3.4.1.7
Efficiency measures how much input energy is usefully transferred
Work, Energy & Power — AQA A-Level Physics
Key Definition
Efficiency — The ratio of useful energy output to total energy input. No units (expressed as a decimal or percentage).
$$\eta = \frac{\text{useful output}}{\text{total input}} \times 100\%$$
$$\text{EfficiencyThe ratio of useful energyThe capacity to do work. Measured in joules (J). output to total energyThe capacity to do work. Measured in joules (J). input, expressed as a fraction or percentage.} = \frac{\text{Useful powerThe rate of energy transfer. Measured in watts (W). output}}{\text{Total powerThe rate of energy transfer. Measured in watts (W). input}} \times 100\%$$
- EfficiencyThe ratio of useful energyThe capacity to do work. Measured in joules (J). output to total energy input, expressed as a fraction or percentage. can also be calculated using energy instead of powerThe rate of energy transfer. Measured in watts (W)..
- No real system is 100% efficient. Some energy is always dissipated as thermal energy or sound.
- Which energy is 'useful' depends on the purpose of the device: light is useful in a bulb but wasted in a motor.
Worked Example
An electric motor with 35% efficiencyThe ratio of useful energy output to total energy input, expressed as a fraction or percentage. lifts a 7.2 kg load through 5 m in 3 s. Calculate the power input.
Show Solution
1
Calculate useful energy output (GPE gained)
$$E_p = mgh = 7.2 \times 9.81 \times 5 = 353.2 \text{ J}$$
2
Calculate useful power output
$$P_{\text{out}} = \frac{353.2}{3} = 117.7 \text{ W}$$
3
Calculate power input using efficiency
$$P_{\text{in}} = \frac{P_{\text{out}} \times 100}{\text{Efficiency}} = \frac{117.7 \times 100}{35} = 336 \text{ W}$$
Answer
Power $input = 336 W$
Examiner Tips and Tricks
- Efficiency can be given as a ratio (0 to 1) or a percentage (0% to 100%).
- If the question asks for a ratio, do not multiply by 100.
- Always check which format is required.