Owen Duffy wrote in
:


You could devise an example with an AC source that had exactly the
same outcome, more complicated mathematics to solve it, but the
exactly the same outcome. But you don't need to do that, the example
is in the set of the general, and stands against a general statement
that power reflected from a source is necessarily dissipated in the
source.


This wasn't so well written, let me move a word around:

You could devise an example with an AC source that had exactly the
same outcome, more complicated mathematics to solve it, but the
exactly the same outcome. But you don't need to do that, the example
is in the set of the general, and stands against a general statement
that reflected power from a source is necessarily dissipated in the
source.

Owen

Owen Duffy wrote:
You could devise an example with an AC source that had exactly the same
outcome, more complicated mathematics to solve it, but the exactly the
same outcome.

But the point is that I can devise an example with an
AC source that has the opposite outcome simply by using
a Norton equivalent source. When the reflection arrives,
the dissipation doubles.
--
73, Cecil, w5dxp.com

Cecil Moore wrote in news:KGhKh.7646$yW.5893
@newssvr11.news.prodigy.net:

Owen Duffy wrote:
You could devise an example with an AC source that had exactly the same
outcome, more complicated mathematics to solve it, but the exactly the
same outcome.

But the point is that I can devise an example with an
AC source that has the opposite outcome simply by using
a Norton equivalent source. When the reflection arrives,
the dissipation doubles.

Your Norton equivalent doesn't pretend to replicate my (real) source in
every way, it only pretends to supply the same voltage and current to its
load... and it does that.

Owen

Owen Duffy wrote:
Your Norton equivalent doesn't pretend to replicate my (real) source in
every way, ...

I thought we were discussing general principles, not
replicating a special case example in every way.
--
73, Cecil, w5dxp.com

Owen Duffy wrote:

Cecil Moore wrote in news:96hKh.483$rj1.92
@newssvr23.news.prodigy.net:


Owen Duffy wrote:

Are the principles that apply to this example inconsistent with the
general case (eg dc, ac, transient, steady state etc)?

Of course, pure DC principles are not adequate to cover
general case distributed network problems. Else, there


The principles are those of the behaviour of transmission lines and the
theory applying to solving the source circuit. In the 'DC' case, the
analysis becomes trivial, but IMHO there were no underlying principles
that do not apply in the general case.

But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.

Best, Dan.

Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122
@news02.roc.ny:

But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.

It semantics Dan, the excitation is part of the scenario or example, it is
not a principle of circuits or transmisssion lines.

Owen

Owen Duffy wrote:

Dan Bloomquist wrote in news:xYhKh.5014$ya1.2122
@news02.roc.ny:


But your underlying principle is that the source steps and doesn't
change. So any general case where the source is at some different level
is now not covered.


It semantics Dan, the excitation is part of the scenario or example, it is
not a principle of circuits or transmisssion lines.

Like saying the impedance of a capacitor is the same no matter what the
frequency......

From here, I'll let you have the last word. I'm not into hand waving....

Owen

Best, Dan.

Owen Duffy wrote in
:

....
When the switch is closed, current flows into the line. Until t=2T
seconds, the current If that flows into the line equals 100V/(100+100
ohms) = 0.5A, during which time the resistor dissipates heat at the
rate of I^2*R = 100W. The voltage of the wave Vf travelling from the
source is I*Zo = 50V. This situation is constant until t=2T seconds.
....

David Ryeburn is awake, even if I wasn't at 5.38 when I wrote this.

should be ...during which time the resistor dissipates heat at the
rate of I^2*R = 25W...

It doesn't materially change anything, but I messed up!

Owen