I2 =-400I 22 +480I~~

160=−⇒=− 20I22 I 8A V1 ⇒ To find the value of h11 = ,the I =-I =8A I1 = ⇒SC 2 V02 Vin 32 equivalent circuit is shown below, Rin = = =4Ω I8SC → 2 P= () I 20 →Now20Ω the 20Ω ckt is from port 2is 32 =20 = 35.55watt 4+ 20

~~Q.14 3 Applying KVL in loop (1),~~

~~VI11−−() II 1 + 2 = 0~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V1= 2I 12 + I~~

~~Applying KVL in loop (2),~~

~~()()II12++ 2II0 12 +=~~

~~3I12+= 2I 0~~

~~From equation (i) and (ii),~~

~~3I V= 2I − 1 112~~

~~V 31 1 =−==2 0.5 I1 22~~

~~Hence,~~

~~V1 h11 = = 0.5 I1 V02 =~~

~~Hence, the correct answer is 0.5~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(IN)~~

~~V Q.3 With 10 V dc connected at port A, Q.1 The DC voltage gain 0 in the V i at port B is following circuit is given by a)3/7Athe current drawn by7Ωb)5/7A connected c)1A d)9/7A [GATE-2012]~~

~~Q.4 For the same network, with 6 V dc~~

R R at port B draws 7/3 A . If 8 V dc is a) Av 2 b) Av 1 RR+ RR+ connected atto port port A, 1ΩA, connectedthe open 12 12 circuit voltage at port B is R 2 c) Av+ R o d) Av a) 6V b) 7V RR12+ c)8V d)9V [GATE-2007] [GATE-2012]

Q.2 For the circuit shown below the Q.5 Considering the transformer to be input resistance ideal, the transmission parameter V ‘A’ of the 2 -port network shown in RR= = 1 | is 11 11 I2= 0 the figure below is I1

a)1.3 b)1.4 a) - c)0.5 d)2.0 [GATE-2013] 3Ω b)[GATE 2Ω -2008] c) 3Ω d) 13Ω Q.6 The output voltage of the ideal Common Data for questions 3 and 4 transformer with the polarities With 10V dc connected at port A in the and dots shown in the figure is linear nonreciprocal two –port network given by shown below, the following were observed: i) at port B draws a current of 3A ii) 1Ω connected current of 2A 2.5Ω connected at port B draws a

~~a) NV1 b) - NV1 c) V1 sin ωtt d) -V1 sin ωt [GATE-2015] sin ω sin ωt~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 2 5 1 2 a) b) 0.2 1 0.5 1 � 5 2� � 1 2� c) d) 0.5 1 0.5 5 � � [GATE� -201� 7]~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ANSWER KEY: 7 1 2 3 4 5 6 (a) (a) (d) (c) (b) (a) (b)~~

~~EXPLANATIONS~~

~~Q.1 (a) R2th = Ω R 2 So, V= 3x2 += 3 9V VV= i th ()RR12+~~

~~R 2 V0V= AV = A V ()RR12+~~

~~Q.2 (d) 9 =+++ + i = V11 I 3I 2 2() I 1 2V 3 V 3 andV 3 ()27+ Ω~~

~~=2() I1 + 2V 3 ⇒= 2I 13 V Q.4 (b) For I2==+++ 0V 11 I 2() I 1 4I 1 2I 1 V ⇒=R1 | = 13Ω 11 I2 = 0 I1~~

~~Q.3 (c) As per the given conditions, we can 7x1 draw the following two figures. So, Vth = 7 / 3x2 + 3 21 = = 7V 3 The Open circuit voltage at port B is 7V.~~

Let Vth and Rth be Thevenin Q.5 (a) voltage & resistance as seen from part B. 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 Vth= 3R th + 3 ……….(1) other coil)

~~Vth= 2R th + 5 ……….(2) Solving (1) &(2) →Using Transformer ratio~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission VN 22= VNi1 N ⇒=V21 V = NV i sin sin ωt 1~~

~~→Vo2 =−=− V NV i sin sin ωt~~

~~Q.7 (a)~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 9 NETWORK SYNTHESIS~~

9.1 RC, RL, LC IMPEDANCE & 4) For the RLC impedance function, the ADMITTANCE FUNCTIONS poles are complex conjugate pair and they are symmetrical wrt the negative 1) For the LC impedance function the real axis. poles and zeros are alternate and lie on the jω axis Note: In the above cases instead of impedance function if admittance function ()()S122++ S3 + e.g. Zs() = is given then they are converted into the sS2S4()()22++ 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 ()S2(S4)++ F(s) .Then the function is ()S++ 1 (S 3) 2) For the RL impedance function the poles and zero are alternate and lie only a) an RL impedance function on the negative real axis and nearest to b) an RC admittance function the origin is zero .It can be at the origin. c) LC impedance function e.g. d) 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 3) For the RC impedance function the poles Hence, F(s) is RC impedance or RL and zeros are alternate and lie only on admittance function. the negative real axis and nearest to the origin is a pole. It can be at the origin. e.g.

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EC)~~

~~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) Rneg≤ Re Z 1 () j ω , ∀ω b) R≤ Zj() ω , ∀ω neg 1~~

c) Rneg≤ Im Z 1 () j ω , ∀ω 3(s+ 3) 2(s+ 3) a) b) d) Rneg≤ ∠ Zj 1 () ω , ∀ω S2 ++ 2S 3 S2 ++ 2S 2 [GATE-2006] c) 3(s− 3) d) 2(s− 3) S2 −− 2S 2 S2 −− 2S 3 Q.5 The RC circuit shown in the figure is [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 pole a) a low–pass filter b) a zero and a zero b) a high-pass filter c) a pole and a pole c) a band–pass filter d) a pole and a zero d)a band-reject filter [GATE-2005] [GATE-2007]

Q.3 The first and the last critical Q.6 Two series resonant filters are as frequencies (singularities) of a shown in the figure. Let the 3-dB driving point impedance function of bandwidth of Filter 1 be B1 and that a passive network having two kinds B1 of elements are a pole and a zero of Filter 2 be B2 .The Value of is B respectively. The above property 2 will be satisfied by a) RL network only b) RC network only c) LC network only d) 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 a) 4 b) 1 be positive real.~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 1 c) d) 2 4 [GATE-2007]~~

The driving point impedance of the Q.7 The value of the load resistance R is following networks is given by L a) R/4 b) R/2 0.2S = c) R d) 2R Zs() 2 S++ 0.1s 2 [GATE-2009] The component values are V (S) Q.9 The transfer function 2 of the V1 (S) circuit shown below is

a) L= 5H,R = 0.5Ω ,C = 0.1F b) L= 0.1H,R = 0.5Ω ,C = 5F c) L= 0.1H,R = 0.5Ω ,C = 5F d) L= 0.1H,R = 2Ω ,C = 5F 0.5s+ 1 3s+ 6 [GATE-2008] a) b) s1+ s2+ s2+ s1+ Q.8 If the transfer functions of the c) d) following network is s1+ s2+ V (S) 1 0 = [GATE-2013] Vi (S) 2+ sCR

~~ANSWER KEY :~~

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

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

Q.1 (b) JωRC = = G(Jω) K(s− z) 222 Zs() = ()1−+ω R C 3JωRC ()SP(SP)−− 12 As ω0→ , GJ()ω=0 K(s+ 3) = As ω → ∞ , GJ()ω=0 ()S1++ j(S1j) +− J K(s+ 3) At ω=11 G() Jω = = = RC, 3 Zs() 2 3J ()s+− 1 j2 Filter is band –pass filter. K(s+ 3) = 2 ∴ ()s1++ 1 Q.6 (d) RRBL12 1 3K B12= ;B = ⇒== Z0|() ω−0 = 0 ⇒ =⇒=3 K2 LL12BL 4 2 21 2(s+ 3) ∴=Zs() S2 ++ 2S 2 Q.7 (a) 1 Z() S= R SL Q.2 (b) SC For stability poles and zero interlace RSL. 1 on real axis. Since its RC, first pole = SC RSL++ R. 1L should come and zero at last. ()SC C RSL Q.3 (b) = SC ()S2 RCL++ R SL Q.4 (a) SC Zs()()= Zs + R SRL 2 1 neg ⇒=ZS() ⇒ =++ 2 Z2()() s() R neg R e Z 1 (s J Im() Z 1 () s ()S RCL++ SL R < &Rneg 0 S. 1 SC For Zs() to be +ve & real, Re = 2 S2 ++S 1 ()RC LC ()Zs() ≥− R ⇒ R() Zs() ≥ R 1 neg e 1 neg 0.2S = 2 ++ Q.5 (c) ()S 0.18 2

1 1 V (S) ()R || /sc =0.2 ⇒= C 5F 0 = C 1 Vi (S) R++1 R || /sc 1 =0.1 ⇒= R 2Ω ()SC () RC V (S) SRC 1 ⇒=0 =⇒=2 L 0.1H 222 LC Vi (S) ()S R C++ 3SRC 1 Q.8 (c) VJ0 ()ω Put SJ=ω, ∴ 1 VJi ()ω VS() ()RL || 0 = SC VS() + 1 i R() RL || SC

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 ()R.L SC 1 ()R L + = SC R+ R.1 ( ()L SC ) + 1 ()R L SC~~

~~R L ()1+ SCR = L R R + L ()1+ SCRL V (S) R 1 ⇒=0 L = ViL (S) R++ R SCRR L() 2 + SCR~~

~~RL = R Satisfies above equation~~

~~Q.9 (d)~~

~~V 2 = ? V1 + 1 VR CS ()RC S+ 1 C 21= = 12 1 + V1++1 () RC 121 S C C R CS 1 CS2 Substituting the values we get V S1+ 2 = V1 S2+~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission GATE QUESTIONS(EE)~~

Q.1 The driving point input impedance Q.2 A major advantage of active filters is that they can be realized without seen from the source Vs of the using a) op-amps b) inductors circuit shown below, in Ω , is ____. c) resistors d) capacitors [GATE-2016]

~~[GATE-2016] ANSWER KEY:~~

~~1 2 20 (b)~~

~~EXPLANATIONS~~

~~Q.1 (20) Vss() 2+1 =I (6+4+48+2)~~

~~The Driving point impedance is Vs 60 nothing but the ratio of voltage to ⇒ = = 20Ω I3s V ⇒ this case it is s current from s the defined port. In Q.2 (b)~~

~~I Inductive coils are bulky in nature.~~

Writing KCL at node x VV −+Ixx − 4V + = 0 s136 Substituting these in Eq(1) V−− 2I V 2I −+Iss − 8I → ss = 0 ss36 11  2 2  Vss +  =I 1+ +8+  36  3 6  ⇒

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

~~Q.1 The v-i characteristic of an element Q.4 Consider the following circuit: is shown in the figure given. The element is~~

Which one of the following statements is correct? a) Passive and linear b) Active and linear a) Non linear, active, non-bilateral c) Passive and non-linear b) Linear, active, non-bilateral d) Active and non-linear c) Non-linear, passive, non-bilateral d) Non-linear, active, bilateral Q.5 Consider the circuits A and B. For what values respectively of I and R, Q.2 The incandescent bulbs rated the circuit B is equivalent to circuit respectively as P1 and P2 for A? 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 d) a)3A, 40Ω b)4 A, 24 Ω Q.6 Forc)1 A,the 100 circuit Ω given in the2 A, figure 100 Ω the power delivered by the 2volt source is given by

~~PP12 22 a) b) PP12+ PP12+~~

~~c) (P12+ P ) d) PP12×~~

Q.3 Consider the circuit as shown above which a current-dependent current a) 4W b) 2W V source has. The value 2 is c) -2W d) -4W V1 Q.7 The current in the given circuit with a dependent voltage source is a) 10 A b) 12 A c) 14 A d) 16 A a) 1 b) 2 Q.8 A certain network N feeds a load 1+α α c) d) resistance R as shown in figure-I. It 2 +α 2 + α

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission consumes a power of ‘P’W. If an identical network is added as shown Q.12 Consider the following circuit: in figure-II, the power consumed by R will be

What is the value of current I in the resistor in the above circuit? a) 0 A b) 2 A c)5 Ω 3 A d) 4 A a) Less than P b) Equal to P Q.13 In the circuit shown above, if the c) Between P and 4P current through the resistor R is d) More than 4P zero, what is the value of I?

~~Q.9 In the circuit shown in Fig., the current I through 2 Ω resistor is~~

~~a) 1A b) 2A c) 3A d) 4A~~

~~Q.14 In the network shown above, what a) – 94.34 mA b) -70.34 mA is the current I in the direction c) 70.34 mA d) 94.34 mA shown?~~

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 a) 0 b) 1/3 A in volts? c) 5/6 A d) 4 resistor of 5Ω, what is the value of V Q.15 The current I in the network in Fig. is

~~a) 5.0 b) 2.5 c) 7.5 d) 3.5 a) 1A b) 3 A Q.11 Consider the circuit in the below c) 5A d) 7A figure. What is the power delivered by the 24 V source? Q.16 The current i in figure is~~

~~a) 96 W b) 144 W a) 0.5 A b) 5/6 A c) 192 W d) 288 W c) 1.5 A d) 2.5 A~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.17 In Fig., the voltage source a) 40V b) 30V c) 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 a) Delivers 200/3 W IC = 6(t – 1) u(t – 1) A, where u(.) b) Absorbs 100 W denotes the unit step function. For c) Delivers 100 W this circuit, the current i at t = - 0.5s d) Absorbs 200/3 W will be

~~Q.18 The nodal voltage V1 in fig. is~~

a) – 9 A b) – 6 A c) – 1 A d) 0 A a) – 13.5 V b) – 6. 75 V Q.23 Match List I (Quantities) with List II c) – 4.5 V d) 0 V (Units) and select the correct answer using the codes given below Q.19 What is the current through the 2 Ω the lists: resistance for the circuit as shown above?

~~a) 5A b) 4A c) 3A d) 2A~~

~~Q.20 What is the value of I for the above Codes: A B C D shown circuit, if V = 2 volts? 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 a) 2 A b) 4 A the value I? c) 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) 10 A b) 6 A c) 3.7 A d) 3 A~~

~~Q.25 If the voltage V across 10 Ω resistance is 10V, what is the voltage~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission E of the voltage source in the circuit will be the ratio of the branch shown above? 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 : 6 c) 6 : 3 : 2 d) 6 : 2 : 4

~~Q.30 For the circuit as shown above, what a) – 50V b) – 10V is the value of I? 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 a) 4A b) 3A the circuit? c) 2 A d) 1 A~~

~~Q.31 In the circuit shown above, when is the power absorbed by the 4 Ω resistor maximum?~~

~~a) 1 MΩ b) 1.5 MΩ c) 3MΩ d) 5MΩ~~

~~Q.27 Consider the following circuit: What is power delivered to resistor a) R = 0 b) R = 2 Ω R in the circuit? c) R = 4 Ω 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 a) -15V values are 18 W, 50W and 98 W. b) 0W When all the sources are acting c) 15 W simultaneously, the possible d) Cannot be determined unless the maximum and minimum values of value of R is known power in R will be Q.28 Consider the following circuit: What is the current I in the circuit?

a) 98W and 18W b) 166W and 18W a) 0A b) 2A c) 450W and 2W c) 5A d) 6A d) 166W and 2W Q.29 Three parallel resistive branches are Q.33 According to maximum power connected across a d.c. supply. What transfer theorem, when is the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission maximum power absorbed by one non-linear, active or passive, time- network from another network? variant or time-invariant? a) The impedance of one of the a) Thevenin theorem networks is half that of the other b) Norton theorem b) The impedance of one is the c) Tellegen theorem complex conjugate of the other d) Superposition theorem c) The impedance of one is equal to that of the other Q.39 In the circuit shown below, if the d) Only the resistive parts of both source voltage Vs = 100 ∠ 53.130 V are equal then the Thevnin’s equivalent voltage in Volts as seen by the load Q.34 In the circuit shown in the given resistance RL is figure, RL will absorb maximum power when its value is

~~a) 100 ∠ 900 b) 800 ∠ 00 c) 800 ∠ 900 d) 100 ∠ 600~~

a) 2.75 Ω b) 7.5 Ω Q.40 Match List –I with List-II and select c) 25 Ω d) 27 Ω the correct answer using the code given below the Lists: Q.35 In a linear circuit, the superposition List – I (Term) principle can be applied to calculate A. Norton equivalent of one port the B. Open-circuit output admittance a) Voltage and power C. Reciprocal network b) Voltage and current D. Transmission parameters c) Current and power List –II (Concept) d) Voltage, current and power 1. Network where loop and node equation have a symmetric Q.36 In the circuit shown in the given coefficient matrix figure, power dissipated in the 5 Ω 2. Hybrid parameter h22 resistor is 3. Parameters where V1 and I1 expressed as functions of V2 and –I2 4. Current source in parallel with the impedance a) Zero b) 80 W Code: A B C D c) 125 W d) 405 W a) 1 3 4 2 b) 4 2 1 3 If two identical 3A, 4Ω Nrton Q.37 c) 1 2 4 3 equivalent circuits are connected in d) 4 3 1 2 parallel with like polarity, the

~~combined Norton equivalent circuit Q.41 Consider the following statements: will be Network NA in figure (A) can be~~

~~replaced by the network NB shown~~

in figure (B), when Ic and Rc, a) 3A, 8Ω b) 6A, 8Ω respectively, are Q.38 c)Which 0A, 2Ω of the followingd) theorems 6A, 2Ω can be applied to any network-linear or 1. 5A and 2Ω 2. 10A and 1Ω

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 1 are the Norton equivalent current IN 3. 15A and Ω 4.30A and Ω 2 5 and resistance RN in the figure (b)?

V RR a) T, TL  RRTL RTL +R Which of these statements given  RTL +R is/are correct? V a) 1 only b) 2, 3 and 4 b) T ,R =R R NT c) 1, 2, 3 and 4 d) 2 and 3 T V T c) ,RNL= R Q.42 Consider the following properties of R T a particular network theorem: d) Noneof theabove 1. The theorem is not concerned with type of elements. Q.45 The terminal volt-ampere 2. The theorem is only based on the conditions of a linear reciprocal two Kirchoff’s laws. network N are shown in the figure 3. The reference directions of the (a). What is the current I branch voltages and currents are corresponding to the terminal arbitrary except that they have conditions shown in the figure (b)? 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 a) – 1 A b) 9 A c) 10 A d) 11 A Q.43 Consider the following circuit: Q.46 What is the value of R required for maximum power transfer in the network shown above?

~~What should be the value of resistance R, in the above circuit it has to absorb the maximum power from the source? a) 2Ω b) 4Ω a) 8/3 ohms b) 3/8 ohms c) d) c) 4 ohms d) 8 ohms 8Ω 16Ω~~

Q.44 A network with independent Q.47 What are the source voltage and sources and resistors shown above source resistance, respectively for in figure (a) has a Thevenin voltage the Thevenin’s equivalent circuit as VT and Thevenin resistance RT. What seen from the terminals indicated in the circuit given above?

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 a) 20V,24Ω b) 20V,48Ω D. Kirchhoffs current law c) 20V,4.8 Ω d) 20V,12Ω List – II (Property) 1. Effects on independent sources Q.48 What is the Thevenin resistance in a linear circuit are addictive seen from the terminals AB of the 2. Law of non-accumulation of circuit shown above in the figure? 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 a) 2Ω b) 4Ω

c) 8Ω d) 12Ω Q.51 For the network shown above I = (0.2V – 2) A, (I = the current Q.49 In the network in fig, the mesh delivered by the voltage source V). current I and the input impedance The Thevenin voltage Vth and seen by the 50V source, resistance Rthfor the network N respectively, are across the terminals AB are respectively

125 11 a) A and Ω 13 8 a) – b) 10 150 13 c) – d) 10 b) A and Ω 13 8 10V, 5 Ω V, 5 Ω 150 11 Q.52 In the10V, circuit, 0.2 Ω S was initiallyV, open.0.2 Ω At c) A and Ω time t=0, S is closed. When the 13 8 current through the inductor is 6A, 125 13 d) A and Ω the rate of change of current 13 8 through the resistor is 6A/s. The value of the inductor would be Q.50 Match List – I with List – II and select the correct answer using the code given below the Lists: List – I (Theorem/Law)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.56 The circuit shown in the given figure is in the steady state with the switch S closed

a) 1H b) 2H c) 3H d) 4H Q.53 In the circuit in fig, the switch S is The current i(t) after S is opened at closed at t = 0. Assuming that there t=0 is is no initial charge in the capacitor, a) A decreasing exponential the current iC (t) for t > 0 is 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 2t 2t 1H and 2H, respectively, are V − V − a) S eARC b) S eARC connected in parallel. At some time, R 2R the currents through L1 and L2 are 1 1 V − V − c) S eARC d) S eARC 1A and 2A, respectively. The current 2R R through R at time t = ∞ will be a) zero b) 1A Q.54 In the circuit shown in the given c) 2A d) 3A figure, the values of i(0+) and I (∞), will be, respectively 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. = − a) zero and 1.5A b) 1.5 A and 3A 1.i(0) 8A di c) 3A and zero d) 3A and 1.5A 2. (t= 0) = 3A / s dt Q.55 In the circuit shown in the given 3.i(∞= ) 4A figure, C1 = C2 = 2F and the capacitor C1 has a voltage of 20V when S is open.

The value of R is a) 0.5 ohm b) 2.0 ohm c) 4.0 ohm d) 12 ohm If the switch S is closed at t = 0, the voltage v will be a Q.59 For the circuit in fig, the switch was c2 a) Fixed voltage of 20V kept closed for a long time before b) Fixed voltage of 10V opening it at time t = 0. The voltage c) Fixed voltage of -10V vL (0+) is d) Sinusoidal voltage

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1. Both make the solution of circuit problems simple and easy. 2. Both are applicable for the study of circuit behavior for t – 3. Both convert differential a) – 10V b) – 1V equations to algebraic equations.α to α. c) 0 V d) 10V 4. Both can be used for transient and steady state analysis. Q.60 For the circuit given, what is the Which of the above statements are expression for the voltage v? 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

a) + b) vvic vc Q.64 If the switch S in the circuit shown dv dv above is opened at t = 0, what are c) RCc − v d) RCc + v dt c dt c the values of V(0+) and d V(0+)/dt, respectively? Q.61 A step voltage is applied to the circuit shown. What is the transient current response of the circuit?

~~a) 100V, 10,000 V/s b) 100 V, - 10,000 V/s~~

a) Undamped sinusoidal c) –100 V, 10,000 V/s b) Overdamped d) –100 V, -10,000 V/s c) Underdamped d) Critically damped Q.65 The response of a linear, time- invariant system to a unit step is s(t) -t/RC Q.62 The circuit shown above is under = (1 – e ) u(t), where u(t) is the steady-state condition with the unit step. What is the impulse switch closed. The switch is opened response of this system? − at t = 0. What is the time constant of a) e t/RC b) e-t/RC u(t) the circuit? c) 1/RC {e-t/RCu(t)} d) δ(t)

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). a) 0.1 s b) 0.2 s dt c) 5 s d) 10 s What is the forced response of the system when x(t) = k (a constant)?

a) k b) k/3 Q.63 Consider the following statements regarding the use of Laplace c) 3k d) 0 transforms and Fourier transforms Q.67 In the circuit shown below, the in circuit analysis: switch in open for a long time and closed at time t = 0. What is the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission current through the switch after the d) 0.25 exp (-6.25 t) mA switch is closed? 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) 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 a) 0A b) 5A initial value of the current through c) 10A d) 20A the capacitor? Q.72 For series R-L-C circuit, the characteristic R1 equation is given as s2 ++= s 0. L LC R 1 a) 0.8 A b) 1.6 A If isdenoted byα and c) 2.4 A d) 3.2 A 2L LC byβ then under thecondition The switch of above circuit was 22 Q.69 ofβ >α , thesystem willbe open for long, and at t = 0 it is closed. What is the final steady state a) Critically damped voltage across the capacitor and the b) Under damped time – constant of the circuit? c) Undamped d) Over damped

Q.73 In the above circuit, the switch has been in position 1 for quite a long a) 0 V and 0.1 sec time. At t = 0 the switch is moved b) 20V and 0.2 sec to position 2. At this position c) 10 V and 0.2 sec what is the time constant? 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) 0.1 s b) 1 s c) 0.11 s d) 1.11 s~~

Q.74 In the above circuit, the switch is a) 0.50 exp (-25 t) mA open for a long time. At time t = 0, b) 0.25 exp (-25 t) mA the switch is closed. What are the c) 0.50 exp (-12.5 t) mA initial and final values of voltages across the inductor?

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission v =, 125e−−50t V,i= 5e 50t A .The value of L will be~~

~~a)0V and 0V b)0V and 80V c)80 V and 0V d)80V and 80V a) 0.005 H b) 0.05 H~~

c) 0.5 H d) 5 H Q.75 The voltage applied to an R-L circuit at t = 0 when switch is closed is 100 In the circuit shown above, switch S cos (100t + 30°). The circuit Q.79 is closed at t = 0. The time constant of the circuit and initial value of 0.6 H (in which initial current is current i(t) are zero).resistance What is 80 isΩ andthe inductance maximum is amplitude of current flowing through the circuit? a) 1A b) 2A c) 5A d) 10A

~~Q.76 The current in the network is 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 -t a) |t – 1 + e | u(t) dv b) |t2 – t + e-t | u(t) at t = 0? dt 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 = a) 0 and 4 b) 4 and 0 0, then v0 for t > 0 is given by 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 V b) 2 + 2(1 – e-100t) c) 2(1 – e-1000t) d) 2 e-1000t V~~

~~Q.78 Voltage and current expressions for a) 0 b) 10 c) 10 e-t d) 10 (1 – e-t)~~

~~the above circuit are given at t ≥ 0 as~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.82 In the circuit shown above, the a) 1 and 2 only b) 2 and 3 only switch is closed after a long time. c) 3 and 4 only d) 1 and 4 only The current is(0+) through the switch is 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) 1A b) 2/3A c) 1/3A d) 0A~~

Q.83 The value of V that would result in a a) 12.5 V b) 5.0 V steady-state current of 1A through c) 10.0 V d) 0.0 V the inductor in the above circuit is Q.87 Initially, the circuit shown in the given figure was relaxed. if the switch is closed at t=0, the values of di2 i(0+), di/dt (0+) and ()0+ will dt2 a) 30V b) 15V respectively be 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) 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 V (s) 4s function G(s) = 0 is V (s) s2 ++ 4s 20 i a) 1/3 A b) 2/3 A when R is 2 ohm. What is the value c) 1 A d) 0 A of L and C?~~

Q.85 Consider the following statements: 1. Voltage across a capacitor cannot change abruptly 2. Voltage across an inductor a) 0.3 H and 1 F b) 0.4H and 0.5F cannot change abruptly c) 0.5H and 0.1F d) 0.5H and 0.01F 3. Current through a capacitor cannot change abruptly Q.89 A unit step u (t-5) is applied to the 4. Current through an inductor RL network. The current i is given cannot change abruptly by Which of these statements are correct?

~~Q.94 For the circuit shown in the given -t a) 1- e figure, if C=20µF, v(0-)=-50V and b) [1-e-(t-5)] u(t-5) dv(0− ) c) (1- e-t) u(t-5) = 500V / S , then R is d) 1-e-(t-5) dt~~

Q.90 The response of a network is i(t)=Kte-α t for t0≥ where α is real positive. The value of‘t’ at which the i(t) will become maximum, is a) 2K b) 3K a) α b) 2α c) 5K d) 10K 2 c) 1/α d) α In the circuit shown in the given -2t Q.95 Q.91 If I = - 10 e , the voltage of the figure, the switch is closed at t = 0. source of the given circuit, Vs is The current through the capacitor given by will decrease exponentially with a time constant:

~~a) -10 e-2t b) -20 e-2t c) 20 e-2t d) -30 e-2t a) 0.5s b) 1s~~

c) 2s d) 10s Q.92 The steady state in the circuit, shown in the given figure is reached A unit step current of 1A is applied with S open. S is closed at t =0. The Q.96 to a network whose driving point current I at t =0+ is V(s) s+ 3 impedance is Z(s) = = ; I(s) (s+ 2)2 then the steady state and initial values of the voltage developed across the source are respectively 3 13 a) V,1V b) V, V 4 44 a) 1 A b) 2 A 3 3 c) 3 A d) 4 A c) V,0V d) 1V, V 4 4 1 Q.93 The system function H(s) = . for The current i in a series R-L circuit s +1 Q.97 an input signal cost, the steady state 20 mH is response is given by i = 2sin 500t A. If v is the with R = 10 Ω and L =

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 a) 20 2∠ 45o b) ∠3020 o o o c) ∠4520 d) ∠30220 Q.98 The input impedance of a series RLC circuit operating at frequency Q.102 In the circuit shown in the above figure, switch K is closed at t=0. The ω=2 ω0 0 being the resonant circuit was initially relaxed. Which frequency, is , ω one of the following sources of v(t) ω0L ω0L a) Rj− Ω b) Rj+ Ω will produce maximum current at 2 2 t=0+ ?

c) (R− j 2 ωΩ0 L) d) (R+ j 2 ωΩ0 L) Q.99 Which one of the following theorems can be conveniently used to calculate the power consumed by Ω the 10 resistor in the network a) Unit step shown in the given figure? b) Unit impulse c) Unit ramp d) Unit step plus unit ramp

Q.103 Consider the following statements: a) Thevenin’s theorem If a network has an impedance of (1- b) Maximum power transfer j) as a specific frequency, the circuit theorem would consist of series c) Millman’s theorem 1. R and C d) Superposition theorem 2. R and L 3. R, L and C Q.100 In the circuit shown in the given Which of these statements are figure, the current supplied by the correct? sinusoidal current source I is 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 a) 28A measured across the terminal 1 and b) 4A 2 was 4H with open terminals 3 and c) 20A 4. It was 3H when the terminal 3 d) not determinable from the data and 4 were short circuited. The given. coefficient of coupling would be

~~Q.101 In the circuit, if the power dissipated in the 6Ω resistor is zero then V is~~

~~a) 1 b) 0.707~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission c) 0.5 power factor of the network is d) indeterminate due to insufficient approximately: data. a) 0.9 b) 0.6 c) 0.3 d) 0.1 Q.105 The circuit shown in the figure, will act as an ideal current source with Q.109 A series R–L circuit is to be respect to terminals A and B, when connected to an a.c. source v(t) = Vm frequency is sin (ωi +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 π a) Zero b) 1 rad/s b) ωt0 = c) 4 rad/s d) 16 rad/s 2 ωL c) ωt0 = tan-1 Q.106 A series LCR circuit with R d) ωt0 has any arbitrary value R=Ω=Ω 10 |XL | 20 , and |Xc| = 20Ω is connected across an ac supply of 200Vrms. The rms voltage across Q.110 A series R – L – C circuit is switched the capacitor is on to a step voltage V at t = 0. What is the initial and final value of the a) 200∠ -90o V b) 200∠ 90o V current in the circuit, respectively? c) o d) o 400∠ 90 V 400∠ -90 V a)V/R, V/R b)Zero, Infinity c)Zero, Zero d)Zero, V/R In the circuit, if |I1| = |I2| = 10A Q.107 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) a) ωCR b) 1/( ωCR) c) ωC/R d) R/( ωC) -188 -1 I12 willlead bytan , I willlag by tan 66 Two coils are coupled in such a way b) Q.112 that the mutual inductance between -166 -1 I12 willlead by tan ,I willlag by tan them is 16mH. If the inductances of 88the coils are 20mH and 80mH c) respectively, the coefficient of −−1188 I12 will lag by tan , I willlead by tan coupling is: 66 a) 0.01 b) 0.4 d) c) 0.1 d) 0.0025 −−1168 I12 willlagbytan ,I willleadbytain 86Q.113 When is a series RLC circuit over damped? Q.108 In a two element series network, the 2 2 R1 R1 voltage and current respectively are a) = b) < 2 given as.V (t) = 50 sin (314 t) + 50 4L LC 4L C R12 R12 sin (942 t) V, i (t) = 10 sin (314 t + c) > d) = 600) + 8 sin (942 t + 450) A, then the 4L C 4C2 LC

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.114 Consider the following circuit: value and frequency is connected. For the above circuit, which one of The circuit will exhibit two the following statements is correct? resonances if The voltage V0 is independent of R, if a) The reactance of the capacitor is the input signal frequency ω 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 1 1 a)is b)is short circuit.  LC 2 LC Q.119 A series R-L-C circuit, excited by a c) is LC d) Has any value 100V variable frequency source, has a resistance of 10 Ω and an Consider the following circuit: For Q.115 inductive reactance of 50 Ω at 100 what value of ω, the circuit shown Hz. If the resonance frequency is exhibits unity power factor ? 500 Hz, what is the voltage across 1 1 a) b) the capacitor at resonance? LC + 22 a) 100V b) 500V LC R C c) 2500V d) 5000V 1 1 c) d) − 22 RC Q.120 In a series RLC, circuit, the locus of LC R C the tip of the admittance phasor in the complex plane as the frequency Q.116 An RLC series circuit has a is varied, is resistance R of 20 Ω and a current a) A semicircle in the upper half of which lags behind the applied the G-B plane having the centre voltage by 45o. If the voltage across 1 1 the inductor is twice the voltage at ,0 and radius across the capacitor, what is the R R value of inductive reactance? b) A circle in the upper half of the a) 10 Ω b) 20Ω G-B plane having the centre at c) 40Ω d) 60Ω 1 1 ,0 and radius ()2R ()R2 Q.117 R and C are connected in parallel c) A semicircle in the bottom half of across a sinusoidal voltage source of the G-B plane having the centre 240V. If the currents through the −11 source and the capacitor are 5A and at ,0 and radius 4A, respectively; what is the value of ()2R() 2R R? d) A semicircle in the upper half of a) 24 Ω b) 48 the G-B plane having the centre c) 80 Ω d) 240 Ω 11 at − ,0 and radius RR Q.118 A parallel circuit consists of two

branches: one with a pure capacitor Q.121 Which one of the following and the other has resistor of 5 Ω in statements is not correct for the series with a variable inductor. To circuit shown at resonant this circuit on ac voltage of fixed frequency?

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 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 a) The current is maximum b) The equivalent impedance is real Q.125 Width of resonance curve in an R – L c) The inductive and capacitive – C network is determined by which reactance are equal in one of the following? magnitude a)R alone b)L alone 1 C c)C alone d)All R, L and C d) The quality factor equals R L Q.126 What is the average power for Q.122 A parallel circuit has two branches. periodic non-sinusoidal voltages In one branch, R and L are in series and currents? and in the other branch, R and C are a) The average power of the in series. The circuit will exhibit fundamental component alone unity power factor when b) The sum of the average powers of the harmonics excluding the L a) R = b) R= LC fundamental C c) The sum of the average powers C L of the sinusoidal components c) R = d) R = L C including the fundamental d) The sum of the root mean square power of the sinusoidal Q.123 x(t): Input voltage y(t): Output voltage components including the Consider the circuit shown above: fundamental What is the natural response of this system? 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 a) A sinusoid with constant c) 8pF d) 10.66pF amplitude b) A growing sinusoid Q.128 Which one of the following relations c) Zero for power is not correct? d) A decaying sinusoid a) P = VIcos ϕ b) P = Re part of [VI*] Consider the following statements: Q.124 c) P = Re part of [V*I] When a series R –L –C circuit is d) P =VIsin ϕ under resonance 1. Current is maximum through R 2. Magnitude of the voltage across Q.129 For the circuit shown in the above L is equal to that across C figure, what is the natural 3. The power factor of the circuit is frequency? unity

a) 1 M rad/s b) 2 M rad/s c) 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 10 1 10 and D=3, then the value of Zm is a)  b)  0 10 01 01 0 10 c)  d)  10 0 10

Q.134 For an ideal step-down (n:1) 12 13 a) Ω b) Ω transformer, which one of the 13 12 following is the ABCD parameter c) 3Ω d) 4Ω matrix? n1 n0 Q.131 The impedance matrices of two, a)  b)  1n 0n two-port networks are given by n0 n 1/n 32 15 5 c)  d)  and  . If these two 0 1/n 1/n 1 23 5 25 What is the expression for h12 in networks are connected in series, Q.135 respect of the network shown: the impedance matrix of the resulting two-port network will be 35 18 7 a) b)  2 25 7 28 15 2 c) d)indeterminate 53 ZZ− ZZ+ a) 21 b) 12 Q.132 A two port network is reciprocal, if ZZ12+ ZZ21− and only if ZZ+ ZZ− c) 12 d) 12 a) = ZZ 2211 ZZ12− ZZ12+ b) BC−=− AD 1

c) YY12= − 21 Q.136 Two two-port networks are connected in parallel. The d) hh12= 21 combination is to be represented as a single two-port network. The Q.133 The input voltage V1 and current I1 for a linear passive network is given parameters of this network are obtained by addition of the by V1 = AV2 + BI2 and I1 = CV2 + individual DI2 Now consider the following network: a) z-parameters b) h – parameter c) y-parameters d)ABCD parameters

Q.137 Two 2-port networks with transmission matrices 12  2 4 TA = and TB =  0.1 4  0.5 3 are connected in cascade. Which is Which one of the following gives the the transmission matrix of the parameters of an equivalent π combination? network shown above? 3 10 36 a) y1 = 4 , y2 = 0, y3 = 1 a)  b)  2.2 12.4 0.2 12.4 b) y1 = 4 , y2 = 4 , y3 = 1 1 10 3 10 c) y1 = 1 , y2 = 1 , y3 = 1 c)  d)  d) y1 = 4 , y2 =0, y3 = 2 2.0 12.0 12.4 2.2   Q.141 Which of the following are the Q.138 Which one of the following is the conditions for a two port passive transmission matrix for the network network to be a reciprocal one? shown in the figure given above? 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 1 1+ yz 1+ yz z a) b)  Q.142 What is the open circuit impedance yz y1 Z’11(s) of the network shown in the 1z 1 1+ yz c)  d)  figure given above? y 1+ yz zy

~~Q.139 What is the value of the parameter h12 for the 2-port network shown in~~

~~the figure given above? 4 a) 10 + 2s b) 10 − s 4 c) 10 + d) 10 – 2s s~~

~~a) 0.125 b) 0.167 Q.143 What is the value of z21 for the c) 0.250 d) 0.625 network shown above?~~

~~Q.140 The currents I1 and I2 at the output of 2-port network can be written as I1 = 5V1 – V2 I2 = - V1 + V2~~

~~a) - 2 Ω b) –1/2 Ω c) 1 Ω d) 2 Ω~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.144 The circuit shown in the figure Q.149 What are the ABCD parameters of above the single element circuit given above?

a) Is reciprocal but not symmetrical b) Is not reciprocal but symmetrical 1Z 11 a)  b)  c) Is both reciprocal and 01 Z0 symmetrical 1Z Z1 d) Is neither reciprocal nor c)  d)  symmetrical 10 11

Q.145 The circuit shown in the above Q.150 For determining the network figure functions of a two-port network, it is 1. is reciprocal required to consider that 2. has Z11 = 2, Z22 = 2 a) All initial conditions remain 3. has Z11 = 4, Z22 = 2 same 4. has Z11 = 0, Z22 = 2 b) All initial conditions are zero Select the correct answer using the c) Part of initial conditions are code given below: 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 a) 1 and 3 b) 1 and 2 correct? c) 1 and 4 d) 3 only a) Z12 = Z21 and Z11 = Z22 Q.146 A reciprocal two-port network is b) Y12 = Y21 and Y11 = Y22 symmetrical if c) AD – BC = 1 and A = D a) ∇ A = 1 b) A = C d) All of the above c) z11 = z22 d) ∆ y = 1 Q.152 Match List I with List II and select Q.147 With respect to transmission parameters, which one of the the correct answer using the code following is correct? given below the lists: a) A and B are dimensionless List-I b) B and C are dimensionless (Network parameter) c) A and D are dimensionless d) B and D are dimensionless A. Z11 B. A Q.148 A Two-port network has z11 = C. C 13/35, z12 = z21 = 2/35, z22 = 3/35. D. Z22 Its y11 and y12 parameters will, respectively, be List-II a) 3, -2 b) 3, 2 (Measure under open-circuit c) 13, -2 d) 13, 2 conditions)

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission A. I1, I2 1. y V2 1) I01 = B. V1, V2 2. z I2 C. I1, V2 3. g~~

V1 D. V1, I2 4. h 2) I02 = V2 Codes: A B C D V1 3) I02 = a) 1 2 3 4 I1 b) 4 2 3 1 I c) 1 3 2 4 4) 1 I0= V 2 d) 4 3 2 1 2 Codes: Q.156 With reference to the above A B C D network the value of Z11 will be 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 a) – 3 b) 3 the transmission matrices of the c) – 1 d) – 5 individual networks, what type of connection is it? Q.157 A two-port network satisfies the a) Series connection following relations: b) Cascade connection 4I1 + 8I2 = 2V1 c) Parallel connection 8I1 + 16 I2 = V2 d) none of the above 1. The network is reciprocal 2. Z11 = 4 and Z12 = 8 Q.154 In the case of ABCD parameters, if 3. Z21 = 8 and Z22 = 16 all the impedances in the network 4. Z11 = 2 and Z12 = 4 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 Which of these relations are correct? d) A and D are unchanged, C is a)1, 2, 3 and 4 b)2 and 3 only doubled and B is halved c)3 and 4 only d)1 and 2 only

Match List I with List II and select Q.158 In the circuit shown, 2-port network Q.155 3 the correct answer using the code N has Z11 = 10 12 21 = - 6 4 given below the lists: 10 22 =10 I Ω, Z = 10Ω, Z List I List II (Two-port 2 gainΩ andis Z Ω. The current (Excitation) parameters) I1

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 0.7− 0.5 0.7 0.5 a) b)  −0.5 0.8 −0.5 0.8 0.8− 0.5 0.7− 0.5 c) d) −0.5 0.7 0.5 0.8 a) -50 b) +50  c) +20 d) -20 Q.163 Assertion A: The fundamental loop Q.159 In the 2-port network shown in the of a linear directed graph contains four twigs and two links figure, the value of Y12 is 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 correct 1 1 explanation of A. a) − mho b) + mho b) Both A and R are individually 3 3 true but R is NOT the correct c) −3mho d) + 3mho explanation of A. c) A is true but R is false. Q.160 The h parameters h11 and h22 are d) A is false but R is true related to z and y parameters as a) h11=z11 and h22=1/z22 Q.164 The number of edges in a compete b) h11= z11 and h22=y22 graph of n vertices is c) h11= 1/y11 and h22=1/z22 − )1n(n d) h11=1/y11 and h22=y22 a) n (n-1) b) 2 c) n d) n-1 Q.161 The lattice has the following impedances ZA = 3 + j4, Zn = 3-j4. Which one of the following is a cut Then the Z- parameters would be Q.165 set of the graph shown in the figure?

3+ j4 0 3− j4 a) 1, 2, 3 and 4 b) 2, 3, 4 and 6 a)b) c) 1, 4, 5 and 6 d) 1, 3, 4 and 5 0 3− j4 − j4 3 3− j4 3 − j4 3 Q.166 The network has 10 nodes and 17 c)d)  3 3+ j4 3+ j4 branches. The number of different node pair voltages would be a) 7 b) 9 Q.162 Which one of the following gives the correct short circuit parameter c) 10 d) 45 matrix Y for the network shown Q.167 The dual of a parallel R-C circuit is a a) Series R-C circuit b) Series R-L circuit c) Parallel R-C circuit d) 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 a) b)~~

Q.169 Which one of the following statements is not correct? a) A tree contains all the vertices of its graph. c) d) 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.173 The graph of a network is shown in Q.170 Consider the following graph: figure above. Which one of the figures shown below is not a tree of the graph?~~

~~Which one of the following is not a tree of the above graph? a) b) a) b)~~

~~c) d) c) d)~~

~~Q.171 What is the total number of trees for the graph shown above?~~

Q.174 A network has 4 nodes and 3 independent loops. Number of branches in the network? a) 5 b) 6 a) 4 b) 8 c) 7 d) 8 c) 12 d) 16 Q.175 Consider the following statements Q.172 For the network graph shown in the with regard to a complete incidence figure given above, which one of the matrix: following is not a tree? 1. The sum of the entries in any column is zero.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2. The rank of the matrix is n – 1 −−1 10 1 01 where n is the number of nodes.  c) 011 d) −−1 10 3. The determinant of the matrix of  a closed loop is zero. 10− 1 0− 11 Which of the statements given above are correct? Q.179 Number of fundamental cut-sets of a) 1 and 2 only b) 2 and 3 only any graph will be c) 1 and 3 only d) 1, 2 and 3 a) Same as the number of twigs b) Same as the number of tree Q.176 What is the number of chords of a branches connected graph G of n vertices and c) Same as the number of nodes e edges? d) Equal to one a) n (n – 1)/2 b) n – 1 c) e – n – 1 d) e – n + 1 Q.180 In a network with twelve circuit elements and five nodes, what is the Q.177 Consider a circuit which consists of minimum number of mesh resistors and independent current equations? sources, and one independent a) 24 b) 12 voltage source connected between c) 10 d) 8 the nodes i and j. The equations are obtained for voltage of n unknown Q.181 The maximum number of trees of nodes with respect to one reference the graph in fig, is node in the form

~~V1 M  V M []∆ 2 =  M M a)16 b)25  c)100 d)125 Vn M~~

What are the elements of the ∆ ? Q.182 Consider the spanning tree of the a) All conductance’s connected graph: What is the b) All resistances number of fundamental cut-sets? c) Mixed conductance’s and constant d) Mixed conductance’s and resistances

Q.178 Consider the directed graph shown a)15 b)16 above: c)8 d)7 What is its incidence matrix? 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:

~~−−1 10 10− 1 a)  b)  0− 11 11 0 −−10 1 0− 11~~

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

~~List X List Y Q.187 For the network shown in the figure Twigs 4,5,6,7 given above, what is the value of Links 1,2,3,8 z(s)? Fundamental cutest 1,2,3,4 Fundamental loop 6,7,8~~

Codes: A B C D a) I II III IV s2 ++ 2s 2 s2+ a) b) b) III II I IV s1+ (s+ 1) 2 c) I IV III II 2 s1+ (s+ 1) d) III IV I II c) d) 2 ++ + s 2s 2 (s 2) Q.184 Driving point impedance Q.188 If Y1 and Y2 are the RC driving point s(s2 +1) Z(s) = is not admittance and impedance 2 + 4s functions, respectively, such that realizable because the Y (s++ 1)(s 3) a) Number of zeroes is more than 1 = then, the value of Y (s+ 2)2 the number of poles 2 b) Poles and zeroes lie on the Y2 can be s3+ s2+ imaginary axis 1. 2. c) Poles and zeroes do not s2+ s1+ alternate on imaginary axis s2+ 3. 4. s+ 2 d) Poles and zeroes are not located (s++ 1)(s 3) on the real axis Which of the above are correct?

~~3 a) 1 and 2 b) 2 and 3 Q.185 The function 2s ++ can be realized c) 3 and 4 d) 1 and 4 s~~

a) Both as a driving point Q.189 A reactive network has poles at ω = impedance and as a driving point 0, 4000 rad/s, and infinity and zeros admittance at = 2000 and 6000 rad/s. The b) As an impedance, but not as an ω impedance of the network is –j 700 admittance Ohm at 1000 rad/s. What is the c) As an admittance, but not as an correct expression for the driving impedance point impedance? d) Neither as impedance nor as (ω−2 4 × 10 62 )( ω− 36 × 10 6 ) admittance a) −ω j(0.1 )22 6 Ohm Q.186 What is the minimum number of ω( ω− 16 × 10 ) elements required to realize a given ω22( ω− 16 × 10 6 ) b) j(0.1ω ) Ohm driving point susceptance function? (ω−2 4 × 10 62 )( ω× 36 × 10 6 ) a) One greater than the total (ω−2 4 × 10 62 )( ω− 36 × 10 6 ) c) ω number of internal poles & zeros j(0.1 )22 6 Ohm ω( ω− 16 × 10 )

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ω22( ω− 16 × 10 6 ) d) −ωj(0.1 ) Ohm (ω−24 × 10 62 )( ω− 36 × 10 6 )~~

Q.190 Consider the following expression for the driving point impedance: ss 22 ++ )9)(1(2 sZ )( = ss 2 + )4( 1. It represents an LC circuit 2. It represents an RLC circuit 3. It has poles lying on the j ω axis 4. It has a pole at infinite frequency and a zero at zero frequency Which of the statements given above are correct? a) 2 and 4 b) 1 and 3 c) 1 and 4 d) 2 and 3

~~Q.191 Which one of the following functions is an RC driving point impedance? s(s++ 3)(s 4) (s++ 3)(s 4) a) b) (s++ 1)(s 2) (s++ 1)(s 2) (s++ 3)(s 4) (s++ 2)(s 4) c) d) s(s++ 1)(s 2) (s++ 1)(s 3)~~

~~Q.192 Laplace transform of sin ( ωt + α ) is s cosα+ω sin α ω a) b) cos α s22+ω s22+ω s ssinα+ω cos α c) sin α d) s22+ω s22+ω~~

~~Q.193 For an RC driving-point impedance function ZRC (s) a) ZRC (0) ≥ ZRC(∞ ) b) ZRC (0) = ZRC( ∞ ) only c) ZRC(0) ≤ ZRC ( ∞ ) d) ZRC (0) > ZRC (∞ ) only~~

~~ANSWER KEY:~~

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)

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

Q.1 (b) Since the independent sources • The given function similar to cancel each other therefore the yx= which is Non linear, circuit will consist of only resistances. Hence the system is • The element is active since it has Linear and passive. two different slopes. One is +ve slope (for i > 0) and other is – ve Q.5 (b) (for i < 0) • For different current directions impedance is different hence non-bilateral

~~Nodal ⇒ V+− 120 V 60 xy+ xy 40 60~~

~~Vxy = 96 Q.2 (c) Rxy = 40 || 60 = 24Ω 2 Let, I be the current P11= IR 2 P22= IR Power Supplied by the source, 2 P= I (R12 + R ) 22 =+=+IR1 IR 2 P 12 P Q.6 (b) Q.3 (c)~~

~~Nodal at V2 ⇒ KCL at node A − + −= VVV− ix 21 0 21+ 2 −αi0 = ------(1) ⇒ RR ix = − 1A~~

VV21− P21= ×−() Also i = −  ------(2) 2v R = −2w(delivered) Substituting (2) in (1) we get Power is delivered since current is V2 1+α = entering from –ve terminal. Pabs due V21 +α to 2V source =Pdel =−−() 2w = 2w Q.4 (a) Q.7 (b)

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V= 2.5v~~

~~Q.11 (d)~~

~~KVL ⇒~~

24−+ I 2Vbb − V − 4I = 0 24−+ 5I V = 0 ------(1) 24 b IR = = 4A Also V= 3I ------(2) 6 b 2I= 8A From (2) and (1) R By KCL I= 12A = = I1R 3I 12A ⇒ Q.8 (a) Power delivered by 24v source =×=24 12 288w Q.9 (a) By applying source transformation Q.12 (a) to the dependent source we get

Nodal at V Nodal at V ⇒ V− 15 V V+ 51V 1 ++x =0 V5VV−− 7 7⇒ 19 1+ 12 −=10 2V− 15 V+ 51V 52 +=x 0 ------(1) 7V−= 5V 20 ------(1) 7 19 12 Nodal at ⇒ V+ 51V V2 = x ------(2) I VVV−− V5 19 21++ 2 2 =0 Vx = 2I ------(3) (by ohm’s law) 2 24 Solving (1) (2) and (3) we get −+2V12 5V = 5------(2) I= − 94.34mA Solving (1) &(2)

~~5V11=⇒= 25 V 5 Q.10 (b) V5− I=1 = 0A 5~~

~~Q.13 (d)~~

~~Here VTh = 5v (when EE= 1 and I is removed) V5 R5Th = = = Ω (For E=0, I=1A and I1 V = 5 volt) KVL in Loop I~~

⇒

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −−2I11 2I −= 16 0~~

~~−4I11 = 16 ⇒=− I 4A KVL in Loop II~~

~~Vx1+ 2I += 0 0 ⇒ =−= 31 Vx1 2I 8v iA= = KVL in Loop III 62~~

~~−−+=Vx 2I 16 0 ⇒ Q.17 (b) ⇒=I 4A~~

~~Q.14 (a)~~

~~Nodal at Vx ⇒~~

~~Vxx− 20 V By applying source transformation +−0.3Vx = 0 we get 1 10 V= 25v 10− 10 x I= = 0A 20− 25 5 I= = − 5A 1~~

Q.15 (c) P20V = 20 ×− 5 =− 100w (Delivered) By finding the Thevenin’s (Since current is entering from –ve equivalent of the circuit across terminal) resistor P=− P =−−() 100w = 100w 1Ω abs del

~~Q.18 (a) Nodal at V1 V9+ V 11++ =⇒ =− 3⇒ 0 V1 13.5v 39~~

~~VTh= VV A − B Q.19 (d) 25 5 20 = −= =10v 22 2 11 R= 11 + 11 =+= 1Ω Th 22 10 I= = 5A 2~~

~~kVL ⇒ V= 6v − 20 −=−5V 20 V= 4volt 4 By ohm’s law, I= = 2A 2 Q.16 (a) Applying source transformation we Q.20 (c) get~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.24 (d)~~

V= 2volt Nodal at V KVL in Loop (I) x V− 10 V −− = ⇒ = xx+ −= Vxx 1 2 0 V 3volt ⇒80 ⇒ 15 By KCL⇒=+=I 2 4 6A 5V−+ 50 V −= 40 0 xx V Q.21 (a) I=x = 3A The Value of R doesn’t affect the 5 value of V1 Hence when R=0, Q.25 (b) V1 = 40v . Since R=0 means circuit

~~hence V21= V = 40v~~

~~Q.22 (c) At t= − 0.5s V= 600() −=− 0.5 300v A V= 10v = V0B = I0C Ixx+− 15 = 0 ⇒ I =− 4 Hence the circuit can be modified as KCL ⇒ E−−() 20 − V = 0 KVL ⇒ E= − 10v~~

Q.26 (b) VV− I2=12 = μA −300 1M I= = − 1A V2 2 300 Rin = = = 1.5MΩ I2μ Q.23 (a) Time constant of RL Ckt = Time Q.27 (b) constant of RC(dimensionally) It is a balanced wheat stone bridge L 30 60 1 1 = RC (sec) = ⇒= R 60 120 2 2 From above equation: Current through R is zero Power R =I2 R = 0w → sec−1 L RC→ sec Q.28 (a) By source Transformation LL =⇒=→R2 R ohm CC 11 2 = ⇒= →rad ωω00() LC LC sec

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The circuit on both sides of 10 is For maximum power across~~

same hence I0= RL= R th = 25 Ω Ω Q.29 (c) Q.35 (b) Let the resistance be 2x, 4x, 6x Superposition theorem can be Since the resistances are connected applied to calculate voltage and in parallel, voltage is same. Currents current only .It can’t be used to find are in the ratio of power. VVV : :⇒ 6:3:2 2x 4x 6x Q.36 (a)

~~Q.30 (d)~~

~~Applying source Transformation we V V− 20 get Nodal −+40xx + = 5 14 Nodal at Vx ⇒ V− 3V Q.37 (d) xx+ −=60 21⇒~~

~~Vxx−+ 3 2V = 12~~

Vx = 5v 53− I= = 1A 2 Q.38 (c) Tellegen theorem is applicable to Q.31 (a) any network linear or nonlinear, When the load ( ) is constant the active or passive, time variant or minimum resistance proved time-invariant maximum power4Ω at load.

~~For resistance to be minimum Q.39 (c) R = Vth is open circuit voltage across R L I0= due to open circuit Q.32 (c) 0Ω 2 2 KVL in input loop~~

P=() PPP123 ±± 3V V−− 3I V =⇒=− 0 V V S 2 S 1L11⇒ L S+ 3 j4 Pmax =() 48 ++ 50 98 = 450W j4 2 = V = −− = S  Pmin ( 98 50 18) 2W 3+ j4 j4 4∟ 90° V= 10V = 10 V = 10 100 Q.33 (b) th L1  S 3+ j4 5 When impedance of one network is = ∟ the complex conjugate of the other 800 90°

~~ZZ= LS Q.40 (b) Norton equivalent: Current source Q.34 (c) in parallel with impedance Rth across RL is Open circuit output admittance: 25Ω~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission I 30 h|= 2 = ×= ×= 22 I1 = 0 I1 4 2.5 4 10A V2 12 Reciprocal network: Network where (By homogeneity principle) loop and equation have a Considering 6v source alone

~~symmetric coefficient matrix I2 = 1A (By reciprocity) Transmission Parameter: By superposition V= AV − BI =+= 1 22 I I12 I 11A I1= CV 22 − DI Q.46 (c)~~

~~Q.41 (b) RR= Th Calculated across R For fig (A) 5× 20 R= 5 || 20 += 4 += 4 8Ω Voltage across 1Ω= 5v Th 25 Current Across 1Ω= 5A 2 10A 1. I= 5A,R = 2Ω, ⇒ I = ×= 5 Q.47 (b) CC 1Ω 33~~

2. ICC= 10A,R =⇒= 1Ω, I 1Ω 5AV 1Ω = 5V 3. I= 15A,R =1 Ω, ⇒ I CC2 1Ω 1 =2 ×=15 5A,V = 5V 3 1Ω 2 = − 1 VTh VV 1 2 4. ICC= 30A,R =Ω, ⇒ I 1Ω 5 40 60 1 = ×−×100 100 100 100 =5 ×=30 5A,V = 5V 1 1Ω =−=−40 60 20v 5 V= 20v Th = + Q.42 (c) RTh 60 | 40 60 | 40 Tellegan’s theorm is applicable for 60× 40 =×=2 48Ω any type of element. It is based on 100 KCL & KVL. Voltage and current direction can be arbitrary. Q.48 (a)

~~Q.43 (a) 8 R= R = 8 || 4 = Ω Th 3~~

~~Q.44 (b)~~

~~4× 12 4 ||12= = 3Ω 16 11 1 R =++ AB 4 6 12 Q.45 (d) 321++ 6 Considering 30v source alone = =⇒=R2AB Ω 12 12~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission −t τ VCC()()()() t= V ∞+ V CC 0 − V ∞ e fort ≥ 0 Q.49 (b) V RC V()() 0 ,V∞=S ,τ = CC 22 −2t VVSSRC VtC () = − e 22 R= 2 ||1 ++ 1 ||1 1 −2t in {}() dVCS V RC 2 itC () ==−− c c e ×21  dt 2 RC = +1 ||1 3 V −2t = S e RC 13 R = Ω 8 50 400 Q.54 (c) IA1 = = Rin 13 1 150 I= ×= IA 5 1 1+ 13 3

+ Q.50 (b) at t= 0 3 i() 0+ = = 3A with shunt resistor 1 Norton’s theorem → current source at t →∞ Q.51 (b) ∞= When V= 0, I = − 2 (short circuit i0() current) (Capacitor gates open circuital) When I= 0, V = 10 (open circuit voltage) Q.55 (d) Since the given circuit is a LC tank V 10 V= 10vR = = =−= 5 5Ω circuit hence the voltage V will be Th Th − C2 I2 sinusoidal.

~~Q.52 (b) Q.56 (a) −t τ itLL()()()()= i ∞+ i0 LL − i ∞ e L i()() 0= 0,i ∞= 18A, τ = = Lsec LL 1 −t τ iL () t= 18 − 18e ------(1) −t τ 6= 18 − 18e (GivenitL () = 6)~~

−t 12 2 e τ = = fort≥ 0 converting the circuit in s- 18 3 domain − t diL () t −1 2 From (1) ⇒=−18e τ  Is() = dt τ 2s+ 18 2 i() t= 2e−2t for t ≥ 0 6= ×⇒= L 2H L3 Q.53 (a) Q.57 (a)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission As t →∞, the inductor gets short Hence, critically damped circuited and current through resistor R would be zero. Q.62 (a)

Q.58 (a) −t τ itLL()()()()= i ∞+ i0 LL − i ∞ e 2 τ= sec R −t τ iL () t= 4 − 12e − iL () 0= 1A − t diL () t −1 = 12e τ  dt τ 0.5V− 10iL −= V 0 KVL ⇒ diL () 0 12R −−= = = 6R 0.5V 10iL 0 dt 2 di −−L = 31 di() 0 10iL 0 R= = = 0.5ΩQ L = 3 dt 6 2 dt −10t  iL () t= Ce Comparing with standard form Q.59 (a) − t τ − i t= Ce = ( L () ) iL () 0 1A −t − τ 1 itLL() = i0e() τ= = 0.1sec 10 L 5m τ= = = 0.5m R 10 Q.63 (c) −2000t ∴=iL () t e for t ≥ 0

~~di Q.64 (b) V() t= LL = − 10e−2000t L dt + VL () 0= − 10V~~

~~Q.60 (d)~~

~~Current across inductor at t0= − is −+= = i0LL()() 0i0 =IR + VC As t →∞~~

dVC i∞= 1A IC= L () dt −t it= i ∞+ i0 − i ∞ eτ dV LL()()()() LL V= RCC + V C − t dt 0.01 L1 =−==1e τ R 100 Q.61 (d) =1e − −100t RC di (t) ξ= (For series RLC ckt) V() t= LL = 100e−100t 2L L dt =1

𝛏𝛏

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission − dv(t) dv (t) 4− 100t = =L =10 e t0 −+ dt dt i0LL()()= 4Ai0 = + dv(0 ) v −+ = −10,000 V0cc()()= 4vV0 = dt s + + = v= () 0 100v (From figure) t0 + 12− 4 8 i0() = = 2.5 2.5 Q.65 (c) = − t 3.2A Step response,s() t=( 1 − eRC ) u(t) given Impulse response, d ht() = st() dt −−tt−1 =1e −RCδ() t +− e RC  u(t) ( ) ( ) RC

~~1 −t = eRC u(t) KCL at node (A) RC + 3.2= 4 + ic 0 () Q.66 (b) + ic () 0= − 0.8A Forced response = particular integral + = ()()D+= 3yt k ic () 0 0.8A k y() t= e0t = k/3 D3+ Q.69 (d)~~

~~Q.67 (a)~~

~~Steady state is achieved as Vc ()∞= 10v t→∞ τ= RC = (10k ||10k) × 20μ =×=5k 20μ 0.1sec~~

~~− Q.70 (a) Vcv() 0= 10 −+ V0cc()()= 0 = V0 The current through the switch is ∞= zero since 0A current flows through Vc () 5v + −t resistor. i(0 )= OA τ Vc () t= 5 − 5e~~

~~−6 Q.68 2Ω (a) τ=() 20kΩ || 20kΩ 4 × 10 =10 ×× 4 1036 × 10−− =× 4 10 2 sec −t τ Vc () t= 5 − 5e cdV() t i() t=c = 4 × 10−−6 125e 25t c dt  −25t = 0.5e mA~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.71 (c) 1 i(t)= V × 50 m I() 0+ = = 10A 100 5 1 =×=100 1A 100~~

~~Q.76 (a)~~

Q.72 (b) R1 ss2 ++= 0 L LC V(s) 1 A B C = = = ++ 2 11 Is() 22 ωωnn= ⇒= = β s1++ s(s1)s s s1 + LC LC 11 1 = −+ R 1R Is() 2 2ξωn = ⇒= ξ LC s s s1+ L 2L −t α i() t=() t −+ 1 e u(t) ξ= <∴ 1( β22 > α ) β Hence, the system is under damped. Q.77 (a)

~~Let iL be the inductor current~~

~~Q.73 (a) −+10 36− i0LL() = = 2Ai0 = () τ= RC =100 ×× 10 10 = 0.1sec 5~~

~~20 i()∞= = 2A Q.74 (c) L 10 −t τ = iLL()()()() t= i ∞+ i L 0 − (i L ∞ e 2 di Vt() = LL = 0 L dt~~

Q.78 (c) −+100 V() t= 125e−50t i0LL() = = 10Ai0 = () 10 −50t + i() t= 5e VL () 0= 100 −= 20 80v L1 R V∞= 0v τ = = ⇒=L L () R 50 50 − V(t) 125e 50t = = = Q.75 (a) R−50t 25 Transfer function, i(t) 5e 25 Is() 1 ∴= = Hs() = = L 0.5H Vs() Zs() 50

~~1 = Q.79 (d) R+ sL 1 Hj()ω| = ω= 100 80+ j60 8 1 −1 =∟ − tan 6 100~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission τ= RC −+ iLL (0 )= 1A = i() 0 36× =×=10 20sec −+68 36+ V033() =×= 4 = V0() 93 + 5 i() 0= = 0.5A −+4 10 V066() = = V0() 3 Q.80 (c) −+ i0LL()()= 1Ai0 = V0()()−+= 2VV0 =

~~So the current~~

~~+ 1 i0s () = A 3~~

~~Q.83 (a) As t →∞, circuit would be in steady state~~

~~at t= 0+ + i0c () = 0 dv() 0+ c0= dt dv() 0+ V= 30v(by kVL) this gives = 0 dt Q.84 (a)~~

~~i0−+= 1Ai0 = Q.81 (c) LL11()() 10 i0−+= 0i0 = Is() = LL22()() 1s+ −+ −t V0cc()()= 4vV0 = i() t= 10e~~

~~Q.82 (c) By KVL + 12− 4 2 i01 ( ) = = A ⇒12 3 By KCL~~

⇒

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission di ()0+ 10− 10i() 0+ −− dt 1 idt = 0 10× 10−6 ∫ di0()+ di02 () ++ i0() − − −= 1025− 0 dt dt 10 ++ (on differentiating) i01s()()+= i0 1 di2 () 0+ = − 2 + 21 2 100A / s i0s () =−= 1 A dt 33

~~Q.88 (c) Q.85 (d) The transfer function of series RLC Voltage across capacitor and current circuit is across inductor can’t change Vo(S) R abruptly. = 1 Vi(S) R++ sL Q.86 (a) SC Nodal at V1~~

⇒

~~R S SRC = = L 2 V− 10 V S LC++ SRC 1 2 R1 11+ −=1.2 0 sS++ 2.5 2.5 L LC Comparing with given transfer 2V−= 10 3 1 function 13 V= = 65v R 1 =⇒=4 L 0.5H 2 L~~

~~1 + =20 ⇒= C 0.1F Vo () 0= 12.5v KVL in Loop (I) ⇒ LC~~

~~Q.87 (a) Q.89 (b)~~

~~e−5s + Is() = i() 0= OA s(s+ 1) + −5s 11 VL () 0= 10v =e  − + s s1+ diL () 0 −− ⇒=L 10A / s =ut()() −− 5 e(t 5) ut − 5 dt −−(t 5) KVL⇒ =−−[1 e ]u() t 5~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.90 (c) −+ V0cc()()= 4vV0 = −αt i() t= kte + t0= To find the condition for maximum Nodal at Vx value di = 0 ⇒ dt ke−−αt+ kte αt () −=α0 t = 1 α To check whether maximum or V −+2x + V − 40 = minimum at t = 1 x α 2 2 3V di x = = negative, hence t = 1 is 6 2 α 2 dt t= 1 α Vx = 4v condition for maximum value. V I() 0+ =x = 2A 2 Q.91 (b) Q.93 (a) 1 Given Hs() = s1+ 1 Hj()ω= I() t= − 10e−2t 1j+ω −10 Is() = s2+ −10 −10S s2+ Input signal r() t= cos cos t Is11() = Is() = s2+ 1 11 s ω = =∟ −°  H(j )ω= 45 1 1j+ 2 I s= Is + I s 21()()() 1 π = − −+10(s 1) ct() cos t  = 2  4  s2+ −+10(s 1) − 10 Vs() = = Q.94 (c) 2 ()s2(s1)s2++ + By KVL 10− 20 Vs()()= Vs −= s2⇒ s2++ s2 −2t dv 0+ Vs () t= − 20e + () i0c () = C dt Q.92 (b) Also, v0()()−+= 50v0 = + + dv() 0 ∴=i0c () C dt =10-2 A S

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 50 ωL R= = 5k Ω (From figure) 0 + =Rj + Ω i0c () 2~~

Q.95 (a) Q.99 (d) τ= RC Two different frequencies are 1 operating on the circuit i.e. =(1||1).1 = = 0.5sec = = simultaneously. 2 (ω 100,ω 200) 1 Since Z =jωL, Z = is frequency Q.96 (c) L c jωC vs()()()= zs&s dependent hence super position s3+ theorem can be used more conveniently. s(s+ 2)2

~~+ Initial value, v() 0= lts→∞ v() s = 0 Q.100 (c) Final value 22 I= IRL += R 144 + 256 + 3 v()∞= v() 0 = lts→∞ sv() s = =400 = 20A 4 steady state value.~~

~~Q.101 (a) Q.97 (a)~~

~~VR = 20∟ − 90°~~

~~VR = 20∟ 0°~~

I0= (Given) Hence V1=V2 Nodal at V1 The angle between V and I is V −∟ +=1  V1 20 0°⇒ 0 −−11VL 20 j1 θ= tan = tan = 45° VR  20 20∟ 0° 20 V12= =∟ 45° = V For RL circuit the power factor is 1j− 2 lagging in nature. Hence i lags v Nodal at V2 by45°. V22 Vv− =+=⇒=⇒0 2V2 v v 55⇒ Q.98 (b) = ∟ 1 20 2 45° zj()ω=++ R jωL jωc Q.102 (b) 1 For unit impulse input zj()ω=+− R j ω.L ωc 1 i0()()+−= i0 + L zj()ω = ω 2ω0 For all other inputs 1 +− =+− i0()()= i0 = 0 Rj 2ωL0 2ωC0 1 Q.103 (b) Q ωL0 = For RC circuit, z= R − jxc ωC0 For series RLC circuit,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission z=+− R j(x x ) x x 20 Lc Q2=L =c = = For parallel RLC circuit, R R 10 ∟ z=+− R j(xCL x ) Vc = 200 − 90° Hence, it can be RC or RLC circuit. Q.107 (c) Q.104 (c) The T-equivalent circuit is given by When 3 & 4 are open circuited

~~22 I2= II RC +~~

~~2 10= IR + 64~~

~~IR = 6A Leq= L p −+= MML p 2 2 I1= I R +−() II LC Lp = 4H~~

~~When 3 & 4 are short circuited IL = 16A~~

~~Leq=() L p − M || M +− L s M~~

~~()Lp − MM 3= +− LMs Lp~~

~~M=⇒= 2 k LLsp 2 1 ⇒==k 0.5 2 −1 IILC− θ1 = tan  IR Q.105 (c) −1 8 For the circuit to behave as an ideal = tan  current source, 6~~

~~−1 IC θ2 = tan  IR~~

−1 8 = tan  6 ZAB = ∞ −1 8 will lead by , I1will lag s1 I2 tan  × 6 16 s = ∞ −1 8 s1 by tan  + 6 16 s s = ∞ Q.108 (b) 2 + s 16 Power Factor = 0.6 s22=−⇒−=−⇒= 16 ω 16 ω 4rad/sec Q.109 (c) Q.106 (a) For transient free response Since, x= x = 20Ω the circuit is at Lc −1 ωL ωto = tan  resonance. R Vc = Qv∟ − 90° Q.110 (c)

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission − = 1 i0() 0 ∴=ω 0 22 + − i0() = 0 LC R c~~

~~i0()∞= Q.116 (c) Z= 20 + j20 (From given condition)~~

~~Z=+− R j(xLC x ) (For serial RLC circuit)~~

~~VLC= 2V (Given)~~

~~Q.111 (a) For parallel circuit Q = ωCR~~

~~Q.112 (b) = M k LL12 ⇒=ZL 2Z CL ⇒= x 2x C M 16× 10−3 16 x−= x 20 (From (1)) k= = = = 0.4 LC −6 −= LL12 20×× 80 10 40 2xCC x 20~~

~~xC = 20Ω Q.113 (c) xL = 40Ω~~

Q.114 (a) Q.117 (c) At resonant frequency L & C will act as short circuit 1 z= R || jωL ++ R L jωc 1 RjωL + jωc 22 zR= + I= IIRC + 1 L Rj++ωL =22 − jωc IRC II at resonance , j- term =0 =5422 −= 3 1 = V 240 ω0 rad / sec R= = = 80Ω LC I3R

Q.115 (c) Q.118 (b) Unity power factor exhibited by the 11 circuit at resonance. Y = + −+jx R jx 11 cL Y = + j R− jx jωL 1 = + L R + 22+ jωc xRxcL j At resonance, j-term =0 R + 1 x −j = L = + ωc 22 x Rx+ ωL 2 1 cL R + 22 22 ωc xxxR0L− cL += j-term = 0 at resonance

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 L1 R −ωLo −= 0 C ωCo LL ⇒RR2 =⇒= CC~~

~~Damping factor, ξ1<− for two resonance x −c <−1 ⇒ x > 2R ⇒>x 10Ω 2R c c~~

Q.119 (c) Q.123 (a) Voltage across capacitor For LC circuit the voltage across At resonance; V= QV ∠− 90° inductor or capacitor is sinusoidal c with constant amplitude. = ωL 50 at 100Hz 1 ⇒L= Hz Q.124 (a) 4π For series R-L-C circuit under 1 2π×× 500 resonance ωL Q=o =4π = 25 1. Current is maximum R 10 2. VL =VC Vc = 2500 ∠− 90° 3. Power factor =cos =1

Q.120 (b) Q.125 (d) ϕ BW of series or parallel R-L-C circuit Q.121 (d) is given by For series RLC, at resonant f frequency BW = o 1. Current is maximum Q 2. Equivalent impedance is R 1 fo = (which is purely real) LC 3. = xxLc 1L Q = for series RLC 1L RC 4. Q = RC C QR= for parallel RLC Q.122 (a) L 11 Hence Bandwidth depends on R, L Y = + and C. + 1 RjωL R + jωc Q.126 (d) j R + Rj− ωL = = + ωc Q.127 (b) 2 22 + 2 1 R ωL R + Let inductance be L and Self ωc22 capacitance be C pF. 0 j-term =0 1 =500 × 103 1 2π L(C+× 36) 10−12 At ω=ωωL ωC o0= 1 2 22 3 + 2 1 =250 × 10 R ωL0 + −12 R 22 2π L(C+× 160) 10 ωc0

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Solving above two equations we get Q.133 (b) C = 5.33 pF 1 10  01 Q.128 (d) Power in option a, b, and c are active V1= AV 22 + BI  ⇒ Given  (1) powers where as P = VI sin is I1= CV 22 + DI  reactive power. ϕ V12= V + 10I 2 Q.129 (d) And from given (2) n/ω = 11 II12  ω = = After comparing (1) & (2) we get LC 400× 10−−6 ×× 100 10 12 = A B  1 10  5Mrad / sec  =  CD  01  Q.130 (a) ABCD parameter equations are ___ Q.134 (c)

~~V1= AV 22 −⇒ BI V 1 n0 ⇒ = − ______(1)  1V22 2I 0 1/n~~

I1= CV 2 −⇒ DI 21 I We have = − (1) =1V22 − 3I _____ (2) V1 AV 22 BI = − (2) V I1 CV 22 DI And Zin = 1 VI12n I1 I02 = ∴ =−= V21 I1 And V22= − 10I So, V12= nV (3) ()−−10I22 2I −−10 2 12 ∴=zin = = 1 ()−10I − 3I −− 10 3 13 And II12= − (4) 22 n 12 After comparing eq (1), (2), (3) & zin = 13 (4) n AB  n 0  Q.131 (b) We get  =  C D  0 1/n  18 7  7 28 Q.135 (a) connected in ZZ− h = 21 series the impedance matrix of the 12 ZZ+ resulting⇒ If two networkstwo port arenetwork will be 12

the addition of the two impedance 1 matrix. = += ⇒Z11 In symmetrical() ZZ 1 2 lattice Z 22 network 2 Q.132 (b) 1 Z12= Z 21=() ZZ 2 − 1 BC−=− AD 1 2 ----- 1 ()ZZ21− Z parameter →=ZZ Z12 21 12 h = = 2 ⇒ Conditions for reciprocity 12 1 Y parameter →=YY Z22 21 12 ()ZZ12+ ABCD parameter →−BC AD 2 − = − ZZ21 1 ∴=h12 ZZ12+ h parameter →=−hh12 21

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.136 (c) Y parameter~~

networks – i)⇒ Combination of two port ii) 1/ y −=I .I parameterSeries connection → z parameter 211 + z iii) seriesParallel– connection → Y y parameter 1 iv) Parallelparallel– connection → h −=II21 parameter 1+ yz v) series connection → g ⇒=+D 1 yz

parameter V12= − zI cascade connection → ABCD ⇒=Bz Q.137 (a) 1z 3 10 So, T =   y 1+ yz 2.2 12.4  Q.139 (c) combination – V ⇒Transmission matrix of the = 1 T= TAB .T h12 V2 = 12  24  I01 T =    8 0.1 4  0.5 3  II32= 88+ 3 10 I I T =  ⇒=I 2 ⇒=−=−IIII 2 2.2 12.4 3 2 4 23 22 I ⇒=I 2 Q.138 (c) 4 2 1Z V= 4I = 2I ----- (1)  132 Y 1+ yz  V2= 4I 24 + 8I ⇒=V AV − BI 1 22 V2= 4I 22 + 4I I= CV − DI 1 22 V22= 8I ------(2) Making O/P port open –circuited i.e. V 2I 1 12= = = 0.250 I02 = V22 8I 4

~~V12= V ⇒= A1~~

~~I12= yV ⇒= c y~~

~~Q.140 (a) Y= 4J ,Y = 0 JJ ,Y = 1 , 123 Now making output port short eq n circuited I1= YV 11 1 + YV 12 2 i.e. V0= ⇒ We have Y parameter 2 I2= YV 21 1 + YV 22 2 Comparing above eq – n~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Y511 = ∴=V1 (2R − 3)I 12 + RI ------(1)~~

~~Y112 = − V2=++ RI 2 R(I 12 I )~~

~~Y121 = − ∴=V21 RI + 2RI 2 ------(2)~~

~~Y122 = We have, z parameter eq - n For network, V1= ZI 11 1 + ZI 12 2~~

~~Y11= YY 1 + 3 V2= ZI 21 1 + ZI 22 2 π Y22= YY 2 + 3 Comparing with eq (1) & (2) We YY= = − Y get n 12 21 3 Z= (2R − 3) So, 11 ZR12 = Y41 = J~~

~~ZR21 = Y02 = J~~

~~Z22 = 2R Y13 = J Condition for reciprocity Q.141 (d) ⇒=ZZ12 21~~

~~1, 2, and 3 Condition for reciprocity Condition for symmetry ⇒=ZZ11 22 – Therefore, the circuit is reciprocal z parameter ⇒=zz12 21 but not symmetrical.~~

~~y parameter ⇒=yy12 21 h parameter ⇒=−hh Q.145 (c) 12 21 1 & 4~~

Q.142 (c) By KVL in mesh (1) ' 4 ⇒V2II1(II) from= given ++ circuit + ____ Z11 () s= 10 + 1 11 12 s VI= ----- (1) ⇒Open ckt impedance 12 By KVL in mesh (2) ' 1 Z11 () s= 10 + V=++ 1I 1(I I ) 1 2 2 21 s = + ----- (2) 4 V21 I 2I 2 4 We have z parameter ' = + Z11 () s 10 = + s V1 ZI 11 1 ZI 12 2

~~V2= ZI 21 1 + ZI 22 2 Q.143 (b) Comparing with eq (1) & (2) We Z21 = − 1/2Ω get n~~

~~This is symmetric lattice network. Z11= 0, Z 12 = 1 13 = = += Z21= 1.Z 22 = 2 ⇒Z11 Z 22 () 21 Ω 22 Condition for reciprocity 11− Z= Z =() 12 −= Ω ⇒=ZZ12 21 =1 12 21 22 Z= 0, Z = 2 11 22~~

~~Q.144 (a) ⇒ and The circuit is reciprocal but not Q.146 (c) symmetrical. Z11 =Z 22 Condition of symmetry for~~

~~V1=−+− 3IRIR(II) 1 1 12 + various parameters are given ⇒ From given circuit, below⇒ -~~

~~© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission z parameter ⇒=ZZ11 22~~

~~y parameter ⇒=yy11 22 h parameter ⇒∆h1 =~~

~~ABCD parameter ⇒=AD Q.150 (b) Q.147 (c) All initial condition are zero. A and D are dimensionless. Equation for transmission Q.151 (d)~~

parameter are V1= AV 22 − BI All of the above ⇒ I= CV − DI ⇒ Condition for reciprocity & 1 22 symmetry. It is clear that A and D are Z parameter = = dimensionless. Z12 Z 21 & Z 11 Z 22 Y parameter Y12= Y 21 & y 11 = y 22 Q.148 (a) 3,-2 ABCD parameter AD-BC =1 & A = D ZZ ⇒=Y 22 = 22 11 ∆− z ZZ11 22 ZZ 12 21 Q.152 (d) 3 / 35 A B C D = 13 3 2 2 ⇒ 3 2 4 1 −. 35 35 35 35 Network Parameter Y3= Measure under open circuit 11 conditions −−ZZ And 12 12 Y12 = = V ∆− V1 1 z ZZ11 22 ZZ 12 21 a) Z11 b) A I1 = V2 = 2 / 35 I02 I02 = 13 3 2 2 I V − c) C 1 d) Z 2 35 35 35 35 22 V2 I0= I2 I0= Y2= − 2 1 12 Q.153 (b) Q.149 (a) Cascade connection 1z AB ABAB    ⇒=aa bb 01    CD CDCDaabb   II12= − ----- (1) is valid for cascade connection i.e.

~~V1−= V 21 IZ = − zI2~~

⇒=−VV12z I 2 We have ABCD parameter___ Q.154 (a) A and D remain unchanged. C is V= AV − BI 1 22 halved and B is doubled = − I1 CV 22 DI ⇒ Relation between ABCD & z – Comparing with equation (1)7(2) parameter of the two port network We get Z A = 11 AB  1z  Z  =  21 CD 01 Z    11Z22 Bz= − 12 Z21

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 Z And V= 8I + 16I ______(2) C = D = 22 21 2 Z21 , Z21 Comparing these equations with standard z parameter equation we Q.155 (a) get – Z211=Ω, Z 12 = 4Ω

~~Z821=Ω, Z 22 = 16Ω~~

~~∴≠ZZ12 21 Given network is not a reciprocal network.~~

Q.156 (d) Q.158 (b) -5 ⇒ From given figure – We have z parameter eq 3 V22=−×× I 10 10 V1= ZI 11 1 + ZI 12 2 n ⇒ For two port network we have V2= ZI 21 1 + ZI 22 2 V2= ZI 21 1 + ZI 22 2 Putting values of V ,Z &Z in above equation we get, 3642 21 22 −×I2 10 × 10 =− 10 I 12 + 10 I

⇒−44 + =− 6 Open circuit output terminal I21() 10 10 10 I V I 106 100 = 1 ⇒=2 = = Z11 4 50 I1 I 2× 10 2 I02 = 1 With I = 0 , draw circuit Apply KVL in loop (1) & (2) Q.159 (a) 2 =++ 1 V1 5I 112 5I 5I − mho V= 10I + 5I (1) 3 1 12 ⇒Short circuit input port. And + += ()10 5 I21 5I 4v z Applying KCL at node V

⇒15I += 5I 4 5 I + I VVVV− 2 2 1() 12 ++ =0 {}∴=+v II 11 1 z 12 ⇒=3V V ______(1) ⇒=−I 3I 2 21 V Net value ofI in eq (1) Also = −I 1 1 V= 10I − 15I n 1 112 ⇒=−vI = − 1 V11 5I Putting value in equation (1)

~~V1 3I()−=12 V ⇒ ⇒=−Z511 I1 = I02 I 1 ⇒==−1 Y mho 12 V32 = Q.157 (c) V01 3 & 4 only~~

~~as ⇒2V Given= 4I + equation 8I can be rewritten 1 12 V112= 2I + 4I ______(1) Q.160 (c)~~

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 11 h= & h = 11yz 22 link forms a closed loop. 11 22 ⇒ (i) In a linear directed graph, a (ii) The fundamental loop of a linear directed graph contains only one VI11 link and a number of twigs. ⇒h11 We= have & y 11 = IV11 v022= v0 = Q.164 (b) Iv h=22 & z = n(n− 1) 22 22 vI22= = I011 I0 2 n(n− 1) ⇒ Number of edges =n = Q.161 (b) C2 2 3j− 4  − j34 Q.165 (d) 1 (1, 3, 4, 5) is cut-set of graph. z= z = (Z + Z ) 11 222 A B 1 1 =(3 + j +− 3 j ) = ()6 2 442

zz311= 22 = 1 z12= z 21 = (Z B − Z A ) 2 Q.166 (d) 1 1 =(3j −44 −− 3j)=( − 2j4 ) Node pair voltage 2 2 10× 9 zz= = − j nCC= 10 = = 45 12 21 4 2221× zz  3− j  ∴=11 12 4  −  Q.167 (b) zz21 22  j3 4  Series R-L circuit

replaced as ⇒RG ↔ In dual circuit, the elements are LC↔ VI↔ Q.162 (a) Series↔ Parallel 0.7− 0.5 Node↔ mesh  −0.5 0.8 Q.168 (c) y=+= y y 0.2 + 0.5 = 0.7 11 A B ⇒=Loop branch – node + 1 = =−=− y12 y 21 y B 0.5 =−+bn1=1161 −+= 6

~~y22=+= y B y C 0.5 + 0.3 = 0.8 Q.169 (d) There are at least two edges in a circuit.~~

~~Q.170 (c)~~

~~Q.163 (d) A is false but R is true~~

a connected graph containing all the Q.179 (a) nodes⇒A tree of isthe a graphconnected but containing sub graph noof Same as number of twigs. loops. -sets of any graph will be same as number Q.171 (d) of⇒ twigs. Number of fundamental cut ⇒ n2− Total number of trees = n 42− = 4 = 16 Q.180 (d) ⇒L =−+ bn1

~~Q.172 (d) Branch() b= 12,node(n) = 5 Minimum number of mesh eq L= 12 −+ 5 1 n L = 8~~

~~Q.181 (d) ⇒A tree never contains a loop. n(n− 2) Where, n is node =5 Q.173 (d) ∴===⇒Then(n−− 2) maximum 5 (5 2) 5 3 number 125 of tree is~~

Q.182 (d) ⇒No. of node =8 No. of fundamental cut-sets = (n 1) =−=(8 1) 7 − a connected graph containing all he nodes⇒ A tree of isthe a connected graph but sub containing graph of no loops. Q.183 (a) ⇒ Twigs → 4,5,6,7 Q.174 (b) Links →1, 2, 3, 8 6 Fundamental cut-set →1,2,3,4 ⇒L =−+ bn1 Fundamental loop → 6,7,8 3=−+ b41 b6= Q.184 (c) For LC circuit, the poles and zeroes Q.175 (d) must be alternate and lie on 1, 2 and 3. imaginary axis.

Q.176 (d) Q.185 (a) e-n+1 3 The function s2++can be realized s Q.177 (a) as driving point impedance or as All conductance. driving point admittance function. If it is impedance function then Q.178 (c) 1 −−1 10 R2=Ω, L = 1H, C = F  3 011 If it is admittance function then  10− 1

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1 G= 2J , L = H, C = 1F 3 Q.190 (b) Since poles and zeroes of z(s) are Q.186 (b) purely imaginary and are alternate The number of elements required hence, the function is an LC circuit. to realize a given driving point susceptance function is equal to Q.191 (d) total number of poles and zeros. For RC driving point impedance function poles and zeroes are Q.187 (c) alternate and must lie on negative 1 real axis, nearest to imaginary axis Admittance, Ys() = ++ s 1 is pole. This is true for s1+ 22 ()s++ 2 (s 1) 1++ (s 1) s + 2s + 2 zs() = = = s++ 1 (s 3) s1++ s1 () 1 s1+ z()s = = y(s) s2 ++ 2s 2 Q.192 (d) L sin ()ωt+ α Q.188 (c) =L[sin ωtcosα + cosωt sinα] ] For RC driving point impedance ωcosα s sinα = + function a pole must be near to the 2222 s ++ωsω imaginary axis and for RC driving s sinα+ ωcos α point admittance function a zero 22 must be near to the imaginary axis. s + ω s3+ 1. Y2 = Zs() = Q.193 (d) s2+ For RC driving point impedance 2 ()()s1s3++ z0zRC()()>∞ RC ⇒=Y1 3 = Y(s) ()s2+ is not possible. s2++ s3 2. Y = Zs() = ⇒= Y Ys() = 21s1++ s2 is not possible. 1 s2+ e.g. zsR() = + 3. Y= Zs() = RC SC 2 ()()s1s3++ z0RC()()=∞ z RC ∞= R 1 ⇒=Y Ys() = is possible. 1 s2+

~~4. Y21= Zs()() =+⇒ s 2 Y = Ys ()s++ 1 (s 3) = is possible. (s+ 2)~~

~~Q.189 (c) For ()()ω2−× 4 10 62 ω − 36 × 10 6 zj()ω= j(0.1ω) ω22() ω−× 16 10 6 z() j1000= − j700~~