Electrical Machines. 1. Transformers 800 V. As the slope is uniform the induced voltage is a square wave. 01. Ans: (b) Sol: 400/200 V 50 Hz

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2 lectricl Mchine. Trnformer 0. An: (b) Sol: 400/00 50 Hz B mx. T 800, 50 Hz liner dimenion ll double B mx? l l b b A l b A 4A B 800 B 400. mx B mx mx 4A A A A f f. B mx. T 4 0. An: (c) 40 Sol: b c.m 40 A net m MF TU An: (d) di Sol: nduced emf M dt di (where i the lope of the wveform) dt A the lope i uniform the induced voltge i qure wve pek voltge 800 ote: A given trnformer i : T/F, the induced voltge on both primry & econdry i me 04. An: () Sol: i(t) 0 in (00 t) A di nduced emf on econdry M dt co(00t) 400 co (00t) 400 in (00t ) When S i cloed, the me induced voltge pper cro the eitive lod pek voltge cro A & B An: () Sol: d dt (where e pq ) e pq 0 (Between 0 & 0.06) e pq 90 (Between 0. & 0.) AC ngineering Publiction

3 : : Potl Coching Solution 06. An: (c) Sol: Core lo core volume 400 W W 6788 W 0. A So w w w w W A ( w i core lo component) eluctnce l A 000 m 4.44 f 000 m 4.44 f contnt; f contnt m m m m mmf e luc tn ce ( i the mgnetizing current of the trnformer) 0 w (.) (0.67).9 A 4.44A 07. An: (b) Sol: ( 0.67) 4.48 A 4.5 A k 0. W 0 0 co 0 W0 W A 400 w o co o (4.44) co , nd in co ( o ) (0.64) 4 ( co(6.0) ( K A ) 4.58ª Power fctor; 4.58 co 0.9 co 6.86 p.f co 0.76 lg k Q 0 w AC ngineering Publiction

4 : 4 : lectricl Mchine 08. An: (c) Sol: Z T (0.8j0.4) nd Z L (4j) line Z Z T L A oltge t the lod, lod ( ) (56.86) And power lo in tr.line ( line ) 0.8 (90.86) W 09. An: (b) Sol: 00, 60Hz, Wh 50W, Wh? W e 90W W e? f W W h h f.6 Wh f f W h When rtio i not contnt f W e v W W W e e e W W i W h W e W 0. An: () Sol: 440 ; f 50Hz ; W i 500 W 0, f 5Hz W i 850 W f f W i Af Bf 500 A 50 B 50 () 850 A 5 B 5 () By olving () & () A 8 ; B 0.64 W e Bf W W h Af W. An: (b) Wi Sol: Wh ; Wi We W W h h Wh Wh W h W h 0.4 W i W W e e Wi W e 0.8 W e 0.8 W e 0.40 W i W i W h W e 0.4 W i 0.40 W i W i 0.8 W i eduction in iron lo i round i.e. 7.% reduction. An: () Sol: At 50 Hz; Given, P cu.6%, P h 0.9%, P e 0.6% AC ngineering Publiction

5 We know tht, P h f -0.6 P P h f 60 h f 50 AC ngineering Publiction Ph %.5 ddy current lo contnt, (ince P e ) nd given totl loe remin ome. P h P P P P P cu e h cu e.% 0.806% Pcu Pcu.694 % 0.6% Pcu i directly proportionl to P P cu cu.08 Output ka.08. An: (d) Sol: 0 ka 0 0/0 50Hz o lod t rted voltge i,e W 0 60Wtt co % % %X % input power output Power Totl lo of power FLcu lo %. %FL cu lo 00 Arting FL cu lo % A rting 0.0 0, Wtt F A rting 0, K lod 85A At 90.9A cu lo 00 W 90.9A : 5 : Potl Coching Solution 85At cu lo? cu lo t 85A Wtt 90.9 Totl lo when 4.96 kw o/p ron lo cu lo t 85A W nput power 4.96 kw4.8w 594.8W 4. An: () Sol: Z 6 Z Z 0.5 X 0 Z co X Z in ectnce t f 5Hz X 5 X 50 X 0.6 Z X Z 0.8 Z (0.06) (0.6) 56.78A 56.65A 0.06 p.f co c lg Z An: () H voltge Sol: L voltge , c

6 : 6 : lectricl Mchine X m f (M) j A 6. An: (d) Sol: Given tht, no lod lo component re eqully divided W h W e 0W nitilly tet i conducted on L ide v 00 ow rtio i f 50 n H ide, pplied voltge i 60; thi voltge on L ide i equl to 80. ow f rtio i contnt, W h f nd W e f. W h W h f f W f 40 We We 0 f 6.4 W 50 Therefore, W W h W e W n SC tet, (H ide) 5A nd lo 5W Current in L ide i k 5 i.e 0A For 0A 5 wtt 5 A? W c 4000 W c W 0 c X m 7. An: (b) ka co Sol: ka co W W Cu W i 60W W Cu A 400 W Cu W 6 W i W Cu 0 W 4K % 00 4K % 8. An: (c) Sol: 98% Let tke ka p.u p.f t full lod : 0.98 W i W Cu W i W Cu () For / full lod W 0.5 i W Cu W i 0.5 W Cu 0.00 () By olving eqution () & () W i W Cu W i 0.5W Cu 0.00 W i ; W Cu 0.06 / (0.75) 98.% 0.06 AC ngineering Publiction

7 : 7 : Potl Coching Solution 9. An: () Sol: mx % lod t which mximum efficiency poible W i Cu 0.06 W ( % mx 0. An: (d) Sol: 0 ka,500/50 OC A, 50W SC 60 A, 45W ron loe 50 W W 0000 ( 4) 4A (rted current) 500 cu lo t A 45W cu lo t 4A? W 9 ka t mx ron lo ka cu lo FL 00 ) 50 0 ka 7.9 ka 80. An: (c) Sol: mx % (50). An: (c) Sol: 0.5 ; 0.04 ron Lo 40 W 0 mx. An: (c) Sol: K ron lo A k k W i W i 5.06W output in kwh lldy output kwh loe kw ka co kw 0 0 kw kwh output 0 40 kw W i kw W Cu S 0 WCu W Cu 7.06 Trnformer i O lod for 0 to hr o W Cu W 40 0 lldy % ll dy 97.9% 97.% An: (*) Sol: Given ron lo.5 kw, co 0.85 Find equivlent reitnce 0 on H. ide k k AC ngineering Publiction

8 : 8 : lectricl Mchine 000 Full lod current on H. ide A Full lod cu lo (9.09) kw fficiency % 5. An: (c) Sol: 00/400v, 500 ka, mx 98% 80% of full lod UPF % Z 4.5% PF mx For min. econdry 0% % %Z ron Lo ron lo W cu lo t 80 % of FL (.8) cu lo of FL FL cu lo W % % FL cu lo FLcu lo A ting % PF mx. % %Z 6. An: (b) Sol: Terminl voltge? % X %Z % ( 4.5) (.7) 4.7% 0.8lg (.7 0.8) ( ) % 4.49% Pu Totl voltge drop on econdry ide PU oltge drop An: () Sol: 0 X 0 X X K (eitnce referred to econdry ide) X k X.4 (0.0 7.) X co X in % eg.7 A eg eg 0.0 % eg.% i me on both ide full voltge 0.0 full Lod 9.6 The voltge pplied cro terminl. % % co %Xin AC ngineering Publiction

9 : 9 : Potl Coching Solution 8. An: (b) Sol: 6600/440 p.u. 0.0 pu p.u.x 0.05 pu 6600 pu % co % Xin % pu oltge drop when with repect to econdry p.u. econdry oltge Terminl voltge An: (b) Sol: f voltge re not nominl vlue % eg will be zero Pu co X pu in 0 tn - (/X).80 p.f co co (.80) 0.98 led 0. An: (c) Sol: pu Convert thi in volt z Turn rtio An: (c) Sol: P co co From given dt, (0.5 j5)68.64 t 0 t efer L ide t t An: (*) Sol: 40 j50 0. j0.5 pu 0.0 X pu , 0.8 lg Tke rted current pu Drop (z) 6.86 (0.0 j0.05) 00 4/.4k P 90kW 0.8pf The equivlent circuit refer to L. ide i AC ngineering Publiction

10 : 0 : lectricl Mchine j quivlent circuit refer to H. ide i 80 j k t ZL 87.4 j A Where voltge pplied cro the trnformer. (0. co 0.5 in) [ ) % egultion %. An: 96.7% Sol: copper loe (.8 0.) (48.9) W 900 % % 4. An: 8.8 Sol: j00 0 X 0 50k 70k r Z L.j.5 5kA nd 8000/0 Z L Trnformer impednce 0 jx A Z L (.788.5) ( ) ow t An: 4.9% t Sol: oltge regultion % 6. An: (*) Sol: Given dt, f 60 Hz, 0 ka, 4000 /0, Z pu 0.04 pu, pu, W 0 00 W, W cu 80 W P 0 0 kw & co 0.8lg AC ngineering Publiction

11 : : Potl Coching Solution 00 Lod current 08. A ted lod current 50 A 0 The copper loe for 08. A i wtt fficiency % The equivlent circuit wrt primry i S Primry rted current 00 P 7.5 A 4000 Given cu loe 80 W P (7.5) Given, Z pu 0.04 Z X (k) MA Z Z 0 X ka 4 k/ Lod current wrt primry i A kw 0.8 pf ecery primry voltge cox in S C A A B C B [ ] An: (b) Sol: c The Poible Connection i Yd 8. An: () 0.4 Sol: X P An: (d) k 0.4 c c b b b 50/ j AC ngineering Publiction

12 : : lectricl Mchine 4. An: (b) Sol: Turn rtio primryinduced voltge ec ondryinduced voltge ec ondryinduced phe voltge ter min lphe voltge % eg % eg % co % X in [Lgging Lod] % ph 0.08 Turn rtio ph ph An: () Sol: P o/p 50 hp kw P o/p of induction motor 6.77 kw P i/p to induction motor (or) power output of trnformer L P o / p kW P L co o 64 A ph 4. An: (c) Sol: A Y B An: (d) Sol: The flux linkge in phe b nd c winding i. Therefore induce voltge i lo become hlf 0 7 / / 0 0 /0 /0 64A ph KL: ph AC ngineering Publiction

13 45. An: (b) Sol: 0 0 Y 0 - blnced lod : : Potl Coching Solution z e z e 500 ( j0.09) j ( j0.08) Y i 0 lgging w.r.t (from ytem) Y 0 And An: () Sol: rted be.00 rted be.00 Under hort circuit, c z e c Since c rted ; z e (0.0)() Or z e 0.0 Short circuit pf co c 0.5, in c n complex nottion, z e 0.0(0.5 j0.968) ( j0.09) pu Similrly z e 0.04(0. j0.95) 0.0 j0.08 pu () When uing pu ytem, the vlue of z e nd z e hould be referred to the common be ka. Here the common be ka my be 00 ka. 500 ka or ny other uitble be ka. Chooing 500 ka be rbitrrily, we get S ka 0.8 S 700co From q. S S z e ze z (7006.9) o ka S (460 )(co6. o ) t pf co6. o lg 7 kw t pf of lg (Check. Totl power kw, lmot equl to 560 kw) 47. An: (d) 45 Sol: Current hred by trnformer 00.5 pu Trnformer i, therefore, overloded by.5%, i.e. 45 ka 460 Current hred by trnformer pu Trnformer i, therefore, under loded by 8%, i.e. 40 ka. oltge regultion, from q. (.40), i given by r co x in For trnformer, the voltge regultion t.5 pu current i.5 ( r co x co ) e o AC ngineering Publiction

14 : 4 : lectricl Mchine.5 ( ).5(0.049) Or Or ( )(400) And: (c) Sol: Here 60, 400 Ze f nd 480 Z e f Ze f Trnformer i loded firt to it rted cpcity, becue ze f h lowet mgnitude. Thu the gretet lod tht cn be put on thee trnformer without overloding ny one of them i, ka Z f f ka... e z ka e f Ze f Z ze f ka The totl lod operte t unity p.f. nd it i nerly true to y tht trnformer i lo operting t unity p.f. 49. An: (c) Sol: Secondry rted current Amp 6.6 Since trnformer i fully loded, it econdry crrie the rted current of 60.6 A. 05 For trnformer, re Full-lod voltge drop for trnformer, re co x e in A A (60.6) (0.85) () 0 50 Secondry terminl voltge An: () Sol: oltge rting of two winding trnformer 600 / 0, 5 KA voltge rting of uto trnformer 600 / 70 from the uto trnformer rting, cn y winding connected in erie dditive polrity. From two winding trnformer rted A rted A n AT, due to erie dditive polrity pry A ting of AT pry pry ka 5. An: (b) Sol: 00 A 0 A A ka 00 A 500 AC ngineering Publiction

15 : 5 : Potl Coching Solution The current through the lod of 050 ka t 500 i 00A 500 The current through the lod of 80 ka t 500 i The ka upplied ka The totl current tken from the upply 0,000 min i 40A 000 Auto trnformer ka rting kA An: () Sol: The rted current of h.v winding i 4 A. Therefore, the current drwn from the upply i 84A. ka trnformed (K) ka AT nd ka conducted ka. 5. An: (b) Sol: From bove olution, current tken by 80 ka lod i 0A 55. An: (d) Sol: An: (c) Sol: The two prt of the l.v. winding re firt connected in prllel nd then in erie with the hv. winding, o tht the output voltge i A 84 A 4 A 84 A 40 A 40 A A 80 A The rted current of l.v. winding i 0,000 40A 50 Totl output current i A Current through 480 winding i A ka rting of uto trnformer MA For two winding trnformer W 0 W 0.79 kw fficiency % 6 AC ngineering Publiction

16 : 6 : lectricl Mchine 56. An: () Sol: A M A rting for 60 lod i A Totllod A Primry current A 400 For reitive lod power fctor i t unity. d 58. An: (c) By eqution 74.69A A Sol: A kw A C 57. An: () Sol: 400v 50Hz 00 A D C c 40 d 0 0Ω 60Ω Lod current mmf j.5 Totl econdry mmf The voltge per turn 4 00 For 80 turn For 60 turn d 0 5. A 60 c 40 A 0 A rting fo 0 lod i 40 c A B 59. An: (b) Sol: Primry current A AC ngineering Publiction

17 : 7 : Potl Coching Solution Sec. mmf (500) [0 0 45] 000[0j 0] 000[ j 0] mmf Primry current A t 0.76 lg 60. An: (b) Sol: From power blnce co co co 0 : : ; 5 0 co co co o 4A p.f co(9.6) 0.77 lg co co co 6. An: () Sol: Given.6, L mh,.44m, f 60Hz, L 9H, c 60k, L m 450 H, P 0 kw, 0 nd co 0.85lg. X fl line X fl m The equivlent circuit i, 60k quivlent circuit referred to H. ide. 00 L 5.88A [.6 co ( )in].6 60k L [ ] nput power cn be clculted by dding loe to the output power. Cu loe:.6 L W Core loe: 600 j H.6 H L 4000/0 450H j Pc P % efficiency P loe 0.44m.6 j j9.55m 0.W 00 L L 00 0kW, 0.85pf 4000, 0kW, 0.85pf AC ngineering Publiction

18 : 8 : lectricl Mchine %. nduction Mchine An: (b) Sol: Given 500, A 00 cm m l 40 c.m m nd r 000 A nduc tn ce L 0 A r H An: (d) Sol: For motoring, the ttor pole nd rotor pole mut be equl. n the bove ce, the ttor winding re wound for 4 pole, where the rotor winding re wound for 6 pole. A the ttor pole nd rotor pole re unequl the torque developed i zero nd peed i zero. 0. An: 4% Sol: The frequency of generted emf by the lterntor i given Ppm 4500 f 50Hz 0 0 The ynchronou peed of nduction motor 4 0f P r % Slip rpm % An:(d) Sol: For 50 Hz, upply the poible ynchronou peed with different pole pole 000 rpm 4 pole 500 rpm 6 pole 000 rpm 8 pole 750 rpm 0 pole 600 rpm pole 500 rpm 0 pole 00 rpm We know tht, the rotor of n induction motor lwy trie to rotte with peed cloer to ynchronou peed, there fore the ynchronou peed cloer to 85 rpm for 50 Hz upply i 00 rpm nd pole re 0 pole. So it 0 pole induction motor 04. An: (d) Sol: Synchronou peed of field i, 0f P rpm 4 Ce (i): When the rotor i rotting in the field direction, AC ngineering Publiction

19 : 9 : Potl Coching Solution r Slip otor frequency f Hz. Ce(ii): When the rotor i rotting in oppoite direction of field. r Slip otor frequency f Hz. 05. An:(d) Sol: Synchronou Mchine: Prime mover peed, pm 0f P rpm The rotor peed of induction motor i fixed t 500 rpm. nduction Mchine: For obtining frequency of 50 Hz t induction motor rotor terminl the rotting field nd rotor mut run in oppoite direction Pin P in P P in P in 8 For obtining frequency of 50 Hz t induction motor rotor terminl the rotting field nd rotor mut run in me direction. in The induction mchine i in generting mode Pin P 500Pin in P in P in An: () Sol: P 4, f 50 Hz, 0.4, L 0 A nd P m 550 W Sttor copper loe /phe W Airgp power P r W nternl torque developed m 07. An: (c) Sol: Slip frequency f Hz Gro mechnicl power outut P G ( )P r W P r AC ngineering Publiction

20 : 0 : lectricl Mchine et mechnicl power output, P net W Pnet % efficiency Pinput % 08. An: (c) Sol: Given induced emf between the lip ring of n induction motor t tnd till (Line voltge), liring 00 For tr connected rotor winding, the induced emf per phe when the rotor i t tntnd till i given by lipring n generl, rotor current, neglecting ttor impednce i 0 X 0 For mller vlue of lip, >>>x0 Then the eqution for rotor current A An:.66 Sol: The ynchronou peed of the motor i 0f rpm P 6 Given, the rotor peed of induction motor, t mximum torque rtmx 940 rpm Therefore, per unit lip t mximum torque, rt mx Tmx We hve, lip t mximum torque i given by Tmx x From thi, x T mx 0. An: () Sol: Given rotor reitnce per phe 0. Stnd till rotor rectnce per phe X 0 7 We hve lip t mximum torque given by 0. Tmx 0.0 X 0 7 The ynchronou peed of the motor i 0f rpm P 4 otor peed t mximum torque i given by rtmx ( ) 500( 0.0) 455 rpm. An: 90 m Sol: T mx 50 -m otor peed t mximum torque, rtmx 660 rpm The ynchronou peed of the motor i 0f rpm P 8 Slip t mximum torque, rt mx Tmx AC ngineering Publiction

21 Operting lip 0.04 T We hve T T 0.6 T mx mx AC ngineering Publiction T mx T mx T m. An: 0.09 Sol: Given rotor reitnce per 0.05 Stnd till rotor rectnce per phe, X 0 0. We hve lip t mximum torque given by Let Tmx for T t 4 Tmx X t T mx TT mx T mx 4 8 T mx T mx 0 Solving for Tmx we hve Tmx ext ext. An: (b) Sol: The ynchronou peed of the motor i 0f rpm P 6 Given T mx 50 -m, lip t mximum torque Tmx 0. 0 ext, : : Potl Coching Solution Given, T mx Tmx Therefore, T mx k Tmx k Tmx T mx nd lo, T fl fl, T fl k fl Full lod net mechnicl power P net 0 kw Mechnicl loe P ml 600 W 0.6 kw P gmd P net P ml kw Pgmd 0.60 otor input, P ri ( ) ( ) T fl P ri 60 fl 0.60 ( ) ( ) fl ( fl) ( fl) Solving for fl, we hve fl rfl ( fl ) 000( ) rpm 4. An: (c) Sol: Given dt P 4, B 00 A, W B B 0 0 kw T t? At trting, otor input otor copper loe. 60 t B Here u rotor reitnce refer to primry ide of mchine 0 Given fl fl fl

22 : : lectricl Mchine t 60 B An: (c) Sol: 400A;k 0. 7 c t, up ply k c m A When tr delt trter i ued, T t time trting torque with DOL trter 5 75 m t time trting current with DOL trter 00 00A 6. An: () Sol: Strting AC ngineering Publiction line current with ttor winding in tr winding in delt Strting line current with ttor Strting line current with ttor winding in delt (DOL) Strting line current with ttor winding in tr 50 50ª 7. An: (d) Sol: Strting current with rted voltge, c 00 A Full lod current, fl 60 A The ynchronou peed of the motor i 0f rpm P 6 Given, the rotor peed of induction motor t full lod r fl 940 rpm Therefore, per unit lip t full lod, rf ST mx Full lod torque, T fl 50 m For DOL trter, we hve T t c 00 Sf T f f 60 T t m 8. An: (b) Sol: The ynchronou peed of the motor i 0f rpm P 4 Given, the rotor peed of induction motor r 440 rpm Therefore, per unit lip, r S The frequency of induced emf in the rotor winding due to negtive equence component i f n ( )f( 0.04)50 98 Hz. Synchronou Mchine 0. An: () Sol: The direction of rottion of conductor i oppoite to direction of rottion of rotor. So by pplying Fleming right hnd rule t conductor we cn get the direction of current.

23 0. An: (c) Sol: A the two lterntor re mechniclly coupled, both rotor hould run with me peed. 0f 0f p p AC ngineering Publiction f f p p p p p :p 0: very individul mgnet hould contin two pole, uch tht number of pole of ny mgnet lwy even number. G : p 0, f 50 Hz 600 rpm G : p, f 60 Hz 600 rpm (or) 0. An: (c) Sol: m lot/pole/phe Slot ngle P 80 0 K d K d m in n n min 0 in 0 in o o An: (b) Sol: Totl umber of conductor : : Potl Coching Solution P 00 0 f 50Hz umber of turn 540 ph (umber of turn (erie) (Phe)) P 80 0 Slot ngle, 0 S nd lot/pole/phe, m 0 in m Then, bredth fctor K b m in in 0 o in0 in 0 0. in0 95 o Hence Ph 4.44 k b f ph An: (d) Sol: For uniformly ditributed -phe lterntor the ditribution fctor m in( ) (K du ) m ( ) 80 Where phe pred m 80 0 for - lterntor in 90 K du The totl induced emf o of turnmf in ech turnk p K du

24 : 4 : lectricl Mchine T k p K du For fullpitched winding K p. T.7T volt 06. An: (b) 48 Sol: ; p 4 48 m lot / pole / phe Slot ngle 5; ( / p) phe pred m Winding fctor K w K p.k d () lot pitch 5 5 m 60 in in K d 5 m.in 4.in 8om7.5 K p co 5 co co (7.5) From eq (), K w co (7.5) 8 in(7.5) cot (7.5) An: (b) Sol: emf/conductor emf / turn 4 Totl turn T T Totl turn / phe For ytem m 60 m 60 Sin in K d m Totl induced mf o.of turn mf in ech turn per phe T K d 4 T 4 4 T 08. An: () Sol: 4 pole, 50 Hz, ynchronou genertor, 48 lot. For double lyer winding o. of coil o. of lot 48 Totl number of turn For -phe winding 480 Turn/phe 60 K p co co K d K d m in min , 60 in in ph 4.44K p K d ftph AC ngineering Publiction

25 : 5 : Potl Coching Solution ph ph L-L An: (c) Sol: ph k d T ph. ph() K K ph() K d() T ph( ) d() d().t.t ph() ph() m 90 in in min 6in ph() ph() ph( ) LL() ph() (Or) Method For phe connection T ph [ m 6] 4 K p 0.95; M (lot / pole / phe) 6 4 K d Sin (90 / ) 6Sin (5/ ) ph LL () Ph An: () Sol: To eliminte n th hrmonic the winding could be hort pitched by (80 0 /n). A the winding i hort pitched by 6 0 fifth hrmonic i eliminted.. An: (66) Sol: MF inductor - connection Kd Tp n 0.5 Kd Tp 0.5 n An: (404, 700 ) Sol: f turn re connected in two prllel pth then Turn/ph Turn / Ph / Pth ph L ph AC ngineering Publiction

26 : 6 : lectricl Mchine. An: (57, 808 ) Sol: f the turn re connected mong two prllel pth for two phe connection 480 Phe Turn/Ph Turn/Phe/Pth 0 Phe L L Phe L L An: (b) Sol: Min field i produced by ttor o it ttionry w.r.t ttor. For production of torque two field (Min field & rmture field) mut be ttionry w.r.t. ech other. So rotor (rmture) i rotting t. But per torque production principle two field mut be ttionry w.r.t ech other. So the rmture field will rotte in oppoite direction to rotor to mke. t peed zero w.r.t ttor flux. 5. An: (d) Sol: Field winding i n rotor, o min field o produced will rotte t w.r.t ttor. Field winding i rotting, field o produced due to thi lo rotte in the direction of rotor. Field produced i ttionry w.r.t. rotor. 7. An: (b) Sol: When tte or diconnected from the upply 0, 0 Without rmture flux, the ir gp flux r m 5mwb With rmture flux, the ir gp flux r m 0mwb So the rmture flux i cuing demgnetizing effect in motor. Hence the motor i operting with Leding power fctor. 8. An: (b) Sol: BD i the field current required to compente drop due to lekge rectnce. 0. An: () Sol: lod ngle in X tn co 5.97 o (0.6) (0.5) (0.8) 0 q o 7. o. An: (b) Sol: q co co(5.97) d in.in(5.97) co q d X d co(7.) 0.588(0) 0.808(0.8).60pu. An: (b) Sol: P.F UPF 0 X d. PU, X q.0 PU, 0 AC ngineering Publiction

27 : 7 : Potl Coching Solution PU, ka PU, PU in X tn co q o. An: () Sol: Given, P.5 MW, co 0.8, L 6.6 k nd 0. mx 96 X d 9.6 min 0 min 90 X q 6 5 ph L mx L P co L L 7.6A ph in X tn co q tn An: (c) Sol: Condition for zero voltge regultion i co ( ) P L Z Z (0.4 j5) Ph co( ) P.f led 5. An: (b) Sol: egultion will be mximum when 85.6 P.f co co(85.4) 0.08 Lg 6. An: (9%) Sol: Mximum poible regultion t rted condition i co in 0 X.9 0 ( ) ( ) % egultion % 7. An: 6.97% Sol: egultion t 0.9 p.f led t hlf rted condition i when AC ngineering Publiction

28 : 8 : lectricl Mchine ( ) ( ).8 % egultion % An: 75 Sol: Given dt, L 00 X 0 nd ph L S ph ph ph 5A 00 X nternl ngle, Tn -, S ka, 90 At mximum voltge regultion,. Therefore, 90 nd co 0. xcittion voltge i co in 0 X 0 50 % egultion % An: () 0 ( ) (00 5 0) Sol: Tht ynchrozing current will produce ynchronizing power. Which will demgnetize the M/C M nd Mgnetize the M/C M 0. An: () Sol: xcittion of M i increed, it nothing but mgnetizing the M. So ynchronizing power will come into picture, it will mgnetize the M/C M men lterntor operting under led p.f nd demgnetize the M/C M men lterntor operting under lgging p.f.. An: (b) Sol: ffect of chnge in tem input (xcittion i kept cont): ffect of chnge in tem input cue only chnge in it ctive power hring but no chnge in it rective power hring. Becue the ynchronizing power i only the ctive power. f the tem input of mchine incree Mchine Mchine ka ka kw kw ka ka p.f p.f Active power hring i depend on the Stem input nd lo depend on the turbine chrcteritic.. An: (b) Sol: xcittion of mchine i increed (Stem input i kept contnt): ffect of chnge in excittion cue only chnge in it rective power hring but no chrge in it ctive power hring, AC ngineering Publiction

29 : 9 : Potl Coching Solution becue the ynchronizing power i only the rective power. f the excittion of mchine incree 9. An: (b) Sol: 4% F(Hz) 5% Mchine Mchine kw kw ka ka ka ka P.f P.f. An: (d) Sol: At perfect ynchroniztion men both ytem h ll the chrcteritic imilr t tht point. o untbility fctor o there i no need for production of ynchronizing power. 4. An: (c) Sol: For ny chnge in field current there will be chnge in rective power of the mchine o there will be chnge in p.f of the mchine kw Without over loding ny one mchine. So here 00 kw i mximum cpcity of mchine. For M/C mximum lod. t cn ber i P P 0 kw Totl lod P P kw 40. An: () Sol: M/C re working t UPF now. For increed f from, inverted curve. We cn find tht there will be chnge in p.f of lterntor A from led to lg. Alterntor nd lgging p.f i over-excited. So it will deliver lgging A to the ytem. 8. An: (d) Sol: te of flickering bet frequency f f Hz 0. Flicker/ec filcker/min 4. An: (c) Sol: For ynchronizing n lterntor, the peed of lterntor need not be me lredy exiting lterntor. 4. An: () Sol: Synchronizing current per phe Z Z given Z Z AC ngineering Publiction

30 : 0 : lectricl Mchine nd mut be of phe quntitie Hz 44. Sol: y.7 y 6.98A. A lope Hz/MW x y xi 5.8 f D 5 f 50 Hz x fig (i) lope Hz/MW y mxc () f x 5.8 x 5 x x 0.8 () x x.8 () From eqution () & () x.6 x.8 MW x MW et frequency (f) x Hz (b) f lod i increed to MW x x.8 MW () x x 0.8 MW (4) From eqution () & (4) x 4.6 x. MW x.5 MW f x 5.8 B x xi C 5 50 Hz x x fig (ii) (c) in prt(b) totl lod x x.8 () t f 50 Hz lod hred by mchine() f x x x.8 MW x.8 x MW for mchine () f x c 50 0 c 50 c Sol: (i) Given dt: G : 00 MW, 4% G : 400 MW, 5% P x P 50x % 5% 00 MW P P 400 MW AC ngineering Publiction

31 : : Potl Coching Solution P x P 80x But, totl lod P P 600 MW () From () 50x 80x x Given, no-lod frequency 50 Hz preent ytem frequency f 50 (50 x %) Hz 00 (ii) Lod hred by M/C i nd M/C i. From bove olution we got x 4.65 P 50 x MW P 80 x MW Here P violte the unit. (iii)mximum lod the et cn upply without overloding ny Mchine i. From bove olution P violted the limit o tke P vlue reference P 00 MW From % egugrph find P P P 0 MW Totl lod P P MW et cn upply. 46. An: (c) Sol: Let power fctor i unity, M/C-A 40 MW nd M/C-B 60 MW m/c-a P 5 x P 5 x 60 5 P 5 x P 85 x 40 5 P P x5 x 80 x P 85 MW Mchine A j P 5 5-x 05 40MW 00 P 5 48MW 47. An: 0.74 Sol: Two prllel connected -, 50 Hz, k, tr-connected ynchronou mchine A & B re operting ynchronou condener. x j Mchine B 50 ka m/c-b L k 50 Hz The totl rective power upplied to the grid 50 MA in in 50 MA 5 P 60MW AC ngineering Publiction

32 : : lectricl Mchine in90 in90 50 ( only rective power pf co 0 90 o ) ( ) A 0 90 X X ,87.8 The rtio of excittion current of mchine A to mchine B i me the rtio of the excittion emf i.e., , An: (b) Sol: L k k ph t 00A, UPF, 0.Z o o 8.94 o xcittion increed by 5% Turbine input kept contnt P P in X in X in in(8.94) o An: () Sol: 0 Z o 90.4 A 50. An: (0.5 lg) Sol: p.f co(58.4) 0.5 lg 5. An: (d) 6600 Sol: X i in % P.U 5%; ph 80 X in i 0.5 Z b (6.6) 0.5 (.) (K) 0.5 MA 9.07 j X n lterntor By ubtituting the vlue P The current ( ) t which the p.f i unity ( 0) 0 ( co ) (in X ) b AC ngineering Publiction

33 5.76 A 4447 (680 0) ( An: (560.9) Sol: j X AC ngineering Publiction 9.07) 6.60 P Ph 80 ; A Ph Ph L P h An: (6.88) Sol: Power ngle (or) An: (b). Sol: P in 0.5 in XS j X X An: () Sol: From bove olution Anwer i An: (0.860 lg) Sol: From bove olution power fctor i p.f co co(0.69) lg 57. An: (0.96 PU) X Sol: ective power (Q) co : : Potl Coching Solution 0.8. co(7.9) 0.96 P.U 58. An: (.05 PU) Sol: The current t which mximum power output i Under mximum output condition Here 90 ( 0) 0 Z PU 59. An: (0.79 led) Sol: Power fctor t mximum power output i p.f co(7.56) 0.79 led 60. An: (.5 PU) Sol: rective power t mximum Q X co Subtitute 90 Q 0.8.co(90).5P. U 6. An:.4 to 4.0 Sol: A non lient pole ynchronou genertor X 0.8 pu, P.0 pu, UPF.pu, 0 P co. 0.9 pu The voltge behind the ynchronou rectnce i.e Z

34 : 4 : lectricl Mchine. j An: Sol: f.pu, X.pu, P 0.6pu,.0pu P Q X S. in 0.6 in. 0.5 X. co. co pu 6. An: () Sol: Motor input L L co W given motor i lo le lectricl power converted to mechnicl power Motor input output W 0f rpm P 4 P T 0.5 m co 65. An: () Sol: We know tht, ynchronou motor lwy rotte only t ynchronou peed but induction motor cn rotte t more or le thn the ynchronou peed. Conider peed of nduction motor, r 750 rpm r lip f r f 50.5 Hz 4 S.M P 8 S 750 rpm.m P 6 S 000rpm 50 Hz 64. An: () Sol: From phor digrm, led the, hence clled Genertor. Here, co > clled over excited genertor. An under excited genertor lwy opertor t lging power fctor. 66. An: (b) Sol: 7. KA kW 60kW KA AC ngineering Publiction

35 : 5 : Potl Coching Solution Totl kw of lod k co P kw ka equirement of lod P tn 60 tn ka KW requirement of ynchronou motor (P ) 0 kw Operting p.f of lod 0.5 led Phe ngle co (0.5) 60 Q P tn 0 0 tn ka (KA upplied by ynchronou motor) Totl lod P P 70 kw Totl KA requirement Overll power fctor tn Q P p.f co 0.74 lg 67. An: 4 A Sol: ka j co(6.87) j40in6.87 j4 A Aume tht the motor drw current j4 A, then overll pf, therefore nwer i 4 A 68. An: (b) Sol: ph , ph P in in X in An: (c) Sol: From the rmture current 7.9. o 9. o i the ngle difference between nd. co co(9. o ) PF Lg 70. An: (d) Sol: 0 Z o ª 7. An: () 500 Sol: ph ph 54.7 Z 0. j o 84.8 o P in co co( ) Z Z AC ngineering Publiction

36 : 6 : lectricl Mchine 8000 (54.7) co(84.8) o.84.8 ( ) co(84.8 ) o Z o o 7. An: (b) Sol: PF co (4.9) led 7. An: (760.9 kw) Sol: Mechnicl power developed * P P Z S co( ) Z S co P co( ) co(84.80) P phe 5.64 kw P kw (Or) P mech P ( 54 0.) P mech 76 kw 75. An: (b) Sol: L 0 0 ph.8 Z 0.6 j o o t 0A, UPF, 0 Z o xcittion i kept contnt 0.9, contnt lod on the motor i,, to 40A (given) Z (0) co co 55.4 o 0 Z o PF Co (7.) lg 74. An: (4.84 m) Sol: (n quetion pole nd frequency not given let tke P 4, F 50) T P/ 4.84 m An: (c) Sol: P Mech P in Copper lo T P mech L L co ( )-(40 0.6).05 kw m AC ngineering Publiction

37 77. An: (b) 6.6 Sol: ph 80.5 P in L L co L A ph ( z ) o o xcittion i contnt, i contnt 5000 P in X AC ngineering Publiction in 9.5 o 78. An: () 0 Sol: Z PF co (.95 o ) 0.9 led 79. An: (*) Sol: Dt given ph , 00 ka, 0. nd X. 000 line phe 44. A 400 Stry loe 4000 W nd power input 75 kw : 7 : Potl Coching Solution Totl cu loe W Totl loe Stry loe Cu loe W % input loe 00 input % 80. An: () Sol: quivlent circuit/ph: j0 Fig A 650 Since problem pecifie tht excittion emf lg the terminl voltge, motor opertion i implied. So, the current received by the motor i pecified A. (Tht i why the reference direction of i hown entering t it terminl. Of coure we could hve revered the current direction in the figure, nd writen the vlue A). (i) Clcultion of excittion emf: 650 [506.87](j0) Solving, 665. (.7, 665.7)

38 : 8 : lectricl Mchine (ii) The mchine, cting motor, i drwing leding current. Hence it i overexcited. By reducing the excittion, it cn be mde to drw current t upf. n tht ce nd will chnge, but will be in phe with. (iii) Clcultion of rmture current nd lod ngle: At upf opertion, let the current per phe drwn by the motor be mp. Auming tht the power input per phe remin unchnged while the excittion i reduced, 65() A. eplcing the current in the circuit of fig. with thi vlue nd pplying KL, (j0). nd cn be clculted from thi eqution by eprtely equting the rel nd imginry prt. co () nd in ().6 4. DC Mchine 0. An: (c) Sol: nduced emf in DC Genertor Z P 60 A An: 0.05 Wb, 0. /turn Sol: Given, P, 00, Z 600, 400 rpm, dt 0.5 ec, /pole? nd emf/turn? (i) ZP 60A / pole Web (ii) flux linkge Wb turn d 50 e. dt 0.5. emf / turn /turn 0. An: (d) Sol: P 6, lp winding, Z 480, 0.06 ext chnged to wre A (prllel pth) AC ngineering Publiction

39 L A w w L w 0.06 A 6 w Sol: () lp winding: P 4, A 4, Z 500 Z Armture AT / pole P A AT / pole Z ATdemgnetiz ing 80 P A AT/pole AT cro-mgnetizing 80 Z 80 P A AT/pole (b) Wve winding P 4; A Z Armture AT/pole. AT / pole P A AT/pole Z AT demgnetizing. 80 P A AC ngineering Publiction : 9 : Potl Coching Solution AT/pole 80 Z AT cro-mgnetizing 80 P A AT/pole 06. An: Sol: t L 00kW L 50A 400 MMF required on ech interpole i Z MMF P g..b P A AT/pole A we know tht interpole winding connected in erie winding of rmture turn 7 turn/pole 07. An: (b) GA xi lo clled bruh xi (or) xi of commuttion which i lwy perpendiculr to the min field flux xi ( m ).Commutting pole re plced in GA xi to eliminte cro mgnetiztion effect under q-xi. umber of commutting pole umber of min field pole for lrge rting motor nd le for mll rting motor.

40 : 40 : lectricl Mchine 08. An: () Sol: Compenting winding & nter pole winding connected in erie with the rmture winding. Compenting flux i ued to nullify the cro mgnetiztion effect under polr xi i.e d-xi only but commutting pole flux ued to compente cro mgnetiztion effect under inter polr region i.e. q-xi nd lo limit rectnce voltge. 09. An:.5 mwb, 5 c.m Sol: Given, P 0, 000 rpm, Z 000, A 0, 400 nd B T Armture copper lo 400 W 00 5 A / (6/5) 46 ZP 60A /pole.5 mwb We know tht, B A.50 Are of pole hoe m Are 5 cm 0. An: (c) Sol: Generted MF g. () lectricl lod, P L. L.8 KW L A f A For genertor L f 9A A 0A From eq(), g 00 (0) (0.4) 04. An :( 49.6 A, 6.4 ) Sol: P 5k 00 f 50A g () P g () g g g f g 5 g () 5 50 Subtitute eqution () in () L L O A D.8kW AC ngineering Publiction

41 : 4 : Potl Coching Solution A An: (d) Sol: 00, L 00 A 0.05, c 0.04 h 00, B.D, g? 00 f.5 A h 00 L f 0.5 A For long hunt compound genertor g L e B.D 00 ( ) (000.04) g An: 550 Sol: t 45 (Grid) 50 A; 800 rpm rpm ; t 45 (Grid)? g t r 45 (50) (0.) 60 g [ g K, g cont] 000 g A A 4. An: (c) Sol: Given externl lod chrcteritic curve i liner, therefore from curve vlue of correponding to 50A i 95 95, L 50A L L Ω Alternte method: At 50 Amp, 00 (50 0.) L. 9 L An: (c) Sol: C 50, 000 rpm C 80 ;? C C rpm An: 8 Sol: Seprtely excited DC motor Flux remin contnt AC ngineering Publiction

42 : 4 : lectricl Mchine Ce (i): At o lod 000 rpm 00 Ω 0 T 0 b Ce(ii): Full lod T T rted Full lod rmture current? 500 rpm b 500 b 000 b b 00 () 00. () Ce(iii): T 50% of T rted 7. An: () Sol: b b b b 080 rpm 8. An:.7 Sol: Motor opertion: 5 Ω A Genertor opertion: 48 A 0.6 Ω. Fig A 40 Supply T 0.5 T rted T T 0.5 T () But b b b b 500 b b 04 b From () & () b 00 (0.5 ) (0.5 ) A A 5 Ω. 5 A 8. Fig Ω A Since the field current i the me in both ce, neglecting rmture rection, flux i contnt in both ce nd (peed genertor)/(peed motor) (8./.) An: (b) Sol: Prcticlly in motor, the torque direction i imilr direction to tht of rotor rottion. 40 Supply AC ngineering Publiction

43 n genertor the torque direction i oppoite direction to tht of rotor rottion. Before Belt Snp (p.m fil) T em L, delivering clled genertor Let h, re poitive h e Hence ct compound genertor T em ( ) ( ) After Belt Snp (p.m fil) PM fil T pm h L T e m L, oborbed clled motor Here h poitive, i egtive h e AC ngineering Publiction h L T pm T em lectricl lod (grid) lectricl upply (grid) : 4 : Potl Coching Solution [ direction chnged] Act differentil motor T em ( ) ( ) We know, fter prime mover fil, T em ( ) ( ) T em (. ). Torque direction i revered, i.e, direction of rotor rottion in differentil compound motor i imilr direction to tht of over compound genertor. 0. An: () Sol: Full lod Armture current ( ) 40A When i inerted in erie with rmture eq.5 Stlling Torque i the Torque t which peed fll to zero b Speed () When A.5 Torque T [ cont] T T T T pm T em 4. An: (d) Sol: 50, 0.5, h 50, b eq

44 : 44 : lectricl Mchine 600 rpm, T L contnt L A, e 50,? h h L h h A 0A A T T h h e contnt 0 h h 40A b 600 b b h h rpm An: (i).778 (ii).498 Sol: Given dt: t rpm, L A f 00, rpm 0 f.a 00 f L t L f. 9.8A (for hunt motor contnt) Ce (i): T L T e nd under tedy tte condition T L T e A t ( ext ) (0. ext ) ext Ce (ii): T L nd T e 9.8 (0.8).67A b t ( ext ) (0. ext ).497. An: (c) Sol: 80 rpm 50 rpm 5 5 AC ngineering Publiction

45 : 45 : Potl Coching Solution h b h b 4. An: (*) Sol: b h h 5 h h h b b h 5 h h h A torque i contnt cont. b 600 e d rpm 5. An: (c) Sol: e e Ampere turn AT e e 5 e e e Ampere (AT p ). e Contnt torque condition T T AT e AT p e e ( i the peed, when erie field winding re connected in prllel) AT AT e e e p By ubtituting 500 rpm 6. An: (b) Sol: P T contnt T T e e 0.5 AC ngineering Publiction

46 : 46 : lectricl Mchine 0.5 pu, pu T pu 0.5 pu 7. Sol: According to motor rting nd neglecting loe b t rpm rted b r / ec A K ; (erie motor) 50 b K K K 0.04 f the two motor re connected in erie nd rotting common hft. A B 5 4 A.5 b A.5 A.5 B b b 50.5 A 8.8 b A 8.8, B. A 8.8 K 5 hft 8.8 (0.04)(5) hft 694 hft A loe re neglected P input P output T (50) (80)( hft ).4 ( hft ) hft rd / ec hft 694 hft rpm (.4)(4.6) 47.40A A 5 hft nd B 4 hft 8. An: (b) Sol: Since dt i not given, for implicity neglect A ll loe including rmture copper loe. 50 Thi men we ume 0. B Problem pecifie tht t ny peed, rmture current h the me vlue, which i equl to the rted vlue. A B 50 Let rted nd rted be the rted vlue. Then A K A rted input rted output rted rted 50 A K (5 hft ); B K (4 hft ) kw. AC ngineering Publiction

47 : 47 : Potl Coching Solution (i). To obtin hlf the peed by rmture voltge control (field current i umed to remin unchnged), rmture voltge mut be mde rted. Thi reult i proved follow. For the hunt motor, K r ince i neglected. Let field excittion be kept contnt. Then neglecting rmture rection, K i contnt, y K nd K r. At rted opertion rted Kr rted. To get r r ; rted mut be rted /. rted Power rted 5 kw (ii). With kept t rted ; r i to be chnged to rrted by field control. But the current i till rted from the given dt. Hence power rted rted 50 kw. Method-: By rmture voltge control we cn control peed below rted vlue. But below rted peed, the mchine ct contnt torque vrible power drive, where P out P P 0.5 P 50k P 5 kw P, T T < rted P > rted By field control method we cn control peed bove rted lue. But bove rted peed, the mchine behve contnt power vrible torque drive P out P rted 50 kw 9. An: 7 Sol: Given, t 400, 0.8 We know tht, Where t n n A, 5.5A, r r Tke logrithm on both ide, n log 0 log0 n The number of reitnce element, n 7. An: (i) d (ii) () Sol: (i) b L h b 40 h A h et voltge L A h b 40 4 AC ngineering Publiction

48 : 48 : lectricl Mchine 474 b (ii) b.5f b b b b. 5. An: Sol: Given dt P 0 0Hp fl 7.5A 0 0.0Wb P 4 Z 666 A 0.67 ottionl loe 600W At, 000 rpm Output torque, T h? b Z 60 P A b b A 665.W Output power hft power P h P gm rottionl loe P h W P h T h T h 60 Output torque T h m 4. An: () Sol: At no lod developed power (P) b b P W Under no lod condition the developed power i ueful to overcome friction & windge Loe. friction & windge Loe 5.7W Under loded condition Armture Cu lo (.5) W Totl loe W nput to motor W input Loe fficiency 00 input % Gro mechnicl power developed P gm P gm b 9.96 AC ngineering Publiction

49 : 49 : Potl Coching Solution 5. Sol: Given dt: 0A A 7A 0 M t 0, fg.5a f A, m 7A g m 0.05 L 0A g m fg f L Armture circuit lo in genertor (67.5) W Armture circuit lo in motor (7 0.05) 66.45W Power drwn from the upply (xcluding the field lo in two mchine) t L fg fm A P lo W o- lod rottionl lo in both the mchine W 0 t r ( g m [(67.5) (7) ] 75.75W G g fg.5 A Z Z o-lod rottionl lo for ech mchine W W For genertor, output t g Wg Totl loe, W g t f gr W g W lo g 00 loe output % For motor, input t ( m fm ) 0(7) 6500W W0 Totl loe, W m mr t fg m loe nput % 6500 AC ngineering Publiction