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Option A: Alternating Currents

Mean power dissipated in a resistor

  • 3 things to know
  • Component 3

Overview

Recall questions on why the mean of an alternating current is zero but the mean power is not, the mean of the square of the current, the factor of a half in the mean power, worked mean power calculations, and why root mean square values are quoted.

What you need to know

Answer each one before you reveal it.

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Key term

Why must rms rather than peak values be used in P = I²R for an alternating current?

Key term

Give the three equivalent expressions for the mean power dissipated in a resistor carrying an alternating current.

Key term

Why does the instantaneous power in a resistor never become negative on an alternating supply?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Give the three equivalent expressions for the mean power dissipated in a resistor carrying an alternating current.

    Model answer

    The mean power is P = IV = I²R = V²/R, in watts. Here I and V are the rms current and rms potential difference, and R is the resistance in Ω.

  2. 2

    Why must rms rather than peak values be used in P = I²R for an alternating current?

    Model answer

    The rms value is defined so that it gives the correct mean rate of energy dissipation. Using peak values would give the peak power. For a sine wave that is twice the mean power.

  3. 3

    How does the peak power dissipated in a resistor on a sinusoidal supply compare with the mean power?

    Model answer

    The peak power is exactly twice the mean power. This follows because the peak values are √2 times the rms values, and power depends on their product.

  4. 4

    A 60 Ω heating element is connected to a 230 V rms supply. Calculate the mean power dissipated.

    Model answer

    P = V²/R = 230²/60 = 52900/60 = 880 W to two significant figures. Working back, √(880 × 60) = √52800 = 230 V.

  5. 5

    A 60 Ω element carries an rms current of 3.8 A. Calculate the mean power dissipated.

    Model answer

    P = I²R = 3.8² × 60 = 14.44 × 60 = 870 W to two significant figures. Working back, √(870/60) = √14.5 = 3.8 A.

  6. 6

    An alternating supply has a peak potential difference of 12 V across a 24 Ω resistor. Calculate the mean power dissipated.

    Model answer

    The rms potential difference is 12/√2 = 8.49 V, so P = V²/R = 8.49²/24 = 72.0/24 = 3.0 W. Working back, √(3.0 × 24) = 8.49 V rms. Multiplying by √2 returns a peak of 12 V.

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