Efficiency
Work, Energy & Power - OCR A-Level Physics
Key Definition
Efficiency
The fraction of the total input energy (or power) that is transferred as useful output. Dimensionless. Usually written as a decimal between 0 and 1, or as a percentage between 0% and 100%.
The fraction of the total input energy (or power) that is transferred as useful output. Dimensionless. Usually written as a decimal between 0 and 1, or as a percentage between 0% and 100%.
$$\text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}} \times 100\%$$
$$\text{efficiency} = \frac{\text{useful output power}}{\text{total input power}} \times 100\%$$
- Both forms are equivalent because power is energy per unit time, so the time cancels in the ratio.
- Real machines are never 100% efficient because some energy is always dissipated to thermal energyThe energy stored in a system due to the random motion of its particles; it is related to temperature. stores via friction, air resistance, or electrical resistance.
- A Sankey diagramA diagram in which the width of each arrow is proportional to the amount of energy it represents. Useful arrows split off from the total input and wasted arrows branch downwards. shows the input energy splitting into useful and wasted streams. The width of each arrow is proportional to the energy.
- Typical values: incandescent light bulb $\sim 5\%$; LED lamp $\sim 40\%$; petrol car engine $\sim 25\%$; electric motor $\sim 90\%$.
Diagram pending
Sankey diagram for an electric motor lifting a load. Input arrow $500 \text{ J}$ electrical energy on the left. Splits into a useful right-going arrow $\Delta E_p \approx 392 \text{ J}$ (gravitational PE gained) and a downward wasted arrow $\sim 108 \text{ J}$ labelled thermal energy (friction in bearings, electrical heating in coils). Arrow widths drawn to scale.
Will be replaced with a GeoGebra/Figma SVG in stream 2.
Common Mistake
MEDIUM
Wrong: Counting total output energy (useful + wasted) in the numerator. By conservation of energy, that always gives 100%.
Right: Only the useful output counts. If a motor uses $1000 \text{ J}$ of electrical energy and produces $800 \text{ J}$ of kinetic energy plus $200 \text{ J}$ of heat, the efficiency is $800/1000 = 80\%$.
Right: Only the useful output counts. If a motor uses $1000 \text{ J}$ of electrical energy and produces $800 \text{ J}$ of kinetic energy plus $200 \text{ J}$ of heat, the efficiency is $800/1000 = 80\%$.
Worked Example [3 marks]
An electric motor uses $500 \text{ J}$ of electrical energy to lift a $4.0 \text{ kg}$ mass through a height of $10 \text{ m}$. Calculate the efficiency. Use $g = 9.81 \text{ m s}^{-2}$.
Show Solution
1
Useful output is the GPE gained
$\Delta E_p = mg\Delta h = 4.0 \times 9.81 \times 10 = 392.4 \text{ J}$.
[1]2
$\text{efficiency} = \frac{\text{useful output}}{\text{total input}} = \frac{392.4}{500}$.
[1]3
$\text{efficiency} = 0.785 = 78.5\%$.
[1]Answer
78.5%
Examiner Tips and Tricks
- Check the units on input and output before dividing. If one is in joules and the other in watts, you have a mistake somewhere.
- An efficiency greater than 1 (or 100%) signals an arithmetic error or that you have confused useful and total energy.