Calculate efficiency using useful energy output
Divide the useful share by the full input and keep consistent units, boundaries and intervals.
Section 1.7.3 · Energy resources
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.7.3. Resource comparison and qualitative efficiency are Core. Solar origins and exceptions, fusion and its electricity research, and numerical efficiency are Supplement. Compare the same useful task, consistent system boundary and stated interval. Written work is self-checked. Human subject-teacher review, supervised practical experience and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Identify the intended useful output.
- Convert kilojoules to joules and use percentages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which fraction represents 40%?
What you will learn
- Recall efficiency = useful energy output / total energy input.
- Convert between fraction and percentage and find missing amounts.
The denominator is the full input
Efficiency as a fraction is useful energy output divided by total energy input. Multiply by 100% for percentage efficiency. For 200 J input and 120 J useful output, the fraction is 0.60 and efficiency is 60%. The remaining 80 J has other conserved destinations; it is not deleted from the input denominator.
Use matching units and the same event, boundary and interval. A 2.0 kJ input is 2000 J, so 500 J useful output gives fraction 0.25 or 25%. Dividing 500 by 2 without converting units gives a meaningless result. If the account includes an additional energy source or an initial store decrease, include it in the full relevant input rather than claiming a conversion exceeds its complete input.
Rearrange a complete account
Useful output = efficiency fraction × total input. At 40% efficiency and 300 J input, useful output is 120 J and other transfers are 180 J in an exhaustive account. If 150 J useful output is required at 30%, total input is 150/0.30 = 500 J. A percentage value of 30 means fraction 0.30, not thirty times the input.
At the same useful task and consistent full boundary, a larger useful fraction means higher efficiency. It need not mean a larger useful amount if input quantities differ. A reported ordinary conversion value above 100% requires checking units, boundaries, omitted inputs or measurements. Heating-system ratios using additional environmental energy cannot be called this complete-input conversion efficiency by ignoring that energy source.
Pause and explain
What denominator belongs in efficiency for 200 J input with 120 J useful and 80 J other output?
Worked example
A device receives 200 J and supplies 120 J to its stated useful task. Find fraction, percentage and the exhaustive other share.
Show the worked solution
- Fraction = 120/200 = 0.60.
- Percentage = 0.60 × 100% = 60%.
- Other destinations receive 200 − 120 = 80 J.
Answer 0.60 or 60%; other share 80 J.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
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 efficiency and the other share for 200 J input and 120 J useful output.
Give me a hint
Use the whole input and then an exhaustive difference.
Compare my reasoning
- Fraction = 120/200 = 0.60.
- Percentage = 60%.
- Other share = 80 J.
60%; 80 J other output.
Look for these in your work
- Use the full denominator.
- Account for the remaining conserved energy.
Find efficiency for 500 J useful from 2.0 kJ input, and useful output from 300 J input at 40%.
Give me a hint
Convert input units and percentage fractions before calculating.
Compare my reasoning
- 2.0 kJ = 2000 J, so 500/2000 = 25%.
- 40% is fraction 0.40.
- Useful output = 0.40 × 300 = 120 J.
25%; 120 J.
Look for these in your work
- Use matching units.
- Multiply by the fraction, not forty.
A device is claimed to supply 150 J usefully at 30% using only 300 J complete input. Check the numbers and find the input actually required by that efficiency.
Give me a hint
Compare the claimed ratio and the rearranged input.
Compare my reasoning
- 150/300 = 50%, not 30%.
- At 30%, input must be 150/0.30.
- The required input is 500 J; the original combined claim is inconsistent.
The claim gives 50%; 30% would require 500 J input.
Look for these in your work
- Check rather than accept inconsistent labels.
- Keep a complete input boundary.
Common mistakes
- Divide by full input, not wasted share.
- Convert units before the ratio.
- Use percentage as a fraction when rearranging.
What can you explain now?
What input supplies 150 J usefully at 30%?
Compare with the explanation
500 J.
150/0.30 = 500 J for the consistent supplied account.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Recall energy efficiency, convert units, find missing input or output and identify a boundary or percentage error.
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.