W5DXP wrote:

Let's say we have two sources each equipped with a circulator and load
resistor. We'll call such a source an SGCR (Signal generator equipped
with a circulator and load). We set up the following experiment.

100W SGCL#1------------50 ohm lossless coax-----------50W SGCL#2

Seems to me that the 100W from SGCL#1 will flow unopposed and
dissipate in SGCL#2's circulator load resistor. Seems to me that
the 50W from SGCL#2 will flow unopposed and dissipate in SGCL#1's
circulator load resistor.

May I modify your experiment slightly? Change SGCL#2 to produce 100W.
Restating your explanation for the new experiment:
"Seems to me that the 100W from SGCL#1 will flow unopposed and
dissipate in SGCL#2's circulator load resistor. Seems to me that
the 100W from SGCL#2 will flow unopposed and dissipate in SGCL#1's
circulator load resistor."

Let's also make the 50 ohm lossless coax long enough that we can
find a voltage maximum. At this voltage maximum the current is 0,
always.
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

The obvious (and probably controversial) answer to this
inconsistency is that the power dissipated in CL#1 originates
in SG#1. Similarly for SGCL#2.

To further the experiment slightly, whenever a conductor in a circuit
has 0 current, the conductor can be cut without changing a thing.

Let's cut the coax at the voltage maximum (where the current is 0).
Nothing changes. The voltage and currents remain the same every
where. The powers remain the same everywhere, and yet there is no
longer a path from SG#1 to CL#2. The power being dissipated in CL#1
must be originating in SG#1.

Comments and corrections invited.

....Keith

wrote:
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

Nope, a directional coupler can still separate out the forward
and reflected waves even at voltage and current nulls. Remember,
those voltage and current nulls are only a 'dl' in length as 'l'
approaches zero. Superposition of two waves traveling in opposite
directions has no effect on the individual waves. There are no
reflections except at an impedance discontinuity point which
doesn't exist in the example.

Let's cut the coax at the voltage maximum (where the current is 0).
Nothing changes.

Of course something changes. You have introduced an impedance discontinuity
where none existed before. The forward wave is 100% reflected at
the cut point and of course, the reflected power is dissipated in
the source's circulator load resistor. If you make the two sources
one Hz different in frequency, you can easily observe what happens
when you cut the line. Otherwise, it's the old steady-state Catch-22
coverup in action. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:

wrote:
Assuming that P = V x I, the power is 0, always.
This seems to be at odds with explanation that SGCL#1's power is
dissipated in SGCL#2's circulator load resistor since there is no
energy flowing at the voltage maximum.

Nope, a directional coupler can still separate out the forward
and reflected waves even at voltage and current nulls.

It is a bit early to move to the complexity of directional couplers.
I am still stuck on how energy can flow through a point on the circuit
where the current or voltage is always 0.

Using instantaneous Power = Vinst x Iinst, the power at such a point
must always be 0, leading to the conclusion that no energy is flowing.

So how does energy flow through a point in the circuit where the voltage
or current is always 0.

Is it that Pinst != Vinst x Iinst?
Or is it that there is no point in the ideal experiment presented where
V or I is always 0?
Or have I missed something in the chain of reasoning above?

....Keith

wrote:
It is a bit early to move to the complexity of directional couplers.
I am still stuck on how energy can flow through a point on the circuit
where the current or voltage is always 0.

There are two voltages and two currents. Sometimes they are 180 degrees
out of phase and their sum is zero. But those waves don't know each
other exists. Someone forgot to tell them that they aren't supposed
to carry any energy so they just keep on carrying energy.

Using instantaneous Power = Vinst x Iinst, the power at such a point
must always be 0, leading to the conclusion that no energy is flowing.

No *NET* energy is flowing but the forward energy and reflected energy
just keep on flowing, carrying an equal amount of energy in each direction.
In a transmission line with reflections, there are always forward and reflected
transactions. Please don't be seduced by the steady-state religion.
--
73, Cecil http://www.qsl.net/w5dxp



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Keith wrote:
"So how does energy flow through a point in the circuit where the
voltage or current is always 0?"

The "0" is the sum of two voltages produced by adding together equal and
opposite voltages in two waves which are passing through each other with
no effect on each other.

Conductors producing the volts and amps from the waves they carry are
actually "bucket brigades". Distributed inductance and capacitance pass
along charges in a travel direction, or in both directions. There is no
problem as charges moving in opposite directions swell and shrink
occupancy at certain spots on the line. They don`t overflow, and charges
moving in each direction, like Ole Man River, they "jes keep movin`
along".

Best regards, Richard Harrison, KB5WZI