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IGCSE · Cambridge (CIE) · 0625

Find operating current or potential difference from electrical power

Rearrange P = IV using non-zero supplied values and check the resulting operating condition.

Electrical energy and power pathway · 4.2.5

Core and Extended; Physics 0625, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked.

Before you begin

  • Use P = IV.
  • Convert watts and kilowatts.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

How many watts equal 1.2 kW?

What you will learn

  • Find current using I = P/V.
  • Find potential difference using V = P/I.
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.

Choose the unknown

If electrical input power P and non-zero component voltage V are known at the same steady condition, current is I = P/V. A component receiving 24 W at 12 V carries 2 A. If P and non-zero I are supplied instead, use V = P/I; 9 W at 0.5 A requires 18 V across that component.

Use the electrical input power. A motor's useful mechanical output is not enough to infer its electrical current unless efficiency or other input information is supplied. Check units before dividing: 1.2 kW is 1200 W, so at a stated steady 240 V the arithmetic current is 5 A.

Check the operating assumptions

A rated power and voltage can identify the rated current under the stated simple model; a different supplied voltage does not guarantee the device retains its rated power. Real components can change their resistance, temperature or control state. Record the stated condition instead of inventing a universal behaviour for the appliance.

Division by zero is not a valid rearrangement. If I and P are both zero, the relation P = IV does not uniquely establish V. An open interval can have a voltage between two points while sustained current is zero. The equation and the measurement context must be read together.

Pause and explain

At 9 W and 0.5 A, what is the voltage?

Put the idea to work

Worked example

A steady component receives 24 W at 12 V. Find its current.

Show the worked solution
  1. Rearrange as I = P/V.
  2. I = 24/12.
  3. I = 2 A.

Answer 2 A.

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 current for a component receiving 24 W at 12 V.

Give me a hint

Divide electrical input power by the non-zero voltage.

Compare my reasoning
  1. I = P/V.
  2. I = 24/12.
  3. I = 2 A.

2 A.

Look for these in your work

  • Use input power.
  • State amperes.
2 · Independent

Find voltage at 9 W and 0.5 A, then current at 1.2 kW and a stated steady 240 V.

Give me a hint

Solve the two cases separately and convert kilowatts first.

Compare my reasoning
  1. V = 9/0.5 = 18 V.
  2. 1.2 kW = 1200 W.
  3. I = 1200/240 = 5 A.

18 V; 5 A.

Look for these in your work

  • Choose the correct rearrangement.
  • Convert power units.
3 · Transfer

A motor delivers 60 W mechanically at 12 V. Is its electrical current necessarily 5 A?

Give me a hint

The useful mechanical output is not the supplied electrical input.

Compare my reasoning
  1. 60/12 would use output power as input.
  2. Other transfers may occur in the motor.
  3. Input power or efficiency is needed to determine the current.

No; electrical input power or efficiency is missing.

Look for these in your work

  • Identify the power boundary.
  • Do not assume perfect efficiency.

Common mistakes

  • Use input power.
  • Convert kW to W before dividing by volts.
  • Zero input and zero current do not uniquely fix voltage.
Recall without your notes

What can you explain now?

How do you find V when P and non-zero I are supplied?

Compare with the explanation

V = P/I.

Rearrange the same operating relation without reversing the quotient.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Rearrange the power relation with matching input data and avoid inferring device behaviour or missing input power.

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 Physics

Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route.

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