On Apr 19, 11:57 pm, Cecil Moore wrote:
Keith Dysart wrote:
Minor changes to the generator for B can give you a very hot one
or one that is exactly the same temperature as the one with
the circulator.

You've been hammering people with the source of your choice
for days but choose to abandon it as soon as someone points out
a logical problem with it.

Yes indeed. But a key realization is that the behaviour on the line
can not depend on arcane details about the construction of the
generator. So when such statements are made, it is instructive
to explore the situation with a different generator. If the outcome
is different, then there is something wrong with the theory. Note
that the standard explanations only require that the impedance
of the source be known. It is not important how this impedance
is achieved. With the hypothesis you are proposing, the internal
details significantly change the behaviour on the line and require
that the line (or the wave) know the internals so that it can
decide to enter or reflect despite the impedance being the same.

The conditions on the lines are indistinguishable and yet you
claim one is reflecting and one is not. How did the line know
whether it should reflect or not? Or is the wave that knows?

One wave encounters a 50 ohm resistor and is dissipated.
The other wave encounters an infinite impedance and is
100% re-reflected - all in accordance with the wave
reflection model.

But how does it know, since the output (source) impedances are
the same for both cases? How does the 50 Ohm output
impedance of generator B become infinite while the 50 Ohm
output impedance of generator A does not? This is the key
issue with your explanation. In fact, there is no difference and
they behave the same. The reverse wave encounters 50 Ohms
and is not reflected in either case. It is the only sensible answer.

Actually Pnet is zero because of basic circuit theory and the
universally accepted understanding that P = VI.

Circuit theory completely falls apart with distributed
network problems. The currents at two points in the
closed loop are often flowing in opposite directions.

This is a misunderstanding. Look at the distributed inductors and
capacitors and make the measurements at a point that makes
sense. This is, of course, how a directional wattmeter works.
It measures the voltage and current at one place on the line.
And this is how a real wattmeter would work if it was placed
at the same point.

In circuits, we measure the power at a particular place.
I certainly want my power company to do this. If the voltage
or current is 0 at a particular place in the circuit then no
energy is flowing at that place in the circuit. If you are disputing
this, I contend that you do not accept that P = VI.

Plenty of energy is flowing - two equal magnitudes
in opposite directions equals zero net energy.

Correction to your P = VI based on DC, not AC/RF:

Net P = V*I*cos(theta) = Pfor - Pref

I think you have made this mistakte before. In my expression P=VI,
V and I are simultaneous instantaneous values and P is the
instantaneous power. If you want average you need to integrate
and divide.

Your expression applies only to sine waves and V and I are peak
voltages. Interestingly, the expression P=V*I*cos(theta) for sine
waves is always derived by starting with Pinst=Vinst*Iinst,
substituting the appropriate functions for Vinst and Inst, integrating
and dividing.

Forget V and I being 0. cos(theta) is always 0 for a
standing wave. There is ZERO net power anywhere in a
standing wave. (It's a lot like my bank account.)

True. But there IS energy flowing at every point that is not a voltage
minimum or maximum. At these points there is NEVER any energy
flowing.

Inventive. But it doesn't fly. P = VI or it doesn't.

Forward waves and reflected waves are completely independent
of each other and do NOT interact. Their powers do NOT
superpose. There is nothing but joules moving at the
speed of light in a transmission line. There's no net
energy but your zero energy assertions are just illusions.
EM waves are incompatible with zero energy.

Well that does leave you with a bit of a conundrum since on
an open ended line, zero energy is flowing at every quarter
wavelength point back from the load. Pinst=Vinst*Iinst.
Vinst or Iinst is for all times zero. Power must be zero for
all times.

At such a point a real instantaneous wattmeter would always
indicate zero. At other points a real wattmeter would show
energy flowing first one way then the other, though the average
would be zero. And the peak magnitude of the energy flow
depends on where you insert the real wattmeter on the line.
An instructive question: Where does the peak energy flow
occur? It is zero at every quarter wavelength point so it
must be highest somewhere else.

....Keith