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When we talk about the effective voltage of AC, we take the Root-Mean-Square, which ends up being $V_{peak} / \sqrt{2}$ for sine wave energy supplies. Why do we not just take the average value of the absolute value of the AC function? When that is done for a simple sinusoidal wave, the result is $2V_{peak}/\pi$, not $V_{peak} / \sqrt{2}$. Am I missing some weird integral property or something of that sort, or is there some reason that taking the mean of $|V(x)|$ does not make sense is this case?

finding the mean of |sin(x)|

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    $\begingroup$Do you agree that the RMS value of the sinusoid produces the same power in a resistor and a pure dc voltage of the same magnitude as the RMS value?$\endgroup$
    – Bob D
    Commentedyesterday

3 Answers 3

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In the case of a sinusoidal alternating voltage $V=V_p\sin x$ and current $I= I_p \sin x$ (with $x=\omega t$), the effective voltage $V_e$ and the effective current $I_e$ are defined as the DC voltage and DC current giving the same mean power $P$ as the AC voltage and current over the AC period:

$$P= V_e I_e=V_p I_p \frac {1}{\pi}\int_0^\pi\sin^2 x dx= V_p I_p\frac{1}{2}=\frac {V_p}{\sqrt 2}\frac {I_p}{\sqrt 2}$$ That's how one arrives at the the RMS voltage and RMS current as the effective AC voltage and effective AC current.

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    $\begingroup$Of the 3 answers, only this one mentions the most important correct reason, which is that we are all lazy and want to reuse the DC power equation $P=VI$ for the AC case, and that is the sole motivation for defining things via RMS. So +1.$\endgroup$
    – naturallyInconsistent
    Commentedyesterday
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The generalized "power" of a signal is proportional to the square of the voltage. In the case of power loss in ohmic materials,

$$ P = \frac{V^2}{R} \propto V^2 $$

This means that to produce the same average power as a sinusoidal wave of peak voltage $V_{peak}$, a constant voltage $V_{RMS} = \sqrt{\frac{1}{T} \int_0^T V(t)^2 dt} = \frac{V_{peak}}{\sqrt{2}}$ is necessary. Since we are normally interested in average power, it is useful to use $V_{RMS}$.

If we were more interested in a quantity that involved averaging out $|V(t)|$, we might use it, but it seems natural to average out $V(t)^2$ since it's proportional to power.

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    $\begingroup$I see. So the case is that V_rms, since it denotes the power output equivalent, is derived from the power being proportional to the square of the voltage, rather than being the average magnitude of the v(t). Quite the thing to wrap my mind around. Thank you.$\endgroup$
    – Igggosha
    Commentedyesterday
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There is no fundamental reason that we could not report $\langle|V(t)|\rangle$, but it turns out that RMS voltage is more useful.

Concretely, we often care about the average power delivered to a circuit element. If this element has resistance $R$, then the power delivered at a given moment is $\frac{V^2}{R}$. Then, the average power becomes $\frac{\langle V^2 \rangle}{R} = \frac{V_{rms}^2}{R}$.

On the other hand, working with $\langle|V(t)|\rangle$ is clunky, since it is rare to want to know the average magnitude of the voltage. We could still report this value and, assuming the voltage is sinusoidal, infer $V_{peak}$ or $V_{rms}$, but at that point, there is little value in calculating $\langle|V(t)|\rangle$ at all.

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  • $\begingroup$Beat me to the same answer by a minute.$\endgroup$
    – ChemiCalChems
    Commentedyesterday
  • $\begingroup$Oops -- the more the merrier!$\endgroup$
    – GusterBuster
    Commentedyesterday

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