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: Idea.
A galvanometer gives full-scale deflection for a very small current \(I_g\) and has a small resistance \(G\); by itself it can stand only a small voltage. To read larger voltages it is converted into a voltmeter by connecting a suitable high resistance \(R\) in series with it.

Step 2: Circuit diagram (described).
The galvanometer G and a high resistance R are joined in series, and this series combination is connected across the two points A and B whose potential difference \(V\) is to be measured:
A \(\longrightarrow\) [ Galvanometer G ] \(\longrightarrow\) [ high resistance R ] \(\longrightarrow\) B, the whole branch placed in parallel with the element.

Step 3: Find R.
For full-scale deflection the same current \(I_g\) passes through G and R when the applied voltage is \(V\): \[ V = I_g (G + R) \] \[ R = \frac{V}{I_g} - G \] Step 4: Why series and high value.
The large series resistance keeps the current through the meter small and gives the voltmeter a very high total resistance, so that when connected across a circuit element it draws negligible current and does not disturb the reading. An ideal voltmeter has infinite resistance. \[\boxed{R = \frac{V}{I_g} - G\ \ (\text{connected in series})}\]