VS=12∠0°=12cos0+j12sin0=12
(2+j4)//-j2=(8-j4)/(2+j2)=[(8-j4)(2-j2)]/[(2+j2)(2-j2)]
=(8-j24)/8=1-j3
12V分壓
12×[(1-j3)/(1-j3+2)]=(12-j36)/(3-j3)=[(12-j36)(3+j3)]/[(3-j3)(3+j3)]
=(144-j72)/18=8-j4
(8-j4)V再分壓
(8-j4)×[2/(2+j4)]=(16-j8)/(2+j4)=(16-j8)(2-j4)/(2+j4)(2-j4)
=-j80/20=-j4=4∠-90°
31 假設交流電壓源 Vs = 12∠0°V ,試求電壓 Vo: ..-阿摩線上測驗