Concept:Using the diode's maximum power rating, calculate the maximum safe forward current \( I_{\text{max}} \), then apply Kirchhoff's Voltage Law to find the required series current-limiting resistor.
Formula:$$P_{\text{max}} = V_D \cdot I_{\text{max}} \implies I_{\text{max}} = \frac{P_{\text{max}}}{V_D}$$
$$R = \frac{V_S - V_D}{I_{\text{max}}}$$
Solution:- Maximum current: \( I_{\text{max}} = \frac{100\text{ mW}}{0.50\text{ V}} = \frac{0.10\text{ W}}{0.50\text{ V}} = 0.20\text{ A} = 200\text{ mA} \).
- Voltage across resistor: \( V_R = V_S - V_D = 1.50\text{ V} - 0.50\text{ V} = 1.00\text{ V} \).
- Required resistance: \( R = \frac{1.00\text{ V}}{0.20\text{ A}} = 5.0\ \Omega \).
Why other options are incorrect:- Option A: 2.5 Ω would allow \( I = 0.4\text{ A} \), resulting in \( P = 200\text{ mW} \) which exceeds the power rating.
- Option C: 10.0 Ω restricts current to 100 mA, which is below the maximum rated capacity.
- Option D: 20.0 Ω limits current to 50 mA.
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