For Woopy: Explanation for why you lose 1/2 energy in a capacitor

[MileHigh]

Okay, so the root answer to explain this problem is deceptively simple.  You know that if you short  a 1000 uF capacitor at 10 volts together with a 1000 uF capacitor at 0 volts you lose half of the energy.  But who is to say that you should lose half of the energy when you short a 1000 uF capacitor to a much smaller capacitor?  If you do the number crunching you will quickly realize that if you short a 1000 uF capacitor at 10 volts to a much smaller capacitor at 0 volts then you lose only a very small amount of energy.  So as the size of the second capacitor goes towards zero the energy loss goes towards zero.

So in the example the first event is a 1000 uF capacitor being shorted with a 500 uF capacitor.  So right away without calculating anything you already know beforehand that you are not going to lose 50% of the energy in the first step.

So now going forward you should be much more confident that the result of the simulation is correct.  There is nothing here that can be worked on or used, it's another case of working with "less losses."   It's very easy to mix up "less losses" with "energy gain" and it happens all the time.

The mechanical analogy is actually very good in this case.  Imagine a 1000 Kg block of putty moving at 10 meters per second on a frictionless surface smashes into a 1000 Kg block of putty that is stationary.  When they hit they deform and stick together and become a 2000 Kg block moving at 5 meters per second.  You know that when two objects hit each other and stick together that you have conservation of momentum.

So if you rework the problem and substitute the capacitors for moving blocks of putty on a frictionless surface the results will be identical.   Obviously it's very intuitive to imagine that when a 1000 Kg block hits another 1000 Kg block that more energy is burned off in the collision as compared to when a 1000 Kg block hits a 50 Kg block.

MileHigh

[MileHigh]

Laurent:

The Falstad sim question was answered above.

For reply number 8 by NerzhDishual the real issue is what happens when you have a coil (which is what the motor looks like) between the two capacitors?   You can remove the resistor also.

With the coil between the two capacitors there is no longer a "crash" when you connect the first capacitor to the second capacitor.  A coil is like a spring in the mechanical analogy, so you can think of a spring between the two masses of putty.  They don't hit each other any more, the spring absorbs the energy and stores it, and then transfers it into the second mass.

So if you ran some simulations with a capacitor connected to a coil connected to the second capacitor, you will see an oscillation when the connection is made.  An oscillation means no energy is being lost (in the ideal case).

So that's what is suggested with the motor.  The motor stores some of the first capacitor's energy in the motor coil and also in the rotation of the mass of the rotor.  As the rotor spins down the motor becomes a generator and turns the mechanical rotational energy back into electrical energy.  So you lose less energy with the motor setup because it resembles an inductor.

MileHigh

[woopy]

Hi MileHigh

thank's for answering this way, and now i feel better regarding my problem with the " energy storing formula "

So it is a different way to see the transfer of energy , between the rigid  approach in the application of the "formula" due to the different mean of discharging a cap into  an other cap !

I mean if i short cut a charged cap in its twin uncharged cap , the formula is valuable, but if i use a different mean (as for instance the motor as per Nerzdishual proposition ) to do it, the formula is not fully applicable.

that's OK for me , now i am feeling much better concerning this problem

thank's very much for your explanation.

Good luck at all

Laurent

[MileHigh]

You are very welcome Laurent and good luck with your experiments.

MileHigh

[Magluvin]

I was thinking the same thing with the motor as an inductor. So the inductance would go down with increased motor load?

As for my lil circuit, and the standard example of cap to cap transfer, are they ever used in any case?  Or is the cap to cap example just used for explanation?

I never considered the 1.67v cap extra as a gain, but just higher eff compared to the standard cap to cap, because we are still at a loss. I think more eff is all I stated. ;]  But I did say that what if we could find more or better ways to decrease losses, it would be good.

M, I can see that we only used a bit of the discharge cap into the 2 series, and then only some more when discharged from 3.33v into the lone 1.67v cap then levels out to 2.5v each. In some way some of the losses are severed in my version.  Your right, calculations will have to be made to see where and why the losses are different. I came up with a few ideas today to try and expand on this. Ya never know. ;] Till ya try ;]

It is an odd situation with my lil circuit, aint it? ;]

Im going to try the motor idea also.

Its funny how an idea can come full circle, as it was only months ago we were on this very subject and it has come alive again.

Mags

lol did anyone see Americas Got Talent last night?  Ooooo, Ouch And they made it to the next round!! What is this world coming to? 

http://www.youtube.com/watch?v=mqxLDoe6BnI

The use of safety goggles is hilarious.