Keith Dysart wrote:
It is not clear what you mean by "through the source".
The source provides or absorbs energy. It does not have
a "through", since it only has a single port.

Good grief! The source is a two terminal device
with one terminal tied to ground. Since it has
a zero resistance, an EM reverse wave can flow
right through it, encounter the ground, and be
100% re-reflected by that ground. I have already
explained that. Did you bother to read it?

If you put an ideal 50 ohm directional wattmeter
between the source and its ground, what will it read?

dir
GND---watt----Vs--Rs-----------------------RL
meter

Did you bother to analyze this configuration?

Rs=50
----50-ohm----/\/\/\/\----50-ohm----
125w-- 100w 50w--
--25w --50w

This explains everything at the average power level.
I suspect it also works at the instantaneous level.

It is not obvious why you want to draw a distinction
between elements that remove energy from a circuit by
heating and those that do so in some other manner.

Could you expand?

We are dealing with an ideal closed system. The
ultimate destination for 100% of the ideal source
power is heat dissipation in the two ideal resistors,
Rs and RL. What happens between the power being sourced
and the power being dissipated as heat is "of limited
utility".

Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?

Heating is the only way that the energy can be removed
from the ideal closed system.

And that you can ignore energy being removed in other ways?

No energy is removed in other ways. There is no other
way for energy to leave the ideal closed system.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

An ideal source has zero source impedance, by definition.
All of the source impedance is contained in Rs, the
ideal source resistor.

And the resistor, could it not also radiate some of
the energy it receives? Perhaps even as visible light?
Would that make it less of a resistor because it was
not 'dissipating' the energy?

God could also suck up the energy or the universe
could end. Why muddy the waters with irrelevant
obfuscations? Please deal with the ideal boundary
conditions as presented.

This is quite a leap.

Nope, it is simple physics through which you must
have been asleep.

How would it violate conservation of energy if it was
dissipated rather than going somewhere else?

It can only be dissipated in the resistances, by definition.
Nothing other than the resistances in an ideal closed system
dissipates power. EE102.

In your model, what things could be done with the energy
that would not violate conservation of energy?

FIVE TMES, I listed three things in previous postings. Since
you avoided reading it FIVE TIMES already, I'm just going to
point you to the "Optics" chapters on "Interference" and
"Superposition".
--
73, Cecil http://www.w5dxp.com