Semiconductor Electronics question for JEE Main (JEE Main 2025 (23 Jan, Shift 2)), with solution

Answer: (B) $0.5\ \text{A}$

Both diodes point the same way and are forward biased by the cell (treat them as ideal, zero resistance in forward bias).

The two branches, each a diode with $20\ \Omega$, are then in parallel:
$$R = \frac{20\times20}{20+20} = 10\ \Omega$$
$$I = \frac{5}{10} = 0.5\ \text{A}$$

Solution by Sreeraj P, M.Sc Physics