DIYbanter - View Single Post

FromTheRafters

After serious thinking wrote :
On Mon, 30 May 2016 20:13:21 -0400, FromTheRafters
wrote:

After serious thinking trader_4 wrote :
On Monday, May 30, 2016 at 2:25:40 PM UTC-4, FromTheRafters wrote:
It happens that formulated :
On Mon, 30 May 2016 10:37:01 -0400, 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.

Yeah that Ohm guy was a moron.

No, *he* knew quite well what he was talking about.

I suppose we should throw out the rest
of electrical engineering too as just a flawed theory.

Ah, engineering, no wonder the math skills are so weak. They use
calculus all day long but few of them know how or why it works.

Show me a definition of "voltage drop" that doesn't involve current
flowing in a circuit. I'll wait, but I haven't got forever so be quick
about it.

V = IR is the formula for voltage drop. Put in zero for I, any finite
resistance for R, you get V = 0. That tells you there is no voltage drop.

Of course there isn't, because there is no current. You can't have a
voltage drop when there is no current. Thanks for finally agreeing with
me.
No voltage drop across the wire does not mean no voltage drop across
the circuit thogh.

You've ( it appears) been arguing both sides of the equation.

To avoid confusion - WHOEVER said there would be a voltage loss in an
unloaded wire is WRONG.

I'm not talking about a voltage loss, but a 'voltage drop'.

Also whoever said there is no voltage drop in an open circuit is ALSO
WRONG.

Then you are calling yourself wrong. There is no current in an open
circuit, and you have agreed several times (correctly) that you need
current to have a 'voltage drop'.

And to top it all off, in order to measure the voltage across an open
circuit, you MUST close the circuit - meaning it is no longer an
"open" circuit, AND

I'm not talking about measurements at all.

In the real world there is no such thing as zero ohms. You can get
REAL close - but "in the wild" it does not exist. - so you are never
REALLY deviding by or multiplying by ZERO when solving ohm's law

You're wrong about that too.

https://en.wikipedia.org/wiki/Superconductivity