《电路》（英文版）9-5 Maximum power transfer in the AC case

89-5 Maximum power transfer in the aC case Zth=rthtjx =p∠0°Z ZL=R+JXL 2 PL IR L XR=V R Zth +ZL (Rn+R1)2+(Xmn+X1)2 The best that X can do to maximize p is to make X+X=0 or The load power becomes P,=v2 R (R +ru) atr= rh p Lmax =卩2/4R t L=RL+jx=rn-jx th

§9-5 Maximum power transfer in the AC case Zth ZL • I − =  + •  V V 0 L L L th th th Z R jX Z R jX = + = + L t h L L L R Z Z V p I R  + = = 2 2 2 The best that XL can do to maximize PL is to make Xth+XL=0 or XL = -Xth. The load power becomes 2 2 ( ) th L L L R R R p V + = at RL Rt h pL V 4Rt h / 2 = max =  L = L + L = th − th = Zth Z R jX R jX 2 2 2 ( R R ) ( X X ) R V t h L t h L L + + + =

Example2:(p175-12) A resistor R, capacitor C, and voltage source Ds =100 cos 300M 150W and the rms capacitor voltage is 10v, determine r andc a are all in series. If the average power dissipated by the resistor R Solution R=V70.72-104=70 100 R702 32702 150 70 3272.143 2.143 714uF 300×10

A resistor R, capacitor C, and voltage source V are all in series. If the average power dissipated by the resistor is 150W and the rms capacitor voltage is 10V, determine R and C. t  s = 100cos300 70.7 10 70 2 2 VR = − = 2.143 32.7 70 I = = C 714uF 300 10 2.143 =  = 32.7 150 702 R = = R C − = +  s 100cos 300tVi Example 2:(p175-12) Solution:

DP:(p175-19)k2300V,=25A,PF=0.815(lag).Find(a)P;(b)Q; (c)s;(d)S;(e)Z. (a):P=46.9kW;(b):Q-333kvar;(c):57.5/35.4VA; (d):S=57.5kVA;(e):z=92235.4° Example: (p176-20)Find l and p (×5+40000021+(400002 =(120002 400kW 12k 007(g)=478(2366) P=5×4782+400000-=4114kW

DP:(p175-19) V=2300V,I=25A,PF=0.815(lag).Find (a)P; (b) Q; (c) ;(d) S; (e) Z. _ S Example:( p176-20) Find I and Ps . 5 0.707( ) 400 lag kW − + 12kV i 2 2 2 2 (12000 ) ( 5 400000) (400000) I I =  + + Ps = 547.8 2 + 400000 = 411.4kW I = 47.8 ( 2366 ) (a):P=46.9kW; (b): Q=33.3kVar; (c): 57.5 35.4 oVA; (d): S=57.5kVA;(e):Z=92 35.4o