Efficiency
Work, Energy & Power - OCR A-Level Physics
Key Definition
Efficiency
The ratio of useful energy (or power) output to total energy (or power) input, often expressed as a percentage.
The ratio of useful energy (or power) output to total energy (or power) input, often expressed as a percentage.
$$\eta = \frac{\text{useful output}}{\text{total input}} \times 100\%$$
$$\text{efficiencyThe ratio of useful energyThe capacity to do work. Measured in joules (J). output to total energyThe capacity to do work. Measured in joules (J). input, expressed as a fraction or percentage.} = \frac{\text{useful output energyThe capacity to do work. Measured in joules (J).}}{\text{total input energy}}$$
- EfficiencyThe ratio of useful energy output to total energy input, expressed as a fraction or percentage. is a dimensionless ratio. It is always between 0 and 1 (or 0% and 100%)
- No real machine is 100% efficient because energy is always dissipated as thermal energyThe energy stored in a system due to the random motion of its particles; it is related to temperature. due to friction, air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω). or electrical resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to current. Measured in ohms (Ω).
- EfficiencyThe ratio of useful energy output to total energy input, expressed as a fraction or percentage. can be calculated using energy OR powerThe rate of energy transfer. Measured in watts (W).: $efficiency = useful output powerThe rate of energy transfer. Measured in watts (W). /$ total input powerThe rate of energy transfer. Measured in watts (W).
- A Sankey diagram visually represents energy transfers; the width of each arrow is proportional to the energy
Worked Example [3 marks]
An electric motor uses 500 J of electrical energy to lift a 4.0 kg mass through a height of 10 m. Calculate the efficiency. Use g = 9.81 $m s^{-2}$.
Show Solution
1
$Useful energy output = gravitational PE gained = mg\Delta h = 4.0 \times 9.81 \times 10 = 392.4 \text{ J}$
[1]2
$Efficiency = \frac{\text{useful output}}{\text{total input}} = \frac{392.4}{500}$
[1]3
$Efficiency = 0.785 = 78.5\%$
[1]Answer
78.5%
Common Mistake
MEDIUM
Students often: It's common to calculate efficiency using total output (including wasted energy) instead of useful output only.
Instead: Only count USEFUL output energy. If a motor converts 1000 J electrical to 800 J kinetic and 200 J heat, efficiency = 800/1000 = 80%, not 1000/1000.
Instead: Only count USEFUL output energy. If a motor converts 1000 J electrical to 800 J kinetic and 200 J heat, efficiency = 800/1000 = 80%, not 1000/1000.