Re: Average, RMS and Power Factor made easy!

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "35045 by way of Terry Fritz <twftesla-at-uswest-dot-net>" <free0076-at-flinders.edu.au>



On Wed, 10 Jan 2001, Tesla list wrote:

> Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>"
<elgersmad-at-fnworld-dot-com>
> 
> Read the spec sheet on your Multimeter, and see if it doesn't just give
> you a readout in RMS.

Cheaper meters (including mine that cost about AU$180) usually read the
peak voltage and are calibrated to display the RMS value assuming that it
is a sinewave. That's the catch of course, since it will be correctly
calibrated for a pure sinewave but not for some other waveform (even just
the mains waveform plus harmonic distortion!).

> 
> Or, use an oscilliscope which reads peak to peak.  Then find the peak
> to peak voltage, divide by two, and multiply by seven tenths, and seven
> hundredths.

It's easier to say 1/sqrt(2) which is approx. 0.707 as you meant to say.
This _only_ works for a very select few waveforms, one of which is the
pure sinewave. It's safe to say that if you aren't looking at a sinewave
then this formula doesn't work.

> 10 volts peak to peak translates to
> 5 volts peak, and 5 volts peak translates to
> 3.535 Volts RMS.
... if you are looking at a pure sinewave.

> 
> The RMS voltage gives you the DC equivalent of AC voltage.  Since, a
> sine wave is not continually on, or off the length of time, and the
> amount of power over time is reduced to a value equal an amount of
> power from a continuous source.
> 


Have fun,
Darren Freeman