What you'll learn
This revision guide covers how electrical appliances transfer energy, how to calculate power and energy consumption, and the costs associated with running household devices. You'll learn the equations needed for exam calculations and how to apply them to real-world contexts that appear frequently in AQA GCSE Physics papers.
Key terms and definitions
Power — the rate of energy transfer, measured in watts (W) or kilowatts (kW); 1 kW = 1000 W
Energy — measured in joules (J) or kilowatt-hours (kWh); the capacity to do work or cause a change
Joule — the standard SI unit of energy; the energy transferred when a force of 1 newton moves an object 1 metre
Kilowatt-hour — the energy used by a 1 kW appliance running for 1 hour; the unit used by electricity companies for billing
Potential difference (voltage) — the energy transferred per unit charge between two points in a circuit, measured in volts (V)
Current — the rate of flow of electrical charge, measured in amperes (A)
Resistance — the opposition to current flow in a component, measured in ohms (Ω)
Efficiency — the proportion of input energy that is usefully transferred, expressed as a percentage or decimal
Core concepts
Calculating electrical power
Power indicates how quickly an appliance transfers energy. The faster the energy transfer, the higher the power rating.
Three power equations you must know:
P = E ÷ t
- P = power in watts (W)
- E = energy transferred in joules (J)
- t = time in seconds (s)
P = I × V
- P = power in watts (W)
- I = current in amperes (A)
- V = potential difference in volts (V)
P = I² × R
- P = power in watts (W)
- I = current in amperes (A)
- R = resistance in ohms (Ω)
The equation you choose depends on the information given in the question. Most commonly, you'll use P = I × V for appliances where current and voltage are known, or P = E ÷ t when dealing with energy transferred over time.
Power ratings on appliances (like "2400 W" on a kettle) tell you the rate of energy transfer when operating normally. A 2400 W kettle transfers 2400 joules of electrical energy every second.
Energy transfer in electrical appliances
When you use an electrical appliance, the power supply does work to move charges through the device. This transfers energy from the power supply to the appliance.
Energy transferred equation:
E = P × t
- E = energy in joules (J)
- P = power in watts (W)
- t = time in seconds (s)
This can also be written as E = I × V × t by substituting P = I × V.
In real contexts:
- A 100 W light bulb transfers 100 J of energy every second
- A 3000 W electric shower transfers 3000 J every second
- A 900 W microwave transfers 900 J every second
Not all of this energy is usefully transferred. For example, in a light bulb, much energy is wasted as heat rather than light. In a motor, some energy is wasted as heat due to friction and resistance in the wires.
The kilowatt-hour as a unit of energy
Electricity companies don't bill in joules because the numbers would be enormous. Instead, they use the kilowatt-hour (kWh).
One kWh is the energy used by a 1 kW (1000 W) appliance running for 1 hour.
To calculate energy in kilowatt-hours:
E = P × t
- E = energy in kilowatt-hours (kWh)
- P = power in kilowatts (kW)
- t = time in hours (h)
Converting between joules and kilowatt-hours:
1 kWh = 3,600,000 J = 3.6 MJ
This is because:
- 1 kW = 1000 W
- 1 hour = 3600 seconds
- Energy = 1000 W × 3600 s = 3,600,000 J
You rarely need to convert between these units, but understanding the relationship helps with context questions.
Calculating the cost of electricity
Electricity bills are calculated using:
Total cost = power (kW) × time (h) × cost per kWh
Or alternatively:
Total cost = energy used (kWh) × cost per kWh
In the UK, electricity costs vary but typically range from 15p to 35p per kWh (this changes with market conditions, so exam questions will always give you the rate).
Step-by-step approach:
- Convert power to kilowatts if given in watts (divide by 1000)
- Convert time to hours if given in minutes (divide by 60)
- Calculate energy: E = P × t (in kWh)
- Multiply by the cost per kWh
Example context: A 2 kW electric heater running for 3 hours when electricity costs 20p per kWh:
- Energy used = 2 kW × 3 h = 6 kWh
- Cost = 6 kWh × £0.20 = £1.20
Power and energy in different appliances
Different household appliances have very different power ratings depending on their function:
High-power appliances (typically 2-3 kW):
- Electric kettles
- Electric showers
- Electric ovens
- Immersion heaters
- Tumble dryers
Medium-power appliances (typically 0.5-2 kW):
- Washing machines
- Dishwashers
- Microwave ovens
- Hair dryers
- Vacuum cleaners
Low-power appliances (typically under 100 W):
- LED light bulbs (5-15 W)
- Mobile phone chargers
- Televisions (modern LED TVs)
- Radios
Understanding these typical values helps you spot calculation errors. If your answer suggests a phone charger uses 3 kW, you've made a mistake.
High-power appliances cost more to run even for short periods. A kettle might only run for 2 minutes but uses significant energy in that time. Low-power devices like phone chargers use very little energy even if left plugged in for hours (though this is still wasteful).
Efficiency and wasted energy
No appliance is 100% efficient. Some input energy is always wasted, usually as heat.
Efficiency equation:
Efficiency = (useful energy transferred ÷ total energy supplied) × 100%
Or in terms of power:
Efficiency = (useful power output ÷ total power input) × 100%
Efficiency can be expressed as:
- A percentage (0-100%)
- A decimal (0-1.0)
Examples of typical efficiencies:
- Filament light bulbs: ~5% (95% wasted as heat)
- LED light bulbs: ~70-90%
- Electric motors: ~80-95%
- Electric heaters: ~100% (all energy transferred to thermal store, though not always where you want it)
Note: Electric heaters are considered 100% efficient because all electrical energy transfers to thermal energy. However, if the heating is where you don't need it (like heating a room when you wanted to heat water), it's not useful.
In exam questions, you might need to:
- Calculate efficiency when given useful and total energy
- Calculate wasted energy: wasted = total input - useful output
- Explain why an appliance isn't 100% efficient
- Compare efficiencies of different appliances
Worked examples
Example 1: Power calculation (3 marks)
Question: A vacuum cleaner operates at 230 V and draws a current of 6.5 A. Calculate the power of the vacuum cleaner.
Solution:
- Use the equation P = I × V [1 mark]
- P = 6.5 × 230 [1 mark]
- P = 1495 W or 1.5 kW [1 mark]
Mark scheme note: Answer must include correct unit. Accept 1500 W if rounded appropriately.
Example 2: Energy and cost calculation (5 marks)
Question: An electric oven has a power rating of 2.5 kW. It is used for 45 minutes. Electricity costs 28p per kWh. Calculate: (a) The energy transferred in kWh [2 marks] (b) The cost of using the oven [2 marks]
Solution:
(a)
- Convert time: 45 minutes = 45/60 = 0.75 hours [1 mark]
- E = P × t = 2.5 × 0.75 = 1.875 kWh [1 mark]
(b)
- Cost = energy × cost per kWh [1 mark]
- Cost = 1.875 × 0.28 = £0.525 or 52.5p (accept 53p) [1 mark]
Mark scheme note: Unit conversion must be shown. Accept answers to 2 significant figures.
Example 3: Efficiency calculation (4 marks)
Question: An electric motor transfers 180 kJ of electrical energy. It does 153 kJ of useful work lifting a load. The rest is wasted as heat.
(a) Calculate the efficiency of the motor. [3 marks] (b) Calculate the energy wasted as heat. [1 mark]
Solution:
(a)
- Efficiency = (useful energy ÷ total energy) × 100% [1 mark]
- Efficiency = (153 ÷ 180) × 100% [1 mark]
- Efficiency = 85% [1 mark]
(b)
- Wasted energy = 180 - 153 = 27 kJ [1 mark]
Mark scheme note: For part (a), accept answer as decimal 0.85 for full marks. Units not required for efficiency.
Common mistakes and how to avoid them
- Forgetting to convert units — always convert power to kW and time to hours when calculating energy in kWh; convert minutes to hours by dividing by 60, not multiplying
- Using the wrong power equation — check what information is given (I and V? Use P = IV. E and t? Use P = E/t)
- Confusing power and energy — power is the rate of energy transfer (joules per second), not the total energy; a high-power device transfers energy quickly but might only run briefly
- Mixing up efficiency formulas — always put useful energy on top, total energy on bottom; efficiency can never exceed 100%
- Incorrect cost calculations — multiply energy (kWh) by cost per kWh; don't forget the time component when calculating energy first
- Not showing working — in calculations worth 3+ marks, you must show the formula, substitution, and answer with units for full marks
Exam technique for "Electrical power and energy transfer in appliances"
- Command word "Calculate" requires you to show the formula, substitute numbers with units, and give a final answer with units; typically worth 2-4 marks with method marks available even if final answer is wrong
- Command word "Explain" needs you to link cause and effect; for efficiency questions, state that energy is wasted (usually as heat due to resistance/friction) and that useful output is less than total input
- Standard form and units — write large numbers in standard form (3,600,000 J = 3.6 × 10⁶ J); always include units with your final answer (W, kW, J, kWh, £, %)
- Context questions often involve comparing running costs of appliances or explaining why LED bulbs are more economical than filament bulbs despite higher purchase price; link calculations to real-world implications
Quick revision summary
Electrical power is the rate of energy transfer (P = E/t, P = IV, P = I²R). Energy used by appliances is calculated in joules (E = Pt) or kilowatt-hours for billing purposes. One kWh equals 3.6 MJ. Electricity cost equals energy in kWh multiplied by the cost per kWh. Different appliances have characteristic power ratings: kettles and showers are high-power (2-3 kW), LED bulbs are low-power (under 15 W). Efficiency equals useful energy output divided by total energy input, expressed as a percentage or decimal. No real appliance achieves 100% useful efficiency; energy is always wasted, typically as heat.