Explain how will you convert moving coil galvanometer into voltmeter?

Answer: By connecting a suitable high resistance \(X=\dfrac{V}{I_g}-G\) in series with the galvanometer.

Let the galvanometer have resistance \(G\) and full-scale deflection current \(I_g\). To measure voltages up to \(V\), a high resistance \(X\) is connected in series with it. [Figure: galvanometer G in series with high resistance X, the combination connected across the points A and B whose potential difference V is to be measured.]

When the full voltage \(V\) is applied, the current through the combination must be \(I_g\):

\(V=I_g(G+X)\quad\Rightarrow\quad X=\dfrac{V}{I_g}-G\).

The scale is then calibrated in volts. The series resistance makes the total resistance of the voltmeter very high, so it draws very little current from the circuit; an ideal voltmeter has infinite resistance.