On Mar 12, 9:49 am, Cecil Moore wrote:
Keith Dysart wrote:
Actually, we have shown that that is not the case.

If the reflected energy is not being dissipated in
the source resistor, where does it go?

Good question.

It is not
dissipated in the load, by definition, and there is
no other source of dissipation in the network besides
the source resistor.

Not true. The voltage source also absorbs energy whenever
its voltage is positive but its current is negative.

There are no reflections and
no average interference to redistribute the reflected
energy back toward the load. The reflected wave
possesses energy and momentum which must be conserved.
Do your reflected waves obey your every whim and just
disappear and reappear as willed by you in violation
of the conservation of energy principle?

It would be nice, but, unfortunately, no.

You might like to visit
http://keith.dysart.googlepages.com/radio6
for a graph that shows the actual power dissipated in the
source resistor along with the power in the reflected
wave. It can be visually seen that the power dissipated in
the source resistor has no relationship to the sum of the
reflected power and the power that would be dissipated in
the source resistor if there was no reflection.

The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?

I have no trouble with the concept, but the exposition you
have offered is weak.

Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

Otherwise... Just handwaving.

The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

I see no reactance that performs this function.

But the actual answer is that it is the voltage source which is
absorbing the energy for part of the cycle and delivering the extra
energy to the source resistor for the other part of the cycle.

So again, it is not the energy in the reflected wave that accounts
for the change of the dissipation in the source resistor.

....Keith