JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Current Electricity

The voltmeter of $400\Omega$ connected across one $100\Omega$ resistor forms a parallel combination of $\frac{400\times 100}{400+100}=80\Omega$. The circuit resistance becomes $100+80=180\Omega$, so the current drawn from the battery is $I=\frac{9}{180}=0.05$ A. The voltmeter reads the drop across the parallel section, $V=0.05\times 80=4$ V. The voltmeter reading is $4$ V.