# Why we take mean of squares in RMS Current instead of taking mean of absolute value of current? [duplicate]

I know $I_{\text{rms}}$ produces the same energy over time as if the current were DC. I am asking the reason why it works.
The main focus of this question is highlighted below which makes it different from similar questions.(why we take mean of square then take root instead of taking mean of absolute value of current.

Let's say the AC current is $I=\sin(t)$ for this case where $t=\omega \tau$ where $\tau$ is time in seconds. This is done to make the question simpler.

Let's say we have graph of $\sin(t)$. It's average is zero so we square: $\sin^2(t)$. Now be careful!
we take average of $\sin^{2}(t)$ and then we take root. Why don't we take the absolute value $|\sin(t)|$ first and then take the mean, which will give us the answer 0.637I not 0.707I.

That will actually mean that in the formula $P=IV$ we will use $I=I_{avr}$ not $I=I_{rms}$ where $I_{avr}=\langle|\sin(t)|\rangle$.

• I think you've answered this yourself already: because $P \neq I_{avr}V$ and $P = I_{rms}V$. – Nick2253 Nov 11 '14 at 15:17
• – Kyle Kanos Nov 11 '14 at 16:38

The formula for power is $$P=IV$$ So one might be tempted to say that the average value of an AC current should produce the same power as a DC current. However, the formula for voltage across a load is $$V=IR$$
Therefore, the power across a load is $$P=I^2R$$
This means that if you want to find the equivalent DC power produced by an AC current, you find the average of the produced AC power, which (across the same load) ends up finding the average of the square of the AC current, $I^2$.
$$I_{DC}=I_{AC~RMS}$$