ENERGY AMPLIFICATION

[XR IX]

The biggest problem is making limitations and not seeing the potential from the larger picture. You see a capacitor, and that's all. But a capacitor can also a coil, and coil can be a capacitor. A plate is also a conductor.

For Example: Look at a Tesla transmitting coil and receiving coil. They are coils, but they are capacitively coupled, ergo two plates of a capacitor. You see the Earth as ground, but it also is a conductor, ergo it can maintain a standing wave (as Tesla said).

We see energy as quantity, but it most likely infinite, as the motion is in the universe. The big bang set everything in motion, and like your hand outside your car window we only harness it when we have the ability to interface the difference (i.e., slow it down).

My Theories anyway...

[pauldude000]

If we charge a 100uf cap to 100v and connect it to a 100uf cap with 0v we end up with 50v in each cap, we lose 50% of the total energy that was in the first cap to start. Calculate the energy total in 2 100uf caps at 50v each. Where did the energy go?
Mags

Very good question, Mag. A full half of the energy is lost and no discernable work has been done, in that it is not dissipated in the form of heat or emr. Half of the energy available within the system jut goes "poof", so to speak. These are the types of questions I like as the electrical energy had to be transformed, but into what? Not electrostatic field, as that is what is being measured. Not electricity, as the voltage has lowered not raised. The available E has become 1/2E no matter how you look at the system.




For those who do not know how to do the math:



E=.5CV^2 (energy in joules stored as an electric field in a capacitor equals one half of the capacitance in farads times the square of the voltage in volts)




The first 100uF (.0001 F) is charged to 100V, so  E(joules) = (.5 X .0001) X (100 X 100) = .00005 X 10000         = .5 Joules total stored energy


The second  has two 100uF charged at 50v each, or 2((.5 X .0001) X (50 X 50)) = 2(.00005 X 2500) = 2 X .125  =  .250 Joules total stored energy




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The problem you have here is that it is not a closed system. The capacitor you charged is system 1. When connected in parallel with the second capacitor thus equalizing charge you have system 2. When they are disconnected from each other to check voltage of each individual you have yet system 3. Something is happening there, no doubt, as it indicates something not identified in which the extra lost energy was stored or converted in system 2. I think the energy was probably stored, but I cannot prove that. It could have been converted into some unidentified force as well. Either way, call it system 4.

[Magluvin]

From capacitor/caps to cups :

You have two cups  in front,not caps  ,one full with water and the other empty !
Now you fill the second = empty cup half-full by the full cup with water !


Now you have two halffull, or one halfempty and an other halffull cups :


I do not think that somebody has to explain where the water from the first cup became lost !?
100% same water content,in two cups now as 100% capacitance in Joule/Farad in 2 capacitors now !

Well that understanding is part of the key to where it went.
Think about it this way....  1 full cup and 1 empty cup. In the end we still have a total amount of water to fill 1 cup....Same with the electron charges in the caps... In this instance we should say, how many more electrons are on the negative plate of the first cap at 100v, then how many are there in total on the 2 caps negative plates?(in total excess electrons compared to a cap with 0v)

Where the work or energy happens is in the action of lifting the full cup to the brim of the empty cup and pouring the water into the empty cup till both are half full.. What useful work did we accomplish? In this case it most likely took more than 50% of the total energy of the stored water of the first cup by having to lift it in the first place, where as a cap, the higher voltage of the first cap has nothing to do with the height physically of the full cap and the empty cap. Its voltage is already at a high point electrically.  So to associate it better, with the water and containers we would need them on level surface with a hose and valve at the bottoms of each container to be equal to the cap function. With that config, we could say that if we ran a small water pump to run a small gen and the valve and hose were inline, then whatever we got out of the generator during the evening out process from one container to the other, then we could say we got some work out of the transfer rather than just moving water and losing energy in the process. So in the cap to cap process we basically wasted the 50% in the process because we did nothing with the transfer other than release the initial pressure into a container 2 times as wide..  And supposedly when we have the 2 containers half full, the amount of work that can be done is 50% less than the 1 full container can do.
Interesting, isnt it?
Mags

[XR IX]

Where the work or energy happens is in the action of lifting the full cup to the brim of the empty cup and pouring the water into the empty cup till both are half full..

This is fun topic. So I think we have to recognize the difference in potential to move to a neutral state. The two 50v caps produced at discharge, doesn't retain the potential the single cap had at 100v. The word "discharge" is the relevant idea. Though I used the example of water, you have to understand the relationships that are and aren't the same. Returning to a lower state is probably flows like water, raising it to a higher level is like winding a spring. If you ever popped a cap, you know what happens when you wind the spring beyond it's breakdown voltage. The capacity is reached when the charges accumulate on the plates, and the plates get full. The charge rate is not linear, because as you place more like charges, they resist being placed there - the action is like compression.

When you discharged the first cap, into the 2nd cap, the energy to bring it to the higher potential was lost, as you can't return the the energy from the 2nd cap into the 1st cap because they are at equal states. You need to wind the spring (build the compression) of either cap to allow it to flow to a lower state capacity.

[lancaIV]

#8552 :
Peak and average !
You have a bottle full with water ,open without cap :
you turn the bottle slowly : water comes out
you turn the bottle fast : no(very low) water comes out
action/reaction time !

Water with 80° ,60°,20°,0° C,0° F,0°K degree(s) : temperature and movement freedom degrees linear/rotatory/translatory( viscosity,Engeler units)

This is in a capacitor(spring : Lord Kelvin analogon) similar

Scaling f.e. Richter or Beaufort : never linear,ever dynamical : scale down plenitude(-x),scale up magnitude(+x)

The (non expected accelerated) movement in nano-tubes !