Is it FREE ENERGY or what?

Do like this:

Put a source of DC or AC current with an electronical millisecond switch and a capacitor of the same power like the source with another switch and finally the load.

When the first switch is on, the other must be off and viceversa, all the time.

If you will make this circuit to run by keeping only one thing in your mind, that is of NEVER PUT THE SOURCE IN DIRECT CONTACT WITH THE LOAD, probably you shall have some FREE ENERGY.

I mean the time till your source will be off its power will be much more greater with this circuit attached than puting in direct contact the source with the load.

It's better to see the whole document here:

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First, I urge you to try it and write your comments after, please.

Reply to
swisscash
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Cannot see the paper. URL is dead.

Sorry, no free energy. The source will charge the capacitor which drives the load. So energy is still being drawn. In fact, you are using more energy than without the circuit because of the losses between the source and the load.

By the way, for very high frequency operation, the switches are not required and the same effect occurs. That is, the charged capacitor will drive a load during the load current drawing period, and recharge on the down swing. This is the primary function of bulk decouplers. Here the capacitors are used even though there is loss, becuase there is too much impedance between the source and load to supply the required power at high frequency.

Reply to
wolfenhawke

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