JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Current Electricity

The voltmeter is in parallel with the $5\Omega$ resistor, so the $2$ V it reads is the voltage across that resistor and the current in it is $i_1=\frac{2}{5}=0.4$ A. The remaining emf, $3-2=1$ V, appears across the $2\Omega$ resistor, so the current drawn from the source is $i=\frac{1}{2}=0.5$ A. The current through the voltmeter is $i-i_1=0.5-0.4=0.1$ A, and with $2$ V across it its resistance is $R=\frac{2}{0.1}=20\Omega$.