Keith Dysart wrote:
Actually, we have shown that that is not the case. If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power. And
we have shown that the instantaneous power is not dissipated
in the source resistor.

Actually, we have shown exactly the opposite. Maybe a
different example will help. We are going to replace
the 23.5+j44.1 ohm load with a *phase-locked* signal
generator equipped with a circulator and 50 ohm load.
The signal generator supplies 18 watts back to the
original source and dissipates all of the incident
power of 50 watts, i.e. all of the forward power from
the original source. The original source sees
*exactly the same 23.5+j44.1 ohms as a load*.

Rs Vg
+----/\/\/-----+----------------+--2---1-----+
| 50 ohm \ / |
| 3 18 watt
Vs 1 wavelength | Signal
100v RMS 50 ohm line 50 Generator
| ohms |
| | |
+--------------+----------------+----+-------+
gnd gnd


The conditions at the original source are identical.
The circulator load resistor is dissipating the 50 watts
of forward power supplied by the original source. The
signal generator is sourcing 18 watts.

Your instantaneous power equations have not changed.
Are you going to try to tell us that the 18 watts from
the signal generator are not dissipated in Rs???? If
not in Rs, where????
--
73, Cecil http://www.w5dxp.com