P-N Junction Diode question from MHT CET 2025 (21 Apr, Shift 2)

Answer: (d)

1. The battery's positive terminal sends current through $20\ \Omega$ to the left junction.
2. $D_1$ points right (forward biased), so the middle $30\ \Omega$ branch conducts; $D_2$ points left (reverse biased), so the top branch carries no current.
3. Total resistance $= 20 + 30 = 50\ \Omega$
4. $I = \dfrac{5}{50} = 0.1$ A
5. So the answer is (d) $0.10$ A.