发布时间:2019-08-08 13:16:05
那位大神会做一题?(要步骤)
电压源单独作用,i1 = 10/( 2 + 1 ) = 10/3 A;
可控源单独作用,i2 = -2i/( 2 + 1 ) = -2i/3 A;
电流源单独作用,i3 = -5 * 1/( 2 + 1 ) = -5/3 A;
i = i1 + i2 + i3 = 10/3 - 5/3 - 2i/3,5i/3 = 5/3,i = 1 A;
u = 1Ω电阻压降 + 2i = 1 * ( i + 5 ) + 2i = 5 + 3 * 1 = 8 v 。