Question

A galvanometer having a resistance of $100\Omega$ has $25$ division. A current of $0.4mA$ deflects the pointer of the galvanometer through one division. To convert this galvanometer into a voltmeter of $0$ to $25$ V, its should be connected with a resistance of

• A. $2000\Omega$ as shunt
• B. $2400\Omega$ as shunt
• C. $2400\Omega$ in series
• D. $2500\Omega$ is series

Answer 1

The correct option is C $2400\Omega$ in series

The resistance of the Galvanometer is low and the resistance of the voltmeter is high, to convert a galvanometer into a voltmeter, we use Resistance $R$ connected in the series.

Given,

$R_g=100\Omega$ have $25$ division

$I=0.4mA$

The total current is shown by the galvanometer

$I_g=25\times 4\times {10}^{-4}=0.01A$

The Resistance $R$ is connected in series to convert Galvanometer into voltmeter $25V$ is,

$R=\frac{V}{I_g}-R_g$

$R=\frac{25}{0.01}-100=2400\Omega$

The correct option is C.