wrote:

Richard Clark wrote:

I stumbled upon the context in this snipe hunt:

Cecil Moore wrote:
What would
Kirchhoff have thought about a coil with 0.1 amp
at the bottom and 0.7 amps at the top? It certainly
doesn't mean that 0.6 amps is flowing sideways.

I think everyone here except Cecil knows where the current goes.

Those are standing wave currents, Tom. What is it about
func(kx)*func(wt) that you don't understand?

Take a look at the standing wave current distribution on a one
wavelength dipole at:

http://www.qsl.net/w5dxp/1WLDIP.GIF

The position in which a coil is installed in the standing wave
environment determines the magnitudes and phases of the currents
at the top and bottom of the coil. No magic displacement current
is required. If magic displacement current is not required in
a transmission line, why is it required in a coil? Hint: because
the lumped-circuit model is flawed.

If Cecil admits to displacement currents, he has to also admit his
argument about reflected waves is incomplete.

We can assume zero displacement current without much changing anything.
In the example at the top of this posting, there sure isn't 0.6 amps
of displacement current. I'm beginning to believe that you don't
understand superposition of forward and reflected waves. That would
explain a lot.

In the above example, the forward and reflected currents superpose to
0.7 amps at the top of the coil. That is simply closer to the standing
wave current maximum point. No displacement current required.

The forward and reflected currents superpose to 0.1 amp at the
bottom of the coil. That is simply closer to the standing wave
current minimum point. No displacement current required.

Exactly the same thing happens along a transmission line with
reflections. There's negligible displacement current between the
0.1 amp point and the 0.7 amp point on a transmission line. For
exactly the same reason, there can be negligible displacement
current in the coil. The forward current and reflected current
superpose in a coil just as they do in a transmission line.

If you would use the proper model and you will not need to resort
to any magic displacement current which is just a patch on a gaping
hole in the flawed lumped-circuit model.
--
73, Cecil http://www.qsl.net/w5dxp