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Old 10-13-2017, 02:15 PM
dragon dragon is offline
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Join Date: Jul 2009
Posts: 964
Once again we jump to the conclusion that the output is higher than the input... I don't believe this is the case. When we charge a cap it takes a certain amount of time, during that time you have a peak energy exchange at the beginning and as the cap fills the voltage rises, reducing the potential difference and the current flow decreases - this gives us an average energy from start to finish.

Then we have the resistive load which is said to be "using" the energy. I tend to question the amount the resistive load actually uses or converts vs what is generally wasted during the neutralizing process of an energy source such as a battery. The ohms law and BTU standards don't always coincide with each other so in order to explain the phenomenon we introduce a COP value that allows us to acceptably bend the rules without really breaking them.

If you take a higher potential and dump it into a lower potential ( thus storing the energy not neutralizing it ) the resistance between them acts as a regulator of time - so then what is really lost?

If you take a fully charged capacitor of a given value and dump it into another capacitor of equal value that is not charged you loose 1/2 of the energy. But... if you charge both caps one higher than the other the losses are reduced drastically and the resistance in line plays a very small role in those losses while balancing the pair.

As an example I'll use the 1F caps - one charged to 24 volts and the other to 12 volts. The 24volt cap contains 288 joules the 12 volt cap 72 joules. If we connect them in parallel they will balance at around 18 volts each. Each now contain 162 joules. We lost 36 joules of energy during the balancing - with or without a resistive load this proves to be accurate. The exchange shows a 10% loss between them, considerably lower than the 50% loss of the first example - one charged one not charged.

Now consider as an (extreme) example the same caps charged to 112 volts and 100 volts using the same potential difference as the first example. We have 6272 joules in the higher potential and 5000 joules in the second a total of 11,272 joules stored. They would balance at around 106 volts each containing 5618 joules or a total of 11,236 joules, again a loss of 36 joules because of the 12 volt difference but the actual loss of energy is less than 1% - with or without the resistive load. 6272 joules - 5000 joules = an exchange of 1272 joules... is it possible to drive a 1200 watt resistive load with only a 1% loss?? This is what I'm trying to find out...

The circuit itself isn't anything special - not overunity nor even unity - simply a circuit that allows us a small view of the possibilities.... if your willing to look beyond what's on the surface and question how and why the accepted rules apply.
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Last edited by dragon; 10-13-2017 at 02:43 PM.
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