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A LevelPhysicsCambridge (CIE)Revision Notes21. Alternating CurrentsCharacteristics of Alternating CurrentsMean Power

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First exams 2025

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Mean Power (Cambridge (CIE) A Level Physics)

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Ashika

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Ashika

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Mean power

  • In mains electricity, current and voltage are varying all the time

  • This also means the power varies constantly, recall the equations for power:

P space equals space I V space equals space I squared R space equals space V squared over R

  • Where:

    • I = direct current (A)

    • V = direct voltage (V)

    • R = resistance (Ω)

  • The r.m.s values means equations used for direct current and voltage can now be applied to alternating current  and voltage

  • These are also used to determine an average current or voltage for alternating supplies

  • Recall the equation for peak current:

I subscript 0 space equals space square root of 2 I subscript r. m. s end subscript

  • The peak (maximum) power and the mean (average) power are given by:

P subscript m e a n end subscript space equals space open parentheses I subscript r. m. s end subscript close parentheses squared R

P subscript p e a k end subscript space equals space I subscript 0 squared R

  • Peak power can be written in terms of r.m.s current as

P subscript p e a k end subscript space equals space open parentheses square root of 2 I subscript r. m. s end subscript close parentheses squared R

  • Therefore, peak power is related to mean power by:

2 open parentheses I subscript r. m. s end subscript close parentheses squared R space equals space 2 P subscript m e a n end subscript

P subscript m e a n end subscript space equals space P subscript p e a k end subscript over 2

  • Therefore, it can be concluded that:

    The mean power in a resistive load is half the maximum power for a sinusoidal alternating current or voltage

Mean power on a graph

Average power graph, downloadable AS & A Level Physics revision notes

Mean power is exactly half the maximum power

Worked Example

An alternating voltage supplied across a resistor of 40 Ω has a peak voltage V0 of 240 V.

Calculate the mean power of this supply.

Answer:

Step 1: Write down the known quantities

  • Resistance, R = 40 Ω

  • Peak voltage, V0 = 240 V

Step 2: Write out the equation for the peak power and calculate

P space equals fraction numerator space V subscript 0 squared over denominator R end fraction

P space equals space fraction numerator open parentheses 240 close parentheses squared space over denominator 40 end fraction space equals space 1440 space straight W

Step 3: Calculate the mean power

  • The mean power is half of the maximum (peak) power

Mean space power space equals space 1440 over 2 space equals space 720 space straight W

Examiner Tips and Tricks

You do not need to remember the derivation for the mean power, but it is useful to know where it comes from. However, makes sure you remember its definition and know how to apply it in questions.

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    Ashika

    Author: Ashika

    Expertise: Physics Project Lead

    Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.