MH's ideal coil and voltage question

[tinman]

Infinite current would flow.

Interesting.

So we have a current flowing through an ideal voltage source,and in an instant,we try to reverse this current flow by inverting the voltage across that ideal source.
What would be the result?


Brad

[tinman]

The mechanical analogy for the two capacitor issue can be quite clear and revealing if you put in the effort in to think about it.

You have a 5 Kg mass going two meters per second on a frictionless surface that hits a stationary 5 Kg mass.  The masses stick together and become a 10 Kg mass.  So what are the energy dynamics?

I am pretty sure that most people know that momentum is conserved when these things happen.

The initial momentum is 10 Kg-meters-per-second.

The final mass is 10 Kg.  Therefore the 10 Kg mass must be moving at one meter per second to give you a momentum of 10 Kg-meters-per-second.

Now, let's look at the energy.

Initial energy:  10 Joules
Final energy:  5 Joules.

Whoops, we lost half of the energy so where did it go?   The answer is that the energy was lost in the hit itself.  It was a perfectly inelastic collision.  When the two 5 Kg masses hit you can imagine a thin layer of putty absorbs the shock and keeps the two masses stuck together.  5 Joules of energy are burnt off in the deforming putty as heat.

This is exactly the same as the two capacitors.

Here is a good question for you MH.
Why can this transfer be made more efficiently between the two caps by way of induction?

I will state for the record that i have no idea as to why,as i have never looked into it.


Brad

[poynt99]

Here is a good question for you MH.
Why can this transfer be made more efficiently between the two caps by way of induction?

I will state for the record that i have no idea as to why,as i have never looked into it.


Brad

Sorry MH, but I'm somewhat shocked that Brad has never seen nor heard of my document on this very subject. Feel free to still answer Brad though.

Brad, here it is: http://overunity.com/downloads/sa/view/down/209/
First posted about 8 years ago. A bit of clarification is in order though:

My reference to the inductor slowing down the transfer process, while true, is not the whole picture. It does that by storing a large portion of the energy, then transferring it to the second cap. A charged ideal capacitor connected to an ideal inductor will oscillate forever because there are no losses, and the energy simply transfers back and forth from one to the other. It is a similar idea with the cap to cap energy transfer. If we insert a high Q inductor (i.e. the same idea as in MH's question where I established a L/R ratio of 50:1 approaches ideal) between the caps, the transfer becomes much more efficient when compared to a pure resistor.

[Magluvin]

Haven't you done your own test with this? What was your result?


I need to correct you here. If you followed what I said, you would understand that the absolute value of resistance is not the deciding factor that establishes the resistance as "close to ideal" for an ideal inductor, it is the RATIO of inductance to resistance, and I established a baseline of 50:1 ratio for a 5% error (from ideal). If you want less error then the ratio must increase.

And yes, 50% of the energy in a transfer through pure resistance between two equal value caps is burned in that resistance, no matter how small the resistance is. Why does the value of resistance make no difference? It is a self-regulating process. We know the power burned in the resistor is P=I²R. We also know that as R decreases, so does tau (tau=RC), and I increases. And energy E is essentially the integral of power burned in the resistor over time. So let's look at two crudely-calculated examples:

1) Source cap voltage is 10V, and R=1 Ohm. C1,2=1uF (ideal). Ipeak=10A, Ppeak=100W, tau=1us.

2) Source cap voltage is 10V, and R=0.1 Ohm. C1,2=1uF (ideal). Ipeak=100A, Ppeak=1000W, tau=0.1us.

In case 1 we have 100W peak and the tau of the transfer process is 1us. In case 2 we have 1000W peak and the tau of the transfer process is 0.1us. In both cases, the total energy burned in the resistor is the same. Case two burns 10 times the power, but for 10 times less time.

This is really general and not exact by any means. The idea is to illustrate how and why the energy burned in the resistor is always the same, regardless of its value.

"Haven't you done your own test with this? What was your result?"

My results were that I get 5v per cap after cap to cap transfer. And What Im saying is that I think that even if they were ideal caps, 10v cap to 0v cap, that we would still have 5v per cap.


Now Ive heard that term 'burn' before. Even before the internet was around.  Id like a detailed explanation of that term here. What is 'it' that is consumed in the 'burn'? Energy? What form?



Here is a couple of questions for you....


We have 2 'Ideal' capacitors of the same value. One is charged at 10v and the other at 0v. 

Lets say just for simplifying the example that the 10v cap has an imbalance of 2000 electrons (pos plate -1000 and neg plate +1000) between the pos and neg plates and the 0v cap has no imbalance.

So we do the cap to cap zap. When all is said and done, we disconnect the caps. What would the voltage be across those caps? And what would the electron imbalance count be for each cap?


Now we will do the same with real world caps..... Same values as described above.....

Cap to cap zap and disconnect. What will the voltages be across those caps? And what would the electron imbalance count be in this example?


Thats as simple as I can put it. So anyone here can understand it.

If you can answer those questions, then I will continue with my explanation.

Mags

[Magluvin]

As to my prev post, Id like to see anyone have a crack at those questions.

Mags