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A Level · Cambridge (CIE) · 9702

Relate mean and maximum power in an AC resistor

Use the sinusoidal current model to explain why a resistive load's mean power is half its maximum power.

Alternating currents · Cambridge International AS & A Level Physics 9702 · 21.1

A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 21. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use period, frequency, sine graphs, current, voltage, power, diodes, capacitors and SI units.

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

Quick readiness check

For a sinusoidal AC in a resistor, how does mean power compare with maximum power?

What you will learn

  • Recall and use mean power = half maximum power for a sinusoidal AC in a resistive load
  • Connect instantaneous resistive power to current squared
  • Distinguish mean power from peak power

Resistive power follows the square of current

For a resistor R, instantaneous power is p = i²R (also p = v²/R). If i = I₀ sin ωt, then p = I₀²R sin²ωt. Power is never negative for an ideal resistor, even when the current reverses, because the current is squared.

The maximum occurs at |i| = I₀, so Pmax = I₀²R. Over a complete sinusoidal cycle, the mean value of sin²ωt is 1/2. Therefore the mean power in a resistive load is Pmean = Pmax/2. This stated half-maximum result applies to a sinusoidal alternating current and a resistive load.

Mean power is the cycle average

Mean power averages the changing instantaneous power over a full cycle. It is not the power at the zero crossings, and it is not the maximum. At current zero crossings, instantaneous resistive power is zero; near either current peak, it is greatest.

If Pmax is given, halve it for Pmean. If the mean is given, double it for Pmax. Check that the question concerns a resistive load and a sinusoidal AC before using this shortcut; a non-resistive load can have a different power relationship.

p = i²R · Pmax = I₀²R · Pmean = ½Pmax

Pause and explain

A sinusoidal AC resistor has maximum instantaneous power 80 W. What is its mean power?

Put the idea to work

Worked example

A sinusoidal current in a resistor produces a maximum instantaneous power of 120 W. Find its mean power over a cycle.

Show the worked solution
  1. Identify the stated sinusoidal current and resistive load.
  2. Use Pmean = Pmax/2.
  3. Substitute the maximum power.

Answer Pmean = 120/2 = 60 W.

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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

A sinusoidal current in a 5.0 Ω resistor has peak value 4.0 A. Calculate its maximum and mean powers.

Give me a hint

Use Pmax = I₀²R, then halve it for the cycle mean.

Compare my reasoning
  1. Pmax = (4.0)² × 5.0 = 80 W.
  2. Pmean = Pmax/2.
  3. Pmean = 40 W.

Maximum power = 80 W; mean power = 40 W.

Look for these in your work

  • Peak power calculated
  • Resistive sinusoidal condition applied
  • Mean distinguished from maximum
2 · Independent

A heater's resistive element receives a sinusoidal current and has mean power 75 W. What is the maximum instantaneous power?

Give me a hint

Reverse the relationship: Pmax = 2Pmean.

Compare my reasoning
  1. Pmean = Pmax/2.
  2. Rearrange to Pmax = 2Pmean.
  3. Pmax = 2 × 75 = 150 W.

Maximum instantaneous power = 150 W.

Look for these in your work

  • Relationship recalled
  • Equation rearranged
  • Power stated
3 · Transfer

A student says an alternating current that averages to zero gives zero mean power in a resistor. Explain the error and state the mean-to-maximum relationship.

Give me a hint

Compare the time dependence of current and i²R.

Compare my reasoning
  1. The current changes sign, so its average over a cycle can be zero.
  2. Resistor power is p = i²R and is non-negative for either current direction.
  3. For sinusoidal current, mean power is half the maximum power.

The current average may be zero, but squaring it gives positive power during both half-cycles; Pmean = Pmax/2.

Look for these in your work

  • Current and power distinguished
  • Square dependence explained
  • Required ratio stated

Common mistakes

  • Averaging current first and then squaring it
  • Allowing negative current to imply negative resistor power
  • Calling peak power the cycle mean
  • Applying the half-maximum result to any load without checking the resistive condition
Recall without your notes

What can you explain now?

For a sinusoidal AC in a resistor, how does mean power compare with maximum power?

Compare with the explanation

It is half the maximum

The cycle mean of sin² is 1/2.

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

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Make it stick

Why can a reversing current have positive mean power in a resistor?

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

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.

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