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Roger Sparks wrote:
I offered the examples of two real sources that will absorb
power when the returning voltage exceeds the output voltage
(a battery and a generator turned into a motor). I think that
we must allow our voltage source to have that same real property.

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy. A motor converts electrical
energy into physical work, i.e. it transforms the electrical
energy. An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

I do understand that when we allow the source to receive
power, then we need to address source impedance.

The series source impedance is zero. It acts like a short
circuit to reflections, i.e. there are no reflections.
However, there seem to be 100% reflection from the GND on
the other side of the source.

Does the idea of source receiving power run counter to what
you were planning to write in Parts 2 and 3?

The source will be shown to adjust its output until an
energy balance is achieved. It will throttle back when
destructive interference occurs at the source resistor
and will gear up when constructive interference requires
more energy.

I am trying to
understand why you have such great reluctance to accept that
the source could receive power for part of a cycle, especially
when it could easily bring the instantaneous power and energy
calculations into balance.

There is no known mechanism that would allow an ideal
source to dissipate or store energy. Consider that the
energy you see flowing back into the source is reflected
back through the source by the ground on the other side
and becomes part of the forward wave out of the source.
That would satisfy the distributed network model and
explain why interference exists in the source.
--
73, Cecil http://www.w5dxp.com

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On Apr 14, 5:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
What is your explanation for this phenomenon?

You first! :-)

I suppose. If you are happy with energy equations that
don't balance.

My energy equations balance perfectly. Yours are the
energy equations that don't balance in violation of
the conservation of energy principle.

"It depends" was no answer - that was just mealy-mouthing.

Except, that it does depend.

Sounds more like religion than anything else.

You have to read more carefully. I did not use the word
dissipation.

You have to read more carefully. I used the word
"dissipation" and you disagreed with me. Please
read it again to verify that fact.

Since we don't know the internals of the source, we do not
know if it is dissipating or not.

Sorry, the source is, by definition, lossless. All of
the source dissipation is lumped in the source resistance
drawn separately on the diagram as Rs.

You really should rethink this a bit. When current flows into
a voltage source, the voltage source is absorbing energy.

But the source is NOT *DISSIPATING* energy because
Rs is not inside the source. Rs is clearly drawn outside
the lossless source so the source generates *ZERO* heat.
I am amazed at the lengths to which you will go to try
to obfuscate the discussion.

You do need to back up a bit and review voltage sources.

When conventional current is flowing out of the positive
terminal of a voltage source, it is usually agreed that
the voltage source is providing energy to the circuit.
This comes from
P(t) = V(t) * I(t)
when V and I are the same sign, P is positive representing
a flow of energy from the source to the other elements
of the circuit.

What do you explain is happening when conventional current
is flowing into the positive terminal of the source?
Is the source still providing energy to the circuit now
that P is negative?
Or is it accepting energy from the circuit, the negative
P representing an energy flow into the source?

...Keith

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Cecil Moore wrote:

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy.

Gee, I'm just a silly old engineer, but I thought batteries converted
chemical energy to electrical energy. Unless you are perhaps speaking
of charging it.

tom
K0TAR

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On Tue, 15 Apr 2008 01:22:11 GMT
Cecil Moore wrote:

Roger Sparks wrote:
I offered the examples of two real sources that will absorb
power when the returning voltage exceeds the output voltage
(a battery and a generator turned into a motor). I think that
we must allow our voltage source to have that same real property.

A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy. A motor converts electrical
energy into physical work, i.e. it transforms the electrical
energy. An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

No, the separation of Vs and Rs was made to better understand why no interference would occur in Figure 1-1. found at http://www.w5dxp.com/nointfr.htm.

Here is a quote from Part 1.
"4. Since the transmission line is 1/8 wavelength (45 degrees) long and the load is purely resistive, the reflected wave incident upon the source resistor will be 2(45) = 90 degrees out of phase with the forward wave at the source resistor. This is the necessary and sufficient condition to produce zero interference at the source resistor."

The problem is that the source and reflected waves behave as two power sources out of time by 90 degrees. As a result, the current flows as the result of two sine waves, and can be described by only one sine wave. The one sine wave description necessarily shows that power *does* flow into the source during part of the cycle. Interference techniques are used to combine the two sine waves into one wave so it would appear that statement 4 is incorrect.

I do understand that when we allow the source to receive
power, then we need to address source impedance.

The series source impedance is zero. It acts like a short
circuit to reflections, i.e. there are no reflections.
However, there seem to be 100% reflection from the GND on
the other side of the source.

It is not the reflections from the source that is the root of the problem. The root is the way two sine waves combine into one wave that runs at a third phase compared to either of the source waves.

Does the idea of source receiving power run counter to what
you were planning to write in Parts 2 and 3?

The source will be shown to adjust its output until an
energy balance is achieved. It will throttle back when
destructive interference occurs at the source resistor
and will gear up when constructive interference requires
more energy.

I am trying to
understand why you have such great reluctance to accept that
the source could receive power for part of a cycle, especially
when it could easily bring the instantaneous power and energy
calculations into balance.

There is no known mechanism that would allow an ideal
source to dissipate or store energy. Consider that the
energy you see flowing back into the source is reflected
back through the source by the ground on the other side
and becomes part of the forward wave out of the source.
That would satisfy the distributed network model and
explain why interference exists in the source.

I can understand a voltage source that throttles up and down but I can't understand why the throttle all has to be on the plus side. What logic prevents the power from returning to the ideal source from whence it just left?

Our real limit is that only one current can flow for only one voltage for each instant at any place in the circuit. This is how we justify a "one sine wave" description. It is why whenever we have a reflection, we also have interference. It is also the reason that we must have power flowing back into the source for part of the cycle.
--
73, Roger, W7WKB

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Tom Ring wrote:
Cecil Moore wrote:
A battery converts electrical energy to chemical energy, i.e.
it transforms the electrical energy.

Gee, I'm just a silly old engineer, but I thought batteries converted
chemical energy to electrical energy. Unless you are perhaps speaking
of charging it.

Here was the preceding comment in context which you trimmed:
"I offered the examples of two real sources that will absorb power when
the returning voltage exceeds the output voltage (a battery and a
generator turned into a motor)."

The context was a battery absorbing power when the
charging voltage exceeds the battery's output voltage.
A battery converts electrical energy to chemical energy
during that charging process.
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
On Tue, 15 Apr 2008 01:22:11 GMT
Cecil Moore wrote:
An ideal source does not dissipate power and there is
no mechanism for storing energy. It seems what you are objecting
to is the artificial separation of Vs and Rs.

No, the separation of Vs and Rs was made to better understand
why no interference would occur in Figure 1-1.

I wasn't talking about my article. I was talking about Vs & Rs
models in general. In the real world, Vs is not separated from
Rs. That only occurs in the ideal model. In the ideal model,
all dissipation is confined to Rs and there is none in Vs.

The problem is that the source and reflected waves behave as
two power sources out of time by 90 degrees.

Not quite correct. The problem is that the forward waves
and reflected waves flowing through the source behave as
two power sources out of time by 90 degrees. The source
wave is the net superposition of the forward wave and
reflected wave. An ideal 50 ohm directional wattmeter
in the circuit will not read the source power. It will
read a forward power which is a different magnitude
than the source power. In any case, only Rs and RL
dissipate power in the system.

I can understand a voltage source that throttles up and
down but I can't understand why the throttle all has to be
on the plus side.

It is not all on the plus side. Whatever energy flows, flows.
Sometimes the flow is forward and sometimes it is backwards.
That's the way AC works. If destructive interference is
present, the source reduces its output power. If constructive
interference is present, the source increases its output
power. But the ideal source does not dissipate power, i.e.
doesn't heat up. All of the heat generated in the entire
system comes from Rs and RL.

Our real limit is that only one current can flow for only one
voltage for each instant at any place in the circuit.

You are, of course, talking about the *net* voltage and the
*net* current after superposition of all the components. But
this discussion is not about net voltage and net current.

This is
how we justify a "one sine wave" description. It is why whenever
we have a reflection, we also have interference. It is also the
reason that we must have power flowing back into the source for
part of the cycle.

I don't know where you got the idea that energy doesn't
flow back into the source for part of the cycle. Since
it is AC, it does flow forward and backward but none is
dissipated, i.e. none is turned into heat in an ideal
source. An equal amount of destructive and constructive
interference occurs during each complete cycle.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
Or is it accepting energy from the circuit, the negative
P representing an energy flow into the source?

Of course there can be an energy flow into the source
and through the source. The point is that the ideal
source doesn't dissipate that energy, i.e. it doesn't
heat up. All of the heating (power dissipation) in the
entire example occurs in Rs and RL because they are
the only resistive components in the entire system.
Any additional heating in the ideal source would
violate the conservation of energy principle.
--
73, Cecil http://www.w5dxp.com

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On Apr 15, 6:16*am, Cecil Moore wrote:
Keith Dysart wrote:
Or is it accepting energy from the circuit, the negative
P representing an energy flow into the source?

Of course there can be an energy flow into the source

Good.

and through the source.

It is not clear what you mean by "through the source".
The source provides or absorbs energy. It does not have
a "through", since it only has a single port.

The point is that the ideal
source doesn't dissipate that energy, i.e. it doesn't
heat up.

It is not obvious why you want to draw a distinction
between elements that remove energy from a circuit by
heating and those that do so in some other manner.

Could you expand?

Is there some reason why you think that it is only
necessary to account for the energy removed from the
circuit by heating?
And that you can ignore energy being removed in other ways?

And how do you know the ideal source does not dispose
of the energy it receives by getting warm? Nowhere
do I find in the specification of an ideal source any
hint of how it disposes of its excess energy. It could
be by heat, could it not?

And the resistor, could it not also radiate some of
the energy it receives? Perhaps even as visible light?
Would that make it less of a resistor because it was
not 'dissipating' the energy?

All of the heating (power dissipation) in the
entire example occurs in Rs and RL because they are
the only resistive components in the entire system.
Any additional heating in the ideal source would
violate the conservation of energy principle.

This is quite a leap. The energy flows into the source.
We have accounted for that energy. We don't know where
it goes from there.

How would it violate conservation of energy if it was
dissipated rather than going somewhere else?

In your model, what things could be done with the energy
that would not violate conservation of energy? What other
things (besides heating) would violate conservation of
energy?

...Keith

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Keith Dysart wrote:
It is not clear what you mean by "through the source".
The source provides or absorbs energy. It does not have
a "through", since it only has a single port.

Good grief! The source is a two terminal device
with one terminal tied to ground. Since it has
a zero resistance, an EM reverse wave can flow
right through it, encounter the ground, and be
100% re-reflected by that ground. I have already
explained that. Did you bother to read it?

If you put an ideal 50 ohm directional wattmeter
between the source and its ground, what will it read?

dir
GND---watt----Vs--Rs-----------------------RL
meter

Did you bother to analyze this configuration?

Rs=50
----50-ohm----/\/\/\/\----50-ohm----
125w-- 100w 50w--
--25w --50w

This explains everything at the average power level.
I suspect it also works at the instantaneous level.

It is not obvious why you want to draw a distinction
between elements that remove energy from a circuit by
heating and those that do so in some other manner.

Could you expand?

We are dealing with an ideal closed system. The
ultimate destination for 100% of the ideal source
power is heat dissipation in the two ideal resistors,
Rs and RL. What happens between the power being sourced
and the power being dissipated as heat is "of limited
utility".

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.

And the resistor, could it not also radiate some of
the energy it receives? Perhaps even as visible light?
Would that make it less of a resistor because it was
not 'dissipating' the energy?

God could also suck up the energy or the universe
could end. Why muddy the waters with irrelevant
obfuscations? Please deal with the ideal boundary
conditions as presented.

This is quite a leap.

Nope, it is simple physics through which you must
have been asleep.

How would it violate conservation of energy if it was
dissipated rather than going somewhere else?

It can only be dissipated in the resistances, by definition.
Nothing other than the resistances in an ideal closed system
dissipates power. EE102.

In your model, what things could be done with the energy
that would not violate conservation of energy?

FIVE TMES, I listed three things in previous postings. Since
you avoided reading it FIVE TIMES already, I'm just going to
point you to the "Optics" chapters on "Interference" and
"Superposition".
--
73, Cecil http://www.w5dxp.com

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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