Energy Loss

[kvholio]

This one humbles me. thanks

Klaas

[SY]

But R is implied when the work word comes up...

To quote mikeks, no.

Look, think of two parallel plates in a vacuum between and a potential between them. The vacuum is perfect and so are the plates.

Now, place a test charge in the space between them. Move the charge toward one of the plates with the same sign potential as the test charge (e.g., for a negative test charge, move it toward the negative plate).

Takes work, doesn't it?


Al, your analogy is quite apt. Of course I mentioned it several pages ago...

[poobah]

How is that relevent to the missing energy? No one is debating how energy is stored in a cap.

[SY]

The energy isn't "missing," it's lost to the work needed to move charges out of a field and into another, along with the changes in the original field due to the charges that cause the field being moved. The motion can be lossless, but if the field that a charge sees changes, the energy changes.

I think I explained this in pretty gory detail in some earlier posts.

[poobah]

Agreed... completely... 100%

So in what form does that energy leave the system? Does not work imply heat?

[kvholio]

moving a charge = work.
or am i oversimplifying ?

Klaas

[poobah]

Charge moved times the the change in voltage equals work.

Power = Volts * Amps = Volts * [Coulombs / second] = Joules / second

So... multiplying power by time to get energy,

Work = Volts * Coulombs

Put 2 Coulombs in a 1 Farad cap... we get 2 volts (Volts = Q/C) and we have 2 Joules of potentail energy in the cap (E = (V^2 *C)/2)

Now do some work... move 1 coulomb through 1 volt, so that can put it on the other cap... this is one 1 Joule of work done.

Now we have 2 caps each with 1 Coulomb and 1 Volt. The energy in each cap is 1/2 Joule.

So:

2 Joules >> 1 Joule of work done + 1/2 Joule of pot. E + 1/2 Joule of pot. E

All is well. The "missing" energy is work done in moving a charge through a field. And we did this (on paper) without a resistor.

Now, where did the energy go? It is not in the system... so it must have fled. By what means did it leave? Radiation.

[ChocoHolic]

Originally posted by SY


To quote mikeks, no.

Look, think of two parallel plates in a vacuum between and a potential between them. The vacuum is perfect and so are the plates.

Now, place a test charge in the space between them. Move the charge toward one of the plates with the same sign potential as the test charge (e.g., for a negative test charge, move it toward the negative plate).

Takes work, doesn't it?


Al, your analogy is quite apt. Of course I mentioned it several pages ago...

Well, in this situation the charge will be forced to accellerate towards one the plates.... gaining mechanical energy, as soon as it reached the plate this charge is reducing the charge difference of the plates, means voltage goes down. The energy of the fast charge particle will trnasfer to electromagnetic radiotion, when hitting hard onto the plate... like the unpleasant Roentgen radiation in TV and monitors So everything is fine, no strange things happening.
If we would push the charge slowly to the other plate, then we would need some force for it and add energy to the system be E=integral F(s) ds. As soon as the charge reaches the plate, the voltage will gp up. The added energy is noe stored in the cap.
Fine also here nothing strange is happenning.
But I think that's not the situation of our discussed issue.

Poobah:
The sentence of de l'hopital is just a tool to find the value
of mathematic expressions like 0/0 or indefinite/indefinite
I.e. which value has y(x) = (x-1) / (x-1) for x = 1 ?
De l'hopital says that this can solved by simply differentiating nominator and denominator in seperate.
In above expamle y(1) = (1-0)/(1-0)=1/1=1.
Originally I was guessing that my energy loss calculation would lead to some experessions like this ..., but before I would have needed such math, integrating the power already eliminated the resistor....
and showed that the losses are independend of R.
I did not need de l'hopital at all.

Discussion about time:

Ohhps good point! Up to now I could live with the model of the dirac pulse in my mind... But if we consider time... Also my brain starts to deliver delirious output, when running at more than 300 000 km/s.
I mean if we guess that time stops at light speed and runs backwards above light speed..
Then our cap discharge system would stabilize more ore less at currents which cause current densities that make the electrons move at light speed. At this level the system is not in trouble with indefinite short times anymore.
Ok.. sounds like we do only need a cap and a supraconductor to desgin our wharp speed engine....

SY:
It is obviously time to open a new forum.
Time has obviously ripen for DIY 'Time Machines and Wharp Engines'

[poobah]

Before the wharp engine... let's take care of the whup engine.

Somehow the destination of the "missing" energy is just conceaaling it's obvious location/destination.

Consider the same exact setup... with one little twist. Put a resistor between the caps to cause the equalization. Charge is conserved. So the caps wind up at the same voltage and energy levels as in the no res. case.

Anybody having an "ah-ha" moment here?

Now, wherever the missing energy was going in the first place, it has now decided to show up as radiated heat in the resistor.

Hmm...

Do the math... exactly 1/2 of the initial energy in the system "burns" in the resistor. The value of R drops out of the solution as we would expect.

So... the value of R isn't needed to understand the energy loss based on the now deposed SY's charge through a field approach. But the R (or L) is needed in any pratical realization.

Therein lies the "trick". No R... no L, NOT real.

Anybody want the math? Integrate the power through the resistor from t = 0 to infinity... exactly half.

[ChocoHolic]

Hi Poobah!
...sounds like you are the first one who has read my attachments to my first postings. I did exactly the approach with R inbetween and originally planned then to move in the resulting math experession the resistor to zero, but when integrating the power then the resistor is dropping out of the equation and the term is independent of the resistor... same thing as we know from hard switching devices, from MosFet gate drives and from switched cap networks.

Of course we cannot make a simple real test set up, because we cannot have a setup without any R and L. But the theoretic model
is still stunning me with losses in zero Ohms. The model with the resistor was/is pushing me to the assumption that we can loose the energy in the zero Ohms connection by an indefinite high current pulse of almost zero duration.
I feel glad that at least you now seem to understand the intesion my first posts.

Not reflecting reality: My horse sense agrees. And because of Einstein I am also not sure if our applied model is valid for the zero time situation with indefinite electron moving speed...
On the other hand we humans tend to say not real or impossible for everthing beyond our imagination. And imagination is at least partially coupled to know how. In the moment we all agree that we have no idea how to make a R-L-free circuit....

Not so long ago the earth was a flat disc.

Originally posted by poobah
Before the wharp engine... let's take care of the whup engine.

Somehow the destination of the "missing" energy is just conceaaling it's obvious location/destination.

Consider the same exact setup... with one little twist. Put a resistor between the caps to cause the equalization. Charge is conserved. So the caps wind up at the same voltage and energy levels as in the no res. case.

Anybody having an "ah-ha" moment here?

Now, wherever the missing energy was going in the first place, it has now decided to show up as radiated heat in the resistor.

Hmm...

Do the math... exactly 1/2 of the initial energy in the system "burns" in the resistor. The value of R drops out of the solution as we would expect.

So... the value of R isn't needed to understand the energy loss based on the now deposed SY's charge through a field approach. But the R (or L) is needed in any pratical realization.

Therein lies the "trick". No R... no L, NOT real.

Anybody want the math? Integrate the power through the resistor from t = 0 to infinity... exactly half.