What you say, Dave, is entirely true and always pragmatically wrong. Most DAQ systems do not sample simultaneously and have an input capacitance that provides averaging. Thus, you will always be reading average current between samples and average voltage. Computing power from average current and average voltage will always be in error if there is any variation. As the sampling period approaches zero, the sampled average current times the sampled average voltage will approach the sampled average power. Unfortunately many DAQ systems have sampling periods of 1-3 seconds to sample many channels and provide best an most accurate readings (of the average) and to filter out 60 Hz and its harmonics. This means that short scale current variation will be averaged (not RMS'ed).
It may be better to create a calibrated analog multiplier and feed that into a DAQ channel. Then you would be reading average power which would be OK. If you had an analog RMS function, that would also be OK to be averaged. Still, for bubble current/voltage noise, this represents only a small error bar in his experiment and does not invalidate the results. Bob On Sun, Oct 26, 2014 at 9:50 PM, David Roberson <[email protected]> wrote: > The total instantaneous power into the system can be calculated by taking > the instantaneous source voltage and multiplying it by the instantaneous > source current. It does not matter whether you want to call it AC or DC > since this is the total that is being delivered. There is no more, > regardless of how the load changes resistance. > > If you then integrate the instantaneous power over the time period of > interest, you get the total energy delivered by that source. The > requirement is that you must accurately measure the voltage and current > waveforms during the period of interest. > > If someone can show that the measuring system used by McKubre was not > capable of following the waveforms then they might have a valid point. I > suspect the Mike knew how to make these measurements in an accurate > manner. The skeptics need to demonstrate otherwise. > > Dave
