The Rest of the Story
 

Cecil Moore wrote:
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.

In case you have forgotten :-), here is what you posted
over the past few days:

When it is sinking current, it is *absorbing* energy.

You will find that for some of the time
energy is being *absorbed* by the source.

But we do know that when the sign of Ps(t) is
negative, the source is *absorbing* energy from
the system,

When current flows into a voltage source, the
voltage source is *absorbing* energy.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

This certainly implies that you consider the dissipation
of "absorbed" energy to be a distinct probability.

The source provides or *absorbs* energy.

When 1.5 amps is flowing into the positive terminal,
the ideal voltage source is *absorbing* 15 joules per
second from the circuit.

The ideal voltage source on the right is *absorbing*
5 joules/second from the circuit.

An ideal source provides or *absorbs* energy to
satisfy its basic function which is to hold the
voltage across its terminals at the desired value.

When it is providing energy we do not know where
this energy comes from and when it is *absorbing*
energy we do not know where this energy goes.

But was this because you have learned that you were
in error and now better understand the behaviour of
sources when they are *absorbing* energy?

The ideal voltage source on the right, after the
circuits settle, will be absorbing 50 joules/s
in both cases.

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

The element *absorbing* energy is an ideal voltage
source, not a resistor.

Despite your protests to the contrary, ideal voltage
sources can, and do, *absorb* energy.

You know, Keith, if you were just ethical enough to
answer my questions, I wouldn't have to treat you
this way.
--
73, Cecil http://www.w5dxp.com

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On Apr 22, 8:11*am, Cecil Moore wrote:
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."

The wordy non-sequitor leaves us with nagging doubts about
whether the fundamentals of DC ideal voltage sources have
been grasped.

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.

Doubts now confirmed.

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.

But since the whole work is predicated on a misunderstanding
of the behaviour of ideal voltage sources, any conclusions
retain no value.

...Keith

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On Apr 22, 8:18*am, Cecil Moore wrote:
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 did use the word 'absorb' to convey the concept of removing
energy from a circuit without knowing where the energy went.
You misinterpreted my use of 'absorb' to mean something else.

These things happen. Most readers, upon learning the intended
meaning of a word was different than expected, would simply
re-read the affected passages to glean the intended meaning.

This seems difficult for you. May I suggest that substituting
'remove' for 'absorb' might help you gain an understanding
of the intent.

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.

Don't worry. Knowing your difficulty with detecting the
intended meaning of words, in future I shall avoid 'absorb'
when I mean "removing energy from a circuit without knowing
where the energy goes."

...Keith

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On Apr 22, 1:12*pm, Cecil Moore wrote:
Cecil Moore wrote:
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.

In case you have forgotten :-), here is what you posted
over the past few days:

With your difficulty using the word 'absorb' to represent the
abstract concept of removing energy without knowing where it
goes, for a better understanding please re-read these passages
substituting 'remove' for 'absorb'.

When it is sinking current, it is *absorbing* energy.

You will find that for some of the time
energy is being *absorbed* by the source.

But we do know that when the sign of Ps(t) is
negative, the source is *absorbing* energy from
the system,

When current flows into a voltage source, the
voltage source is *absorbing* energy.

And how do you know the ideal source does not dispose
of the energy it receives by getting warm?

This certainly implies that you consider the dissipation
of "absorbed" energy to be a distinct probability.

It certainly is one of the many possible things that might
happen to the energy that is removed. As I wrote previously,
"In practice, devices which are designed to approximate
ideal voltage sources do simply dissipate the energy they
remove from the circuit."

Still, since we do not *know* what happens to the energy removed
by an ideal voltage source, we can make no assumptions.

The source provides or *absorbs* energy.

When 1.5 amps is flowing into the positive terminal,
the ideal voltage source is *absorbing* 15 joules per
second from the circuit.

The ideal voltage source on the right is *absorbing*
5 joules/second from the circuit.

An ideal source provides or *absorbs* energy to
satisfy its basic function which is to hold the
voltage across its terminals at the desired value.

When it is providing energy we do not know where
this energy comes from and when it is *absorbing*
energy we do not know where this energy goes.

But was this because you have learned that you were
in error and now better understand the behaviour of
sources when they are *absorbing* energy?

The ideal voltage source on the right, after the
circuits settle, will be absorbing 50 joules/s
in both cases.

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

The element *absorbing* energy is an ideal voltage
source, not a resistor.

Despite your protests to the contrary, ideal voltage
sources can, and do, *absorb* energy.

You know, Keith, if you were just ethical enough to
answer my questions, I wouldn't have to treat you
this way.

Perhaps. But I doubt that it would actually alter your
behaviours.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Your challenge is to prove that a single source
removes an average amount of energy from the
network.

But since the whole work is predicated on a misunderstanding
of the behaviour of ideal voltage sources, any conclusions
retain no value.

Keith, if you can prove that a single source removes
an average amount of energy from the network during
steady-state conditions, I will worship at your feet.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
I did use the word 'absorb' to convey the concept of removing
energy from a circuit without knowing where the energy went.
You misinterpreted my use of 'absorb' to mean something else.

No, I assumed the standard IEEE Definition of the word.
You mistakenly neglected to state the non-standard definition
that you were using.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
Perhaps. But I doubt that it would actually alter your
behaviours.

Well, let's try one more time and see. Assuming a 50 ohm
environment (by definition) in the following example, what
will an ideal 50 ohm directional wattmeter indicate at
point 'x' for forward power (Pf1) and reverse power (Pr1)?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----short
| --Pr1 Rs --Pr2=50w
| 100w
Vs=100v
|
Gnd

It's a simple question with easily calculated and
quantifiable answers.
--
73, Cecil http://www.w5dxp.com

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On Wed, 23 Apr 2008 07:58:49 -0500
Cecil Moore wrote:



Well, let's try one more time and see. Assuming a 50 ohm
environment (by definition) in the following example, what
will an ideal 50 ohm directional wattmeter indicate at
point 'x' for forward power (Pf1) and reverse power (Pr1)?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----short
| --Pr1 Rs --Pr2=50w
| 100w
Vs=100v
|
Gnd

It's a simple question with easily calculated and
quantifiable answers.

Your circuit is incomplete. The circuit parameters are not fully defined.

Is this the same circuit that you wish to analyze?

Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----+
| --Pr1 Rs --Pr2=50w |
| 100w |
Vs=100v |
| |
Gnd Gnd

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

What reflection characteristics should we assume for the voltage source Vs?
--
73, Roger, W7WKB

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Roger Sparks wrote:
Pf1-- 50 ohm Pf2=50w--
+----x-----/\/\/\/\-------45 deg 50 ohm-----+
| --Pr1 Rs --Pr2=50w |
| 100w |
Vs=100v |
| |
Gnd--------------------------Braid------------+

The coax is shorted at the end, and Gnd is 45
degrees away from the short.

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

Assume that Rs is 100% of the source impedance.

What reflection characteristics should we assume for the voltage source Vs?

Assuming that Rs is 100% of the source impedance
should answer that question.
--
73, Cecil http://www.w5dxp.com

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On Apr 23, 12:03*pm, Cecil Moore wrote:
Roger Sparks wrote:
* * * Pf1-- * 50 ohm * * Pf2=50w--
* *+----x-----/\/\/\/\-------45 deg 50 ohm-----+
* *| *--Pr1 * * Rs * * * --Pr2=50w * * * * * |
* *| * * * * * *100w * * * * * * * * * * * * * |
*Vs=100v * * * * * * * * * * * * * * * * * * * |
* *| * * * * * * * * * * * * * * * * * * * * * |
*Gnd--------------------------Braid------------+

The coax is shorted at the end, and Gnd is 45
degrees away from the short.

What is the impedance of the voltage delivered from Vs before the wave encounters the resistor Rs?

Assume that Rs is 100% of the source impedance.

What reflection characteristics should we assume for the voltage source Vs?

Assuming that Rs is 100% of the source impedance
should answer that question.

Is this a standard ideal voltage source which provides energy
to the circuit when the product of its voltage and current is
positive and removes energy from the circuit when the product
of its voltage and current is negative?

Or have you assigned it some other non-standard definition?

...Keith