Question

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

• A. $2500\,\Omega$ as a shunt
• B. $245\,\Omega$ as a shunt
• C. $2550\,\Omega$ in series
• D. $2450\,\Omega$ in series

Answer 1

Answer: (D)

$2450\,\Omega$ in series

Answer 2

To convert a galvanometer into voltmeter, high resistance should be connected in series with it. Let $R$ is the resistance connected in series with the galvanometer. Galvanometer current $i_{g}=\frac{V}{G+R}$ or $R=\frac{V}{i_{g}}-G$ Given, $G=50 \,\Omega$, $i_{g}=25 \times 4 \times 10^{-4}=10^{-2} A , V=25 \,V$ $\therefore R=\frac{25}{10^{-2}}-50$ $=2500-50=2450\, \Omega$