JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Current Electricity

The diodes $D_1$ and $D_3$ are forward biased and conduct, while $D_2$ is reverse biased and carries no current. The branch through $D_3$ places $10\Omega$ directly across the battery, while the branch through $D_1$ together with the $10\Omega$ arm on the right offers $10+10=20\Omega$; these two conducting paths are in parallel, so $R_{eq}=\frac{10\times 20}{10+20}=\frac{20}{3}\Omega$. The current drawn from the battery is $I=\frac{V}{R_{eq}}=\frac{10}{20/3}=1.5\text{A}$.