Physics · Semiconductor and Electronic Devices

JEE Main 2026 — 6 April, Morning Shift — Question 10

The maximum rated power of the LED is $2\mathrm{mW}$ and it is used in the circuit with input voltage of $5\mathrm{V}$ as shown in the figure below. The current through resistance $R_s$ is $0.5\mathrm{mA}$ . The minimum value of the resistance of $R_s$ to ensure that the LED is not damaged is ______ kΩ.

Option A:
1
Option B:
2
Correct
Option C:
3
Option D:
4

Answer: B

Step-by-step solution

LED is in reverse bias, so no current through it. Power rating: $P = V_{LED} I$ ? Actually maximum power dissipation = $2$ mW, and current through Rs is 0.5 mA, so voltage across LED = $P/I = 2\times10^{-3} / 0.5\times10^{-3} = 4$ V. Then voltage across Rs = $5-4=1$ V. Rs = $1 / (0.5\times10^{-3}) = 2000 \Omega = 2$ kΩ.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Semiconductor and Electronic Devices
Topic: p-n Diode and its Applications