A galvanometer having resistance of `50 Omega` requires a current of `100 micro A` to given full scale deflection. How much resistance is required to convert it into an ammeter of range of 10 A ?

Explanation

(a) to convert galvanometer intio ammeter a shunt has to be connected in parallel to the galvanometer.
Thus `2xx10^(-3)R_(G)=(0.302xx10^(-3))r_(s)`

`r_(s)=(2xx10^(-3)xx30)/0.3=0.2Omega`
(b) To convert galvanometer into voltmeter a resistance has to be connected in series
`implies2xx10^(-3)(R_(G)+R)=0.2`
`implies30+R=100`
`impliesR=70Omega`