# Homework Help: AC voltage measurement - Vrms

1. Apr 24, 2005

### StonieJ

I feel like I'm so close to the correct answer for this problem, but I just can't seem to bring it all together.

Code (Text):

To find the V[sub]RMS[/sub] of an AC sine wave, you use the following
forumla, where V[sub]max[/sub] is the maximum amplitude:

V[sub]RMS[/sub] = V[sub]max[/sub] / sqrt(2)

To find the V[sub]RMS[/sub] of an AC triangle wave, you use:

V[sub]RMS[/sub] = V[sub]max[/sub] / sqrt(3)

In one sentence, give a good qualitative reason why sqrt(3) is appropriate
for the triangle wave.

I have a bunch of vague and un-elegant ideas, but not really one good sentence. I've been Googling on the subject and coming up with crest factor (peak / RMS), which is sqrt(3) for triangle waves and sqrt(2) for sine waves. But I have yet to bring it all together. Any help is appreciated.

2. Apr 24, 2005

### OlderDan

You need to integrate the square of the wave form over one period and divide by the integration interval to find the mean square. Then you take the square root.

3. Apr 24, 2005

### vincentchan

because in the triangle wave, V is proportional to t... and V^2 is proportional to t^2, when you do the RMS, you need to take the squreroot of $$\int V^2 dt$$, this is where the root 3 come from

4. Apr 24, 2005

### chroot

Staff Emeritus
Hint:

The method for calculating the "crest factor" is simple:

1) Square the function under consideration. (That's the "square" part of the term "root mean squared.")

2) Average that squared function over time; one period is enough. (That's the "mean" part.)

3) Take the square root of the result. (That's obviously the "root" part.)

For the sine wave, the RMS multiplier is thus:

$$\sqrt{\frac{\int_0^{2 \pi} \sin^2 x dx}{2 \pi}} = \frac{1}{\sqrt{2}}$$

To come up with an elegant sentence, first, recognize that $1/\sqrt{3}$ is smaller than $1/\sqrt{2}$. Next, graph the squared triangle wave superimposed over the squared sine wave. Notice that the squared sine wave is always equal to or greater than the squared triangle wave. Thus, its time average is correspondingly larger.

- Warren
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