Compute resistor power: When 8 mA flows through a 12 kΩ resistor, what is the resulting power dissipation?

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:

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