8.1.7 Efficiency & Power | Edexcel GCSE Physics Revision Notes 2018

Efficiency & Power

The efficiency of a system is a measure of the amount of wasted energy in an energy transfer

The ratio of the useful energy output from a system to its total energy output

If a system has high efficiency, this means most of the energy transferred is useful
If a system has low efficiency, this means most of the energy transferred is wasted

$efficiency = \frac{useful energy transferred by the device}{total energy supplied to the device}$

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

An electric motor has an efficiency of 35%. It lifts a 7.2 kg load through a height of 5 m in 3 s.

Calculate the power of the motor.

Step 1: List the known quantities

- Efficiency = 35%
- Mass, m = 7.2 kg
- Height, h = 5 m
- Time, t = 3 s
- Gravitational field strength, g = 10 N/kg

Step 2: Write down the efficiency equation

$efficiency = \frac{useful power output}{total power input} \times 100 %$

Step 3: Rearrange to make power input the subject

$useful power input = \frac{power output \times 100}{efficiency}$

Step 4: Calculate the power output

$P = \frac{E}{t}$

- Energy was transferred from the kinetic store of the motor to the gravitational potential store of the load

$∆ G P E = m \times g ∆ h$

$∆ G P E = 7.2 \times 10 \times 5$

$∆ G P E = 360 J$

$P = \frac{360}{3}$

$P = 120 W$

Step 5: Substitute values into power input equation

$useful power input = \frac{power output \times 100}{efficiency}$

$useful power input = \frac{120 \times 100}{35}$

$useful power input = 343 J$

Efficiency can be in a ratio (between 0 and 1) or percentage (between 0 and 100%)

If the answer is required as a percentage, remember to multiply the ratio by 100 to convert it: if the ratio = 0.25, percentage = 0.25 × 100 = 25 %

Remember that efficiency has no units (just the % sign)