| ChocoHolic |
Hi Ho !
... just strolled through some of my old records and found a nice phenomenon, which might make fun to discuss here...
Imagine two caps.
Both have the capacitance.
One is charged (U1), the second is uncharged. The energy in the first is evidently E=1/2 x C x U1^2.
If we now connect ideal the second cap, then the charge will split half and half on both caps.... balancing at U2=1/2 x U1.
If we now calculate the energy, we will find that half of the energy is lost.. ... somehow.. each cap has just a quarter of the original energy.. Curious thing, only ideal and lossfree assumed components involved and still energy is lost.
Some years back I started some calculation on this.
I started with an Resistor in between both caps and then planned to move the R mathematically to zero.... hoping to get reasonable results with the sentence of de l'hopital or similar...
Aehem well, of course before I could move R to zero.... I came to the well know result from capacitve switching devices , that the losses are independent from impedance of the switching device. As soon as you calculate the dissipated power the resistor drops out of the calculation....
Sorry for poor scan quality. In the last millenium my docu-system was not really high end :( |
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| ChocoHolic |
...so for me this charging-changing-event looks similar to a Dirac pulse.
Means: Indefinite high current pulse, indefinite short time, well defined energy.
E = t x P = t x I^2 x R = 0 x (indefinite)^2 x 0
:clown: |
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| ChocoHolic |
In case of two identical caps:
E_loss = 0^2 x (indefinite)^2 = 1/4 x C x U^2 |
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| poobah |
Dude... I was planning on sleeping tonight... good one!
:D |
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| ChocoHolic |
...Arizona... must be something like midnight...
OK: Access to bed granted.
And dream with your wife :yes: , not about caps :Ohno: |
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| SY |
| Here's an analogy that (at least for me) clarifies things. Imagine your first situation, a cap with a voltage V and a charge Q. Now, move one of the plates closer to the other to increase the capacitance... |
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| poobah |
Choc,
aaaarrggghhhhhhh!
Something about no resistance, infinite current, no inductance bugs me.
Consider the problem chronologically reversed... where does the energy COME from.
:xeye: |
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| SY |
poobah, you can still conserve energy and charge with idealized conductors. Try answering my question first: what happens when you move capacitor plates closer together to double the capacitance while conserving charge?
BTW, 2nd Law makes the reversal unphysical. |
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| rpapps |
Sy is on the right track.
If you connect two equal capacitors in parallel the charge will be shared equally (Law of conservation of energy).
Now V=Q/C so, in a perfect world, if you double C, V halves.
No violation.
In the real world, some energy is lost due to thermal dissipation and electromagnetic radiation in the connecting wires so the end result is slightly less than V/2.
Physics 101 people!
Reference:
Physics for Scientists and Engineers
Fourth Edition
Raymond A. Serway |
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| ChocoHolic |
| quote: | Originally posted by rpapps
Sy is on the right track.
If you connect two equal capacitors in parallel the charge will be shared equally (Law of conservation of energy).
Now V=Q/C so, in a perfect world, if you double C, V halves.
No violation.
In the real world, some energy is lost due to thermal dissipation and electromagnetic radiation in the connecting wires so the end result is slightly less than V/2.
Physics 101 people!
Reference:
Physics for Scientists and Engineers
Fourth Edition
Raymond A. Serway |
rpapps:
Our trouble in the moment is that double C and half voltage DOES violate the law of conservation of energy. Half of the original energy is lost.
Which is also plausible, because if split two double C and half V the unchanged amount of charge is placed on lower potential. Less energy.
The funny thing that we are loosing the energy even with a theoretically lossfree charge changing model.
If you want to split the energy onto two caps without energy loss, you would need an addiotinal ideal choke and some ideal switches.
...resulting in a nice energy transfer and then the voltage in each cap would be something like 70% of the original volatge.
This resonant event would nicely match to the law of conservation of energy, but would somehow violate the law keeping the amount of charge constant.... charge is then more afterwards, but we did not supply external charge... strange world ! |
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| rpapps |
Sorry chocoholic, I don't see it and I can't read your scans to verify the math.
Cheers
Rob |
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| ChocoHolic |
Our discussion is based on the following phenomenon:
Imagine two caps, both 1F.
In the beginning Cap1 is charged to U1=2V.
Cap2 is uncharged:
Total energy in the beginning E1+E2 = 1/2 x 1F x (2V)^2 + 0 = 2J
After connecting the second in parallel, both caps are charge to half of the voltage:
Total energy in the end: E1+E2 = 1/2 x 1F x (1V)^2 + 1/2 x 1F x (1V)^2 = 1 J
Half of the original energy is lost. |
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| ChocoHolic |
| quote: | Originally posted by SY
Here's an analogy that (at least for me) clarifies things. Imagine your first situation, a cap with a voltage V and a charge Q. Now, move one of the plates closer to the other to increase the capacitance... |
Hi SY,
I think in your example things are less strange.
Due to opposite charge on both plates there is mechanic force, which is pulling together both plates.
If we now make one plate movable (and free of friction), we could use this force to pull something up. Mechanical energy can be calculated by increased mechanical potential energy or by integral of F (ds). |
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| rpapps |
Hmmm
I shall have to think about this.
Hopefully not for 7.5 million years.
Bedtime here, back tomorrow.
Cheers
Rob |
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| Joules |
Dude --- You keep assuming the voltage will be cut in half by adding a second capacitor in parallel, but there is no law to prove that it will.
The law of consevation of energy is real. so based on that, the problem is
> 1 capacitor - .5(1F)*(2v)^2 = 2 joules
> 2 capacitors - 2 joules/(.5(2F) = sqrt of 2 or 1.414 Volts shared on bothe capacitors.
a simple reverse check > .5(2F)*(1.414v)^2 = 2 joules
:D :D :D |
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| SY |
Y'all have to work harder on this one...
edit: ok, I'll give you yet another way to think about the problem. Your capacitor is charged, then we insert a chunk of dielectric with k = 2. |
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| ChocoHolic |
| quote: | Originally posted by SY
Y'all have to work harder on this one...
edit: ok, I'll give you yet another way to think about the problem. Your capacitor is charged, then we insert a chunk of dielectric with k = 2. |
How about answers, not questions?
I would guess that in this case the voltage will drop to 70.7%, but that's just a guess.
Let us know your view.
P.S.
No comment on my 'integral F ds' from your previous question?
At least your own answer to this question might help us.
Joules:
Besides conservation of energy, there is also a law of constant charge... Both laws seem to contradict here. |
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| poobah |
Alright SY... poop rolls downhill and only downhill.
rpapps... the rub here is that V = Q / C, & E = Q^2 / (2 * C)
In this example charge must be conserved, energy as well, no one said the enregy had to stay in the caps though. |
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| BWRX |
You're assuming the voltage will become halved by connecting an equal capacitor in parallel with the first. What you must assume is that the energy will be shared between the caps. From that you can determine what the voltage across the caps will be.
First situation with one cap charged:
U1=0.5*C*V1*V1
Second situation where an uncharged capacitor of equal value is added in parallel to the first:
U2=0.5*2*C*V2*V2
Assuming U1=U2 (conservation of energy), it follows that V2=V1*sqrt(0.5)
So if you had 2V across a single cap and added another uncharged cap of equal value in parallel the new voltage across those caps would be 1.414V. |
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| poobah |
Sorry Brian,
V = Q / C |
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| BWRX |
Yes, and your point?
Let me rewrite the equation.
0.5*C*(Q1/C)^2 = 0.5*(2C)*(Q2/(2C))^2
What does that boil down to? |
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| poobah |
Well... if the charge is halved... will not the voltage be halved?
Conservation of charge of charge and energy must both hold.
:) |
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| BWRX |
Here's what it boils down to.
(Q1^2)/C = (Q2^2)/(2*C)
which can be simplified to
Q2 = Q1*sqrt(2) |
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| BWRX |
So now we have the following equations:
Q2 = Q1*sqrt(2)
and
V2=V1*sqrt(0.5)
C=Q1/V1, right?
from the equations 2C = Q2/V2 = (Q1*sqrt(2))/(V1*sqrt(0.5)) = 2*(Q1/V1)
Is that not right, or am I missing something? |
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| [Ro]Sharky |
| The energy is lost moving the electrical charge from the source (the charged cap.) to destination (the fully discharged cap) |
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| BWRX |
| How is that possible if the capacitors are ideal? In reality you are correct, some energy will be lost to resistances and some to heat and light because there will be a nice spark if you connect a discharged cap across a charged cap :) |
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| poobah |
Hey BWRX,
No one is screwing up the math here... what has to screwed up is the assumptions being made.
Consider this: a lake and a dock at exactly the same height. Pull a bucket of water from the lake and set it on the dock. Now... we have 1 bucket-mass (B) moved an average distance of 1/2 the bucket's height (H)... ie work or potential energy... B x 1/2 H amount. Now... do it again... twice the energy (this is the final state in the cap quandry... we have 2B x 1/2H. Then... put one bucket atop the other the other. The end result is we have 2B x H (average height of mass) worth of potential energy.
Now it is easy to see why we had the add energy to put the 2nd bucket on top... and how the total energy doubled when we did so. Now return the top bucket to the dock... it gives up energy when doing so.
So in the cap world we lost half the energy... makes perfect sense. Nothing can be violated here... we just aren't seeing where the energy in our cap went.
:xeye: |
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| sawreyrw |
I don't think anyone has this exactly correct. It can be shown that the charge shared between the 2 capacitors is conserved, but energy is lost because the circuit has some finite resistance. Assume initially C1 is charged to Uinit and C2 is charged to 0. After the switch connects the two caps in parallel, the voltage across them is Ufinal.
Before the switch is closed, Q1=C1*Uinit.
After the switch is closed, Qfinal=(C1+C2)*Ufinal.
From these equations: Ufinal=Uinit*C1/(C1+C2).
The initial energy is Einit=.5*C1*Uinit^2
The final energy is Efinal=.5*(C1+C2)*Ufinal^2
The energy lost is Elost=.5*C1*Uinit^2-.5(C1+C2)*Ufinal^2.
The energy is lost in the resistance; although low it will never be zero. |
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| poobah |
Mind you... all the classical equations puke with divide by zero errors. This does imply infinity squared over zero (l'hopital and all that) which is undefined? Who remembers that stuff?
BTW... :king:SY is in Austria I believe... he's drunk already.
:) |
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| BWRX |
So basically one of the conservation rules has to break down for this to work with ideal capacitors, right? Isn't that what we showed with our equations? Just take your pick whether it be conservation of charge or energy.
Otherwise we can obviously see where the energy goes if we short a discharged cap across a charged cap.
If you don't buy that then maybe this is how dark matter is created ;) |
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| sawreyrw |
poobah,
I'm not sure who your last post was directed too, but dividing zero is avoided by assuming some finite resistance. Notice that the resistor value doesn't even appear in these equations. You caneasily solve the differential equations of the equivalent RC circuit. It's easier to solve these, if you recall that a cap with an initial voltage can be represented by a voltage source is series with the cap. If you don't like the math, try a simulation.
Rick |
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| poobah |
sawrey,
We are the same page here. The quandry is about zero resistance... which ain't real. Half the energy MUST leave the system... we are just not seeing where. I'm thinking we can't assume zero resistance.
You can't put the second bucket back down on the dock without removing energy from it.
Now, zero resistance is possible... this would leave inductance as the place for the "missing" energy. I know of nothing with zero inductance.
:xeye:
thinking out loud here... |
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| sawreyrw |
poobah,
If you wish to solve this problem assuming R and L are both zero, proceed as follows:
A capacitor (C1) with an initial voltage can be modeled and an IMPULSE of current (of amplitude Uinit*C1) in parallel with C1. Now if at time zero, this impulse is applied to the parallel combination of C1 and C2 you get Ufinal=Uinit*C1/(C1+C2). I don't care much for impulses, but the fact is that it is very esay to imagine a circuit where a very fast pulse 'look like' an impulse to a circuit with a long time constant. So if you replace the impulse above with a narrow large current spike and assume dU/dt =I/C you can easily compute Ufinal.
Rick |
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| poobah |
BINGO!
I was gettin' there mo...
:D |
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| BWRX |
| Yes, thanks for posting a link to that article moamps. |
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| poobah |
You know...
That paper makes alot of sense; and it seems to say that you can't have both zero inductance and resistance... |
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| burnedfingers |
| You guys have way too much time on your hands. |
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| poobah |
Joe,
We simply have no time here for meta conversation.
You are not required to post here.
If you make a topical reply I will find it and respond.
:D |
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| burnedfingers |
| I guess I got told...:bawling: |
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| BWRX |
Hey, some people get all charged up when they get a chance to use their mental capacity to do work! ;)
| quote: | Originally posted by poobah
Joe,
We simply have no time here for meta conversation.
You are not required to post here.
If you make a topical reply I will find it and respond.
:D |
:drink: Joe that wasn't aimed at you in particular. It was more of a joke about another person, another thread, another forum. |
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| poobah |
Sorry,
My inner child got out of control...
:smash:
:clown: |
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| poobah |
Hey I just remebered something...
"The voltage in a cap cannot change instantaneously"
Implies mandatory resistance... SOMEWHERE. |
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| BWRX |
| Tis true: i = C (dV/dt) because if voltage could change instantaneously then current would be infinite. |
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| burnedfingers |
| Just kidding:D |
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| Giaime |
Hey, my phisics professor asked me this paradox at the exham... :D
(but I already knew it) |
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| poobah |
Yep!
And all this leads NO zero R AND L... which is the premise of the paradox. And also where the equations freak-out.
I remember now calculating the energy dissipated by a resistor charging a cap... equal to the final energy in the cap, and 1/2 of that applied to entire circuit.
So... no circuit without some R or L. |
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| poobah |
I've been calculating charge and energy in caps for 25 years. It is a convenient way of solving steady state equations for switchers.
You guys suppose Choco is just hosing us???
:mad:
I was about to employ the "chewbacca defense".
Giaime,
Next time... punch your professor in the nose. |
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| sawreyrw |
| Reread posts 30 and 35. The answer is there. |
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| poobah |
I'm with you on post 30. Which alludes to the "no R or L" fallacy.
You lost me on post 35 though.
I'm not really interested in calculating anything, been doing that forever, correctly it would seem... just explaining the paradox. As far as the missing energy is concerned... you gotta burn it (R and other things) or store it (L with some E & H into the ether). When you have neither, the paradox, which is BS, arises.
parlor tricks...
now
here's a parlor trick!
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| SY |
One person has gotten this right- the Shark. The R and L stuff is a total red herring.
I'll push you a little further- charge is conserved. Energy isn't because it's not a closed system. In order to move charge around in the presence of fields, work has to be done. The sum of the cap energy and the work is equal to the original energy, so there's no 1st Law violation. All this (to me) is much clearer in my equivalent problems, but either way, work done in moving charges in a field MUST be accounted for.
poobah- I'm perfectly sober at the moment. I cannot say the same about last night. ;) |
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| ChocoHolic |
Hi Poobah !
No I am not tricking you. I am serious in this discussion.
But I do disagree a lot with the doug-smith paper.
His postualte that moving charge needs energy is correct, but this energy is then available in the charge and not lost.
We can move charge without losses. Just imagine a simple ideal resonant circuit. BTW, using a choke and two switches would allow us two transfer the energy lossless from C1 to C2... may be I will prepare a simulation.
Also the doug smith paper tells that the unit of capacitances would be a length. ...just hoping that this guy does not really believe all the stuff, which he is saying..... His formula is slightly strange from geometrics anyway and especially it is missing the natural electric field constant which is 8.859 E-12 F/m .
I do also not fully agree to post 30, because in theory we can have zero resistance. And in theory we can have indefinite high currents for indefinite short times. Normally we solve such division by zero or multiplication by indefinite terms by several math tools (i.e. de l'hopital). But in our case it is much easier, because the R is droping from the equation. For me this means the losses do happen, no matter which resistance we are putting there.
Let's make it simpler. Just one cap which is charged in the beginning and then short circuited by zero Ohms. We will again get the same phenomenon of lost energy.
Also forget about thinking if we loose the energy in some inductance, which we might suspect instead of a resistance.
If we would apply an ideal inductance instead of our zero Ohms, then we will have an ideal resonant circuit, which is ringing at its resonant frequency and is loosing no energy at all and does never stop to ring.
Burnedfingers:
Trying to understand theory is sometimes helpful to avoid burning fingers. :D So we are not wasting our time. |
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| SY |
| quote: | | We can move charge without losses. |
Only in the absence of a field. Not the case here. |
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| ChocoHolic |
SY, no.
Without field we can easily move charge without changing the potential energy of the charge.
If there is a field, then the might be a change of potential energy depending on the direction in which we move the charge. Furtheron not only the potential energy may change but also the mechanical energy (i.e. brown tube... speed of electrons).
Look at an simple resonant circuit, there we can move the charge without any losses. Theoretically an ideal resonant circuit will never stop to ring. Sometimes you will have all the energy in the cap (high voltage), sometimes you will have all the energy in the choke (high current). But we don't loose it.
Even an lossy resonant circuit allows to transfer much more charge than the Doug Smith paper would allow.
Look to the attached simulation. It is lossy due to switches, which have Ron= 1 Ohm, Roff = 1meg Ohm and further switching losses, because this primitive circuit does not work all with soft switching.
Even in this lossy arangement we are able to transfer the energy from one C1 to C2 in a way that is ending up in almost perfect match to the ideal result of law of preserving energy. |
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| ChocoHolic |
Simulation results:
(If you do not believe the results, pick the old school books, invest some hours and calculate by hand) |
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| ChocoHolic |
...uhps, now with results..
Hm, now I have to go on with my class D project... |
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| SY |
A resonant circuit is not an apt analogy because you're exchanging energy back and forth in a closed system. This is purely a matter of doing work, and the 1st Law demands that the final energy equal the original energy minus work.
Interestingly, you can also consider this problem from an entropy standpoint; think of an isothermal expansion... |
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| ChocoHolic |
Resonant Circuit:
Do we move charge within a field? Yes.
Do we have losses? In an ideal resonant circuit, No.
So we can move charge in a field without losses. |
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| SY |
| Do not confuse losses and work. |
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| ChocoHolic |
| quote: | Originally posted by SY
A resonant circuit is not an apt analogy because you're exchanging energy back and forth in a closed system. This is purely a matter of doing work... |
I do disagree again.
It would be a matter of work, if we would have losses and would generate heat (lowest form of work). But in an ideal resonant circuit... there is no work done! If we look to p(t) in each component it has equal areas in negative and positive direction. Resulting work is zero. |
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| ChocoHolic |
| quote: | Originally posted by SY
Do not confuse losses and work. |
You are right. Losses and heat are not work.
For work we would need to introduce an ideal DC motor...., the consumed energy and done work would look for the resonant circuit like lost energy. |
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| ChocoHolic |
| For the resonant circuit ist does not change anything if we would have work or heat. As soon as there is real power, not only appearant power, then we would shift energy of the choke-cap system into heat or work. But a cap and a choke do not make work, nor heat. |
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| SY |
That's correct. And that's exactly why it's a poor analogy. There's no mass and springs here, only springs. Consider a massless, lossless ideal spring, contrained on one end. Compress it a distance X. What is its potential energy? 1/2 k x2. Now butt up an identical uncompressed spring. Release the first one so that in equilibrium, each spring is now compressed 1/2 x. The energy of each spring is then 1/2 k (x/2)2, or 1/8 k x2. Two springs, so the total energy is 1/4 k x2.
What happened to the energy? Work was done compressing the second spring even though there's no mass or friction. And that work is subtracted from the initial energy. |
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| ChocoHolic |
Did we loose our original topic? :clown:
Losses in a zero Ohms connection due to indefinite high current? |
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| ChocoHolic |
| quote: | Originally posted by SY
That's correct. And that's exactly why it's a poor analogy. There's no mass and springs here, only springs. Consider a massless, lossless ideal spring, contrained on one end. Compress it a distance X. What is its potential energy? 1/2 k x2. Now butt up an identical uncompressed spring. Release the first one so that in equilibrium, each spring is now compressed 1/2 x. The energy of each spring is then 1/2 k (x/2)2, or 1/8 k x2. Two springs, so the total energy is 1/4 k x2.
What happened to the energy? Work was done compressing the second spring even though there's no mass or friction. And that work is subtracted from the initial energy. |
Your model without mass would ring with indefinite high frequency.
Normally the resonant circuit is a nice analogy to a spring-mass system.
A ideal spring mass system does not make work, if we calculate the full period.... But I agree, within one half period there is positve work and in the other one is negative work.
Analogue we could consider the same in the resonant circuit. Means
moving charge up or down , positive area of p(t) and negative area of p(t). I agree to this way of modelling. But still no losses. |
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| ChocoHolic |
...hm, would your system ring with indefinite frequency?...
No, I think it would not really ring, but it would balance in indefinite short time, with indefinite high velocities and start and stop with indefinite high accellerations...
Hey that's a cool analogy to our original system ! |
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| SY |
This is a problem in statics. Release it as slowly as you want.
edit: for you p-chem veterans, think of an isothermal expansion of an ideal gas with volume increasing from V to 2V. |
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| ChocoHolic |
... release it slowly, means keeping it in hand until balanced situation?
With this we would not loose energy, but would get well defined mechanical work out of it by integral F(s) ds .
That's boring. Would be a analogy to a mechanically loaded DC motor between the caps.
Interesting is the situation of simply releasing it and let it move freely to balanced situation. There you would loose the energy and would get similar indefinite values like in our cap-cap circuit. |
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| SY |
Yes, but that wasn't the question. :D
In a dynamical situation where you're interested in the time behavior, there's nothing to do but to set up the diffy qs and the boundary conditions and flail away. If you get singularities, it means you've picked an inappropriate model- massless, lossless springs, frictionless pulleys, and infinitely rigid rods seem to be out of stock at McMaster-Carr these days...
| quote: | | ... release it slowly, means keeping it in hand until balanced situation? |
Yep, that's one way to think about it. |
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| ChocoHolic |
It is really true, if we release it slowly, then the spring system will deliver work to the external releasing mechanism. It will always put some force (unbalance of F1 and F2) to the system. Resulting in a function F(x). Force time distance is work. Or better if force is not constant: E = F(x) dx
Nothing lost.
We only get a strange thing if we simply release it.
Do singularities really always mean that the chosen model is not valid? Why?
I really think that your model is a perfect analogy to the original question of the two caps.
Energy in starting situation is clear, balanced end situation and energy in the balanced situation are are clear. But during the transition from unbalanced to balanced we loose some energy and do not really know where. Also both systems make somehow sense, because balanced static situation is usually the lower energy situation. (Different to the ideal resonant circuit which has no balanced static situation and the system keeps ringing).
Furtheron the double spring system and the C-C system show singularities if start to solve it with the classic diffy qs.
If you look to your spring-spring system, there is a force inbalance in the beginnig. The difference of the force will start to move the system. But with zero mass you will get indefinite accelarations and velocity.
I love it !!! |
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| ChocoHolic |
| quote: | Originally posted by ChocoHolic
Or better if force is not constant: E = F(x) dx
Nothing lost.
I love it !!! |
uhps... lost the integral...
E = Integral (F(x) dx) |
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| SY |
| quote: | | Do singularities really always mean that the chosen model is not valid? Why? |
Not always, but when you're trying to calculate something simple and everything blows up, it's probably a safe assumption that you're trying to calculate something too simple.
edit: STRONG book recommendation: Shive and Weber, "Similarities in Physics." Really excellent. |
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| EC8010 |
| Why are people getting in a tizzy about this one and making incorrect assumptions? The charged capacitor has a voltage across it. The uncharged capacitor has no voltage across it. Therefore, when we connect the two together, a charge will flow from the charged capacitor into the uncharged capacitor until both voltages are equal. The charge lost by one capacitor must be equal to the charge gained by the other. The final voltage will be dependent on the relative capacitances. It is not necessary to invoke differential equations, Dirac pulses, EM radiation or even L and R. I would scribble this all down properly with equations but I no longer have paper or pen (moving house in nine days). |
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| SY |
| But that doesn't answer the original question: why is the ending energy different than the starting energy? That took some explaining. |
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| ChocoHolic |
| quote: | Originally posted by EC8010
Why are people getting in a tizzy about this one and making incorrect assumptions? The charged capacitor has a voltage across it. The uncharged capacitor has no voltage across it. Therefore, when we connect the two together, a charge will flow from the charged capacitor into the uncharged capacitor until both voltages are equal. The charge lost by one capacitor must be equal to the charge gained by the other. The final voltage will be dependent on the relative capacitances. It is not necessary to invoke differential equations, Dirac pulses, EM radiation or even L and R. I would scribble this all down properly with equations but I no longer have paper or pen (moving house in nine days). |
As you say: The charge lost by the first must be equal to the charge gained by the other.
And we made it even easier. We picked two identical caps. This leads to halving the voltage, if we follow the charge approach.
But if we calculate the energy in the beginning and the energy in the end, we loose half of the energy.
If we would chose the energy approach then the resulting voltages would be 70.7% of the original and funny thing the charge would have increased to 141%. That's the discussion.
Two fundamental approaches, which both usually would be easily accepted by most people, but here they do not match together. |
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| poobah |
Guys,
This is WAY to much much fun, but, I actually have to work today.
Keep in mind, "moving charge through a field" can mean a resistor... that's where the "missing energy" is going. This doesn't apply to caps... the field is between the plates, and there ain't 'posed to be no 'lectrons there. "But, but, but what about?" they say with theit hands flying in the air. Consider a cathode and anode... charge moving through a field... when those electrons land with their mv^2/2 the plate gets hot... hmmm. "How hot they ask", [Va-k * I] says me.
It is clear that energy leaves the system, then in what form?
Place a resistor across the two caps; you will find the missing energy. Now, choose any value of resistance; the amount of energy will stay the same. You will have to settle for 99.9% something becuase in this case the caps will take forever to charge. Somebody that can still manage that l'Hopital junk can do something more elegant. They should be able to prove that as R approaches zero the energy lost will always be 1/2 of the initial.
Go back to the basic RC charge scenario... e^(-t/RC) etc... Charge the cap... wait for eternity. The energy in the cap is exactly half of the energy applied to the RC system... sound familiar?
After 25 years, I couldn't do a differential equation to save my butt. EE's are tought diff. eq. for punishment only... it's then on to LaPlace (which I can do)
The spring thing is a suitable analogy as well as buckets on the dock. In the course of letting the springs move to the relaxed state 1/2 the initial energy is available to do work outside the system.
No R... no L... no reality... no unreal supporting math.
:) |
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| poobah |
| quote: | | If you get singularities, it means you've picked an inappropriate model- massless, lossless springs, frictionless pulleys, and infinitely rigid rods seem to be out of stock at McMaster-Carr these days... |
Frictionless... hmm... no R?
Massless... no L?
:cool: |
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| janneman |
| quote: | Originally posted by SY
[snip]poobah- I'm perfectly sober at the moment. I cannot say the same about last night. ;) |
Stop bragging, you only had a bottle and a half :cool:
Jan Didden |
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| poobah |
Looks like SY didn't finish his greens...
:smash: |
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| moamps |
Jan:
The thread is about energy loss and not energy gain. :clown:
Regards,
Milan |
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| SY |
| quote: | Originally posted by poobah
Looks like SY didn't finish his greens...
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30 seconds after this was taken, they vanished. I never let arugula go to waste. |
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| poobah |
Alright then... you can have cheesecake.
:) |
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| ChocoHolic |
Janneman in Bandung....
SY in Europe...
Is there a growing DIYAUDIO tourism? :cool: :cool: :cool:
...cheese cake, beer and gaining energy...
Hm, may be gaining energy is easier than expected and loosing energy would be better than I thought! *seriouslylookingintothemirror* :clown: |
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| ChocoHolic |
...crepe & red wine....
Envy! |
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| Pafi |
QUOTE]We picked two identical caps. This leads to halving the voltage, if we follow the charge approach. But if we calculate the energy in the beginning and the energy in the end, we loose half of the energy. If we would chose the energy approach then the resulting voltages would be 70.7% of the original and funny thing the charge would have increased to 141%. That's the discussion. [/QUOTE]
ChocoHolic!
Laws of phisics tell something about real world. In real world there is no ideal switch, so there is no violation of charge and energy conservation.
And charging method counts very much!
If you use 2-pole device between two cap, then charge conservation works because of continuity, but there is no lossless 2-pole element can equalize voltage of caps. Here is no electrical energy conservation.
If you use 3-pole network (eg. made of several switches, and an inductance) one pole connected to common pole of caps, then Q1+Q2 will change by the amount of charge flowed into common point. This doesn't violate charge conservation rule, because Q1 of a real capacitor means +Q1 in positive electrode and -Q1 in negative electrode, so sum of all charge is 0 at all time.
Generally: charge conservation only applies to electrically isolated domains, and energy conservation only applies to whole energy (motion+electrical+nuclear) of a closed system. |
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| poobah |
OK choco,
Can we really have a circuit with no R or L? That's a keystone here I think...
:xeye: |
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| kvholio |
First of all: i dont even understand half of what's been said in this thread, so please forgive my ignorance..
| quote: | | Can we really have a circuit with no R or L? |
this implies discharging/charging takes place in ZERO time.
Do the laws of physics used here still apply in this (theoretical) situation ?
Klaas |
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| poobah |
Yep... good point!
Would it not require infinite energy to move electrons (with mass) at infinite speed (speed of light... whatever)? |
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| kvholio |
afaik you're right, but It's the TIME that gives me a headache, Poobah.
What happens to time when objects travel at infinite speed ?
btw, am i still making sense :xeye:
With kind regards,
Klaas |
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| poobah |
uh... time stops when you hit the speed of light?
:xeye: |
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| kvholio |
When you exceed the speed of light the clock starts ticking backwards right ?
(btw, i must be a masochist for gettin into this :D )
Klaas |
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| poobah |
hmmmm... I don't know... my understanding stops just before the speed of light.
:xeye: |
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| kvholio |
LOL mine too Poobah.
my point is perhaps, that most people in this thread try to calculate from a real-world perspective, a situation that is "out of this world".
I tried to walk backwards from the situation that was given (didnt get far sofar :rolleyes: )
Klaas |
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| poobah |
Yep...
If you conserve charge, then you violate energy. If you conserve enegry, then you violate charge.
I think you must have R or L... :smash:
Choco was talking about L'Hopital... wonder if he has some math for us to try and remember?
Every equation I know falls apart with "divide by zero" issues.
:xeye: |
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| kvholio |
I've got two fr*ggin' wikipedia open, one with Einsteins theory of relativity, the other with the rule of L'Hopital.
the L'hopital rule.....well goes way beyond my understanding of maths (hated calculus back in school)
By the time i would understand it and know how to implement it, it would be 2046 ;)
Nevertheless curious about how this turns out.
i need a beer after this one :cheers: (think i'll take a Heineken)
With kind regards,
Klaas |
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| poobah |
2046... :D
Let's see if Choco can show us some L'Hopital...
Giaime,
What's your opinion here? |
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| anatech |
Relativity requires adjustment. Since everything is based on time, an infinetly small slice tending towards zero blows up everything.
It could be the universe's way of saying, not possible. I refuse to get a headache trying to think on this one. BTW, the plane takes off! :D
-Chris ;) |
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