A galvanometer has resistance ' \(\mathrm{G}\) ' and range ' \(\mathrm{V} g\) '. How much resistance is required to read voltage upto ' \(\mathrm{V}\) ' volt?

Given: Resistance of the galvanometer \(=\mathrm{G}\)
Range of the galvanometer \(=\mathrm{V}_{\mathrm{g}}\)
The series resistance value to be used for converting the galvanometer into a voltmeter of range 0 to \(\mathrm{V}_{\mathrm{g}^{\prime}}\) is,
\(\mathrm{R}=\frac{\mathrm{V}_{\mathrm{z}}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G}\)
Also,
\(\mathrm{I}_{\mathrm{g}}=\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}\)
To increase the measuring range to \(\mathrm{V}\), the new
resistance value \(\mathrm{R}^{\prime}=\frac{\mathrm{V}}{\left(\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{G}}\right)}-\mathrm{G}\)
\(=\frac{\mathrm{VG}}{\mathrm{V}_{\mathrm{s}}}-\mathrm{G}=\mathrm{G}\left(\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{g}}}-1\right)\)