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FromTheRafters

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.

Also, in another post, you started writing about semiconductors and I
am familiar with forward voltage drop, and it requires a current.

Excerpted from Wikipedia:

"In a small silicon diode operating at its rated currents, the voltage
drop is about 0.6 to 0.7 volts."

Notice the word "currents" in there? There is no 'voltage drop' when no
current is present.

I'm still waiting for "voltage drop" with no reference to current. You
see, devices don't dissipate power when there is no current through
them, so how can there be any 'voltage drop' with no current?