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\(\Omega\).

Since the reverse-biased diode branch acts as an open circuit, the entire current through the series resistor \(R_S\) flows through the LED:
\(I_{\text{LED}} = I_{R_S} = 0.5\text{ mA}\)

At the boundary condition where the LED dissipates its maximum rated power of \(2\text{ mW}\):
\(P_{\text{LED}} = V_{\text{LED}} \cdot I_{\text{LED}}\)

\(V_{\text{LED}} = \dfrac{P_{\text{LED}}}{I_{\text{LED}}} = \dfrac{2\text{ mW}}{0.5\text{ mA}} = 4\text{ V}\)

Applying Kirchhoff's voltage law to the loop:
\(V_{\text{input}} - V_{R_S} - V_{\text{LED}} = 0\)

\(5 - (I_{R_S} \cdot R_S) - 4 = 0\)

\(I_{R_S} \cdot R_S = 1\text{ V}\)

\(R_S = \dfrac{1\text{ V}}{0.5 \times 10^{-3}\text{ A}} = 2000\ \Omega = 2\text{ k}\Omega\)

Hence, the correct option is \((2)\ 2\).