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) $2$
Concept. The branch with the reverse-biased diode carries no current, so all $0.5$ mA flows through the LED. Its voltage follows from its power rating.
Step 1. LED voltage. $V=\dfrac{P}{I}=\dfrac{2\times10^{-3}}{0.5\times10^{-3}}=4$ V.
Step 2. Resistor. $R_S=\dfrac{5-4}{0.5\times10^{-3}}=2$ k$\Omega$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
