Help explain free energy in my house light circuit?

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jollyrgr
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Post by jollyrgr » Thu May 18, 2006 7:30 pm

I think you are picking up stray power by inductance. While you might have anything from 0.1V to 58V my guess is that as soon as you attempted to draw power you'd see next to zero current flow.

With the compact fluorescent you will have a somewhat steady impedance that will allow for some voltage to build up. But the incandscent bulb is (when cold or off) nearly a dead short. Do this...

Pull out all bulbs from the circuit and measure the voltage.

Using a pair of clip leads on a light socket, attach one clip to ground. Connect the other lead to where you are measuring the voltage. Likely you will see the voltage drop to zero but the light will not light.
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positronicle
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Post by positronicle » Thu May 18, 2006 7:56 pm

--Edited by Positronicle--

Robert Reed
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Post by Robert Reed » Thu May 18, 2006 8:47 pm

Posi-
If these weird bulbs you guys are experimenting with have any circuitry containing capacitance, the amount of charge on them at exact time of opening the circuit would depend on the level of the ac sine wave at that same moment. Seems it could be anwhere from zero to +/- 170volts and statically read back as DC on your meter. The reading would be different every time you did the test owing to the probability that you would never break the circuit at the same AC sinewave level.
And speaking of capacitance, has anyone considered 60 Hz capacitive coupling on a non energized traveler wire via it companion wires in the same romex cable?

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Chris Smith
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Post by Chris Smith » Thu May 18, 2006 9:42 pm

Yes, a circuit is very simple.

You start off with one feed wire.

This wire will carry all the usable current to, and when it reaches the target it powers the circuit but only IF, the return out wire or ground returns this current/ voltage properly to ground, and thus this return wire has in essence ZERO volts, but only IF the return/ ground does its job fully.

Most circuit grounds are actually about 99.8% perfect meaning that you will always read a .01 volt or so.

IF the ground fails to carry ALL of the flow back to ground, then it will become a resistor and voltage divider dropping some of the value [voltage] but not all, thus you will see the leak when you place the ground under test. 20 v

Like a sewer pipe, you flush it but only MOST of the fluid makes its way to where it belongs, and the remaining POTENTIAL leaks out some where else, as tested by you with your value of 20 V.

When you double ground the return wire, in essence you place your leaky pipe within another sealed pipe and both have only one direction to go. Ground.

The actual ground.

Thus when you check for leaks, even though the inner pipe is leaking, the outer pipe catches this drip and sends it to the proper place any way and thus you will read NO voltage or leak on the grounds any where within the whole system.

AND, if this voltage of yours happens to just be stray inductance, RF induced power, or ghost signals of no value or actual workable current, it will also disappear because the extra lead to ground will in essence place a load on this inductance and force it as well to become at ground level.

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Edd
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Post by Edd » Sat May 20, 2006 12:30 am

.


Hey, IT’S A MIRACLE ! the images of the wiring layout are up now…….I just had dos red equis’ displayed at the initial posting time.

Some delay in getting back, as my posts graphics were residing separately on my lap tops hard drive.

Colorful isn’t it ….(your wiring schema)…but I also agree on the abortive color coding scheme assignment…until you clarified. I zeroed in on the main three color codes of house wiring even prior to my teens where I thought over in my mind, a comparative color analogy for myself where I could use it for making a mental association. In my case, I then utilized:

Black - the color of "death" if you possibly contact with that “ hotâ€

rshayes
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Post by rshayes » Sat May 20, 2006 10:03 pm

Most DVMs have an input impedance of 10 megohms. A 30 megohm reactance in series with the meter would attenuate 125 volts to about the 28 volts that you are reading.

At 60 Hertz, a capacitance of of 88 picofarad would have a reactance of 30 megohms. The parallel wires in an 8 or 10 foot run of romex could have this much capacitance, especially since the run to a three-way switch may have two hot wires and one floating wire when the switches are in the off position.

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