Keith Dysart wrote:
Regardless, the average voltage is zero.

No, that contradicts what you said before which
was, if it ever wasn't zero instantaneously, it
cannot possibly average out to zero.

I'm sorry, Keith. Your assertions have gotten
just as irrational as my 96 year old aunt's.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.
I see no reactance that performs this function.

Huh??? When the instantaneous source voltage is zero,
the instantaneous source power is zero. Yet, there
is instantaneous power dissipation in every resistive
element in a circuit with a reactive component. Where
do you reckon that power is coming from? It is apparent
that you are now just trying to pull my leg with your
ridiculous assertions. That's too bad. For awhile, I
thought you were serious.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
On Mar 13, 7:02 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.
Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?

Agree with what? That it is the same logic? You will have to
expand on your question.

Regardless, the average voltage is zero.

...Keith

Keith;

Put your fingers across the bare wires and then tell me the results, if
you are still around that is. ;^)

Dave N

Keith Dysart wrote:
So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,

Sorry Keith, that's not the way circulators work. The
power incident upon port 2 does NOT change when the
18w generator is turned on.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
Cecil Moore wrote:
The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?

Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

I don't have time to teach you AC circuit theory which
you really do need to understand. The power function
for the reactance is of the same form as it is for a
resistance, i.e. V(t)*I(t). I haven't seen you include
that reactance power function anywhere. When you do,
you will find your "missing" power and balance your
energy equations.

The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

I see no reactance that performs this function.

Please go take an AC circuits course. That's exactly
what a reactance does. The source terminals see +j44.1
ohms of reactance looking into the transmission line.
--
73, Cecil http://www.w5dxp.com

On Mar 13, 10:33*pm, Cecil Moore wrote:
Keith Dysart wrote:
Regardless, the average voltage is zero.

No, that contradicts what you said before which
was, if it ever wasn't zero instantaneously, it
cannot possibly average out to zero.

If you could kindly point out which of my previous
writings you misinterpreted to be say this, I will
gladly correct your misunderstanding.

...Keith

On Mar 13, 11:48*pm, "David G. Nagel"
wrote:
Keith Dysart wrote:
On Mar 13, 7:02 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.
Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?

Agree with what? That it is the same logic? You will have to
expand on your question.

Regardless, the average voltage is zero.

...Keith

Keith;

Put your fingers across the bare wires and then tell me the results, if
you are still around that is. ;^)

Dave N- Hide quoted text -

- Show quoted text -

The average of a sine wave is 0. Positive half the time,
negative half the time, sums to 0.

Perhaps you are confusing average with root-mean-square?

...Keith

On Mar 13, 10:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

You've got to be kidding - EE201.

I see no reactance that performs this function.

Have you no ideal what the +j44.1 ohms means?

This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

Since you have not done so, I conclude that you can't
find it either.

...Keith

On Mar 13, 11:22*pm, Cecil Moore wrote:
Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.
I see no reactance that performs this function.

Huh??? When the instantaneous source voltage is zero,
the instantaneous source power is zero.

So true.
Yet, there
is instantaneous power dissipation in every resistive
element in a circuit with a reactive component. Where
do you reckon that power is coming from? *

Fairly obviously, the equations I have presented show
that it comes from energy stored in the line.

Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.

It is apparent
that you are now just trying to pull my leg with your
ridiculous assertions. That's too bad. For awhile, I
thought you were serious.

Completely serious, I am.

...Keith

On Mar 14, 12:49*am, Cecil Moore wrote:
Keith Dysart wrote:
So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,

Sorry Keith, that's not the way circulators work. The
power incident upon port 2 does NOT change when the
18w generator is turned on.

You need to read more carefully. I made no statement about
the energy incident upon port 2, only about the energy
flowing into port 2, which, after the 18 W generator is
turned on, is
Pcp2(t) = 32 + 68cos(wt)

...Keith