[R-390] 6080 in place of 6802 - RMS ???

[twleiper@juno.com]

On Wed, 9 Jan 2002 17:41:15 -0500 "Barry L. Ornitz" <ornitz@tricon.net>
writes:
> But since the RMS seems to be confusing people, let us say
> we have 25.2 volts DC and a switch that opens and closes
> very rapidly with a 50 percent duty cycle.  The result is the same.

Really? I beg to differ. The area under the half square wave is
equal to the area under the FULL sine wave, so you actually
have to switch your DC on at a 25% duty cycle (relative to the
full AC cycle) if you want to achieve the same effect as the diode.
Remember, a square wave has twice the power of a sine wave
of equivalent peak voltage. On the other hand, the peak AC
voltage is much greater than the RMS value, so why doesn't the
filament nuke out on the AC peaks? Scrap all this crap.

Let's try a simpler explanation than even the switch example:

However we want to calculate it, RMS, ohm's law and P=EI aside,
we should be able to agree that the power consumed is directly
proportional to the quantity of electrons flowing...regardless
of direction and potential. Furthermore, in the AC circuit we are
talking about, we know that half the electrons flow one direction
and half the other, and at equivalent and opposite rates. If we put in
a diode, we eliminate half the total flow of electrons at various
potentials, thus half the power delivered. If one agrees that the circuit
WITHOUT the diode consumes four times the normal power (120
watts instead of 30), then WITH the diode it will be half of the 120
watts
or 60 watts...twice the original.

If diodes could do the job, why do the utilities use transformers?
Why not just put a giant triac dimmer switch on the pole?

More grist for our ever grinding mill...

Tom