《电路》（英文版）9-4 Complex power

§94 Complex power A sinusoidal voltage: V=V29 A sinusoidal current: I=I9 We define the complex power S=vI=v∠9I∠-91 V1∠9-1=W1∠9=c0s8+jsin8=P+Q P=Vcos 9 -active(average)power 0=vIsin 9 --reactive power S=vI s2=P2+g2 (功率三角形) P

§9-4 Complex power A sinusoidal voltage: =  • V V A sinusoidal current: i I = I • =VI −i =VI =VI cos + jVI sin = P + jQ We define the complex power i S =V I =V I −  _ • • P =VI cos --active(average) power Q =VIsin --reactive power S =VI 2 2 2 S = P + Q (功率三角形) S Q P 

0=vIsin 9=vIsin(9-9) 9=9-9>00>0 inductive load 9=9-9

sin sin( ) Q =VI  =VI  −i  = −i  0 Q  0 inductive load  = −i  0 Q  0 capacitive load I cos − −active current I sin − −reactive current • V • I I cos Isin  The complex power delivered to several interconnected loads is the sum of the complex power delivered to each of the individual loads no matter how the loads are interconnected. ( ) ( ) ( ) 1 1 2 2 _ 2 _ 1 _ _ N N N S = S + S ++ S = P + jQ + P + jQ ++ P + jQ P = P1 + P2 ++ PN and Q = Q1 +Q2 ++QN ( S S S S )  1 + 2 ++ N

P174-6(a)Which of the element A, B, C, D must be source (b How mu nuch average is each of the sources supplying?:s +40∠150° 4.984∠116°A 156∠-7.37° 120∠0°A4 D 2∠110°4 3∠120°4 24=J0×、-II0。=-85I-22 2B=-10℃I0。×-II0。=-QI3-12I寸 =-I2Q一3山。×寸081-IIQ。=1yI+!Q寸o3 2D=I-3×3-I0。=-8I-13y

−  − +  156 7.37 A B C D −  +  120 0 A  2110 A  3120 +  −  40 150 . A  4 984116 P174-6(a) Which of the element A,B,C,D must be source? (b) How much average is each of the sources supplying? SA = 120 2 − 110 = −82.1 − j225.5 − SB = −40150  2 − 110 = −61.3 − j51.4 − S C = −156 − 7.37  4.984 − 116 = 427.1 + j649.3 − S D = 156 − 7.37  3 − 120 = −284.1 − j372 −

We raise pF V=200∠0° Z1--l0d,P=1000W, cos 9=0.8 lagging). ff PF=0.95(lagging, find C. B=1000-S1=1000/0.8=125014 c0s1=08→>91=36.9° O,=s sin 9=1250sin 36.9=750var S1=P+j01=1000+j750VA at cos=0.95→9=18.2°S 1000 ∠182°=1000j329A 0.95 Sc=S-S1=-j42arSc=H2→l2=421/200=2.107A4 2.107 → 33.5F a314×200

Z1 1 • I • I − + • V C 2 • I We raise PF 0.95( ), . cos 0.8( ). , 1000 , 200 0 1 1 1 If PF lagging find C lagging Z load P W V V = = − − = =  •   P1 = 1000W → S1 = 1000/ 0.8 = 1250VA cos = 0.8 → = 36.9 1 1 Q S sin 1250sin36.9 750Var 1 = 1 1 = =   S1 P1 jQ1 1000 j750VA _ = + = + at cos = 0.95 →  = 18.2 S 18.2 1000 j329VA 0.95 1000 _ =  = +  S C S S1 j421Var _ _ _ = − = − SC =VI2 → I2 = 421/ 200 = 2.107A F V I I C C V    33.5 314 200 1 2 2.107 2 =  = → = =

fr g9 2=11 cos 9 tan 91-11 cos 9 tan 9 9 =11 cos 9(tan 9-tan 9) 1 cos 9(tan 9-tan 9 1 tan v1 tan JPF=0.8→>0.95 Pi(tan 91-tan 9) 1000(tan 36.90-tan 18.20) 33.5∠F 2 314×200 2

• V • I 1 • I 2 • I 1  1 I 1 cos I 2 = I 1 cos1 tan1 − I 1 cos1 tan VV VI V I C      cos (tan tan ) 2 1 1 1 − = = 2 1 I C V  = If PF = 0.8 → 0.95 F V P C     33.5 314 200 (tan tan ) 1000(tan 36.9 tan18.2 ) 2 2 1 1 =   −  = − = 2 1 1 (tan tan ) V P   −  = cos (tan tan ) = I 1 1 1 − 