JEE Main 2026 Apr 6 Shift 1, Physics Q14: Energy, Power And Heating Effect

JEE Main2026Apr 6, Shift 1Physics

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
$2$
C
$4$
D
$5$

Solution

In the configuration shown, the LED is reverse-biased so no current flows through it; the $0.5$ mA flows entirely through the $1$ k $Ω$ branch in parallel.

Voltage across the parallel branch: $V_{R} = I_{R} \cdot R$ . With the LED dissipating its rated power $P = V_{LED} I_{LED}$ at the same branch voltage, the constraint gives $V_{branch} = 4$ V (matching $P / I = 2 mW /0.5 mA = 4$ V).

By KVL: $5 = V_{R_{S}} + 4 \Rightarrow V_{R_{S}} = 1$ V.

$R_{S} = \frac{V _{R_{S}}}{I} = \frac{1}{0.5 \times 1 0 ^{- 3}} = 2$ k $Ω$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q14: Energy, Power And Heating Effect

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