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Author Topic: Increase the potential energy without any energy  (Read 17580 times)

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #75 on: September 26, 2017, 01:07:21 PM »
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

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Re: Increase the potential energy without any energy
« Reply #75 on: September 26, 2017, 01:07:21 PM »

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #76 on: September 26, 2017, 01:18:46 PM »
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

Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #77 on: September 26, 2017, 02:55:22 PM »
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 :D


Vidar

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Re: Increase the potential energy without any energy
« Reply #77 on: September 26, 2017, 02:55:22 PM »
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Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #78 on: September 26, 2017, 05:22:47 PM »
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.

Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #79 on: September 26, 2017, 06:36:52 PM »
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

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Re: Increase the potential energy without any energy
« Reply #79 on: September 26, 2017, 06:36:52 PM »
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Offline activ25

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Re: Increase the potential energy without any energy
« Reply #80 on: September 26, 2017, 07:13:06 PM »
Low-Q, webby1, Magluvin: please, speak about my device because if you mix electricity and mechanics (or other example in mechanics) it becomes impossible to read. I'm agree the energy is conserved with standard capacitors and Magluvin didn't understand something. But anyway, come back to my device, I count all the energies. And again, the theorem of Noether cannot be applied here because the function is not continuous. Could you be precise, where in my thinking, I'm wrong ? The energies X, Y and W are known and without a white disk the energy is conserved, the springs lost a potential energy but the walls give an energy, I calculate with integrals and the sum of energy is conserved without the white disk. But with the white disk, the sum of energy is not zero, it is 'e'. Now, be specific, and try to explain where I'm wrong please:

1/ move in/out spheres need an energy ?
2/ rotate the disk around the magenta point needs an energy ?
3/ I can't recover the energy from the difference of length L1/L2 ?
4/ the potential energy is not decrease by the same value 'd' at start and at final ?

Because, there are no other possibility.

Like there is no mass, there is no delay to transmit the pressure, so for you if the question 1/ and 2/ need and energy, could you explain ?
I don't know why the question 3/ could be true.
I calculate the sum of length for all springs at start and at final and it is the same (even it is logical with geometry), so I don't see why the question 4/ could be true.

NB: change the length of a spring don't need nor give an energy because it is like a molecule of water inside a glass of water on Earth, the molecule is attracted by gravity but others molecules around give Archimede force, so the sum if zero. It is the same with the springs and spheres. It is verified by calculation, without the white disk the sum of energy is well conserved and I take in account only the potential energy of the springs and the work from walls.
 

Offline activ25

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Re: Increase the potential energy without any energy
« Reply #81 on: September 26, 2017, 08:30:31 PM »
You might be surprised at the connection between mechanical and electrical.
No, I'm not.

This is wrong,, you keep the same quantity of spheres. 
And what is the problem ? The container keeps the same volume, the number of spheres is the same, the white disk keeps constant its volume. All volumes are constant.

Same quantity within a smaller volume simply means more pressure,, same quantity larger volume less pressure.
  The volumes are constant.

webby1: the springs have no volume, again, to simplify the calculations.


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Re: Increase the potential energy without any energy
« Reply #81 on: September 26, 2017, 08:30:31 PM »
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Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #82 on: September 26, 2017, 09:10:11 PM »
I must figure this out. I can't sleep at night with a puzzle like that riding my brain all night :D


Vidar
]

The cap to cap is a well known exercise that has been misconstrued for some time claiming that the 50% loss is due to heat by way of resistance.

Here is the capacitor energy calculator....
http://www.calctool.org/CALC/eng/electronics/capacitor_energy

10uf 10v is .5mJ   

10uf 5v is ,125mJ

2 10uf caps(20uf) at 5v is .25mJ  Total of 1/2 the original energy in the single 10uf cap at 10v .5mJ

We lose half the energy dumping a full cap into an identical empty cap of the same capacitance value.

Now how I started thinking about it to come to my conclusion that the resistance plays no part in the 50% energy loss as many were claiming, was when they went ahead and said that if the caps were ideal, zero resistance, superconductive per say, that we would end up with 7.07v in each cap after the cap to cap and the voltage was leveled out between the 2 caps.  If we did the electron count, it would be impossible to get 7.07v in each cap from a dump of 10v in the initial cap. Cant happen just like we can get 2 buckets of water of 7.07 gal each from a single 10 gal bucket.

1 10uf 7.07v is .2499mJ   not .25 exactly because the 7.07 is rounded off from the actual figure. But you should get the drift.

20uf(2 10uf caps in parallel) at 7.07v is .499mJ  again would be .5mj if the .707 were not rounded off

So resistance or zero resistance, we still lose 50% of the original total energy of the 10uf 10v cap by dumping half of the charge into another same value cap.

Mags

Offline Low-Q

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Re: Increase the potential energy without any energy
« Reply #83 on: September 26, 2017, 10:01:40 PM »
]

The cap to cap is a well known exercise that has been misconstrued for some time claiming that the 50% loss is due to heat by way of resistance.

Here is the capacitor energy calculator....
http://www.calctool.org/CALC/eng/electronics/capacitor_energy

10uf 10v is .5mJ   

10uf 5v is ,125mJ

2 10uf caps(20uf) at 5v is .25mJ  Total of 1/2 the original energy in the single 10uf cap at 10v .5mJ

We lose half the energy dumping a full cap into an identical empty cap of the same capacitance value.

Now how I started thinking about it to come to my conclusion that the resistance plays no part in the 50% energy loss as many were claiming, was when they went ahead and said that if the caps were ideal, zero resistance, superconductive per say, that we would end up with 7.07v in each cap after the cap to cap and the voltage was leveled out between the 2 caps.  If we did the electron count, it would be impossible to get 7.07v in each cap from a dump of 10v in the initial cap. Cant happen just like we can get 2 buckets of water of 7.07 gal each from a single 10 gal bucket.

1 10uf 7.07v is .2499mJ   not .25 exactly because the 7.07 is rounded off from the actual figure. But you should get the drift.

20uf(2 10uf caps in parallel) at 7.07v is .499mJ  again would be .5mj if the .707 were not rounded off

So resistance or zero resistance, we still lose 50% of the original total energy of the 10uf 10v cap by dumping half of the charge into another same value cap.

Mags
You haven't lost it. You just havent spent all yet as useful energy.
I can agree that it is a confusing experiment.
1. What is the initial PE?
2. What is the loss of PE after both have the same charge?
3. What is the difference from 2. after emtying both caps?


What is the sum of the answer in question 2 and 3?
The sum is the same as initial PE in the first cap.


Energy is conserved.


Vidar (sorry for polluting your thread active25)...

Free Energy | searching for free energy and discussing free energy

Re: Increase the potential energy without any energy
« Reply #83 on: September 26, 2017, 10:01:40 PM »
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Offline activ25

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Re: Increase the potential energy without any energy
« Reply #84 on: September 27, 2017, 06:37:27 AM »

If you place the sphere back in at L2, so that you can see\use the change in energy of the spring as compared to L1 , you would need to elevate that sphere up and over the white disk raising the spring potential higher than at L1 and then allow it to reduce down to L1 as the wheel rotates.



I recover an energy from the springs because L1>L2, is it ok for you ? for all spheres I need to move in/out.

When I move out a sphere of the container, the pressure is exactly the same than the sphere I move in. Look at the identical lines of pressure, they are always perpendiculary to the springs, and the springs change their orientation.


If then you were to use that energy difference between L1 and L2 you would need to replace that energy in order to move the sphere around and back to the "place" it started from.
No, you misunderstand the potential energy stored in the spring and the energy I recover/need to move out/move in  a sphere. The length of the spring is a potential energy recovered. I recover an energy when I move out a sphere, exactly the same when I recover an energy when I move out an object from a side of a container full of water under gravity. And I lost an energy when I move in the object. But here, the pressure are symmetric.

Offline activ25

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Re: Increase the potential energy without any energy
« Reply #85 on: September 27, 2017, 07:33:02 AM »
I think you need to look into this part deeper.
Explain how you see the pressure please, because the springs simulate gravity. So, the lines of pressure are perpendiculary to the force of attraction with gravity, tou are ok with that ? So, with the springs it is the same.

You are not using the correct reference frame,, you do not have all of the energy in and out accounted for,, mainly that that is being put into your system, the deformation and or rotation is an input.
Your sentence is like, you're wrong, I don't know where, but you're wrong.


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Re: Increase the potential energy without any energy
« Reply #85 on: September 27, 2017, 07:33:02 AM »
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Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #86 on: September 27, 2017, 07:34:34 AM »
You haven't lost it. You just havent spent all yet as useful energy.
I can agree that it is a confusing experiment.
1. What is the initial PE?
2. What is the loss of PE after both have the same charge?
3. What is the difference from 2. after emtying both caps?


What is the sum of the answer in question 2 and 3?
The sum is the same as initial PE in the first cap.


Energy is conserved.


Vidar (sorry for polluting your thread active25)...

You tell me. ;) Ive laid out what I claim. I dont have 2 months to convince you like I had to do when I last made the claim. It took that long for that session to sink in with other top guys here. Once they actually took notice they finally agreed and it has been settled here. I made the claim several times before that and was ignored over bout a 2 year stint. Reread what I wrote here about this and do some tests and calculations yourself. If you find fault in what I claimed then just simply prove me wrong. I put it to you here in a more detailed manner than I had done to convince others before because its ingrained in my mind and can describe it in better detail now than then. So put your thinking cap on and work it out, and if you can, provide convincing argument that what I said is not correct. Then Ill respond with a rebuttal.

You want to question me on Pe without giving me your version of that. Show me how and what I said is wrong and give me your numbers and how you came to arrive at those numbers and I will respond.

Mags

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #87 on: September 27, 2017, 07:41:09 AM »
P.S.


The water in the 2 tanks has the same weight as the first tank,, but now it can only fall 1\2 the distance that it could of fallen from the first tank.

To correctly use the water tank analogy for the cap to cap, the tanks need to be on the same level, not one above the other. Doing so with tanks at different levels offsets the balance that would be inherent with 2 like caps. The water needs to balance to equal parts by leveling out just like the caps. Having one water tank at a different height, that divided balance cant be had.

mags

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #88 on: September 27, 2017, 09:19:58 AM »
You tell me. ;) Ive laid out what I claim. I dont have 2 months to convince you like I had to do when I last made the claim. It took that long for that session to sink in with other top guys here. Once they actually took notice they finally agreed and it has been settled here. I made the claim several times before that and was ignored over bout a 2 year stint. Reread what I wrote here about this and do some tests and calculations yourself. If you find fault in what I claimed then just simply prove me wrong. I put it to you here in a more detailed manner than I had done to convince others before because its ingrained in my mind and can describe it in better detail now than then. So put your thinking cap on and work it out, and if you can, provide convincing argument that what I said is not correct. Then Ill respond with a rebuttal.

You want to question me on Pe without giving me your version of that. Show me how and what I said is wrong and give me your numbers and how you came to arrive at those numbers and I will respond.

Mags

One more thing.  My actual claim is that resistance is not the cause for the loss. But you are arguing the basic fact that there is no loss in the cap to cap. So you have a ways to go with understanding the basic point of the exercise. We lost it by reducing pressure that resulted in no work being done during the cap to cap connection of a charged cap into an identical cap at 0v.  A very similar case would be to have 2 12v car batteries. It may not be a 50% loss as a battery is a bit different than a cap.
If we have a fully charged 12v battery and a fully drained battery of the same make and model then we can calculate how much work can be done with the fully charged battery and use it till it is all the way drained to get the result, then we take the fully charged battery and connect it to the fully drained battery, you will not end up with the same amount of total energy of the 2 batteries combined as you could get out of the single fully charged battery.

There are a lot of sites and articles that will give you the same results for the cap to cap 50% loss. Most all will say the "unanimously agreed upon loss" is is the fault of resistance. On your point of there being no loss, that is an argument that just about anyone that works out the numbers will argue against you till you break. And if you dont end up realizing it for what it really is, then you must not agree with the capacitor energy calculators, nor any of the equations of formula that all arrive at the real and actual lossy conclusion. When we first discussed this when it was brought up in a big discussion about caps, most of us were just like you are now. Denial without full understanding. But once actual test were done, and the energy calculations were made over and over, the loss became very real. So like I said, show me your numbers and how exactly you arrived at those numbers that you say prove your reasons for believing that there is no loss and I will respond with where you are making your mistakes and correct them for you.

in the real world, doing the cap to cap as described is not a function used in any electronic devices. Its not a useful circuit in the least. it is a losing proposition. It is only a learning tool.  We do not want this circuit in any products because of the inherent loss.  One more time.......



If we have a 10uf cap at 10v and we use that to do work till its 0v, then we can measure and calculate the total energy used and the amount of work that can be done.



1)  If we used the energy from the 10uf 10v cap only till it is drained down to 5v and we disconnect out load, then we did a certain amount of work with that usage. Now we are left with a 10uf cap at 5v.

2)  So now say we have 2 10uf caps and each has 5v. Now lets just use one of them to do work till it is 0v and leave the other one at 5v.

In both cases 1 and 2, we end up with 1 cap with 5v. Are you going to tell me that the work we did with the first cap beginning at 10v till it was drained to the 5v level then disconnect the load, that the amount work that the cap provided during that drain down to 5v is not more work done than the second case where we did work from 10uf cap at 5v starting and drained down to 0v???  If that is your argument then you need to work on your basics of V*A=P.   If the load were a resistor and we repeated the work done tests 1 and 2 and we measured the heat generated by the resistor in each case, do you believe that in both cases the resistor produces the same 'amount' of heat for each case???  P=V*A is not a credible equation here? The higher voltage during the case 1 drain is not always throughout the test more than the voltage in case 2 during its drain into the resistor? ??? Think man think!

Mags
« Last Edit: September 27, 2017, 01:04:23 PM by Magluvin »

Offline Magluvin

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Re: Increase the potential energy without any energy
« Reply #89 on: September 27, 2017, 05:14:09 PM »

Well,, I did not say they were,, BUT the WATER is!


mgh,,, interesting formula,, CoM of the water column is, when uniformly shaped, at the midpoint in the height of the column,, so a 1m tall tank has its CoM at 0.5m,, if you take the water above the 0.5m mark and dump it into the second tank :)


The P.S. part is after all is said and done the CoM is now lower and if it is mgh the same m with less h at the same g means less stored.


Funny thing that I can balance things together that are not even close to being in the same system except for the shared force to balance,,,,,  :)


Do you realize there is the same kind of loss if you charge the cap from a fixed DC voltage?  unless you charge it my way anyhow :)

"Do you realize there is the same kind of loss if you charge the cap from a fixed DC voltage?  unless you charge it my way anyhow"

Well i might not agree there. If the 2 caps are the same value, then there is a total 50% loss. If 1 cap is say 10uf 10v and the other is 1uf at 0v and then we do cap to cap. We get 9v each. Now calculate the energy left in the 10uf 9v and calculate the energy in the 1 uf 9v then add them together. The loss is significantly less. So if we think about the power supply as a very very large cap 10v and we charge a 1uf 0v to the ps 10v, wouldnt that in comparison to what I just explained be very similar with not that much of a loss? Would seem to me that the power supply lost nearly as much as the cap gained.  The loss happens hard in the identical cap to cap because the source cap lost a lot of pressure(V) resulting in part of the whole loss, where the power supply if solid would not. The 10u 10v to 1uf 0v cap to cap, the 10uf did not lose a lot of pressure so its energy level is still quite high, and then we add that to what energy level in in the 1uf 9v to find the end total.

I really cant see 50% loss in charging a cap from a power supply that simulates basically an infinitely large cap compared to the receiving empty cap. And I might say that the loss in charging a very tiny cap in comparison that the losses would be very minuscule.

Mags

 

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