Q 12-14-038JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
The maximum rated power of the LED is $2$ mW and it is used in the circuit with input voltage of $5$ V as shown in the figure below. The current through resistance $R_S$ is $0.5$ mA.
The minimum value of the resistance of $R_S$, to ensure that the LED is not damaged is ______ k$\Omega$.
Answer: (B) $2$
The diode in the middle branch is reverse biased, so no current flows through the $1\ \text{k}\Omega$ branch; all of the $0.5$ mA goes through the LED.
LED power $\le 2$ mW: $V_{\text{LED}} \le \dfrac{2\ \text{mW}}{0.5\ \text{mA}} = 4$ V.
So $R_S$ must drop at least $5 - 4 = 1$ V: $R_S \ge \dfrac{1}{0.5 \times 10^{-3}} = 2\ \text{k}\Omega$.
Solution by Sreeraj P, M.Sc Physics