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TK
Thanks for the article.
I guess the difference between electricity and electronics rears its ugly head once more.
For the present, I'm willing to accept that this is a property of "Electrical Systems". I've got several books handy regarding electrical theory and I promise I will do "Due Diligence" to better understand. Are you open to more questions regarding this topic or would you prefer me to let it alone because it is an accepted fact, period.
Don, of course I am open to questions to this or any topic.
The smarter I get you and any others interested will make my life easier in answering questions accurately.
Believe me, I much rather be just answering posters questions than correcting answers or clearing up concepts.
TKRussel, Yeah, I have problems explaining that one to a layman too.
Don, Current must be looked at in the context of the whole circuit, not just one conductor or point on a conductor. If you look at any given point on the hot conductor you can measure current without having any voltage change there. Voltage drop occurs wherever work is done in the circuit. Current is the flow of electrons. It is the same throughout the whole circuit. Voltage is how we measure the potential difference between two points.
Take any point on EITHER current carrying conductor and apply the P=V*A (using common abbreviations) equation. Voltage change at that point for all intents and purposes is 0. Amps can be whatever you want. Which gives 0 power at that POINT. There is no work being done there. The equation holds with V/R=A too. If you have 0 volts change divided by 0 ohms, the Amp side of the equation is completely indeterminant. It can be anything you want at that POINT. Of course the work done in the complete circuit will determine the magnitude of the current.
Example: Take a 10 foot section of current carrying conductor in a 120v circuit. Lets make it #10 copper and put 10A of current on it. The resistance of that section will be somewhere around .01 ohms, and so the voltage measured from end to end of that section will be lost in the noise of your meter, and yet you will easily be able to measure 10 amps of current. I think the confusion comes when we try do divide the circuit into discrete parts and only think of the voltage as the supply voltage rather than the voltage measured across work done in that section. For the sake of the discussion here we don't have work done (and therefore voltage drop) on the current carrying conductors, and yet there is current there.
Of course when you run 600 feet of 16 gauge extension cord, and wonder why your saw doesn't work right, then we can talk about voltage drop (work done/heat developed) on the current carrying conductors!
EPMiller
I have to disagree, but certainly does not mean a reddie is in order:
Quote:
Originally Posted by EPMiller
Well done!
Would this happen if you used 12/3 wire for example to feed two circuits sharing a common neutral? Like many kitchens are...
Codyman,
The neutral conductor in a correctly implemented shared neutral circuit situation only carries the differential current between the two halves of the circuit. In other words, If the one breaker for the one half of the circuit is supplying 14 amps and the other breaker is supplying 9 amps, the neutral will be carrying 5 amps.
This is why you MUST run those circuits on a double pole breaker. (There are other reasons too, but we aren't discussing that here.) If you split it up on 2 single pole breakers and get them both on the same leg of the service (an even number of spaces separating them in the panel) the neutral conductor carries the SUM of the current in both halves of the circuit. So from the above example, the neutral would be carrying 23 amps. That's too much even for a 20 amp kitchen circuit. And it could go as high as 40 amps in that situation. I've corrected that on more than a few panels where the homeowner knew just enough to be dangerous.
Let's not go into 3-phase here. That's even more fun.
EPM
EPM,
Right because you are using two phases with the two poll breaker. I do understand the proper installation of such a circuit as I have read past posts on it. Although given I just bought 250' of 12/2 romex for a ridiculously cheap price I doubt I will use this setup when I (eventually) re-wire the kitchen. It will be more cost effective to buy additional breakers than the wire I would need.
But would this explain why you might read zero voltage but still pass current? Do the phases run unversed to each other meaning when one was at +120V the other would be at -120V on the neutral wire? That would mean the neutral wire would be passing amps but read at 0V if tested on the grounding wire?
I am sure that I am missing something or many things thanks just want to discuss.
- JC
Read my post #20 above for an explanation on why a neutral carries current but is at earth potential of zero.
I guess it's my turn:
A couple of things:
The two 120 volts "phases" are derived from a transformer, whose center tap is at the 0 volt potential. The 120 V is really 155 * sin (wt) so it really goes through a peak and valley of +-155 volts. Meaning 120 VAC is equivalent to about 310 Vpeak to peak. (Vp-p)
So, you statement of negative and positive is correct except you have to be talking about p-p values. In a real world the voltages won't be exactly "identical".
-----------------> 120 (L1)
*
*
*---------------->0, (N), (Earth), L0 (Just a made up term)
*
*
------------------>120 (L2)
The "*"'s are transformer windings.
If you have ONLY 240 V loads, there is no current on L0
If you have 10 A on L1 with respect to L0 (120 v load)
and 10 A on L2 with respect to L0 (120 V load)
The currents cancel and there would be none on L0
If you have an unbalanced load, then (N) caries the difference. The sign may be different depending on your reference.
Now back to the age old problem of current with no voltage.
Depending on where a voltmeter is inserted and it's reference will determine if there is appreciable voltage.
A stupid example:
Say you had a 12 V battery and a non-contact DC amp probe.
If you measure from the - terminal to the +terminal you will have 12 V.
If you measure from the + to the -, you will have -12V
The current will change sign based on the orientation of the AMP probe.
In AC there is no sign change in terms of the measured value because RMS or Root-Mean_Square means the square root of the square of the average. RMS values of voltages have the same "heating" value as a DC voltage of the same potential and ohm's law works for resistive loads. Not motors.
Cheap meters are RMS reading, average responding. I'll stop here before it gets to complicated. This means that the sinusoid is flipped, so that both half cycles are positive, averaged and multiplied by a constant.
If you measure the current with a probe close to the reference, the voltage at the same spot would be nearly zero.
KISS,
Wow okay I was with you up until the battery/motor part. Ahh, I am just going to stop here before my head explodes. You guys sure as hell know your S***
For any detail freaks out there, let me state that I generally am ignoring the small resistance of the copper wire and its accompanying details that are not necessary for this type of explanation. I am also ignoring counter EMF and reactive loads.Quote:
Originally Posted by codyman144
Codyman,
The neutral conductor in a correctly implemented shared neutral circuit only carries the differential current between the two hot sides of the circuit (restating my first post for emphasis). This is a statement about current, it has NOTHING to do with voltage. Reread that earlier post after you are done with this one.
In a residential electrical service, the neutral conductor (grounded conductor, not to be confused with the bare grounding conductor) has 0 volts on it everywhere. Think about it. It is copper the whole way from the low side of the load to the ground stake.
Voltage is always measured across a power source or across the load (power sink) in a circuit. Since the wiring from the panel to the load does not consume any power, there is no voltage change on the wires. So you would be able to measure 120 volts anywhere along the hot wire right up to the load. And you would be able to measure 0 volts anywhere on the neutral wire too. The voltage is dropped across the load. If you could measure from the exact middle of the load to ground you would get 60 volts.
Now on the other hand current requires a complete circuit to "travel" through and therefore can be measured anywhere in that circuit regardless of the voltage potential at the point of measurement.
You can think of voltage as potential and current as movement. Wrap your head around Amps = Voltage / Resistance and Watts = Voltage * Amps. Using basic algebra, or that table that gets posted every so often around here, you can figure any parameter out if you have two known parameters.
It took me a while to see the whole picture too. I am self taught, there's hope for others too. Get a book on basic electricity and read it. It will use electronics to teach you, but the fundamentals are the same everywhere. Residential electricians are just working with the special case of a center grounded 60 Hz AC transformer power supply with enough wallop to kill you without any warning.
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