On Mon, 30 May 2016 17:26:52 -0400, FromTheRafters
wrote:

brought next idea :
On Mon, 30 May 2016 11:45:57 -0400, 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.
You are burying yourself deeper.
Quit while you are ahead. If there is no semiconductor in the "open
circuit" there will be no voltage drop. As soon as you put a meter on
to check the voltage it IS a circuit.. A smiconductor has a "forward
voltage drop" that behaves differently than a resistance - but we are
not talking about semiconductor physics here.

We're not talking about completing a circuit with a meter either are
we? Show me how a semiconductor has a voltage drop without any current
flowing and maybe I'll take your comments seriously.
The voltage drop across a perfect semiconductor is not affected by
current - other than requiring the current to be NON zero. - and it is
vityually impossible to read voltage anywhere in a circuit without
completing the circuit - in a DC circuit requiring a resistive load -
an an AC circuit either a resistive, inductive or capacitive load. -
which means you can NOT measure voltage in a totally unloaded or
"open" circuit.