W5DXP wrote:
wrote:
In reality there is not zero volts in the incident wave or in the
reflected wave. There`s full voltage coming and going. The volts just
happen to be out-of-phase at this point.
Yes, indeed. But there is no power.
Power is the same as irradiance in optics. When total V=0, it is simply
the result of destructive interference. Perhaps this quote from _Optics_,
by Hecht, will enlighten you. "The principle of conservation of energy
makes it clear that if there is constructive interference at one point,
the "extra" energy at that location must have come from elsewhere. There
must therefore be destructive interference somewhere else."
My knowledge of optics is insufficient to comment on any analogies you
choose to draw. Fortunately, a knowledge of optics is unnecessary to
understand circuits and transmission lines.
The voltage goes to zero because two voltage waves are engaged in destructive
interference. The current goes to maximum because two current waves are engaged
in constructive interference. The momentum in the voltage waves simply transfers
to the current waves and they just keep on rolling along. There is no mechanism
of physics existing at that point to change the momentum of the waves. Believing
that no energy crosses a superposed V=0 boundary is just a wet dream.
This puts you in
group a) P(t) is not always equal to V(t) * I(t); or
group c) "double think".
Care to think about which and comment?
The current
is at an absolute maximum point so plenty of charge carriers are crossing that
boundary.
Yes indeed, but current by itself is not energy. Remember
P(t) = V(t) * I(t) [unless you choose option a)]
Both volts and amps are simultaneously necessary for power.
....Keith