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 + + + =