Idea about capacitors charge and amperage?

[Magnethos]

I have seen in youtube that anyone can charge a capacitor with pure voltage. That means that you don't destroy (discharge) the dipole. I have seen also a circuit to charge a capacitor in a microsecond (200 Volts I dont know the Farards). And I have thought... when you charge a capacitor using the common method (here I'm not talking about free energy principles)... The battery draws current? Or only voltage? I think batteries draw current + voltage in common conditions to charge a capacitor, but I have to ask it.

QUESTION B
Imagine that the dipole only draws voltage to load the capacitor... and the capacitor charge is 4000VDC 30uF.
When I discharge the capacitor.... The capacitor only draws current+volt or only voltage?

* If the device only consumes 12 Volts... Can I build a device to discharge the 4000 Volt capacitor little by little and have the energy for more time? I mean if I have a 4000 Volt Capacitor loaded and I want use only 12 VDC... I will have energy for more time?

[amigo]

Here's my semi-educated take on it...

My understanding is that when capacitor is connected to the power source, current "flows" strongly at first but slows down as voltage catches up with it (current is leading voltage 90 degrees). I put "flows" because current does not really flow through a capacitor, the capacitor plates are statically charged and the current which appears to flow is a displacement current going back and forth.

Theoretically, to slow down the discharge of the capacitor you would need to affect the movement of the electrons from the negative plate, affecting the path they travel towards neutralizing the charge on the positive plate. If electrons are current, then putting an inductor in their path creates that opposition to change in current flow (until magnetic field is built up in the inductor). But then you have a series resonant tank and are producing oscillations...plus you will have energy stored in the magnetic field of the inductor in the form of a collapsing field.

Regarding the 4000V rated cap, the voltage indicates a breakup point of the dielectric makeup of the capacitor. When the rated voltage is reached the dielectric will break and capacitor will short.

What you seem to be looking for is high capacitance (Farads) not high voltage, so that enough energy (electrons) is stored in the electric field between the capacitor plates to last sufficiently enough to drive your 12V load. This will all be governed by the capacitor discharge rate, which exponentially drops the current as voltage drains away.

Here's a page explaining charge/discharge in simple terms: http://www.coilgun.info/theorycapacitors/capacitorcharging.htm but I'm sure you can find more complex explanations yourself.

[Magnethos]

uhmm... ok.
So I have understood that when you charge a capacitor, the capacitor loads with pure voltage/static electricity. So, we're not drawing amperage from the battery. And we know that a device can work with the load from a cap. So, we could draw only voltage from a battery to load the cap and then discharge the cap into a device. Using this technique we could get unlimited free energy. I have seen some videos in youtube explaining and showing this technique and it works!
Of course, they use a special circuit.

[gyulasun]

uhmm... ok.
So I have understood that when you charge a capacitor, the capacitor loads with pure voltage/static electricity. So, we're not drawing amperage from the battery. And we know that a device can work with the load from a cap. So, we could draw only voltage from a battery to load the cap and then discharge the cap into a device. Using this technique we could get unlimited free energy. I have seen some videos in youtube explaining and showing this technique and it works!
Of course, they use a special circuit.

Hi,

Respectfully, the text above  I changed red colour is simply not correct.  When you charge a capacitor that has no any charge initially, the current draw from the source is determined by the source inner resistance and the ESR of the capacitor (ESR= equivalent series resistance).  Normally a source's inner resistance like a battery is well in the some ten or hundred milliOhm range and a good high uF capacitor also has an ESR in the same order.  Hence the current in the very first moment you start charging up the capacitor can be in the order of several Ampers!  And of course this peak current will decrease exponentially to near zero as the capacitor takes up the charge and finally the voltage across it will be the same as that of the source. 

See many pages on this like this one here: http://www.coilgun.info/theorycapacitors/capacitorcharging.htm

And see this thread from Sandy showing a COP of 1.2 merely on discharging/charging capacitors in an interesting way (that is a special circuit if you like):
http://www.overunity.com/index.php?topic=4419.0

rgds,  Gyula

[CTG Labs]

Hi all,

I think many people have thought about this!  Some caps will charge in what appears to be an instant, thats just beacuse the time constant is short and of course current must flow from the battery.  Voltage lags current in to a capacitor, but it can be so fast that it appears like nothing has happened, but really of course electrons have flowed from the negative battery plate in to the cap, then some of the other side of the cap have travelled to the positive plate of the battery and will continue to do so until the cap reaches the battery voltage at approx 5 x tc.

Of course Bearden would have us believe that you can potentialise a circuit then switch the source away before current flows from the source.  But the relaxation time is something like 0.00000000000015 seconds, far to fast for even the most cutting edge high speed mosfets and switching circuits.

If you charge a cap in 100mS and you measure energy stored in the cap, then that is how much energy you have taken from the battery plus some resistive loss from the wire, no FE here?!


Regards,

D.