A multirange current meter can be constructed by using a galvanometer circuit as shown in figure. We want a current meter that can measure `10mA`, `100mA` and `1A` using a galvanometer of resistance `10Omega` and that produces maximum deflection for current of `1mA`. Find `S_1, S_2` and `S_3` that have to be used.

Explanation

`I_(G).G=(I_(1)-I_(G))(S_(1)+S_(2)+S_(3))` for `I_(1)=10mA`
`I_(G)(G+S_(1))=(I_(2)-I_(G))(S_(2)+S_(3))` for `I_(2)=100mA`
and `I_(G)(G+S_(1)+S_(2))=(I_(3)-I_(G))(S_(3))` for `I_(3)=1A`
gives `S_(1)=1W,S_(2)=0.1W`
and `S_(3)=0.01W`