Calculate electrical power from current and potential difference
Use P = IV at a stated operating condition and distinguish electrical input power from useful output power.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence 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
- Read current in amperes and voltage in volts.
- Define power as energy transferred per time.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does one watt represent?
What you will learn
- Calculate power from amperes and volts.
- Distinguish rate of transfer from total energy and useful output.
Multiply matching operating values
Electrical power is P = IV: current in amperes multiplied by potential difference in volts gives power in watts. Use I and V for the same component at the same operating condition. A 3 V component carrying 0.6 A receives electrical power of 1.8 W, meaning an input rate of 1.8 J each second under constant conditions.
Convert prefixes first: 250 mA is 0.250 A. A component at 12 V carrying 250 mA receives 3 W, not 3000 W. The voltage belongs across the component whose power is being calculated; the source voltage cannot automatically replace a smaller component voltage.
Read the boundary of the power
P = IV gives the component's electrical input rate using the stated passive operating convention. It does not by itself establish how much of that input becomes useful light, mechanical output or another chosen output. A lamp receiving 12 W electrically may deliver less than 12 W of visible light.
These simple calculations use stated steady d.c. conditions. A nameplate value describes a specified rated condition, not a promise that the appliance always uses that power. For varying values, the average of simultaneous power readings is needed; multiplying unrelated averages of I and V is not generally equivalent.
Pause and explain
At 12 V and 250 mA, what is the input power?
Worked example
A component has 3 V across it and carries 0.6 A steadily. Find its electrical input power.
Show the worked solution
- Use P = IV.
- P = 0.6 × 3.
- P = 1.8 W.
Answer 1.8 W.
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 power at a steady 3 V and 0.6 A.
Give me a hint
Multiply the matching current and voltage with their units.
Compare my reasoning
- P = IV.
- P = 0.6 × 3.
- P = 1.8 W.
1.8 W.
Look for these in your work
- State watts.
- Use matching operating values.
Find input power for a component at 6 V carrying 2 A, and another at 12 V carrying 250 mA.
Give me a hint
Convert the prefixed current before treating the two cases separately.
Compare my reasoning
- First: 6 × 2 = 12 W.
- Second current: 250 mA = 0.250 A.
- Second: 12 × 0.250 = 3 W.
12 W; 3 W.
Look for these in your work
- Calculate each component.
- Convert the prefix.
A lamp receives 12 W electrically. A student claims it produces 12 W of useful light. Is that conclusion established?
Give me a hint
Check whether a useful-output measurement or efficiency is supplied.
Compare my reasoning
- The electrical input is known.
- Heating and other output pathways can also occur.
- Useful light power is not separately specified.
No; 12 W is the input, and useful light power needs further information.
Look for these in your work
- Keep input and output distinct.
- Identify missing output data.
Common mistakes
- Convert mA to A.
- Use the component voltage.
- Electrical input is not automatically useful output.
What can you explain now?
What are the units in P = IV?
Compare with the explanation
Watts, amperes and volts.
Use unprefixed consistent units for this direct calculation.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Calculate electrical input power with consistent units and explain what the answer does and does not measure.
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.