Calculate efficiency using power and complete stage chains
Use matching power rates and multiply stage fractions without adding efficiencies.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Distinguish watts from joules.
- Use efficiency fractions and matching units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many watts is 0.80 kW?
What you will learn
- Recall efficiency = useful power output / total power input.
- Find power inputs and outputs and combine consistent successive stages.
Compare rates over a consistent account
Efficiency can also be useful power output divided by total power input, multiplied by 100% for a percentage. A 1000 W input and 350 W useful output give 35%. Convert kilowatts to watts before comparing: 0.80 kW input and 240 W useful output give 240/800 = 30%. Both values must describe the relevant same boundary and operating interval.
An energy amount in joules cannot be divided by a power in watts and called an efficiency without first obtaining consistent quantities. If 600 J reaches a desired output in 20 s, useful power is 30 W. With total input power 50 W over that same interval, efficiency is 60%; equivalently the input energy is 1000 J and 600/1000 is the same fraction.
The second stage receives only the first output
For a 500 W input, a first stage with fraction 0.60 supplies 300 W onward. A second stage with fraction 0.40 of that supplied amount delivers 120 W usefully. Overall fraction is 0.60 × 0.40 = 0.24, or 24%, rather than their sum 100%. The second stage does not receive the original 500 W again.
The multiplication requires a defined chain in which the first useful output is the second stage's input and no unaccounted additional source enters. If another input is supplied, include it in a new complete account instead of applying the same two-stage product automatically. Higher efficiency does not alone imply greater useful power when input rates differ.
Sources for this lesson
Pause and explain
For consistent 60% and 40% consecutive stages, is overall efficiency 100%?
Worked example
A 500 W input passes through 60% then 40% stages, with the first output the only second input. Find final useful power and overall efficiency.
Show the worked solution
- First output = 0.60 × 500 = 300 W.
- Second output = 0.40 × 300 = 120 W.
- Overall efficiency = 120/500 = 24%.
Answer 120 W; overall 24%.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find final power and overall efficiency for 500 W through 60% and then 40% stages with no added input.
Give me a hint
Use the first output as the second input.
Compare my reasoning
- First stage supplies 300 W.
- Second stage supplies 120 W.
- 120/500 = 24%.
120 W; 24% overall.
Look for these in your work
- Use the stage boundary correctly.
- Multiply fractions.
Find efficiency for 240 W useful from 0.80 kW input; then for 600 J useful in 20 s with input 50 W over that interval.
Give me a hint
Convert the first input and derive the second useful rate.
Compare my reasoning
- 240/800 = 30%.
- Second useful power = 600/20 = 30 W.
- 30/50 = 60%.
30%; 60%.
Look for these in your work
- Match units and intervals.
- Convert energy into a consistent rate when needed.
A 35%-efficient device supplies 350 W. Find its required input, then explain why another device's 500 W useful output alone cannot establish higher efficiency.
Give me a hint
Rearrange the ratio, then identify the missing input for the comparison.
Compare my reasoning
- Required input = 350/0.35 = 1000 W.
- Another useful output amount does not state its input.
- Compare each useful fraction using the same intended task and a complete power account.
1000 W; the other device's input is needed to compare efficiency.
Look for these in your work
- Use the fraction correctly.
- Do not rank efficiency from useful power alone.
Common mistakes
- Use watts with watts or energy amounts with energy amounts.
- Multiply successive stage fractions, not percentages added together.
- Include any additional supplied energy source.
What can you explain now?
What input power supplies 350 W usefully at 35%?
Compare with the explanation
1000 W.
350/0.35 = 1000 W for the complete specified account.
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Make it stick
Use power efficiency, compare energy and rate accounts and explain a two-stage product under its stated no-extra-input condition.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.