Electric power and the electricity bill

A 60 W bulb left on for 1,000 hours uses 60 kWh; at 12 cents per kWh that is $7.20.

Power: energy per second

A bulb's rating in watts tells you how fast it turns electrical energy into light and heat. Power is energy per second, and in a circuit it has a neat form. Voltage is energy per coulomb and current is coulombs per second, so multiplying them gives energy per second: P = I V. One ampere at one volt is one watt. A car's power outlet rated at 20 A on a 12 V system can deliver up to 20 × 12 = 240 W. Combine P = I V with Ohm's law and you get two more forms for resistors: P = I² R and P = V² ÷ R.

The three forms tell different stories. P = V² ÷ R says that on a fixed voltage, a lower resistance means more power. That is why a 60 W bulb made for 120 V has a lower resistance (120² ÷ 60 = 240 Ω) than a 25 W bulb for the same voltage (120² ÷ 25 = 576 Ω). Because V is squared, doubling the voltage on the same resistance makes four times the power, which is why a bulb made for 120 V burns out quickly on 240 V.

Which bulb is brighter?

In the lab, a bulb's glow follows its power. Here is a puzzle that catches many people. Take a 6 Ω bulb and a 24 Ω bulb. Side by side in parallel on 6 V, each has the full 6 V, and P = V² ÷ R gives 6 W for the 6 Ω bulb and 1.5 W for the 24 Ω one, so the low-resistance bulb is brighter. Now put the same two in series on 6 V. The current is 6 ÷ 30 = 0.2 A through both, and P = I² R gives 0.24 W for the 6 Ω bulb and 0.96 W for the 24 Ω bulb. Now the high-resistance bulb is four times brighter. In series the same current flows through both, so the bigger resistance takes more power; in parallel both have the same voltage, so the smaller resistance does.

From watts to the bill: the kilowatt-hour

You pay for energy, not power. Energy is power multiplied by time, E = P t. A joule is far too small for household use, so electricity meters count kilowatt-hours (kWh): the energy used by 1,000 W running for one hour. One kWh is 1,000 W × 3,600 s = 3.6 × 10⁶ J. To estimate the cost of running an appliance, change its power to kilowatts, multiply by the hours it runs to get kWh, and multiply by the price per kWh on your bill. A 2,000 W heater used for half an hour takes 2 × 0.5 = 1 kWh. A 10 W LED bulb on 5 hours a day for 30 days uses 10 × 5 × 30 = 1,500 Wh, which is 1.5 kWh.

OpenStax works through a comparison. At 12 cents per kWh, a 60 W incandescent bulb costing 25 cents and lasting 1,000 hours uses 60 kWh, which costs $7.20 for the electricity, $7.45 in all. A compact fluorescent bulb gives the same light at a quarter of the power, 15 W, so its electricity for the same 1,000 hours costs $1.80. It costs $1.50 but lasts 10,000 hours, so only a tenth of its price, 15 cents, belongs to those 1,000 hours. Its total is $1.95. LED bulbs use even less. Prices per kWh vary by country, by supplier and often by how much you use in a month, so the price on your own bill is the one to use.