MH's ideal coil and voltage question

[Magluvin]

Yeah, the ideal cap with infinite capacitance is a neat 'ideal' idea. But lets say it is at 0v. It would never charge to any particular voltage above 0v. It would be an infinite ideal load. It would be far from falling under a definition of a capacitor in any form.

So if we could buy one, it would need to be precharged.    

But I understand the comparison to an ideal voltage source. I also understand that if we have an ideal voltage source, we would be just considering it it at its face value of being a source and as to where its power comes from doesnt matter. That would be another hangup for me being the screw ball that I am. Its all sorta fun to think about for a bit, but I dont believe spending great amounts of time on the subject is necessary to 'advancing' any projects here. I just find it to be a huge distraction. How many pages on that alone here? Tires me to think about it any longer.

Im just perplexed as to any need to think of things 'ideally' and that it would help me with any of my projects. For me it doesnt help me understand inductance any better. It doesnt help me understand caps any better. Im just not getting it.   Just like I dont get if we have 2 ideal caps, one at 10v and 1 at 0v, that the result of going cap to cap would be 7.07v per cap, ideally. I still stand by what I believe on that, until someone can physically show me that the results would be such. On paper only, and according to 'laws' just doesnt get me there in the least. And none of what has been posted here has shown me any different.

When I say we just lost 'pressure', basically that is what we lost.  50% doing the ideal cap to cap.  I find that the 50% loss due to resistance(of any value above 0ohms) is a prick in the science as to conservation of energy and resistance was given as the culprit by way of heat loss. Especially when it is 'claimed here' that 1uohm is 'seamless' to being ideal, by Mh and backed up by Poynt, but also claiming that .000001 picoohm is still responsible for the 50% loss from cap to cap. Just doesnt work for me.... Like why wouldnt there be some ledge of very very slight resistance that would give us say 6v in each cap? Or 5.1v??? Or 7.06v? ?

Ideal. No resistance. No heat.  So 100% of the 'ideal energy' used to charge the cap to 10v was 'converted' into 'pressure' of opposite charge in the caps plates measured in voltage.  So if we decide to diversify or better yet divide that 10v pressure of that cap between 2 caps of the same value, then we have lost pressure, as Ive said, stupidly. Ideally or not.

Mags

" I find that the 50% loss due to resistance(of any value above 0ohms) is a prick in the science as to conservation of energy and resistance was given as the culprit by way of heat loss."


We couldnt have anyone knowing that heat may be had for free in some way, could we.

Mags

[verpies]

And what would happen if this charged ideal capacitor of infinite capacitance,has an inverted voltage placed across it,from an ideal voltage source?.

Infinite current would flow.

[verpies]

Yeah, the ideal cap with infinite capacitance is a neat 'ideal' idea. But lets say it is at 0v. It would never charge to any particular voltage above 0v. It would be an infinite ideal load.

More like an ideal short.

Ideal. No resistance. No heat.  So 100% of the 'ideal energy' used to charge the cap to 10v was 'converted' into 'pressure' of opposite charge in the caps plates measured in voltage. 

Not 100% because of EM radiation.

So if we decide to diversify or better yet divide that 10v pressure of that cap between 2 caps of the same value, then we have lost pressure,

Yes, but neither pressure nor voltage alone are energy.

[poynt99]

Just like I dont get if we have 2 ideal caps, one at 10v and 1 at 0v, that the result of going cap to cap would be 7.07v per cap, ideally. I still stand by what I believe on that, until someone can physically show me that the results would be such. On paper only, and according to 'laws' just doesnt get me there in the least. And none of what has been posted here has shown me any different.

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

When I say we just lost 'pressure', basically that is what we lost.  50% doing the ideal cap to cap.  I find that the 50% loss due to resistance(of any value above 0ohms) is a prick in the science as to conservation of energy and resistance was given as the culprit by way of heat loss. Especially when it is 'claimed here' that 1uohm is 'seamless' to being ideal, by Mh and backed up by Poynt, but also claiming that .000001 picoohm is still responsible for the 50% loss from cap to cap. Just doesnt work for me.... Like why wouldnt there be some ledge of very very slight resistance that would give us say 6v in each cap? Or 5.1v??? Or 7.06v? ?

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.

[MileHigh]

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.