A galvanometer of resistance `5 Omega` can be converted into a voltmeter of range 0 - V voltmeter of range 0 - V volts by connecting a resistance `R Omega` in series with it. How much resistance will be required to charge its range to `0 - V/2` volts?

Explanation

As galvanometer of resistance G can be converted into a voltmeter of range 0 - V volts by adding a series resistance `R Omega`, hence
`R = V/(I_g) - G implies I_(g) = V/(R + G)`
Let a resistance `R. Omega` be joined in series with galvanometer to convert it into a voltmeter of range `V. = V/2` volts , then
`R. = (V.)/(I_g) - G = (V//2)/(V//R + G) - G = (R + G)/(2) - G = (R - G)/(2)`.