正孩稳态电路的计算 例4. 已知:is=4L90°A,Z1=Z2=-j302 is白zzz Z3=302,Z=452 求:i. 方法一:电源变换 Z1∥Z3= 30(-j3 30-j3 2=15-j1592 Z IIZ Z,/1Z,)j4(15-j15) Z,1iZ,+乙,+7=15-15-j30+45 亿1112) =5.657∠45 5∠-36.9° =1.13∠81.9°A 例4. . 3 0 , 4 5 4 9 0 A , j3 0 3 1 2 o S I Z Z I Z Z 求 : 已 知 : Ω Ω Ω = = = = = − 方法一:电源变换 = − − − = 15 15 30 30 30( 30) // 1 3 j j j Z Z Z2 S I Z1 Z3 Z I 1 3 S ( / / ) Z Z I& Z2 Z1 Z3 Z I + - 1 1 3 2 S 3 / / ( / / ) I Z Z Z Z I Z Z · = + + & 15 15 30 45 4(15 15) − − + − = j j j j o o 5 - 36.9 5.657 45 = A o = 1.1381.9