Re: Power Costs


"Peter van Merkerk" <merkerk@xxxxxxxxxxxx> wrote in message
news:42fb96bf$0$807$3a628fcd@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Clockmeister wrote:
> > "Rick Balkins" <nospam.rickbalkins@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote
in
> >>If you have a wattage counter, you can test the equipment.
> >
> > Or just use ohms law like normal people, and multiply the current draw
by
> > the voltage at the outlet.
>
> This is one of the rare cases that Rick's response is actually more
> correct than yours (assuming one uses a decent wattage counter).
>
> Your method gives yields the VA rating of the equipment, which is not
> necessarily the same as Watts. Only for resistive loads the VA rating
> equals Watts. With non-resistive loads (like for example transformers,
> motors, a typical power supply...etc) connected to an AC source the
> current is not in phase with the voltage, so simply multiplying voltage
> with current won't do.
>
> To be able to compute the amount of watts consumed by the equipment with
> the method you suggested, one would have to know the phase difference
> (phi) between the voltage and the current. After that the outcome of
> your formula should be multiplied by cos(phi):
>
> P = V x I x cos(phi)
>
> For resistive loads phi would be 0, so cos(phi) would be 1, so in that
> particular case the cos(phi) factor can be omitted.
>

Thanks for the clarification, it's been a long time since I've played with
A/C voltages so I should do a refresher I guess, it does ring a bell now
that I think about it ;-)

> (Rick is probably going to pretend he knew all of this...I'm still
> wondering who the hell he thinks he is kidding)
>
> > Try and get the basics right first Rick.
>
> Don't be cruel Clocky, we all know that even the basics are way beyond
> his mental capacity.

Maybe I should concentrate on getting the basics right first then :-)



.

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