Efficiency = useful output ÷ total input. Multiply by 100 to give a percentage. The numerator and denominator must be the same kind of quantity, both energy or both power, and in the same unit.
This lesson follows tracking energy stores, because you must know what is useful and what is wasted. Later, power and energy resources uses the same idea for devices.
How do I calculate efficiency step by step?
- Identify the total input. This is the energy (or power) supplied to the device.
- Identify the useful output. This is the energy (or power) in the form you wanted.
- Make the units match. Convert kJ to J, or kW to W, before dividing.
- Divide useful output by total input.
- Convert to a percentage by multiplying by 100, then check that the answer is between 0% and 100%.
Because the two quantities are in the same unit, the efficiency itself has no unit. It is a ratio, often written as a percentage.
Worked example
A winch lifts a 40 kg load through 6.0 m. It takes 3.2 kJ of electrical energy. Take g = 10 N/kg. (Invented example data.)
Step 1, useful output. The useful energy is the gain in gravitational potential energy: mgh = 40 × 10 × 6.0 = 2400 J.
Step 2, total input. 3.2 kJ = 3200 J.
Step 3, efficiency. 2400 ÷ 3200 = 0.75.
Step 4, percentage. 0.75 × 100 = 75%.
Step 5, wasted energy. 3200 − 2400 = 800 J went to the thermal store and sound.
The power version works the same way. If a motor takes in 800 W and gives 200 W of useful mechanical power, the efficiency is 200 ÷ 800 = 0.25, or 25%.
What mistake should I watch for?
The most frequent slip is to divide numbers that are in different units.
Mistaken answer: efficiency = 2400 ÷ 3.2 = 750%.
The student divided joules by kilojoules.
The correction is to write 3.2 kJ as 3200 J first, giving 2400 ÷ 3200 = 75%. The result of 750% should have rung an alarm, because efficiency cannot exceed 100%. A second slip is to divide the input by the output, which also gives a value above 100%.
Check yourself
Try these without a calculator, then open each answer.
1. A lamp takes in 60 J of electrical energy and gives 6 J as light. What is its efficiency?
Show answer
6 ÷ 60 = 0.10, so 10%.
2. A machine has an input power of 800 W and an efficiency of 25%. What is its useful output power?
Show answer
Useful power = 0.25 × 800 = 200 W.
3. A heater takes in 2.4 kJ and gives 1.8 kJ of useful thermal energy. Find the efficiency and the wasted energy.
Show answer
Efficiency = 1.8 ÷ 2.4 = 0.75, so 75%. Wasted energy = 2.4 − 1.8 = 0.6 kJ (600 J).
Where this leads next
Once you can calculate efficiency, explain an energy loss without saying energy disappears teaches the words that earn the explanation marks. The bounds and rounding explainer helps you check how many significant figures an answer should carry.
Efficiency questions combine unit handling, choice of useful output and percentages in a few lines. A teacher in online one-to-one Physics tuition can watch which of these three steps slips first and tighten it.