Is it possible to combine a voltage source with a current source ?

[fxeconomist]

I have done these tests without the diodes. I will be repeating these tests with the diodes to block power from going to the other source to see if the results change.

So, without diodes, what were the results ?

[floodrod]

Edit-  My setup was different..  I had the resistor between the 2 hot leads.-  In that configuration- Without diodes, current will not flow out of the power supply till voltage is higher than the 12V batteries voltage. And Current will calculate as if we are only sending the differential, but in reality we are sending the full monte..  Verified with wall watt meters several different ways. I will be testing the way I proposed next..  But I am pretty sure the diodes will be making a big difference.

My point was though, you present complex questions and chain of events, which is great to tackle.. But I recommend bench testing each piece.  No one is going to be able to instruct you how to accomplish this task.

"We collect the EMF along with the homopolar amps"

Ok, so "Collect the amps"..  So Homopolar generator out lead to a positive of a capacitor or battery?  Amps go into the storage device-->  then what comes out of the negative of the storage device?   We collected amps, but now the same amount of amps came out..  Where do those amps go?  It won't collect anything with an open circuit.

Do we send the negative of the collection device to another positive terminal?  If so, again- amps come out of that negative. Eventually the collected amps need to connect to the negative of the source battery..  SO in the end, any amps we "collect" result in the exact same amount of amps going to the Negative terminal of the source..

And what happens when we put amperage into the negative terminal?  The exact same amount come out the positive.  because amps are always equal in a closed circuit.  Basically what I am saying, you may get some charge in a collection device, but by robbing peter.  Ultimately, it all came out of the source.

I personally "believe" there are reactions that cause a circuit to make extra potential. (and some pretty easy to create)..  And I believe my roadblock preventing me from OU is basically what I explained above. Along with other snafu's. 

[fxeconomist]

Do we send the negative of the collection device to another positive terminal?  If so, again- amps come out of that negative. Eventually the collected amps need to connect to the negative of the source battery..  SO in the end, any amps we "collect" result in the exact same amount of amps going to the Negative terminal of the source..

And what happens when we put amperage into the negative terminal?  The exact same amount come out the positive.  because amps are always equal in a closed circuit.  Basically what I am saying, you may get some charge in a collection device, but by robbing peter.  Ultimately, it all came out of the source.

I personally "believe" there are reactions that cause a circuit to make extra potential. (and some pretty easy to create)..  And I believe my roadblock preventing me from OU is basically what I explained above. Along with other snafu's.

But if storage doesn't work, we can simply provide this to a runtime, working load. After all, one of the reasons of targeting free energy is to avoid battery reliance, right ?

[floodrod]

Lol, I had to try with the diodes..

It can be balanced and split..  If the Supply is putting out the same amount as the battery, the current splits. 1/2 from the battery, 1/2 from the supply.  As you raise the supply, the battery puts out less amps and the supply puts more.  Raise supply high enough and the supply takes over and battery amps go to zero.

Resistor still received wattage dictated by Ohms Law, And it is possible to split that wattage between 2 sources in varying amounts.

[fxeconomist]

Lol, I had to try with the diodes..

Resistor still received wattage dictated by Ohms Law, And it is possible to split that wattage between 2 sources in varying amounts.

I don't understand how it applies, because we have two sides, with two different voltages. How much was the voltage past the resistor, how many ohms was the resistor and how much did the source on right gave, given the battery was 12 V ? Say in the example with 16 V on the right.