Semiconductor Electronics question for JEE Main (JEE Main 2024 (4 Apr, Shift 1)), with solution

Answer: (A) $6\ \Omega$

The left terminal ($-6\ \text{V}$) is at the higher potential, so current tends to flow from left to right.

The diode in the $10\ \Omega$ branch points left to right: forward biased, it conducts. The diode in the $5\ \Omega$ branch points right to left: reverse biased, no current.
$$R = \frac{15\times10}{15 + 10} = 6\ \Omega$$

Solution by Sreeraj P, M.Sc Physics