Power in a resistor from a battery: A 68 Ω resistor is connected directly across a 3 V battery. What is the resistor’s power dissipation?

Introduction / Context:
Computing power dissipation is essential for part selection and thermal design. Given voltage and resistance, you can directly compute power without finding current separately by using P = V^2 / R.

Given Data / Assumptions:

• Voltage V = 3 V (DC source).
• Resistance R = 68 Ω.
• Purely resistive load, steady state.

Concept / Approach:

Use the relation P = V^2 / R, which derives from combining Ohm’s law (V = I * R) with P = V * I. This is often the fastest way when V and R are directly given.

Step-by-Step Solution:

Verification / Alternative check:

Find current first: I = V / R = 3 / 68 ≈ 0.0441 A. Then power P = I^2 * R ≈ (0.0441)^2 * 68 ≈ 0.132 W. Both methods agree.

Why Other Options Are Wrong:

13.2 mW is 10× too small. 22.6 mW does not match any correct derivation. 226 mW is too large. 6.8 mW is far too small.

Common Pitfalls:

Rounding too early; confusing mW and W; forgetting to square the voltage in P = V^2 / R.

Final Answer:

132 mW