正孩稳态电路的计算 Z=Z +Z, =92.11-j289.13+10+j157 =102.11-j132.13 =166.99∠-52.3°2 100∠0 ΓZ-166.99∠-52. 。=0.6∠52.3°A→1=0.6V2c0s(314t+52.3)A 1/R i=.=0.181∠-20A→5=0.181V2c0s(314t-20)A 1/R+j@C 132 j0Ci,=.=0.57∠70A→4，=0.57N2c0s(3141+70)A 1/R+joC Z1 Z2 U + R 2 _ R 1 1 I 2 I 3 I C j  − 1 jL =  −  = − = − + + = +  166.99 52.3 102.11 132.13 92.11 289.13 10 157 1 2 j j j Z Z Z 1 2 1 2 1 1 / 0.181 20 A A 0.181 2 cos(314 20 ) 1 / R I I t R j i C  = = = + = →  − − 3 1 3 1 0.57 70 A A 0.57 2 cos(314 70 ) 1 / j jC I I i C t R    = = + =  → = + 1 1 100 0 0.6 52.3 A A 0.6 2 cos(314 52.3 ) 1 . 66 99 52.3 i U I t Z    = =  = =  → +  −