1.1.15 Efficiency | AQA GCSE Physics Revision Notes 2018

Efficiency

The efficiency of a system is a measure of the amount of wasted energy in an energy transfer
Efficiency is defined as:

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

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

$efficiency = \frac{useful output energy transfer}{total input energy transfer}$

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

Since power is the energy transferred per unit time, power can also be used to calculate efficiency

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.

Answer:

Step 1: Write down the efficiency equation

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

Step 2: Rearrange to make power input the subject

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

Step 3: Calculate the power output

$power output = \frac{E}{t}$

ΔE is equal to the change in gravitational potential energy as the load is lifted

$∆ E_{P} = m g ∆ h$

$∆ E_{P} = 7.2 \times 9.8 \times 5$

$∆ E_{P} = 352.8 J$

Therefore, $power output = \frac{352.8}{3}$

$power output = 117.6 W$

Step 4: Substitute the values into the power input equation

$power input = \frac{117.6}{0.35}$ OR $power input = \frac{117.6 \times 100}{35}$

$power input = 336 W$

Efficiency can be given in a ratio (between 0 and 1) or percentage format (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