P-N Junction Diode question from JEE Main 2026 (21 Jan, Shift 2)

Answer: (c)

1. In steady state no current flows in the capacitor branch.
2. The battery makes the left node positive, so the $4\ \Omega$ branch diode is forward biased and the $3\ \Omega$ branch diode is reverse biased.
3. Loop current: $I = \frac{2.5}{1 + 4} = 0.5$ A.
4. Voltage between the two nodes $= 2.5 - 0.5 \times 1 = 2$ V, and this is the capacitor voltage (no drop across $2\ \Omega$).
5. $Q = CV = 5\ \mu\text{F} \times 2 = 10\ \mu\text{C}$. So the answer is (c).