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AQA GCSE Physics · Topic 4.2 Electricity
Electrical Power and Energy Transfers
In this lesson you'll learn to: explain that power is the energy transferred every second, use
P = I V and P = I²R, calculate the energy transferred with E = P t (so E = I V t) and
E = Q V, and work out how the energy from a battery is shared between components in series and
parallel circuits.
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Prefer to watch? Scroll to the bottom for the video on this topic
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A video walkthrough is coming soon — it will appear at the bottom of this page
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Start here — have a play
Don't read anything yet. Close the switch and watch each coulomb hand its red joules to the bulb. The gold box beside the bulb counts every joule it receives. Leave it for 10 seconds, then check the tally against the three calculations underneath. Then try two bulbs in parallel and two in series: which one makes the battery work harder?
1 coulomb of charge1 joule of energygold boxes = joules transferred so far
Battery
Cells are 2 V. Bulbs are 2 Ω (1 Ω on the series tab). The stopwatch and tallies go up every tenth of a second.
▲ Each ball is 1 coulomb and each red dot is 1 joule, so the tally is literally E = Q V.
Watched the joules pile up? Now calculate them.
Four equations, one idea: energy transferred = how many coulombs × how many joules each. The notes show which equation to use when, then work through series and parallel.
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Revision notes
4.2.4.1 Power
⭐ Definition
Power is the energy transferred per second. 1 watt (W) = 1 joule per second (J/s).
P
P = I V
power = current × potential difference
P watts (W) · I amps (A) · V volts (V)P
P = I² R
power = current² × resistance
Use it when you know the current and resistance but not the p.d.Why P = I V works: the current tells you how many coulombs pass each second, and the p.d. tells you how many joules each coulomb transfers. Multiply them and you get the joules per second.
4.2.4.2 Energy transfers in everyday appliances
Everyday appliances transfer energy from batteries or the mains to other stores: the kinetic energy of a motor (a fan, a drill, a food mixer) or the thermal energy of a heater (a kettle, a hairdryer, a toaster). The energy transferred depends on how long the appliance is on and its power.
E
E = P t
energy transferred = power × time, so E = I V t
E joules (J) · t seconds (s)E
E = Q V
energy transferred = charge flow × potential difference
Q coulombs (C) · V volts (V)💡 Exam tip — units first
Time must be in seconds (2 minutes = 120 s), current in amps (300 mA = 0.3 A) and power in watts (2 kW = 2000 W). Big answers are often given in kJ: 14 400 J = 14.4 kJ.
Series and parallel: where does the battery's energy go?
| Series | Parallel | |
|---|---|---|
| Current | The same through the battery and every component | Splits: the battery current is the sum of the branch currents |
| p.d. | Shared: each component gets part of the battery's voltage | Each branch gets the full battery voltage |
| Energy | The energy transferred by the battery = the total energy transferred by all the components | |
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How to write it in the exam
Every calculation here is worth marks for the same four steps. Write all four.
The four-line calculation
1. Equation — P = I V, P = I²R, E = I V t or E = Q V
2. Convert — minutes → s, mA → A, kW → W
3. Substitute — put the numbers in
4. Answer + unit — W or J (or kJ)
2. Convert — minutes → s, mA → A, kW → W
3. Substitute — put the numbers in
4. Answer + unit — W or J (or kJ)
equation & answerconvertsubstitute
Worked example · power and energy · 4 marks
A kettle: 230 V, 10 A, on for 3 minutes. Find its power and the energy transferred.
- P = I V = 10 × 230 = 2300 W
- t = 3 × 60 = 180 s
- E = P t = 2300 × 180 = 414 000 J
Worked example · P = I²R · 2 marks
Find the power of the heating element.
- P = I² R = 2.0² × 12
- P = 48 W
Watch out: square the current first.
Worked example · E = Q V · 2 marks
50 C flows round the circuit. Find the energy transferred.
- E = Q V = 50 × 9.0
- E = 450 J
Exam-style question · series · 4 marks
Identical bulbs, on for 1 minute. Find the energy from the battery and from bulb A.
- Battery: E = I V t = 0.50 × 12 × 60 = 360 J
- Each bulb gets half the voltage: 6.0 V
- Bulb A: E = 0.50 × 6.0 × 60 = 180 J
Check: 180 + 180 = 360 J
Exam-style question · parallel · 4 marks
Identical bulbs, on for 2 minutes. Find the energy from the battery and from bulb A.
- Battery current = 0.20 + 0.20 = 0.40 A; t = 120 s
- Battery: E = 0.40 × 6.0 × 120 = 288 J
- Bulb A gets the full 6.0 V: E = 0.20 × 6.0 × 120 = 144 J
Check: 144 + 144 = 288 J
📘 Now do it in your workbook
Complete the power and energy equations, make up your own circuit, answer the heater practical question, then work through the series and parallel energy questions.
AQA GCSE Physics Workbook · ELECTRICITY · pages 13 & 14The full workbook covers the whole of Topic 4.2 Electricity, with exam-style questions and worked answers.
✅ Can you do it? Tick as you go
0% complete🎉 Nice work! You've ticked off every objective for this spec point.
Don't forget to hit “Mark complete” at the bottom of the lesson.
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Prefer to watch? Here's the whole thing
Video walkthrough — coming soon
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it. The video is on its way.
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Video coming soon
The walkthrough for this lesson is being filmed and will appear right here. Until then, the animation and the notes above cover every mark.
Now put it into practice
Pages 13 and 14 of the workbook have the equations, the heater practical question and the series and parallel energy questions, with full answers at the back.
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