Minimum Series Resistance for LED Circuit Protection

To find the minimum value of the series resistance $R_S$ to protect the LED from damage, we analyze the given circuit diagram and parameters step-by-step:

1. Analysis of the Circuit Branches:

   • The DC source provides an input voltage of $V_{\text{in}} = 5\text{ V}$.
   • The diode in the parallel branch with resistance $R = 1\text{ k}\Omega$ has its anode connected to the ground (lower potential) and its cathode connected to the positive terminal (higher potential). Thus, this diode is reverse-biased and acts as an open circuit (no current flows through this branch).
   • Therefore, the entire current flowing through $R_S$ passes through the LED: $I_{\text{LED}} = I = 0.5\text{ mA} = 0.5 \times 10^{-3}\text{ A}$

2. Maximum Allowable Voltage Across the LED:

   • The maximum rated power of the LED is $P_{\text{max}} = 2\text{ mW} = 2 \times 10^{-3}\text{ W}$.
   • For the LED to remain safe and not be damaged, the electrical power dissipated by it must satisfy: $P_{\text{LED}} = V_{\text{LED}} \cdot I_{\text{LED}} \le P_{\text{max}}$
   • Solving for the maximum operating voltage across the LED: $V_{\text{LED, max}} = \frac{P_{\text{max}}}{I_{\text{LED}}} = \frac{2 \times 10^{-3}\text{ W}}{0.5 \times 10^{-3}\text{ A}} = 4\text{ V}$

3. Determining the Resistance $R_S$:

   • Applying Kirchhoff's Voltage Law (KVL) around the loop: $V_{\text{in}} = V_{R_S} + V_{\text{LED}}$ $V_{R_S} = V_{\text{in}} - V_{\text{LED}}$
   • To keep $V_{\text{LED}} \le 4\text{ V}$, the minimum voltage drop across the resistor $R_S$ must be: $V_{R_S, \text{min}} = 5\text{ V} - 4\text{ V} = 1\text{ V}$
   • Using Ohm's Law, the minimum value of $R_S$ is: $R_{S, \text{min}} = \frac{V_{R_S, \text{min}}}{I} = \frac{1\text{ V}}{0.5 \times 10^{-3}\text{ A}} = 2000\text{ }\Omega = 2\text{ k}\Omega$

Correct Answer: B (2)