On Mar 11, 7:18*pm, Cecil Moore wrote:
Keith Dysart wrote:
So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

Of course there is, Keith. That's what reactances do.
A reactance stores energy during part of the cycle
and gives it back during a different part of the cycle.
The energy stored during part of the cycle is the
destructive interference energy that you are missing
from you equation. It is delivered back during the
next part of the cycle and dissipated 90 degrees
later.

When the signs of the two superposed voltages are opposite,
there is "excess" energy available which is stored in the
transmission line. 90 degrees later, when the voltages
have the same sign thus requiring constructive interference
energy, that "excess" energy is delivered back to the source
resistor to be dissipated.

Since we already know that the interference energy averages
out to zero, the energy imbalance that you discovered is
obviously energy being displaced in time by the reactance
whether it is from a coil or from the transmission line.

I can see that words describe a somewhat plausable conjecture,
but to be convincing, a mathematical exposition is needed.

When is the energy stored where? Energy stored in a capacitor,
for example, should correlate with the voltage on the
capacitor. Similarly for current in an inductor.

To be convincing, the various functions of time need to
align appropriately.

So far, the only convincing expressions I have seen
a
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)
and, of course, all the corresponding voltage and
current functions line up as expected.

To show that the reflected power is dissipated in
the resistor will require the derivation of some
X(t) such that
Prs(t) = 50cos(2wt) + Pr.g(t) + X(t)
and X(t) needs to be in terms of some known
physical quantities of the circuit.

...Keith