Why is RMS used for P=VI (where there are no squares)?

I understand why RMS values are used for

$$P = {\bar{v^2} \over R}$$

because $$\\bar{v^2}\$$ would be the "MS" so to speak, and then by square rooting we get the "RMS" ($$\\sqrt{\bar{v^2}}\$$), which can then be squared to give $$\\bar{v^2}\$$, meaning the above equation is equivalent to: $$P_{av} = {V^2_{rms} \over R}$$

However, why are RMS values used for calculating the average power using $$\P = VI\$$, i.e

$$P_{av} = \bar{V}\bar{I}$$

In a textbook I see this equation for power for AC written as:

$$P_{av} = V_{rms}I_{rms}$$

Why is the mean of the square values used here when there are no squares in the equation? Surely it should just be the (non squared) average of the absolute voltage and current?

• So, you do not know whay we introduce the RMS value for a voltage/current?
– G36
Oct 27 '19 at 20:05
• @G36 I understand why it is used for power formula where there is a square value, I don't understand why it is used for P=VI. Oct 27 '19 at 20:06
• Because only V_rms x I_rms give us average power equal to the DC power if a DC voltage is equal to V_rms and the DC current is equal to I_rms . allaboutcircuits.com/textbook/alternating-current/chpt-1/…
– G36
Oct 27 '19 at 20:08
• It's kind of the other way around. We use RMS because it makes the power work out nicely like that. Oct 27 '19 at 20:17
• @G36 But why? For a formula with squares it obviously makes sense to take the mean of the square values. But for P = VI there are no squares. So why use the mean of the squares? Surely we should use the average of V and I, not the average of the square values of V and I? Oct 27 '19 at 21:08

The instantaneous power is product of the instantaneous voltage $$\v(t)\$$ and the instantaneous $$\v(t)\$$ current:

$$p(t) = v(t)i(t)$$

Therefore, the average power is the area underneath the $$\v(t)i(t)\$$ curve divided by time. In other words, it's the average of the instantaneous power over time. For a periodic signal, this is:

$$P_{avg} =\frac{1}{T}\int{p(t)dt}= \frac{1}{T}\int_{-T/2}^{T/2} {v(t)i(t)dt}$$

These are, as far as I know, the fundamental definitions for electrical power from which all others arise. Everything else is simplifications and generalizations since our monkey brains aren't able to sit there and calculate power from a bunch of instantaneous values on the fly.

The definition of RMS for a periodic signal is:

$$RMS = \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {[f(t)]^2dt}}$$

Now this is going to be an empirical example since it's just faster but hopefully it will demonstrate things.

Let's assume the following voltage and current waveforms, which covers all resistors values with any voltage waveform being applied across them: $$v(t)=v(t)$$ $$i(t)=\frac{v(t)}{R}$$

Therefore, the average power is: $$P_{avg} = \frac{1}{T}\int_{-T/2}^{T/2} {\left( v(t) \times \frac{v(t)}{R}\right)dt} =\boxed{ \frac{1}{RT}\int_{-T/2}^{T/2} {[v(t)]^2dt}}$$

Now let's compare this result to what happens when you multiply the RMS of the voltage and current waveforms together.

First we take:

$$V_{RMS} = \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {[v(t)]^2dt}}$$

$$I_{RMS} = \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {[i(t)]^2dt}} = \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {\frac{[v(t)]^2}{R^2}dt}}$$

Then multiply them together:

$$V_{RMS} \times I_{RMS} = \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {[v(t)]^2dt}} \ \times \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {\frac{[v(t)]^2}{R^2}dt}}$$

$$= \sqrt{\frac{1}{T}\int_{-T/2}^{T/2} {[v(t)]^2dt}} \ \times \sqrt{\frac{1}{R^2T}\int_{-T/2}^{T/2} {[v(t)]^2dt}}$$

$$= \sqrt{\frac{1}{R^2T^2}\int_{-T/2}^{T/2} {[v(t)]^2dt} \ \times \int_{-T/2}^{T/2} {[v(t)]^2dt}}$$

$$= \sqrt{\frac{1}{R^2T^2} \left(\int_{-T/2}^{T/2} {[v(t)]^2dt} \right)^2}$$

$$\boxed{= \frac{1}{RT} \int_{-T/2}^{T/2} {[v(t)]^2dt}}$$

So you can see that the calculation of power from instantaneous values is equal to the result of multiplying the RMS values together.

This can be generalized to all functions but was a lot easier to just use example functions (much stronger calculus skills are required to solve for the multiplication of those two integrals when the function inside the integral, $$\v(t)\$$ and $$\i(t)\$$ are both arbitrary. I feel that it would involve using Taylor expansions to represent the arbitrary functions which would give you something that would still be arbitrary but have enough substance to be mathematically manipulated. Either that, or representing the functions as generic fourier series which would put into a form similar to what I have done here with sine (or cosine), but with an infinite number of terms.

So I guess the answer to your question is that the squaring is there because it arises from the fundamental definition of instantaneous power, and the average power being time average of the instantaneous power.

The definition for RMS builds in the squared values into the definition, which is then used to determine $$\V_{rms}\$$ and $$\I_{rms}\$$. That is why if you go $$\P_{avg}=V_{rms}I_{rms}\$$ you don't need any squares since it's already accounted for.

And when you use $$\P = I^2R = \frac{V^2}{R} \$$ you are only ever working with instantaneous power and instantaneous current/voltages, or average power and constant DC values where the squaring has not happened yet so you must handle the squaring manually for yourself.

It just happens to be a coincidence and special case that for constant DC signals, RMS and AVG are the same, but you don't ever actually care about the average amplitude of voltage or current as far as average power is concerned. I think @Hearth put it best when he said the average of the product is not the product of the averages. It might be unintuitive to see why the squaring needs to be there, but with that quote it should be intuitive to see why just multiplying the averages doesn't work. Gotta be careful when combining averages...the average of the a bunch of averages isn't the same as the average of everything either, especially weighted averages.

• Thanks for taking the time to write up this proof, I can see how both are equivalent now - although I haven't had the time to fully understand everything. Oct 27 '19 at 23:11
• @JShorthouse Fixed some errors in the last couple of lines. Oct 27 '19 at 23:47

In short: Using RMS values for AC power yields a simpler expression that is analogous to the expression for DC power of $$\P = IV\$$.

The expression for AC power, using peak values for current and voltage, is: $$P = \frac{1}{2} \cdot I_{pk} \cdot V_{pk}$$

RMS values simplify this by absorbing the $$\\frac{1}{2}\$$ factor into the current and voltage values:

\begin{align} P &= \frac{1}{2} \cdot I_{pk} \cdot V_{pk}\\ P &= \frac{1}{\sqrt{2}}I_{pk} \cdot \frac{1}{\sqrt{2}}V_{pk}\\ P &= (\frac{I_{pk}}{\sqrt{2}}) \cdot (\frac{I_{pk}}{\sqrt{2}})\\ P &= (I_{RMS}) \cdot (V_{RMS})\\ \end{align}

Note that the equations above are only representing real power across a purely resistive load. Of course, with a reactive load, you'll need to work with complex power, but from your question it seems you haven't started studying that just yet.

• This makes sense, but wouldn't your first formula using the peaks only work for a square wave? By my own calculations shouldn't the power of a sine be: $P_{av} = {4\over\pi}V_pI_p$ ? Oct 27 '19 at 20:27
• Based on the formula $V_{av} = {2\over\pi}\times V_p$ Oct 27 '19 at 20:34
• Power is defined as the product of current and voltage. Instantaneous power is always the product of current and voltage at any point in time. For a resistive load, the current and voltage waveforms are in-phase, so for a sinusoidal input, the average power turns out to be simply half the product of the peak sinusoidal values. Assume $V(t) = A \cdot sin(t)$ and $I(t) = B \cdot sin(t)$. Calculate the average value of the product and see what you get. Hint: you should arrive at a value of $\frac{AB}{2}$. Oct 27 '19 at 20:38
• Could you please explain where the half is coming from? The average of a half cycle sine wave is given by ${2\over\pi} \cdot Peak$, correct? So therefore $V_{av} = A \cdot {2\over\pi}$ and $I_{av} = {B}\cdot{2\over\pi}$ meaning $P_{av} = {4AB\over\pi}$? Oct 27 '19 at 20:52
• @JShorthouse The average of the product is not the same as the product of the averages. Oct 27 '19 at 23:44