A galvanometer of resistance 50 Ω shows full scale deflection for a current of 3 mA. To convert it into a voltmeter to measure 15 V maximum potential difference, the resistance required to be connected in series with it will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4950 Ω
To make a voltmeter, a high resistance R in series is chosen so that the full-scale current flows when the maximum voltage is applied across the combination.
\( \displaystyle I_g (G + R) = V \)
\( \displaystyle R = \frac{V}{I_g} - G \)
\( \displaystyle R = \frac{15\,\mathrm{V}}{3\times10^{-3}\,\mathrm{A}} - 50\,\Omega \)
\( \displaystyle R = 5000\,\Omega - 50\,\Omega = 4950\,\Omega \)