On Jun 22, 5:49*am, Keith Dysart wrote:
What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

Reflected "power" doesn't know anything. Reflected voltages and
currents simply respond to the known laws of physics involving wave
reflection and wave superposition, and obey the conservation of energy
principle.

In 1, it doesn't stay out of the generator. It is redistributed from
the generator back toward the load by superposition associated with
*destructive interference) which happens when two superposed coherent
waves are between 90 deg and 180 deg out of phase. The entire
experiment is set up in a perfect 50 ohm environment so power =
V^2/50. What you guys are missing is that step which carries the
energy along with the voltage, i.e. the voltage (and current) require
a certain energy level which is being ignored.

In 2, the source wave never encounters the reflected wave so there is
no superposition or interference at the source resistor.

You can add a couple of more configurations. Number 3 could be when
the interference between the source wave and the reflected wave occurs
and they are less than 90 deg out of phase. More power than the
average reflected power plus average load power will be dissipated in
the source resistor because of superposition associated with
*constructive interference*.

Number 4 could be the special case when the source wave and the
reflected wave are 90 degrees out of phase. This is the condition of
zero interference (neither destructive nor constructive) and 100% of
the average reflected power is dissipated in the source resistor. This
is the easiest case to understand because there are no reflections and
no interference.
--
73, Cecil, w5dxp.com