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

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, 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

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:
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

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

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

On Apr 21, 12:26*pm, Cecil Moore wrote:
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".

You need to read more carefully. Not the first time at all.

Have you changed your mind about energy being "absorbed",
by the source, i.e. turned into heat?

What I have said is that an ideal voltage source removes energy
from a circuit and that we do not know what it does with the
energy it removes. In practice, devices which are designed to
approximate ideal voltage sources do simply dissipate the energy
they remove from the circuit.

But that is not part of the definition of an ideal voltage source,
for which no statement about where the energy removed goes is
made.

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.

No. We do not know what an ideal voltage source does with
energy it removes. We can not say that it stores and then
returns it, though a particular implementation might do
so. Another implementation might not. So this can not be
used as an explanation.

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.

I am sorry that the occasional use of the word 'absorb' so
mislead you. I avoided 'dissipate' for that reason. It is
not so obvious why you were mislead by the use of 'remove'.

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"?

No. Because the definition of an ideal voltage does not specify
where the energy goes. Therefore we had better not care when
we use an ideal voltage source.

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?

No. My ideal voltage source just obeys the definition of an
ideal voltage source. It provides energy to the circuit when
the circuit conditions demand that it do so and similarly it
removes energy when the circuit conditions demand that it do
so. The definition does not tell us where the energy it
provides comes from, nor does it tell us where the energy
it removes goes.

A fairly simple defintion: The voltage at the terminals is
maintained at the desired value, regardless of the current
flow needed to do so.

No magic in the definition whatsoever. And no need to obey
whims.

...Keith

Keith Dysart wrote:
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?

It follows that it is futile to try to track any
movement of instantaneous energy. You have convinced
me that you are correct - "We don't care where the
(instantaneous) energy goes."

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.

Since there is an instantaneous leak in the closed
system, it is useless to proceed. You say you don't
care what happens to the energy. I said a couple of
months ago that I didn't care what happens to
instantaneous power. And indeed, you have convinced
me that any attempt to track instantaneous power is
doomed to failure.

My part 1 article based on a single source and
*AVERAGE* powers doesn't have those conceptual
problems and stands as written. Here's the second
paragraph from that article:

"Please note that any power referred to in this paper
is an AVERAGE POWER. Instantaneous power is beyond the
scope of this article, irrelevant to the following
discussion, and "of limited utility" according to
Eugene Hecht. [4]

Your challenge is to prove that a single source
removes an average amount of energy from the
network.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
What I have said is that an ideal voltage source removes energy
from a circuit ...

Sorry, you specifically said that an ideal voltage source
"absorbs" energy, i.e. irreversibly converts energy to
another form, the most common form of which is heat.

I am sorry that the occasional use of the word 'absorb' so
mislead you. I avoided 'dissipate' for that reason.

The IEEE Dictionary says that, in this context, "absorb"
and "dissipate" are virtual synonyms.
--
73, Cecil http://www.w5dxp.com