Question

A galvanometer of $50\Omega$ resistance has $25$ divisions. A current of $4\times {10}^{-4}\text{A}$ gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of $5\text{V}$, it should be connected with a resistance of

• A. $2450\Omega$, in parallel
• B. $2550\Omega$. in series
• C. $2450\Omega$, in series
• D. $2550\Omega$, in parallel

Answer 1

The correct option is C $2450\Omega$, in series

Galvanometer has $25$ divisions and current of $4\times {10}^{-4}\text{A}$ gives a deflection of $1$ division.

$\Rightarrow$ the maximum current that can pass through galvanometer is$I_g=25\times 4\times {10}^{-4}={10}^{-2}\text{A}$
Range of the voltmeter is, $V=25\text{V}$ and $G=50\Omega$
So, to convert it into a voltmeter, say, a resistance $R$ must be added in series.
$\Rightarrow R=\frac{V}{I_g}-G$
$=\frac{25}{{10}^{-2}}-50=2450\Omega$

Hence, $\left(C\right)$ is the correct answer.