Question

A galvanometer of $50$ ohm 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 $25$ $\text{V}$, it should be connected with a resistance of

​​​​​​​$[$1 Mark$]$

• A. $245\Omega$ in parallel
• B. $2550\Omega$ in series
• C. $2450\Omega$ in series
• D. $2500\Omega$ in parallel

Answer 1

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

Galvanometer resistance, $R_g=50\Omega$

Full-scale deflection current, $I_g=25\times 4\times {10}^{-4}\text{A}$

$I_g={10}^{-2}\text{A}$

To convert this galvanometer into a voltmeter having a range $\text{V}$, a suitable high resistance should be connected in series with the galvanometer such that when a potential difference of $\text{V}$ is applied, full-scale deflection current passes through galvanometer,


$V=I_g(R_g+R)$

$25={10}^{-2}(50+R)$

$R=2450\Omega$

$\therefore$ option $\left(c\right)$ is correct