Increase the potential energy without any energy

[Magluvin]

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

"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."

Absolutely not.  You are missing it.

The 10v down to 5v has a great advantage over 5v down to 0v. 

The 10v to 5v still has the potential of the beginning output of the 5v when it gets down to 5v, And the 10v to 5v is working on potential above 5v throughout the use, where the 5v only has 5v down to 0v potential during its use. Run the numbers.  10uf 10v holds 2 times the energy of a 20uf cap at 5v. Yep.

In the cap to cap, we lost 50% doing the cap to cap.  There are many online calculators for caps. Run the calculator to calc the energy of a 10uf cap at 10v, then run it for 20uf at 5v(equal to total 2 10uf at 5v) You will see the loss. You can do more work with a 10uf at 10v than 20uf at 5v. Check it out for yourself. Its not what you think. Its the same for water tanks and gravity and the same as air pressure tanks. You would need to end up with 7.07v in each cap for the 2 caps combined energy to equal the energy of 1 cap at 10v. Run that on the energy in a cap calculator online or work out the equation yourself. Well by the electron imbalance count we cannot end up with 7.07v each. Just like we cannot end up with 2 buckets of water with 7.07gal from 1 bucket that had 10gal.

Its been a big discussion on and off here. It is what it is.  Even Woopy said Opah! Couldnt believe the loss. Its been through the ringer.  In the back of my mind I couldnt see the resistance as the loss, considering the electron count evidence.  If you could count how many electrons you take from one plate of the cap and shove it into the other plate, you can determine the resulting voltage charge of a particular cap value, every time. More electrons taken from one plate and forced into the other plate is stored pressure(voltage) If you know the exact capacitance and the voltage charge on it, you can use Coulombs law to figure the electron imbalance of the plates. And that number will be the exact same if the cap is charged to exactly 10v after discharging it to 0 and charge back to exactly to 10v. Each cap value will have its own electron count imbalance across its plates for 10v charge.  A tiny cap will need less electron count differential than a large cap, just like an air tank, less air molecules to get a small tank to 100lb than a large tank to get to 100lb.

Some claim that charging a cap say in a output of a power supply has this loss. Its not true. If it were then we would not have power supplies that approach 100%eff. We are mostly just topping off an output cap in a supply, not replenishing from 0v.  Only at start up of the supply where the cap is actually 0v is there a possibility of these losses, and even then the new slow start supplies take care of that issue with step charging till optimum output is achieved.

Mags

[Magluvin]

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 ?

The energy loss in the cap to cap scenario can be associated to the energy it took to move the electrons/water/air from 1 container to the other one till leveled out. That is the only thing I can think of as to where the energy went. But, we could have used that energy in the transfer from cap to cap to do something and still end with 5v in each cap.

But the saying that the energy is lost in the cap to cap is due to resistance and heat is not true. It is only a valve set to a particular position between the 2 tanks to slow the transfer down, but in the end all of the electrons leveled out just like water or air and we lost the energy by not using the energy that occurred in the transfer to do other work in the process. We just released potential pressure into a tank that is basically 2 times as large. We lost it stupidly as I say, unless the goal was to divide that pressure/water/air into 2 containers for what ever reason. That is a lossy goal.

Mags

[Low-Q]

The energy loss in the cap to cap scenario can be associated to the energy it took to move the electrons/water/air from 1 container to the other one till leveled out. That is the only thing I can think of as to where the energy went. But, we could have used that energy in the transfer from cap to cap to do something and still end with 5v in each cap.

But the saying that the energy is lost in the cap to cap is due to resistance and heat is not true. It is only a valve set to a particular position between the 2 tanks to slow the transfer down, but in the end all of the electrons leveled out just like water or air and we lost the energy by not using the energy that occurred in the transfer to do other work in the process. We just released potential pressure into a tank that is basically 2 times as large. We lost it stupidly as I say, unless the goal was to divide that pressure/water/air into 2 containers for what ever reason. That is a lossy goal.

Mags

I must figure this out. I can't sleep at night with a puzzle like that riding my brain all night


Vidar

[Low-Q]

So if we have two tanks. 1x1x1m each. One filled with 1000 l of fluid.
That full tank have a potential energy of 500 units with reference to the ground because center of the mass is 0.5 meter above ground.
Then we open the valve so 500 liter fills the other tank.
As the waterlevel rises in the other tank, the remaining potential energy in the first tank drop both because the level is decreasing, and the initial reference level for the other tank will rise.


In reality, the potential energy in the first tank is only 250 units because we choose to fill the other tank, and not poor the fluid on the ground.


The remaining potential energy in each tank will finally reach 125 units each. 250 units combined. The same as the initial potential energy in the first tank. Energy is conserved.


Does this make sense? Will this apply to the capacitors as well?


Vidar.

[Low-Q]

I just claimed that no energy is lost during the filling of the other tank.
The initial scenario is filling up a 2.nd cap with voltage.
While doing that from a 10V charged cap, the PE of the first cap will decrease. Not only because the voltage drops in that cap, but also because the reference voltage is rising in the other cap that is being charged.
The average PE of the first cap is therefor 1/2 of the PE that could exist if we discharged that 10V cap to zero. But we are not. It is discharged to 5V while the second cap is charged to 5V. The sum of discharged PE and the remaining PE is the same as the fully 10V charged single cap.


I just confirmed that energy is conserved.


Vidar