ENE 104 Electric Circuit Theory

Transcription

1 Elecric Circui heory Lecure 0: AC Power Circui Analysis (ENE) Mon, 9 Mar 0 / (EE) Wed, 8 Mar 0 : Dejwoo KHAWPARSUH hp://websaff.ku.ac.h/~dejwoo.kha/

2 Objecives : Ch Page he insananeous power he average power he rs value of a ie-varying wavefor coplex power: average and reacive power he power facor, how o iprove.

3 nsananeous Power: p = v i() Page 3 he device is a resisor: p = i R = v R he device is enirely inducive: p = Li di d = v() v d L he device is enirely capaciive: p = Cv dv d = i() i d C

4 nsananeous Power: Page 4 Consider he series RL circui, i = V 0 R e R L u() he oal power delivered by he source p = v i = V 0 R e R L u()

5 nsananeous Power: Page 5 Consider he series RL circui, i = V 0 R e R L u() he power delivered o he resisor is p R = i R = V 0 R ( e R L ) u()

6 nsananeous Power: Page 6 Consider he series RL circui, i = V 0 R e R L u() o deerine he power absorbed by he inducor, firs hen, v L () = L di() d = V 0 e R L u + LV 0 R = V 0 e R L u p L () = v L i() = V 0 e R L du() d R e R L ( e R L ) u()

7 nsananeous Power: Page 7 Consider he series RL circui, p = p R + p L ()

8 nsananeous Power: Page 8 Power Due o Sinusoidal Exciaion: he response is Where i( ) cos( ) R V L an L R

9 Pracice:. Page 9 A curren source of cos(000) A., a 00-oh resisor, and a 0.-H inducor are in parallel. Assue seady-sae condiions exis. A = s., find he power being absorbed by he (a) resisor (b) inducor (c) sinusoidal source

10 Pracice:. Page 0

11 Average Power: Page ENE 04 For Periodic Wavefors: ) ( d p P ) ( ) ( f f d p P ) ( x x x d p P ) (

12 Average Power: Page For Periodic Wavefors: P n x n x p( ) d n,,3,... P n n n p( ) d P li n n n n p( ) d

13 Exaple: Page 3 ENE 04 find he average power delivered o a resisor R by he periodic sawooh curren wavefor i ), ( ) ( i 0, ) ( R v R i i v p ) ( ) ( ) ( ) ( ) (

14 Exaple: Page 4 ENE 04 i ), ( ) ( i 0, ) ( R p 0, ) ( R d R P 0 3 R p, ) ( ) (

15 Average Power: Page 5 n he Sinusoidal Seady Sae v( ) V cos( ) i( ) cos( ) he insananeous power is Or p( ) V cos( )cos( ) p( ) V cos( ) V cos( ) By inspecion: P V cos( )

16 Exaple: Page 6 Given he ie-doain volage V., find boh he average power and an expression for he insananeous power ha resul when he corresponding phasor volage V = 4 0 o V. is applied across an ipedance Z = 60 o oh. v 4cos / 6 Sol:

17 Exaple: Page 7 v 4cos / 6 V. V = 4 0 o V. Z = 60 o oh. Sol: he phasor curren is V/Z = -60 o So he average power is: P i V cos( ) (4)() cos(60) cos / 6 60 W

18 Exaple: Page 8 v 4cos / 6 i cos / 6 60 he insananeous power is: p( ) V cos( )cos( ) V cos( ) V cos( ) 8cos( )cos( 60) 4cos( 60) 6 6 3

19 Exaple: Page 9 p( ) 4cos( 60) 3 v 4cos / 6 i cos / 6 60

20 Pracice:. Page Given he phasor volage V = V. across an ipedance Z = o oh., obain an expression for he insananeous power, and copue he average power if = 50 rad/s. Sol: p( ) V cos( )cos( ) p( ) V cos( ) V cos( )

21 Pracice:. Page

22 Average Power: Page Absorbed by an deal Resisor: Or P R V P R cos(0) R V R V Absorbed by Purely Reacive Eleens: P x 0

23 Exaple: Page 3 Find he average power being delivered o an ipedance Z L = 8-j oh. by a curren = 5 0 o A. Sol: P R R W.

24 Pracice:.3 Page 4 Calculae he average power delivered o he ipedance 6 5 Ω by he curren = + j5 A.

25 Pracice:.4 Page 5 Copue he average power delivered o each of he passive eleens. Sol:

26 Pracice:.4 Page 6 For he circui in he figure below, copue he average power delivered o each of he passive eleens. Verify your answer by copuing he power delivered by he wo sources.

27 Pracice:.4 Page 7

28 Pracice:.4 Page 8

29 Maxiu Power ransfer: Page 9 An independen volage source in series wih an ipedance Z h delivers a axiu average power o ha load ipedance Z L which is he conjugae of Z h, or Z L =Z * h

30 Exaple.5: Page 30 f we are assured ha he volage source is delivering axiu average power o he unknown ipedance, wha is is value? Sol: Z * Z 500 3? h j

31 Pracice:.5 Page 3 f he 30-H inducor of Exaple.5 is replaced wih a 0-μF capacior, wha is he value of he inducive coponen of he unknown ipedance Z? if i is known ha Z? is absorbing axiu power?

32 Pracice:.5 Page 3

33 Average Power: Page 33 For Nonperiodic Funcion: for exaple i( ) sin sin he average power delivered o a oh resisor: P li (sin Wa sin sin sin ) d

34 Average Power: Page 34 For i( ) cos cos... N cos N he average power delivered o a resisor R: P... N R

35 Pracice:.6 Page 35 A volage source v s is conneced across a 4-Ω resisor. Find he average power absorbed by he resisor if v s equals (a) 8sin00 (b) 8sin00 6cos (00 45 ) V. (c) 8sin00 4sin00 V. (d) 8sin00 6 cos sin V.

36 Pracice:.6 Page 36

37 Effecive Values: Page 37 f he resisor received he sae average power in par (a) and (b), hen he effecive value of i() is equal o eff, and he effecive value of v() is equal o V eff

38 Effecive Values: Page 38 of a Periodic Wavefor: he average power delivered o he resisor, P 0 i he power delivered by he direc curren is: d R 0 i d P eff R

39 Effecive Values: Page 39 of a Periodic Wavefor: R 0 i d eff R we ge eff 0 i d ENE 04 noe: he effecive value is ofen called he roo-eansquare value, or siply he rs value.

40 Effecive Values: Page 40 of a sinusoidal Wavefor: i( ) cos( ) eff 0 i d ENE 04 which has a period o obain he effecive value eff 0 0 cos [ ( ) d cos( )] d

41 Effecive Values: Page 4 ENE 04 of a sinusoidal Wavefor ) cos( ) ( i eff d i 0 ] [ 4 )] cos( [ ) ( cos eff eff d d

42 Effecive Values: Page 4 Use of RMS Values o Copue Average Power eff P R eff R V eff R P V cos( ) V eff eff cos( )

43 Effecive Values: Page 43 wih Muliple-Frequency Circuis eff P eff eff... Neff R eff eff eff... Neff

44 Effecive Values: Exaple Page 44 Ex Page 384: find he effecive value of: eff 0 i d

45 Pracice:.7 Page 45 Calculae he effecive value of each of he periodic volages: (a) 6cos5; (b) 6cos5 + 4sin ( ) V. (c) 6cos5 + 5cos (5) V. (d) 6cos5 + 5sin V.

46 Pracice:.7 Page 46

47 Pracice:.7 Page 47

48 Pracice:.7 Page 48

49 Apparen power and Power facor: Page 49 he sinusoidal volage v V cos( ) ENE 04 is applied o he nework and he resulan sinusoidal curren is he average power delivered o he nework P V he apparen power: he power facor: cos( i eff eff eff eff cos( ) ) Veff eff cos( ) V PF V P

50 Exaple: Page 50 Calculae values for he average power delivered o each of he wo loads, he apparen power supplied by he source, and he power facor of he cobined load.

51 Exaple: Page 5 Calculae values for he average power delivered o each of he wo loads, he apparen power supplied by he source, and he power facor of he cobined load. P V eff eff cos( angv ang) PF V eff eff P V eff eff

52 Exaple: Page 5 S ( j) ( j5) A. rs P S V eff eff (60) () cos[0 43 W. cos( angv ang) ( 53.)]

53 Exaple: he apparen power supplied by he source: Page 53 V eff eff ( 60) () 70 VA. he power facor of he cobined load. PF P V eff eff 43 (60) () P () 0.6 P ()

54 Pracice:.8 Page 54 For he circui of he figure below, deerine he power facor of he cobined loads if Z L = 0Ω.

55 Pracice:.8 Page 55

56 Coplex Power: Page 56 he average power, P V eff eff cos( ) express as, P V eff eff Re j( ) j j * e Re V e e Re V eff eff eff eff he coplex power, S V eff V e * j( ) eff eff eff P jq Q V eff eff sin( )

57 Power Measureen: Page 57 he coplex power delivered o several inerconneced loads is he su of he coplex power delivered o each of he individual loads. S = V * = V( + ) * = V( * + * ) = V * +V *

58 Exaple: Page 58 An indusrial consuer is operaing a 50-kW inducion oor a a lagging PF of 0.8. he source volage is 30 Vrs. n order o obain lower elecrical raes, he cusoer wishes o raise he PF o 0.95 lagging. Specify a suiable soluion.

59 Exaple: Page 59 A purely reacive load us be added o he syse, in parallel, since he supply volage us no change. he coplex power supplied o he oor us have a real par of 50-kW and an angle of cos - (0.8)=36.9 o Hence, S j37. 5 kva

60 Exaple: S = o achieve a PF of 0.95, he oal coplex power us becoe: S cos j6. 43 hus he correcive load is 50 (0.95) kva j37. 5 kva Page 60 S = j. 07 kva

61 Exaple: Page 6 S = j. 07 kva We selec a phase angle of 0 o for he volage source, and he curren drawn by Z is Or herefore, * S j070 j9.6 A. V 30 j9.6. A Z V j9.6 j

62 Pracice:.9 Page 6 Find he coplex power absorbed by he (a) -oh R (b) -j0 C (c) 5+j0 Z (d) source

63 Pracice:.9 Page 63

64 Pracice:.9 Page 64

65 Pracice:.9 Page 65

66 Pracice:.0 Page 66 A 440-Vrs source supplies power o a load Z L = 0 + j Ω hrough a ransission line having a oal resisance of.5ω. Find (a) he average and apparen power supplied o he load; (b) he average and apparen power los in he ransission line; (c) he average and apparen power supplied by he source; (d) he power facor a which he source operaes.

67 Pracice:.0 Page 67

68 Exaple: Final /47 Page 68 จากวงจรตามร ป ให หา ) he average power ท จ ายโดย แหล งจ าย (source) ) ค า power facor ท แหล งจ าย (source) 3) ขนาดของ ต วเก บประจ (capacior) ท เม อนาไปต อ ขนานก บ แหล งจ าย ทาให ค า power facor = 4 oh 0 Vrs 60Hz j6 oh oh

69 Exaple: Final /47 Page 69

70 Exaple: Final /47 Page 70 4 oh 0 Vrs 60Hz j6 oh oh

71 Exaple: Final /47 Page 7

72 Ex: Page 7 5 Ω µf 50 µh V g 7.5 Ω Find he average power, he reacive power, and he apparen power supplied by he volage source in he circui if v g = 50 cos 0 5 V.

73 Ex: Page 73 5 Ω µf 50 µh V g 7.5 Ω

74 Suary: Page 74

75 Hw: Page 75

76 Reference: Circui Analysis and Elecrical Engineering Page 76 W.H. Hay, Jr., J.E. Keerly, S.M. Durbin, Engineering Circui Analysis, Sixh Ediion. Copyrigh 00 McGraw-Hill. All righs reserved.