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Physics · Ch 4 — Electromagnetic Induction and Alternating Current

Mean or Average Value of AC

4.7.2

Mean or Average Value of AC

Because a symmetric alternating current spends exactly as much time (and reaches exactly the same peak magnitude) in its positive half-cycle as in its negative half-cycle, the SIMPLE average of current over one FULL cycle is always exactly zero -- the positive and negative contributions cancel completely -- making a full-cycle average useless as a way to characterise an AC waveform's typical size. The average (or mean) value of alternating current is therefore instead defined over just ONE HALF cycle (positive or negative), as the average of all the current values over that half-cycle.

Deriving this value: with i=Imsin⁡θi=I_m\sin\theta, the sum of all current values over the positive half-cycle is represented by the AREA under the curve from θ=0\theta=0 to θ=π\theta=\pi, found by integrating an elementary strip i dθi\,d\theta: Area=∫0πImsin⁡θ dθ=Im[−cos⁡θ]0π=Im(−cos⁡π+cos⁡0)=2Im\text{Area}=\int_0^{\pi}I_m\sin\theta\,d\theta = I_m[-\cos\theta]_0^{\pi}=I_m(-\cos\pi+\cos0)=2I_m. Dividing this area by the half-cycle's base length π\pi gives the average value,

Iav=2Imπ≈0.637 Im(4.32)I_{av} = \dfrac{2I_m}{\pi} \approx 0.637\,I_m \qquad (4.32) …

Figure 4.36Sine wave of an alternating current, with an elementary strip

What this figure shows. A single positive half-cycle of a sinusoidal current wave i=Imsin⁡θi=I_m\sin\theta is drawn against the angle axis from 00 to π\pi, with a thin vertical elementary strip of angular width dθd\theta marked at some intermediate angle, whose mid-ordinate height represents the instantaneous current i at that point. The figure supports the integration used to derive the average value: treating the area under this half-cycle curve as the sum of many such thin strips (i dθi\,d\theta each), integrating Imsin⁡θI_m\sin\theta from 00 to π\pi gives the total area as 2Im2I_m, and dividing by the half-cycle's base length π\pi gives the avera …