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the RMS (root mean square) value of $f(x)$ is defined as:

$$f(x)_{rms}=\sqrt{\frac{\int^b_a (f(x))^2dx}{b-a}}$$

Why do we do this very specific thing of taking the square, the mean, and then the square root of the function? For an AC circuit, why does this tell us the power consumption and not something like the expression below?

$$\frac{\int^b_a|f(x)|dx}{b-a}$$

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Consider the instantaneous current $i(t)$ through a resistance $R$. The instantaneous power dissipation is $$P(t) = Ri^2(t).$$ The average power dissipation during a sufficiently long time $T$ after $t=0$ is $$P_{avg}=\frac{1}{T}\int\limits_0^T{P(t)dt}=R\frac{1}{T}\int\limits_0^T{i^2(t)dt}=R\sqrt{\frac{1}{T}\int\limits_0^T{i^2(t)dt}}^2=Ri_{rms}^2.$$ This illustrates why RMS values are useful: you can use them to calculate the average power from e.g. the RMS voltage and current in the exact same way you would use the instantaneous values to calculate the instantaneous power (in AC you do need to take phase angles into account). Other ways of averaging quantities, like the mean absolute value as you propose, do not have this property.

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  • $\begingroup$ The important assumption is, that the restistance is Ohmic (so no depency of R on U or I) $\endgroup$ – lalala Jun 15 '20 at 6:33
  • $\begingroup$ RMS is an average that is useful when applied to the so-called field or root-power quantities such as voltage, current and electric/magnetic field which are proportional to the square root of power. Ignoring the absolute value for the moment, the integral you are talking about is nothing but the time average. You could apply it to instantaneous power to get the average power, but this is a trivial thing to do. The problem is that power is not a field quantity. Also the absolute value is no good: the average power of a capacitor is 0, but the absolute value would yield a positive power. $\endgroup$ – Puk Jun 15 '20 at 16:30
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By definition the rms, also called the effective or heating value of AC, is the equivalent of DC with respect to resistance heating. The reason for taking the square is because both positive and negative values of current equally produce resistance heating.

Hope this helps.

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  • $\begingroup$ good observation on the positive-and-negative bit! $\endgroup$ – niels nielsen Jun 14 '20 at 22:35
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    $\begingroup$ His question is why not take the absolute instead of the square root? I think he understands that both positive and negative produce resistance heating. Look at his proposed equation $\endgroup$ – slebetman Jun 15 '20 at 4:17
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it is the ac voltage source which is creating the trouble since the direction of voltage is reverse according to its time period so if we have a dc source then power dissipated through r will be $$P=i^2r$$, so here we have to replace this I with the average current because the net current is zero in the total time.

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I think your definition of the mean value of $f(x)$ over an interval $ab$, $\frac{\int^b_a|f(x)|dx}{b-a}$, is just as valid as $f(x)_{rms}=\sqrt{\frac{\int^b_a (f(x))^2dx}{b-a}}$. The values of both definitions though are different and have different physical units. For an AC-circuit only the second definition gives the right value and unit for power consumption.

For the current $i(t)$ we can write: $$i(t)_{rms}=\sqrt{\frac{\int_0^T (i(t))^2dt}{T}},$$ giving $${i(t)_{rms}}^2={\frac{\int_0^T (i(t))^2dt}{T}}$$

Now $$P(t)=R{i(t)}^2,$$ so $$P(t)_{rms}=R({i(t)_{rms}})^2,$$ which means The RMS-value of $P(t)={P(t)}_{rms}$ is equal to $R$ times the the squared value of The RMS-value of $i(t)$ squared, which is equal to

$$P(t)_{rms}=\sqrt{\frac{\int_0^T (P(t))^2dt}{T}}$$

The RMS-value is most convenient for periodic functions.

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  • $\begingroup$ Can you clarify your point about units? I think both $f_{avg}$ and $f_{RMS}$ have the same units as $f(x)$. $\endgroup$ – jacob1729 Jun 15 '20 at 16:50
  • $\begingroup$ @jacob1729 Do you mean with $f_{avg}$ the definition in the question (with the absolute value)? (sorry for reacting a bit late; I went for a walk with the dog and didn't notice yet your comment) :) $\endgroup$ – Deschele Schilder Jun 15 '20 at 19:47
  • $\begingroup$ Of course, they have the same units :) $\endgroup$ – Deschele Schilder Jun 17 '20 at 14:01

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