Question

A moving coil galvanometer has a resistance of $10\, O$ and full scale deflection of $0.01\, A$ . It can be converted into voltmeter of $10\, V$ full scale by connecting into resistance of :

• A. $9.90 \,O$ in series
• B. $10 \,O$ in series
• C. $990 \,O$ in series
• D. $0.10 \,O$ in series

Answer 1

Answer: (C)

$990 \,O$ in series

Answer 2

Let $G$ be resistance of galvanometer and $i_g$ the current through it. Let $V$ is maximum potential difference, then from Ohm?? law

$i_{g} = \frac{V}{G + R}$
$\Rightarrow \,R = \frac{V}{i_{g}} - G$
Given, $G = 10\,\Omega, i_{g} = 0.01\,A$
$V = 10$ volt
$\therefore \, R=\frac{10}{0.01}-10=990\, \Omega$
Thus, on connecting a resistance $R$ of $990 \Omega$ in series with the galvanometer, the galvanometer will become a voltmeter of range zero to $10\, V$.
For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.