The secret to Overunity

[Tajerek]

We can also charge a capacitor inductively by applying a potential to one
plate.
The other will induce an equal and opposite charge
as a function of the distance between them and change in voltage.


Charging a capacitor without a load has no measurable current.
it is so small it can only be calculated,
You can scope it with an adequately high-frequency oscilloscope.
and we see a sharp spike at the front, and a decrease in current
inversely proportionate to the voltage.
Current at 0 charge is 'infinite', current at max charge is 0.
Internal resistance to a single capacitor plate increases with charge.


The large spike (across a the charging time) of current is overwhelmed
by the increasingly low current as it approaches max charge.
This will continue until the potential between the charged plate and the
induced plate equals the potential between the source and our reference.
Resulting in almost no current at all.
Current is a function of time.
The time is so short it destroys the equation.
It is approximately 0 current.


Hope this helps you understand.

That's what I was trying to say. And explained in my previous message with formulas that you need current from the source only in proportion to the circuit resistance. If you remove the resistance component you literally don't need current to charge a cap. It is the voltage pressure across the plates that causes the charges that are already on the plates to be displaced, when the charges are displaced there appears current as a consequence. Current is not the cause of the charging cap but the consequence. The current doesn't come from the source but is created by the cap when it is charging, and the cause is the voltage.
Obviously we cannot have zero resistance that's why small current is still needed, but in theory a perfect cap doesn't need current to charge but only voltage.

As example Take a high voltage cap of 1uF and ESR=0.2 ohm  it will charge up in 5*R*C as known by time constant formulas (this is science not me claiming) . It means you fully charge it up within 1 micro second ! And you charge it up to 63% in 1/5th of a microsecond. The charges are displaced from one plate to the other so fast that you see large spike of current but that's not really drawn from the source but displacement of charges from 1 plate to the other. Again it is the resistive component of your circuit and cap that's consuming current/power.

Now anyone who tells me that cap doesn't charge fully in 5*R*C then they don't know science...

The current across the cap is i=C*dv/dt  That means it is resulting from the rate of voltage change across the capacitor it is not the source current.
I hope this clears the confusion for Void who has wrong assumptions thinking that the current C*dv/dt is required from the source, while as formula clearly says it comes as consequence of the voltage increase across the cap as I just explained that is in fact the secret to OU

Hi Tajerek. It has already been pointed out to you that you are making major mistakes in
your assumptions and interpretations. The total charge on a capacitor at a given point in time
is equal to C x V, yes, but that in no way means or implies that to build up that charge on
the capacitor that no current has to flow. You are showing that you have no understanding
of what the formulas represent.

There is no possible way that you have tested this, because if you did do some actual testing
you would quickly realize that what you have been saying here is false. To charge a capacitor
requires a flow of current (flow of charges). The formula which you yourself have posted for the capacitor
charge current clearly shows that the higher the applied voltage, the higher the initial capacitor charging current will be.

[Void]

That's what I was trying to say. And explained in my previous message with formulas that you need current from the source only in proportion to the circuit resistance. If you remove the resistance component you literally don't need current to charge a cap. It is the voltage pressure across the plates that causes the charges that are already on the plates to be displaced, when the charges are displaced there appears current as a consequence. Current is not the cause of the charging cap but the consequence. The current doesn't come from the source but is created by the cap when it is charging, and the cause is the voltage.
Obviously we cannot have zero resistance that's why small current is still needed, but in theory a perfect cap doesn't need current to charge but only voltage.

Hi Tajerek my friend. You keep saying the same thing, but unfortunately you seem to be unwilling
to give due consideration to what people are telling you in response. You ignoring where people
are pointing out how you are wrong is not helping your situation at all.

Based on your circuit diagram, you appear to be talking about charging a two plate type capacitor
from an rectified DC input voltage source. The applied voltage is the 'pressure' (AKA electromotive force) which
causes the charges to start to move from one cap plate to the other.  This movement of charge from one plate to the
other is a current. Packing electrons into one plate from the other plate takes an expenditure of energy. This power
consumption is represented as the applied voltage times the capacitor current at any given point in time. The formula for capacitor
current which you posted clearly shows that the higher the applied voltage, the higher the initial capacitor charge current
will be for a given circuit. There is simply no denying that. Just look at the capacitor charge current formula.

This capacitor charge current will fall off in an exponential curve which is accounted for in the cap current formula
by the exponential component in the formula. Nothing you have shown or described in your circuit diagram
would seem to do anything to bypass this normal type of capacitor charge behavior. Anyway, you have made it clear
that you are not interested in listening to reason about this, as this has been pointed out to you several times already,
so I'll leave it at that.

In order to have a chance of seeing OU, it would seem a setup would have to be doing something very out of the ordinary.
I see nothing out of the ordinary in your circuit setup, and your basic premise is clearly at odds with the formula
for capacitor charge current. In other words, you are not making sense.

All the best...

[Tajerek]

Hi Void , you missed the point you are thinking in the traditional way of conservation of energy which leads to "overunity doesn't exist".
All your assumptions are based on a false premise that "if there's work it means there's equivalent expenditure which it has to come from the source" that negates any chance of believing in OU every existing. I am wondering what you are doing in this OU forum then 

I will explain where you make the mistake

The applied voltage is the 'pressure' (AKA electromotive force) which
causes the charges to start to move from one cap plate to the other.  This movement of charge from one plate to the other is a current.

This is correct , I agree with you here. that there is current and I said it serveral times and in my last message. I used the term 'displacement' current. I say it doesn't have to come from the source.

Packing electrons into one plate from the other plate takes an expenditure of energy. This power
consumption is represented as the applied voltage times the capacitor current at any given point in time.

This is a misunderstanding. It applies to resistive loads not capacitors. Capacitors do not perform any work and don't consume power. Do your research. https://www.crazyengineers.com/threads/how-to-calculate-power-in-capacitor.12638/

The formula for capacitor
current which you posted clearly shows that the higher the applied voltage, the higher the initial capacitor charge current
will be for a given circuit. There is simply no denying that. Just look at the capacitor charge current formula.

that formula says i=c dv/dt and it doesn't say the higher the voltage the higher the current  it links the current in the cap directly to the rate of change of the voltage of the source. This current is NOT the current of the source like you think it is.
If it was like you think it means the source voltage and the source current will have to be linked by this formula which is absurd. to explain the absurdity i will take example of a linear rising voltage at a rate of Const*t. since we are using very high voltage to Const is very high as voltage will jump from 0 to high value across the cap when it is charged. such as 3000v in 1 sec. Then Const = 3000
take a cap of 5 uF as example.

i=C*dv/dt  = 0.005* d(Const*T)/dt = 15A

This means when you charge a 5uF cap with 3kv it draws 15A. But what if you are using a voltage source of 3kv and 100ma ?? does it make sense that the cap will draw 15A from 100ma source ?? that's absurd.

The real answer is that your source remains 3kv 100ma and STILL you will see the current shooting up to 15A but it is NOT related to that 100ma of the source. That's the secret of excess power i've been trying to explain

This capacitor charge current will fall off in an exponential curve which is accounted for in the cap current formula
by the exponential component in the formula. Nothing you have shown or described in your circuit diagram
would seem to do anything to bypass this normal type of capacitor charge behavior.

current will go down exponentially so what? that's not what my circuit is trying to show. my circuit is an illustration that you can draw small power to store high energy into a cap, then covert that electric energy into useful form.

[blueplanet]

I agree.

There have been a lot of real overunity devices around. There have been a lot of great ideas posted on the internet. Unfortunately, these real devices attract no attention. Whenever you disclose a self-running device that has been running for several decades, the ordinary people will look for hidden wires underneath the carpet. The credibility of free energy science has gone down to almost zero.

Instead, the devices which never fail to attract public attention are those overunity spark gaps, 5MW devices, overunity resistor, abnormal supercap, etc, etc. Some individuals are even trying to release disinformation to the public in a way to discredit the science of free energy.   The self-running benchmark will certainly clear a lot of doubts.

Hi blueplanet. Yes, being able to demonstrate a self-sustaining setup is pretty much the
benchmark for OU claims now, given all the many false claims and setups showing
very obvious improper measurements over the years. It is easy to make claims, but it is
a whole different kettle of fish to be able to back up those claims with a self-sustaining setup. 
This quickly separates the real from the fantasy.

P.S. If someone can show that they get a very much larger output power than input power
then that is not anything to sneeze at, but the next logical step would be to try to loop it and
make it self-sustaining if possible, as a truly self-sustaining setup is something that is pretty
hard to argue against, outside of the possibility of plain fraud.

All the best...

[Void]

Hi Void , you missed the point you are thinking in the traditional way of conservation of energy which leads to "overunity doesn't exist".
All your assumptions are based on a false premise that "if there's work it means there's equivalent expenditure which it has to come from the source" that negates any chance of believing in OU every existing. I am wondering what you are doing in this OU forum then 

Hi Tajerek. Your statement that I am thinking "overunity doesn't exist" is a complete fabrication on your part.
I have never said anything even remotely like that, and I have pointed out that I have done experimentation
in this area for years. Please do not put words into my mouth that I have clearly never said.

This is a misunderstanding. It applies to resistive loads not capacitors. Capacitors do not perform any work and don't consume power. Do your research.

Good grief man. You are mixing apples and oranges. You posted a link to a capacitor in an AC circuit (sinewave) 
which does not consume power, that is correct, but I was clearly talking about the power   
consumed from the high voltage DC power source when charging the first capacitor in your circuit
diagram to a DC voltage.  Very obviously I wasn't talking about a capacitor in an AC circuit.
Sorry, but for you to mix up the difference between charging up a capacitor with a DC voltage (which is what
we have been talking about all along based on your circuit diagram) and how a capacitor behaves in an AC circuit
with a sinewave applied to it shows that you have no idea what you are talking about.

that formula says i=c dv/dt and it doesn't say the higher the voltage the higher the current 

The formula I am referring to is the formula for the charge current of a capacitor when a DC voltage source
is applied to it to charge up the capacitor, which you yourself have posted previously:
I = (V/R)  x e^-t/RC
This is the formula for determining the charge current in a capacitor that is being charged with a DC voltage V.
As you can see this formula contains Tau (RC). This is the formula you must use for determining the capacitor charge current
during the 5 Tau charge period when charging from a DC power source V. That you did not realize something this basic again
shows that you do not know what you are talking about.

As you can see from the formula, the lower 'R' is (in your circuit there is no actual R, but there is a diode which
has some losses plus the ESR of the capacitor), the higher the initial charge current will be, and this charge current
will fall exponentially from the maximum as the capacitor charges. So, definitely, the higher the source DC voltage
is, the higher the initial capacitor charge current will be, and it will fall off exponentially from there over the period of 5 x Tau.

By the way, when you are charging a capacitor from a high voltage DC power supply, which usually have
fairly low current capacity, typically what happens is the output voltage of the high voltage DC power supply
drops when it is charging the capacitor due to the high internal resistance of the high voltage power source, so
the actual cap charge current in this situation will be limited by the internal resistance of the power supply. However, 
clearly power is consumed from the high voltage DC power supply when charging the capacitor. It doesn't make any difference
what the actual DC supply voltage is, power will still be consumed when charging a capacitor to a DC voltage.
This is very, very basic stuff.

Tajerek, sorry man, but you clearly have no idea what you are talking about. For absolute certain power is consumed from
the high voltage DC power source when the capacitor is being charged up with this DC voltage applied, regardless of the
applied voltage.  Anyone with even the most basic understanding of capacitors should be aware of this. 
Enough with the nonsense already.