JEE Main 2026 — Current Electricity Question with Solution

Let $G$ be the resistance of the galvanometer and $I_g$ be the current required for full scale deflection.

When the galvanometer is shunted with $S=2\Omega$, it acts as an ammeter of range $I=500\text{ mA}=0.5\text{ A}$. The potential difference across the galvanometer and the shunt is equal:

$I_{g}G=(I-I_g)S$

$I_{g}G=(0.5-I_g)\times 2$

$I_{g}G+2I_g=1$

$I_g=\frac{1}{G+2}$

When a resistance $R=470\Omega$ is connected in series, it acts as a voltmeter of range $V=10\text{ V}$. The total voltage is the sum of the voltage drops:

$V=I_g(G+R)$

$10=I_g(G+470)$

$I_g=\frac{10}{G+470}$

Equating the two expressions for $I_g$:

$\frac{1}{G+2}=\frac{10}{G+470}$

$G+470=10(G+2)$

$G+470=10G+20$

$9G=450$

$G=50\Omega$

Answer: $50$