On Thursday, October 24, 2002, at 09:00 PM, der Mouse wrote:
> When the load is not simply resistive, multiplying RMS amps times RMS
> volts will mis-estimate actual power consumed, sometimes drastically.
> (For example, if you have a purely capacitive or inductive load,
> current will be 90 degrees out of phase with voltage, and power
> consumed will be zero, even though current is flowing.  Power consumed
> is the integral of instantaneous current times instantaneous voltage;
> with an inductive or capacitive load, RMS volts times RMS amps will
> overestimate actual power drawn.
>
> Now, when the load is a switching power supply, current drawn consists
> of a spike at the peak of each voltage half-cycle, rather than anything
> even close to a sine wave.  In this case, actual power is something
> more like RMS amps times peak (not RMS) volts; measuring the current
> and multiplying by the nominal voltage will *under*estimate power
> consumed....
>
> Or, of course, I could be burbling nonsense.  Any power engineers here?

   Well I've spent a good portion of the last year designing switching 
power supplies...does that count? ;)  *poke*

   Of course it won't be purely resistive, and won't be terribly 
accurate...but it's a *lot* more accurate than comparing electric 
bills. :-)

         -Dave

--
Dave McGuire          "PC users only know two 'solutions'...
St. Petersburg, FL       reboot and upgrade."    -Jonathan Patschke