Semiconductor Electronics question for JEE Main (JEE Main 2023 (15 Apr, Shift 1)), with solution

Answer: (D) $1.5\ \text{A}$

The cell's positive terminal is at the right, so current leaves the bottom-right corner and returns to the left wire.
- $D_3$ points from the bottom-right corner towards the left wire: forward biased. Path: $10\ \Omega$.
- Through the right $10\ \Omega$ to the top-right corner, then $D_1$ (pointing left): forward biased. Path: $10+10=20\ \Omega$.
- $D_2$ points from the left wire to the top-right corner: reverse biased, no current.

The two conducting paths are in parallel: $R=\dfrac{10\times20}{30}=\dfrac{20}{3}\ \Omega$.
$$I=\frac{10}{20/3}=1.5\ \text{A}$$

Solution by Sreeraj P, M.Sc Physics