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“Subject To losses”?That’s your explanation?Power is power. V*IWhatever is lost after that is subtracted from the total.It doesn’t change the total energy involved.At what size does a static-electric generatorEqual to the power of a conventional steam-electric generator?And how much “energy” has to be put into each to achieve this power?“Static” is only static until you allow it to move.In the same manner that the potential energy of a battery is “static”.The same can be said of any capacitor.The condenser can dump 100% of its energy at one time.The ‘actual current’ is a large spike. The ‘average current over time’Is the value commonly used, this is why it is often micro amps in quantity.Instantaneous current is all induction cares about.This is an important distinction when dealing with high frequencies.You can melt steel with a ‘static’ discharge.To do this with a conventional generator connected to an arc welderRequires exponentially greater quantities of energy.The math is available for both of these situations.If we take an absolute value for the energy required, it becomes apparent thatOur conventional generator and arc welder are less than 1% efficient.
Formula 2) The Charge on capacitor Q = C* V_source*[1-Exponential(-t/RC)] This is again great secret for those who can see. Notice how the charge (ie. the energy) overtime is defined in relation only the voltage of the source , R , and C . No mention of current here because it doesn't affect the charge on the cap. Only voltage does. You don't believe it ? look again at the formula. The overunity secret is right there.
I have tamed the static.Some of it anyway. http://www.youtube.com/watch?v=f-aP7sk48jw
You are right if you are talking about a closed circuit. The total energy should be from the source if the environment factor remains normal.
Would be nice if it so easy, but u are wrong.Formula 4) The time taken to fully charge a cap from a voltage source is 5*R*CYes, formula is true but in your schematics R is very low like a short circuit. So I=U/R, that means a huge amount of current.....
The charge TIME is to be consider for the power transfer.. Look at scopes shot of a disruptive high voltage trough a spark gap and the RAISING TIME occupy most of the screen.. The relation between the L1 and L2 is directly related to raise and surge of L1 because L2 has it`s capacitance and is acting 180 degree trough the curent cycle... This will give the L2 a more sinusoidal wave given a full amplitude.
You lower the charge time by increasing the voltage (even way beyond the rating of the cap) and reducing the R. this means you have to pick low ESR cap. I won't go into the formulas and calculations to prove it but you can test and see.
You lower the charge time by increasing the voltage (even way beyond the rating of the cap) and reducing the R. this means you have to pick low ESR cap. I won't go into the formulas and calculations to prove it but you can test and see.You will probably ask how come I raise voltage way beyond the rating of my cap ? won't that bend the plates of the cap and it fries and becomes a deadshort ?! well No as long as you are discharging it as soon as it fills up. That's what the spark gap does.
Applying too high of a voltage to a cap will NOT bend the plates. What is does is punch a hole in the dielectric and short out the cap. If the current is high enough it can cause the dielectric to burst into flames and make the cap explode with a very loud bang. And the voltage applied can damage the cap before it fills up. As has already been pointed out to you at least a couple of times you really need to do some basic research before you start expounding theories about OU. Then you would realize that you don't really understand the formulas.
Ok your a skeptic who only belies what an expert official tells you right!