应用叠加定理求各支路电流I1 I2 I3~
1个回答

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 .