Power rating adequacy: A 10 Ω resistor carries 500 mA continuously. Is selecting a 2 W resistor adequate for safe operation?

Introduction / Context:
Selecting an appropriate resistor wattage rating is critical for reliability and safety. The rating must exceed the expected power dissipation, typically with margin to account for temperature rise, airflow, enclosure, and tolerance.

Given Data / Assumptions:

• Resistance R = 10 Ω.
• Current I = 0.5 A (500 mA), steady-state.
• Continuous operation; ambient conditions not specified (assume typical lab/room temperature).

Concept / Approach:
Power in a resistor is computed using P = I^2 * R = V^2 / R = V * I. After calculating the actual dissipation, choose a resistor with a power rating comfortably above that value. Many designers target 1.5× to 2× margin depending on environment and derating guidelines.

Step-by-Step Solution:

Verification / Alternative check:
Using P = V * I: first compute V = I * R = 0.5 * 10 = 5 V. Then P = 5 V * 0.5 A = 2.5 W, confirming the result.

Why Other Options Are Wrong:

• “Adequate — 2 W” (option a) is below required 2.5 W; the resistor will overheat.
• “Adequate — 1.5 W” (option c) miscalculates power.
• “Not adequate — 5 W dissipation” (option d) overstates the load; required dissipation is 2.5 W, not 5 W.

Common Pitfalls:
Ignoring derating for temperature, using nameplate ratings at 25 °C without considering enclosure heating, and rounding current/voltage before squaring in P = I^2 * R.

Final Answer:
Not adequate — actual dissipation is 2.5 W; select at least a 3 W resistor (with margin).