Keith Dysart wrote:
Ahhh, but there is then agreement that energy flows (power) must
be in balance to satisfy conservation of energy.

You can probably answer your own question by figuring
out how the light energy in an interference pattern
gets from the dark ring to the bright ring some
distance away. Do you think it happens faster than
the speed of light?

It is just that it comes up short since there is no
explanation of where the energy goes. Were an adequate
explanation to be offered, I would quite accept it.

I am satisfied with the destructive/constructive
interference explanation. That you have come up
short in tracking those component energies is not
unexpected given your prejudices.

If you told me your car disappeared from existence
because you cannot find it, I wouldn't believe you
either.

You should tread back through the posts, the question was
answered.

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

In the special case presented in Part 1, there are only
two sources of power dissipation in the entire system,
the load resistor and the source resistor.

Three! The source can also take energy from the system.

The Vs source has zero resistance rendering dissipation
impossible. I repeat: There are only two sources of
power *DISSIPATION* in the entire system, the load
resistor and the source resistor.

The source can certainly throttle back its output when
there is destructive interference in the source resistor
and increase its output when there is constructive
interference in the source resistor, but it CANNOT
dissipate any power.

Since you have overlooked the source, the rest of your
post is quite flawed in its conclusions.

Nope, you are confused. The source can adjust its output
but the source cannot *DISSIPATE* power. As I said, the
only sources of power *DISSIPATION* are the two resistors.
No amount of obfuscation is going to change that.

Perhaps adding a circulator to my Part 1, Fig. 1-1 will
allow you to see things in a clearer light. Of course,
using light would be even better.

Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL
\ /
3
|
50 ohms
|
GND

How much power is dissipated in the circulator
resistor?

How much power does the source have to supply to
maintain 50 watts of forward power on the transmission
line?
--
73, Cecil http://www.w5dxp.com

On Apr 14, 1:36*pm, Cecil Moore wrote:
Keith Dysart wrote:
Ahhh, but there is then agreement that energy flows (power) must
be in balance to satisfy conservation of energy.

You can probably answer your own question by figuring
out how the light energy in an interference pattern
gets from the dark ring to the bright ring some
distance away. Do you think it happens faster than
the speed of light?

What is your explanation for this phenomenon?

It is just that it comes up short since there is no
explanation of where the energy goes. Were an adequate
explanation to be offered, I would quite accept it.

I am satisfied with the destructive/constructive
interference explanation. That you have come up
short in tracking those component energies is not
unexpected given your prejudices.

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

If you told me your car disappeared from existence
because you cannot find it, I wouldn't believe you
either.

You should tread back through the posts, the question was
answered.

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

Except, that it does depend.

In the special case presented in Part 1, there are only
two sources of power dissipation in the entire system,
the load resistor and the source resistor.

Three! The source can also take energy from the system.

The Vs source has zero resistance rendering dissipation
impossible. I repeat: There are only two sources of
power *DISSIPATION* in the entire system, the load
resistor and the source resistor.

You have to read more carefully. I did not use the word
dissipation. This is because there are ways other than
dissipation to remove energy from the system. Just as
the source provides energy and we do not care where it
comes from, it can remove energy and we do not care where
it goes.

Examine Ps(t). You will find that for some of the time
energy is being absorbed by the source. This occurs
when the sign of Ps(t) is negative.

The source can certainly throttle back its output when
there is destructive interference in the source resistor
and increase its output when there is constructive
interference in the source resistor, but it CANNOT
dissipate any power.

Since we don't know the internals of the source, we do not
know if it is dissipating or not. But we do know that when
the sign of Ps(t) is negative, the source is absorbing
energy from the system, exactly analagous to it providing
energy when the sign of Ps(t) is positive.

Since you have overlooked the source, the rest of your
post is quite flawed in its conclusions.

Nope, you are confused. The source can adjust its output
but the source cannot *DISSIPATE* power. As I said, the
only sources of power *DISSIPATION* are the two resistors.
No amount of obfuscation is going to change that.

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

Perhaps adding a circulator to my Part 1, Fig. 1-1 will
allow you to see things in a clearer light. Of course,
using light would be even better.

Gnd--1---2---Vs---Rs-----45 deg 50 ohm----------RL
* * * *\ /
* * * * 3
* * * * |
* * *50 ohms
* * * * |
* * * *GND

How much power is dissipated in the circulator
resistor?

This circulator, while often used, does not in any way add
clarity.

Changing the circuit changes the results, especially when
you add a circulator which alters rather dramatically the
energy flows.

How much power does the source have to supply to
maintain 50 watts of forward power on the transmission
line?

This depends on the design of the generator and the length
of the line. With a shorted line that is 90 degrees long,
and a generator constructed using the circuit of Fig 1-1, the
source supplies no energy to maintain an imputed forward
power of 50 watts on the line.

...Keith

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.

Changing the circuit changes the results, ...

When you get more serious about the technical facts
of physics than you are about saving face, please get
back to us, but please, not before. Everyone is
tired of your delusions of grandeur where the laws
of physics obey your every whim.
--
73, Cecil http://www.w5dxp.com

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