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 .

Step-by-step solution

Since the reverse-biased diode branch acts as an open circuit, the entire current through the series resistor R_S flows through the LED: I_ LED = I_ R_S = 0.5 mA At the boundary condition where the LED dissipates its maximum rated power of 2 mW : P_ LED = V_ LED I_ LED V_ LED = P_ LED I_ LED = 2 mW 0.5 mA = 4 V Applying Kirchhoff's voltage law to the loop: V_ input - V_ R_S - V_ LED = 0 5 - (I_ R_S R_S) - 4 = 0 I_ R_S R_S = 1 V R_S = 1 V 0.5 10⁻³ A = 2000 = 2 k Hence, the correct option is (2) 2 .