A galvanometer of resistance $G$ has voltage range $V_g$. Resistance requirec to convert it to read voltage…
- $\frac{G \cdot V_g}{V}-G$
- $\left(\frac{G+V_g}{V}\right) \cdot G$
- $\left(\frac{V-V_g}{V}\right) \cdot G$
- $G \cdot\left[\frac{V}{V_g}-1\right]$
Solution
Consider the following, net potential drop across the voltmeter:
$I_g R+l_g G=V$
On re-writing, $R=\frac{V}{I_g}-G$, and we know, $V_g=I_g G$.
Therefore, $R=G\left(\frac{V}{V_g}-1\right)$Asked in: MHT CET 2022 (05 Aug Shift 1)