# Re: Amps drawn

```> >From P=VI, that means that power is proportional to voltage and also
> proportional to current.  Keeping power constant, if you increase
> voltage, then current has to go down.  I suppose it is possible that with
> higher voltage at the bulbs, that the power rating of the bulb goes up a
> little, which would allow a little more current.

Whoa... power isn't constant, it's going to depend on the voltage across the
bulb (V*V/R where R is the resistance of the bulb, which I'm assuming to
be fairly constant over the couple of volts range we are talking about).
So, going from 10V at the bulb to 12V at the bulb is going to increase
the power dissipation by a factor of 12*12/10*10, 1.44, a 44% increase!

OK, so the resistance of the bulb probably increases a little with being
hotter (hotter = brighter), so you probably won't get all that 44%.

Hmm, maybe I could connect up a bulb to my 30A 0-15V power supply and
plot power against voltage.  Then we would get a good idea of what really
happens as the voltage is raised.

> > Relays work so well because you're bypassing all the shitty (sorry, but
> > it's the truth) Audi Factory wiring with [presumably] high-guage direct-
> > wiring with only the relay between the load (lights) and the generator
> > (alternator, battery), and using a good [presumably] high-current relay,
> > so that overall the resistance of the wiring is much less, therefore
>
> That's what I was thinkin' too.

Very true.

Orin.

```

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