For a 470 Ω resistor carrying 1 mA of current, determine a suitable power (wattage) rating from the given choices, based on calculated dissipation and standard derating practice.

Introduction / Context:
Selecting an appropriate resistor wattage rating prevents overheating and improves reliability. The rating should exceed the expected power dissipation with ample margin, considering ambient temperature and enclosure conditions.

Given Data / Assumptions:

• Resistance R = 470 Ω.
• Current I = 1 mA = 0.001 A.
• Continuous operation; standard ambient conditions; typical derating margin applied.

Concept / Approach:
Resistor power dissipation is P = I^2 * R or P = V^2 / R. Compute actual dissipation, then choose a commercial wattage rating comfortably above that value. Common preferred ratings are 1/8 W (0.125 W), 1/4 W (0.25 W), 1/2 W (0.5 W), etc.

Step-by-Step Solution:
Compute power: P = I^2 * R = (0.001)^2 * 470 = 1e-6 * 470 = 0.00047 W.Convert to milliwatts: 0.00047 W = 0.47 mW.Compare with ratings: even 1/8 W = 125 mW greatly exceeds 0.47 mW.Choose a standard rating: 1/4 W is a very common minimum stocked value for through-hole parts, providing abundant margin.

Verification / Alternative check:
Voltage check: V = I * R = 0.001 * 470 = 0.47 V; P = V * I = 0.47 * 0.001 = 0.00047 W, matching the earlier calculation.

Why Other Options Are Wrong:
Higher ratings (1/2 W, 1 W, 2 W) are unnecessary here and larger/costlier without benefit.

While 1/8 W would also be sufficient electrically, 1/4 W is the more common stocking option in many labs and offers extra robustness; among the provided typical selections, 1/4 W is the best fit.

Common Pitfalls:
Choosing a rating too close to the calculated power can cause overheating under elevated ambient temperatures or tolerances; include adequate margin.

Final Answer:
1/4 watt