Definition of rms voltage

Question

I know that $V_{rms}$ is the equivalent DC voltage that causes the same heating affect in a wire as an AC voltage supply does. But another definition I've read and also how the eqn of $V_{rms}=\frac{V_{peak}}{√2}$ is derived is that it is the square root of the average of the squares, but you can't simply divide by 2 to take an average of a non-linaer graph, so I am confused how the equation has come about. But I understand the need for the equation.

Answer 1

First, rms stands for root mean square. To find the rms value of a periodic function:

(1) square the function

(2) calculate the mean over a period

(2) take the square root

---

So, for example, let $v(t) = V_{peak}\cos(\omega t)$ and follow the steps above:

(1) $v^2(t) = V^2_{peak}\cos^2(\omega)t = \frac{V^2_{peak}}{2}\left[1 + \cos(2\omega t)\right]$

(2) the mean of $\cos(2\omega t)$ over a period is zero so the mean of $v^2(t)$ over a period is just $\frac{V^2_{peak}}{2}$

(3) the square root is then $\frac{V_{peak}}{\sqrt{2}}$

Thus, for a sinusoidal voltage, $v_{rms} = \frac{V_{peak}}{\sqrt{2}}$

---

Why is this the equivalent DC voltage required to heat a resistor?

The instantaneous power delivered to a resistor is $P = \frac{v^2(t)}{R}$.

The average power delivered over a period is just the mean of $P$ over one period, i.e., the mean of the square of the voltage divided by $R$.

Can you take it the rest of the way?

Answer 2

$$\langle V_\text{peak}^2\cos^2\omega t\rangle=\frac{1}{2}V_\text{peak}^2$$ because it is well know that the average of $\cos^2$ is $1/2$.

$$\frac{1}{T}\int_0^T\cos^2\omega tdt=\frac{1}{\omega T}\int_0^{2\pi}\cos^2\omega td(\omega t)$$ $$=\frac{1}{2\pi}\int_0^{2\pi}\frac{1+\cos 2\omega t}{2}d(\omega t)$$ $$=\frac{1}{2\pi}\left[\frac{\omega t}{2}+\frac{\sin 2\omega t}{4}\right]_0^{2\pi}$$ $$=\frac{1}{2\pi}\left[\frac{2\pi - 0}{2}+\frac{\sin4\pi-\sin 0}{4}\right]$$ $$=\frac{1}{2}$$