Define power as energy transfer per unit time
Use watts as joules per second and distinguish the transferred amount from its rate.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 1.7.4. Power is work done or energy transferred per unit time: P = W/t = ΔE/t. A whole-interval quotient gives average power unless a constant rate is stated. Keep system, units and interval consistent. Written practice 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
- Measure work and energy in joules.
- Measure elapsed time in seconds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which unit is a joule per second?
What you will learn
- Define power as work done or energy transferred per unit time.
- Recall P = W/t and P = ΔE/t.
A rate includes time
Power describes how quickly energy is transferred or work is done. P = ΔE/t, or P = W/t for a specified work transfer. A watt is one joule per second. Transferring 450 J in 15 s gives average power 30 W; transferring 900 J in 30 s also gives 30 W. The energy amounts differ, but their interval rates agree.
An energy of 450 J alone does not establish a power until its transfer interval is supplied. A larger energy transfer can have a smaller power if it takes sufficiently longer. Do not replace a joule amount by the same numerical watt value or treat 'power' as the amount permanently stored in a device.
Average and constant rates need distinct claims
Dividing a complete interval's transferred amount by elapsed time gives its average power. In the original straight energy-time model E = 30t from 0 to 30 s, each equal interval transfers at the same constant 30 W rate, so the whole-interval average also describes that constant rate. The graph is of cumulative energy transferred, not the remaining absolute store inside a battery.
A device can vary its instantaneous transfer rate during an interval while still averaging 30 W. The same total and duration do not prove equal peak power or an identical time profile. For zero duration a finite interval-average quotient is not defined; specify a real positive interval before using the simple amount/time calculation.
Pause and explain
Does 450 J alone imply 450 W?
Worked example
Find average powers for 450 J in 15 s and 900 J in 30 s.
Show the worked solution
- First P = 450/15 = 30 W.
- Second P = 900/30 = 30 W.
- The amounts differ but energy transferred per second is the same.
Answer 30 W in both intervals.
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 average power for 450 J transferred in 15 s.
Give me a hint
Divide the amount by its supplied positive duration.
Compare my reasoning
- Energy transferred is 450 J.
- Elapsed time is 15 s.
- P = 450/15 = 30 W.
30 W.
Look for these in your work
- Use energy per time.
- Report the rate in W.
Compare 900 J in 30 s with 600 J in 10 s.
Give me a hint
Calculate both quotients before comparing amount and rate.
Compare my reasoning
- 900/30 = 30 W.
- 600/10 = 60 W.
- The smaller energy amount has the larger rate in this comparison.
30 W and 60 W; 600 J is transferred faster on average.
Look for these in your work
- Use each supplied interval.
- Do not rank power by energy alone.
Two devices each transfer 450 J in 15 s. Does that prove identical peak powers and time profiles? Explain.
Give me a hint
Identify exactly what the interval quotient establishes.
Compare my reasoning
- Both average powers are 30 W.
- Transfer can be distributed differently during each interval.
- The totals do not establish peaks or instantaneous profiles.
Same 30 W average; peaks and profiles are not determined.
Look for these in your work
- Retain the verified average.
- State the missing time-profile evidence.
Common mistakes
- Energy amount and power rate have different units.
- Use a stated positive elapsed interval.
- Average power does not determine peak power.
What can you explain now?
How many joules per second is 30 W?
Compare with the explanation
30 J/s.
A watt is defined as one joule transferred per second.
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
Define power and watts, compare two interval rates and explain the limits of an average.
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.