A galvanometer can pass a small current of 1 mA. It is converted to a voltmeter which can read 10 V in full scale deflection by connecting a high resistance of 9900 ohm. The resistance of the galvanometer is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 100 ohm
A galvanometer is converted into a voltmeter by connecting a high resistance R in series. At full-scale deflection the full-scale current \( I_g \) flows through both, and the total voltage across them is the range V.
\( \displaystyle V = I_g (G + R) \)
\( \displaystyle G = \frac{V}{I_g} - R \)
\( \displaystyle G = \frac{10\,\mathrm{V}}{1\times10^{-3}\,\mathrm{A}} - 9900\,\Omega \)
\( \displaystyle G = 10000\,\Omega - 9900\,\Omega = 100\,\Omega \)