counting neutral conductor as ccc

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jeepx

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Maybe im confused but if you have a twenty amp single phase two wire circuit and check the amperage on the neutral at the panel you should read almost zero on a neutral sharing three circuits you will read the imbalance between the three phases so with that said if you have a@5amps b@5amps c@7amps then the neutral should read the difference of roughly 2 amps and should be considered a CCC where as in reality the neutral of a single circuit really carries no amps it is all being used up at what ever the circuit is powering up engineering and diagrams look good on paper as well as theory so am I wrong
 
There cannot be a neutral of a single circuit. The grounded conductor of a single pole circuit is very much a ccc. It carries every bit as much as the ungrounded conductor. As for the unbalanced current in your example, I dont believe the code considers that neutral a ccc, even with the two (approx) amps it might be carrying. Now, drop one of those phases out of the equation and the code specifically says the neutral is a ccc.
 
arcsnsparks98 said:
There cannot be a neutral of a single circuit. The grounded conductor of a single pole circuit is very much a ccc. It carries every bit as much as the ungrounded conductor. As for the unbalanced current in your example, I dont believe the code considers that neutral a ccc, even with the two (approx) amps it might be carrying. Now, drop one of those phases out of the equation and the code specifically says the neutral is a ccc.
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In an unbalanced condition, every amp that is added to the neutral is an amp that is subtracted from the phase conductors, so as long as all the phase conductors are in there with the neutral, the net result is the same. You count the phase conductors and you don't count the neutral.
 
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arcsnsparks98 said:
There cannot be a neutral of a single circuit. The grounded conductor of a single pole circuit is very much a ccc. It carries every bit as much as the ungrounded conductor. As for the unbalanced current in your example, I dont believe the code considers that neutral a ccc, even with the two (approx) amps it might be carrying. Now, drop one of those phases out of the equation and the code specifically says the neutral is a ccc.
Click to expand...
Arcs, what you are missing is that for there to be two amps of unbalanced current, one of the hot leads must be carrying two amps less than the other, reducing the heat contributed by that conductor by the same amount as the heat produced in the neutral.
This effect justifies not counting the neutral even when the imbalance is the full ampacity of the circuit.
 
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GoldDigger said:
Arcs, what you are missing is that for there to be two amps of unbalanced current, one of the hot leads must be carrying two amps less than the other, reducing the heat contributed by that conductor by the same amount as the heat produced in the neutral.
Click to expand...

Great minds think alike. :D
 
GoldDigger said:
Arcs, what you are missing is that for there to be two amps of unbalanced current, one of the hot leads must be carrying two amps less than the other, reducing the heat contributed by that conductor by the same amount as the heat produced in the neutral.
This effect justifies not counting the neutral even when the imbalance is the full ampacity of the circuit.
Click to expand...

Hmmm...I believe you summed up my point. Not sure I missed it.
 
jeepx said:
. . . if you have a twenty amp single phase two wire circuit and check the amperage on the neutral at the panel you should read almost zero . . . .
Click to expand...
jeepx said:
. . . the neutral of a single circuit really carries no amps it is all being used up at what ever the circuit is powering up . . . .
Click to expand...
I am afraid that both of these statements are wrong. Current needs a complete path, from source to load and back to source, in order to flow at all. The current is not "used up" by the load. Rather, it travels to the load along the black wire (the "ungrounded conductor"), then travels through the load, and then travels back to the source along the white wire (the "grounded conductor," also traditionally - rightly or wrongly - called the "neutral wire"). If you measure 10 amps of current in the black wire, then a measurement of the white wire is also going to read 10 amps. That is why both are considered CCCs in a single phase, two-wire circuit.

What is used up by the load is some of the energy that the current possessed. That is why the voltage becomes lower as the current travels through the load.
 
GoldDigger said:
. . . for there to be two amps of unbalanced current, one of the hot leads must be carrying two amps less than the other, reducing the heat contributed by that conductor by the same amount as the heat produced in the neutral.. . . .
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That is close to the truth, but is not quite true. I once did the math to prove to myself that it was acceptable, from a physics standpoint, to disregard the neutral wire in a three phase circuit. (Let's ignore harmonic loads for this discussion.) What I discovered is that the heat generated by a balanced current in the three ungrounded conductors (i.e., therefore with a zero neutral current) is always greater than, or equal to, the heat that would be generated by current flowing in all four wires in an unbalanced situation. The heat generated by current flowing in the neutral will never be the same amount by which the heat generated in the ungrounded conductors is reduced, unless the imbalance is created by turning off one of the phases. That is, if you create the imbalance by having the Phase A current equal the Phase B current and having the Phase C current be zero, then the total heat generated in the neutral wire will exactly offset the heat that is no longer being generated by the Phase C wire. In any other unbalanced condition, the total heat generated will be lower.

The reason things work out this way is that the amount of heat being generated in a wire is related to the square of the current. So if you keep Phases A and B the same and you reduce Phase C by some small amount, thus causing a small amount of current to flow in the neutral, then the calculation of heat will involve the squares of three slightly smaller but still large numbers being partially offset by the square of a much smaller number.
 
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