(a) Briefly explain how a galvanometer is converted into a voltmeter.
(b) A voltmeter of a certain range is constructed by connecting a resistance of 90 `Omega` in series with a galvanometer. When the resistance of `470 Omega` is connected in series, the range gets halved. Find the resistance of the galvanometer.

Explanation

A galvanometer can be converted in a voltmeter by connecting a large resistance in series to the galvanometer.
`V=I_(g)G+I_(g)R`
`V=I_(g)(G+R)`
Hence value of high Resistance in series

`R=(V)/(I_(g))-G`
(b) `V=I_(g)(G+R)`
`V=I_(g)(G+980)" ....(1)"`
`(V)/(2)=I_(g)(G+470)" ....(2)"`
Solving equation (1) and (2) `G=40 Omega.`