Gene Fuller wrote:

Cecil,

Nice try.
Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

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?

73, Jim AC6XG