A moving coil galvanometer has a resistance of \( 40\,\Omega \). It gives a full scale deflection with a current of 2 mA. How is it to be modified to convert it to a voltmeter of 5 V range by connecting some resistance?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2460\,\Omega \) connected in series with galvanometer
A galvanometer is converted into a voltmeter by connecting a high resistance in series with it, chosen so that the full-scale current flows when the full range voltage is applied.
\( \displaystyle I_g\,(G + R) = V \;\Rightarrow\; R = \frac{V}{I_g} - G \)
\( I_g = 2\,\mathrm{mA} = 2\times10^{-3}\,\mathrm{A} \)
\( G = 40\,\Omega \)
\( V = 5\,\mathrm{V} \)
\( \displaystyle R = \frac{5}{2\times10^{-3}} - 40 = 2500 - 40 = 2460\,\Omega \)