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Author Topic: Silly question about capacitors  (Read 17229 times)

Offline Magluvin

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Re: Silly question about capacitors
« Reply #60 on: February 16, 2014, 10:30:42 PM »
Energy is lost as heat. Condesators heat up dramatically at high current. Look at induction heating - condesator  in LC tank heats up dramaticaly. Polyester condesators with no electrolyte heats up! Currents in that very moment are enormous and so energy losses are. Your setup looks like Colpitt oscillator with very low inductance(and straight wire have inductance). Current jumps back and forth maybe and million times(some miliseconds) before come to stop.

If you read my previous post with the analogy of 2 air tanks, then consider this....


2 caps of the same size, 1000uf with 5v and another 1000uf cap with 0v. We connect them together till each has 2.5v, and we have lost half the energy, and of the loss, it is said to be lost in heat.

Now lets take the same 2 caps, one with 5v and one with 0v. Now we set it up with a very low ohm switch that is controlled by a circuit that opens the switch when a voltage gets to 2.5v or lower.
So between our 2 caps, we include our switch device and an inductor of say 10mh(which can be had with pretty low ohm resistance) in series with the switch. 

The circuit that detects a voltage at or below 2.5v will monitor the source cap. We close the switch and when the circuit opens the switch, we will find that the 2 caps have very close to 2.5v each, with near 0v across the inductor just before the time of switch off. But the inductor has a stored amount of energy that is near equal to what looks like we have lost during the transfer from on cap to the next.

In the end, the 2 caps have the same voltage using the switch and inductor, as we did if we just connected the caps directly.  Now, there is resistance in all things, the caps, the coil, the switch, the wires. Well where are our heat losses when using the inductor? Sure the inductor slows down the transfer, so less current, but over a longer period of time using the inductor. So there must be heat losses while using the inductor, right?

Well after all that, say the 2 caps have no resistance, and say no inductance, just to avoid ringing, what voltage would you say would be in each cap when we connect them together? More than 2.5v each????  No losses would mean what? 3.5v in each?  I cannot agree.  Lets say we consider electron charge.  2 empty caps, 0v each. The have say an equal amount of electrons on each plate. We take 1 cap and we pull electrons from one plate and push them onto the other plate. For simplified example we say 5v has pulled 5000 electrons from one side of the cap to the other. So now 1 plate is missing 5000 electrons and one plate is over loaded with 5000 electrons. When we connect the empty cap to the charges cap, how do you think that imbalance of electrons is going to be divided up between the 2 caps? Do we lose any electrons in the exchange?? Would we gain any electrons using an inductor?????? ::)   Can we not see that in order for there to be a balance in the end, the empty cap, one plate will take on 2500 electrons from the overfilled plate of the source cap, and the other empty cap plate will give 2500 electrons to the source caps depleted plate. So now each cap has a plate with an over flow of 2500electrons and the other 2 plates have a depletion of 2500 electrons, giving each cap 2.5v. Why is that so hard to comprehend? Any other way would have to add or take way electrons from the caps from an outside source. But no, we end up with half the voltage in each once the imbalance, balances out.

And my contention is that we lose that energy between cap to cap by simply not putting it to work during the transfer. Heat, what ever. Lol, that heat is a spike, just like the spike that many claim is just about nothing when it comes to a field collapse of an inductor, when we consider the time period that it happens. So why is this spike such a great loss, when they consider the field collapse to be such a little thing. ;)   

And the reason why there is less loss with a direct transfer from a tiny cap to a large cap is the fact that once things balance out, the amount of energy in the tiny cap is mostly all in the large cap, and can be used, as such from the large cap, just a different voltage range. Less current, longer time.

Along with the large cap to the smaller cap, the large cap doesnt lose much because the tiny cap only took little.  Same can be said with the air tanks. ;)

The only heat that I would consider the loss would be the heat generated by charging that source cap to full, then wasting it by depressurizing into another container of equal size without taking advantage of the action of the transfer by using it to do something of value. ;) ;D In the end, there is no more or less total air in the tanks(total), nor is there in the caps(total), at any given point. ;)

Mags

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #60 on: February 16, 2014, 10:30:42 PM »

Offline Vladokv

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Re: Silly question about capacitors
« Reply #61 on: February 16, 2014, 11:15:08 PM »
I understand what you say and will try to elaborate where you are wrong.
I don't like air tank analogy because is lossy, so I will stick to electronic.
First assume that LC tank have zero resistance. Gives implication ->
1. Then we can not avoid ringing. Not possible.
2. Where is energy? -> 25% one cap, 25% another cap and 50% in inductor field.
 both caps have 2,5V and there is present magnetic field in and around inductor.
3. What happens when inductor return field to zero? -> one cap have 5V and another 0.
 After full cycle again first cap have 5V and another 0.
  Also to mention when caps have equal voltage it is 2,5V and field of inductor




Offtopic note: inductor will slowly lose energy by emitting radio wave, if everything else is from superconductive material

Offline Farmhand

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Re: Silly question about capacitors
« Reply #62 on: February 16, 2014, 11:42:08 PM »
No one mentioned charge till I did.

And since the total charge is conserved if the action were to conserve energy rather than charge then the total charge would need to increase. As Mags I think pointed to. Just because the first post was talking of energy being lost does not mean someone cannot point out that the charge is conserved. Just because the first post was about lost energy does that mean some people seem to think that mentioning that the charge was conserved is making some claim or something ?

I think the fact the charge was conserved is relevant as for the reasons I gave above. If the action described in the first post did anything less than lose half the energy then charge would have to be increased I think and that is a relevant factor towards explaining why it must be that way in that example.

I still say if you want half the charge in each cap then just charge the two caps, if you want the potential energy in one cap then do that.

Like the brick analogy, if we make one big stack then decide we want two small ones we can't just push over half the stack and hope it lands in a neat stack half the height, bricks will break, so we must remove them and make the second stack by hand the same way the first on was made, and it takes energy to undo it, then re-stack the second small stack.

So the moral of the story is if you want two caps charged to 8 volts then just do that. Don't push stuff up hill then carry it back down again.

Cheers

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #62 on: February 16, 2014, 11:42:08 PM »
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Offline Farmhand

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Re: Silly question about capacitors
« Reply #63 on: February 17, 2014, 12:32:18 AM »
sorry, I was not clear enough. Take two non-polar capacitors in paralell (maybe would be good to use two electrolytic to form non-polar to have bigger capacitance from start and lower leakage to air), charge them to the same voltage  (because of being parallel), then disconnect and connect them in series and measure. what is going on ? energy rise ????

To explain it without all the hoopla, if we have one 10000 uf cap at 16 volts it has 160 Microcoulombs of charge and 1.28 Joules of energy.

Now to the question I think Forest asked.

A 10000 uf cap at 8 volts has 80 Microcoulombs of charge and 320 Millijoules of energy.

If we put two 10000 uf caps in series the capacitance is then halved so we end up with a 5000 uf capacitor charged to 16 volts, which has 80 Micocoulombs of charge and 640 millijoules of energy. 

Charging the two caps in parallel then putting in series conserves both charge and energy because no actual transfer was done, the connection is assumed to cost us no energy, as in it cost no energy to just put them in series. However the cap is half the size, double the caps to 20000 uF so we end up with a 10000 uF capacitor charged to 16 volts and we need to double the work we need to do to charge them.

A 20000 uF capacitor charged to 8 volts has 160 Microcoulombs of charge and 640 Millijoules of energy, two in series is a 10000 uf Cap charged to 16 volts and has the original 160 Micocoulombs of charge and 1.28 Joules of energy.

Cheers

I guess it is always very good to know the formula for things, but I have to say that programs like the "Electronics Assistant", are powerful tools if used intuitively, they save time and don't usually lie. They can even teach us if we use them and practice the calculation till we get it right.

EDIT: Fixed typo.
..

Offline forest

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Re: Silly question about capacitors
« Reply #64 on: February 17, 2014, 09:50:35 AM »
sorry, I'm not good with formulas I tried but without assuming  both laws of conservation (energy and charge) I got silly results. When you assume energy is conserved then something weird is going on with charge when such capacitors are re-arranged from parallel to series connection according to maths IF you do not assume conservation of charge but simply the capacitance values halved , probably the other way is what Farmhand presented - if you do not assume energy must be conserved but charge is conserved then you got weird results with energy stored and so on..... only when you assume both and disregard possible capacitance change you got the same charge and energy in both situation. Sorry, it's probably due to my lack of deep understanding of electrical science....


Please , can you find a bug in my conclusion ... :
Let Ci be a capacitance of one capacitor (both are identical)


When connected in parallel and charged to U1:
Q =Ci*U1 (of one capacitor)
C=2*Ci
E=0.5*(2*Ci)*U1^2=Ci*U1^2
Q=2*Ci*U1 (total charge stored)




when you change connection to series  capacitance is changed and I only assume that energy is the same and look for the final charge :


now capacitance is:


1/C=1/Ci + 1/Ci => C=0.5*Ci
Energy is the same so


0.5*(0.5*Ci)*U2^2=Ci*U1^2 => U2=2*U1 (as we can predict voltage rised 2 times)


Q=(0.5*Ci)*U2 =(0.5*Ci)*2*U1=Ci*U1




but originally that was Q=2*Ci*U1


where is the error ?  :(




There must be an error somewhere ...  ???

Free Energy | searching for free energy and discussing free energy

Re: Silly question about capacitors
« Reply #64 on: February 17, 2014, 09:50:35 AM »
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Offline MarkE

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Re: Silly question about capacitors
« Reply #65 on: February 17, 2014, 12:00:10 PM »
Forest, what is throwing you off is that when you connect the two capacitors in series, charge does not transfer unless there is a current path around the loop.  So, if the first was charged before, it retains that charge and voltage difference, and ditto for the second capacitor, including for 0V charge on either.

 

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