量程1mA:Rg与(R1+R2)并联,总电流 I=1mA,总电压为U
U=Ug=IgRg =(I-Ig)(R1+R2)
得到:
(R1+R2)=IgRg/(I-Ig)=0.2mA×100Ω/(1mA-0.2mA)=25Ω ①
量程10mA:R1与(Rg+R2)并联,总电流 I=10mA,总电压为U
U=Ug=Ig(Rg+R2) =(I-Ig)R1
得到:0.2mA(200Ω+R2)=(10mA-0.2mA)R1
200Ω+R2=49R1 ②
①与② 联立起来解得:
R1=4.5Ω
R2=20.5Ω