戴维宁定理，电源的转换方法，求支路电流电压未知量，尽量详细谢谢！

一、用戴维南定理求I2开路R3；Is + i1 = i3，即i1 = i3 - 3uab = U1 - i1 * R2 = U2 + i3 * R4代入数值，6 - ( i3 - 3 ) * 3 = 10 + i3 * 2，解得 i3 = 1A；戴维南等效电压 uab =10 + 1 * 2 = 12v；短路电压源，开路电流源，戴维南等效电阻rab = R2//R4 = 3//2 = 6/5Ω接入R3；I2 =uab/( R3 + rab ) = 12/( 2 + 6/5 ) = 3.75 A ；二、求支路电流及电流源的端电压Uab = I2 * R3 = 2 * 3.75 = 7.5 V；I1 = ( U1 - Uab )/R2 = ( 6 - 7.5 )/3 = -0.5 A；I3 = ( Uab -U2 )/R4 = ( 7.5 - 10 )/2 = -1.25 A；负号表示电流实际方向与标注方向相反。Us = Is * R1 + Uab = 3 * 10 + 7.5 = 37.5v；三、用电源转换方法求解I2U1变换为 2A电流源与R2并联；【 6v/3Ω = 2A 】U2变换为 5A 电流源 与 R4 并联；【 10v/2Ω = 5A 】3个电流源总电流i= 3 + 2 + 5 = 10A；总电阻r = R2//R3//R4 = 3//2//2 = 3//1 = 3/4ΩUab = i *r = 10 * 3/4 = 7.5vI2 = Uab/R3 = 7.5/2 = 3.75v四、判断各元件的工作状态各电阻都消耗功率；电流源输出功率；I1 实际方向上至下，U1吸收功率；I3 实际方向下至上，U2输出功率 。