P-N Junction Diode question from JEE Main 2024 (4 Apr, Shift 1)

Answer: (b)

1. The left end ($-6$ V) is at higher potential than the right end ($-8$ V), so current flows left to right.
2. The $15\ \Omega$ branch conducts; the $10\ \Omega$ branch diode points right (forward) and conducts; the $5\ \Omega$ branch diode points left (reverse) and is open.
3. $R_{net} = \dfrac{15\times 10}{15 + 10} = 6\ \Omega$
4. So the answer is (b) $6\ \Omega$.