Originally Posted By: hgordon This post was automatically imported from our archived forum.
Josh, I agree with Patrick. What caused you to do this? Were you doing an exhaustive inspection?
Remember, if you take the time to do stuff like that, and lets say that you don't on a home that, God forbid, burnt down...realize that a good attorney will ask your previous clients if you checked that in their homes and if the answer is yes then the judge will look at you and...dude! And to say nothing about the fact that if you do this type of work, you are now stepping out of the basis of a visual inspection and now what else did you not check the amps on?
Like Patrick said, not trying to give ya a hard time...just wondering where your coming from.
Originally Posted By: jpeck This post was automatically imported from our archived forum.
Bob Badger wrote:
jrabanus wrote:
I came across a electrical breaker panel that had a neutral wire and ground that had some amps in it but no voltage. Does anyone know what this means?
This is not possible.
A difference of potential (voltage) would have to be present in order to get a reading of current flow (amps).
Bob,
Yes (technically) and no (practically speaking).
Yes, it needs voltage to serve as the electromotive force to drive the current through the conductor, however, if the neutral is connected at the panel, and at the ground rod, you will have current on it (even though the neutral is (practically speaking) at -0- potential to ground. This is because there is technically (and it is measurable) voltage drop across the neutral, meaning there is voltage at the driving end end as compared to the other end, voltage which one might not readily measure unless one was looking for that small of a voltage.
My guess, as to the original question, is that there is a problem somewhere (where I don't know?), which has caused current to flow on the ground, this is indicating a ground fault somewhere in the electrical system.
Originally Posted By: jpeck This post was automatically imported from our archived forum.
Bob Badger wrote:
Anyway you want to slice it you can not have current without voltage.
If you want to argue that it is possible I will finally realize you just want to argue. 
Bob
I'm not presenting the argument that it is possible, only that the voltage was so small (in relation to what was being expected) that it was not measured (very small deflection on analog multi-meter) which was discounted.
If you are stating that such occasions are not possible, I suspect it may be you who wants to argue, in which case, you may have found a suitable person to carry-on such an argument with. Hopefully, I will make a worthy debater as your opponent. (Either that, or look foolish by making a math mistake, which I have done before.)
Let us assume a #6 awg conductor was used for the ground (just as an example to put the voltage required for current flow into perspective).
#6 awg, uncoated, has a Direct-Current Resistance at 75?C (167?F) of 0.491 ohm/ kFT. 0.491 ohm/ kFT divided by 1,000 equals a Direct-Current Resistance at 75?C (167?F) of 0.00491 ohm/ FT. Now, let's assume the length of that ground conductor was, oh, let's say ... 10 feet. That's reasonable, right? Okay, at 10 feet the Direct-Current Resistance at 75?C (167?F) of 0.0491 ohm/ 10 FT or 0.0491 ohms.
Now, using Ohm's Law, where E (voltage) = I (current) times R (resistance), thus, if we assume the current was 10 amps (a healthy amount and easily measurable by almost any clamp-on ammeter), then we has the following:
E (unknown) = I (10 amps) times R (0.0491 ohms for 10 FT of conductor)
Let us now solve for the voltage necessary to push 10 amps through 10 feet of #6 awg copper. I have simplified this by using the dc resistance, even though this is an ac circuit, but still, how far off can that be? 100%? Okay, lets say that is 100% off.
E = 10 X 0.0491
E = 0.491 volts - round off to 0.5 volt (wait, don't forget we are assuming that is 100% off, so we will correct for that)
E = 0.5 X 2 (100% correction factor)
E = 1 volt.
Lets double check my math (I haven't done this in many, many years).
I = E/R
I (10 amps) = E (.5 volt) divided by R (0.0491 ohm - rounded off to 0.5 ohms)
10 = 0.5 / 0.05, right, and the 100% correction factor makes it 1 volt.
Did I do the math right? (Like I said, it has been many years.)
If I did, and you used an analog meter, would you even see 1 volt deflection on a 150 volt scale? Maybe, if you looked very closely.
Would you consider it an accurate reading with a digital meter when it read '1'? " '1' volt? I guess I must be connected wrong."
And that is reading that '1' volt across the length of the ground wire itself, no telling what you would get with a little extra resistance measuring any other way.
Added with edit: I know I made an error, by my eyes are glazing over, I think found it and corrected it (changed some numbers, just not the answer). Am I right this time?
Originally Posted By: jrabanus This post was automatically imported from our archived forum.
Thanks for the info guys. I stumbled across this when I was checking the panel to see if it was hot. I checked both amps and volts, and I was curious as to where it came from. Thanks for the legal advice as well. I appreciate anyone who is looking for me.
