A moving coil galvanometer has 100 equal divisions. Its current sensitivity is 20 divisions per mA and voltage sensitivity is 4 divisions per mV. To convert it into a voltmeter in which each division reads 1 V, the resistance needed to be connected in series with the coil will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 19995\,\Omega \)
A galvanometer becomes a voltmeter when a large resistance is joined in series with it, so that the full-scale current flows at the full-scale voltage. The two sensitivities give the full-scale current and the coil resistance.
\( \displaystyle I_{g} = \frac{100\ \text{div}}{20\ \text{div/mA}} = 5\,\mathrm{mA} = 5\times10^{-3}\,\mathrm{A} \)
\( \displaystyle V_{g} = \frac{100\ \text{div}}{4\ \text{div/mV}} = 25\,\mathrm{mV} \)
\( \displaystyle G = \frac{V_{g}}{I_{g}} = \frac{25\,\mathrm{mV}}{5\,\mathrm{mA}} = 5\,\Omega \)
Each of the 100 divisions must read 1 V, so the range is \( V = 100 \times 1 = 100\,\mathrm{V} \).
\( \displaystyle R = \frac{V}{I_{g}} - G = \frac{100}{5\times10^{-3}} - 5 = 20000 - 5 = 19995\,\Omega \)