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Don't read anything yet. Drag the two sliders and watch the equation fill itself in. Then try to push the useful output above the total input — the widget will not let you, and the reason why is the most important idea on this page.
Set the energy going in, and how much of it does the job you wanted
Notice what you cannot do. Push the useful output above the total input and the widget stops you. Efficiency can never be more than 1 (100%), because you cannot get more energy out than you put in — energy is never created.
Pick a device
Both AQA equations, same idea. efficiency = useful output energy transfer ÷ total input energy transfer, and efficiency = useful power output ÷ total power input. Joules with joules, watts with watts — never mix them.
Physics With Kate · physicswithkate.com — AQA GCSE Physics 4.1.2.2
The notes below give you the equation in both of its AQA forms, how to rearrange it, when to answer as a decimal and when as a percentage, and the four ways of making a device more efficient that examiners ask for by name.
Every device is given energy to do a job. Some of that energy does the job — that is the useful output. The rest is wasted, and it almost always ends up transferred to the thermal store of the surroundings, which spreads out and cannot be got back.
Efficiency is simply the fraction of the energy that did the job you wanted.
The same energy transfer can be useful in one device and wasted in another. A filament bulb heating the room is wasted energy, because you wanted light. An electric heater heating the room is useful, because heating is the job.
So never learn a list of "useful" and "wasted" transfers. Ask what the device is for.
efficiency = useful output energy transfer ÷ total input energy transfer
AQA also gives it in terms of power, and it works exactly the same way:
efficiency = useful power output ÷ total power input
Use joules with joules, or watts with watts. Never mix the two.
Because you are dividing joules by joules (or watts by watts), the units cancel — so efficiency is never measured in joules, watts, or anything else. What you write after the number depends on which form you give the answer in, and there are only two:
As a decimal — nothing at all. Write 0.08 on its own. No J, no W, and no % sign, because 0.08 is not eight per cent.
As a percentage — the % sign is required. Write 8%. Leaving the % off turns your answer into 8, which is a hundred times too big and cannot be right, since efficiency can never be above 1.
So "efficiency has no unit" means no J and no W — it does not mean you can drop the per cent sign.
The equation always gives a decimal between 0 and 1. To turn it into a percentage, multiply by 100.
| Calculation | As a decimal | As a percentage |
|---|---|---|
| 250 J useful out of 1000 J in | 0.25 | 25% |
| 45 J useful out of 60 J in | 0.75 | 75% |
| 18 W useful out of 20 W in | 0.9 | 90% |
Read the question. If it says "give your answer as a percentage", the decimal on its own is not the answer. If it does not say, either is accepted — but say which you have given.
Two of the three quantities are always given, so you need the equation all three ways round:
If you are working from a percentage, turn it into a decimal first — 80% becomes 0.8. Multiplying by 80 instead of 0.8 is the single most common slip on this topic.
Your efficiency must come out between 0 and 1 (or 0% and 100%). If it does not, you have divided the wrong way round — the useful output goes on top.
Energy is never created or destroyed, so you can never get more out than you put in. But you always get less useful energy out, because some is dissipated — spread out into the surroundings where it cannot be used.
The usual culprits are:
The one near-exception is an electric heater, which is about 100% efficient — not because it is cleverly built, but because the thermal store is the useful output, so almost nothing counts as waste.
The trick is always the same: reduce the wasted transfer. Learn one method for each cause, and say what it reduces — the mark is in the reason, not the word.
| Method | What it reduces | Example |
|---|---|---|
| Lubrication — oil or grease the moving parts | friction between surfaces, so less energy is dissipated as heating | oiling a bicycle chain or a motor's bearings |
| Insulation — lag or double-glaze | the rate of energy transfer by heating out of the system | loft insulation, a lagging jacket on a hot water tank |
| Streamlining — smooth the shape | air resistance, so less energy is wasted pushing air aside | the shape of a car or a train |
| Better components | the waste built into the device itself | an LED bulb instead of a filament bulb |
What you do NOT needSankey diagrams are not required for AQA GCSE Physics. You will not be asked to draw one, label one, or read values off one, and there are none anywhere on this page.
If you have seen them in a textbook, on a poster, or from a friend on a different course, you are not missing a topic — they belong to other specifications, such as Edexcel International GCSE. For AQA, everything on efficiency comes from the one equation above.
A lamp is supplied with 500 J of energy and usefully transfers 40 J as light. Calculate its efficiency.
efficiency = useful ÷ total = 40 ÷ 500 = 0.08 — a decimal, so nothing
is written after it.
As a percentage that is 0.08 × 100 = 8% — and here the % sign is needed.
Either form is accepted, but "0.08%" and a bare "8" are both wrong.
A motor is 75% efficient. It is supplied with 1200 J. How much energy is usefully transferred?
Turn the percentage into a decimal first: 75% = 0.75.
useful = efficiency × total = 0.75 × 1200 = 900 J
So 300 J is wasted, mostly heating the motor through friction.
A kettle usefully transfers 180 kJ to the water and is 90% efficient. How much energy was supplied to it?
total = useful ÷ efficiency = 180 ÷ 0.9 = 200 kJ
Check it looks right: the total must be bigger than the useful output. 200 kJ > 180 kJ ✓
A device has a total power input of 60 W and a useful power output of 21 W. Calculate the efficiency as a percentage.
efficiency = 21 ÷ 60 = 0.35, so 35%.
Watts divided by watts — the power equation works exactly like the energy one. The question
asked for a percentage, so the % sign must be there; no W and no J ever appear after an
efficiency.
A student calculates an efficiency of 2.5. Explain why this must be wrong.
Efficiency can never be greater than 1, because that would mean more energy coming out than went in, and energy cannot be created. The student has divided total by useful instead of useful by total.
Work through the efficiency calculations yourself — forwards, then both rearrangements — and check them against the answers at the back of the book.
AQA GCSE Physics Workbook · ENERGY · page 14Free sample = the pages for this lesson. The full workbook covers the whole Energy topic for AQA, with exam-style questions and worked answers.
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.
Video coming soon
The walkthrough for this lesson is being filmed and will appear right here. Until then, the notes and the free workbook pages above cover every mark.
Watching someone else rearrange an equation is not the same as rearranging it yourself. The workbook pages give you the calculations all three ways round, with worked answers at the back.
Spec-aligned revision resources, group courses, and 1:1 tutoring for GCSE and A Level Physics — built by an experienced teacher and examiner.