No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, photocopying, or otherwise without the prior permission of the author.GATE SOLVED PAPERElectronics & Communication Network AnalysisCopyright By NODIA & COMPANYInformation contained in this book has been obtained by authors, from sources believes to be reliable.However, neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Nodia and its authors are supplying information but are not attempting to render engineeringor other professional services.NODI A AND COMPANYB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, J aipur 302039Ph :+91 - 141 - 2101150www.nodia.co.inemail : enquiry@nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount*GATE SOLVED PAPER - ECNETWORK ANALYSI S NODIA & COMPANY2013ONE MARKQ. 1Consideradeltaconnectionofresistorsanditsequivalentstarconnectionas shown below. If all elements of the delta connection are scaled by a factor k, k 0 >, the elements of the corresponding star equivalent will be scaled by a factor of(A) k2(B) k(C)/ k 1(D)kQ. 2The transfer function V sV s12^^hh of the circuit shown below is(A) .ss105 1++(B) ss23 6++(C) ss12++(D) ss21++Q. 3A sourcecos v t V t 100sp = ^ hhas an internal impedance ofj 4 3 W + ^ h . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be(A) 3(B) 4(C) 5(D) 72013TWO MARKSQ. 4Inthecircuitshownbelow,ifthesourcevoltage100 53.13 V VSc + = thenthe Thevenins equivalent voltage in Volts as seen by the load resistance RL isGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) 100 90c + (B) 800 0c +(C) 800 90c + (D) 100 60c +Q. 5Thefollowingarrangementconsistsofanidealtransformerandanattenuator whichattenuatesbyafactorof0.8.Anacvoltage100V VWX1= isapplied acrossWXtogetanopencircuitvoltage VYZ1acrossYZ.Next,anacvoltage 100V VYZ2=is applied across YZ to get an open circuit voltage VWX2 across WX. Then,/ V VYZ WX 1 1,/ V VWX YZ 2 2 are respectively,(A) 125/ 100 and 80/ 100(B) 100/ 100 and 80/ 100(C) 100/ 100 and 100/ 100(D) 80/ 100 and 80/ 100Q. 6Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency is(A) q q1 2+(B)/ / q q 1 11 2+ ^ ^ h h(C)/ qR q R R R1 1 2 2 1 2+ + ^ ^ h h (D)/ qR q R R R1 2 2 1 1 2+ + ^ ^ h hQ. 7Three capacitors C1, C2 and C3 whose values are 10 F m , 5 F m , and 2 F mrespectively, have breakdown 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,andthecorrespondingtotalchargeinC m storedintheeffective capacitance across the terminals are respectively,(A) 2.8 and 36(B) 7 and 119(C) 2.8 and 32(D) 7 and 80GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inCommon Data For Q. 8 and 9:Consider the following figureQ. 8The current IS in Amps in the voltage source, and voltage VS in Volts across the current source respectively, are(A), 13 20 -(B), 8 10 -(C), 8 20 -(D), 13 20 -Q. 9The current in the 1W resistor in Amps is(A) 2(B) 3.33(C) 10(D) 122012ONE MARKQ. 10Inthefollowingfigure, C1and C2areidealcapacitors. C1hasbeenchargedto 12V before the ideal switch S is closed at. t 0 =The current( ) i tfor all t is(A) zero(B) a step function(C) an exponentially decaying function(D) an impulse functionQ. 11Theaveragepowerdeliveredtoanimpedance(4 3) j W - byacurrent 5 (100 100) cos A t p+is(A) 44.2W(B) 50W(C) 62.5W(D) 125WGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 12In the circuit shown below, the current through the inductor is(A)Aj 12+(B)Aj 11+-(C)Aj 11+(D) 0A2012TWO MARKSQ. 13Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is(A) 0.8W(B) 1.4W(C) 2W(D) 2.8WQ. 14If6V V VA B- =then V VC D-is(A)5V - (B)V 2(C)V 3 (D)V 6Common Data For Q. 15 and 16 :With 10V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed :(i)1W connected at port B draws a current of 3A(ii)2.5W connected at port B draws a current of 2AGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 15With 10V dc connected at port A, the current drawn by 7W connected at port B is(A) 3/ 7A(B) 5/ 7A(C) 1A(D) 9/ 7AQ. 16For the same network, with6V dc connected at port A, 1W connected at port B draws 7/ 3 . AIf 8V dc is connected to port A, the open circuit voltage at port B is(A) 6V(B) 7V(C) 8V(D) 9V2011ONE MARKQ. 17In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is(A) 6.4 . j 48 - (B) 6.56 . j 787 -(C) 10 0 j + (D) 16 0 j +Q. 18In the circuit shown below, the value ofRL such that the power transferred to RL is maximum is(A) 5W(B) 10W(C) 15W(D) 20WQ. 19The circuit shown below is driven by a sinusoidal input ( / ) cos v V t RCi p= . The steady state output vo isGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) ( / 3) ( / ) cos V t RCp(B) ( / 3) ( / ) sin V t RCp(C) ( / 2) ( / ) cos V t RCp(D) ( / 2) ( / ) sin V t RCp2011TWO MARKSQ. 20In the circuit shown below, the current Iis equal to(A) 1.4 A 0c + (B) 2.0 A 0c +(C) 2.8 0 A c + (D) 3.2 0 A c +Q. 21In the circuit shown below, the network N is described by the following Ymatrix:....SSSSY0100100101=-> H. the voltage gain VV12 is(A) 1/ 90(B) 1/ 90(C) 1/ 99(D) 1/ 11Q. 22In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time0 t = . The current( ) i tat a time t after the switch is closed is(A)( ) 15 ( 2 10 ) exp A i t t3#= -(B)( ) 5 ( 2 10 ) exp A i t t3#= -(C)( ) 10 ( 2 10 ) exp A i t t3#= -(D)( ) 5 ( 2 10 ) exp A i t t3#=- -GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in2010ONE MARKQ. 23For the two-port network shown below, the short-circuit admittance parameter matrix is(A)S4224 --> H (B) ..S105051 --> H(C) ..S105051> H (D)S4224> HQ. 24For parallel RLCcircuit, which one of the following statements is NOT correct ?(A) The bandwidth of the circuit decreases if R is increased(B) The bandwidth of the circuit remains same if L is increased(C) At resonance, input impedance is a real quantity(D) At resonance, the magnitude of input impedance attains its minimum value.2010TWO MARKSQ. 25In the circuit shown, the switch S is open for a long time and is closed at t 0 = . The current( ) i tfor t 0 $+ is(A)( ) 0.5 0.125 A i t et 1000= --(B)( ) 1.5 0.125 A i t et 1000= --(C)( ) 0.5 0.5 A i t et 1000= --(D)( ) 0.375 A i t et 1000=-Q. 26The current Iin the circuit shown is(A)1A j - (B)1A j(C) 0A(D) 20AGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 27In the circuit shown, the power supplied by the voltage source is(A) 0 W(B) 5 W(C) 10 W(D) 100 WGATE 2009ONE MARKQ. 28In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power.Which of the following can be the value of the current source I ?(A) 10 A(B) 13 A(C) 15 A(D) 18 AQ. 29If the transfer function of the following network is ( )( )V sV sio sCR 21=+The value of the load resistance RL is(A) R4(B) R2(C) R(D)R 2Q. 30A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in(A) 220 J (B) 12 kJ(C) 13.2 kJ (D) 14.4 JGATE 2009TWO MARKQ. 31AnACsourceofRMSvoltage20Vwithinternalimpedance( ) Z j 1 2sW = +feeds a load of impedance( ) Z j 7 4LW = +in the figure below. The reactive power consumed by the load is(A) 8 VAR(B) 16 VAR(C) 28 VAR(D) 32 VARQ. 32The switch in the circuit shown was on position a for a long time, and is move to position b at time t 0 = . The current( ) i tfor t 0 >is given by(A) 0.2 ( ) e u tt 125 - mA(B) 20 ( ) e u tt 1250 - mA(C) 0.2 ( ) e u tt 1250 - mA(D) 20 ( ) e u tt 1000 - mAQ. 33In the circuit shown, what value of RL maximizes the power delivered to RL?(A). 24W(B) 38W(C) 4W(D) 6WGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 34The time domain behavior of an RL circuit is represented byLdtdiRi + (1 ) ( ) sin V Be t u t/ Rt L0= +-.For an initial current of( ) iRV00= , the steady state value of the current is given by(A)( ) i tRV0" (B)( ) i tRV 20"(C)( ) ( ) i tRVB 10" + (D)( ) ( ) i tRVB210" +GATE 2008ONE MARKQ. 35In the following graph, the number of trees ( ) Pand the number of cut-set ( ) Qare(A), P Q 2 2 = = (B), P Q 2 6 = =(C), P Q 4 6 = = (D), P Q 4 10 = =Q. 36Inthefollowingcircuit,theswitchSisclosedatt 0 = .Therateofchangeof current( )dtdi0+ is given by(A) 0(B) LR Is s(C) ( )LR R Is s+(D) 3GATE 2008TWO MARKSQ. 37TheTheveninequivalentimpedanceZthbetweenthenodesP andQinthe following circuit is(A) 1(B)ss11+ +(C)ss21+ + (D) s ss s2 1122+ ++ +GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 38The driving point impedance of the following network is given by( ) Z s..s ss01 2022=+ +The component values are(A)5 , 0.5 , 0.1 L R C H F W = = =(B). , 0.5 , L R C 01 5 H F W = = =(C)5 , , 0.1 L R C 2 H F W = = =(D). , , L R C 01 2 5 H F W = = =Q. 39Thecircuitshowninthefigureisusedtochargethecapacitor Calternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows:For( ) nT t n T 2 2 1 # # + , ( , , ,..) n 0 1 2 =S1 to P1 and S2 to P2For ( ) ( ) , n T t n T 2 1 2 2 # # + +( , , ,...) n 0 1 2 =S1 to Q1 and S2 to Q2Assume that the capacitor has zero initial charge. Given that( ) u tis a unit step function , the voltage( ) v tc across the capacitor is given by(A)( ) ( ) tu t nT 1 nn 1- -3=/(B)( ) ( ) ( ) u t u t nT 2 1 nn 1+ - -3=/(C)( ) ( ) ( )( ) tu t u t nT t nT 2 1 nn 1+ - - -3=/(D) . . e e T 05 05( ) ( ) t nT t nTn2 21- + -3- - - -=6 @/Common Data For Q.40 and 41 :The following series RLC circuit with zero conditions is excited by a unit impulse functions( ) t d .GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 40For t 0 > , the output voltage v tC ^ h is(A)e e32 t t2123--^ h(B)te t3221 -(C)cos e t3223 t21 -c m(D)sin e t3223 t21 -c mQ. 41For t 0 > , the voltage across the resistor is(A)e e31 t t2321--_ i(B)cos sin e tt233123 t21--c c m m= G(C)sin et3223 t21 -c m(D)cos e t3223 t21 -c mStatement for linked Answers Questions 42 and 43:A two-port network shown below is excited by external DC source. The voltage andthecurrentaremeasuredwithvoltmeters, V V1 2andammeters., A A1 2(all assumed to be ideal), as indicatedUnder following conditions, the readings obtained are:(1) S1 -open, S2 - closed0 4.5 , 1.5 1 A V V A , V V,1 1 2 2= = = =A(2) S1 -open, S2 - closed4 6 , 6 A V V A 0 A, V V,1 1 2 2= = = =Q. 42The z-parameter matrix for this network is(A) ....15451515= G(B) ....15154545= G(C) ....15154515= G(D) ....45151545= GQ. 43The h-parameter matrix for this network is(A) .313067--= G(B) .331067- -= G(C) .313067= G(D) .331067 - -= GGATE 2007ONE MARKQ. 44An independent voltage source in series with an impedance Z R jXs s s= +delivers a maximum average power to a load impedance ZL when(A) Z R jXL s s= + (B) Z RL s=(C) Z jXL s= (D) Z R jXL s s= -GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 45The RC circuit shown in the figure is(A) a low-pass filter(B) a high-pass filter(C) a band-pass filter(D) a band-reject filterGATE 2007TWO MARKSQ. 46Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be. B2 the value BB21 is(A) 4(B) 1(C) 1/ 2(D) 1/ 4Q. 47For the circuit shown in the figure, the Thevenin voltage and resistance looking into X Y -are(A), 2 V34W(B) 4 , V32W(C)V,3432W(D) 4 , 2 V WQ. 48In the circuit shown, vC is 0 volts at t 0 =sec. For t 0 > , the capacitor current ( ) itC, where t is in seconds is given by(A). ( ) exp t 050 25 -mA (B). ( ) exp t 025 25 -mA(C) 0.5 ( 12.5 ) exp t 0 -mA (D). ( . ) exp t 025 625 -mAGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 49In the ac network shown in the figure, the phasor voltage VAB (in Volts) is(A) 0(B) 5 30c +(C). 125 30c + (D) 17 30c +GATE 2006TWO MARKSQ. 50A two-port network is represented by ABCD parameters given by VI11= G AC BDVI22=-= = G GIf port-2 is terminated by RL, the input impedance seen at port-1 is given by(A) C DRA BRLL++(B) BR DAR CLL++(C) BR CDR ALL++(D) D CRB ARLL++Q. 51In the two port network shown in the figure below, Z12 and Z21 and respectively(A) re andr0b (B) 0 andr0b -(C) 0 androb (D) re andr0b -Q. 52Thefirstandthelastcriticalfrequencies(singularities)ofadrivingpoint impedance function of a passive network having two kinds of elements, are apole and a zero respectively. The above property will be satisfied by(A) RL network only(B) RC network only(C) LC network only(D) RC as well as RL networksQ. 53A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is(A) 0.5 A(B) 2.0 A(C) 1.0 A(D) 0.0 AGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 54In the figure shown below, assume that all the capacitors are initially uncharged. If( ) ( ) v t u t 10i=Volts,( ) v to is given by(A) 8e/ . t 0004 - Volts(B) 8(1 ) e/ . t 0004-- Volts(C)( ) u t 8Volts(D) 8 VoltsQ. 55AnegativeresistanceRnegisconnectedtoapassivenetworkNhavingdriving point impedance as shown below. For( ) Z s2 to be positive real,(A)( ), Re R Z jneg 16 # w w(B)( ) , R Z jneg 16 # w w(C)( ), Im R Z jneg 16 # w w(D)( ), R Z jneg 16 + # w wGATE 2005ONE MARKQ. 56The condition on, R L and C such that the step response( ) y tin the figure has no oscillations, is(A) RCL21$ (B) RCL$(C) RCL2 $ (D) RLC1=Q. 57The ABCD parameters of an ideal: n 1 transformer shown in the figure are nx 00> HThe value of x will beGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A) n(B) n1(C) n2(D) n12Q. 58InaseriesRLCcircuit,2 R kW = ,L 1 = H,andC4001F m = Theresonant frequency is(A) 2 104#Hz(B)101 4p#Hz(C) 104 Hz(D) 2 104p #HzQ. 59ThemaximumpowerthatcanbetransferredtotheloadresistorRLfromthe voltage source in the figure is(A) 1 W(B) 10 W(C) 0.25 W(D) 0.5 WQ. 60ThefirstandthelastcriticalfrequencyofanRC-drivingpointimpedance function must respectively be(A) a zero and a pole(B) a zero and a zero(C) a pole and a pole(D) a pole and a zeroGATE 2005TWO MARKSQ. 61For the circuit shown in the figure, the instantaneous current( ) i t1 is(A)90210 3cA(B) 210 390c -A(C) 5 60cA(D) 5 60c -AQ. 62Impedance Z as shown in the given figure is(A)j29W(B)j9W(C)j19W(D)j39WGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 63For the circuit shown in the figure, Thevenins voltage and Thevenins equivalent resistance at terminals a b -is(A) 5 V and 2W(B) 7.5 V and. 25W(C) 4 V and 2W(D) 3 V and. 25WQ. 64If R R R R1 2 4= = =and. R R 113=in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is(A). 0238 V(B) 0.138 V(C). 0238 -V(D) 1 VQ. 65The h parameters of the circuit shown in the figure are(A) ....01010103 -= G(B) .1011005-= G(C) 30202020= G(D) .1011005 -= GQ. 66A square pulse of 3 volts amplitude is applied to C R -circuit shown in the figure. The capacitor is initially uncharged. The output voltage V2 at time t 2 =sec is(A) 3 V(B)3 -V(C) 4 V(D)4 -VGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inGATE 2004ONE MARKQ. 67Consider the network graph shown in the figure. Which one of the following is NOT a tree of this graph ?(A) a(B) b(C) c(D) dQ. 68The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is(A) L L M1 2+ +(B) L L M1 2+ -(C) L L M 21 2+ +(D)L L M 21 2+ -Q. 69The circuit shown in the figure, with, R L3141H W = =and3 C F =has input voltage( ) sin v t t 2 = . The resulting current( ) i tis(A)( . ) sin t 5 2 531c +(B)( . ) sin t 5 2 531c -(C)( . ) sin t 25 2 531c +(D)( . ) sin t 25 2 531c -GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 70Forthecircuitshowninthefigure,thetimeconstantRC 1 = ms.Theinput voltage is( ) sin v t t 2 10i3= . The output voltage( ) v to is equal to(A)( ) sin t 10 453c - (B)( ) sin t 10 453c +(C)( ) sin t 10 533c - (D)( ) sin t 10 533c +Q. 71FortheR L - circuitshowninthefigure,theinputvoltage( ) ( ) v t u ti= .The current( ) i tisGATE 2004TWO MARKSQ. 72For the lattice shown in the figure, Z j2aW =and Z 2bW = . The values of the open circuit impedance parameterszzz zz11211222=6=@G are(A) jjjj1111-+++= G(B) jjjj1111-- ++-= G(C) jjjj1111+-+-= G(D) jjjj1111+- +- ++= GGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 73Thecircuitshowninthefigurehasinitialcurrent( ) i 0 1L=-Athroughthe inductorandaninitialvoltage( ) v 0 1C=--Vacrossthecapacitor.Forinput ( ) ( ) v t u t = , the Laplace transform of the current( ) i tfor t 0 $is(A) s ss12+ +(B) s ss122+ ++(C) s ss122+ +-(D) s s 112+ +Q. 74The transfer function( )( )( )H sV sV sio=of an RLCcircuit is given by( ) H ss s 20 10102 66=+ +The Quality factor (Q-factor) of this circuit is(A) 25(B) 50(C) 100(D) 5000Q. 75Forthecircuitshowninthefigure,theinitialconditionsarezero.Itstransfer function( )( )( )H sV sV sic=is(A) s s 10 1012 6 6+ +(B) s s 10 10102 3 66+ +(C) s s 10 10102 3 63+ +(D) s s 10 10102 6 66+ +Q. 76Consider the following statements S1 and S2S1 : At the resonant frequency the impedance of a series RLC circuit is zero.S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor.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 FALSEGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inGATE 2003ONE MARKQ. 77The minimum number of equations required to analyze the circuit shown in the figure is(A) 3(B) 4(C) 6(D) 7Q. 78A source of angular frequency 1 rad/ sec has a source impedance consisting of 1 W resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is(A) 1 W resistance(B) 1 W resistance in parallel with 1 H inductance(C) 1 W resistance in series with 1 F capacitor(D) 1 W resistance in parallel with 1 F capacitorQ. 79AseriesRLC circuithasaresonancefrequencyof1kHzandaqualityfactor Q 100 = . If each of, R L and C is doubled from its original value, the new Q of the circuit is(A) 25(B) 50(C) 100(D) 200Q. 80The differential equation for the current( ) i tin the circuit of the figure is(A)( ) sindtd idtdii t t 2 222+ + = (B)( ) cosdtd idtdii t t 2 222+ + =(C)( ) cosdtd idtdii t t 2 222+ + = (D)( ) sindtd idtdii t t 2 222+ + =GATE 2003TWO MARKSQ. 81Twelve 1 W resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is(A) 65W(B) 1W(C) 56W(D) 23WGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 82The current flowing through the resistance R in the circuit in the figure has the formcos P t 4where Pis(A) ( . . ) j 018 072 + (B) ( . . ) j 046 190 +(C)( . . ) j 018 190 - + (D)( . . ) j 0192 0144 - +Common Data For Q. 83 and 84 :Assume that the switch S is in position 1 for a long time and thrown to position 2 at t 0 = .Q. 83At t 0 =+, the current i1 is(A) RV2-(B) RV -(C) RV4-(D) zeroQ. 84( ) I s1and( ) I s2aretheLaplacetransformsof( ) i t1and( ) i t2respectively.The equations for the loop currents( ) I s1 and( ) I s2 for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t 0 = , are(A) ( )( )R LsLsLsRI sI s 0CsCssV 1112+ +--+= > = = H G G(B) ( )( )R LsLsLsRI sI s 0CsCssV 1112+ +--+=-> = = H G G(C) ( )( )R LsLsLsR LsI sI s 0CsCssV 1112+ +--+ +=-> = = H G G(D) ( )( )R LsLsCsR LsI sI s 0CsCssV 1112+ +--+ += > = = H G GQ. 85Thedrivingpointimpedance( ) Z s ofanetworkhasthepole-zerolocationsas shown in the figure. If( ) Z 0 3 = , then( ) Z sisGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.in(A) ( )s ss2 33 32+ ++(B) ( )s ss2 22 32+ ++(C) ( )s ss2 23 32+ ++(D) ( )s ss2 32 32- --Q. 86Aninputvoltage( ) 10 ( 10 ) 10 ( ) cos cos v t t t 2 5 2 10 c c = + + + Visapplied toaseriescombinationofresistanceR 1W = andaninductanceL 1 = H.The resulting steady-state current( ) i tin ampere is(A) 10 ( 55 ) 10 (2 10 2) cos cos tan t t1c c + + + +-(B) 10 ( 55 ) 10 (2 55 ) cos cos t t23c c + + +(C) 10 ( 35 ) 10 (2 10 2) cos cos tan t t1c c - + + --(D) 10 ( 35 ) (2 35 ) cos cos t t23c c - + -Q. 87The impedance parameters z11 and z12 of the two-port network in the figure are(A). 5 z 2711W =and0. 5 z 212W = (B)3 z11W =and. z 0512W =(C) z 311W =and0. 5 z 212W = (D)2.25 z11W =and0.5 z12W =GATE 2002ONE MARKQ. 88The dependent current source shown in the figure(A) delivers 80 W(B) absorbs 80 W(C) delivers 40 W(D) absorbs 40 WQ. 89In the figure, the switch was closed for a long time before opening at t 0 = . The voltage vx at t 0 =+ is(A) 25 V(B) 50 V(C)0 5 -V(D) 0 VGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inGATE 2002TWO MARKSQ. 90In the network of the fig, the maximum power is delivered to RL if its value is(A) 16 W(B) 340W(C) 60 W(D) 20 WQ. 91If the 3-phase balanced source in the figure delivers 1500 W at a leading power factor 0.844 then the value of ZL (in ohm) is approximately(A). 90 3244c + (B). 80 3244c +(C). 80 3244c +- (D). 90 3244c +-GATE 2001ONE MARKQ. 92The Voltage e0 in the figure is(A) 2 V(B)/ 4 3 V(C) 4 V(D) 8 VQ. 93IfeachbranchofDeltacircuithasimpedanceZ 3 ,theneachbranchofthe equivalent Wye circuit has impedance(A) Z3(B)Z 3(C) 3 Z 3 (D) Z3Q. 94The admittance parameter Y12 in the 2-port network in Figure is(A). 002 -mho(B) 0.1 mho(C)0.05 -mho(D) 0.05 mhoGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inGATE 2001TWO MARKSQ. 95The voltage e0 in the figure is(A) 48 V(B) 24 V(C) 36 V(D) 28 VQ. 96Whentheangularfrequencywinthefigureisvaried0to3,thelocusofthe current phasor I2 is given byQ. 97In the figure, the value of the load resistor RL which maximizes the power delivered to it is(A). 1414W(B) 10 W(C) 200 W(D). 2828WGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 98The z parameters z11 and z21 for the 2-port network in the figure are(A); z z116111611 21W W = = (B); z z11611411 21W W = =(C); z z116111611 21W W = =- (D); z z11411411 21W W = =GATE 2000ONE MARKQ. 99The circuit of the figure represents a(A) Low pass filter(B) High pass filter(C) band pass filter(D) band reject filterQ. 100In the circuit of the figure, the voltage( ) v tis(A) e eat bt- (B) e eat bt+(C) ae beat bt- (D) ae beat bt+Q. 101In the circuit of the figure, the value of the voltage source Eis(A)16 -V(B) 4 V(C)6 -V(D) 16 VGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inGATE 2000TWO MARKSQ. 102Use the data of the figure (a). The current iin the circuit of the figure (b)(A)2 -A(B) 2 A(C)4 -A(D) 4 AGATE 1999ONE MARKQ. 103IdentifywhichofthefollowingisNOTatreeofthegraphshowninthegiven figure is(A) begh(B) defg(C) abfg(D) aeghQ. 104A 2-port network is shown in the given figure. The parameter h21 for this network can be given by(A)/ 1 2 - (B)/ 1 2 +(C)/ 3 2 - (D)/ 3 2 +GATE 1999TWO MARKQ. 105The Thevenin equivalent voltage VTH appearing between the terminals A and B of the network shown in the given figure is given byGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.in(A)( ) j j 16 3 4 - (B)( ) j j 16 3 4 +(C)( ) j 16 3 4 + (D)( ) j 16 3 4 -Q. 106The value of R (in ohms) required for maximum power transfer in the network shown in the given figure is(A) 2(B) 4(C) 8(D) 16Q. 107A Delta-connected network with its Wye-equivalent is shown in the given figure. The resistance, R R1 2 and R3 (in ohms) are respectively(A) 1.5, 3 and 9(B) 3, 9 and 1.5(C) 9, 3 and 1.5(D) 3, 1.5 and 9GATE 1998ONE MARKQ. 108A network has 7 nodes and 5 independent loops. The number of branches in the network is(A) 13(B) 12(C) 11(D) 10Q. 109The nodal method of circuit analysis is based on(A) KVL and Ohms law(B) KCL and Ohms law(C) KCL and KVL(D) KCL, KVL and Ohms lawQ. 110Superposition theorem is NOT applicable to networks containing(A) nonlinear elements(B) dependent voltage sources(C) dependent current sources(D) transformersQ. 111The parallel RLC circuit shown in the figure is in resonance. In this circuit(A)I 1 R L+mA(C)I I 1 R C+mAGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inQ. 112The short-circuit admittance matrix a two-port network is // 01 21 20-> HThe two-port network is(A) non-reciprocal and passive(B) non-reciprocal and active(C) reciprocal and passive(D) reciprocal and activeQ. 113The voltage across the terminals a and b in the figure is(A) 0.5 V(B) 3.0 V(C) 3.5 V(D) 4.0 VQ. 114A high-Q quartz crystal exhibits series resonance at the frequency swand parallel resonance at the frequency pw . Then(A) swis very close to, but less than pw(B)>s pw wGATE 1997ONE MARKQ. 115The current i4 in the circuit of the figure is equal to(A) 12 A(B)12 -A(C) 4 A(D) None or theseQ. 116The voltage Vin the figure equal to(A) 3 V(B)3 -V(C) 5 V(D) None of theseGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inQ. 117The voltage Vin the figure is always equal to(A) 9 V(B) 5 V(C) 1 V(D) None of the aboveQ. 118The voltage Vin the figure is(A) 10 V(B) 15 V(C) 5 V(D) None of the aboveQ. 119In the circuit of the figure is the energy absorbed by the 4 W resistor in the time interval ( , ) 0 3is(A) 36 J oules(B) 16 J oules(C) 256 J oules(D) None of the aboveQ. 120In the circuit of the figure the equivalent impedance seen across terminals, , a b is(A) 316Wb l(B) 38Wbl(C)j3812 W +b l(D) None of the aboveGATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.inGATE 1996ONE MARKQ. 121In the given figure,, A A1 2 and A3 are ideal ammeters. If A2 and A3 read 3 A and 4 A respectively, then A1 should read(A) 1 A(B) 5 A(C) 7 A(D) None of theseQ. 122The number of independent loops for a network with n nodes and b branches is(A) n 1 -(B) b n -(C) b n 1 - +(D) independent of the number of nodesGATE 1996TWO MARKSQ. 123Thevoltages, , V VC C 1 2and VC3acrossthecapacitorsinthecircuitinthegiven figure, under steady state, are respectively.(A) 80 V, 32 V, 48 V(B) 80 V, 48 V, 32 V(C) 20 V, 8 V, 12 V(D) 20 V, 12 V, 8 V***********GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inSOLUTI ONSSol . 1Option (B) is correct.In the equivalent star connection, the resistance can be given asRC R R RR Ra b cb a=+ +RB R R RR Ra b ca c=+ +RA R R RR Ra b cb c=+ +So, if the delta connection components Ra, Rb and Rc are scaled by a factor k thenRAl kR kR kRkR kRa b cb c=+ +^ ^ h h kk R R RR Ra b cb c2=+ +kRA=Hence, it is also scaled by a factor kSol . 2Option (D) is correct.For the given capacitance, C F 100m =in the circuit, we have the reactance.XC sc1=s 100 1016# #=- s104=So, V sV s12^^hh 1010s ss10101044444=+ ++ ss21=++Sol . 3Option (C) is correct.For the purely resistive load, maximum average power is transferred whenRLR XTh Th2 2= +where R jXTh Th+is the equivalent thevenin (input) impedance of the circuit. Hence, we obtainRL4 32 2= +5WSol . 4Option (C) is correct.For evaluating the equivalent thevenin voltage seen by the load RL, we open the circuit across it (also if it consist dependent source).The equivalent circuit is shown belowAs the circuit open across RL soI20 =or, 40 jI20 =GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free**Maximum Discount* www.nodia.co.ini.e., the dependent source in loop 1 is short circuited. Therefore, VL1 jjV4 34s=+^ hVThV 10 L1= 100 53.13j j4 340c =+ .100 53.135 531340 90ccc =800 90c =Sol . 5Option (C) is correct.For the given transformer, we have VVWX .1125=Since,VVYZ. 08 =(attenuation factor)So, VVWXYZ. . 08 125 1 = = ^ ^ h hor,VYZVWX=at VWX1100V = ; VV100100WXYZ11=at VWZ2100V = ; VV100100YZWX22=Sol . 6Option (C) is correct.The quality factor of the inductances are given byq1 RL11w=andq2 RL22w=So, in series circuit, the effective quality factor is given byQ RXR RL LeqLeq1 21 2w w= =++ R RR RLR RL1 12 11 211 22w w=++ R RRqRq1 121222 1=++ R RqR q R1 21 1 2 2=++Sol . 7Option (C) is correct.Consider that the voltage across the three capacitors C1, C2 and C3 are V1, V2 and V3 respectively. So, we can write VV32 CC23= ....(1)Since, Voltage is inversely proportional to capacitance GATE SOLVED PAPER - ECNETWORK ANALYSI S www.nodia.co.inBuy Online all GATE Books: shop.nodia.co.in*Shipping Free* *Maximum Discount* www.nodia.co.inNow, given thatC1F 10m = ;10V Vmax1= ^hC2F 5m = ;5V Vmax2= ^hC3F 2m = ;2V Vmax3= ^hSo, from Eq (1) we have VV32 52=forVmax3 ^h 2 =We obtain,V20.8volt52 25