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

On Mon, 22 Nov 2004 10:13:00 -0800, 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?

Hi Jim,

In the vein of shouldaaskedquestions: If there are no current paths in
shunt, how do you get current? (Which leads us back to Reg's
observation about R and G).

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Mon, 22 Nov 2004 10:13:00 -0800, 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?


Hi Jim,

In the vein of shouldaaskedquestions: If there are no current paths in
shunt, how do you get current? (Which leads us back to Reg's
observation about R and G).

73's
Richard Clark, KB7QHC

Roger that.

ac6xg

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

Jim,

I have conceded. I simply do not know how to deal with someone who uses
models to prove that models are misleading. Cecil's "reality", wherever
that comes from, trumps all.

I have not measured transmission lines in the vacuum of free space, so I
must rely on models. Since Cecil claims some sort of devine connection
to model-less reality he clearly must be correct. (Probably explains his
intimate knowledge of reality in the vacuum of free space.)

73,
Gene
W4SZ


Jim Kelley wrote:



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

Gene Fuller wrote:
Since Cecil claims some sort of devine connection
to model-less reality he clearly must be correct.

On the contrary - I don't believe in divinities at all.
That's exactly why I don't worship math models (like some
people I know) as if they possessed supernatural powers.

It is ridiculous to try to force a distributed network
problem to obey the rules of circuit analysis. The
distributed network analysis model was developed
specifically because distributed networks do NOT obey
the boundary conditions necessary for the circuit analysis
model to yield valid results.

When the only tool you have is a hammer, everything
looks like a nail. Math models are tools. Please
choose a valid one for whatever problem you are
trying to solve.
--
73, Cecil http://www.qsl.net/w5dxp