How can a galvanometer be converted into a voltmeter to read a maximum potential difference of V ? Support your answer with related mathematical expression.

Explanation

To convert a galvanometer into a voltmeter, we join a suitable high resistance in series of galvanometer. If the galvanometer gives full scale deflection for a current `I_g` and has a resistance `R_G`, then to measure a potential difference up to V volt we must join a resistance R in series, so that
`V = I_g (R_G + R)`
`:. " " R = V/(I_g) - R_G`.