Can I take the RMS of DC and AC, or only AC? And why isn't it exactly equal to Peak/sqrt(2)? My teacher said there is a pretty big margin for error when calculating RMS, but he didn't state why.

Printable View

- April 30th 2007, 03:16 AMDivideBy0RMS value
Can I take the RMS of DC and AC, or only AC? And why isn't it exactly equal to Peak/sqrt(2)? My teacher said there is a pretty big margin for error when calculating RMS, but he didn't state why.

- April 30th 2007, 03:30 AMCaptainBlack
- April 30th 2007, 03:31 AMtopsquark
You CAN take an RMS value for a DC source. But since the current (potential, etc.) is constant for a DC source it makes no sense to talk about an RMS value for it. RMS stands for "Root-Mean-Square" which takes the average value of the square of the signal (mean), then takes the square root of that. If the signal is constant this gives you nothing new to work with.

-Dan - April 30th 2007, 03:34 AMDivideBy0
If you had a fully rectified AC wave, then you could take an RMS of the DC, couldn't you? Because it's still fluctuating up and down? And if you have a half-rectification, would the RMS be half that of fully rectified AC?

- April 30th 2007, 05:28 AMtopsquark
Of course you can. But as CaptainBlack stated, the RMS value is simply equal to the magnitude of the signal. ie if you are looking for the RMS voltage, it would simply be |V|, the absolute value of the voltage.

If the signal were half-rectified, then the square of the signal appears as a constant value: V^2. So the average of this would still be V^2, and upon taking the square root we would get |V| again.

-Dan - April 30th 2007, 06:07 AMDivideBy0
Alrighty, thanks guys. I have the physics final exam tomorrow! XD

- April 30th 2007, 07:33 AMCaptainBlack