[Test-Equipment] rms voltmeter?

[brianclarke01 at optusnet.com.au]

Forster's information is misleading to say the least.

If you take a sinewave, divide each loop, positive and negative, into two halves and divide one of each of the two halves by ten, square and integrate, the difference from sqaring and integrating the original sinewave is immitesticle.

In other words, unless you need to know to the 5th electronics degree, the waveform matters little. 

What is far more important, and is completely missing from the original proponent's and critic's messages, is the phase between the voltage and the current.

73 de Brian, VK2GCE.



> J Forster <jfor at quik.com> wrote:
> 
> Only, if you KNOW the waveform is a pure sine wave or square wave, with 
> an
> exception if you are using a digital scope with sampling and computation
> capabilities for doing a point by point power measurement and 
> integrating it
> over a cycle.
> 
> Best,
> -John
> 
> 
> 
> Ron Youvan wrote:
> 
> > [snip]
> >    Accurate power measurements can be made by putting a "T" at the
> > input of the load and viewing * the peak to peak (sine wave) voltage
> > across the load and mathematically determine the power level.
> >
> > * Using a 'scope that is flat beyond the observed frequency if
> > you have or can borrow one
> 
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