It happens that formulated :
On Mon, 30 May 2016 17:28:23 -0400, FromTheRafters
wrote:

pretended :
On Mon, 30 May 2016 14:18:28 -0400, FromTheRafters
wrote:

After serious thinking trader_4 wrote :
On Monday, May 30, 2016 at 11:46:09 AM UTC-4, FromTheRafters wrote:
trader_4 brought next idea :
On Monday, May 30, 2016 at 10:37:10 AM UTC-4, FromTheRafters wrote:
on 5/30/2016, Al Gebra supposed :
On 05/30/2016 12:53 AM, 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.

Voltage drop is represented by the formula E=I*R
Seems to me if the current flow is zero, then the voltage would be
zero as well.

That's a good theory, but IMO it is wrong.

No theory, it's correct from electricity 101. With an open circuit, no
current flow, the voltage across the wires is the same at both ends.

That I can agree with. There is no "voltage drop" because there is no
circuit in which current can flow. Voltage drop is defined in terms of
a circuit with current flowing from the source through a load and back
to the source and voltage drop is across all of the resistances in the
circuit including the internal resistance of the source.

Think of it this way:

Distance equals rate times time (D=RT) and you have two trains on a
railroad track. Detroit to Chicago on one end and Chicago to Detroit on
the other. Neither train is moving. Does the distance between them drop
to zero?

No, it doesn't.

Voltage drop has no meaning in an 'open circuit', which isn't actually
a circuit at all, just as 'rate' has no meaning for objects which
aren't moving.


From Wikipedia:

"An electronic circuit is composed of individual electronic components,
such as resistors, transistors, capacitors, inductors and diodes,
connected by conductive wires or traces through which electric current
can flow."

If current can't flow, it stands to reason that it is not a circuit,
and 'voltage drop' has no meaning.

It does have meaning, it's ZERO. Zero in physics has meaning. It can
be measured. The voltage on the ends of that 100 ft of wire is exactly
the same.,

No argument on that point.

because the voltage drop across the wire with no current
flowing is zero.

There is no such thing as a 'voltage drop' when there is no current
flowing.

https://www.youtube.com/watch?v=ggKnH-95ty0

It's like saying a velocity of zero has no meaning.

D=RT, so when the rate (velocity) is zero then D=zero*T and since
zero*T equals zero then D equals zero, so you are saying that the
distance is zero. This is obviously not the case.

Consider this:

What is the amperage (A) when you put zero ohms (R) resistance across
an ideal 12 volt (V) source?

Answer:

Undefined! Because A=V/R and you can't divide by zero.

This is simple math, why are you struggling with it?
You are over your head -- WAY over.

Says you. Do you divide by zero for a living?
Deviding by zero is not involved because, as stated before, by
measuring the voltage you are introducing a non-zero value to both
resistance and current.

Who said anything about measuring? It is well known that the act of
detecting a thing affects the thing being detected. We were talking
about the existence of 'voltage drop' when there is zero (by
definition) current.

[snipped the what if almost zero scenarios]