Jim Kelley wrote:
So Gene, you should have asked the question: if one has a circuit with a
thousand resistors all connected in series and no current paths in
shunt, how does one arrive at a different current through R1000 than
through R1?

I've already explained that twice, Jim. Your circuit example above
bears no resemblance to the discussion of distributed networks
involving EM wave transmission lines with a fixed Z0. The Z0 of
a transmission line forces the ratio of voltage/current to be
equal to Z0. The voltage cannot change without a corresponding
change in the current. Even if there is no physical shunt path for
current, there is always a path for the H-field to supply energy
to the E-field and vice versa. If it will make you feel any better,
feel free to consider the Z0-fixing of the H-field and E-field ratio
to be a shunt path for the energy involved.

If your idealized circuit above is a transmission line with a constant
Z0, modeled as 1000 resistors in series, zero resistance in shunt,
and carrying EM waves, the source of the attenuation of the current
is clear. The voltage and current will be attenuated by exactly the
same percentage. That's why the attenuation factor is identical for
voltage and current in the exponential transmission line equations.

1. The series resistance causes a drop in voltage along with a
corresponding decrease in the amplitude of the E-field.

2. Since the ratio of the voltage to current is fixed by Z0, the
H-field decreases by exactly the same percentage as the E-field.

3. Since the H-field decreases by the same percentage as the E-field,
the current decreases by the same percentage as the voltage. That's
why the attenuation factor is identical for the voltage and current
for the exponential transmission line equations.

For the Nth time, the moral is: Don't be seduced by a circuit model
solution if it yields invalid results for a distributed network problem.

If your above idealized configuration is a circuit and not a distributed
network, then by all means, use circuit theory on it. But using a circuit
model solution for a distributed network problem is like using a DC ohm-meter
to measure the feedpoint impedance of an antenna. I trust you know enough
not to do that. :-)
--
73, Cecil http://www.qsl.net/w5dxp