例5用叠加定理计算电流i2 已知:Us=100∠45V Is=4∠0°A Z1=Z3=50∠30°9, Z,=50∠-30°g (2)Us单独作用Is开路): U 解(1)单独作用U短路): Z+Z 100∠45° 1.155∠-135°A 2 十 3 50 3 50∠30° =4∠0° 50∠-30°+50∠30° 200∠30° =2.31∠300+1.155∠-135° 2.31∠300A 50 √3 123∠-159A例5 用叠加定理计算电流 2 I  Z2 S I  Z1 Z3 2 I  S • U + - 解 (1) ( ): I  S单独作用U  S短路 2 3 3 S ' 2 Z Z Z I I +  =  o o o o 50 30 50 30 50 30 4 0  − +   =   2.31 30 A 50 3 200 30 o o =   = 2 3 " S 2 Z Z U I + = −   o o " 2 ' 2 2 = 2.3 13 0 + 1.155 − 135 I = I + I    1.155 135 A 50 3 100 45 o o =  − −  = 1.23 15.9 A o =  − (2) S ( S ) : 单独作用 开路 • • U I ' 2 I  " 2 I  : 100 45 V o S 已 知 U =  5 0 3 0 . 5 0 3 0 , 4 0 A, o 3 o 1 3 o S Ω Ω =  − = =  =  Z Z Z I 