A galvanometer has an internal resistance of `50Omega` and current required for full scale deflection is 1 mA. Find the series resistances required [ as shown fig] to use it as ameter with different ranges, as indicated in Fig.

Explanation

a) For range 1 volt, `I_(g)=V/((G+R_(1)))`
`therefore10^(-3)=1/((50+R_(1)))` or `50+R_(1)=1000`
`R_(1)=1000-50=950Omega`
b) For range 10 volt, `10^(-3)=10/((G+R_(1)+R_(2)))`
`thereforeG+R_(1)+R_(2)=10/(10^(-3))=10xx10^(3)`
or `R_(2)=10000-(50+950)=9000=9kOmega`
c) and for range 100V, `10^(-3)=100/((G+R_(1)+R_(2)+R_(3)))`
`thereforeG+R_(1)+R_(2)+R_(3)=100/(10^(-3))=100xx10^(3)`
or `R_(3)=100xx10^(3)-(G+R_(1)+R_(2))`
`=100xx10^(3)-10xx10^(3)=90xx10^(3)=90kOmega`