Back-solve current from power: A 3.3 kΩ resistor is dissipating 0.25 W under steady DC conditions. What current is flowing through the resistor?

Introduction / Context:Given any two of power, resistance, and current for a resistor, you can find the third. This is particularly useful for verifying whether a resistor is within safe operating limits or diagnosing abnormal heating.

Given Data / Assumptions:

• Power P = 0.25 W.
• Resistance R = 3.3 kΩ = 3300 Ω.
• DC or RMS AC steady operation.

Concept / Approach:

Use the formula P = I^2 * R. Solving for current gives I = sqrt(P / R). Convert units carefully to avoid errors.

Step-by-Step Solution:

Verification / Alternative check:

Check via P = V * I and V = I * R: If I ≈ 8.7 mA, then V ≈ 0.0087 * 3300 ≈ 28.7 V. Power P ≈ 28.7 V * 0.0087 A ≈ 0.25 W. Consistent.

Why Other Options Are Wrong:

87 mA would give P ≈ (0.087)^2 * 3300 ≈ 25 W (far too high). 8.7 µA is 1000× too small. 8.7 A is unrealistic (thousands of watts). 0.87 mA yields negligible power relative to 0.25 W.

Common Pitfalls:

Forgetting the square root; mis-converting kΩ to Ω; mixing mA and A.

Final Answer:

8.7 mA