+i2g0.125F 解 m-(2+ j0.125o0.5g2 10∠0° 2+j4 4.165H 2∠-60° =(0.5+j8.33)2N0 i + - u (b) 2 0.125F 解： = +  − + −    = = − + (0.5 j8.33) 2 j4 2 60 10 0 ) j0.125 1 (2 m m 0 I  U Z   4.165H 0.5