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 🙁
... 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 🙁
Attachments
...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
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
...Arizona... must be something like midnight...
OK: Access to bed granted.
And dream with your wife
, not about caps 
OK: Access to bed granted.
And dream with your wife
, not about caps 
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...
Choc,
aaaarrggghhhhhhh!
Something about no resistance, infinite current, no inductance bugs me.
Consider the problem chronologically reversed... where does the energy COME from.
aaaarrggghhhhhhh!
Something about no resistance, infinite current, no inductance bugs me.
Consider the problem chronologically reversed... where does the energy COME from.
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.
BTW, 2nd Law makes the reversal unphysical.
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
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 said:
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 !
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.
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.
SY said:
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).
Hmmm
I shall have to think about this.
Hopefully not for 7.5 million years.
Bedtime here, back tomorrow.
Cheers
Rob
I shall have to think about this.
Hopefully not for 7.5 million years.
Bedtime here, back tomorrow.
Cheers
Rob
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
😀 😀 😀
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
😀 😀 😀
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.
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.
SY said:
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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