IES MASTER GATE Session-1. Electrical Engineering Questions and Detailed Solution

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2 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION. The equivalent resistane between the terminals A and B is. A B Sol. (3 ) Simplifying the iruit A B A B Combining resistanes 3, 6, and as these are parallel 3 6 Ciruit redues to or.8 6 eq eq GATE 7 Eletrial Engineering Questions and Detailed Solution Session-. Consider an eletron a neutron and a proton initially at rest and plaed along a straight line suh that the neutron is exatly at the enter of the line joining the eletron and proton. At t, the partiles are released but are onstrained to move along the same straight line. Whih of these will ollide first? () Sol : The partiles will never ollide All will ollide together Proton and neutron Eletron and neutron e n p d Mass of eletron Kg Mass of proton Kg Eletrostati fore will exist between eletron and proton only. Let say fore is F then by relation F ma, where m is mass of partile and a is aeleration. Sine mass of eletron is lesser than proton so aeleration of eletron will be more than proton. By equation As u so s s d ut at s distane travelled u inital speed at To travel distane d eletron will take lesser time so eletron will ollide with neutron first. egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

3 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 3. The slope and level detetor iruit in a CO has a delay of ns. The start-stop sweep generator has a response time of 5 ns. In order to display orretly a delay line of Sol. 5 ns has to be inserted into the y-hannel 5 ns has to be inserted into the x- hannel () 5 ns has to be inserted into both x and y hannels ns has to be inserted into both x and y hannels In a CO during the sweep time the beam moves left to right aross the CT, during the retrae time the beam quikly moves to the left side of the CT sreen as shown in figure below. volts t s sweep time t retrae time t s t r t s t r Given data slope and level detetor has delay time (t d ) ns response time (t re ) 5 ns so total time taken for one sweep yle of x-plate (t re t d t s t r ) 5 ns (t s t r ) In order to display orretly signal to y- hannel has to be applied after a delay of 5 ns. r t. The following measurements are obtained on a single phase load V V ± %. I 5A ± % and W 555 W ± %. If the power fator is alulated using these measurements the worst ase error in the alulated power fator in perent is (Give answer up to one deimal plae) Sol. (%) Sine, So, W W V V ± % I 5A ± % W 555W ± % W. Hene in worst ase (p.f.) p.f. VIos V I (os ) V I os (p.f.).. p.f.... or in perent % or.% 5. A losed loop system has the harateristi equation given by s 3 Ks (K )s 3. For this system to be stable, whih one of the following onditions should be satisfied? Sol. < K <.5.5 < K < () < K < K > By outh Hurwitz Criteria 3 s k s k 3 s k(k ) 3 k s 3 For stability k > (i) k(k ) 3 (ii) k egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

4 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 3 (k )(k 3) k Either k > ; k > 3 or k < ; k < 3 From equation (i) and (iii) (iii) k > hene option is the orret answer. 6. The matrix 3 A 3 has three distint eigenvalues and one of its eigenvetors is. Whih one of the following an be another eigenvetor of A? () Sol. () A 3 3 () To find the eigen values of A, det(a I ) i.e., ( ) ( ) 3 ( ) 3 ( ) or ( )( )( ) Hene eigen values of A are, and. Now, let X be an eigen vetor of A assoiated to, then So, AX X 3 3 On solving it; Thus for by taking [A I ]X 3 x z y x z or x z y Hene option () satisfies. x X y z egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

5 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 7. A -bus power system onsists of four generator buses indexed as G, G, G3, G and six load buses indexed as L, L, L3, L, L5, L6. The generator-bus G is onsidered as slak bus and the load buses L3 and L are voltage ontrolled buses. The generator at bus G annot supply the required reative power demand and hene it is operating at its maximum reative power limit. The number of non-linear equations required for solving the load flow problem using Newton-aphson method in polar form is. Sol. () Given data Total number of buses (N) Number of PV buses (x ) (i.e. G 3 and G ) Number of voltage ontrolled buses (x ) (i.e. L 3 and L ) Number of slak buses (i.e. G ) Number of load buses 5 The total number of equations to be solved [N (x x )] [() ( )] The size of the jaobian matrix [N (x x )] [N (x x )] 8. A 3-bus power system is shown in the figure below, where the diagonal elements of Y bus matrix are : Y j pu, Y j5 pu and Y 33 j7 pu Bus- jr jq jp Bus-3 Bus- The per unit values of the line reatanes p, q and r shown in the figure are Sol. p., q., r.5 p., q., r.5 () p 5, q, r p 5, q, r From bus diagram y y jq y 3 y 3 jr y 3 y 3 jp Sine diagonal elements So, Y y y y 3 j jq jr (i) q r Similary for Y j5 or j5 jq jp 5 (ii) q p For Y 33 j7 or j7 jr jp 7 (iii) r p On solving equations (i), (ii) and (iii) 5 p ; q ; r Hene, p., q., r.5 egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

6 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 5 9. For the power semiondutor devies IGBT, MOSFET, Diode and Thyristor, whih one of the following statement is TUE? () Sol. All the four are majority arrier devies All the four are minority arrier devies IGBT and MOSFET are majority arrier devies, whereas Diode and Thyristor are minority arrier devies MOSFET is majority arrier devie, whereas IGBT, Diode, Thyristor are minority arrier devies MOSFET is the only majority arrier devie among MOSFET, DIODE, Thyristor and IGBT. In majority arrier devies ondution is only beause of majority arriers whereas in minority arrier devies ondution is due to both majority and minority arriers.. Consider the unity feedbak ontrol system shown. The value of K that results in a phase margin of the system to be 3 is. (Give the answer up to two deimal plaes.) Sol. (.5) U(s) Ke s For unity feedbak system with s Y(s) Ke G(j )H(j ) s s Phase margin is given by where, P.M. 8 G(j ).H(j ) g and G( j ).H( j ) j Sine, e g So, k So, k g at g Now, G( j )H( j ) at k g k PM k 5 k.7 Upto two deimal plaes k.5. The transfer funtion of a system is given by V (s) s V (s) s i Let the output of the system be v (t) V sin( t ) for the input, v i (t) m V sin( t). Then the minimum and m maximum values of (in radians) are respetively Sol. and and () and and For transfer funtion V (s) V (s) i i V (j ) V (j ) i s s V (j ) V (j ) tan Here, v i (t) Vm sin( t) v (t) Vm sin( t ) egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

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8 6 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION So, for to tan varies from 8 to Hene, option is the orret answer.. For the iruit shown in the figure below, assume that diodes D, D and D 3 are ideal. v(t) sin (t) V v D D D 3 v The DC omponents of voltages v and v, respetively are () Sol. V and V.5 V and.5 V V and.5 V V and V During positive half yle D ON, D and D 3 will be OFF During negative half yle D and D 3 ON but D OFF v(t) v v sin t sin (t) sin t sin ( t) V (avg) sin t d t sin td t.5v sin t d t os os V (avg.).5 3. The power supplied by the 5V soure in the figure shown below is W Sol. (5) 5 V 5 V I 7 V I 7 V A A a.i.i Using Kirhoffs urrent law at node a. A Power supplied by 5V soure; P 5V A P 5 watt. A three-phase, 5 Hz, star-onneted ylindrial-rotor synhronous mahine is running as a motor. The mahine is operated from a 6.6 kv grid and draws urrent at unity power fator (UPF). The synhronous reatane of the motor is 3 per phase. The load angle is 3. The power egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

9 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 7 deliver to the motor in kw is.(give the answer up to one deimal plae). Sol. (838.3 kw) Given that, V 6.6 kv 3 P.f (UPF) Synhronoes reatane (X s ) 3 P VEf sin For unity P.f. for synhronous motor. 3 I X From above phasor diagram, E os V or, E f f s V os 6.6kV os3 7.6 kv V E f jix s Hene, P sin3 MW MW or, P kw 5. A soure is supplying a load through a - phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and urrent in phase-a are v an sin(t)v and i a sin(t)a, respetively. Similarly for phase-b, the instantaneous voltage and urrent are V bn os( t)v and i b os( t)a Soure a b i a v anib n v bn a b n Load The total instantaneous power flowing from the soure to the load is () Sol. W sin (t)w W sin( t) os( t)w Instantaneous power; P i P an ia bn ib sin( t) sin( t) os( t) os( t) sin ( t) os ( t) P W 6. For a omplex number z, Sol. z lim zi 3 is z z i(z ) i i () i i z lim zi 3 z z i z This is form, so on differentiating both numerator and denominator z lim zi 3z zi egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

10 8 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION i i i 3 i i i 3 7. Let I xy dxdy, where is the region shown in the figure and 6. The value of I equals. (Give the answer up to two deimal plaes). Sol. (.99) I y xy dx dy y 5 x 5 x egion is bounded by y and y x I 5 x 5 5 xy 3 xy dy dx 3 x 8 x dx 3 dx x I.99 (upto two deimal plaes) 8. A pole indution mahine is working as an indution generator. The generator supply frequeny is 6 Hz. The rotor urrent frequeny is 5 Hz. The mehanial speed of the rotor in PM is Sol. () () 95 5 For pole, 6 Hz indution mahine synhronous speed; N s f P s 6 8 r.p.m. f f r s 5 6 For indution generator slip is negative, So, N s N N s r 8 N r 8 r 8 N N r Nr r.p.m. 9. A 3-phase voltage soure inverter is supplied from a 6 V DC soure as shown in the figure below. For a star onneted egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

11 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 9 resistive load of per phase, the load power for devie ondution, in kw, is 6 V Sol. (9 kw) In devie andution mode and star onneted load: At any instant only IGBTs will ondut so, when IGBT and 6 are onduting in 6 yle, equivalent kt an be given as Vs 6 V Vs 6 V So power, P or, a 3 6 V s C a V s O V s b 5 O b P V s (6) 9 watt 9 kw. A solid iron ylinder is plaed in a region ontaining a uniform magneti field suh that the ylinder axis is parallel to the magneti field diretion. The magneti field lines inside the ylinder will () Sol. Bend loser to the ylinder axis Bend farther away from the axis emain uniform as before Cease to exist inside the yliner The magneti field lines will bend loser to the ylinder axis to find a minimum relutane path.. Consider the system with following inputoutput relation y[n] ( () n )x[n] where, x[n] is the input and y[n] is the output. The system is () Sol. Invertible and time invariant Invertible and time varying Non-invertible and time invariant Non-invertible and time varying y[n] ( () n ) x[n] A system is said to be invertible if there is a one-to-one orrespondene between its input and output signals. for n, for n, for n 3, y[] ( () ) x[] y[] ( () ) x[] y[3] ( () 3 ) x[3] Here for odd values of n output will always be zero so system is non-invertible. To hek time invariany For delayed input, y[n n ] ( () n ) x[nn ]...(i) For delayed response, egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

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13 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION nn y[n n ] ( ( ) )x[n n ]...(ii) For time invariant system output for delayed input should be equal to delayed response. Hene this system is time varying.. Consider g(t) t t, t, where t t t, otherwise Here, t represents the largest integer less than or equal to t and t denotes the smallest integer greater than or equal to t. The oeffiient of the seond harmoni omponent of the Fourier series representing g(t) is Sol. (.796) g(t) t t, t t t, otherwise g(t) 3 3 t T T g(t) t t g(t) jnt Gne n T jn t g t e T T G n jnt t e j nt t e dt G n t e jnt dt dt dt jnt jnt e e t dt jn jn jn j n G n jn G j G.796 jn e 3. The boolean expression AB AC BC simplifies to () Sol. BC AC AB AC B AB AC AB BC Using k-map, f AB AC BC BC B A A C f AC BC. Let z(t) x(t)*y(t), where " " denotes onvolution. Let be a positive real-valued onstant. Choose the orret expression for z(t) () Sol..x(t) y(t) x(t) y(t).x(t) y(t).x(t) y(t) z(t) x(t) * y(t) taking fourier transform Zj Xj Y j () egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

14 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION z(t) j Z Also, byusing eq. () j Z j j X Y j Z X j Y j Multiplying and dividing.h.s. by j Z j j X Y x t * y t z(t) () 5. In the onverter iruit shown below, the swithes are ontrolled suh that the load voltage v (t) is a Hz square wave. V S S LOAD v (t) S 3 S The MS value of the fundamental omponent of v (t) in volts is Sol. (98.69) For single phase full bridge inverter, Volts n n, 3, 5... Vs sinn n t MS value of fundamental omponent is given by: Vs Volts. 6. The positive, negative and zero sequene reatanes of a wye-onneted synhronous generator are. pu,. pu and. pu respetively. The generator is on open iruit with a terminal voltage of pu. The minimum value of the indutive reatane, in pu, required to be onneted between neutral and ground so that the fault urrent does not exeed 3.75 pu if a single line to ground fault ours at the terminals is (assume fault impedane to be zero). (Give the answer up to one deimal plae). Sol. (. pu) Positive sequene reatane X. pu Negative sequene reatane X. pu Zero sequene reatane X. pu Fault urrent i f 3.75 pu For single line to ground fault, 3E i f X X X 3Xn Where X n is the retane onneted between neutral and ground. Therefore,for E pu On solving, i f X X n. pu n Two passive two-port networks are onneted in asade as shown in figure. A voltage soure is onneted at port I I I 3 Two port Two port V network V network V 3 Part Part Part 3 Given V A V B I I C V D I V A V 3 B I 3 I C V 3 D I 3 A, B, C, D, A, B, C and D are the generalized iruit onstants. If the Thevenin equivalent iruit at port 3 onsists of a voltage soure V T and an impedane Z T, onneted in series, then () V V V T T T V A B B D, ZT A A A A B C V A B B D, ZT A A B C A A V A B B D, ZT A A A A egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

15 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Sol. V T V A B B D, ZT A A B C A A B C For asaded network, equivalent ABCD parameters are given by, A C B D A B A B C D C D AA BC AB B D C A DC CB DD Therefore, V A B V 3 C D 3 Now, open iruit voltage V T V 3 when. 3 So, V AV 3 V T V T AV T V A A A V B C To alulate Thevenin equivalent impedane Z voltage soure is short T iruited. So for voltage soure of V 3 feeding urrent (I 3 ) Z T V 3 I 3 AV 3 BI 3 Z T B A Z T A B B D A A B C Hene, option is the orret answer. 8. A V DC series motor runs drawing a urrent of 3 A from the supply. Armature and field iruit resistanes are. and., respetively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature urrent. To redue the speed of the motor by 5%, the resistane in ohms the should be added in series with the armature is. (Give the answer up to two deimal plaes). Sol. (.75 ) 3A. f. Bak E.m.f. E 3(..) 5V Torque, a {where flux, a armature urrent} a as in series motor I a a...(i) Also, N...(ii) here N is the speed of motor From eq. (i) & (ii) or a N...(iii) a N Therefore to redue speed by 5%, I a will redue to 5% i.e., 5A. Now bak emf will hange to E 5(..) Where is the external resistane added in series with armature. Sine, E N So, E N E N N N egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

16 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 3 Thus, 5 N 5(.5) N dividing eq. (v) by eq. (iv), 5(.5) 5 N N 5(.5) 5 on solving,.75...(iv)...(v) / N / N 9. The logial gate implemented using the iruit shown below where, V and V are inputs (with V as digital and 5 V as digital ) and V OUT is the output, is Sol. V k 5 V k V OUT Q V k Q NOT NO () NAND XO From the given iruit it an be dedued that Q will be ON when V is high, Q is ON when V is high Truth Table V Q V Q Vout High ON High ON Low High ON Low OFF Low Low OFF High ON Low Low OFF Low OFF High This is the truth table of NO gate. 3. A funtion f(x) is defined as f(x) x e, x, ln x ax bx x where x. Sol. Whih one of the following statement is TUE? () f(x) is NOT differentiable at x for any values of a and b. f(x) is differentiable at x for the unique value of a and b. f(x) is differentiable at x for all values of a and b suh that ab e. f(x) is differentiable at x for all values of a and b f(x) x e, x ln x ax bx, x f(x) f Left hand derivative lim x x where f() ln () a() b () ab so, Left hand derivatives (LHD) lim e a b x x this limit exists if e a b then e LHD lim e x ight hand derivative (HD) x x f(x) f() lim x x () ln x ax bx a b lim x x this limit exist if a b ab or a () Hene ax b HD lim x a b x LHD will be equal to HD if a and b e Hene, f(x) is differentiable at x for unique values of a and b. 3. The swith in the figure below was losed for a long time. It is opened at t. The egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

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18 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Sol. urrent in the indutor of H for 6 5 V e t 5e t ().5e.5t 5e.5t t, is H For t iruit an be represented as below 5V H Indutor an be taken as short iruit at steady state. So urrent in indutor at t will be i L ( ) i L.5 A On opening of swith at t, i L an be given by i L (t) i L () e t/ where L/ is equivalent aross L to alulate equivalent, Hene i L (t) 3 L 8 t.5 e / or i L (t).5e t L H eq 8 3. A three-phase, three winding / / Y (. kv/6.6 kv/ V) transformer is energized from AC mains at the. kv side. It supplies 9 kva load at.8 power fator lag from the 6.6 kv winding and 3 kva load at.6 power fator lag from the V winding. The MS line urrent in ampere drawn by the. kv winding from the mains is. (Give the answer up to one deimal plae). Sol. (65. A) B C A Given (.kv / 6.6kV V) Tr. Primary Seondary a b Let us assume V a"n" as referene, n Tertiary a 3 V a"n" 3 Load at tertiary 3 kva,.6 Pf lag I a " A V 3 AB..763 V / 3 a "n " V AB V a "n " are in phase IAB " orresponding to I A We know, IA IAB ICA A Now Load at seondary 9kVA,.8 p.f. lag V AB. V i.e. a' b' V 6.6 kv a'b' Ia' b' A IAB orresponding to I a'b' 7.73 A AB AC I I I a b egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

19 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 5 IA orresponding Ia' b' A Total line urrent from supply. I I (orresponding to load) L A I A(orresponding load) MS line urrent 65. A 33. Consider the line integral Sol. I x iy dz, where z x iy. The line is shown in the figure below y i (, i) (, ) The value of I is () i 3 i C x Given integral I i 3 i 5 x iy dz, z x iy the line C is a straight line passing through origin and its equation is given by y x y i (, i) x i i dx x i dx i x dx i x dx 3 x i 3 i 3 3 i 3 3. The figure shows the single line diagram of a power system with a double iruit transmission line. The expression for eletrial power is.5 sin, where is the rotor angle. The system is operating at the stable equilibrium point with mehanial power equal to pu. If one of the transmission line iruits is removed, the maximum value of, as the rotor swings, is. radian. If the expression for eletrial power with one transmission line iruit removed is P,max sin, the value of P max, in pu is (Give the answer up to three deimal plaes.) Sol. (.pu) With double iruit transmission line P ei.5 sin with single line P eii P max sin x substituting y x in given integral we get I x ix dx idx P m.5 P max pu A A P.5 sin ei P eii P max sin m egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

20 6 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Here, m For stability sin.8.5 or.797 radian. radian or A A m Pmax sin d max os P max P P sin d os m max m P max os m os m substituting values and solving for P max P max (os os.8 ) P max. pu 35. The approximate transfer harateristi for the iruit shown below with an ideal operational amplifier and diode will be () Sol. V in V ss V ss D V V V For V in < V in V in V V V in V in Output of operational amplifier will be negative hene due to presene of diode (reversed biased) in this ase output will be zero For V in > Diode will be forward biased so V in V Thus transfer harateristis V V in 36. A load is supplied by a 3 V, 5 Hz soure. The ative power P and the reative power Q onsumed by the load are suh that Sol. kw P kw and kva Q kva. A apaitor onneted aross the load for power fator orretion generates kva reative power. The worst ase power fator after power fator orretion is.7 lag.77 lag ().89 lag For worst ase power fator P kw, Q kva After addition of apaitor for power fator orretion Q beomes kva new or P.f ostan P Q os tan os 5 P.f.77 lag 37. A separately exited DC generator supplies 5 A to a 5 V DC grid. The generator is running at 8 PM. The armature resistane of the generator is.. If the speed of the generator is inreased to PM, the urrent in amperes supplied by egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

21 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 7 the generator to the DC grid is. (Give the answer up to one deimal plae). Sol. (55A) f. a I 5A a 5V V t N 8 rpm (speed of generator) Sine bak emf E V t I a a and E N...(i)...(ii) For separately exited generator remains onstant so E N...(iii) For N 8 rpm E Using equation (iii) V E N or (iv) For N rpm E On solving equation (iv) and (v) E V Thus 5 I a. or I a I a A 38. Consider the differential equation Sol....(v) dy (t 8) 5ty sin(t) dt with y(). There exists a unique solution for this differential equation when t belongs to the interval (, ) (, ) () (, ) (, ) Given differential equation is dy t 8 5ty sin t () dt initial ondition y() Converting the given equation into standard form dy 5t y dt t 8 this is of the form dy py dt Q sin t t 8 5t sin t where P,Q t 8 t 8 we know integrating fator (IF) e e 5t dt t 8 5 t t ln t 8 ln x () e dt e e x IF 5 t 8 y(if) Q IF dt y(t 8) 5/ y sin t t 8 dt t sin t t 8 5 t 8 5 dt t 8 y 5 sin t t 8 dt t 8 and solving from the options by verifying initial ondition we get unique solution If t ±9 then solution is not unique hene range (, ), (, ), (, ) an be eliminated, then left option is (, ) 39. The input voltage V DC for the buk-boost onverter shown below varies from 3 V to 7 V. Assume that all omponents are ideal, indutor urrent is ontinuous and output voltage is ripple free. The range of duty ratio D of the onvetor for whih the egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

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24 8 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION magnitude of the steady-state output voltage remains onstant at 8 V is the readings of the wattmeters in watts will be: Sol. V DC 3 D 5 5 S L C V 3 D 3 () D D 3 3 For buk-boost onverter V Vs where is duty yle of onverter V s V supply voltage output voltage For V s 3 V and V 8 V For V s 7 V and V 8 V The load shown in the figure is supplied by a V(line-to-line), 3-phase soure (YB sequene). The load is balaned and indutive, drawing 36 VA. When the swith S is in position N, the three wattmeters W, W and W 3 read W eah. If the swith is moved to position Y, Sol. Y B N v, 3-phase soure w w w 3 Y S N LOAD W 73 and W W 3 W, W 73 and W 3 () W 866, W, W W W and W 3 73 Apparent power 36 VA eal power W 73.5 Watts P.f. eal power Apperent power P.f..5 When swith is moved to position Y Voltage aross potential oil of wattmeter two is zero so W For YB phase sequene V BY V Y V Y I B VB 3 3 V I V Y Voltage aross potential oil of wattmeter one is V Y. Voltage aross potential oil of wattmeter two is V BY. So W V I os3 Y egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

25 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 9 W V I os 3 BY For P.f.5; os (.5) 6 Thus W V I os 3 6 Y W 3 V I os W BY B. The bus admittane matrix for a power system network is j39.9 j j j j39.9 j pu j j j39.9 There is a transmission line, onneted between buses and 3, whih is represented by the iruit shown in figure Suseptane is.5 pu B eatane is.5 pu Suseptane is.5 pu If this transmission line is removed from servie, what is the modified bus admittane matrix? () j9.9 j j j39.9 j pu j j9.9 j39.95 j j j39.9 j pu j j39.9 j9.95 j j j39.9 j pu j j9.95 j9.95 j j j39.9 j pu j j j9.95 Sol. () Given y-bus [Y] 3 3 j39.9 j j j j39.9 j j j j39.9 Converting the given transmission line parameters into Y parameters we get Bus y j.5 y j 3 Bus 3 y 33 j.5 whenever we remove the transmission line between Bus and Bus 3 the parameters Y, Y 3, Y 3, Y 33 will get affeted Y j39.9y y 3 j39.9 (j.5)(j) j39.9 j.5 j j 9.95 Y 3 j y 3 j j Y 3 j y 3 j j Y 33 j39.9 y 33 y 3 j39.9 j.5 (j) j9.95 New Y bus matrix [Y] 3 3 j9.95 j j j39.9 j j j9.95. In the system whose signal flow graph is shown in the figure, U (s) and U (s) are inputs. The transfer funtion U /L /s k k U Y(s) U (s) /J /s Y is egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

26 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION k JLs Js k k V () Sol. k JLs Js k k k U ( sl) JLs (J U L)s k k U k U (sl ) JLs (J U L)s k k U Using mason s gain formula T.F. P T.F. P k k k k L s J s JLs kk Ls JLs k k Ls JLs JLs Js k k JLs k JLs JLs Js k k JLs T.F. k JLs Js k k 3. The iruit shown in the figure uses mathed transistors with a thermal voltage V T 5 mv. The base urrents of the transistors are negligible. The value of the resistane in k that is required to provide A bias urrent for the differential amplifier blok shown in. (Give the answer up to one deimal plae). ma Sol. (7.69 k ) V Given data V T 5 mv given I I I IB C C B ma A Differential Amplifier A µa (Bias urrent) Applying the KVL for the given iruit we get I B C ma V V BE V BE V BE BE E DA V V I V V IE BE BE I we get µa I C I E () IE I C µa () substituting eq () in eq () we get (µ) VBE V BE (3) egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

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28 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION V BE I VT ln I C s () IC V BE VT ln I (5) s substituting equations () and (5) in eq. (3) we get (µ) 7.69 k I I VT ln C C VT ln I s I s I C VT ln I C m 5mln. For a system having transfer funtion G(s) s s a unit step input is applied at time t. The value of the response of the system at t.5 se (rounded off to three deimal plaes) is Sol. (.55) G(s) s s For unit step input (s) s So output y(s) (s). G(s) y(t) at L L s ss s s s e t t t e dt t t e e y(t) e t t.5 se y(.5) e or y(.5) Two parallel onneted, three-phase, 5 Hz, kv, star-onneted synhronous mahines A and B are operating as synhronous ondensers. They together supply 5 MVA to a kv grid. Current supplied by both the mahines are equal. Synhronous reatanes of mahine A and mahine B are and 3, respetively. Assuming the magneti iruit to be linear, the ratio of exitation urrent of mahine A to that of mahine B is. (Give the answer up to two deimal plaes). Sol. (.7) As the mahines works at same voltage and same urrent so they supply same reative power. As mahines are operating as synhronous ondenser so they will work as overexited synhronous motor. Total urrent, I T I A 5 MVA 3 KV.6 ka I B IT.6 ka.3ka As motor is working as synhronous ondenser, urrent taken by motor will be leading Also, I A 9 E A Vt jiaxs V t egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

29 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Similarly Hene, kv j.3 9 ka kV 3 E kv j.3 9 ka 3 B 3 I I A B kv 6. Let the signal Sol. () EA E.86 B k k x(t) ( ) t k be passed through an LTI system with frequeny response figure below H( ), as given in the H( ) 5 5 The Fourier series representation of the output is given as os(t) os(t) os(t) os(t) () os(t) A os(t) Given funtion in time domain x(t) k k t k This funtion looks like f(tt) delayed by time T here k t is ompared with t kt Where T the values of x(t) for k,,,... are k,, 3,... x(t) t for k x(t) t for k x(t) t for k Drawing the funtion x(t) for various values of k we get, (t) t T k k k t T T T t T k k t T t t The figure shown above passess even half wave symmetry with time period T T T In the ase of even half wave symmetry b n and onsists of only odd harmonis of a n. The frequeny omponents are,3... i.e.,6 and is the only frequeny available in the above range or 5 to 5 a n T T / T / f t osn tdt egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

30 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 3 a The output T T T T / T / T / f tosn tdt tos dt tdt T y(t) os os (3 t)... os t os 6 t... Hene, os t is in the range of 5 to The output expression for the Karnaugh map shown below is () Sol. CD AB BD BCD BD AB BD ABC BD ABC f BD ABC CD C AB A 8. The transfer funtion of the system Y(s)/ U(s) whose state-spae equations are given below is: D B () Sol. x (t) x (t) u(t) x (t) x (t) x (t) y(t) [ ] x (t) (s ) (s s ) (s ) (s s ) (s ) (s s ) (s ) (s s ) Transfer funtion T.F C(sI A) B sia s s si A s(s ) s s s [si A] s s s s [si A] B s s s CsI A B s s s s s s s 5 s T.F. s s 9. The magnitude of magneti flux density (B) in miro Teslas ( T), at the enter of a loop of wire wound as a regular hexagon of side length m arrying a urrent (I A) and plaed in vauum as shown in the figure is. (Give the answer up to two deimal plaes). egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

31 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Sol. (.69) Magneti field due to finite length of urrent arrying ondutor is given by I os os aˆ.. () H 6 I In ase of regular hexagon a a a a a where a 3 m Using formula in eq-(i) magneti field intensity at entre due to one side of regular Hexagon H os6 os H/m Magneti field intensity due to all six sides of regular hexagon will be H 6 H H/m Magneti flux density B Hene B 7 H H (in vaum) Tesla or B.698 T B.69 T upto two deimal plaes 5. The figure below shows an unontrolled diode bride retifier supplied from a V, 5 Hz, -phase a soure. The load draws a onstant urrent I A. The ondution angle of the diode D in degrees (rounded off to two deimal plaes) is V, 5 Hz Sol. (.7 ) Ls mh D D 3 D D I A For single phase ontrolled bridge retifier effet of soure indufane will modify the average output voltage as, V os os m V where is overlap angle But for diode (unontrolled) bridge, so V Also V V m Vm os L s I...()... () where L s soure indutane From eq. () and eq (). V m L s I Vm os egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

32 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 5 Substituting all the values in above equation 3 5 os Solving for os os Condution angle for diode will be 8 Hene ondution angle upto two deimal plaes. 5. In the iruit shown below, the maximum power transferred to the resistor is W 5 V Sol. (3.5 W) V P max aross will be given by P max Th V Th 5 A Where V Th Thevenin s voltage aross Th Thevenin s resistane aross To Calulate th Short irauiting all voltage soures and open iruiting all urrent soures, iruit redues to Hene Th To alulate V th Using superposition theorem Taking 5V soure only Ciruit redues to 5 V 5 V 5.5 V 5 Taking 6V soure only, iruit redues to 5 V 3V V 3 6 V Taking A urrent soure only, V 3 5V 5 V A V Th V V V V P max or P max 3.5 W 5. A 375 W, 3 V, 5 Hz, apaitor start single-phase indution motor has the following onstants for the main and auxiliary windings (at starting): Z (.5 j5.75) (main winding), m Z a (.5 j.75) (auxiliary winding). Negleting the magnetizing branh, the value of the apaitane (in F) to be added in series with the auxiliary winding to obtain maximum torque at starting is egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

33 6 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION Sol. (98.87 F ) Capaitor start single phase mdution motor V M X M I A Swith A otor I M X A V I Torque will be maximum when 9 between urrents of auxiliary winding and mains winding. 3 I M.5 j5.75 M 5.75 tan.5 Taking X as reatane of apaitor C s 3 I A.5 j.75 jx A.75 XC tan.5 Taking m 9 A C s.75 X tan 9.5 tan tan X tan Taking tan on both sides X X.5.5 tan X.5.5 Solving for X X 3.9 also X C s or C s C X s F Consider a ausal and stable LTI system with rational transfer funtion H(z), whose orresponding impulse response begins at n. Furthermore, H() 5. The poles of H(z) are p k (k ) exp j for k,, 3,. The zeros of H(z) are all at z. Let g[n] j n h[n]. The value of g[8] equals. (Give the answer up to three deimal plaes). Sol. (.98) P k k exp j, k,, 3, P P j e os jsin j e j3 j 3 3 os jsin egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

34 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 7 P 3 P e j5 j e 5 5 os jsin j7 j 7 7 os jsin System is ausal so order of numerator an not be greater than order of denominator. Therefore, K Z H(z) Z P Z P Z P Z P 3 K Z Z Z Z Z j j j j H(z) H() 5 K K KZ Z H(z) H(z) K 6 5 Z 6 Z 8 Z Z n n n h[n] h[8] h[8] g[n] n j h n g[8] j 8 h 8 h[8].98 g Let a ausal LTI system be haraterized by the following differential equation, with initial rest ondition d y dy dx(t) 7 y(t) x(t) 5 dt dt dt where, x(t) and y(t) are the input and output respetively. The impulse response of the system is [u(t) is the unit step funtion] () Sol. e t u(t) 7e 5t u(t) e t u(t) 7e 5t u(t) 7e t u(t) e 5t u(t) 7e t u(t) e 5t u(t) Differential equation d y dy dx t 7 y t x(t) 5 dt dt dt Taking Laplae on both sides (initial rest ondition) s Y(s) 7sY(s) Y(s) X(s) 5s X(s) H(s) Y s X s 5s s 7s Impulse response h(t) L H(s) egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

35 8 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 5s L h(t) s s 5 7 L s s 5 h(t) e t (t) 7e 5t (t) 55. Only one of the real roots of f(x) x 6 x Sol. () lies in the interval x and bisetion method is used to find its value. For ahieving an auray of., the required minimum number of iterations is In bisetion method, the minimum number of iterations is given by where a: lower limit of interval b: upper limit of interval : Error in approximation n : Number of iteration Thus n n > n. Aptitude b a n. The probability that a k-digit number does NOT ontain the digits, 5 or 9 is Sol. ().3 k.6 k ().7 k.9 k k- digit number digit 3 k k Exluding digits, 5 or 9 Probability P C C C... C C C C... C k times... k times or P (.7) k (k times) (k times). ahul, Murali, Srinivas and Arul are seated around a square table. ahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Whih of the following pairs are seated opposite eah other? () Sol. () ahul and Murali Srinivas and Arul Srinivas and Murali Srinvas and ahul ahul Srinivas Murali Arul 3. esearh in the workplae reveals that people work for many reasons () Sol. money beside beside money money besides besides money Beside next to Besides Exept. After ajendra Chola returned from his voyage to Indonesia, he to visit the temple in Thanjavur. () was wishing is wishing wished had wished egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

36 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION 9 Sol. () 5. Find the smallest number y suh that y 6 is a perfet ube Sol. 7 () 3 36 y 6 as perfat ube to make it perfet ube y 3 3 or y The expression Sol. () (x y) x y is equal to The maximum of x and y The minimum of x and y None of the above Given expression is x y x y we know modulus of any number should be a positive value Case (): x > y (here y is minimum) then x y (x y) positive value then x y x y x y x y y y (minimum of x and y) Case (): y > x (here x is minimum) then x y y x (positive value) then x y x y x y y x x x (minimum of x and y) In both the ases we get the minimum of x and y and the orretion option is. 7. The hold of the nationalist inagination on our olonial past is suh that anything inadequately or improperly nationlist is just not history. Whih of the following statements best reflets the author s opinion? () Sol. Nationalists are highly imaginative History is viewed through the filter of nationalism Our olonial past never happened Nationalism has to be both adequately and properly imagined 8. A ontour line joins loations having the same height above the mean sea level. The following is a ontour plot of a geographial region. Contour lines are shown at 5 m intervals in this plot. If in a flood, the water level rises to 55 m. Whih of the villages P, Q,, S, T get submerged? () Sol. () P, Q P, Q, T, S, T Q,, S 5 S Height above mean sea level for P H P 575 m Q H Q 55 m Q T egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org

37 3 GATE7 ELECTICAL ENGINEEING SESSION QUESTION AND DETAILED SOLUTION H 75 m S H s 75 m T H T 5 m if water level in a flood is 5 m then,s,t will be submerged. 9. Six people are seated around a irular table. There are at least two men and two women. There are at least three righthanded persons. Every woman has a lefthanded person to her immediate right. None of the women are right-handed. The number of women at the table is 3 () Sol. Cannot be determined Total perosns6 Conditions:. Atleast two men and two women. Atleast 3-right handed persosn 3. Every women has a left handed person to her immediate right and all women are left handed. Let us hoose at least two women (minimum) then total left handed persons ( man is immediate right of one woman when both woman are sitting together) 3 emaining three will be right handed Hene orret answer is.. Arun, Gulab, Neel and Shweta must hoose one shift eah from a pile of four shirt oloured red, pink, blue and white respetively. Arun dislike the olour red and Shweta dislikes the olour white. Gulab and Neel like all the olours. In how many different ways an they hoose the shirts so that no one has a shirt with a olour he or she dislikes? Sol. 8 () 6 Colour ed, Pink, Blue, White Arun ed Shweta White CaseArun hooses pink shirt then Shweta will have two options ed and blue so number of ways n C C C C Case Arun hooses bule shirt, Shweta will have two options ed and Pink, so n C C C C Case3 Arun hooses white, then Shweta will have three options, so n 3 C 3 C C C 6 So total number of ways 6 egd. offie : F-6, (Upper Basement), Katwaria Sarai, New Delhi-6 Phone : -36 Mob. : 89955, ies_master@yahoo.o.in, info@iesmaster.org