ENERGY AMPLIFICATION

[Magluvin]

One more thing about the resistive load circuit.  If the energy used by the resistive load is near equal to what was taken from the source, then how did we get the same amount of energy into the cap, which is 'also' near equal to what was taken from the source, if there is sooo much 'supposed losses' in the 'resistive' load? For the first cycle of the circuit we produced heat from the source and stored 'that same energy' into the cap. 

Mags

[Magluvin]

Over 1100 reads in 24 hours. 

Mags

[forest]

Resistance is like having a lot of small holes in pipe. The way to ovecome it is by having a high pressure sharp water pulse so water have no time to escape. What do you think about this idea ?
Sure,the better is not to have any resistance....

[Magluvin]

Hmm, I just ran the circuit I described in sim and what I posted is wrong.
The peaks are the same, but the long term power is not. I have to find those files. looking right now.  There was something back then that was interesting.

Mags

[Magluvin]

Resistance is like having a lot of small holes in pipe. The way to ovecome it is by having a high pressure sharp water pulse so water have no time to escape. What do you think about this idea ?
Sure,the better is not to have any resistance....

Holes in the pipe.

Lets look at a simple circuit. A source and a resistor.  We will compare 2 values of resistor, 1 ohm and 10 ohm.

With the 1 ohm we have 10 amps of current with a dissipation of 100 watts.

With the 10 ohm resistor we have 1 amp of current and a dissipation of 10 watts.

Voltage is pressure. Current is amount of electron flow.

If the resistor is like a pipe with a lot of holes in it, then that would mean that current flow on one side of the resistor is less than the other, no? But that isnt the case.

I think if one were to think of an analogy of electrical properties, air pressure is a better choice than liquid because air is compressible, and so is electrical charge.

In the case of the resistors above, the the higher the resistance value, current flow will be less as well the source will also be less strained with heavy output.

The resistor is like a valve or better yet a generic form of a pressure regulator.

Like my nitrous bottle in my Fiero can have 1000 psi or more when it is full and at temperature. The pressure is released by a solenoid valve the to a nozzle with a preset jet that you choose that provides the added HP level increase that you desire. I run a .065 which provides near 80 HP increase for my car.  If I want less HP increase I get a smaller jet, with a smaller hole.

But as the bottle(source) with 10lb of NO is used and gradually is not able to produce 1000psi, the HP increase will be less also. Less pressure, less flow, less current. Just like the circuit above, if the source voltage is decreased, less current will flow through the resistor(jet) thus less power output in heat. So ideally a gas is a better analogy for electrical comparison. One can also consider atmospheric pressure as 0 volts where by one could simulate positive and negative pressures or +(pressure) and -(vacuum) of a source. All just for visualization purposes of simple pressures and current flows of course, as there are many electrical functions that would be difficult to emulate.

So we have to concede that the resistor is not losing electrons when current flows, it is just limiting the speed that they can flow. depending on the pressure of the source. Its like a loosely regulated power supply vs a fully regulated supply, where a loosely regulated power supplies output is not locked down to a particular voltage output. Its output can vary some when loaded or when there are differences in input voltage, more in more out.

This is where I have some conflict with heat losses as they are described.

It is shown that if we charge a cap from a source directly or through a resistance that the source delivered 2 times the amount of energy than what ends up in the fully charged cap.  Did half of the number of electrons not make it back to the + side of the source than left the - side? Did less current flow on one side of the cap than the other? Did less current flow on one side of the resistor then the other?

If we charged the cap to full from a 10v source, through a resistor, to 10v, where was half of the energy from the source really lost and how if no electron left the circuit and 'less' current flows because of the resistor yet the cap is still brought to full charge, just not instantaneously, did energy leave the system via heat.?  So what does that mean? 

It is said that due to resistance if we have 2 caps same value, 1 is full say 10v, and the other empty, then we connect the 2 caps each will have 5v but we will have lost 1 half of the energy in the process.

So what would the voltage be in the 2 caps if resistance did not exist? 5v each from 10v?

So if I took 2 10lb tanks of NO and one is full and one is empty, and I connect them together so that eventually each has 5lb, did I lose half of the energy that was held in the full tank to begin with? Well you can be sure that neither tank will reach 1000lb of pressure at temperature, There is still just as much NO total in the 2 tanks as was in the full one, but pressure is way down. The same goes for the 2 caps. the total amount of electrons in the caps, whether less on one side of the cap than the other from being fully charged to equal amounts with no charge, when connected together, we dont lose any electrons, weve just moved them and lost pressure stupidly by letting the contained pressures of one full tank expand into a container of twice the size.

Its what I call stupid losses. We didnt do anything with the pressure while it was expanding to the other tank. We just let it depressurize. If we were to use that flow from one tank to the other via turbine, pump, we would be gathering that stupidly lost energy, and still get 5lb into each tank in the end.

If there were no resistance, superconductive, and zero heat produced, would there be more electrons or charge in the caps after connecting because of no heat losses? No, there would not be.  The pressures would level out and be half of what the first cap started with because we released that total pressure into a container twice as large.

It would be a superconducting super loss.

Resistance losses in a circuit are due to voltage divisions across those resistances where the load we want to provide power to suffers from the drop because it is not getting the actual potential of the source. So we can say that oh we are not getting full power to the load because of voltage drops due to restrictions in the line of current. But in the case of full cap connected to an empty cap, I believe the losses would be the same whether there be resistance or not. 10v to 5v and 10lb to 5lb 

Now, if we use an inductor in the circuit and apply timing, we can stop the initially full cap at near 7v and use a freewheel diode with the inductor to get near 7v into the other cap.
How does that happen? There is still 'resistance' in the circuit, so called losses all over, but when we calculate the total energy in the 2 caps now with 7v each, we come real close to no loss at all. 


Mags