ES D: 5 CE Engineering cdemy Hyderd Delhi hopl Pune huneswr engluru Lucknow Ptn Chenni ijywd iskhptnm irupti Kuktplly Kolkt H.O: 4, Floor, hmn Plz, Opp. Methodist School, ids, Hyderd-5, Ph: 4-33448, 4-33449, 4-3344, 4-475437 ESE- 8 (Prelims) - Offline est Series est- 9 ELECONCS & ELECOMMUNCON ENGNEENG SUJEC: MEL SCENCE NEWOK HEOY SOLUONS. ns: (). ns: (c) Sol: he dots indicte tht when current entering Sol: Given circuit, the dotted terminl of one coil, voltge induced is positive t the dotted terminl of other coil. Given circuit, i 5 i Here current entering the dotted terminl of coil. induced voltge is positive t the dotted terminl of coil. = M i di M dt ut, = = L L Coil Coil di M dt 4 pplying KL, 5( ) 4 = = 5i 3. ns: (c) Sol: Given grph () () 3 (4) 6 4 5 (3) 7 Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
= Numer of possile trees = [ r ][ r ] Where, [ r ] is the reduced incidence mtrix ncidence mtrix, for the given grph is 3 4 5 6 7 () () (3) (4) y tking the (4) node s reference, educed incidence mtrix, [ r ] = [ r ][ r ] 4 = 3 3 4 [ r ][ r ] = 3 3 = 4(9 ) (3) (3) = 3 4 4 = 4 Numer of possile trees = 4 : : E & E 4. ns: () Sol: Given networks, N 5 N Fig. i N 3 Fig. pplying reciprocity theorem to Fig () N y using linerity principle, N 3 Fig.3 3 = 3 With respect to the terminls nd the short circuit current SC = = 3 Similrly with respect to the terminls nd the Norton s equivlent resistnce N is N Fig.4 N Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 3 : ESE - 8 (Prelims) Offline est Series Fig () is the energized version of Fig. (4) 5 ssume voltge source of vlue o hving frequency of 4 rd/sec is pplied cross the circuit then N = 5 = 4 he Notron s equivlent of Fig. (3) with respect to terminls nd is N x j.5 j4 3 x hen the Fig. () ecomes i 4 i = SC 4 4 i = 3 = 6 N =4 sc =3 N =4 sc =3 x = x pplying KL in the loop, o x j3.5 3 x = o j3.5 3 = o = =.884.85 o 4 j3.5 Circuit power fctor = cos(4.85) 6. ns: () Sol: Given network, =.755 lgging 5. ns: () Sol: Given network, H x 3 x i j 5 i When nd terminls re open circuited th /F i = hen the equivlent network ecomes Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 4 : E & E i= j 5 i s there is no source to drive the ove circuit th = 7. ns: () Sol: Given network, th Power consumed P s 3 N 3N Y Ps. () s () = () Y Y = 8. ns: (d) Sol: Given network, j5 Y j5 Here the power is consumed only in the resistor nd its vlue is D P D Y.. () Now connect lnced str lod, ech of resistnce vlue s shown elow pplying KL, Y N ( ) = = Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt D =..
: 5 : ESE - 8 (Prelims) Offline est Series 9. ns: () Sol: Given network, Convert into Lplce domin C 5F F F F (s) s (s) Cs Cs nitil chrge in C = 5C nitil voltge cross cpcitor C is q C 5 5 he equivlent, circuit fter the switch is closed is C =5F = 5F /s (s) Cs (s) s C nd (s) (s) Cs. Cs s C Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 6 : E & E Stedy stte vlue of v(t) is = = s Lt s(s) s Lt s. s(cs ) = 5 Stedy stte voltge cross F cpcitor = 5 Chrge in the F cpcitor q = C = 5 = 5 C rms rms / rms. ns: () (4) Sol: Given circuit dt 6 / 8. ns: (c) Sol: Given network 6Ω 8 3Ω Ω F /5 3H he ove circuit cn e redrwn s he dul of the ove network is H 6 3 8 5 3F 8 3 6. ns: () Sol: he wve form is s shown elow 8 8 4 / 3/ t otl power delivered y the sources = 8 8 = 34W Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 7 : ESE - 8 (Prelims) Offline est Series 3. ns: (d) m Sol: t the cut off frequency L C L L C C t the lower cut off frequency, L C L C Net rectnce = C E 5. ns: () C C 6 5mJ Sol: We know, when two, -port networks re connected in prllel, their individul Y- prmeters gets dded. First, we need to convert given CD prmeters to Y-prmeters. 4. ns: () For -port network Sol: i C C Y-prmeters re Y Y () Y Y Under stedy stte condition c = = nd i(t) = e t/c Energy dissipted y the resistor E i dt CD prmeters () C D Mke = to find Y & Y Eqution () ecomes = E e t / C. t / C e / C dt Y D = D Y = Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 8 : E & E Similrly, =, to find Y & Y From eqution () = = Y D C D = C D C D = C D C D = Y Y ut = C D = C D (3) Y C D C D nd = eqution (3) C D C D = Y = put this vlue in Y-prmeter mtrix in terms of CD Y or Y eq = Y Y = D Y or Y eq = C D D D C D ` C D Now, gin connect ck to CD prmeters D C D D Mke = = = 4 prmeter is Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 9 : ESE - 8 (Prelims) Offline est Series =.5 D C D = D D / D / Mke =, eqution (4) ecomes = & = = D C D D C D = C = C = C C D C.5 D Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : E & E 6. ns: (c) Sol: Given network function s Fs s s 3 he pole zero pttern for the ove network function is s shown elow s we cn see from ove pole zero plot poles nd zeros lternte on negtive rel xis nd nerest to the origin is pole. t cn e relized s C impednce function or L dmittnce function. 7. ns: () Sol: ssume the circuit is in stedy stte efore the switch opens. he inductor will cts s short circuit. hen the circuit t t= is 3 i 3Ω 3Ω j i x ( ) Ω i L ( ) 8. ns: () Sol: Y he equivlent network ecomes Equivlent resistnce seen y the port is = = 3 6 = 6 Y = 6 / /3 / /3 /4 6 Equivlent resistnce seen y the voltge source is = 3 3 = = 7.5 4 4 i = 7.5 3 i x ( ) = i 3 4 = 4 3 = 3 4 9. ns: (d) Sol: Given network 3 s we re unle to write interms of nd therefore Y prmeters doesn't exist. Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : ESE - 8 (Prelims) Offline est Series. ns: () Sol: For prllel resonnt circuit, dmittnce, Y jc L Susceptnce = C L he plot of susceptnce is s shown elow C L t t = is = L di = di dt dt = di dt t t = is = /sec.. ns: () Sol: For L C driving point impednce function C driving point dmittnce function L driving point impednce function L driving point dmittnce function. ns: (c) Sol: ssume the circuit is in stedy stte efore the switch is chnged from to inductor will cts s short circuit. hen the circuit is i( ) i( ) =. nd initil voltge cross the cpcitor is C ( ) = t t= the equivlent circuit is i L. 3. ns: () Poles nd zeros should lternte only on negtive rel xis s (s 3) Sol: Given Z(S) = s (s ) he pole zero pttern is s shown elow 3 s we cn see the poles nd zeros lternte on negtive rel xis nd nerest to the origin is pole. t cn e relized s C driving point impednce j s 3 Z(S) = = s(s ) 3/ s / s Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : E & E he network for the ove impednce function is s shown elow wo energy storge elements re present. 4. ns: (d) Sol: t t= the circuit is 4 F F 3 5. ns: () Sol: he equivlent network t t= is i L ( ) = = t t= the network is 5 i L ( ) i(t) F c ( ) = 5 s ( ) Convert into Lplce domin S ( t= ) = 5 = 5 C (s) 5 s s 6. ns: (d) Sol: t t = the network is C (s) 5 s 5 = s v C (t) = 5 u(t) t t= u(t= ) = s v C ( ) = 5 i L ( ) = = t t= the network is H 3 i L ( ) H Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
Convert the ove circuit into Lplce domin pplying KCL t node '' (s) (s) (s) = s 3 s 3 (s) s 3 s 3 (s) = v (t) = 3 e 4.5 t 9 s s t (t) = 7. ns: () Sol: he equivlent cpcitnce cross, is clculted y simplifying the ridge circuit s shown in Fig. to Fig. 5. s c Note: he ridge is lnced nd the nswer is esy to get. c Fig. (s) c is eliminted. 3 Fig. 3 s.. d. 3 d. 3 : 3 : ESE - 8 (Prelims) Offline est Series 8. ns: (d) Sol: i(t) i(t) v(t) d t, L (s) (s s Ls ) he inductor in time domin nd trnsform Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt L t domin re shown in Fig. nd Fig.. 9. ns: (c) Sol: L = 3 3 H, i(t) Fig. v(t) t 5 3 d 3 i(t) 5 d t / s, d i(t) v(t) L 5 d t = s 3 t (s) p(t) = v(t) i(t) = 55 3 t W = 75t 3. ns: () Sol: su-network N cn e replced y its hevenin equivlent even if the su-network N contins two-terminl element which is non-liner. 3. ns: (c) Sol: f network hs ll liner elements except for few non liner ones, then superposition theorem my hold on creful selection of element vlues, source wveform nd response. Note: he reder my see this importnt sttement nd n exmple in comment (3), pge 66, of ext ook sic Circuit heory y Desoer & Kuh. Ls Fig. (s)
: 4 : E & E 3. ns: (d) Sol: he figure ccompnying the question specifies voltge cross C s in fig.. Polrity of c is therefore specified s in fig.. t t =, c would hve cquired vlue of volts. For t, the circuit cn e redrwn s in fig.3. o C c C i(t) Fig: C SW C C Fig: C (t) C t. / C (t) i dt e t c C Check: t t = ; c (t) must equl, which if does. t t =, c (t) must equl, which gin it does. Note: n the question if the polrity given for is reversed, then (d) would e the nswer. 33. ns: (d) Sol: n the circuit given elow, (t) t = C C (t) t t = ; (t) = nd (t) =. t t = (immeditely fter the switch is closed) the two cpcitnces must cquire some common voltge. Since this mens voltges cross cpcitnces chnging instntneously, there must e n impulse current through the loop. C (t= )= C (t= )= Fig: 3 34. ns: () Sol: KL: i C t i dt Solving, i(t) = e t/c. t 4H F F t t =, = or = thus i(t) = t / C e Fig. he given circuit is shown in Fig. Z = Z Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 5 : ESE - 8 (Prelims) Offline est Series j j where, Z = j4 j j j4 = j j j4 4 = j j4 For circuit to e resonnt i.e., = 4 = =.5 rd/ sec resonnce =.5 rd/sec 35. ns: () Sol: 36. ns: () Sol: Output voltge = t g p = 3.55.5 6 = 37.5 37. ns: () Sol: onic polriztion tkes plce y displcement of ctions nd nions. t is independent of temperture i = e m onic Polrizility is inversely proportionl to the squre of nturl frequency () 3 (6j) 38. ns: () Sol: Polriztion (P) = ( r )E = 8.854 (8 ) 3 = 6. 7 P 3 cos ph L cos P L 6 3 L 3 Ph Z.5( 6 3.5 6 3) 9.9 6 3(3)(9.9)(.8) 6348 = 6.34 kw 4.8 5 39. ns: (c) 4. ns: () 4. ns: () Sol: ne ne 8 9 6.6.54 = 6.76 3 m /-s 8 Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 6 : E & E 4. ns: (c) Sol: H C = H C 8 5 = 5 C 5 5 5 5 8 c 8 c = 64.5 c c C =.3 K 43. ns: () Sol: J c = 64.5 ic H C H C 44. ns: () Sol: sed on F.London nd H.London reserch, mgnetic flux density is llowed y super conductor up to some lyers from the surfce. London penetrtion depth: t is the depth from the surfce of super conductor upto which flux density is decresed y 63%. X L = e = Flux density t depth X = flux density t surfce of super conductor L = London penetrtion depth 45. ns: (d) Sol: Hll ngle tn H = =.4.=.4 H = tn (.4) H =.349 Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 7 : ESE - 8 (Prelims) Offline est Series 46. ns: () 47. ns: (c) Sol: otl mgnetic moment = N i = 5 3 ( ) = 5 3 4 4 = 4 3 -m 48. ns: () Sol: Mgnetic flux density () = 88 4 W/m Field strength (H) = /m = 4 H = 88 4 7 = 7 49. ns: () Sol: he hysteresis loop of ferromgnetic mteril depends on. emperture. Crystllogrphic imperfection 3. Cold working 5. ns: () Sol: M [] [] [] H 5. ns: () Sol: Mgnetorheologicl mterils hs the ility to increse viscosity drsticlly with pplied field. 5. ns: (c) 53. ns: (d) 54. ns: () Sol: Wiess-Domin heory: sed on Wiess Domin theory in domin ll the dipoles re ligned in prticulr direction. f the mgnetic field is pplied Domin growth tkes plce in the field direction nd t higher field, inside domin dipole reltion tkes plce. 55. ns: () Sol: eson: he no. of toms per unit cell in CC Crystl = he no.of toms per unit cell in FCC crystl = 4 So, the difference is 4-= 56. ns: (d) Sol: G-s compound is zinc lende structure 57. ns: () Sol: 4 4 d h k 9 4.8 = o Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 8 : E & E 58. ns: (d) Sol: X (,, ) Y,, Z 59. ns: (d) Sol: he ullet proof jocket is mde up of rmid fier reinforced polymer (F) with reinforcement phse is rmid nd mtrix phse is polymer. 64. ns: (c) Sol: f the temperture of metl increses, the lttice virtion in the crystl structure increses. Hence collision frequency increses nd relxtion time decreses. Due to tht resistivity of metl increses. 65. ns: (c) Sol: Cermics hve high melting point nd cn with stnd high temperture. 6. ns: () 6. ns: (c) 6. ns: (c) 63. ns: (d) Sol: he onding in cermics is predominntly ionic, it consists of nions nd ctions. he crystl structure of cermics is influenced y the rdius rtios of the ions. he coordintion numer depends on the rdius of the onding ions. Cermics re clssified ccording to their crystl structure s X, X, X 3 nd X 4 types. Exmples under ech type re given elow: X : NCl, CsCl, Zns X : SiO, CF, PuO, ho X 3 : io 3, SrZrO 3, SrSnO 3 X 4 : Mgl O 4, Fel O 4 66. ns: () Sol: f the rdius X =.55, more stle configurtion is possile with three nions onding with ction. his form stle structure only upto n X vlue of.5 for.55 < X <.5 the nions do not touch ech other. 67. ns: () 68. ns: () Sol: io 3 crystl is Ferroelectric mteril upto o C, due to non-centro symmetric (symmetric) structure. ut ove C it ecome centro-symmetric nd hence it looses its ferroelectric chrcter. 69. ns: () Sol: he ntiferro mgnetic mteril depends on Neel s lw Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
= C N : 9 : ESE - 8 (Prelims) Offline est Series 7. ns: (d) Sol: Equivlent network otined from - Y trnsformtion reltion is vlid for ny frequency. So sttement () is flse. nti ferro pr N 73. ns: () 7. ns: (d) Sol: mpressed voltge = ( j9) Current = (3 j4) Complex power, S = * = ( j9) (3 j4) = (66 j3) el power = 66 W Sttement () is flse. 7. ns: () 74. ns: () Sol: Conversion to equivlent NW nd ppliction of hevenin s heorem hve no reltion. 75. ns: () Sol: Poly crystl mterils re stronger thn single crystl mteril ecuse they require more stresses to initite slip nd yielding. Poly crystlline mterils there re mny preferred plnes nd direction for different grins due to their rndom orienttion. Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : E & E Hyderd Delhi hopl Pune huneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt