用节点电压法求图示电路中9A电流源的功率

更新时间：2019-01-08 04:05:23 115次访问

I10 = U10，I12 = 2U12，I13 = 12 - I10 - I12 = 12 - U10 - 2U12；

I23 = U23 = 9 - I13 = 9 - ( 12 - U10 - 2U12 ) = U10 + 2U12 - 3；

U12 + U23 = U13，U12 + U10 + 2U12 - 3 = I13/2 = 6 - U10/2 - U12，

8U12 + 3U10 = 18；①

U02 = U12 - U10 = 5；②

① + ②*3，11U12 = 33，U12 = 3 v；U10 = U12 - 5 = -2 v；

U23 = U10 + 2U12 - 3 = 1 v；

U03 = U23 + 5 = 6 v；

9A电流源输出功率  6 * 9 = 54 w 。