I mean, If I connect a device to the + pole of a battery and then to the - pole of the battery the device works. But... Why that not happens when I connect the + pole and the - pole from different batteries? ???
PS The red points are leds
Quote from: Magnethos on October 18, 2008, 05:51:58 PM
I mean, If I connect a device to the + pole of a battery and then to the - pole of the battery the device works. But... Why that not happens when I connect the + pole and the - pole from different batteries? ???
PS The red points are leds
That is simple for you have what is called an open which means no circuit. you need a closed circuit to do anything with your batteries and batteries have to be in line with the circuit. ::)
Quote from: AB Hammer on October 18, 2008, 10:09:32 PM
That is simple for you have what is called an open which means no circuit. you need a closed circuit to do anything with your batteries and batteries have to be in line with the circuit. ::)
Tom Bearden says that if we connect the two poles in a closed circuit, we will kill the battery. He says something about the open circuits. Anyone knows more about that?
Quote from: Magnethos on October 18, 2008, 05:51:58 PM
I mean, If I connect a device to the + pole of a battery and then to the - pole of the battery the device works. But... Why that not happens when I connect the + pole and the - pole from different batteries? ???
PS The red points are leds
both cant work
1. open circuit ( lik eswitch is opened)
2. not (like switch closet, but 1.5 volts to low - to light and LED)
G.Pese
Possible the meaning of Newman is an "other one" that you (we) have (also) not understand
Electrons need a path, the 2 battery system won't work unless you connect the other ends to a common wire, then you close the loop and electrons can flow.
Your first pic. shows an OPEN... circuit is not closed and electrons cannot flow.
your second pic shows a completed circuit, CLOSED
the other variation is a SHORT, since electrons follow the path of least resistance, if you took another wire and connected it to both ends of the same battery in the second pic, the LED would get dimmer or even go out... (pese is correct about the low voltage from a 1.5v cell being too low to even light that LED)
Hi Magnethos,
I think that before believing in Bearden it may help to learn and understand normal forms of electricity rather than jump in trying to find cold electricity which is essentially a an unproven idea.
When Bearden is talking about not destroying the dipole he means that you have a closed loop and you switch away before current flows. Ie once you close a circuit, there is a small time after voltage is applied before current actually flows, this time gap however is extremely fast, faster than anything we have can switch at. If you could switch that fast, you could switch away a potentialised circuit before you have any current flow and prevent the battery from running down. Unfortunately it is not possible without slowing down this time period to something we can switch at. But bear in mind, this is all just theory from Bearden, you must take it with a pinch of salt!
Regards,
Dave.
Your post got me brainstorming a bit... Isn't it interesting how if you need to jumpstart your car battery, you can hook the black jumper cable to the frame of the car and still jumpstart it?!?!?! Wouldn't that technically be an open circuit, since you only have the positive pole of the battery in the circuit?
Second thought... Since you can jump start a car with only the positive cable connected, wouldn't that mean you could theoretically charge a capacitor using only the positive, and not have a need for a diode? I'm almost positive I have tried this with a capacitor and it does not work, but now I'm wondering why it doesn't when thinking of the concepts of jump starting your car... Or does it work, and I did not do it right?
Just curious... Any answers?
the car's battery is also grounded to the frame so it is in actuality a closed circuit.
Good point... So doesn't that mean theoretically you could charge a capacitor if the negative was hooked into an earth ground?