Special Purpose Diodes question from JEE Main 2026 (6 Apr, Shift 1)

Answer: (b)

1. The diode in series with $R = 1\text{ k}\Omega$ is reverse biased, so that branch carries no current; all $0.5$ mA flows through the LED.
2. Largest allowed LED voltage: $V_{LED} = \frac{P}{I} = \frac{2\text{ mW}}{0.5\text{ mA}} = 4$ V
3. Drop across $R_S$ must be at least $5 - 4 = 1$ V.
4. $R_{S,min} = \frac{1}{0.5 \times 10^{-3}} = 2\text{ k}\Omega$, so the answer is (b) $2$.