trader_4 presented the following explanation :
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.

They're the same? Tell these guys then, 'cause they have it all wrong.

http://ecmweb.com/electrical-testing...s-voltage-drop

https://appauto.wordpress.com/2008/0...-the-question/