On Monday, May 30, 2016 at 6:01:38 PM UTC-4, FromTheRafters wrote:
wrote on 5/30/2016 :
On Mon, 30 May 2016 06:53:52 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

You are out of your area of expertise. With no current flow there is
NO voltage loss.

It almost looks like you are agreeing with me now, except you said
'voltage loss' instead of 'voltage drop' which are *not* the same
thing.


He's right, you're in way over your head. This silliness over Ohm's law started over the voltage drop over a 100 ft of #12 wire. It's also referred
to as voltage loss. Everyone here in the thread at least understands that.

And again, V= IR. With a current of 14 amps, a wire resistance of .16 ohms
you get a voltage loss of 2.2 Volts. Now put in a current of zero, and what
do you get? Voltage loss of Zero. And it has meaning, with no current
flowing the voltage loss is zero, we have the full supply voltage at the far
end of wire. Note: No division by zero was done here, no electrons were
harmed either.