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IGCSE · Cambridge (CIE) · 0654

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.

Section P4.2.5 · Electrical energy and electrical power

Electrical energy and electrical power · Cambridge IGCSE 0654 · P4.2.5

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

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.
Original steady d.c. input example uses 0.6 A at 3 V to give 1.8 W. Power is an energy-transfer rate, and the electrical input is not automatically the useful output. Current and voltage must describe the same operating state.Matching steady input valuesI = 0.6 A · V = 3 VP = I V = 1.8 W1.8 J each secondAn energy-transfer rateElectrical input powerNot automatically useful output
Original steady d.c. input example uses 0.6 A at 3 V to give 1.8 W. Power is an energy-transfer rate, and the electrical input is not automatically the useful output. Current and voltage must describe the same operating state.

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.

Sources for this lesson

Pause and explain

At 12 V and 250 mA, what is the input power?

Put the idea to work

Worked example

A component has 3 V across it and carries 0.6 A steadily. Find its electrical input power.

Show the worked solution
  1. Use P = IV.
  2. P = 0.6 × 3.
  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.

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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.

1 · Guided

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
  1. P = IV.
  2. P = 0.6 × 3.
  3. P = 1.8 W.

1.8 W.

Look for these in your work

  • State watts.
  • Use matching operating values.
2 · Independent

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
  1. First: 6 × 2 = 12 W.
  2. Second current: 250 mA = 0.250 A.
  3. Second: 12 × 0.250 = 3 W.

12 W; 3 W.

Look for these in your work

  • Calculate each component.
  • Convert the prefix.
3 · Transfer

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
  1. The electrical input is known.
  2. Heating and other output pathways can also occur.
  3. 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.
Recall without your notes

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.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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