Compute resistor power: When 8 mA flows through a 12 kΩ resistor, what is the resulting power dissipation?
Electrical Engineering
Energy and Power
Difficulty: Easy
Choose an option
-
A7.68 mW
-
B768 mW
-
C7.68 W
-
D76.8 W
Answer
Correct Answer: 768 mW
Explanation
Introduction / Context:Translating current through a known resistance into power helps ensure components operate within safe limits. This is commonly needed when verifying resistor wattage in bias networks and load resistors.
Given Data / Assumptions:
- I = 8 mA = 0.008 A.
- R = 12 kΩ = 12,000 Ω.
- DC or RMS values implied; use standard power formulas.
Concept / Approach:Use P = I^2 * R. Alternatively, compute voltage first via V = I * R and then apply P = V * I. Both yield the same result when done consistently in SI units.
Step-by-Step Solution:
Compute P = I^2 * R = (0.008)^2 * 12,000.(0.008)^2 = 0.000064; multiply: 0.000064 * 12,000 = 0.768 W.Convert to milliwatts: 0.768 W = 768 mW.Verification / Alternative check:Find V: V = I * R = 0.008 * 12,000 = 96 V. Then P = V * I = 96 * 0.008 = 0.768 W. Both methods agree.
Why Other Options Are Wrong:
- 7.68 mW: Off by a factor of 100; likely from moving the decimal incorrectly.
- 7.68 W and 76.8 W: Too large; would require far higher current or resistance.
Common Pitfalls:
- Forgetting to convert kΩ to Ω and mA to A before substituting.
- Dropping zeros when squaring the current.
Final Answer:768 mW