GATE SOLVED PAPER - EC

0 ONE MARK Q. Conider a delta connection of reitor and it equivalent tar connection a hown below. If all element of the delta connection are caled by a factor k, k > 0, the element of the correponding tar equivalent will be caled by a factor of (A) k (B) k (C) /k (D) k V ^h Q. The tranfer function of the circuit hown below i V ^ h (A) 05 ṡ + + (B) 6 + + (C) + + (D) + + Q. A ource v ^th Vco 00pt ha an internal impedance of ^4+ jh W. If a purely reitive load connected to thi ource ha to extract the maximum power out of the ource, it value in W hould be (A) (B) 4 (C) 5 (D) 7 0 TWO MARKS Q. 4 In the circuit hown below, if the ource voltage V S 00+ 5.c V then the Thevenin equivalent voltage in Volt a een by the load reitance R L i

(A) 00+ 90c (B) 800+ 0c (C) 800+ 90c (D) 00+ 60c Q. 5 The following arrangement conit of an ideal tranformer and an attenuator which attenuate by a factor of 0.8. An ac voltage V WX 00 V i applied acro WX to get an open circuit voltage V YZ acro YZ. Next, an ac voltage V YZ 00 V i applied acro YZ to get an open circuit voltage V WX acro WX. Then, V / VWX, V / VYZ are repectively, YZ WX (A) 5/00 and 80/00 (B) 00/00 and 80/00 (C) 00/00 and 00/00 (D) 80/00 and 80/00 Q. 6 Two magnetically uncoupled inductive coil have Q factor q and q at the choen operating frequency. Their repective reitance are R and R. When connected in erie, their effective Q factor at the ame operating frequency i (A) q + q (B) ^/ qh+ ^/ qh (C) ^qr + qr h/ ^R+ Rh (D) ^qr + qr h/ ^R+ Rh Q. 7 Three capacitor C, C and C whoe value are 0 m F, 5m F, and m F repectively, have breakdown voltage of 0 V, 5V and V repectively. For the interconnection hown below, the maximum afe voltage in Volt that can be applied acro the combination, and the correponding total charge in m C tored in the effective capacitance acro the terminal are repectively, (A).8 and 6 (B) 7 and 9 (C).8 and (D) 7 and 80

Common Data For Q. 8 and 9: Conider the following figure Q. 8 The current I S in Amp in the voltage ource, and voltage V S in Volt acro the current ource repectively, are (A), - 0 (B) 8, - 0 (C) -8, 0 (D) -, 0 Q. 9 The current in the W reitor in Amp i (A) (B). (C) 0 (D) 0 ONE MARK Q. 0 In the following figure, C and C are ideal capacitor. C ha been charged to V before the ideal witch S i cloed at t 0. The current it () for all t i (A) zero (B) a tep function (C) an exponentially decaying function (D) an impule function Q. The average power delivered to an impedance (4 - j) W by a current 5 co(00p t + 00) A i (A) 44. W (B) 50 W (C) 6.5 W (D) 5 W

Q. In the circuit hown below, the current through the inductor i (A) (C) A + j (B) - A + j A + j (D) 0A 0 TWO MARKS Q. Auming both the voltage ource are in phae, the value of R for which maximum power i tranferred from circuit A to circuit B i (A) 0.8 W (C) W (B).4 W (D).8 W Q. 4 If V - V 6V then V - V i A B C D (A) (C) - 5V (B) V V (D) 6 V Common Data For Q. 5 and 6 : With 0 V dc connected at port A in the linear nonreciprocal two-port network hown below, the following were oberved : (i) W connected at port B draw a current of A (ii).5 W connected at port B draw a current of A

Q. 5 With 0 V dc connected at port A, the current drawn by 7 W connected at port B i (A) /7 A (B) 5/7 A (C) A (D) 9/7 A Q. 6 For the ame network, with 6V dc connected at port A, W connected at port B draw 7/ A. If 8V dc i connected to port A, the open circuit voltage at port B i (A) 6V (B) 7V (C) 8V (D) 9V 0 ONE MARK Q. 7 In the circuit hown below, the Norton equivalent current in ampere with repect to the terminal P and Q i (A) 6.4 - j 48. (B) 6.56 - j 787. (C) 0 + j0 (D) 6 + j0 Q. 8 In the circuit hown below, the value of R L uch that the power tranferred to R L i maximum i (A) 5 W (C) 5 W (B) 0 W (D) 0 W Q. 9 The circuit hown below i driven by a inuoidal input vi Vpco(/ t RC). The teady tate output v o i

(A) ( Vp /) co( t/ RC) (B) ( Vp /) in( t/ RC) (C) ( Vp /) co( t/ RC) (D) ( Vp /) in( t/ RC) 0 TWO MARKS Q. 0 In the circuit hown below, the current I i equal to (A).4+ 0c A (B).0+0c A (C).8+ 0c A (D).+ 0c A Q. In the circuit hown below, the network N i decribed by the following Y matrix: 0. S -00. S Y > 00. S 0. S H. the voltage gain V i V (A) /90 (B) /90 (C) /99 (D) / Q. In the circuit hown below, the initial charge on the capacitor i.5 mc, with the voltage polarity a indicated. The witch i cloed at time t 0. The current it () at a time t after the witch i cloed i (A) it ( ) 5 exp( - # 0 t) A (B) it () 5 exp( -# 0 t) A (C) it ( ) 0 exp( - # 0 t) A (D) it () -5 exp( -# 0 t) A

00 ONE MARK Q. For the two-port network hown below, the hort-circuit admittance parameter matrix i 4 - - 05. (A) > S - 4 H (B) > S -05. H (C) > 05. S 05. H (D) 4 > 4 H S Q. 4 For parallel RLC circuit, which one of the following tatement i NOT correct? (A) The bandwidth of the circuit decreae if R i increaed (B) The bandwidth of the circuit remain ame if L i increaed (C) At reonance, input impedance i a real quantity (D) At reonance, the magnitude of input impedance attain it minimum value. 00 TWO MARKS Q. 5 In the circuit hown, the witch S i open for a long time and i cloed at t 0. The current it () for t $ 0 + i -000t (A) it ( ) 0.5-0.5e A -000t (B) it ( ).5-0.5e A -000t (C) it ( ) 0.5-0.5e A -000t (D) it ( ) 0.75e A Q. 6 The current I in the circuit hown i (A) - ja (B) ja (C) 0A (D) 0 A

Q. 7 In the circuit hown, the power upplied by the voltage ource i (A) 0 W (C) 0 W (B) 5 W (D) 00 W GATE 009 ONE MARK Q. 8 In the interconnection of ideal ource hown in the figure, it i known that the 60 V ource i aborbing power. Which of the following can be the value of the current ource I? (A) 0 A (B) A (C) 5 A (D) 8 A Q. 9 If the tranfer function of the following network i Vo () V() + CR i The value of the load reitance R L i (A) R 4 (C) R (B) R (D) R Q. 0 A fully charged mobile phone with a V battery i good for a 0 minute talktime. Aume that, during the talk-time the battery deliver a contant current of A and it voltage drop linearly from V to 0 V a hown in the figure. How much energy doe the battery deliver during thi talk-time?

(A) 0 J (C). kj (B) kj (D) 4.4 J GATE 009 TWO MARK Q. An AC ource of RMS voltage 0 V with internal impedance Z ( + j) W feed a load of impedance ZL ( 7+ 4j) W in the figure below. The reactive power conumed by the load i (A) 8 VAR (C) 8 VAR (B) 6 VAR (D) VAR Q. The witch in the circuit hown wa on poition a for a long time, and i move to poition b at time t 0. The current it () for t > 0 i given by -5t (A) 0. e u( t) ma -50t (B) 0 e u( t) ma -50t (C) 0. e u( t) ma -000t (D) 0 e u( t) ma Q. In the circuit hown, what value of R L maximize the power delivered to R L? (A) 4W. (C) 4 W (B) 8 W (D) 6 W

Q. 4 The time domain behavior of an RL circuit i repreented by L di / + Ri V ( Be Rt L in t) u( t) dt 0 + -. For an initial current of i() V0 0, the teady tate value of the current i given R by (A) it () " V 0 (B) it ()" V 0 R R (C) it () V0 " ( + B) (D) it () V0 " ( + B) R R GATE 008 ONE MARK Q. 5 In the following graph, the number of tree ( P ) and the number of cut-et ( Q ) are (A) P, Q (B) P, Q 6 (C) P 4, Q 6 (D) P 4, Q 0 Q. 6 In the following circuit, the witch S i cloed at t 0. The rate of change of current di ( 0 + ) i given by dt (A) 0 (C) ( R+ R ) I L (B) RI L (D) GATE 008 TWO MARKS Q. 7 The Thevenin equivalent impedance Z th between the node P and Q in the following circuit i (A) (B) + + (C) + + (D) + + + +

Q. 8 The driving point impedance of the following network i given by Z () 0. + 0. + The component value are (A) L 5 H, R 0.5 W, C 0. F (B) L 0. H, R 0.5 W, C 5F (C) L 5 H, R W, C 0.F (D) L 0. H, R W, C 5F Q. 9 The circuit hown in the figure i ued to charge the capacitor C alternately from two current ource a indicated. The witche S and S are mechanically coupled and connected a follow: For nt # t # ( n + ) T, ( n 0,,,..) S to P and S to P For ( n+ ) T # t # ( n+ ) T, ( n 0,,,...) S to Q and S to Q Aume that the capacitor ha zero initial charge. Given that ut () i a unit tep function, the voltage vc () t acro the capacitor i given by / (A) (-) n tu( t -nt) n / (B) ut () + (-) n ut ( -nt) n / (C) tu( t) + (-) n u( t -nt)( t -nt) n -( t-nt) -( t-nt) (D) / 605. - e + 05. e -T@ n Common Data For Q.40 and 4 : The following erie RLC circuit with zero condition i excited by a unit impule function d () t.

Q. 40 For t > 0, the output voltage vc ^th i t t - (A) - ^e - e h (B) te t (C) - e co t c m (D) - e in c Q. 4 For t > 0, the voltage acro the reitor i t m t t - - (A) _ e e i - (B) e t co t in t c m- c mg (C) (D) e e - t - t inc coc t m t m Statement for linked Anwer Quetion 4 and 4: A two-port network hown below i excited by external DC ource. The voltage and the current are meaured with voltmeter V, V and ammeter. A, A (all aumed to be ideal), a indicated Under following condition, the reading obtained are: () S -open, S - cloed A 0, V 4.5 V, V.5V, A A () S -open, S - cloed A 4A, V 6 V, V 6V, A 0 Q. 4 The z -parameter matrix for thi network i 5. 5. 5. 45. (A) 45. 5. G (B) 5. 45. G 5. 45. 45. 5. (C) 5. 5. G (D) 5. 45. G Q. 4 The h-parameter matrix for thi network i - - - (A) - 067. G (B) 067. G (C) 067. G (D) - -067. G GATE 007 ONE MARK Q. 44 An independent voltage ource in erie with an impedance Z R+ jx deliver a maximum average power to a load impedance Z L when (A) ZL R+ jx (B) ZL R (C) Z jx (D) Z R -jx L L

Q. 45 The RC circuit hown in the figure i (A) a low-pa filter (C) a band-pa filter (B) a high-pa filter (D) a band-reject filter GATE 007 TWO MARKS Q. 46 Two erie reonant filter are a hown in the figure. Let the -db bandwidth of Filter be B and that of Filter be B. the value B i B (A) 4 (B) (C) / (D) /4 Q. 47 For the circuit hown in the figure, the Thevenin voltage and reitance looking into X- Y are (A) 4 V, W (B) 4 V, W 4 (C) V, W (D) 4 V,W Q. 48 In the circuit hown, v C i 0 volt at t 0 ec. For t > 0, the capacitor current i () t, where t i in econd i given by C (A) 050. exp( - 5t) ma (B) 05. exp( - 5t) ma (C) 0.50 exp( -.5 t) ma (D) 05. exp( - 65. t) ma

Q. 49 In the ac network hown in the figure, the phaor voltage V AB (in Volt) i (A) 0 (B) 5+ 0c (C). 5+ 0c (D) 7+ 0c GATE 006 TWO MARKS Q. 50 A two-port network i repreented by ABCD parameter given by V A B V I G C D G -I G If port- i terminated by R L, the input impedance een at port- i given by (A) A+ BRL (B) ARL + C C + DR BR + D L L (C) DRL + A BR + C L (D) B+ AR D + CR L L Q. 5 In the two port network hown in the figure below, Z and Z and repectively (A) r e and b r 0 (B) 0 and -br 0 (C) 0 and b r o (D) r e and -br 0 Q. 5 The firt and the lat critical frequencie (ingularitie) of a driving point impedance function of a paive network having two kind of element, are a pole and a zero repectively. The above property will be atified by (A) RL network only (B) RC network only (C) LC network only (D) RC a well a RL network Q. 5 A mh inductor with ome initial current can be repreented a hown below, where i the Laplace Tranform variable. The value of initial current i (A) 0.5 A (C).0 A (B).0 A (D) 0.0 A

Q. 54 In the figure hown below, aume that all the capacitor are initially uncharged. If vi () t 0u() t Volt, vo () t i given by (A) 8e -t/ 0. 004 Volt (B) 8( - e -t/ 0. 004 ) Volt (C) 8 ut () Volt (D) 8 Volt Q. 55 A negative reitance R neg i connected to a paive network N having driving point impedance a hown below. For Z () to be poitive real, (A) R # Re Z ( jw), w (B) R # Z ( jw), w neg 6 neg 6 (C) R # Im Z ( jw), w (D) R # + Z ( jw), w neg 6 neg 6 GATE 005 ONE MARK Q. 56 The condition on RL, and C uch that the tep repone yt () in the figure ha no ocillation, i (A) R $ L C (B) R $ (C) R $ L C (D) R Q. 57 The ABCD parameter of an ideal : n 0 > 0 x H L C LC n tranformer hown in the figure are The value of x will be

(A) n (B) n (C) n (D) n Q. 58 In a erie RLC circuit, R kw, L H, and C mf The reonant 400 frequency i (A) # 0 4 Hz (B) 4 0 p # Hz (C) 0 4 Hz (D) p # 0 4 Hz Q. 59 The maximum power that can be tranferred to the load reitor R L from the voltage ource in the figure i (A) W (C) 0.5 W (B) 0 W (D) 0.5 W Q. 60 The firt and the lat critical frequency of an RC -driving point impedance function mut repectively be (A) a zero and a pole (B) a zero and a zero (C) a pole and a pole (D) a pole and a zero GATE 005 Q. 6 For the circuit hown in the figure, the intantaneou current i () t i TWO MARKS (A) 0 90c A (B) 0 (C) 5 60c A (D) 5-60c A Q. 6 Impedance Z a hown in the given figure i - 90c A (A) j9 W (C) j9 W (B) j9 W (D) j9 W

Q. 6 For the circuit hown in the figure, Thevenin voltage and Thevenin equivalent reitance at terminal a- b i (A) 5 V and W (B) 7.5 V and 5W. (C) 4 V and W (D) V and 5W. Q. 64 If R R R4 R and R. R in the bridge circuit hown in the figure, then the reading in the ideal voltmeter connected between a and b i (A) 0. 8 V (B) 0.8 V (C) - 0. 8 V (D) V Q. 65 The h parameter of the circuit hown in the figure are. (A) 0 0. -0. 0. G (B) 0-005. G (C) 0 0 0 0 G (D) 0-005. G Q. 66 A quare pule of volt amplitude i applied to C- R circuit hown in the figure. The capacitor i initially uncharged. The output voltage V at time t ec i (A) V (B) - V (C) 4 V (D) - 4 V

GATE 004 ONE MARK Q. 67 Conider the network graph hown in the figure. Which one of the following i NOT a tree of thi graph? (A) a (B) b (C) c (D) d Q. 68 The equivalent inductance meaured between the terminal and for the circuit hown in the figure i (A) L+ L+ M (B) L+ L-M (C) L+ L+ M (D)L + L -M Q. 69 The circuit hown in the figure, with R W, L H and C F 4 ha input voltage vt ( ) in t. The reulting current it () i (A) 5 in( t + 5. c) (B) 5 in( t - 5. c) (C) 5 in( t + 5. c) (D) 5 in( t - 5. c)

Q. 70 For the circuit hown in the figure, the time contant RC m. The input voltage i v ( t) in 0 t. The output voltage vo () t i equal to i (A) in( 0 t - 45c) (B) in( 0 t + 45c) (C) in( 0 t - 5c) (D) in( 0 t + 5c) Q. 7 For the R- L circuit hown in the figure, the input voltage vi () t u() t. The current it () i GATE 004 TWO MARKS Q. 7 For the lattice hown in the figure, Za j W and Zb W. The value of the open z z circuit impedance parameter 6 z @ z z G are - j + j - j + j (A) + j + j G (B) - + j - j G + j + j + j - + j (C) - j - j G (D) - + j + j G

- Q. 7 The circuit hown in the figure ha initial current il ( 0 ) A through the - inductor and an initial voltage vc ( 0 ) - V acro the capacitor. For input vt () ut (), the Laplace tranform of the current it () for t $ 0 i (A) + + (C) - + + (B) + + + (D) + + Vo () Q. 74 The tranfer function H () of an RLC circuit i given by V() H () 0 + 6 0 + 0 The Quality factor (Q-factor) of thi circuit i (A) 5 (B) 50 (C) 00 (D) 5000 i 6 Q. 75 For the circuit hown in the figure, the initial condition are zero. It tranfer Vc () function H () i V() i (A) (C) + 0 + 0 6 6 0 + 0 + 0 6 (B) (D) 6 0 + 0 + 0 6 6 0 + 0 + 0 6 6 Q. 76 Conider the following tatement S and S S : At the reonant frequency the impedance of a erie RLC circuit i zero. S : In a parallel GLC circuit, increaing the conductance G reult in increae in it Q factor. Which one of the following i correct? (A) S i FALSE and S i TRUE (B) Both S and S are TRUE (C) S i TRUE and S i FALSE (D) Both S and S are FALSE

GATE 00 ONE MARK Q. 77 The minimum number of equation required to analyze the circuit hown in the figure i (A) (B) 4 (C) 6 (D) 7 Q. 78 A ource of angular frequency rad/ec ha a ource impedance coniting of W reitance in erie with H inductance. The load that will obtain the maximum power tranfer i (A) W reitance (B) W reitance in parallel with H inductance (C) W reitance in erie with F capacitor (D) W reitance in parallel with F capacitor Q. 79 A erie RLC circuit ha a reonance frequency of khz and a quality factor Q 00. If each of RL, and C i doubled from it original value, the new Q of the circuit i (A) 5 (B) 50 (C) 00 (D) 00 Q. 80 The differential equation for the current it () in the circuit of the figure i (A) di + di + it ( ) in t (B) di + di + it ( ) co t dt dt dt dt (C) di + di + it ( ) co t (D) di + di + it ( ) in t dt dt dt dt GATE 00 TWO MARKS Q. 8 Twelve W reitance are ued a edge to form a cube. The reitance between two diagonally oppoite corner of the cube i (A) 5 W (B) W 6 (C) 5 6 W (D) W

Q. 8 The current flowing through the reitance R in the circuit in the figure ha the form Pco 4 t where P i (A) ( 08. + j07. ) (B) ( 046. + j90. ) (C) -(. 08+ j90. ) (D) -(. 0 9 + j0. 44) Common Data For Q. 8 and 84 : Aume that the witch S i in poition for a long time and thrown to poition at t 0. Q. 8 At t 0 +, the current i i (A) -V R (C) -V 4R Q. 84 I () (B) -V R (D) zero and I () are the Laplace tranform of i () t and i () t repectively. The equation for the loop current I () and I () for the circuit hown in the figure, after the witch i brought from poition to poition at t 0, are R+ L+ C -L I () V (A) > -L R + H I () G G 0 R+ L+ C -L I () V (B) > -L R + H - C I() G G 0 R+ L+ C -L I () V (C) > -L R+ L+ H - C I() G G 0 R+ L+ C -C I () V (D) > -L R+ L+ H I () G G 0 C C Q. 85 The driving point impedance Z () of a network ha the pole-zero location a hown in the figure. If Z() 0, then Z () i

(A) (C) ( + ) + + ( + ) + + (B) (D) ( + ) + + ( - ) -- Q. 86 An input voltage vt ( ) 0 co( t+ 0 c) + 0 5co( t+ 0c) V i applied to a erie combination of reitance R W and an inductance L H. The reulting teady-tate current it () in ampere i - (A) 0 co( t+ 55 c) + 0 co(t+ 0c+ tan ) (B) 0 co( t+ 55 c) + 0 co(t+ 55 c) - (C) 0 co( t- 5 c) + 0 co(t+ 0c-tan ) (D) 0 co( t- 5 c) + co(t- 5 c) Q. 87 The impedance parameter z and z of the two-port network in the figure are (A) z 7. 5 and z 0. 5 (B) W and z 05. W W z W (C) z W and z 0. 5 W (D) z.5 W and z 0.5 W GATE 00 Q. 88 The dependent current ource hown in the figure ONE MARK (A) deliver 80 W (C) deliver 40 W (B) aborb 80 W (D) aborb 40 W Q. 89 In the figure, the witch wa cloed for a long time before opening at t 0. The voltage v x at t 0 + i (A) 5 V (B) 50 V (C) - 50 V (D) 0 V

GATE 00 TWO MARKS Q. 90 In the network of the fig, the maximum power i delivered to R L if it value i (A) 6 W (C) 60 W (B) 40 W (D) 0 W Q. 9 If the -phae balanced ource in the figure deliver 500 W at a leading power factor 0.844 then the value of Z L (in ohm) i approximately (A) 90+. 44c (B) 80+. 44c (C) 80+ -. 44c (D) 90+ -. 44c GATE 00 ONE MARK Q. 9 The Voltage e 0 in the figure i (A) V (B) 4/ V (C) 4 V (D) 8 V Q. 9 If each branch of Delta circuit ha impedance Z, then each branch of the equivalent Wye circuit ha impedance (A) Z (B) Z (C) Z (D) Z Q. 94 The admittance parameter Y in the -port network in Figure i (A) - 00. mho (B) 0. mho (C) - 0.05 mho (D) 0.05 mho

GATE 00 TWO MARKS Q. 95 The voltage e 0 in the figure i (A) 48 V (C) 6 V (B) 4 V (D) 8 V Q. 96 When the angular frequency w in the figure i varied 0 to, the locu of the current phaor I i given by Q. 97 In the figure, the value of the load reitor R L which maximize the power delivered to it i (A) 4. 4 W (B) 0 W (C) 00 W (D) 8. 8 W

Q. 98 The z parameter z and z for the -port network in the figure are (A) z ; z 6 6 W W (B) z ; z 6 4 W (C) z ; z 6 6 W - W (D) z ; z 4 4 W W W GATE 000 Q. 99 The circuit of the figure repreent a ONE MARK (A) Low pa filter (C) band pa filter (B) High pa filter (D) band reject filter Q. 00 In the circuit of the figure, the voltage vt () i (A) e at (C) ae bt - e (B) e + e at bt - be (D) ae + be Q. 0 In the circuit of the figure, the value of the voltage ource E i at at bt bt (A) - 6 V (B) 4 V (C) - 6 V (D) 6 V

GATE 000 TWO MARKS Q. 0 Ue the data of the figure (a). The current i in the circuit of the figure (b) (A) - A (B) A (C) - 4 A (D) 4 A GATE 999 ONE MARK Q. 0 Identify which of the following i NOT a tree of the graph hown in the given figure i (A) begh (C) abfg (B) defg (D) aegh Q. 04 A -port network i hown in the given figure. The parameter h for thi network can be given by (A) - / (B) +/ (C) - / (D) +/ GATE 999 TWO MARK Q. 05 The Thevenin equivalent voltage V TH appearing between the terminal A and B of the network hown in the given figure i given by

(A) j6( - j4) (B) j6( + j4) (C) 6( + j4) (D) 6( - j4) Q. 06 The value of R (in ohm) required for maximum power tranfer in the network hown in the given figure i (A) (B) 4 (C) 8 (D) 6 Q. 07 A Delta-connected network with it Wye-equivalent i hown in the given figure. The reitance R, R and R (in ohm) are repectively (A).5, and 9 (B), 9 and.5 (C) 9, and.5 (D),.5 and 9 GATE 998 ONE MARK Q. 08 A network ha 7 node and 5 independent loop. The number of branche in the network i (A) (B) (C) (D) 0 Q. 09 The nodal method of circuit analyi i baed on (A) KVL and Ohm law (B) KCL and Ohm law (C) KCL and KVL (D) KCL, KVL and Ohm law Q. 0 Superpoition theorem i NOT applicable to network containing (A) nonlinear element (B) dependent voltage ource (C) dependent current ource (D) tranformer Q. The parallel RLC circuit hown in the figure i in reonance. In thi circuit (A) IR < ma (B) IR+ IL > ma (C) I + I < ma (D) I + I > ma R C R C

Q. 0 - / The hort-circuit admittance matrix a two-port network i > / 0 H The two-port network i (A) non-reciprocal and paive (B) non-reciprocal and active (C) reciprocal and paive (D) reciprocal and active Q. The voltage acro the terminal a and b in the figure i (A) 0.5 V (C).5 V (B).0 V (D) 4.0 V Q. 4 A high-q quartz crytal exhibit erie reonance at the frequency w and parallel reonance at the frequency w p. Then (A) w i very cloe to, but le than wp (B) w << wp (C) w i very cloe to, but greater than wp (D) w >> w p GATE 997 Q. 5 The current i 4 in the circuit of the figure i equal to ONE MARK (A) A (B) - A (C) 4 A (D) None or thee Q. 6 The voltage V in the figure equal to (A) V (B) - V (C) 5 V (D) None of thee

Q. 7 The voltage V in the figure i alway equal to (A) 9 V (C) V (B) 5 V (D) None of the above Q. 8 The voltage V in the figure i (A) 0 V (C) 5 V (B) 5 V (D) None of the above Q. 9 In the circuit of the figure i the energy aborbed by the 4 W reitor in the time interval ( 0, ) i (A) 6 Joule (C) 56 Joule (B) 6 Joule (D) None of the above Q. 0 In the circuit of the figure the equivalent impedance een acro terminal ab,, i (A) 6 b W l (B) b 8 W l (C) 8 b + j W l (D) None of the above

GATE 996 ONE MARK Q. In the given figure, A, A and A are ideal ammeter. If A and A read A and 4 A repectively, then A hould read (A) A (C) 7 A (B) 5 A (D) None of thee Q. The number of independent loop for a network with n node and b branche i (A) n - (B) b- n (C) b- n+ (D) independent of the number of node GATE 996 TWO MARKS Q. The voltage VC, VC, and V C acro the capacitor in the circuit in the given figure, under teady tate, are repectively. (A) 80 V, V, 48 V (C) 0 V, 8 V, V (B) 80 V, 48 V, V (D) 0 V, V, 8 V ***********

SOLUTIONS Sol. Sol. Sol. Sol. 4 Option (B) i correct. In the equivalent tar connection, the reitance can be given a R RR b a C Ra+ Rb+ Rc R RR a c B Ra+ Rb+ Rc R RR b c A Ra+ Rb+ Rc So, if the delta connection component R a, R b and R c are caled by a factor k then ^krbh^krch R A l k RR b c kr kra+ krb+ krc k R a+ R b+ R A c Hence, it i alo caled by a factor k Option (D) i correct. For the given capacitance, C 00mF in the circuit, we have the reactance. X C c - 6 0 4 # 00 # 0 So, 4 0 4 V ^h + 0 V ^ 4 4 h 0 4 + 0 + 0 + + Option (C) i correct. For the purely reitive load, maximum average power i tranferred when R L RTh + XTh where RTh + jxth i the equivalent thevenin (input) impedance of the circuit. Hence, we obtain R L 4 + 5 W Option (C) i correct. For evaluating the equivalent thevenin voltage een by the load R L, we open the circuit acro it (alo if it conit dependent ource). The equivalent circuit i hown below A the circuit open acro R L o I 0 or, j40i 0

Sol. 5 i.e., the dependent ource in loop i hort circuited. Therefore, ^j4hv V L j4 + j40 40 90c V Th 0V L 00 5.c 00 5.c j4 + 5 5. c 800 90c Option (C) i correct. For the given tranformer, we have V V 5. WX Sol. 6 Sol. 7 Since, V YZ 08. (attenuation factor) V So, V V YZ ^08. h^5. h WX or, V YZ V WX at VYZ V WX 00 V; 00 V WX 00 at VWX V WZ 00 V; 00 V 00 Option (C) i correct. The quality factor of the inductance are given by q R L w and q R L w So, in erie circuit, the effective quality factor i given by XLeq Q L L w w Req R+ + R w RR L w RR L q q + + R R qr+ qr + + R+ R R R R R Option (C) i correct. YZ Conider that the voltage acro the three capacitor C, C and C are V, V and V repectively. So, we can write V V C...() C Since, Voltage i inverely proportional to capacitance

Sol. 8 Now, given that C 0 mf ; ^V h max 0V C 5 mf ; ^V h max 5V C mf ; ^V h max V So, from Eq () we have V V 5 for ^V h max We obtain, V # 0.8 volt < 5 5 i.e., V < ^V hmax Hence, thi i the voltage at C. Therefore, V volt V 0.8 volt and V V+ V.8 volt Now, equivalent capacitance acro the terminal i C CC eq + C 5# + 0 80 m F C + C 5 + 7 Equivalent voltage i (max. value) V max V 8. So, charge tored in the effective capacitance i Q C eq V max 80 b 8. 7 l # ^ h mc Option (D) i correct. Sol. 9 Sol. 0 At the node, voltage i given a V 0 volt Applying KCL at node I V V S + + - 0 I 0 0 S + + - 0 I S - A Alo, from the circuit, VS - 5# V & V S 0 + V 0 volt Option (C) i correct. Again from the hown circuit, the current in W reitor i I V 0 0 A Option (D) i correct. The -domain equivalent circuit i hown a below.

Sol. Sol. I () vc()/ 0 vc() 0 + + C C C C I () CC b ( V) C+ C l C eq v C (0) V Taking invere Laplace tranform for the current in time domain, it () C eq d() t (Impule) Option (B) i correct. In phaor form, Z 4- j 5-6. 86cW I 5 00c A Average power delivered. P avg. I Z co 5 5 co 6.86 # c 50 W Alternate Method: Z (4 - j) W, I 5 co(00pt + 00) A P avg Re$ I Z. Re () 5 ( 4 j) # " # -, 00 50 W # Option (C) i correct Applying nodal analyi at top node. V+ 0c V 0c + + 0c j V( j + ) + j+ 0c j V - + j V 0 c j Current I + - + + j j j ( jj ) + + j A

Sol. Option (A) i correct. We obtain Thevenin equivalent of circuit B. Thevenin Impedance : Z Th R Thevenin Voltage : Now, circuit become a V Th 0c V Current in the circuit, I 0 - + R Power tranfer from circuit Ato B P ( I ) R+ I 0-0 : R R + D + : + - R D or P 4 + 70R ( + R) dp ( + R) 70- ( 4+ 70R) ( + R) 0 dr 4 ( + R) ( + R)[( + R)70 -(4+ 70 R)] 0 & R 0.8 W Sol. 4 Option (A) i correct. In the given circuit VA- VB 6V So current in the branch will be I AB A 6 We can ee, that the circuit i a one port circuit looking from terminal BD a hown below

For a one port network current entering one terminal, equal the current leaving the econd terminal. Thu the outgoing current from A to B will be equal to the incoming current from D to C a hown i.e. I DC I AB A Sol. 5 The total current in the reitor W will be I + I DC (By writing KCL at node D) + 5A So, V CD # (- I) -5V Option (C) i correct. When 0 V i connected at port A the network i Now, we obtain Thevenin equivalent for the circuit een at load terminal, let Thevenin voltage i V Th, 0 V with 0 V applied at port A and Thevenin reitance i R Th. I L R For R L W, I L A V R V Th,0 V Th + R Th,0 V Th + L...(i)

For R L.5 W, I L A VTh,0 V RTh + 5. Dividing above two RTh + 5. R + RTh + R Th Th R + 5 W Subtituting R Th into equation (i) Th...(ii) V Th,0 V (+ ) 9V Note that it i a non reciprocal two port network. Thevenin voltage een at port B depend on the voltage connected at port A. Therefore we took ubcript V Th,0 V. Thi i Thevenin voltage only when 0 V ource i connected at input port A. If the voltage connected to port A i different, then Thevenin voltage will be different. However, Thevenin reitance remain ame. Now, the circuit i a hown below : Sol. 6 VTh,0 V For R L 7 W, I L 9 A + R + 7 L Option (B) i correct. Now, when 6V connected at port A let Thevenin voltage een at port B i V Th,6 V. Here R L W and I 7 L A V Th, 6 V R 7 7 Th # + # 7 7 7V # + Thi i a linear network, o V Th at port B can be written a V Th V a+ b where V i the input applied at port A. We have V 0 V, V Th,0 V 9V ` 9 0a+ b...(i) When V 6V, V Th, 6 V 9V ` 7 6a+ b...(ii) Solving (i) and (ii) a 05., b 4 Thu, with any voltage V applied at port A, Thevenin voltage or open circuit voltage at port B will be

Sol. 7 So, V Th, V 05. V + 4 For V 8V V,8 V Th 05. # 8+ 4 8 Voc (open circuit voltage) Option (A) i correct. Replacing P- Q by hort circuit a hown below we have Sol. 8 Uing current divider rule the current I c i I 5 SC (6 0) (6.4 -j4.8) A 5 + 5 + j0 Option (C) i correct. Power tranferred to R L will be maximum when R L i equal to the Thevenin reitance. We determine Thevenin reitance by killing all ource a follow : Sol. 9 R TH 0 # 0 + 0 5 W 0 + 0 Option (A) i correct. The given circuit i hown below For parallel combination of R and C equivalent impedance i R $ j w C Z p R R + + jwrc j w C Tranfer function can be written a R Vout Zp + jwrc Vin Z + Zp R + + R j w C + j w RC jwrc jwrc+ ( + jwrc) j Here j + ( + j) w RC

Sol. 0 Vout j V in ( + j) + j V Thu v out b lco(/ trc) Option (B) i correct. From tar delta converion we have p Thu R. RR 66 Ra+ Rb+ Rc 6 + 6 + 6 Here R R R W Replacing in circuit we have the circuit hown below : a b W Sol. Sol. Now the total impedance of circuit i ( + j4)( -j4) Z + 7W ( + j4)( - j4) Current I 4+ 0c + 0c 7 Option (D) i correct. From given admittance matrix we get I 0.V- 0.0V and...() I 0.0V+ 0.V...() Now, applying KVL in outer loop; V -00I or I - 00. V...() From eq () and eq () we have - 00. V 00. V+ 0. V - 0. V 00. V V V - Option (A) i correct. Here we take the current flow direction a poitive. At t 0 - voltage acro capacitor i

- - Q VC ( 0 ) - - 5. # 0-6 -50 V C 50 # 0 + Thu V C (0 ) -50 V In teady tate capacitor behave a open circuit thu V( ) 00 V Now, VC () t VC( ) + ( VC(0 + )-VC( )) e - / 00 + (-50-00)e - 00 50 e ( # - 0t) Now ic () t C dv dt -t - 0# 50# 0 6 trc Sol. - 50 0 6 50 0 - e # # # # # 0 t A -# 0 5e t ic () t 5 exp( - # 0 t) A Option (A) i correct. Given circuit i a hown below Sol. 4 Writing node equation at input port I V V V + 05. 05-4V- V...(). Writing node equation at output port I V V V + 05. 05 - - V+ 4V...(). From () and (), we have admittance matrix 4 - Y > - 4 H Option (D) i correct. A parallel RLC circuit i hown below : Input impedance At reonance So, Thu (D) i not true. Z in wl Z in + + jwc R jwl wc /R R (maximum at reonance)

Sol. 5 Furthermore bandwidth i w B i.e w R B \ and i independent of L, Hence tatement A, B, C, are true. Option (A) i correct. Let the current - it () A+ Be t t " Time contant When the witch S i open for a long time before t < 0, the circuit i At t 0, inductor current doe not change imultaneouly, So the circuit i Current i reitor (AB) i() 0 075. 0.75 A Similarly for teady tate the circuit i a hown below Sol. 6 i( ) t 5 R L eq 0.5 A - 5 # 0 0 0 + ( 0 0) it () A+ Be - # 0 A + Be -00t - Now i() 0 A+ B 0. 75 and i( ) A 05. So, B 0. 75-0. 5-0. 5-000 t Hence it () 0. 5-0.5e A Option (A) i correct. Circuit i redrawn a hown below t - ec Where, Z j L j 0 - w # # 0 # 0 0j Z R X C X C -6-0j jwc j # 0 # 50 # 0

Z ( - 0j) - 0j R W Sol. 7 Voltage acro Z -0j c V Z - 0j m Z : 0 0 : 0 Z + Z 0j c0j - - 0j m (- 0 j) c : 0 -j 0j+ 400-0j m Current in reitor R i V I Z j - -j A R Option (A) i correct. The circuit can be redrawn a Sol. 8 Applying nodal analyi VA- V 0 + A + - 0 0 VA - 0+ 0 V 4 4V Current, I 0 4 A - Current from voltage ource i I I - 0 Since current through voltage ource i zero, therefore power delivered i zero. Option (A) i correct. Circuit i a hown below Since 60 V ource i aborbing power. So, in 60 V ource current flow from + to - ve direction So, I+ I I -I

Sol. 9 Sol. 0 Sol. Sol. I i alway le then A So, only option (A) atifie thi condition. Option (C) i correct. For given network we have ( RL XC) Vi V 0 R + ( RL XC) RL V0 () RL + C RL Vi () R RL + R + RRLC + R + RL C RL R + RRLC + RL + R + RC RL But we have been given V0 () TF.. Vi () + CR Comparing, we get + R & R R L R L Option (C) i correct. The energy delivered in 0 minute i t t E # VIdt I # Vdt I# Area 0 0 # ( 0 + ) # 600. kj Option (B) i correct. From given circuit the load current i I L V 0+ 0c 0+ 0c Z + ZL ( + j ) + ( 7+ 4j ) 8 + 6 j ( 8-6 j) 0+ 0c - + - f where f tan 5 0+ f 4 The voltage acro load i V L IZ L L The reactive power conumed by load i P r VI L L * * IZ L L# IL ZL IL (7 4 j) 0+ 0c # ( 7 + 4j) 8 + 6j 8+ 6j Thu average power i 8 and reactive power i 6. Option (B) i correct. At t 0 -, the circuit i a hown in fig below : L - V (0 ) 00 V Thu V (0 + ) 00 V At t 0 +, the circuit i a hown below

+ I( 0 ) 00 0 ma 5k At teady tate i.e. at t i I( ) 0 t + - Now it () I(0 ) e u( t) RC eq Sol. (. 05m+ 0. m). 0m C eq 0.6 mf 05. m+ 0. m+ 0. m -6 50 RCeq 5 # 0 # 0. 6 # 0 - it () 0 e 50t u( t) ma Option (C) i correct. For P max the load reitance R L mut be equal to thevenin reitance R eq i.e. R L R eq. The open circuit and hort circuit i a hown below Sol. 4 Sol. 5 The open circuit voltage i V oc 00 V From fig I 00. 5 A 8 V x - 4#. 5-50 V I 00 Vx + 00-50. 5 A 4 4 I c I+ I 5 A R Voc th 00 4 W Ic 5 Thu for maximum power tranfer R L R eq 4 W Option (A) i correct. Steady tate all tranient effect die out and inductor act a hort circuit and forced repone act only. It doen t depend on initial current tate. From the given time domain behavior we get that circuit ha only R and L in erie with V 0. Thu at teady tate it ()" i( ) V 0 R Option (C) i correct. The given graph i

There can be four poible tree of thi graph which are a follow: There can be 6 different poible cut-et. Sol. 6 Option (B) i correct. - Initially i (0 ) 0 therefore due to inductor i (0 + ) 0. Thu all current I will flow in reitor R and voltage acro reitor will be I R. The voltage acro inductor will be equal to voltage acro R a no current flow through R. Sol. 7 + Thu vl ( 0 ) IR + + di( 0 ) but vl ( 0 ) L dt di( 0 + + ) vl( 0 ) Thu IR dt L L Option (A) i correct. Killing all current ource and voltage ource we have, Z th ( + ) ( + ) ( + )( + ) [ + + + ] ( + ) + ( + ) + + + or Z th Alternative : Here at DC ource capacitor act a open circuit and inductor act a hort circuit. Thu we can directly calculate thevenin Impedance a W

Sol. 8 Sol. 9 Option (D) i correct. Z () R C L C + RC + We have been given Z () 0. + 0. + Comparing with given we get 0. or C 5 F C 0. or R W RC or L 0. H LC Option (C) i correct. Voltage acro capacitor i t V c # C idt 0 Here C F and i A. Therefore t V c # dt 0 LC For 0 < t < T, capacitor will be charged from 0 V # V c dt t 0 t At t T, Vc T Volt For T < t < T, capacitor will be dicharged from T volt a # V c T- dt T-t At t T, V c 0 volt For T < t < T, capacitor will be charged from 0 V # t T V c dt t -T T t At t T, Vc T Volt For T < t < 4T, capacitor will be dicharged from T Volt # V c T- dt 4T-t At t 4 T, V c 0 Volt For 4T < t < 5T, capacitor will be charged from 0 V # t t T V c dt t -4T 4T At t 5 T, Vc T Volt Thu the output waveform i

Sol. 40 Sol. 4 Sol. 4 Sol. 4 Only option C atify thi waveform. Option (D) i correct. Writing in tranform domain we have V () c V () ^ + + h ( + + ) Since V() t d() t " V() and Vc () ( + + ) or Vc () ( + ) + 4 G Taking invere Laplace tranform we have V t e - t inc t m Option (B) i correct. Let voltage acro reitor be v R VR () VS () ( + + ) ( + + ) Since v d() t " V() we get VR () ( + + ) ( + ) + ( + ) ( + ) + 4 - ( + ) + or vr () t e co t e in t - - - # t e - co t in t - G Option (C) i correct. From the problem tatement we have z v 6 5. W i i 0 4 z v. 45 45. W i i 0 z v 6 5. W i i 0 4 z v. 5 5. W i i 0 Thu z -parameter matrix i z z 5. 45. z z G 5. 5. G Option (A) i correct. From the problem tatement we have h v. 45 v i 0 5. h i 067. v 5. From z matrix, we have i 0 4 4

Sol. 44 Sol. 45 If v 0 then v zi + zi v zi + zi i z - - 5. - i z 5. or i -i Putting in equation for v, we get v i v 0 v ( z -z ) i Hence h -parameter will be h h h h G - - 067. G h z - z.. h 5-45- Option (D) i correct. According to maximum Power Tranform Theorem * Z L Z ( R-jX) Option (C) i correct. At w ", capacitor act a hort circuited and circuit act a hown in fig below Here we get V V 0 i At w " 0, capacitor act a open circuited and circuit look like a hown in fig below 0 Here we get alo V V 0 i So frequency repone of the circuit i a hown in fig and circuit i a Band pa filter. 0 Sol. 46 Option (D) i correct. We know that bandwidth of erie RLC circuit i R. Therefore L Bandwidth of filter i B R L Bandwidth of filter i B R R 4R L L/ 4 L Dividing above equation B B 4

Sol. 47 Option (D) i correct. Here V th i voltage acro node alo. Applying nodal analyi we get Vth Vth Vth + + - i But from circuit i V V th th Therefore Vth Vth Vth Vth + + - or V th 4 volt From the figure hown below it may be eaily een that the hort circuit current at terminal XY i ic A becaue i 0 due to hort circuit of W reitor and all current will pa through hort circuit. Sol. 48 Therefore R Vth th 4 W i Option (A) i correct. The voltage acro capacitor i At t 0 + +, Vc ( 0 ) 0 At t, VC ( ) 5 V c The equivalent reitance een by capacitor a hown in fig i 0 0 0kW R eq Time contant of the circuit i t Req C 0k# 4m 0. 04 Uing direct formula -t Vc () t VC( ) -[ Vc( ) -Vc(0)] e / t -t/ t -t/ t -t VC ( )( - e ) + VC (0) e 5( -e /. 004 ) or Vc () t 5( -e -5t )

dvc () t Now IC () t C dt - 4 0 6 - # #(-5# 5 e 5t ) 0.5e -5t ma Sol. 49 Sol. 50 Option (D) i correct. ( 5- j) # ( 5+ j) Impedance (5 - j) ( 5+ j) 5- j+ 5+ j () 5 - ( j) 5 + 9 4. 0 0 V AB Current # Impedance 5+ 0c# 4 7+ 0c Option (D) i correct. The network i hown in figure below. Sol. 5 Now V AV- BI...() and I CV- DI...() alo V - IR L...() From () and () we get Thu V AV BI - I CV- DI Subtituting value of V from () we get Input Impedance Z A IRL BI in - # - - C# IRL - DI or Z ARL in + B CRL + D Option (B) i correct. The circuit i a hown below. Sol. 5 At input port V ri e At output port V r ( I - bi) - r bi + r I Comparing tandard equation V zi + zi V zi + zi z 0 and z -r 0 b Option (B) i correct. For erie RC network input impedance i Z in + R + RC C C 0 0 0

Sol. 5 Sol. 54 Thu pole i at origin and zero i at - RC For parallel RC network input impedance i Z in C R C + R + RC C Thu pole i at - and zero i at infinity. RC Option (A) i correct. We know v Ldi dt Taking Laplace tranform we get + V () LI() -Li( 0 ) A per given in quetion + -Li( 0 ) - mv Thu i( 0 + ) mv 0.5 A mh Option (B) i correct. At initial all voltage are zero. So output i alo zero. + Thu v0( 0 ) 0 At teady tate capacitor act a open circuit. Thu, v 0 ( ) 4 # v 4 i # 0 8 5 5 The equivalent reitance and capacitance can be calculate after killing all ource R eq 4 0.8 kw C eq 4 5 mf t Req Ceq 0.8kW # 5mF 4 m + - v0 () t v0( ) -[ v0( ) -v0(0 )] e t/ t - 8 -(8-0)e t/. 0 004

Sol. 55 Sol. 56 Sol. 57 Sol. 58 v0 () t 8( -e -t/. 0004 ) Volt Option (A) i correct. Here Z () Rneg + Z() or Z () Rneg + Re Z() + jim Z() For Z () to be poitive real, Re Z () $ 0 Thu Rneg + Re Z() $ 0 or Re Z () $- R neg But R neg i negative quantity and - R neg i poitive quantity. Therefore Re Z () $ R neg or R neg # Re Z( jw ) For all w. Option (C) i correct. Tranfer function i Y () C LC U () R L + + LC + cr + + R + C L LC Comparing with + xwn+ wn 0 we have Here xw n R, L and w n LC Thu x R LC R C L L For no ocillation, x $ Thu R C $ L or R $ L C Option (B) i correct. For given tranformer I V n I V or I and V nv n Comparing with tandard equation V AV+ BI I CV+ DI A B C D G n 0 0 n G Thu x n Option (B) i correct. We have L H and C -6 0 400 # I Reonant frequency f 0 p LC p # # 0 400-6

Sol. 59 Sol. 60 Sol. 6 Sol. 6 Sol. 6 4 0 # 0 0 Hz p p Option (C) i correct. Maximum power will be tranferred when RL R 00W In thi cae voltage acro R L i 5 V, therefore P max V 5# 5 05. W R 00 Option (C) i correct. For tability pole and zero interlace on real axi. In RC erie network the driving point impedance i Z in R + + RC C C Here pole i at origin and zero i at - / RC, therefore firt critical frequency i a pole and lat critical frequency i a zero. For RC parallel network the driving point impedance i R Z inp C R R + + RC C Here pole i - / RC and zero i at, therefore firt critical frequency i a pole and lat critical frequency i a zero. Option (A) i correct. Applying KCL we get i() t + 5+ 0c 0+ 60c or i () t 0+ 60c- 5+ 0c 5+ 5 j -5 or i () t 5 + 90c 0 + 90c Option (B) i correct. If L j5w and L jw the mutual induction i ubtractive becaue current enter from dotted terminal of jw coil and exit from dotted terminal of j5w. If L jw and L jw the mutual induction i additive becaue current enter from dotted terminal of both coil. Thu Z L - M + L + M + L - M + M j5+ j0+ j+ j0+ j- j0+ j0 j9 Option (B) i correct. Open circuit at terminal ab i hown below Applying KCL at node we get Vab Vab + - 0 5 5 or V ab 75. Vth

Short circuit at terminal ab i hown below Sol. 64 Sol. 65 Sol. 66 Short circuit current from terminal ab i I c + 0 A 5 Thu R Vth. th 75.5 W I c Here current ource being in erie with dependent voltage ource make it ineffective. Option (C) i correct. Here Va 5 V becaue R R and total voltage drop i 0 V. Now V R b # 0. # 0 5. 8 V R+ R4. V Va- Vb 5-5. 8-0. 8 V Option (D) i correct. For h parameter we have to write V and I in term of I and V. V hi + hv I hi + hv Applying KVL at input port V 0I + V Applying KCL at output port V 0 I+ I or I I V - + 0 Thu from above equation we get h h 0 h h G - 005. G Option (B) i correct. Time contant -6-4 RC 0. # 0 # 0 0 ec Since time contant RC i very mall, o teady tate will be reached in ec. At t ec the circuit i a hown in fig. V c V V - V c - V

Sol. 67 Sol. 68 Sol. 69 Sol. 70 Sol. 7 Option (B) i correct. For a tree there mut not be any loop. So a, c, and d don t have any loop. Only b ha loop. Option (D) i correct. The ign of M i a per ign of L If current enter or exit the dotted terminal of both coil. The ign of M i oppoite of L If current enter in dotted terminal of a coil and exit from the dotted terminal of other coil. Thu L eq L+ L-M Option (A) i correct. Here w and V + 0c Y + R j w C + jwl + j# + j4 + j # 4-5+ tan 4 5+ 5. c I V* Y ( + 0c)( 5+ 5. c) 5+ 5. c Thu it () 5 in( t + 5. c) Option (A) i correct. vi () t in 0 t Here w 0 rad and V i + 0c jwc Now V 0. V t R j CR V i + + w j w C + 0c + - j # 0 # 0-45c v0 () t in( 0 t -45c) Option (C) i correct. Input voltage vi () t ut () Taking Laplace tranform Vi () Impedance Z () + Vi () I () + ( + ) or I () ; - + E Taking invere Laplace tranform t it () ( -e ) u( t) At t 0, it () 0 At t, it () 0. At t, it () 05. Graph (C) atifie all thee condition.

Sol. 7 Option (D) i correct. We know that V zi + zi V zi + zi where z V I I 0 z V I I 0 Conider the given lattice network, when I 0. There i two imilar path in the circuit for the current I. So I I Sol. 7 For z applying KVL at input port we get V IZ ( a+ Zb) Thu V I( Za+ Zb) z ( Za+ Zb) For Z applying KVL at output port we get V Z I Z I a - b Thu V I( Za-Zb) z ( Za-Zb) For thi circuit z z and z z. Thu R V SZa+ Zb Za- Zb z z z z G S Z Z Z Z S a- b a+ b S T X Here Za j and Zb W z z j j Thu z z G + - j - + j G Option (B) i correct. Applying KVL, Ldi() t vt () Ri() t + + # itdt () dt C 0 Taking L.T. on both ide, + + I () vc ( 0 ) V () RI() + LI() - Li( 0 ) + + C C vt () ut () thu V ()

Sol. 74 Sol. 75 Sol. 76 Sol. 77 Sol. 78 Hence + or I () I () I() + I() - + - I () 6 + + @ + + + Option (B) i correct. Characteritic equation i + 0+ 0 6 0 Comparing with + xwn+ wn 0 we have 6 w n 0 0 xw 0 Thu x 0 00. 0 Now Q 50 x 00. Option (D) i correct. V0 () H () C Vi () R L + + LC + CR + C - -4-4 4 ( 0 # 0 ) + ( 0 # 0 ) + 6 0 6 6 6 0 + + - + 0 + 0 Option (D) i correct. Impedance of erie RLC circuit at reonant frequency i minimum, not zero. Actually imaginary part i zero. Z R+ j wl ` - wc j At reonance wl - 0 and Z R that i purely reitive. Thu S wc i fale Now quality factor Q R C L Since G, Q C R G L If G - then Q. provided C and L are contant. Thu S i alo fale. Option (B) i correct. Number of loop b- n+ minimum number of equation Number of branche b 8 Number of node n 5 Minimum number of equation 8-5+ 4 Option (C) i correct. For maximum power tranfer Z L Z S * R-jX Thu Z L -j

Sol. 79 Sol. 80 Sol. 8 Option (B) i correct. Q L R C When RL, and C are doubled, Q ' L L Q R C R C Thu Q ' 00 50 Option (C) i correct. Applying KVL we get, di() t in t Ri() t + L + () dt # C itdt di() t or in t it () + + # itdt () dt Differentiating with repect to t, we get di() t dit () co t + + it () dt dt Option (A) i correct. For current i there i imilar path. So current will be divide in three path Sol. 8 Sol. 8 o, we get V i i ab -b # l-b6 # l-b #l 0 V ab R eq + + 5 W i 6 6 Option ( ) i correct. Data are miing in quetion a L& L are not given. Option (A) i correct. At t 0 - circuit i in teady tate. So inductor act a hort circuit and capacitor act a open circuit.

At t 0 -, i ( 0 ) i ( 0) 0 - v ( 0 ) V c - - At t 0 + the circuit i a hown in fig. The voltage acro capacitor and current in inductor can t be changed intantaneouly. Thu Sol. 84 Sol. 85 At t 0 +, i i V - R Option (C) i correct. When witch i in poition, a hown in fig in quetion, applying KVL in loop (), RI () V + + () [ () ()] C I + L I - I 0 or I () R 8 + + L -I () L c B -V zi + zi V Applying KVL in loop, L[ I ( ) I ( )] RI ( ) - + + ( ) C I 0 ZI + ZI V or - LI () R L + 8 + + I () c B 0 Now comparing with Z Z I V Z Z G I G V G we get R R L V S + + -L W C I () V S -L R+ L+ W I() G > - H S C W 0 T X Option (B) i correct. Zero - Pole - + j Pole - -j Z () K ( + ) K ( + ) ( + + j)( + - j) ( + ) - j From problem tatement Z() 0 w 0 Thu K and we get K ( + ) Z () + + K ( + ) ( + ) +

Sol. 86 Sol. 87 Option (C) i correct. vt () 0 co( t+ 0c) + 0 5 co( t+ 0c) 44444444 44444444 Thu we get w and w Now Z R+ jw L + j Z R+ j L + j it () w v v() t v() t 0 co( t + 0c) 0 5 co( t + 0c) + + Z Z + j + j 0 co( t+ 0c) 0 5 co( t+ 0c) + - - + + tan + tan 0 co( t+ 0c) 0 5 co( t+ 0c) + - - + tan 45c 5tan - it () 0 co( t- 5 c) + 0 co(t+ 0c-tan ) Option (A) i correct. Uing - Y converion v R # 05. + + 4 R # 05. + + 4 R # 05. + + Now the circuit i a hown in figure below. Sol. 88 Now z V + 0. 5+ 0. 5. 75 I I 0 z R 05. Option (A) i correct. Applying KCL at for node, V V V + - V 5 5 5

Sol. 89 or V V 0 V Voltage acro dependent current ource i 0 thu power delivered by it i PV V # 0 # 0 80 W 5 5 It deliver power becaue current flow from it +ive terminal. Option (C) i correct. - When witch wa cloed, in teady tate, i ( 0). 5 A L Sol. 90 + - At t 0 +, il(0 ) il(0 ).5 A and all thi current of will pa through W reitor. Thu V x -.5 # 0-50 V Option (A) i correct. For maximum power delivered, R L mut be equal to R th acro ame terminal. Applying KCL at Node, we get 05. I V + I 0 th or Vth + 0I 0 but I Vth 50-40 Thu V Vth 50 th + - 0 4 or V th 0 V For I c the circuit i hown in figure below. I c 05. I - I -05. I but I - 50-5. A 40 I c -0. 5 #-. 5 0. 65 A

Sol. 9 Sol. 9 Sol. 9 R th Vth 0 6 W I 0. 65 Option (D) i correct. IP, VP " Phae current and Phae voltage IL, VL " Line current and line voltage Now V VL P c m and IP IL So, Power VI P L co q 500 VL c ( IL)coq m alo I VL L c Z m c L 500 VL VL c coq mc Z m Z L A power factor i leading ( 400) (. 844) 90 W 500 So, co q 0. 844 " q. 44 A phae current lead phae voltage Z L 90+ - q 90+ -. 44c Option (C) i correct. Applying KCL, we get e0- + e0 + e0 0 4 4 + or e 0 4 V Option (A) i correct. The tar delta circuit i hown a below L Sol. 94 Here Z AB ZBC ZCA Z and Z ZABZCA A ZAB + ZBC + ZCA Z ZABZBC B ZAB + ZBC + ZCA Z ZBC ZCA C Z + Z + Z AB BC CA Now Z A Z Z Option (C) i correct. B C Z Z Z Z+ Z+ Z

Sol. 95 y y y y y y y G + - -y y + y G y -y y - - 005. mho 0 Option (D) i correct. We apply ource converion the circuit a hown in fig below. Sol. 96 Sol. 97 Sol. 98 Now applying nodal analyi we have e0-80 + e0 + e0-6 0 0 + 6 or 4e 0 e 0 8 V 4 Option (A) i correct. I Em + 0c jwc Em + 0c R + + jwcr + I + 90c - + tan wcr I EmwC - + (90 c - tan wcr ) CR + w At w 0 I 0 E m jwc and at w, I R Only figure given in option (A) atifie both condition. Option (A) i correct. X wl 0 W For maximum power tranfer R L R + X 0 + 0 4.4 W Option (C) i correct. Applying KVL in LHS loop E I+ 4( I+ I) -0E or E 6I 4I + Thu z 6 Applying KVL in RHS loop E 4( I+ I) -0E 4( I I ) 0 6I 4I + - + 6I 4I c m - + Thu z 6 -

Sol. 99 Option (D) i correct. At w 0, circuit act a hown in figure below. V0 RL V RL+ R At w, circuit act a hown in figure below: (finite value) V0 RL (finite value) V RL+ R At reonant frequency w circuit act a hown in fig and V0 0. LC Sol. 00 Sol. 0 Thu it i a band reject filter. Option (D) i correct. Applying KCL we get i L e at bt + e Now Vt () v L dil L L d [ e + e ] ae + be dt dt Option (A) i correct. Going from 0 V to 0 V at bt at bt 0 + 5 + E + 0 or E - 6 V

Sol. 0 Sol. 0 Option (C) i correct. Thi i a reciprocal and linear network. So we can apply reciprocity theorem which tate Two loop A & B of a network N and if an ideal voltage ource E in loop A produce a current I in loop B, then interchanging poition an identical ource in loop B produce the ame current in loop A. Since network i linear, principle of homogeneity may be applied and when volt ource i doubled, current alo double. Now applying reciprocity theorem i A for 0 V V 0 V, i A V - 0 V, i - 4 A Option (C) i correct. Tree i the et of thoe branch which doe not make any loop and connect all the node. abfg i not a tree becaue it contain a loop l node (4) i not connected Sol. 04 Option (A) i correct. For a -port network the parameter h i defined a h I I V 0( hort circuit) Applying node equation at node a we get Va- V + Va- 0 + Va- 0 0 R R R V a V & V a V Sol. 05 Now I V V a V R R R - - and I V V 0 a R R R V - - - Thu I V / R h I V R - - V 0 / Option (A) i correct. Applying node equation at node A Vth - 00( + j0) Vth + 0-0 4j

Sol. 06 or 4jV - 4j00 + V 0 th or Vth ( + 4j) 4j00 V 4j00 th + 4j By implifying 4j00-4j V th + 4j # - 4j th V th 6j( -j4) Option (C) i correct. For maximum power tranfer R L hould be equal to R Th at ame terminal. o, equivalent Reitor of the circuit i R eq 5W 0W+ 4W Sol. 07 Sol. 08 Sol. 09 Sol. 0 Sol. Sol. Sol. R eq 50. + 4 4+ 4 8 W 5 + 0 Option (D) i correct. Delta to tar converion R RabRac 5# 0 50 W Rab + Rac + Rbc 5 + 0 + 5 50 R RabRbc 5# 5 5. W Rab + Rac + Rbc 5 + 0 + 5 R RacRbc 5 # 0 9 W R + R + R 5 + 0 + 5 ab ac bc Option (C) i correct. No. of branche n+ l- 7+ 5- Option (B) i correct. In nodal method we um up all the current coming & going at the node So it i baed on KCL. Furthermore we ue ohm law to determine current in individual branch. Thu it i alo baed on ohm law. Option (A) i correct. Superpoition theorem i applicable to only linear circuit. Option (B) i correct. Option (B) i correct. For reciprocal network y y but here y -! y. Thu circuit i non reciprocal. Furthermore only reciprocal circuit are paive circuit. Option (C) i correct. Taking b a reference node and applying KCL at a we get Vab - V + ab or V - + V 6 ab ab