《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<00<0 capacitive load Icos e d sine I cos 6--active current Ising--reactive current 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 S=S1+S2+…+SN=(1+jg1)+(2+jQ2)+…+(P+jQNy) P=B+2+…+Pmdg=g1+g2+…+QN S≠S1+S,+…+S) N

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 − 