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)?
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A150 mW
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B4.8 W
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C15 mW
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D480 mW
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E75 mW
Answer
Correct Answer: 150 mW
Explanation
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:
Compute P = I^2 * R = (0.025)^2 * 120 = 0.000625 * 120 = 0.075 W.Required minimum is 0.075 W (75 mW).Choose the next sensible standard rating above 75 mW → 150 mW (0.15 W) or 1/4 W (250 mW). Among provided options, 150 mW is the smallest that exceeds the need.Therefore, select 150 mW.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