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JEE Main 2026 · 6 Apr · Shift 1, Physics question 39

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 RSR_S is 0.5 mA.

The minimum value of the resistance of RSR_S, to ensure that the LED is not damaged is ________ kΩ.

A 5 V cell feeds R_S in series; after R_S two parallel branches return to the cell: one with R = 1 kΩ in series with a diode, the other with the 2 mW LED.
Choose one option
Answer and worked solution

Official answer: B

The R branch. Its diode is reverse-biased, so it carries no current. The 0.5 mA through RSR_S all goes through the LED.

LED voltage. At its maximum power, VLED=2 mW0.5 mA=4V_{LED}=\dfrac{2\text{ mW}}{0.5\text{ mA}}=4 V.

Resistor.
RS=5−40.5×10−3=2 kΩ.R_S=\frac{5-4}{0.5\times10^{-3}}=2\text{ k}\Omega.

(B)

Topics:Special diodes (LED, Zener, photodiode, solar cell)Semiconductors and p-n junction
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