Increase the potential energy without any energy

[Low-Q]

Well, that is not completely true. If you charge a 1000 l tank with water. The tank is a capacitor. This tank is connected to an equal tank at the same level. The connection is a tube with a ball valve at the bottom.
Now, open the ball valve, and let the water flow into the other tank. After a while there is 500 liters in each. Now you have the same total storage of potential water.
What you suggest, is that both tanks should have 707 liters in each, and asks where did the lost 414 liters of water go, and conclude with the following state: There is lost energy on the way.


The same example applies to a capacitor. It is not only the Voltage (pressure at the bottom of the tank), but also the potential energy storage. You end up with 20uF capacitors with 5V in each. That is the exact same potential energy storage as one 10uF capacitor at 10V. No energy is lost.


Vidar

[Magluvin]

Well, that is not completely true. If you charge a 1000 l tank with water. The tank is a capacitor. This tank is connected to an equal tank at the same level. The connection is a tube with a ball valve at the bottom.
Now, open the ball valve, and let the water flow into the other tank. After a while there is 500 liters in each. Now you have the same total storage of potential water.
What you suggest, is that both tanks should have 707 liters in each, and asks where did the lost 414 liters of water go, and conclude with the following state: There is lost energy on the way.


The same example applies to a capacitor. It is not only the Voltage (pressure at the bottom of the tank), but also the potential energy storage. You end up with 20uF capacitors with 5V in each. That is the exact same potential energy storage as one 10uF capacitor at 10V. No energy is lost.


Vidar

You need to look at it as energy potential(pressure) and the size of the container.  If you open the valve of the water tank that is 1000L there is going to be an amount of work that it can produce if you apply that output. Once you release some of that pressure into another identical tank and each is at 500L, there is a 50% loss. You cannot do the same amount of work with the total of 2 tanks with 500l than you could with the initial tank of 1000L  Here is why....


If you emptied the 1000l tank into another identical tank till they were both 500L, you used that high pressure(due to gravity) of the 1000l to only transfer from tank A to tank B till the equal. No work done other than the transfer from tank to tank. We could have run a gen during the transfer, but we didnt. We lost it, 50%.

So if we do work with the 1000l, the pressure is much greater for the first half of the 1000l, and the second half is the same as either 500l tank.  You get more work done with the initial 1000l.  Same as 100lb of air pressure. You will get more work done during the first 50lb used because it is still higher pressure than 2 identical tanks with 50lb.  We use half of the 100lb and we got more work done at the higher pressure, and we are left with 50lb, then compare what amount of work you get from 1 of the 2 identical tanks with 50lb ea, and are still left with 1 tank at 50lb.  Get it??? 

And to what you say of the caps, no. We dont end with 2 20uf caps.  We have 2 caps period and each are 10uf.  If we combine them we end with basically 1 capacitance at 20uf.    10uf at 10v and another 10uf at 0v. Connect them together and the total is 20uf. Disconnect them and we have a 10uf cap at 5v and another 10uf at 5v.  Its very simple. And my conclusions are validated here on this forum.

Mags

[Magluvin]

And that is how you do the electron count. 100gal of water div into 2 containers, you end with 50gal in each. The resistance is the valve and it only changes the time of the transfer to leveling out, the valve does not consume energy nor take or add water to the system. Same with the electron count, we end up with the same electron count and the resistance only changes the time it takes for each cap to be equal and the resistance doesnt add or take away electrons from the system.

So if we use the energy stored in the 10uf 10v cap only till it is down to 5v, we got more done with that first half of the 10v down to 5v than we can get from either pair of caps at 5v each


10uf 10v use energy till you are down to 5v.   So we still have 5v left

Now have 2 10uf 5v ea  and use the energy of 1 of them.  We still have 1 10uf 5v left just like above.


So the energy from the 10uf at 10v will do more work down to the 5v cutoff than 10uf 5v can do and in both situations we are left with 10uf 5v.

Mags

[Low-Q]

And that is how you do the electron count. 100gal of water div into 2 containers, you end with 50gal in each. The resistance is the valve and it only changes the time of the transfer to leveling out, the valve does not consume energy nor take or add water to the system. Same with the electron count, we end up with the same electron count and the resistance only changes the time it takes for each cap to be equal and the resistance doesnt add or take away electrons from the system.

So if we use the energy stored in the 10uf 10v cap only till it is down to 5v, we got more done with that first half of the 10v down to 5v than we can get from either pair of caps at 5v each


10uf 10v use energy till you are down to 5v.   So we still have 5v left

Now have 2 10uf 5v ea  and use the energy of 1 of them.  We still have 1 10uf 5v left just like above.


So the energy from the 10uf at 10v will do more work down to the 5v cutoff than 10uf 5v can do and in both situations we are left with 10uf 5v.

Mags

Actually not. From 10V to 5V, or 5V to 0V in the same capacitor provides the same amount of energy. The difference in both is 5V at the same capacity.
The difference is time.
If you load the capacitor with a 1kOhm resistor, it takes shorter time to discharge from 10 to 5V, and longer time to discharge from 5V to 0V - actually, it takes forever to discharge from 5V to 0V if the capacitor is perfect. The energy in both cases is equal. When you charge the capacitor through the same resistor, it takes shorter time to charge from 0V to 5V, than from 5V to 10V. In the meantime you have lost energy through the resistor during charging. The energy input is therfor higher than the potential energy in the fully charged capacitor.


Lets say for the cause of simplicity, you have a mass of 1kg you want to lift 1 meter. Lets say you spend 1 second the first half meter. Then 1 hour the last half meter. What is the potential energy between 0m and 0.5m, and between 0.5m and 1m? Not surprisingly they are equal.
In this particular case, the altitude is the charge, and the mass is the capacity.


Vidar

[activ25]

Mags: look at this link: http://www.smpstech.com/charge.htm

But I'm agree with you Mags, it is possible to create the energy (or destroy).

Low-Q: my document is not clear ?