You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
This lesson covers electrical power and energy transfer in circuits, including the key equations you need for the AQA GCSE Physics exam (4.2.3). These equations are essential for calculation questions and for understanding how energy is transferred in domestic and industrial contexts.
Power is the rate of energy transfer — it tells you how much energy is transferred per second.
Power is measured in watts (W).
1 watt = 1 joule per second (1 W = 1 J/s)
Larger powers are often measured in kilowatts (kW): 1 kW = 1000 W.
A 100 W light bulb transfers 100 joules of energy every second. A 2000 W kettle transfers 2000 joules of energy every second.
There are three important power equations you must know for the GCSE exam:
P = E / t
Or rearranged: E = P x t
Where:
P = I x V
Where:
P = I^2 x R
Where:
This equation is derived by substituting V = I x R into P = I x V: P = I x (I x R) = I^2 x R
| Equation | Variables | When to Use |
|---|---|---|
| P = E / t | Power, energy, time | When you know energy and time |
| P = I x V | Power, current, p.d. | When you know current and p.d. |
| P = I^2 x R | Power, current, resistance | When you know current and resistance |
Exam Tip: All three power equations are on the AQA equation sheet, but you MUST be able to rearrange them confidently. Practise rearranging each equation for every variable. For example, from P = I x V: I = P/V and V = P/I.
When charge flows through a component, energy is transferred. The amount of energy transferred depends on the charge that flows and the potential difference across the component.
E = Q x V
Where:
This equation tells us that 1 volt means 1 joule of energy transferred per coulomb of charge.
E = P x t
This is a rearrangement of P = E / t.
Where:
Exam Tip: When a question mentions "energy transferred" or "work done", these mean the same thing in physics. Energy transferred = work done. Use the appropriate equation based on which quantities you are given.
When current flows through a component, electrical energy is transferred into other forms:
graph LR
A[Electrical Energy from supply] --> B[Component]
B --> C[Useful Energy Output]
B --> D[Wasted Energy - usually thermal]
| Component | Useful Energy Transfer | Wasted Energy |
|---|---|---|
| Filament lamp | Light (and heat) | Heat to surroundings |
| LED | Light | Very little heat |
| Electric motor | Kinetic energy | Heat (friction, resistance) |
| Electric heater | Heat to surroundings | None (all heat is "useful") |
| Speaker | Sound | Heat |
| Resistor | Heat (sometimes useful) | Heat to surroundings |
In any circuit, the total energy transferred by the power supply equals the total energy transferred by all the components. Some of this energy is usefully transferred and some is dissipated (wasted), usually as thermal energy (heat).
The energy dissipated by a resistor heats the resistor and its surroundings. This is sometimes useful (e.g., in an electric heater) but is often wasted.
In domestic electricity, energy is measured in kilowatt-hours (kWh), not joules. This is because a joule is a very small unit — a typical household uses millions of joules per day.
1 kWh is the energy transferred by a 1 kW appliance in 1 hour.
To convert: 1 kWh = 1000 W x 3600 s = 3,600,000 J = 3.6 MJ
E = P x t
Where:
Cost = Energy (kWh) x Price per kWh (in pence)
| Step | What to Do |
|---|---|
| 1 | Convert power to kW (divide watts by 1000) |
| 2 | Convert time to hours (divide minutes by 60, or seconds by 3600) |
| 3 | Calculate energy: E = P x t (in kWh) |
| 4 | Calculate cost: Cost = E x price per kWh |
Exam Tip: The most common mistake in energy cost calculations is forgetting to convert units. Power MUST be in kW (not W) and time MUST be in hours (not minutes or seconds) when calculating energy in kWh. Always check your units before substituting into the equation.
A kettle draws a current of 10 A from the 230 V mains supply. Calculate its power.
P = I x V = 10 x 230 = 2300 W (2.3 kW)
A current of 4 A flows through a 5-ohm resistor. Calculate the power dissipated.
P = I^2 x R = 4^2 x 5 = 16 x 5 = 80 W
A 2000 W heater is used for 3 hours. Calculate the energy transferred in kWh.
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.