# How is a moving coil Galvanometer converted into Voltmeter? Explain giving the necessary circuit diagram and the required mathematical relation used.

Let

*G*= resistance of galvanometer

*n*= number of scale divisions in the galvanometer

*K*= figure of merit of galvanometer

Current which produce full deflection in the galvanometer,

*I*g =

*nK*

Let

*V*be the potential difference to be measured by galvanometer.

If

*R*be the resistance connected in series. Then

Total resistance of galvanometer =

*R*+

*G*

From ohm's law

${I}_{\mathrm{g}}=\frac{V}{R+G}\phantom{\rule{0ex}{0ex}}\Rightarrow R+G=\frac{V}{{I}_{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{V}{{I}_{\mathrm{g}}}-G$

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