Electric Power
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
Power is the rate a component uses electrical energy: P = IΔV, which can also be written I²R or ΔV²/R. A bulb’s brightness is its power. Pick the form that uses whatever the components share.
1. Which form to use
2. Brightness puzzles
Series: same current, so P = I²R. The larger resistance uses more power.
Parallel: same voltage, so P = ΔV²/R. The smaller resistance uses more power. That’s why household appliances, all in parallel at 120 V, draw more power when their resistance is lower.
A bulb rated “60 W at 120 V” uses 60 W only at 120 V. At a different voltage, use its resistance: R = ΔV²/P = 240 Ω.
Power companies bill in kilowatt-hours: 1 kW for 1 hour.
3. Example, then try it
A bulb is rated 60 W at 120 V. Find the current through it and its resistance.
I = P/ΔV = 60 ÷ 120 = 0.50 A.
R = ΔV/I = 120 ÷ 0.50 = 240 Ω (or ΔV²/P = 14,400 ÷ 60).
A 1500 W space heater runs on 120 V for 2.0 hours a day for 30 days. Find its current and resistance, the energy it uses, and the cost at $0.15 per kWh.
Current: I = 1500 ÷ 120 = 12.5 A. Resistance: R = 120 ÷ 12.5 = 9.6 Ω.
Energy: 1.5 kW × 60 h = 90 kWh (that’s 3.2 × 10⁸ J).
Cost: 90 × $0.15 = $13.50.
Answer: 12.5 A, 9.6 Ω, 90 kWh, $13.50.
4. Common mistakes
The voltage across a resistor is cut in half. What happens to the power it uses?
P = ΔV²/R. Halving ΔV makes ΔV² a quarter as large, so P is ¼ as large.
Ready to practice?
4 questions on Electric Power, each with a hint and an explanation for every wrong answer.