Efficiency

Every device is given energy to do a job, but not all of it reaches the useful store. Some is always dissipated (spread out and wasted) to the surroundings, usually as thermal energy. Efficiency tells you what fraction of the input actually does the useful job.

Efficiency = useful output energy transfer ÷ total input energy transfer.

A device with high efficiency wastes little of its input; a low-efficiency device wastes a lot. Efficiency is a ratio of two energy values, so it has no units. As a decimal it lies between 0 and 1, and as a percentage between 0% and 100%.

Consider a lamp supplied with 100 J of electrical energy that transfers 20 J usefully as light. The other 80 J is dissipated as thermal energy. The useful fraction, 20 out of 100, is what efficiency captures.

The energy form of the equation compares the useful output with the total input.

$\text{efficiency}=\frac{\text{useful output energy transfer}}{\text{total input energy transfer}}$

This equation is not on the Physics equation sheet. You must recall and apply it.

The total input is always the larger number, so a correct efficiency is never greater than 1 (or 100%). To turn a decimal into a percentage, multiply by 100.

Worked example — an electric motor is supplied with 300 J of electrical energy and transfers 240 J usefully to the kinetic energy store of a load. Find its efficiency.

$\text{efficiency}=\frac{240}{300}=0.8$

As a percentage: $0.8\times 100=80\%$.

So 80% of the input does the useful job and the remaining 20% (60 J) is dissipated, mostly as thermal energy from friction and electrical resistance.

Exam questions often want the answer as a percentage. Divide as before, then multiply by 100.

Worked example — a kettle transfers 90 000 J of electrical energy. Of this, 63 000 J is usefully transferred to the thermal energy store of the water. Find the efficiency as a percentage.

$\text{efficiency}=\frac{63000}{90000}=0.7$

$0.7\times 100=70\%$

The kettle is 70% efficient. The wasted 30% (27 000 J) heats the kettle body, the element and the surrounding air rather than the water.

The units of the two energy values must match before you divide. If one is in joules and the other in kilojoules, convert first, or the ratio will be wrong by a factor of 1000.

Because power is the rate of energy transfer, you can also compare power values instead of energy values.

Efficiency = useful power output ÷ total power input. This form is also not on the equation sheet, so recall it.

$\text{efficiency}=\frac{\text{useful power output}}{\text{total power input}}$

Power is measured in watts (W), where 1 W = 1 J/s. The equation works identically because both quantities are rates measured over the same time.

Worked example — a filament lamp has a total power input of 60 W and a useful power output (as light) of 9 W. Find its efficiency.

$\text{efficiency}=\frac{9}{60}=0.15=15\%$

Only 15% of the electrical power becomes light; the other 51 W is dissipated as thermal energy. This is why filament lamps have been replaced by LED lamps, which waste far less power as heat.

A Sankey diagram shows the split visually: the input arrow enters on the left, and its width divides into useful and wasted branches in proportion to their energy values. The wider the wasted branch, the lower the efficiency.

For the 80% efficient motor above, out of every 100 J of input, 80 J is transferred usefully and 20 J is wasted:

The total energy is conserved: the useful and wasted branches always add back up to the input, because energy cannot be created or destroyed. Reading efficiency straight off the diagram is quick, the useful branch here is 80 out of 100, giving 0.8.

The same equation finds a missing energy value when the efficiency is known. Rearrange it to make the unknown the subject.

Worked example — a device is 25% efficient and is supplied with 2000 J of energy. How much energy is usefully transferred?

First write the efficiency as a decimal: $25\%=0.25$. Rearranging to make useful output the subject:

$\text{useful output}=\text{efficiency}\times \text{total input}$

$\text{useful output}=0.25\times 2000=500\text{J}$

So 500 J is transferred usefully and 1500 J is wasted.

Worked example — a pump usefully transfers 450 J and is 60% efficient. Find the total energy input.

$\text{total input}=\frac{\text{useful output}}{\text{efficiency}}=\frac{450}{0.6}=750\text{J}$

Check: $\frac{450}{750}=0.6=60\%$. The rearrangement is confirmed.

(Higher Tier only) You must be able to describe ways to increase the efficiency of an intended energy transfer.

The strategy is always the same: reduce the unwanted energy transfers so a larger share of the input reaches the useful store. Common methods:

Wasted transfer	How to reduce it	Effect on efficiency
Friction between moving parts	Lubricate with oil or grease	Less energy wasted as heat and sound, so efficiency rises
Heat loss from a hot object	Add thermal insulation	Less thermal energy dissipated, so more stays useful
Air resistance on a moving object	Streamline the shape	Less energy wasted overcoming drag
Resistance in wires	Use lower-resistance conductors	Less electrical energy wasted as heat

No real device can reach 100% efficiency, because some energy is always dissipated during a transfer. The aim is to push efficiency as high as practical, not to eliminate waste entirely.

1. Getting an efficiency above 1 or above 100%

The total input is always the largest energy value, so a correct efficiency never exceeds 1 (or 100%). If your answer is bigger, you have divided the wrong way round. Put the useful output on top and the total input underneath.

2. Forgetting to convert a decimal to a percentage

Dividing gives a decimal such as 0.7. If the question asks for a percentage, you must multiply by 100 to get 70%. A bare 0.7 written as "0.7%" loses the mark.

3. Mixing up useful and total input

The denominator is the total energy supplied, not the wasted part. If a question gives useful and wasted values, add them together to find the total input before dividing.

4. Not matching the units

Both energy values must be in the same unit before you divide. Convert kilojoules to joules (or vice versa) first, otherwise the ratio is out by a factor of 1000.

5. Claiming a device can be 100% efficient

Some energy is always dissipated to the surroundings, so a real device cannot be perfectly efficient. Reducing friction and heat loss raises efficiency but never removes waste completely.