Question

A galvanometer has a resistance of $30\Omega$ and a current of $2\text{mA}$ is needed for a full-scale deflection. What is the resistance needed to convert galvanometer into a voltmeter of $0.2\text{V}$ range ?

• A. $24\Omega$
• B. $70\Omega$
• C. $10\Omega$
• D. $48\Omega$

Answer 1

The correct option is B $70\Omega$

Given:
$G=30\Omega ;I_g=2\text{mA};V=0.2\text{V}$

Let $R$ be the required resistance to convert galvanometer into a voltmeter.

We know that, $R=\frac{V}{I_g}-G$

$R=\frac{0.2}{2\times {10}^{-3}}-30$

$\therefore R=70\Omega$

Hence, option (b) is the correct answer.

Key Concept: To convert a galvanometer into a voltmeter of certain range say $V$, a high resistance $R$ is connected in series with the galvanometer so that current passing through the galvanometer of resistance $G$ becomes equal to its full scale deflection value $\left(I_g\right)$. This is possible only if $V=I_g(G+R)i.e.R=\frac{V}{I_g}-G$