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:
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

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

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

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

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

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

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

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