Re: Simple electricity, and yet.....!


jimp@xxxxxxxxxxxxxxxxxxxx wrote:
xray4abc <lemhenyil@xxxxxxxx> wrote:

Randy Poe wrote:
xray4abc wrote:
Consider an ideal metallic conductor, connected,
in a closed electrical circuit, to a non-ideal DC
power source.
E=0 inside the conductor; E and B are perpendicular
to the conductor outside the conductor and at its surface too.

What is, then, causing the electrical current through
the conductor? What is pushing the electrons along
the conductor?

If E is zero everywhere inside the conductor, including where
it attaches to the source, then grad V is also zero. That is, there
is no voltage difference across the conductor and no current.

Sorry, you are wrong!
In a simple DC circuit where e= I(R+r), if you have R=0
you still get I= e/r , where r is the internal resistance/ impedance
of the DC source and "e" is the open circuit voltage of the source.
As such you can have current through a conductor even if there
is no voltage difference across it!


On the other hand, if you claim there is indeed a different
voltage at the two ends of the conductor, then obviously
E is not 0 everywhere.

Your initial conditions violate E = - grad V.

- Randy

I would reconsider the answer to the question in
your place now.

LL

If you want an answer in terms of voltage pushing current, it is the
voltage across the internal resistance of the source, just the same
as if the internal resistance were external and connected with ideal
wire.

--
Jim Pennino

Remove .spam.sux to reply.

The issue is a lot more subtle than it appears to be.
The standard answer I was expecting could be something
referring to Poynting's vector though it would still be questionable.
LL

.

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