电压源独立作用:i'1=25/(100+2000//500)=25/500=50mA
i'2=50mA x 500/2500 =10mA
i'3=50mA x 2000/2500=40mA
电流源独立作用:暂将2k和500欧并联看成一个400欧电阻,
i"1 = 125mA x 400/500 = 100mA 反方向所以 -100mA
i"2=(125mA-100mA) x 500/2500 =25mA x 1/5 =5mA
i"3 = 25mA x 2000/2500 =20mA
i1=i'1+i"1=50-100=-50mA
i2=i'2+i"2=15mA
i3=i'3+i"3=60mA .