Question

A galvanometer of resistance $50\Omega$ is connected to a battery of $3V$ along with a resistance of $2950\Omega$ in series. A full scale deflection of $30$ divisions is obtained in the galvanometer. In order to reduce this deflection to $20$ divisions the resistance in series should be:

• A. $4450\Omega$
• B. $5050\Omega$
• C. $5550\Omega$
• D. $6050\Omega$

Answer 1

The correct option is A $4450\Omega$

A galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.
Let $V$ be the potential difference to be measured by galvanometer.
To do so, a resistance $R$ of such a value is connected in series with the galvanometer so that if a potential difference $V$ is applied across the terminals $A$ and $B$, a current $I_g$ flows through the galvanometer.
Now, total resistance of voltmeter $=G+R$
From Ohm's law, $I_g=\frac{V}{G+R}$
or $G+R=\frac{V}{I_g}$
or $R=\frac{V}{I_g}-G$
Here,
The current through the galvanometer
$=\frac{3}{2950+50}={10}^{-3}A$
To reduce the deflection from 30 divisions to 20 divisions, the required current
$=\frac{20}{30}\times {10}^{-3}=\frac{2}{3}\times {10}^{-3}A$
The required resistance $R$ is given by
$\frac{3}{R+50}=\frac{2}{3}\times {10}^{-3}$
or $R=4450\Omega$