Efficiency (Oxford AQA IGCSE Physics): Revision Note

An electric motor turns the blades of a fan. In one second, 300 J of energy is transferred electrically from the mains power supply. 85 J of energy is wasted due to friction and sound.

Calculate the efficiency of the motor.

 Answer:

Step 1: List the known quantities

• Total energy in = 300 J

• Total wasted energy = 85 J

Step 2: State the equation

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{energy}\mathrm{out}}{\mathrm{total}\mathrm{energy}\mathrm{in}}\times 100\%$ 

Step 3: Determine the total energy out

• Due to the conservation of energy:

$\mathrm{total}\mathrm{energy}\mathrm{in}=\mathrm{total}\mathrm{energy}\mathrm{out}$

$\mathrm{total}\mathrm{energy}\mathrm{out}=300\mathrm{J}$

Step 4: Calculate the useful energy out

• Due to the conservation of energy

$\mathrm{total}\mathrm{energy}\mathrm{out}=\mathrm{useful}\mathrm{energy}\mathrm{out}+\mathrm{wasted}\mathrm{energy}$

$\mathrm{useful}\mathrm{energy}\mathrm{out}=\mathrm{total}\mathrm{energy}\mathrm{out}-\mathrm{wasted}\mathrm{energy}$

$\mathrm{useful}\mathrm{energy}\mathrm{out}=300-85$

$\mathrm{useful}\mathrm{energy}\mathrm{out}=215\mathrm{J}$ 

Step 5: Substitute these values into the efficiency equation 

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{energy}\mathrm{out}}{\mathrm{total}\mathrm{energy}\mathrm{in}}\times 100\%$

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

$\mathrm{efficiency}=0.72$

$\mathrm{efficiency}=72\%$

Efficiency can be given as a ratio (between 0 and 1) or as a percentage (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 %)