Efficiency (Edexcel GCSE Combined Science: Physics): Revision Note

The blades of a fan are turned by an electric motor. In one second, 300 J of energy is transferred electrically from the mains supply to the fan. 85 J is wasted due to friction and sound.

Calculate the percentage efficiency of the motor.

Answer:

Step 1: List the known quantities

• Total energy supplied to the device (input) = 300 J

• Total wasted energy = 85 J

Step 2: State the equation

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}\left(\mathrm{output}\right)}{\mathrm{total}\mathrm{energy}\mathrm{supplied}\mathrm{to}\mathrm{the}\mathrm{device}\left(\mathrm{input}\right)}\times 100\%$

Step 3: Determine total energy transferred by the device (output)

• Due to the conservation of energy:

$\mathrm{total}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}\left(\mathrm{output}\right)=\mathrm{total}\mathrm{energy}\mathrm{supplied}\mathrm{to}\mathrm{the}\mathrm{device}\left(\mathrm{input}\right)$

• Therefore, total energy transferred by the device (output) = 300 J

Step 4: Calculate the useful energy transferred by the device (output)

• Due to conservation of energy

$\mathrm{total}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}=\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}+\mathrm{wasted}\mathrm{energy}$

$\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}=\mathrm{total}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}-\mathrm{wasted}\mathrm{energy}$

$\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}=300-85$

$\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}=215\mathrm{J}$

Step 5: Substitute these values into the efficiency equation 

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{energy}\mathrm{transferred}\mathrm{by}\mathrm{the}\mathrm{device}\left(\mathrm{output}\right)}{\mathrm{total}\mathrm{energy}\mathrm{supplied}\mathrm{to}\mathrm{the}\mathrm{device}\left(\mathrm{input}\right)}\times 100\%$

$\mathrm{efficiency}=\frac{215}{300}\times 100\%$

$\mathrm{efficiency}=72\%$

Efficiency can be in a ratio (between 0 and 1) or percentage format (between 0 and 100%)If the question asks for efficiency as a ratio, give your answer as a fraction or decimal.

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 (only %)