ESE (Prelims) - Offline Test Series ELECTRICAL ENGINEERING SUBJECT: Power Electronics & Drives, Power Systems and Signals & Systems SOLUTIONS

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1 TEST ID: 8 ESE- 9 (Prlims) - Offlin Tst Sris ELECTRICAL ENGINEERING Tst-5 SUBJECT: Powr Elctronics & Drivs, Powr Systms and Signals & Systms SOLUTIONS. Ans: (b) Powr diod rcovry dpnds on th charg stord in th diod during forward condition. QR IRRtrr. RR I dpnds on di. So, as dt frquncy incrass, tim dcrass and slop of currnt tim graph incrass, so charg stord incrass and thus frquncy plays an important rol in transint rcovry.. Ans: (b) In a powr transistor with larg junction ara, undr crtain conditions of currnt and voltag, th collctor currnt concntrats in a small spot of th bas-mittr junction. This oftn lads to th dstruction of transistor. This is calld scondary brak down of BJT.. Ans: (b) To dsign a bas driv circuit, bas currnt and bas voltag ar rquird and th rlation btwn bas currnt and bas voltag for diffrnt valus of bas currnt is rquird. So, w us I B vs V BE charactristics. 4. Ans: (c) Any switch is opratd in hard saturation stat or in cut off stat. For xampl, if th V c, sat =.7 V for a transistor, it is not saf to oprat at V c =.7V (at th vrg of saturation). Thr is a scop of transistor going into linar mod. So, it is nsurd that it is opratd in th hard saturation stat. BJT can t oprat at high frquncy, sinc it is majority carrir dvic. So, if opratd at high frquncy activ mod and soft saturation losss will b mor. Hydrabad Dlhi Bhopal Pun Bhubanswar Lucknow Patna Bngaluru Chnnai Vijayawada Vizag Tirupati Kukatpally Kolkata

2 5. Ans: (a) During rvrs bias, J and J ar rvrs bias and J is forward biasd. So, gat currnt can t ffct th J,J junction status, whras tmpratur ffct th rvrs saturation currnt through th junction. 6. Ans: (b) Invrtr grad thyristors ar usd in invrtr applications. In a invrtr switchs oprat at mor than fundamntal output frquncy. In PWM invrtrs th switching frquncy is much mor compard to fundamntal frquncy. So, thyristors ar dsignd for lss turn-off. Convrtr grad thyristors ar usd in rctifir circuits. So, switching frquncy is comparabl to sourc voltag frquncy. So, thy hav high turn off tims. 7. Ans: (d) For any rctifir, if it is a n-puls convrtr thn th sourc sid harmonics ar givn by nk harmonics and load sid harmonics ar givn by nk. Sourc sid contains odd harmonics and load sid contains vn harmonics. So, lowst harmonic in sourc currnt is = th harmonic. 8. Ans: (c) V(t) = sin ( t) : : Elctrical Enginring V rms V V P R 5V rms 9. Ans: (b) m 5.5W MOSFET oprats in thr rgions of opration.. Cut-off rgion: VGS V GS th. Triod rgion/ ohmic rgion: V DS V GS V GS th. Saturation rgion: VDS VGS VGSth. Ans: (b) Diod is in ON condition, in th first half positiv cycl th inductor absorbs nrgy and in th scond half ngativ cycl inductor rlass nrgy. So, diod conducts th voltag sourc for 6. So, th whol sourc voltag drops across th inductor. V d = V M = ; V M = V rms. Ans: (d) =4.4V Th conduction of lin currnt to load taks plac from 6 + to +. So, from this w can infr that if > 6 thn as it is R-load, voltag bcoms zro discontinuity occurs.

3 : : ESE - 9 (Prlims) Offlin Tst Sris. Ans: (d) In -, half wav rctifir, Vml V V =.4775 V ml. Ans: (d) Bcaus of sourc inductanc th switch currnts cannot bcom zro abruptly. So, thr is ovrlap angl whr switch currnts ovrlap and this xtnds th commutation tim of th switchs. This ovrlap is calld commutation ovrlap and th ovrlap angl is dnotd by. 4. Ans: (c) I L DT T T t r t f t t f 5. Ans: (c) Vrms V Rippl factor V 6. Ans: (c) V rms DV dc dc dc D.Vdc D.V D.V = D =.49. I rms = 4V V = D.V dc I dc = 4 4A = (.49). = 98V A dc DT V. T DT V V dc t. V r = dc t r tf t r t r. Vdc tf 7. Ans: (c) Circuit turn of tim for main SCR, t c = 4. = 8s CV I S

4 : 4 : Elctrical Enginring Circuit turn of tim for auxiliary SCR = LC 46 = 4 =.56 s. Ans: (a) I L Vdc.D D.T L ms mh = (.6) =.48A 8. Ans: (b) In th abov fig. b, th input currnt is continuous. So, th answr cannot b Buck convrtr or a buck-boost convrtr. From fig. a, output voltag is lss than input voltag. So, th answr cannot b boost convrtr. So, th answr is boost-buck convrtr or C uck Convrtr. 9. Ans: (c) V dc = V; V = 4V T.ms f D V. D D D V dc D T on = DT D s 5 = 8s. Ans: (d) E = 7V V ml 44 V o V 44 I o = Amp R=.5 I o E V R cos =.. Ans: (d) o cos As th siz of conductor incrass kping distanc btwn conductors constant th valu of inductanc dcrass and capacitanc rmains sam, so th valu of surg impdanc rducs. As th distanc btwn conductors is fixd charging currnt also rmains constant.

5 : 5 : ESE - 9 (Prlims) Offlin Tst Sris As th inductiv ractanc rducs Frranti ffct incrass. So statmnt is incorrct.. Ans: (d) By using synchronous condnsr it is not conomical to rgulat ractiv powr flow individually in ach phas. By using static compnsator w can asily rgulat ractiv powr individually in ach phas. 4. Ans: (a) C s =.6 C s = 4. F C s + C c = 9 C c = 9 4. = 4.7 C c =.6 F 5. Ans: (b) Load angl = sin = o.5 ~ V Q Q g Q C V o j.5 SD = p.u V Q = V V cos X o = cos.5 =.68 p.u Capacitor, Q c = Ans: (d) dp EV cos d X 7. Ans: (d) E =. pu, v =. pu P max P max P max Ev x q...6 =.68 p.u.6.. =. Pmax =.667 P.6 max 8. Ans: (b) Fault currnt x (during fault ) Q g + Q c = Q Q c = Q Q g I I in f I I sub transint synchronou s X X syn sub trans.5 = 5.5

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7 : 7 : ESE - 9 (Prlims) Offlin Tst Sris 9. Ans: (b) Undr rach: whn impdanc sn by th rlay du to prsnc of arc rsistanc, th impdanc sn by th rlay appars to b mor than th actual valu of th impdanc up to th fault point and th rlay tnds to undr rach. Ovr rach: Whn impdanc sn by th rlay is lss than th st valu th distanc rlay is pron to ovr rach on a transint fault consisting of a dc offst. All high spd distanc rlay tnds to s mor currnt du to th prsnc of dc offst.. Ans: (a) A = cos hrl B = Z c sin hrl if l is rducing, A is incrasing B is dcrasing. Ans: (d) For a n bus powr systm, w considr on bus as slack bus and rmaining (n ) buss ar ithr gnrator bus or load bus for which activ powr should b spcifid. Only for gnrator and slack bus voltag should b spcifid for gnrator bus voltag is not spcifid so it is a wrong statmnt. 4. Ans: (c) Zro squnc ntwork for th givn systm is quivalnt impdanc as sn from nod is X g =. n g X T X L X g =. n Equivalnt impdanc as run from nod is X q. Ans: (a). Ans: (c) EV P = sin( ) X = sin( ). sin( ) =., ar vry small =.. X q X q =.... X q = =. pu..

8 : 8 : Elctrical Enginring 5. Ans: (b) Th lin is normally oprating with ngativ polarity as th corona loss and th radio intrfrnc ar rducd 6. Ans: (c) I = I +I g Vc = Vc + Vc g c = c +c g c = (+)F c = F 9. Ans: (c) Gnrator buss = Load buss = 9 5 load buss ar convrtd into gnrator buss total gnrator buss = 5 Load buss = 85 Gnrator quations = (5 - ) = 4 Load bus quations = 85 = 7 Total numbr of quations = Ans: (d) V = z c =. = 8. Ans: (c) = 5 km/s LC L = C L 5 = 8 LC C L int 8 L int L int r L L int int C = 8 F/Km = 8 C = F/km =. F/km r 8 r r r 7 H / m 8 Lint Ans: (b) 4. Ans: (a) Slf GMD = r ' SS whr r = imaginary radius du to intrnal as wll as xtrnal turn linkags. r r r S = distanc btwn two conductors (cntr to cntr) Slf GMD =.7788 r r r r.7788 r r

9 : 9 : ESE - 9 (Prlims) Offlin Tst Sris 4. Ans: (d) If any machin ovr xcitd thn it always dlivrs lagging vars 4. Ans: (d) Elctrostatic prcipitator is usd to collct th dust particls from flu gass. 44. Ans: (d) Givn x(t) = u(t ) & h(t) = u(t ) W know that u(t) u(t) = r(t) u(t t ) u(t t ) = r(t t t ), u(t ) u(t ) = r(t ) 45. Ans: (b) = r(t ) = (t )u(t ) Givn x(t) = m(t) y(t).. () and Y() = X( c ).. () Apply IFT to quation () y(t) x(t) x(t) jc t jct y(t).. () Comparing () & (), thn m(t) = 46. Ans: (b) j ct dw(t) Givn y(t) x(t).. () dt dy(t) w(t). () dt Apply Laplac transform to quation () sw(s) = Y(s) + X(s).. () Apply Laplac transform to quation () sy(s) = W(s).. (4) From () & (4) s[ sy(s)] = Y(s) + X(s) s Y(s) Y(s) = X(s) H(s) Y(s) X(s) s Apply Invrs Laplac Transform h(t) = sin(t)u(t) 47. Ans: (c). y(n).4y(n ) = x(n) Apply z-transform Y(z).4 z Y(z) = X(z) H(z) Y(z) X(z).4z Pol =.4, lis insid th unit circl. So, it is stabl. Statmnt () is fals.. All pols of FIR filtr lis insid th unit circl. So, FIR filtrs ar always stabl. Statmnt () is tru.. h(n) (.4) n So, it is stabl. n n,.4 Statmnt () is fals.

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11 : : ESE - 9 (Prlims) Offlin Tst Sris 48. Ans: (b) Givn x(n) = (n + ) + (n ) X( j ) = j + j and H( j ) = 4 j Y( j ) = X( j ) H( j ) = 8 j j 4 4j Apply IDTFT y(n) = 8(n + ) 4(n) + 8(n ) 4(n 4) y(n) = {8,, y( ) = 49. Ans: (b) 4, 8,, 4} (n) x cos n 4 6 is not in th rang of 4 So, y (n) = y(n) = y (n) + y (n) y (n) 5. Ans: (b) y(t) = x(t)*h(t) y (t) 4k sin (n ) 4 y ( ) x( )h(t ) d x( )h( ) d to. Acos( n+) Asin( n+) Givn x(n) = Considr 4 sin n cos n (n) x 4sin n 4 lis in th rang of So, (n) y H( j ) j A H( j ) cos[ n++ H( ) ] j A H( j ) sin[ n++ H( ) 4k sin n 4 ] h( ) 5 y ( ) x() 5..d 5 h() h( ) (n) y 4k sin (n ) 4 y( ) 5

12 : : Elctrical Enginring 5. Ans: (a) Assum dx(t) y(t) dt From diffrntiation in tim domain proprty dx(t) dt jk a k Assum y(t) cofficints is b k thn b k = k = k = k othrwis So, b = 5. Ans: (d) x (t) X (s), ROC R x (t) X (s), ROC R x (t) x (t) X (s)x (s), ROC R R 5. Ans: (a) Rctangular window main lob width is N 4 Hanning window main lob width is N 8 Hamming window main lob width is N 8 Blackmann window main lob width is 54. Ans: (b) y(n) = x(n) + nx(n ) y (n) = x (n) + nx (n ) N y (n) = x (n) + nx (n ) y (n) = [x (n) + x (n) ] + n[x (n ) + x (n )] y (n) = y (n) + y (n) So, linar systm y (n) = x(n k) + nx(n k ) y(n k) = x(n k) + (n k)x(n k ) y (n) y(n k) So tim variant. 55. Ans: (c) Th pak sid lob in th cas of Hanning window has a valu of db. 56. Ans: (b) z Y(z) z z Y(z) z z n ( ) u(n) So, y(n) = 57. Ans: (c) If n () u(n) x(n) = z n y(n) = z n H(z) x(n) = ( ) n y(n) = ( ) n H(z) z H(z) z Y(z) X(z)

13 : : ESE - 9 (Prlims) Offlin Tst Sris H (z) Z So, y(n) = ( ) 58. Ans: (b) X (k) 7 n x(n) X() = x (n) X(4) = 7 n n j nk 8 = x()+ x()+ x() + x() + x(4) 7 n + x(5) + x(6) + x(7) x(n)( ) n = x() x() + x() x() + x(4) x(5) + x(6) x(7) X() + X(4) = x() + x() + x(4) + x(6) n n 59. Ans: (c) Sgn x(n) X() X(4) 6 x(n) 8 t jf From Tim rvrsal proprty Sgn t Sgn t jf jf So, x(t) = Sgn(t) x(t) = Sgn(t) = t > 6. Ans: (a) X () = X () > E X ( ) 6. Ans: (b) X(s) = t < = < 4 X ( ) E 4 = E x(t) d X( ) d st dt Convrgnc condition is x(t) 6. Ans: (b) t dt Th ncssary and sufficint condition for a priod signal x(t) can xpandd by Fourir T sris is x(t) dt < 6. Ans: (b) t u(t) s

14 : 4 : Elctrical Enginring (t) (t) t 64. Ans: (a) s u(t ) s u(t) s (s) u(t ) s Phas dlay 65. Ans: (c) t p ( ) ( ) =.5msc 4 Th maximum frquncy m = 8, f m = 4Hz Nyquist intrval = 66. Ans: (a) x(n) = {,,,} h(n) = {,,4,8} f m So, y(n) = {, 6,, 6} =.5 msc 67. Ans: (a) Condition for stability of LTI Systm is h t dt 68. Ans: (a) Givn y(n + ) 5y(n + ) + 6y(n) = x(n) Apply z-transform Y(z) = = z X z 5z 6 Xz z z Charactristic quation is (z )(z ) = Pols ar,. Pols ar lis outsid th unit circl. So, it is unstabl systm. So, statmnt (I) is corrct. A systm is unstabl if th roots of th charactristic quation lis outsid th unit circl. So, statmnt (II) is corrct. 69. Ans: (d) Statmnt I is not corrct and statmnt II is corrct 7. Ans: (d) For LLG fault positiv squnc currnt is givn by I a = Ea ZZ Z Z Z

15 : 5 : ESE - 9 (Prlims) Offlin Tst Sris And for LG fault positiv squnc currnt is givn by I a = Z Ea. Z Z So, I a(lg) < I a(llg) Zro squnc in synchronous matrix dpnds upon th chording and bradth factors. Howvr, zro squnc impdanc is much smallr than positiv and ngativ squnc impdanc. 7. Ans: (a) Both corrct and (R) is th corrct xplanation of (A). 7. Ans: (d) Th output rippl frquncy of - full bridg rctifir is 6f and output rippl frquncy of - full bridg rctifir is f. 7. Ans: (a) In a Full wav rctifir, if firing angl incrass it draws ractiv powr from ac supply du to lagging powr factor. Th natur of currnt in th supply lin would b non sinusoidal. 74. Ans: (a) In forward blocking stat of SCR, junction J is rvrs biasd and acts lik a capacitor whras junction J and J ar forward dv biasd. So, if larg is applid to th SCR dt a larg charging currnt flows and turns ON th SCR. So, Statmnt-I validats Statmnt-II. 75. Ans: (a) In CSI drivs, th sourc is currnt sourc. So, a larg inductanc is prsnt on sourc sid which stabiliss th currnt in th sourc. As, th sourc is currnt stiff lmnt load has to b voltag stiff lik capacitor. So, inductiv loads ar not prfrabl. Hnc Statmnt-II validats Statmnt-I.

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