Question

如图电路中,电源电动势为15伏特,内电阻为5欧姆,电阻R1为25欧姆,当K闭合后,伏特表的读数是9伏特,试求：

Answer 1

(1) K断开时

I=E/(r+R1)=15/(5+25)=0.5A

U=I*R1=12.5V

(2) K闭合时

I'=(E-U')/r=(15-9)/5=1.2A

(3)

I1'=U'/R1=9/25=0.36A

I2'=I'-I1'=0.84A

R2=U'/I2'≈10.7Ω