>
> > > In any event, when you run any current through a
> > > wire, it will heat up. How much it heats up is a factor of how much
>current
> > > you put through it. Your lighting switch carries all the current for the
> >
> > Actually, heat is the result of the voltage drop due to the resistance
> > of the wire. There shouldn't be a voltage drop across the switch.
> >
> Really actually, the voltage drop is due to the resistance _times_
> the current; the heat is due to the current squared times the resistance.
> There should be no voltage drop across an ideal switch or ideal wire; in
> reality both have some resistance, and therefore heat up when current
> flows
>
> Len Bugel
> '51 TD
> '57 MGA
I was trying not to confuse anyone. Evidently, it didn't work. If you
substitute "equal to" for "due to", then your statements would be
correct.
However, the voltage drop and heat are still "due to" to resistance in
the wire. While it is reasonable to expect a voltage drop across ten or
twelve feet of non-ideal wire, there should not be enough resistance
between switch contacts to cause even a non-ideal switch to get hot.
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