B處電流=U/Rb
總電流=U/Ra+U/Rb
總電阻=U/(U/Ra+U/Rb)=1/(1/Ra+1/Rb)
即:1/總電阻=1/Ra+1/Rb
同理,將原推導(dǎo)擴(kuò)展,有n條支路時(shí):
1處電流=U/R1
2處電流=U/R2
3處電流=U/R3
4處電流=U/R4
……
n處電流=U/Rn
總電流=U/R1+U/R2+U/R3+U/R4+……+U/Rn
總電阻=U/(U/R1+U/R2+U/R3+U/R4+……+U/Rn)=1/(1/R1+1/R2+1/R3+1/R4+……+1/Rn)
即:1/總電阻=1/R1+1/R2+1/R3+1/R4+……+1/Rn
![](http://h.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=30a34a63252dd42a5f5c09ad330b778d/dbb44aed2e738bd45495ad3ba18b87d6277ff900.jpg)