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On Tue, 15 Apr 2008 14:40:28 -0700 (PDT)
Keith Dysart wrote:

On Apr 15, 12:47*pm, Cecil Moore wrote:
Keith Dysart wrote:
Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?

Heating is the only way that the energy can be removed
from the ideal closed system.

And that you can ignore energy being removed in other ways?

No energy is removed in other ways. There is no other
way for energy to leave the ideal closed system.

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

An ideal source has zero source impedance, by definition.
All of the source impedance is contained in Rs, the
ideal source resistor.

I see we need to back up a bit further.

Consider a 10 VDC ideal voltage source.

When 2 amps are flowing out of the positive terminal, the
ideal voltage source is delivering 20 joules per second
to the circuit.

Q1. Where does this energy come from?

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

Q2. Where does this energy go?

Answer to both. We do not know and we do not care. An
ideal voltage source can deliver energy to a circuit
and it can remove energy from a circuit; that is part
of the definition of an ideal voltage source. It does
not matter how it does it.

But just as easily as it can supply energy, it can
remove it.

Without understanding these basics of the ideal voltage
source, it will be impossible to correctly analyze
circuits that include them.

Perhaps this has been the root cause of the misunderstandings.

...Keith

This WIKI article mentions the ability of an ideal voltage source to absorb power.

http://en.wikipedia.org/wiki/Voltage_source
--
73, Roger, W7WKB

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Keith Dysart wrote:
Q1. Where does this [source] energy come from?

An ideal source simply supplies a fixed voltage
devoid of any concern for efficiency or where
the energy comes from. This results in an
average steady-state number of joules being
supplied to the closed system per second.

Q2. Where does this [reverse] energy go?

Once introduced into a closed system, it obeys
the laws of physics including the conservation
of energy/momentum principles and the principle
of superposition of forward and reverse
electromagnetic fields. You can use an ideal
directional wattmeter to track the forward and
reverse energy flows through the ideal source.

Answer to both. We do not know and we do not care.

I have been telling you that your concepts violate
the conservation of energy principle and now you
have essentially admitted it.

But just as easily as it can supply energy, it can
remove it.

I'm afraid you will find that once the ideal source
has supplied the energy to a closed system, that energy
cannot be destroyed. If you are allowed to willy-nilly
suspend the conservation of energy principle, then any
magical event is possible and there is no valid reason
to even try to track the energy. It is one thing to
assume an introduction of steady-state power to a
closed system from an ideal source. It is another
thing entirely to allow destruction of that energy
once it has been introduced.
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power.

It says: "A primary voltage source can supply (or absorb)
energy ..."

That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.
If we define an RF source as a coherent RF battery, anything
is possible (at least in our minds). Which of the following
makes more sense?

1. Destructive interference energy is stored somewhere
in the system and delivered back to the system 90 degrees
later in the cycle just as it is by a physical inductor
or capacitor.

2. An RF battery inside the ideal source stores the extra energy
in coherent RF form and delivers it back to the system as needed.

http://en.wikipedia.org/wiki/Voltage_source

It also says: "The internal resistance of an ideal voltage
source is zero;" So exactly how does something with an
internal resistance of zero absorb any power?
--
73, Cecil http://www.w5dxp.com

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On Wed, 16 Apr 2008 09:31:32 -0500
Cecil Moore wrote:

Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power.

It says: "A primary voltage source can supply (or absorb)
energy ..."

That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

If we define an RF source as a coherent RF battery, anything
is possible (at least in our minds). Which of the following
makes more sense?

1. Destructive interference energy is stored somewhere
in the system and delivered back to the system 90 degrees
later in the cycle just as it is by a physical inductor
or capacitor.

No, the problem here is that the individual switched batteries that would make up this system would not be equally discharged. Some would actually gain charge, and never deliver power to the system. So I don't like this choice.

2. An RF battery inside the ideal source stores the extra energy
in coherent RF form and delivers it back to the system as needed.

This sounds identical to the 1st choice. You can make such a circuit with many individually switched batteries so it becomes a good test of the theory. Again, some of the batteries would actually gain energy.

http://en.wikipedia.org/wiki/Voltage_source

It also says: "The internal resistance of an ideal voltage
source is zero;" So exactly how does something with an
internal resistance of zero absorb any power?

How do we deal with the I x E power when the reflected voltage is greater than the forward voltage from the source? One way is to ignore it and only deal with averages. Another is to allow the ideal voltage source to absorb power. A third way is to allow the energy to be stored in constructive and destructive interference some place in the system. A fourth way is to allow reflections from the ideal voltage source with the result that power will build internally and the phase shift of the resultant system wave will shift farther away from the driving voltage, until some stable power input to disipation rate is reached.

I like your RF battery idea. It could actually be built to a close approximation of an actual sine wave.
--
73, Roger, W7WKB

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Roger Sparks wrote:
Cecil Moore wrote:
That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.

Unless we allow it absorb energy with the same ease that it emits energy.

How can a device with a zero impedance, i.e. zero resistance,
zero capacitance, and zero inductance, absorb energy? We can
certainly allow it to magically absorb energy but of what
use is that?

A third way is to allow the energy to be stored in constructive
and destructive interference some place in the system.

As you know, this is the one I prefer. Another thing that
neither you nor Keith has done is to account for the
reverse-flowing energy through the source. I suspect if
that was done, every thimbleful of energy would be
accounted for. So far, net energy calculations have been
used on one side of Rs and component energy calculations
on the other. That would work only if Rs was not dissipating
power.
--
73, Cecil http://www.w5dxp.com

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On Apr 16, 10:04*am, Cecil Moore wrote:
Keith Dysart wrote:
Q1. Where does this [source] energy come from?

An ideal source simply supplies a fixed voltage
devoid of any concern for efficiency or where
the energy comes from. This results in an
average steady-state number of joules being
supplied to the closed system per second.

Q2. Where does this [reverse] energy go?

You have morphed the questions. Let us try again.

Try two ideal voltage sources arranged in the
circuit below.
5 ohms
+----------\/\/\/\/-----------+
+| +|
Vsl=10 VDC Vsr=5 VDC
| |
+-----------------------------+

Using the circuit analysis technique of your
choice you should find that 1 amp is flowing
through the resistor.

The ideal voltage source on the left is providing
10 joules/second to the circuit.

Q1. Where does this energy come from?

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

Q2. Where does this energy go?

Answer to both. We do not know and we do not care. An
ideal voltage source can deliver energy to a circuit
and it can remove energy from a circuit; that is part
of the definition of an ideal voltage source. It does
not matter how it does it.

But just as easily as it can supply energy, it can
remove it.

Without understanding these basics of the ideal voltage
source, it will be impossible to correctly analyze
circuits that include them.

This has been the root cause of the misunderstandings.

...Keith

The Rest of the Story
 

On Apr 16, 10:31*am, Cecil Moore wrote:
Roger Sparks wrote:
This WIKI article mentions the ability of an ideal voltage source to absorb power.

It says: "A primary voltage source can supply (or absorb)
energy ..."

That's easy to comprehend for a battery source. Not so
easy for an ideal RF source with a zero series impedance.
If we define an RF source as a coherent RF battery, anything
is possible (at least in our minds). Which of the following
makes more sense?

Neither. See 3.

1. Destructive interference energy is stored somewhere
in the system and delivered back to the system 90 degrees
later in the cycle just as it is by a physical inductor
or capacitor.

2. An RF battery inside the ideal source stores the extra energy
in coherent RF form and delivers it back to the system as needed.

3. 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.

http://en.wikipedia.org/wiki/Voltage_source

It also says: "The internal resistance of an ideal voltage
source is zero;" So exactly how does something with an
internal resistance of zero absorb any power?

Unknown. Just as we do not know where the energy that an ideal
voltage source sometimes provides comes from.

But you are invited to speculate on the many ways that it might
be done. Just as you might speculate on where the ideal voltage
source gets its energy when it is providing energy to the
circuit.

Just as there are real devices which approximate ideal voltage
sources delivering energy to a circuit (commonly called power
supplies), there are devices which approximate ideal voltage
sources that remove energy from a circuit. Check the Agilent
catalog for "DC Loads". Perhaps the schematic will provide
what you seek. Of course power supplies and DC loads are
typically one quadrant devices, but with a little ingenuity
you can combine them to form a two quadrant device which
would be an even better approximation to an ideal voltage
source.

...Keith

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Keith Dysart wrote:
Q1. Where does this energy come from?
Q2. Where does this energy go?

Answer to both. We do not know and we do not care.

Well, that certainly puts an end to this
discussion.
--
73, Cecil http://www.w5dxp.com

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On Apr 16, 11:56*pm, Cecil Moore wrote:
Keith Dysart wrote:
Q1. Where does this energy come from?
Q2. Where does this energy go?

Answer to both. We do not know and we do not care.

Well, that certainly puts an end to this
discussion.

An intriguing response.

It was such a good example that it just stopped you in
your tracks.

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?

Or was it because you have realized the contradictions
inherent in your previous explanations but can not
work out how to resolve them?

Or was it because you have detected just a hint of the
contradictions to come and want to give up before having
to realize their full impact?

...Keith

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This discussion about ideal sources and power absorption would be rather
amazing if I hadn't seen so much of the long sad history of this whole
thread. It's just another diversion to deflect the discussion away from
some of the sticky problems with alternative theories.

It looks like Cecil could benefit from momentarily abandoning his power
waves, virtual reflections, photons, s parameters, and constructive
interference, and go back to basic first or second semester electric
circuit theory. Connect an ideal voltage source to a series RL or RC --
all elements lumped, not distributed. Find the power P(t) at each of the
three nodes. (It's easy. Just calculate v(t) and i(t) and multiply the
two.) Remember that the sign of P(t) shows the direction of energy flow.
If I had the patience and self-control to participate in this endless
thread (and I admittedly have neither), I'd refuse to say another word
about it until Cecil shows that he knows what the power waveforms are at
those three nodes. Why argue with someone about distributed networks who
hasn't mastered lumped circuit analysis?

Roy Lewallen, W7EL