Q39. The maximum rated power of the LED in the circuit shown in the figure is 2 mW, and the input voltage is 5 V. The current through the series resistance Rₛ is 0.5 mA. What is the minimum value of Rₛ required to ensure that the LED is not damaged?

Question

G1 Physics · Accepted Answer

From the polarity shown in the circuit, the branch containing the diode and the 1 kΩ resistor is reverse biased, so no current flows through that branch. Therefore the current through the LED is the same as the current through Rₛ: I = 0.5 mA. At the maximum safe LED power, P(LED) = V(LED)I. Thus, V(LED) = 2 mW ÷ 0.5 mA = 4 V. The supply voltage is 5 V, so the voltage across Rₛ is V(Rₛ) = 5-4 = 1 V. Therefore, Rₛ = 1 V ÷ 0.5 mA = 2000 Ω = 2 kΩ. Correct option: B.