ESE (Prelims) - Offline Test Series ELECTRICAL ENGINEERING SUBJECT: Electrical Machines & Systems and Signal Processing SOLUTIONS

Transcription

1 TEST ID: 30 ESE- 09 (Plims) - Offlin Tst Sis ELECTRICAL ENGINEERING Tst-3 SUBJECT: Elctical Machins & Systms and Signal Pocssing SOLUTIONS 0. Ans: (c) Sol: With hot-olld stl laminations max [B m ] A. A A max. With CRGO laminations max.6 A A max.6 wight of thcrgo (W ) Wight of th hot olld co W Co volum Dnsity Co volum W W % savings 00 W % Ans: (d) Sol: Th cofficint of coupling of th tansfom can b incasd by. Incasing th window hight.. Aanging th pimay and sconday windings concntically 3. Sandwiching th pimay and sconday windings 4. Using shll typ constuction A (Hight of lim b) A Hight of lim b max.6 max. 3 W W Ans: (c) Sol: 44 A 500 V + 40 A V L O 4 A A 750 V D

2 : : Elctical Engining I L atd A I H atd 4 A 500 kva auto tansfom (40 + 4) 00 k + 0 k 0 kva kva tansfomd kva kva conductd 0 K 0 K 00 kva 04. Ans: (b) 000 Sol: Tansfomation atio K Voltag Rgulation of auto tansfom ( K) voltag gulation of two winging tansfom 05. Ans: (b) ( 0.8) 0 % Sol:. Th quivalnt lakag impdanc in ohms should b invsly popotional to thi spctiv kva atings.. Th tansfom with gat lakag 06. Ans: (a) impdanc angl opats at low pf as compad to oths with low lakag impdanc angl. 0K Sol: Full load cunt A Ohmic losss at full load cunt P SC watts Th voltag applid to HV winding to gt 4 4 A cunt though it V 3 Maximum pcntag gulation 80 Pcntag Z % 5 Pcntag sistanc 07. Ans: (a) Sol: X 0.5 Pi P Cu % Efficincy Full load copp losss 00 kva ating K 0.8% X kva ating pf 00 X kva ating pf losss 0.5 0k k

3 : 3 : ESE - 09 (Plims) Offlin Tst Ans: (d) Sol: P i P h + P 4k % 4.k K f + K f P i K + K f f K K 50.. () 40 K + K 60.. () 60 () () 0K P i K (40) + K (40) 00 W 09. Ans: (d) K 0 and K Sol: Th following a ffcts of amatu action. Distotion of main fild flux. Incasd ion losss 3. Dlayd commutation 4. cost of th fild winding incass 0. Ans: (b) Sol: I I a f I a I L Vt R I f f 50 A A Total AT p pol Z P Ia A P m Dmagntizing AT p pol. Ans: (b) Total AT p pol Sol: Th following a asons to faild to build- up of voltag at no load in shunt gnato () High fild cicuit sistanc () Fild connction vsd (3) No sidual magntism (4) Spd is Lss than citical spd.. Ans: (a) Sol: Gnating Mod I a I L + I f 48+50A E g V t + I a R a V Motoing Mod I a I L I f 5 50A

4 : 4 : Elctical Engining E b V t I a R a V N m N g E E 3. Ans: (c) b g Sol: Citical fild sistanc is dictly popositional to spd X Ans: (a) Sol: Cas X 600 pm I a I L I f E g V t I a R a T constant K a I a I a constant Cas II Eg E g N N N Eg Eg N E g V t I a (R a +R xt ) ( 0.5+R xt ) 50 R xt V 5. Ans: (c) Sol: At No load Input pow W constant losss + No load copp loss I a0 I L I F 6 4A No load copp losss (I a0 ) R a W Constant loss At Full Load Input pow (50+) KW I a 50A Load Copp Losss I a R a W Total Losss Ans: (c) 74 W P out 3k W 058 Efficincy % Sol: P m E b I a (V t I a R a ) I a P I m a 0

5 : 5 : ESE - 09 (Plims) Offlin Tst-3 V t I a R a 0 V I a R a t 5 3. Th ngy stod in th mchanical systms E b V t I a R a V 7. Ans: (d) Sol: chaactistics of th svo moto. Toqu spd chaactistics a lina. High toqu/intia atio 3. Fast spons 4. Quick vsal 5. Cannot ovloadd 8. Ans: (c) Sol: Stp angl 9. Ans: (d) 360 Numb of stps p volution Sol: In th actual pocss of convting lctical ngy to mchanical ngy (o vic vsa) is indpndnt of. Th loss of ngy in ith th lctical o mchanical systms. Th ngy stod in lctic o magntic filds which a not in common to both systms. 0. Ans: (a) Sol: y b C K coil sids p pol K odd P C numb of coil sids P numb of pols K numb (intg o faction) y b 40 K 6 40 K y b y f y f 3 fo pogssiv y f 3+ 5 fo togssiv. Ans: (b) Sol: Th p phas quivalnt cicuit and phaso diagam cosponding to th poblm a shown in figs. and spctivly. E (EI ) + I jx s f jx s E V IjX s Fig. V + I jx c

6 : 6 : Elctical Engining Assum that th tminal voltag V is always kpt constant, by changing th fild xcitation whn ncssay. With I 0 (X c ), th inducd mf du to th fild, E, quals V. With I 0, phaso diagam of fig. indicats that th inducd mf du to th fild, E, dcass to lss than V. Thfo, fild xcitation must hav dcasd. But th voltag acoss th load must still b V. So th diffnc (V-E) must hav bn inducd by th amatu mmf. Hnc amatu mmf aids th fild mmf. This is calld magntizing action of th amatu action.. Ans: (c) Sol:. Shot-cicuit atio: Th shot-cicuit atio of a 3-phas altnato is dfind as fild xcitation fo atd voltag V on OC sc fild xcitation fo atd cunti onsc Ifoc I fsc I V jx c E Fig. IjX s V Sinc it is th atio of two fild cunts, it is dimnsionlss.. Synchonous actanc: Fo som fild cunt I f, opn-cicuit voltag V I f Ifoc (assuming a lina OCC). Fo th sam fild cunt, shot-cicuit cunt I f I. Ifsc (SSC is always naly lina). X s (unsatuatd) sc X / V s Xsin pu / I V If I I I I foc V I fsc f Fom th givn data, X s in p unit 7.6 sc 3.4 sc 3 7.6

7 : 7 : ESE - 09 (Plims) Offlin Tst-3

8 3. Ans: (c) Sol: V and Invtd V cuvs: I a, p.f 0 Q-absobs Q-dlivs UPF Lading p.f Lagging p.f U.E N.E O.E Fig. I a pf I f : 8 : Elctical Engining 5. Ans: (a) Sol: Diction of fild cunt I f on th oto winding shows (by ight hand ul) that th fild pols a as in fig blow. oto N s s s N f I f Sf s s S s Fig. A synchonous gnato is opating with lading pow facto, if fild cunt incass, thn amatu cunt will dcass to minimum valu and thn incass 4. Ans: (a) Sol: X s.5, R 0 At no load, 0 and E V. P sy 3EV cos Xs P m 3V P X s 6 6 ad kw. Th oto pols do otat at s /s w..t stato. (Thy a, howv, stationay w..t stato pols). (i) is tu. N s of stato and N f of oto pl ach oth. Hnc th dvlopd toqu is in acw diction. Hnc th oto is bing divn by a pim mov a against this dvlopd toqu. This psnts gnato action. (ii) is fals. 6. Ans: (c) Sol: i. VI cos input to th moto I copp losss W hav VI cos I P m. shaft output of moto + mchanical and ion losss in th moto. Not: This is also calld th mchanical pow dvlopd by moto. Mchanical

9 : 9 : ESE - 09 (Plims) Offlin Tst-3 pow dvlopd dos not includ copp losss. ii. Dividing both sids of th quation by (sistanc/ph) and witing I I cos + I sin, w gt, v Icos I cos I sin P Multiply both sids by ( ); and add V to both sids; 4 V V V Pm, 4 4 Icos Icos Isin Lt I cos x, I sin y. Thn locus of th point B(x,y) fo a constant P m is a v cicl with cnt at x, y 0 and adius of v 4 Pm. (This is calld a pow cicl). 7. Ans: (a) Sol: Pow dvlopd in synchonous machin EV P.sin X P P max.sin Pmax s EV X s P max E, V P max I f and V [ E I f ] m 8. Ans: (d) Sol: Fo th gnato, w hav E IjX V. s i. Phaso diagam fo a puly capacitiv load: Fom th diagam, w can wit V X s I + E. V I cuv is a staight lin, with a slop X s, and an intcpt on th V axis of E. (E th inducd mf is constant sinc xcitation and spd a constant). This cosponds to cuv. (whn I 0, V E. As I incass, V incass). (ii) Phaso diagam fo a puly inductiv load: Fom th diagam, w hav V I X s E VI cuv is a staight lin with slop ( X s ), and I achs a maximum of (shot-cicuit). I I E V IjX s E whn V 0 X s This cosponds cuv (whn I 0, V E. As I incass, V dcass). V IjX s E

10 : 0 : Elctical Engining 9. Ans: (c) Sol:. Initial opation as synchonous moto: Pow civd by th moto/phas P EV V sin sin Xd Xq X d Synchonous pow luctan c pow Mchanical losss of th machin. Sinc th mchanical losss a likly to b small compad to th atd pow of th machin, will b vy small.. Excitation ducd to zo: E bcoms zo (sidual magntism of oto pols is nglctd). But th luctanc pow is not zo. (It dos not dpnd on E). will now incas till th luctanc pow just supplis th mchanical losss. Th machin is likly to un as a luctanc moto at synchonous spd. [Not: is th angl btwn E & V. Whn E is zo, th qustion aiss as to how is to b dfind. W can tak it as th angl btwn th stato mmf axis and th oto axis]. 30. Ans: (b) Sol: Ods of th slot hamonics of stato (o (No.of slots) oto) a givn by. No.of pols H, numb of slots of th stato ; Thus slot hamonics of stato mmf a of ods 7 and 9. (Th 9 th hamonic obtaind by using th positiv sign, otat at N s in th sam diction as th 9 fundamntal componnt of stato mmf, i.., in clockwis diction. Th hamonic obtaind by using th ngativ sign otats in th opposit diction i., anti-clockwis). 3. Ans: (c) Sol:. Sinc th oto has 40 slots, slot hamonics of th oto mmf a of ods 40 and 9. 4 Th 9 th hamonic (which is obtaind by using th ngativ sign) otats at 9 (synchonous spd cosponding to fquncy of oto cunts in th anticlockwis diction wt th oto).. If th stato supply fquncy is f, fquncy of oto cunts is Ns N f. N Fo a fquncy f, th synchonous spd is N s in clockwis. Fo a fquncy Ns N f, th synchonous spd N s s

11 : : ESE - 09 (Plims) Offlin Tst-3 cosponding to oto cunts is (N s N) (cw). 3. Thus spd of th 9 th hamonic of oto 3. Ans: (d) mmf wt oto anticlockwis diction. N s N, in th 9 Sol:. Th nomal spd dop du to incas in load is not calld cawling. Stady unning at a spd much small than th synchonous spd (usually at about 7 th th synchonous spd) is calld cawling. Statmnt is fals.. Cawling can b poducd by both hamonic induction toqus and hamonic synchonous toqus. Statmnt is tu. 3. If th cawling is du to hamonic induction toqu, spd can vay with load. If it is du to hamonic synchonous toqu, spd will main constant. Statmnt (3) is fals. 4. With oto slots stato slots, two hamonic synchonous toqus of naly qual magnitud and opposit dictions can b poducd, which can lad to cogging. Statmnt (4) is tu. 33. Ans: (d) Sol: Slip of th machin is gat than on mans induction machin is woking in baking mod. In this mod, th machin will daw lctical pow fom th supply mains. 34. Ans: (c) Sol: T T st FL I I st FL s FL Fom data T st T FL ; I I st FL Ist 4I fl 35. Ans: (d) Sol: As th oto sistanc incas, hnc total cicuit sistanc incasd. So stato cunt will dcas. Fo constant toqu load, slip R As R incass, slip also incass. 36. Ans: (d) Sol: Suitability of th chaactistics of th moto fo a givn application, cost, and siz a som of th impotant considations in slcting a moto fo a givn application. Th usual pactic is listd blow: Domstic pumps us -phas induction motos. Stpp motos a usd in comput

12 : : Elctical Engining numical contol. Univsal motos a usd in a hand-dilling machins. Bushlss dc motos a usd in fast-acting svo divs. Thus, th coct paiing would b A- 4, B-3, C-, D-. [Usually x is substantially lag than in induction machins and hnc is nglctd in th dnominato of ()]. Thus a high stating toqu is obtaind by incasing th oto sistanc. 39. Ans: (b) 37. Ans: (c) Sol: y (t) t x( ) h(t ) d Sol: X m, th magntizing actanc p phas of an induction moto flux linkag / phas du to aigap flux. phas cunt Magntizing actanc can b shown to b popotional to N R g a. Wh N th numb of tuns/phas, R th adius of th stato bo, a th axial lngth, and X() h() h(t) g th adial lngth of th ai-gap. Fo a machin of givn atings, N R a can b considd appoximatly constant. Hnc X m is influncd most by g. t < 0 t t dt y (t) t t t t 38. Ans: (a) Sol: In a 3 ph induction moto, T st Stating toqu 3V x s 3V s x... () 40. Ans: (c) Sol: y(0) x(0) h(0) h(0) h(0) y() x(0) h() + x() h(0) h() + 0 h(0) h() h()

13 : 3 : ESE - 09 (Plims) Offlin Tst-3

14 : 4 : Elctical Engining 4. Ans: (d) Sol:. Th pow signals has finit pow and infinit ngy. Th ngy signal has finit ngy and zo pow 3. All piodic signals a pow signals 4. All infinit duation signals a may not 4. Ans: (c) pow signals Sol:. x (t) y (t) t x (t) + x (t) y (t) t x (t) + x 3 (t) a x (t) + b x (t) y 3 (t) t(a x (t)+b x (t)) + ay (t)+ by (t) Hnc non-lina. y(t t 0 ) (t t 0 ) x(t t 0 )+ 43. Ans: (d) Sol: Givn x (t) x(t t 0 ) y (t) t x(t t 0 ) + y(t t 0 ) Hnc tim vaiant y (n) y ( ) n k k x(k) x(k) Psnt output dpnds on futu input. So, systm is non-causal. Fo a boundd input u(n) systm poducs unboundd output(amp). So, systm is unstabl. 44. Ans: (b) Sol: 4 ( at b) a b t a cos( t) (t )dt 4 Fom SIFTING popty t cos( t) t dt x(t) t t 0 dt x(t 0 ) ; t t 0 t t cos(t) (t )dt Ans: (c) 0 othwis cos( t) cos( ) t Sol: A signal is calld ngy signal if ngy is finit and pow is zo. A signal is calld pow signal if pow is finit and ngy is infinit. In th givn statmnt, and 3 a tu

15 : 5 : ESE - 09 (Plims) Offlin Tst-3 But statmnt (4) is fals. Bcaus t u (t )u( t) ct 4 is an ngy signal. K jk sin K ;K 0 ;K Ans: (a) cosn Sol: Givn Cn. (n) C n cosn C (n) C n C n. So, x( t) x(t) So, x(t) is vn signal. 47. Ans: (b) Sol: Givn signal is odd signal hnc contains only sin tms 48. Ans: (d) Sol:. If th signal is al thn C K n C K. If th signal is vn thn C K C K 3. If th signal is odd thn C K C K 4. Avag pow of th signal C K K 49. Ans: (c) Sol: x(t) C K jk x(t ) C K x(t ) C K 50. Ans: (c) Sol: A. R{x(n)} X ( j ) B. jimg{x(n)} X 0 ( j ) C. x (n) R{X( j )} D. x 0 (n) jimg{x( j )} 5. Ans: (a) Sol: Foui tansfom pai is t Invs Foui tansfom is x(t) x(0) X( ) X j t ( ) d d Assum X(), x(t) t d x(0) 5. Ans: (a) d 0 Sol: Foui tansfom of impuls signal is Constant. Foui tansfom of ctangula function is Sinc function

16 : 6 : Elctical Engining Foui tansfom of constant is impuls function. Foui tansfom of Gaussian puls is also Gaussian puls. 53. Ans: (c) Sol: A systm is calld minimum phas systm, if all pols and zos a lis on th lft sid of s-plan. A systm is calld stabl if all pols lis on th lft sid of s-plan. So, Minimum phas systms a always stabl and hav smallst goup dlay. So, statmnt is tu. A systm is calld mixd phas systm, if on o mo zos lis on th ight sid of s-plan. So, statmnt is fals. A systm is calld maximum phas systm, if all zos a lis on th ight sid of s- plan. So, statmnt 3 is tu Fo systm to b both causal & stabl, all pols hav ngativ al pats in th s-plan. So, statmnt 4 is tu 54. Ans: (a) Sol: Xj T T x(t) jt jt dt dt j jt T j sin 55. Ans: (b) T T T sin T T T sin c Sol: Givn signal is th piodic signal and Foui sis cofficint C and C k 0 Fo K 0 0. Fo th piodic signal Foui tansfom K X(j) C K 56. Ans: (c) ( 0 ) Sol: Using pasval s thom X 57. Ans: (d) Sol: x(n) K d x t (n) 3(n 4) 0 dt n (n 3) y(n)

17 : 7 : ESE - 09 (Plims) Offlin Tst-3 () Numb of buttflis in ach stag x(n) n h(n) (n) 3(n 4) (n 3) h(n) 3 h(3) (0) 3 h(3) Ans: (a) n (n 3) 3 ( 4) n 4 (n 7) Sol: A. x(n) {,,, }. It is a finit duation both sidd squnc. So, ROC is 0 < z <. B. It is a finit duation ight sidd squnc, So, ROC is z > 0. C. It is a finit duation lft sidd squnc, So, ROC is z <. D. Th squnc is x(n) (n) and th z- tansfom qual to and ROC is nti z-plan. 59. Ans: (a) (n)+3(n 4) Sol: () Numb of stags in flow gaph a log N. So, statmnt () is wong n (n 3) y(n) a N. (3) Numb of complx multiplications a N log N. (4) Numb of complx additions a Nlog N. (5) Inputs a bit vsal od, outputs 60. Ans: (b) a nomal od. So, statmnt (5) is fals Sol: Th is a non-lina lation ship btwn analog and digital fquncy in bilina tansfomation mthod. So, statmnt () is fals, maining all a tu. 6. Ans: (b) Sol: L[x(t)] T 0 x(t) St ST St 0 4S dt dt S 4S S S S

18 : 8 : Elctical Engining 6. Ans: (a) Sol: Givn cicuit is Apply Foui tansfom to abov cicuit. Thn Y X j Y H ( ) X j Givn x(t) t -t u(t) X( ) ( j) Y( ) X( )H( ) ( j) t u n at x(t) + y(t) t t 6. Ans: (b) t n! n j a u(t) Sol: Fom multiplication of xponntial in tim domain popty. x(n) + X() X(z) H j + y(t) + Y() 3 n (a) x(n) X(z / a) n (j) x(n) X(z / j) Fom th givn data k k X(z) 3j 3j z z z z k X j z j 9 4 k z k 9 z 4 z 9 Pols of X a z 0, z j Ans: (c) jn jn Sol: S : u(n) So, Fo an LTI systm if input is thn output is of th fom S is not LTI systm S : S 3 : n j 3 j n AH ( 3 j 0n A j0 0 ) j n, input and output fquncis a not sam, so it is not LTI systm. n j 5 j n j n, input and output fquncis a sam, so it is an LTI systm. 65. Ans: (b) Sol: Assum Y(k) X (k) Apply IDFT.

19 : 9 : ESE - 09 (Plims) Offlin Tst-3 y(n) x(n) cicula convolution x(n) y (n) y(n) {,,, } 66. Ans: (a) 0 0 Sol: Th fild winding of a singl-phas synchonous machin (acting as a moto o as a gnato) civs a dc xcitation cunt, just lik a 3-phas synchonous machin. Whn th oto is otating, a singl-phas voltag is inducd in th singl phas winding psnt. Th phas winding thn dlivs an ac cunt (gnato action) o civs an ac cunt (moto action). In ith cas, th phas cunt flowing though th winding poducs two otating mmfs, otating in opposit dictions at synchonous spd (just lik in a singl phas induction moto). Th oto itslf is otating at synchonous spd. Th mmf componnt otating in th sam diction as th oto is stationay w..t th oto and dos not induc any voltag in th fild winding. Th mmf componnt otating in opposit diction to th oto howv, causs a doubl fquncy ac componnt of cunt in th fild. Both statmnts a coct. Statmnt-II xplains statmnt-i. 67. Ans: (c) Sol: Givn induction machin with a 4-pol stato and a -pol wound oto. Sinc th oto winding is wound fo pols, ach oto phas can b psntd by on full-pitchd coil with a coil span of 80 o mch (which is th sam as 80 o lc fo pols). Diction of mf inducd in coil-sid is givn by that of (a z a y ) a x (fom Lontz's foc law). Diction of mf inducd in coil-sid is givn by that of (a z a y ) a x again. s N a y oto -a z S Ths two mfs cancl ach oth and th is no cunt. This is tu at vy instant fo vy oto phas and so th is no toqu dvlopd, and th oto dos not otat. a z N -a y

20 : 0 : Elctical Engining Howv, if th oto is placd by a cag oto, th will b nomal otation sinc cag winding automatically adjusts itslf to hav th sam numb of pols as th oto. 'Statmnt-I' is tu and 'Statmnt-II' is claly fals. 68. Ans: (b) Sol: Statmnt I: 3 point stat is not usd fo th abov atd spd contol of th DC shunt moto Statmnt II: 3 point stat is usd to potct th DC shunt moto fom accidntal opn cicuit of fild winding Statmnt II is not coct xplanation of statmnt I 69. Ans: (a) Sol: DC sis moto should not opat at light and no load condition, bcaus sis moto und no load condition will opat dangously at vy High spd. 70. Ans: (a) Sol: Statmnt I: Concntic and intlaving winding will b usd in th constuction of th co typ tansfom Statmnt II: By using th concntic and intlaving winding w can duc th lakag flux, amount of th insulation quid and amount of th copp quid. Both statmnts a coct and statmnt II is coct xplanation of statmnt I. 7. Ans: (b) Sol: Statmnt I: Th spd of th slf xcitd DC shunt moto is almost constant by changing th tminal voltag. Bcaus flux is popotional to tminal voltag. Statmnt II: Th spd of th DC shunt moto can b incasd by incasing th fild cicuit sistanc. Bcaus flux dcass, back mf is almost constant hnc spd incass. 7. Ans: (a) Sol: Statmnt I: In th paalll opation of th tansfoms, th tansfom which has low pu impdanc on th common bas will sha mo amount of th load. Statmnt II: In th paalll opation th impdanc of th tansfom should b invsly popotional to kva ating of tansfom. Both statmnts a coct and statmnt II is coct xplanation of statmnt I. 73. Ans: (b) Sol: Both a diichlt conditions fo xistnc of Foui sis. But th is no lation btwn ach oth.

21 : : ESE - 09 (Plims) Offlin Tst Ans: (c) Sol: FIR filts a always stabl, bcaus all pols lis insid th unit cicl. So, Statmnt (I) is tu. FIR filts hav lina phas only whn it is symmtic/anti symmtic. i., h(n) h(nn) o h(n) h(nn), So, Statmnt (II) is fals. 75. Ans: (b) Sol: Sampling in on domain maks th signal to b piodic in th oth domain. So, statmnt I is coct. Accoding to multiplication in tim domain popty, multiplication in on domain is th convolution in th oth domain. So, statmnt II is coct.

22 : : Elctical Engining