A galvanometer coil has a resistance of 10 Ω. It shows full-scale deflection for a current of 3.0 mA. To convert it into a voltmeter of range 15 V, we must connect a resistance R of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4990\ \Omega \) in series with the galvanometer
A voltmeter is a galvanometer with a large resistance \( R \) in series. At full scale the current \( I_g \) flows through both the coil and \( R \), so \( V=I_g(G+R) \). A resistance in parallel would only take current away from the coil. It cannot raise the voltage range.
Given:
\( G=10\ \Omega \)
\( I_g=3.0\ \mathrm{mA}=3.0\times10^{-3}\ \mathrm{A} \)
\( V=15\ \mathrm{V} \)
\( \begin{aligned} R &= \frac{V}{I_g}-G \\ &= \frac{15}{3.0\times10^{-3}}-10 \\ &= 5000-10 \\ &= 4990\ \Omega \end{aligned} \)
Therefore, a resistance of \( 4990\ \Omega \) must be connected in series with the galvanometer.
NCERT (2026-27):
The Moving Coil Galvanometer – “… To ensure this, a large resistance R is connected in series with the galvanometer. …”