Watt's law

Updated November 8, 2026 · Written for the 2023 NEC

What Watt's law is

Watt's law describes how fast electrical energy is converted into heat, light or motion. One watt is one volt pushing one ampere. A 1,500-watt heater turns electrical energy into heat 15 times faster than a 100-watt lamp turns it into light and heat.

On the exam, power formulas show up in load calculations, heater and lamp problems, conductor heating and questions about running equipment at the wrong voltage. They pair with Ohm's law, and together they make up the Ohm's law wheel.

The power formulas

The three forms of the power formula:

Rearranged to find the other values:

How to pick the right one:

• Know voltage and current? Use P = E × I.
• Know current and resistance? Use P = I² × R. This is the formula for heat lost in conductors.
• Know voltage and resistance? Use P = E² ÷ R. This is the formula for how a fixed resistance responds to a change in voltage.

All three give the same answer for the same circuit. The second and third forms come from substituting E = I × R into P = E × I.

Watts vs volt-amperes

Watts measure true power, the power that actually does work. Volt-amperes (VA) measure apparent power, which is simply volts times amps. For a purely resistive load such as a heater element or incandescent lamp, current and voltage are in step and watts equal volt-amperes.

For inductive loads such as motors and transformers, current lags the voltage, so some current flows back and forth without doing work. The load draws more volt-amperes than watts. The ratio of watts to volt-amperes is the power factor:

This is why transformers are rated in kVA rather than kW, and why NEC load calculations in Article 220 are done in volt-amperes. A transformer has to carry the full current whether or not that current is doing useful work. More on this under power factor.

Energy in kilowatt-hours

Power is a rate. Energy is power multiplied by time. Utilities bill energy in kilowatt-hours (kWh):

A 1,000-watt load running for one hour uses 1 kWh. So does a 100-watt lamp running for 10 hours.

Worked examples

Example 1: heater current and resistance. A 4,500-watt, 240-volt water heater element is operating at its rated voltage. What current does it draw, and what is its resistance?

1. I = P ÷ E = 4,500 ÷ 240 = 18.75 A
2. R = E² ÷ P = 240² ÷ 4,500 = 57,600 ÷ 4,500 = 12.8 Ω
3. Check with P = I² × R: 18.75² × 12.8 = 4,500 W

Example 2: lamp. A 60-watt, 120-volt incandescent lamp is on at rated voltage. What are its current and hot resistance?

1. I = P ÷ E = 60 ÷ 120 = 0.5 A
2. R = E² ÷ P = 14,400 ÷ 60 = 240 Ω

Example 3: heater on lower voltage. The 4,500-watt, 240-volt element from Example 1 is installed on a 208-volt supply. How much power does it produce?

The resistance of the element stays at 12.8 ohms. Only the voltage changes, so use P = E² ÷ R:

1. P = 208² ÷ 12.8 = 43,264 ÷ 12.8 = 3,380 W
2. Shortcut: P = 4,500 × (208 ÷ 240)² = 4,500 × 0.751 = 3,380 W
3. Current also falls: I = 208 ÷ 12.8 = 16.25 A

The voltage dropped by about 13%, but the power dropped by about 25%. Power in a fixed resistance changes with the square of the voltage. The same rule works in reverse: a 120-volt, 1,500-watt heater wrongly connected to 240 volts would try to produce 1,500 × 2² = 6,000 W.

Example 4: conductor heating. A branch circuit's two conductors have a combined resistance of 0.3 ohm and carry 20 amperes. How much power is lost as heat in the wire?

1. P = I² × R
2. P = 20² × 0.3 = 400 × 0.3
3. P = 120 W

Because current is squared, doubling the current on the same wire makes four times the heat. This is the reason conductor ampacity is limited.

Example 5: energy cost. The 4,500-watt element from Example 1 runs for 3 hours. At $0.14 per kWh, what does that cost?

1. kW = 4,500 ÷ 1,000 = 4.5 kW
2. kWh = 4.5 × 3 = 13.5 kWh
3. Cost = 13.5 × $0.14 = $1.89

Example 6: watts vs volt-amperes. A single-phase motor draws 10 amperes at 240 volts with a power factor of 0.80. What are its apparent and true power?

1. VA = E × I = 240 × 10 = 2,400 VA
2. W = VA × PF = 2,400 × 0.80 = 1,920 W

Using P = E × I alone would overstate the true power by 480 watts.

Common exam mistakes

• Assuming power changes in step with voltage. For a fixed resistance, power changes with voltage squared. A 10% voltage drop cuts power by about 19%, not 10%.
• Forgetting to square. In I²R and E²/R, square the current or voltage before you multiply or divide.
• Treating VA as watts on motor loads. Multiply by the power factor to get true power.
• Mixing watts and kilowatts. Divide watts by 1,000 before calculating kilowatt-hours.
• Using resistance from the wrong voltage. Find the element's resistance from its rated voltage and wattage, then apply the new voltage.

For more formula practice, see the exam calculations guide, and you can check current and resistance values with the Ohm's law calculator.

Practice questions

Frequently asked questions