14.1 Electrical power

Power is the rate of energy transfer. For an electrical component, P = IV. A device operating at 12 V and 2.0 A transfers 24 J each second, so its power is 24 W.

Power rating states the intended rate of energy transfer at the rated voltage. A 60 W lamp transfers 60 J s⁻¹ when operating normally. It does not necessarily use 60 J in total; energy depends on time.

14.2 Alternative power equations

Combining P = IV with V = IR gives P = I²R and P = V²/R for an ohmic component. Choose the equation matching known quantities. P = I²R is useful for cable heating, while P = V²/R is useful for a resistor across a fixed voltage.

These forms assume resistance applies under the operating conditions. For a filament lamp, resistance changes with temperature, so use measured operating values rather than a cold resistance unless directed.

14.3 Electrical energy

Energy transferred in time t is E = Pt. Combining with P = IV gives E = IVt, and since Q = It, E = QV. Use watts and seconds for joules.

A high-power device can use less energy than a low-power device if operated for a much shorter time. Energy comparisons require both power and duration.

14.4 Kilowatt-hour

Electricity suppliers commonly measure energy in kilowatt-hours, kWh. One kilowatt-hour is the energy used by 1 kW for 1 hour. It equals 3.6 × 10⁶ J.

To calculate kWh, convert power to kilowatts and time to hours. Do not insert watts and seconds into a kWh calculation without conversion.

14.5 Cost of electricity

Cost = energy in kWh × tariff per kWh. A fixed standing charge may be added if stated. Tariffs can vary by time or usage, so use the exact information supplied.

A meter reading difference gives energy consumed directly in kWh. Subtract the earlier reading from the later one, then multiply by the tariff.

14.6 Appliance current and fuse choice

At a known supply voltage, I = P/V estimates operating current. A 2300 W appliance on 230 V draws about 10 A. This helps select cable and fuse ratings.

A fuse rating should be slightly above normal current but below a dangerous current. Choosing a much larger fuse removes effective protection; choosing one below operating current causes nuisance melting.

14.7 Efficiency and useful output

Efficiency = useful energy output/total energy input, or useful power output/total power input. It may be expressed as a decimal or percentage. Waste is often transferred as unwanted heating or sound.

An efficient appliance provides the required useful output with less input energy, reducing running cost and environmental impact. However, purchase price, lifetime and use pattern may also matter.

Worked examples

Practical focus

Examination guidance

Check your understanding

1. A 2.0 kW kettle runs for 3.0 min. Find energy in kWh.
2. Find current of a 1150 W appliance on 230 V.
3. Convert 2.5 kWh to joules.