Question

How can a galvanometer be converted into a voltmeter? Explain with the help of a circuit diagram.

Hint:

Put a high resistance \(R\) in series with the galvanometer so that \(V=I_g(G+R)\), giving \(R=\dfrac{V}{I_g}-G\).

Answer 1

Step 1: Starting point.
Let the galvanometer have coil resistance \(G\) and let \(I_g\) be the current that deflects its pointer fully. Across it this corresponds to only a tiny voltage \(I_gG\), far too small to serve as a voltmeter.

Step 2: Add a series multiplier resistance.
To measure up to a chosen voltage \(V\), put a large resistance \(R\) (called the multiplier) in series with the coil, and connect this series pair in parallel with the component whose voltage is wanted.

Step 3: Apply Ohm's law to the series pair.
When \(V\) appears across the pair, the current is \(I_g\) and the total resistance is \((G+R)\): \[I_g=\frac{V}{G+R}\ \Rightarrow\ G+R=\frac{V}{I_g}\ \Rightarrow\ R=\frac{V}{I_g}-G.\]
Step 4: Consequence for measurement.
Because \(R\) is large, the converted meter has a high resistance and draws almost no current from the circuit, so it reads the true potential difference. Choosing a larger \(R\) raises the voltage range of the same galvanometer.
Circuit: element in parallel with (G in series with R). \[\boxed{R=\frac{V}{I_g}-G}\]