Re: Is it FREE ENERGY or what?
- From: wolfenhawke@xxxxxxxxx
- Date: Sun, 3 Feb 2008 19:17:34 -0800 (PST)
On Jan 31, 4:07 am, swisscash <tiberiugrab...@xxxxxxxxx> wrote:
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:
http://www.cheniere.org/techpapers/Final%20Secret%209%20Feb%201993/in...
First, I urge you to try it and write your comments after, please.
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.
.
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