Resistor power rating: A 120 Ω resistor must safely carry a maximum current of 25 mA. What minimum power rating should be specified (choose the next sensible standard rating above the calculated dissipation)?

Introduction / Context:Choosing an appropriate resistor power rating prevents overheating and premature failure. The rating must exceed the expected worst-case dissipation, ideally with safety margin and considering ambient conditions.

Given Data / Assumptions:

• Resistance R = 120 Ω.
• Maximum current I_max = 25 mA = 0.025 A.
• Assume continuous operation and pick at least the next higher standard rating.

Concept / Approach:

Resistor power dissipation is P = I^2 * R. After computing the required wattage, select a commercially available rating above that value (e.g., 1/8 W, 1/4 W, 1/2 W, etc.).

Step-by-Step Solution:

Verification / Alternative check:

Temperature rise and tolerances suggest margin. 150 mW provides 2× nominal headroom over 75 mW dissipation, a reasonable conservative choice.

Why Other Options Are Wrong:

15 mW is below requirement; would overheat. 4.8 W and 480 mW are overkill compared to the actual need; while safe, they are not the 'minimum' practical choice here. 75 mW equals the exact dissipation with zero margin, not advisable as a rating.

Common Pitfalls:

Choosing a rating equal to calculated power with no margin; mis-converting milliamps to amps; forgetting squared dependence on current.

Final Answer:

150 mW