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 $\mathrm{R}_{\mathrm{S}}$ is 0.5 mA. The minimum value of the resistance of $R_{s}$, to ensure that the LED is not damaged is $\_\_\_\_$ $\mathrm{k} \Omega$.

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Current Electricity

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 Ω

(A)6
(B)2✓ Correct
(C)4
(D)5

Answer: (B)

Solution

The diode in the middle branch is reverse biased and carries no current, so the whole

0.5

mA passes through the LED. At rated power the LED drops

V_{L E D} = \frac{P}{i} = \frac{2 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}} = 4

V. The remaining voltage lies across

R_{S}

V_{R_{S}} = 5 - 4 = 1

V, so

R_{S} = \frac{V _{R_{S}}}{i} = \frac{1}{0.5 \times 1 0 ^{- 3}} = 2 \times 1 0^{3} Ω

. The minimum series resistance is

2 kΩ

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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