Electrical Power, Energy, and the Kilowatt-Hour

Electrical power is the rate of transfer of electrical energy, P = VI = I2R = V2 ÷ R, measured in watts, and the energy transferred is E = Pt; electricity is billed in kilowatt-hours, where 1 kWh = 3.6 × 106 J is the energy used by a 1 kW appliance in one hour. In brief: to find the cost of using an appliance, multiply its power in kilowatts by the time in hours to get kWh, then multiply by the price per kWh. A 2 kW heater running for 3 hours uses 6 kWh. Current heats a wire because of collisions of charge carriers with the atoms of the lattice (Joule heating).

Electricity bills, appliance ratings and the sizes of wires and fuses all depend on electrical power and energy. This guide gives the power formulas, explains the kilowatt-hour and the cost calculation, compares appliances in a table, works through numerical examples and corrects common mistakes.

How is electrical power calculated?

Power is energy per second. Since V = W ÷ Q and I = Q ÷ t, the energy transferred in time t is W = QV = VIt, so P = VI. With Ohm’s law (V = IR) two other forms follow: P = I2R and P = V2 ÷ R. Which one to use depends on the data: I2R shows that for a given current, power loss in a wire rises with the square of the current (important for power transmission), and V2 ÷ R shows that for a given voltage, a lower resistance gives more power (a 1,000 W heater has a lower resistance than a 100 W lamp at the same mains voltage). The appliance rating, such as “230 V, 2,000 W”, gives the power at the stated voltage.

What is electrical energy and the kilowatt-hour?

The energy transferred is E = Pt = VIt, in joules when P is in watts and t in seconds. The joule is small, so electricity companies use the kilowatt-hour (kWh): the energy transferred by a power of 1 kW in 1 h. 1 kWh = 1,000 W × 3,600 s = 3.6 × 106 J = 3.6 MJ. The meter at home records the number of kWh used (the “units”). The cost = energy in kWh × price per kWh. The prices vary between countries and over time, and the values below are only for illustration.

Why do wires heat up?

As charge carriers move through a conductor, they collide with the ions of the lattice and transfer energy to them, which raises the temperature. This is called Joule heating (resistive heating), and the heating power is P = I2R. It is useful in kettles, toasters and electric heaters, where wires with high resistance are used, but it is wasteful in power lines. For this reason electricity is transmitted at a high voltage: for a given power P = VI, a higher V means a lower current and so the loss I2R is smaller.

Power lost in a line of 5.0 ohms against current 0 5 10 15 20 0 500 1000 1500 2000 Current in line (A) Power lost (W) 500 W at 10 A
Heating power P = I2R for the 5.0 Ω power line from the worked example. The loss rises with the square of the current: 500 W at 10 A, and four times as much (2,000 W) at 20 A. This is why lower currents, and so higher voltages, are used for transmission.

Appliances compared

Typical power ratings and current at 230 V
Appliance Power Current at 230 V (A) Energy in 1 hour
LED lamp10 W0.040.010 kWh
Laptop60 W0.260.060 kWh
Television100 W0.430.100 kWh
Microwave oven1,000 W4.31.0 kWh
Electric kettle2,300 W102.3 kWh

Worked example: current, energy and cost

  1. Power from V and I. A motor on 230 V draws 4.0 A: P = 230 × 4.0 = 920 W.
  2. Resistance of a heater. A 2,300 W kettle on 230 V: I = P ÷ V = 10 A; R = V ÷ I = 23 Ω (or V2 ÷ P = 52,900 ÷ 2,300 = 23 Ω).
  3. Energy in joules. The kettle runs for 3.0 minutes (180 s): E = Pt = 2,300 × 180 = 414,000 J = 0.414 MJ. In kWh: 2.3 kW × 0.050 h = 0.115 kWh.
  4. Cost. A 100 W television is used for 5.0 h per day for 30 days: E = 0.100 kW × 5.0 h × 30 = 15 kWh. At an illustrative price of 0.25 per kWh, the cost is 15 × 0.25 = 3.75.
  5. Comparing lamps. A 60 W filament lamp and a 10 W LED lamp with similar light output, each used for 1,000 hours: 60 kWh and 10 kWh. The saving is 50 kWh, or 12.50 at 0.25 per kWh.
  6. Transmission loss. A line of resistance 5.0 Ω carries 10 A: loss = I2R = 100 × 5.0 = 500 W. At 100 A (a lower voltage for the same power): 10,000 × 5.0 = 50,000 W, a hundred times larger.

Where do the units come from?

James Prescott Joule published his law of heating in 1841, showing that the heat produced in a wire is proportional to I2R. The watt was named after James Watt, and the kilowatt-hour came into use with the first electricity supply companies in the 1880s. The first electricity meters, developed by Elihu Thomson and others in the 1880s and 1890s, recorded the energy used by customers.

Which mistakes are common?

Mistake 1: Treating the kilowatt-hour as a unit of power

It is a unit of energy: power multiplied by time.

Mistake 2: Forgetting to convert watts to kilowatts

Divide by 1,000 before multiplying by hours to get kWh.

Mistake 3: Using the wrong form of the power formula

Use P = I2R or V2 ÷ R only when R is the resistance of the part in question, and V and I are those for that part.

Mistake 4: Thinking that a higher resistance always means more heat

At constant voltage, a higher resistance gives less power. At constant current, a higher resistance gives more.