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Option A: Alternating Currents

Mean power and energy in a.c. circuits

  • 3 things to know
  • A2 Unit 4 Section B

Overview

Recall questions on Mean power and energy in a.c. circuits.

What you need to know

Answer each one before you reveal it.

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Key term

Explain why a sinusoidal alternating current has a mean value of zero over a whole cycle yet still heats a resistor.

Key term

Explain why the mean power dissipated in a pure inductor over a whole cycle is zero.

Numbers

A heating element of resistance 100 Ω is connected to a 230 V rms supply. Calculate the mean power and the peak instantaneous power.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Explain why a sinusoidal alternating current has a mean value of zero over a whole cycle yet still heats a resistor.

    Model answer

    The current is negative for exactly as long as it is positive, so the values cancel and the mean is zero. The heating rate depends on the square of the current, and a square is positive whichever way the current flows, so energy is dissipated throughout the cycle.

  2. 2

    Explain what the rms value of an alternating current means in terms of the energy dissipated each cycle.

    Model answer

    The rms value is the steady direct current that would transfer the same energy to the same resistor in the same time. An alternating current of rms value 3 A therefore dissipates in one cycle exactly what a steady 3 A would dissipate over that time.

  3. 3

    State the mean power dissipated in a resistor in terms of the peak current and peak voltage.

    Model answer

    The mean power is P = ½V₀I₀, where V₀ is the peak potential difference in V and I₀ the peak current in A.

  4. 4

    How does the peak instantaneous power in a resistor compare with the mean power dissipated in it?

    Model answer

    The peak instantaneous power is exactly twice the mean power, because the mean of a squared sinusoid over a whole cycle is half of its maximum.

  5. 5

    An alternating current of rms value 4.0 A flows through a 12 Ω resistor from a 50 Hz supply. Calculate the energy dissipated in one cycle.

    Model answer

    The mean power is P = Iᵣₘₛ²R = 16 × 12 = 192 W, and one cycle lasts 1 ÷ 50 = 0.020 s, so the energy is 192 × 0.020 = 3.84 J. Substituting back, 3.84 ÷ 0.020 returns a mean power of 192 W.

  6. 6

    Explain why the mean power dissipated in a pure inductor over a whole cycle is zero.

    Model answer

    The current is a quarter of a cycle out of step with the potential difference, so the instantaneous power is positive for one quarter cycle and negative for the next. Energy taken from the supply to build the magnetic field is returned to the supply as the field collapses, and nothing is dissipated.

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