Q.Explain the conversion of galvanometer into voltmeter.

A galvanometer becomes a voltmeter of a chosen range $V$ by adding a large series resistance $R=(V/I_g)-G$, so that the maximum voltmeter reading $V$ drives exactly the galvanometer's full-scale current $I_g$.

Working

Let the galvanometer have resistance $G$ and require a current $I_g$ for full-scale deflection.

To use it as a voltmeter reading up to a maximum voltage $V$, a high resistance $R$ is connected in series with the galvanometer, and this series combination is connected across the points whose potential difference is to be measured.

For the combination to show full-scale deflection exactly when the voltage across it is $V$, the current through the series circuit at that moment must equal $I_g$:

$V=I_g(G+R)$

$R=\dfrac{V}{I_g}-G$

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