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Why do we use the power of $2$? What is the relation between this and having the same heat energy in both AC and DC?

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  • $\begingroup$ Why would we use root mean cube? it gives you a bunch of annoying (-) signs everywhere. But I don't fully understand what you're asking about heat energy. $\endgroup$ – user191954 Oct 10 '18 at 18:27
  • $\begingroup$ @Chair RMS of AC generates the same heat energy as DC whey they are equal. $\endgroup$ – Alsaraha Oct 10 '18 at 18:29
  • $\begingroup$ @Chair So why not use fourth root mean quad? $\endgroup$ – Alsaraha Oct 10 '18 at 18:30
  • $\begingroup$ @Alsaraha It is not a question of " why not use fourth root mean quad?", but a question of "why would you?".Sure you could use it, but RMS already solves the problem of averaging positive and negative numbers, why make it more complicated? $\endgroup$ – Hugo V Oct 10 '18 at 18:39
  • $\begingroup$ The cube root mean would be 0. $\endgroup$ – Michael Oct 10 '18 at 19:01
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what is the relation between this and having the same heat energy in both AC and DC?

First, the context here is the voltage across and current through a resistor. The power delivered to a circuit element is the product of the voltage across and current through. So, for a resistor (only), the instantaneous power is

$$p_R(t) = v_R(t) \cdot I_R(t) = \frac{V^2_R(t)}{R} = I^2_R(t)\cdot R$$

So you can already see why we're interested in the square of a voltage across (or current through) and not the cube or higher powers.

For a periodic voltage across with period $T$, we can ask what the average (mean) power is over the period and find

$$\langle p_R(t)\rangle = \frac{1}{T}\int_0^T\mathrm{d}t\,\frac{v^2_R(t)}{R}$$

Note that the average power (over a period) is proportional to the mean of the square of the voltage across.

Now we can ask what constant voltage across $V_R$ would give the same power as the average power of the periodic, time varying voltage.

The answer is, by inspection, the root of the mean of the square of the periodic, time varying voltage:

$$V_R = \sqrt{\frac{1}{T}\int_0^T\mathrm{d}t\,v^2_R(t)} = V_{rms}$$

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This particular example happens to have a main reason. The power and hence energy of a signal is proportional to the square of the amplitude (provided other variables like load resistance is constant). So if you take the RMS, you have just summed up the total energy, average it over time and you get average power

RMS also removes the negative sign, which is a nice bonus. But you can also use absolute value functions for this. It is also easy to integrate, differentiate, and a number of other things (compared to absolute values).

Unrelated to the current example, the exponential functions have a property of exaggerating differences between the 'highs' and 'lows'. This is one of the factors affecting your decision when picking one (such as with an error function like MSE), how much do you want to emphasize bigger values?

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The root mean square can be derived from something more general.

First lets look at the space of T-periodic, real functions. Its inner product is

$$\langle x,y\rangle = \int_0^T x(t)y(t) \mathrm{d}t.$$

Induced from this inner product, we can define a norm on this space:

$$\lvert\lvert x \rvert\rvert_2 = \sqrt{\langle x, x \rangle} = \sqrt{\int_0^T x(t)^2 \mathrm{d}t}.$$

We already see some similarities between the definition of the norm and the definition of the RMS.

Now if our function $x$ has the unit $\lbrack V \rbrack$, the inner product will have the unit $ \lbrack V^2s \rbrack$ and the norm $\lbrack Vs^\frac12 \rbrack$. As we want the RMS to express a quantity per time division, we have to divide the inner product by the period $T$ to not only match the desired unit $\lbrack V \rbrack$ again but also to relate it to a moment in time and not a timespan (the period e.g.).

So our RMS is really just $\frac{\lvert\lvert x \rvert\rvert_2}{\sqrt{T}}$.

To read more about why the 2-norm is used in a lot of places, I invite you to have a look into 1 and 2.

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Average power $\left <\rm power \right > = \dfrac{\left <\rm \rm voltage^2 \right >}{R} = \left <\rm \rm current^2 \right > R$ so it is the mean of the values squared which you need to use.

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In the example above you will see that the areas $A$ and $B$ above and below the $\left <\rm \rm i^2 \right > = \left <\rm \rm current^2 \right >$ line are equal so the extra energy delivered during the time period of $A$ is exactly compensated for by the reduced amount of energy delivered during the time period of $B$. (Think of it as levelling some ground which originally had a lot of bumps on it.)
Thus a steady current of magnitude $\rm i_{dc} = i_{\rm rms} = \sqrt{\left <\rm \rm current^2 \right >} $ will, over a cycle, dissipate the same amount of electrical energy as the alternating current does over a cycle.

If you wanted the mean of the power squared then you would raise the quantities to the power of four but note that in general $\left <\rm power^2 \right >\ne \left <\rm power \right >^2$.

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