A Level · Cambridge (CIE) · 9702

Electrical power and energy

Distinguish a rate of energy transfer from the total energy transferred.

What you will learn

Use P=VI and E=Pt.
Derive alternative power formulas for resistive components.

Power is a rate

Potential difference is energy transferred per unit charge and current is charge per unit time. Multiplying them gives energy per unit time, which is power. The watt is a joule per second.

For a constant power, multiply by elapsed time to calculate energy. Use seconds with watts if the result is needed in joules. Energy in kWh uses kilowatts and hours instead.

Use the quantities you know

Combining P=VI with V=IR gives P=I²R or P=V²/R for a resistive component. Do not use these substitutions blindly for complex alternating-current components; the simple starter calculations concern direct-current circuit quantities.

P=VI; E=Pt; P=I²R=V²/R

Put the idea to work

Worked example

A component has 10 V across it and carries 0.6 A for 2 minutes. Find power and energy transferred.

Show the worked solution

Power is 10×0.6=6 W.
Convert two minutes to 120 seconds.
Energy is 6×120.

Answer 6 W and 720 J

Common mistakes

A watt is not a unit of energy.
Remember to square the current in I²R.

Recall without your notes

What can you explain now?

A 5 Ω resistor carries 2 A. Find its power.

Compare with the explanation

20 W

P=I²R=4×5=20.

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.

Practise this idea Back to Today

Keep learning Physics

Check the official syllabus 2025–2027

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.