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Unipolar Electricity: What happens here?

Started by Magnethos, December 30, 2008, 03:10:15 PM

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0 Members and 2 Guests are viewing this topic.

sm0ky2

Quote from: Magnethos on December 31, 2008, 05:25:11 AM
One wire doesn't works for me if I want to run a motor.

the key is the battery configuration, because I think this is Cold Electricity. Yes, in cold electricity a capacitor has inductance, so the electrical properties in cold electricity are different. So, if your battery acts like a diode... I'm almost 100% sure that this kind of electricity is Cold Electricity.

We can use both 2 positive or 2 negative poles to run any device in cold electricity. Maybe this is the easiest way to produce radiant/cold electricity.

You will need also more voltage and less current, because cold electricity is the opposite of hot electricity. Hot works in low volt-high amps. Cold works in low amps-high volt.
This is why my the brightness of my light bulb is low when I use cold electrical current.

This is all wrong....    You do indeed have two like poles.  but what you have is 1 pole with 3v, and 1 pole with 1.5v    together they make a 1.5v circuit, and would be the same as connecting ONE battery straight to the lightbulb,, but cooincidently will last Three times as long.  The battery is not acting as a diode, but rather as a 1-way resistor. because it has twice the current potential in one direction, vs. in the other.

the electrical properties are 'different' because you're dealing with the other side of a circuit and trying to compare those results to a  circuit with a '0 ground'.

your ground is positive, 1.5v.  once you account for this things wil start making a lot more sense.
I was fixing a shower-rod, slipped and hit my head on the sink. When i came to, that's when i had the idea for the "Flux Capacitor", Which makes Perpetual Motion possible.

TCadd

Your configuration of batteries has this effect:

1.5v + 1.5v = 3v (These are the two in normal series, they become, effectively, a single 3v battery)
This 3v is then added to a 1.5v battery in opposition (hooked up backwards)

so....

3v - 1.5v = 1.5v

You have 1.5v powering that light bulb in the configuration described above. The "larger" battery wins, but is reduced by the voltage of the opposing battery.

Magnethos

Quote from: CrazyEwok on January 01, 2009, 08:53:53 PM
Your all kidding me right? for electricity to work you don't need a Positive and a Negative current. You need a higher and a lower potential. In other words electricity is a FLOW it tries to balance the charge by flowing from something that has more charge to something that has less charge. If you have 2 "Positive" terminals one that has either more current available to it or one that has more charge available to it then the other the electricity will flow to try and balance out the connected charges. You want something the ponder on rather than trying to "Force" charge into a battery (thats what your doing on your lower potential side BTW, and that can blow up that battery if your not careful) is if you have power source "A" which is 9v+ and has a potential current of 2amp connected to "B" a 12v+ 1.5amp which way will the charge flow? both are the same Watts but have different atributes.

Very usefull comment, thanks

Magnethos

Quote from: spinner on January 01, 2009, 10:27:53 PM
Unipolar electricity, Cold electricity...? I'd just like to know, when will Magnethos stop spamming this board with countless threads about his electricity misconceptions and delusions?


I'm just asking, this section is called HALF-BAKED ideas. So I can ask it.

If you have a brilliant idea, post it here, because I will learn from it and I will stop "spamming" this board. Meanwhile I will ask all my doubts here.  ;)

Magnethos

Quote from: TCadd on January 01, 2009, 11:45:36 PM
Your configuration of batteries has this effect:

1.5v + 1.5v = 3v (These are the two in normal series, they become, effectively, a single 3v battery)
This 3v is then added to a 1.5v battery in opposition (hooked up backwards)

so....

3v - 1.5v = 1.5v

You have 1.5v powering that light bulb in the configuration described above. The "larger" battery wins, but is reduced by the voltage of the opposing battery.

Yes, that is the info I was looking for.