ES D: 5 CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr engluru Lucknow Ptn Chenni ijywd iskhptnm irupti Kuktplly Kolkt H.O: 4, Floor, hmn Plz, Opp. Methodist School, bids, Hyderbd-5, Ph: 4-33448, 4-33449, 4-3344, 4-475437 ESE- 8 (Prelims) - Offline est Series est-9 ELECCL ENGNEENG SUJEC: ELECC CCUS ND FELDS MEL SCENCE SOLUONS. ns: () Sol: he dots indicte tht when current entering ut, = = di M dt the dotted terminl of one coil, voltge induced is positive t the dotted terminl of. ns: (c) other coil. Sol: Given circuit, Given circuit, i i M i 5 L L Coil Coil 4 Here current entering the dotted terminl of coil. induced voltge is positive t the dotted pplying KL, 5( ) 4 = = 5i terminl of coil. = di M dt CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : Electricl Engineering 3. ns: (c) Sol: Given grph 6 () 4 () 3 (4) 5 (3) 7 Number of possible 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 Number of possible trees = 4 4. ns: () Sol: Given networks, 5 N Fig. i N 3 Fig. pplying reciprocity theorem to Fig () N y using linerity principle, N 3 b Fig.3 CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 3 : ESE - 8 (Prelims) Offline est Series 3 = 3 With respect to the terminls nd b the short circuit current SC = = 3 Similrly with respect to the terminls nd b the Norton s equivlent resistnce N is 4 i = SC 4 i = 5. ns: () 4 3 = 6 Sol: Given network, H N b N Fig.4 b N x 3 x Fig () is the energized version of Fig. (4) N = 5 = 4 he Notron s equivlent of Fig. (3) with respect to terminls nd b is hen the Fig. () becomes b 5 b N N =4 sc =3 i N =4 b sc =3 ssume voltge source of vlue o hving frequency of 4 rd/sec is pplied cross the circuit then x = x /F x j.5 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) j4 3 x =.755 lgging CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 4 : Electricl Engineering 6. ns: () Sol: Given network, When nd terminls re open circuited i = hen the equivlent network becomes i= i CE Engineering cdemy j 5 j 5 i i th th s there is no source to drive the bove circuit th = 7. ns: (b) Sol: Given network, Here the power is consumed only in the resistor nd its vlue is P D Y.. () Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt Y j5 j5
: 5 : ESE - 8 (Prelims) Offline est Series Now connect blnced str lod, ech of resistnce vlue s shown below Power consumed P s 3 N 3N Y Ps. () s () = () Y = N Y Y pplying KL, ( ) = = D =. 9. ns: () Sol: Given network,. C 5F F F F 8. ns: (d) Sol: Given network, nitil chrge in C = 5C nitil voltge cross cpcitor C is q C 5 5 he equivlent, circuit fter the switch is closed is = D C =5F 5F CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 6 : Electricl Engineering Convert into Lplce domin (s) s (s) Cs Cs. ns: (c) Sol: Given network F /5 3H /s (s) Cs (s) s C nd (s) (s) Cs. Cs s C 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 he dul of the bove network is. ns: (b) Sol: he wve form is s shown below 4 rms rms / 3/ / rms (4) H dt 6 / 8 5 3F t CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 7 : ESE - 8 (Prelims) Offline est Series. ns: (b) Sol: Given circuit 6Ω b Ω 8 3Ω he bove circuit cn be redrwn s L C L C t the lower cut off frequency, L C L C Net rectnce = 4. ns: () 6 3 b 8 Sol: i C C 8 3 6 Under stedy stte condition c = = 8 8 nd i(t)= e t/c Energy dissipted by the resistor otl power delivered by the sources = 8 8 = 34W 3. ns: (d) m Sol: t the cut off frequency L C E E E i e dt t / C t / C e / C C. dt C 6 5mJ C CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 8 : Electricl Engineering 5. ns: () 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. For -port network Y-prmeters re Y Y () Y Y CD prmeters () C D Mke = to find Y & Y Eqution () becomes = Y = D D Y = 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 prmeter is Y D C D Similrly, =, to find Y & Y From eqution () = Y = Y = C D C D = C D C D = ut = Y or Y eq = Y Y = D C D D C D ` CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 9 : ESE - 8 (Prelims) Offline est Series Y or Y eq = D C D Now, gin connect bck to CD prmeters D C D D Mke = = = D =.5 D / D / 4 Mke =, eqution (4) becomes = & = = C = C = = D C D D C D D C D = C C D C 6. ns: (c) Sol: Given network function F s s s s 3.5 D he pole zero pttern for the bove network function is s shown below j 3 s we cn see from bove pole zero plot poles nd zeros lternte on negtive rel xis nd nerest to the origin is pole. t cn be relized s C impednce function or L dmittnce function. 7. ns: () Sol: ssume the circuit is in stedy stte before the switch opens. he inductor will cts s short circuit. hen the circuit t t= is CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : Electricl Engineering i i x ( ) he equivlent network becomes 3Ω Ω / 3Ω i L ( ) /3 Equivlent resistnce seen by the voltge source is = 3 i = 4 7.5 3 i x ( ) = i 3 4 8. ns: () Sol: Y = = 4 3 = 3 4 3 = 7.5 4 Equivlent resistnce seen by the port is = = 3 6 = 6 9. ns: (d) Sol: Given network 6 Y = 6 3 / s we re unble to write interms of nd therefore Y prmeters doesn't exist. /3. ns: () /4 Sol: For prllel resonnt circuit, dmittnce, Y jc L Susceptnce = C L he plot of susceptnce is s shown below CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : ESE - 8 (Prelims) Offline est Series L t t = is = C L di = di dt dt = di dt t t = is = /sec. L. ns: (b) Sol: For. ns: (c) Sol: ssume the circuit is in stedy stte before the switch is chnged from to b inductor will cts s short circuit. hen the circuit is C driving point impednce function C driving point dmittnce function L driving point impednce function L driving point dmittnce function Poles nd zeros should lternte only on negtive rel xis i( ) 3. ns: (b) s (s 3) Sol: Given Z(S) = s (s ) he pole zero pttern is s shown below i( ) =. nd initil voltge cross the cpcitor is C ( ) = t t= the equivlent circuit is b i L. j 3 s we cn see the poles nd zeros lternte on negtive rel xis nd nerest to the origin is pole. t cn be relized s C driving point impednce CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : Electricl Engineering s 3 Z(S) = = s(s ) 3/ s / s he network for the bove impednce function is s shown below 4 F F 3 wo energy storge elements re present. 5. ns: (b) Sol: he equivlent network t t= is i L ( ) = = t t= the network is 5 i L ( ) 4. ns: (d) Sol: t t= the circuit is s ( ) i(t) F c ( ) = 5 S ( t= ) = 5 = 5 Convert into Lplce domin 6. ns: (d) Sol: t t = the network is C (s) 5 s s i L ( ) C 5 (s) s s v C (t) = 5 u(t) t t= u(t= ) = v C ( ) = 5 5 = s i L ( ) = = t t= the network is CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 3 : ESE - 8 (Prelims) Offline est Series H 3 H 8. ns: (b) Sol: Q Convert the bove circuit into Lplce domin pplying KCL t node '' (s) (s) (s) = s 3 s (s) (s) = 3 s 3 s 3 9 s s t (t) = 7. ns: () v (t) = 3 e 4.5 t Sol: Sttement- is correct becuse in cse of chrged bodies of rbitrry shpe, it is difficult to find the ctul distnce between them. s (s) 3 s Sttement- is incorrect s Coulomb s lw is vlid only when the point chrges re t rest. Consider string, the net force on string is given s, F F F C cos Where, F F C F C Q 4 nd = 6 Q F cos6 4 Q 4 3Q.. () 8 For string to brek due to electrosttic force, F 3 From eqution -, 3Q Q C Q F Fig.() 3 Q 8 8 or, Qmin 8, CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 4 : Electricl Engineering 9. ns: (d) Sol: Given dt, H = /m Let, be the current in the loop nd be the rdius of the loop, then H t the centre of the loop is given s, H H H / m H H Fig.() & H H 4H 3. ns: () H Sol: Given dt, J kx ˆ 5y ˆ x 4 8 / m y hen for stedy current i.e, chrge entering nd leving cross-section of the conductor to be equl t ny time,. J {Continuity eqution,.j t but v = constnt} â x x â y k5 = k = 5 y â z y. v kx â 5y â x y CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 5 : ESE - 8 (Prelims) Offline est Series 3. ns: (c) Sol:..( ) (correct) i.e, divergence of curl of vector is lwys zero.. (.) (correct) i.e; curl of grdient of sclr is lwys zero. ds.d (correct) 3. s L he bove expression is the Stoke s theorem which sys tht closed line integrl of vector is equl to the surfce integrl of the curl of tht vector. dv.ds (incorrect) 4. s Guss s diversion theorems is given s,. dv.ds v 3. ns: (b) s Which sys tht closed surfce integrl of vector is equl to the volume integrl of the diversion of tht vector. Sol: Electric flux: ccording to Guss s lw, the surfce integrl of the electric field intensity gives the mount of electric flux. q E.ds electricl 33. ns: (c) Sol: Sttement- is correct For liner dielectric Polriztion (P) E (Electric field) P = e E e is susceptblity (con stnt for liner dielectric) cpcitor with liner dielectric hs cpcitnce independent of the chrge on the pltes nd their potentil difference, rther the cpcitnce is equl to the rtio of the chrge on the plte to tht of the potentil difference between them. C = Q Sttement- is correct s the cpcitor is connected to bttery, hence voltge is fixed. Let Q d nd Q be the chrge on the pltes for dielectric nd ir cse, then, Qd Cd Cd r d Q C C d = r Q d = Q r = ut, r = e = e = Coulombs unit of Coulombs / Newton meter e = Newton-meter /Coulombs. CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 6 : Electricl Engineering 34. ns: (b) Sol: pplying Guss s lw, D.ds Q enclosed 4 D(4r) 3 D = 3 3r 3 t the surfce of the sphere, r = D = 35. ns: (d) 3 O Sphericl Gussin surfce Fig. Sol: Sttement - is correct s ngle between r nd r is 9 o, therefore, r.( r) = Sttement - is correct s.r ˆ ˆ ˆ. xˆ x x x yˆ zˆ.r 3 x y z x y z Sttement 3 is correct s,.(r.r) ˆ ˆ ˆ x y z. x y z x x x ˆ ˆ ˆ = x x y y z z r = ( xˆ x yˆ ˆ y z z ) = r Sttement-4 is correct s, ngle between nd r is 9 o herefore,.( r ) = 36. ns: (d) Sol: s the sphere is conducting nd hs potentil on its surfce (grounded) Hence, = = Where, nd re the potentil t points nd due to q nd q, hence For = q q 4(d ) 4( b) For =, q q 4(d ) 4( b) Eqn () q = Eqn () q = b b d d ( b) q (d ) ( b) q (d ) d bd b = d bd b = bd = bd b = d (3) = () =.. () CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 7 : ESE - 8 (Prelims) Offline est Series hen, q= q = 37. ns: () q d q d d d d q (4) d Sol: onic polriztion tkes plce by displcement of ctions nd nions. t is independent of temperture i = e m onic Polrizbility is inversely proportionl to the squre of nturl frequency () 38. ns: (b) Sol: Polriztion (P) = ( r )E = 8.854 (8 ) 3 = 6. 7 39. ns: (b) Sol: f the rdius X =.55, more stble configurtion is possible with three nions bonding with ction. his form stble structure only upto n X vlue of.5 for.55 < X <.5 the nions do not touch ech other. 4. ns: (c) 4. ns: () 4. ns: (b) Sol: ne ne 6 43. ns: (c) Sol: H C = 8.6 9 = 6.76 3 m /-s H C.54 8 5 = 5 C 5 5 5 5 8 c 8 c = 64.5 c c C =.3 K 44. ns: (b) Sol: J c = 45. ns: () 64.5 ic H C H Sol: sed on F.London nd H.London reserch, C 8 mgnetic flux density is llowed by super conductor up to some lyers from the surfce. CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 8 : Electricl Engineering London penetrtion depth: t is the depth from the surfce of super conductor upto which flux density is decresed by 63%. = e X L = Flux density t depth X = flux density t surfce of super conductor L = London penetrtion depth 5. ns: () Sol: he hysteresis loop of ferromgnetic mteril depends on. emperture. Crystllogrphic imperfection 3. Cold working 5. ns: () Sol: M 46. ns: (d) Sol: Hll ngle tn H = =.4.=.4 H = tn (.4) H =.349 [] [] [] H 47. ns: () 48. ns: (c) Sol: otl mgnetic moment = N i = 5 3 ( ) = 5 3 4 4 = 4 3 -m 49. ns: () Sol: Mgnetic flux density (b) = 88 4 Wb/m Field strength (H) = /m = H 4 88 = 4 7 5. ns: (b) Sol: Mgneto rheologicl mterils hs the bility to increse viscosity drsticlly with pplied field. 53. ns: () 54. ns: (c) 55. ns: (d) 56. 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 = 7 tkes plce in the field direction nd t higher field, inside domin dipole reltion tkes plce. CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: 9 : ESE - 8 (Prelims) Offline est Series 57. ns: () Sol: Output voltge = t g p = 3.55.5 6 = 37.5 58. ns: (d) Sol: G-s compound is zinc blende structure 59. ns: (b) 4.8 9 = o 6. ns: (d) Sol: X (,, ) Y,, Z Sol: 4 4 d h k 6. ns: (d) Sol: he bullet proof jocket is mde up of rmid fiber reinforced polymer (F) with reinforcement phse is rmid nd mtrix phse is polymer. CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : Electricl Engineering 6. ns: (d) Sol: he bonding in cermics is predominntly ionic, it consists of nions nd ctions. he crystl structure of cermics is influenced by the rdius rtios of the ions. he coordintion number depends on the rdius of the bonding ions. Cermics re clssified ccording to their crystl structure s X, X, X 3 nd X 4 types. Exmples under ech type re given below: 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 63. ns: (c) Sol: f the temperture of metl increses, the lttice vibrtion in the crystl structure increses. Hence collision frequency increses nd relxtion time decreses. Due to tht resistivity of metl increses. 64. ns: (c) Sol: Cermics hve high melting point nd cn with stnd high temperture. 65. ns: (b) Sol: Lplcin eqution () ut we know, from Poisson s eqution herefore Lplcin eqution is true for chrge free region where = Every physicl problem must contin tlest one conducting boundry but my contin more thn one. Solution of Lplce s eqution with two different methods (vlid methods) led to sme solution. Since E field is hrmonic (conservtive). 66. ns: (d) Sol: Sttement is incorrect s S. ds Mxwell eqution:.ds i.e., net flux leving closed surfce is zero. When surfce is open.ds m weber Sttement is correct s ubes of mgnetic flux hve no source (or) sink i.e. monopoles do not exist in cse of mgnetic field. N S CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
: : ESE - 8 (Prelims) Offline est Series 67. ns: () So sttement () is flse. Sol: When C field is pplied to dielectric mteril, then dielectric constnt of mteril is no longer rel t is hving both rel prt s well s imginry prt r = r j r j r prt (mginry prt) of r is due to power loss in mteril. When C field is pplied to dielectric 7. ns: (b) 73. ns: (b) Sol: Poly crystl mterils re stronger thn single crystl mteril becuse 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. mteril, r becomes complex quntity s result power loss in dielectric mteril. 68. ns: (b) Sol: he ntiferro mgnetic mteril depends on Neel s lw = C N nti ferro pr 74. ns: () Sol: io 3 crystl is Ferroelectric mteril upto o C, due to non-centro symmetric (symmetric) structure. ut bove C it become centro-symmetric nd hence it looses its ferroelectric chrcter. 75. ns: (d) 69. ns: (d) N Sol: mpressed voltge = ( j9) Current = (3 j4) Complex power, S = * = ( j9) (3 j4) = (66 j3) Sol: Sttement is incorrect becuse when there is no chrge inside the conductor the electric field inside conductor is zero not infinity. Sttement is correct s el power = 66 W Sttement () is flse. 7. ns: (b) 7. ns: (d) Sol: Equivlent network obtined from - Y trnsformtion reltion is vlid for ny frequency. Guss lw: Electric flux leving ny closed surfce is equl to the chrge enclosed. n cse of conductor s the chrge enclosed by ny closed surfce inside conductor is zero hence there should not be ny electric field inside the conductor. CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt
CE Engineering cdemy Hyderbd Delhi hopl Pune hubneswr Lucknow Ptn engluru Chenni ijywd izg irupti Kuktplly Kolkt