# How to calculate the peak-to-peak voltage of a sine wave given RMS voltage

How do I calculate the peak-to-peak voltage of a sine wave given RMS voltage?

\begin{equation} \frac{V_{p-p}}{V_{rms}} = 2\sqrt{2} \end{equation}

### The long answer, or how to derive the above:

As noted on the Wikipedia page for root mean square, the RMS of a sine wave is equal to its amplitude divided by the square root of two. (You can also derive this by doing the integral over a sine wave yourself.)

\begin{equation} V_{rms} = \frac{a}{\sqrt{2}} \end{equation}

The peak-to-peak voltage is twice the amplitude of the wave, since it's measuring from the tip of a peak to the tip of a trough.

\begin{equation} V_{p-p} = 2a \end{equation}

We can rearrange these two equations:

\begin{equation} V_{rms} \cdot \sqrt{2} = a \end{equation} \begin{equation} V_{p-p} = 2 \cdot (V_{rms} \cdot \sqrt{2}) \end{equation} \begin{equation} V_{p-p} = V_{rms} \cdot 2\sqrt{2} \end{equation}

Thus, we multiply the RMS voltage by twice the square root of two - a factor of about 2.828 or so:

\begin{equation} 2\sqrt{2} = 2.8284271247.... \end{equation}

Of course, the process also works in reverse - if you can measure the peak-to-peak voltage, dividing that by the same factor will give you the RMS voltage assuming a perfect sine wave.

• I hope we get LaTeX equation support here - this would be a great use case for it! – Dan Oct 24 '13 at 0:11
• @Dan yep, hoping it gets turned on soon. – Amber Oct 24 '13 at 0:28
• Can you explain a little more? I'm still not entirely following. – Dan Oct 28 '13 at 20:17

In a pure sine wave $V_{RMS} \times 1.414$ will give you peak. $V_{Peak} \times .707$ will give you RMS.

• Note, however, that "peak" and "peak to peak" are not the same thing. – Amber Oct 29 '13 at 22:37
• That formula will work for "peak "or "peak to peak" depending on what you feed it. Of course peak × 2 also works for" p to p. – Paul Stiles Oct 29 '13 at 23:57