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

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

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

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

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

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

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

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

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