On Apr 18, 10:48*am, Cecil Moore wrote:
Keith Dysart wrote:
So you are saying that ideal voltage sources work differently
in distributed networks than they do in lumped circuits.

"Work differently" is a loaded expression. Since both
lumped circuit and distributed network models exist,
it is safe to say that the lumped circuit model and
the distributed network model indeed "work differently".
If they didn't "work differently", there would be no
need for both of them to exist.

Your previous error is obvious. You were using the
lumped circuit model on the left side of Rs and
using the distributed network model on the right
side of Rs. If it is necessary to use the distributed
network model for part of the network, then it is
absolutely necessary to be consistent for all of
the network.

When you switched to the lumped circuit model on
the right side of Rs, the energies balanced. When
you switch to the distributed network model on
the left of Rs, the energies will also balance.

The lumped circuit model is a subset of the distributed
network model. See:http://www.ttr.com/corum/andhttp://w...1-MASTER-1.pdf

Here's some quotes: "Lumped circuit theory fails because
it's a *theory* whose presuppositions are inadequate.
Every EE in the world was warned of this in their first
sophomore circuits course. ... Lumped circuit theory
isn't absolute truth, it's only an analytical *theory*.
... Distributed theory encompasses lumped circuits and
always applies."

It is a good thing I checked the original references, otherwise
I would have had to assign Corum and Corum immediately to the
flake bucket where they could join some of the other contendors
on this group. But no, it turns out they have the appropriate
qualifications on all their statements about when it is
appropriate to use a lumped analysis and when it is not. And
when is lumped okay, when the physical dimensions of the
elements can be measured in small fractions of a wavelength.
Nothing new there, most of us know that.

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

Firstly, there are no inductors to cause any sort of difficulty.
Secondly, it is constructed of ideal components, which have the
luxury of being infinitely small.
And thirdly, it is a DC circuit so the two points above do not
matter any way.

So we are back to the question you keep dodging...

Where does the energy being absorbed by these ideal voltage
sources go?

0+j0 ohms cannot absorb energy.

Now that is a non-sequitor. The element absorbing energy is an
ideal voltage source, not a resistor.

As Eugene Hecht said:
"If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

It is a DC circuit. This means the instantaneous value is
the average value.

(But poor Hecht, here he is saying power is more useful than
cumulative energy, and you misinterpret him to be comparing
instantaneous to average. And is it the distribution over
the area that is being averaged, or the distribution over
time?)

You seem to have discovered that "limited utility".
You have an instantaneous power calculated over an
infinitesimally small amount of time being dissipated
or stored in an impedance of 0+j0. Compared to that
assertion, the Virgin Birth seems pretty tame. :-)

Again, it is a DC circuit since we have had to go back
to learning the fundamentals of ideal voltage sources.

But back to the simple question...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

...Keith

Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?
--
73, Cecil http://www.w5dxp.com

On Apr 19, 11:23*pm, Cecil Moore wrote:
Keith Dysart wrote:
So we are back to the question you keep dodging...

I'm not dodging anything. I am ignoring irrelevant
examples. The context under discussion is configurations
of single source systems with reflections where the
average interference is zero. So please explain how
your example applies to what we are discussing. Where
are the reflections? Where is the average interference
equal to zero?

Your explanations are predicated on a misunderstanding of
the behaviour of ideal voltage sources. Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

The discussion needs to digress to address this fundamental
misunderstanding since this misunderstanding renders all
else moot. And so...

Let me remind you of the circuit at hand:

50 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=100 VDC Vsr=50 VDC
| |
+-----------------------------+

And we are back to the question you keep dodging...

Where does the energy that is flowing into the ideal
DC voltage source on the right go?

If no energy is flowing into the ideal voltage source
on the right, where is the energy that is being
continuously provided by the ideal voltage source on
left going? Of the 100 W being provided by the ideal
source on the left, only 50 W is being dissipated in
the resistor. Where goes the remaining 50 W, if not
into the ideal voltage source on the right?

Once it is agreed that ideal DC voltage sources can indeed
absorb energy, we can discuss the same for ideal AC voltage
sources. Once that is agreed, we can return to the
discussion of your 'interference free' circuit.

...Keith

Keith Dysart wrote:
Where does the energy that is flowing into the ideal
DC voltage source on the right go?

I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.
--
73, Cecil http://www.w5dxp.com

On Apr 21, 1:23*am, Cecil Moore wrote:
Keith Dysart wrote:
Where does the energy that is flowing into the ideal
DC voltage source on the right go?

I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

...Keith

Keith Dysart wrote:
Cecil Moore wrote:
I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

No, I can quit wasting my time thinking about it
because "we do not care". We not caring takes away
any reason or purpose for continuing the discussion.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

The discussion is moot unless you can prove that an
ideal source in a single-source system can absorb
AVERAGE power. So there's your challenge. If you
cannot do that, my article about average power
stands as written.
--
73, Cecil http://www.w5dxp.com

On Apr 21, 12:11*pm, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
I'll just quote you on that one:

Where does this energy go?
We do not know and we do not care.

Finally, you have got the correct answer.

Having accepted that ideal voltage sources can remove energy
from a circuit, you can now re-evaluate your explanations.

No, I can quit wasting my time thinking about it
because "we do not care". We not caring takes away
any reason or purpose for continuing the discussion.

With an ideal voltage source being capable of removing
energy, it should no longer be necessary for you to
propose strange notions of energy flowing through an
ideal voltage source and bouncing off the ground.

The discussion is moot unless you can prove that an
ideal source in a single-source system can absorb
AVERAGE power. So there's your challenge. If you
cannot do that, my article about average power
stands as written.

The best analogy I can think of is someone saying:
"Until you prove the earth is flat, I will not consider
any evidence that it is round".

Another question remains, since it is difficult to
discern from your writings: Have you grasped the
behaviour of an ideal voltage source when current
is flowing into the source?

You now understand that it removes energy from the
circuit?

If you have that for DC ideal voltage sources, we
can move on to discover what happens with AC ideal
voltage sources. After that, we can go back to what
is happening in your circuit.

...Keith

Keith Dysart wrote:
Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

From the "IEEE Dictionary":

"Absorption: (2) A general term for the process by which
incident flux is converted to another form of energy,
usually and ultimately to heat. (4) The irreversible
conversion of the energy of an EM wave into another form
of energy as a result of wave interaction with matter."

By what mechanism does an ideal source with an impedance
of 0+j0 manage to dissipate heat? Since the energy absorption
by the ideal source is *irreversible*, where is the heat
(or other form of energy) stored and for how long?
--
73, Cecil http://www.w5dxp.com

On Apr 21, 3:25*am, Cecil Moore wrote:
Keith Dysart wrote:
Despite your
protests to the contrary, ideal voltage sources can, and
do, absorb energy.

*From the "IEEE Dictionary":

"Absorption: (2) A general term for the process by which
incident flux is converted to another form of energy,
usually and ultimately to heat. (4) The irreversible
conversion of the energy of an EM wave into another form
of energy as a result of wave interaction with matter."

By what mechanism does an ideal source with an impedance
of 0+j0 manage to dissipate heat? Since the energy absorption
by the ideal source is *irreversible*, where is the heat
(or other form of energy) stored and for how long?

Just as ideal voltage sources can provide energy to a circuit,
they can also remove energy from a circuit.

Feel free to substitute the word of your choice for 'remove'.

Dissipate is not a good choice since it usually implies
conversion to heat.

Absorb is not a good word for you, since you can find absorption
in the IEEE dictionary and it also suggests conversion to heat.

The thesaurus (http://thesaurus.reference.com/ suggests 'consume',
'assimilate', 'digest', 'imbibe',
'take up', 'sop up', and 'devour'.

Pick the word that you find least confusing.

Recalling that an ideal voltage source can provide (deliver,
furnish, supply, transfer) energy to a circuit, we need a
non confusing word to describe the concept that an ideal
voltage source can also remove energy from a circuit.

A word that gives no hint about where this energy goes would
be best, since, just as we do not know where the energy
that an ideal voltage source delivers to a circuit comes
from, we do not know where the energy that an ideal voltage
source removes from a circuit goes.

...Keith

Keith Dysart wrote:
Feel free to substitute the word of your choice for 'remove'.

That's the first time you have used the word "remove".
Have you changed your mind about energy being "absorbed",
by the source, i.e. turned into heat?

Dissipate is not a good choice since it usually implies
conversion to heat.

Whoa there Keith, "absorb" is equally not a good choice
since it usually implies conversion to heat as in the IEEE
definitions. If the source only removes energy, then that
is a plus for my side of the argument. If the source has
the ability to remove the destructive interference and
supply it back 90 degrees later as constructive interference,
the entire mystery of where the reflected power goes is
solved. When I previously offered that as a solution, you
turned it down flat. Now you seem to be agreeing with it.

Absorb is not a good word for you, since you can find absorption
in the IEEE dictionary and it also suggests conversion to heat.

That's why I have been arguing loud and long against the
absorption of energy by the source. It would imply that
the source is heating up or has an infinite ability to
"irreversibly convert the energy of an EM wave into another
form of energy". That irreversible energy conversion is
what I have been objecting to. There is no way an impedance
of 0+j0 can cause an irreversible energy conversion.

A word that gives no hint about where this energy goes would
be best, ...

So you can sweep it under the rug and "not care where it
went"? As I said, further discussion is pointless. You
have a magic source that obeys your every whim. Why didn't
you just say that in the first place?
--
73, Cecil http://www.w5dxp.com