Efficiency

Efficiency measures how much of the energy supplied to a device does the useful job. It is short, formulaic and worth easy marks. Yet answers above 100% appear in every exam series, an instant giveaway of an inverted fraction. Get the fraction the right way up and this subtopic is nearly free.

What is the efficiency equation?

In words: $\text{efficiency} = \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100\%$. The same works with power: $\text{efficiency} = \dfrac{\text{useful power output}}{\text{total power input}} \times 100\%$. In symbols:

Quantity	Symbol	Unit
Efficiency	none	% (or a decimal, no unit)
Useful energy output	$E_{\text{useful}}$	J
Total energy input	$E_{\text{total}}$	J
Useful power output	$P_{\text{useful}}$	W
Total power input	$P_{\text{total}}$	W

Efficiency has no unit. It can never exceed 100%, because conservation of energy forbids a device giving out more useful energy than it takes in. The “missing” energy is not destroyed. It is dissipated, usually to the internal (thermal) store of the surroundings by heating or by sound waves.

Why is no device 100% efficient?

Friction in moving parts and resistance in wires transfer some input energy to the surroundings as heating. A typical petrol engine runs near 25-30% efficient; an electric motor can exceed 80%; an LED lamp beats a filament lamp roughly five-fold for the same light output. Sankey diagrams show this visually: arrow widths are proportional to energy, the useful arrow continues forward, and the wasted arrow bends away. Reading values off a Sankey diagram is a standard Core question.

Worked Exam Question

An electric motor lifts a 30 N load through 2.0 m. The motor is supplied with 90 J of energy.

(a) Calculate the useful work done on the load. [2] (b) Calculate the efficiency of the motor. [2] (c) State what happens to the rest of the input energy. [1]

Solution (a). Equation: $W = Fd$. Substitute: $W = 30 \times 2.0$. Answer: $W = \textbf{60 J}$.

Solution (b). Equation: $\text{efficiency} = \dfrac{\text{useful energy output}}{\text{total energy input}} \times 100\%$. Substitute: $\text{efficiency} = 60 \div 90 \times 100\%$. Answer: efficiency = 67% (accept 66.7% or 0.67).

Solution (c). It is dissipated to the internal (thermal) store of the motor and surroundings (by heating, plus some sound).

Mark scheme

• M1: $W = Fd$ used with $30 \times 2.0$; A1: 60 J.
• M1: $60/90 \times 100$ (their (a) over 90 gains the method mark).
• A1: 67% (no unit other than % accepted).
• B1: dissipated as heating / to thermal store of surroundings.

Common Mistakes

• Inverting the fraction. $90/60$ gives 150%, which is impossible. Fix: useful goes on top; if your answer exceeds 100%, flip it.
• Mixing energy and power in one fraction. Fix: use J over J, or W over W, never one of each.
• Writing joules or watts as the unit of efficiency. Fix: efficiency is a percentage or a plain decimal.
• Calling wasted energy “lost”. Fix: say it is dissipated to the surroundings; energy is conserved overall.
• Misreading Sankey diagrams. Width represents energy, so a half-width branch is half the energy. Fix: add the branch values and check they equal the input.

Exam Technique Tip

Sanity-check every efficiency answer against two limits: it must sit between 0% and 100%, and a real machine should look plausible (a motor near 67% is believable; 2% or 99.9% probably is not). This five-second check converts the most common wrong answer in this subtopic into full marks, because the only repair needed is usually flipping the fraction.

How This Is Examined

Efficiency runs through every written paper, Core and Extended alike, using the same equations. Papers 1 and 2 (MCQ) give input and useful output and ask for the percentage, or test Sankey-diagram reading. Papers 3 and 4 chain it after work or power calculations, exactly as in the worked question, so a slip earlier costs only that part, thanks to error carried forward. It also appears inside Energy Resources questions: comparing power stations often hinges on efficiency figures. Paper 6 may ask why a measured output is below the input in a ramp or motor experiment; “energy dissipated by friction as heating” is the expected line.