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Thread: How does the neutral wire prevent MWBCs from operating at 240v?

  1. #21
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    Quote Originally Posted by ActionDave View Post
    Roger's diagrams, Larry's written explanation, and the video Jamesco posted all do a good job of telling us what the neutral does but none explain why. How do the electrons on an unbalanced MWBC coming from phase A know they are supposed to turn right and head down the neutral path, and the electrons from phase B know they are supposed to turn left and head down the same neutral? Why don't they go wherever they want on whatever copper wire they like the best?
    Tell me how this explanation works for you. Think of small beads. And the wire paths are 2" pipes. one pipe has a narrow part that is 1/2" in diameter, and the other has no restriction 2" all the way. With a 2" representing the neutral going back from the center. If you push beads in the end with 1/2" restriction, my common sense tells me that I will get an equal number of beads coming out of the "neutral" and the other "line" pipe. If I then, (for the other half of the cycle) push beads in the end that has 2" the whole way, my brain conceives that more beads will come out of the "neutral" than the 1/2" restricted other "line" In fact I am guessing about a 3 to 1 ratio.


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  2. #22
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    Quote Originally Posted by romex jockey View Post
    If i'm reading you right it's because a 120V load uses one side of the cycle , not both
    No. It works with DC, too. The direction of current in a DC version of a 3-wire circuit depends on which half of the circuit has more current.
    Code references based on 2005 NEC
    Larry B. Fine
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  3. #23
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    Quote Originally Posted by LarryFine View Post
    No. It works with DC, too. The direction of current in a DC version of a 3-wire circuit depends on which half of the circuit has more current.
    Correct, put two batteries in series and connect two lamps from each end to mid point. If both lamps are same resistance the "neutral" current is zero. If lamps are different resistance - neutral will carry imbalanced current.

    By connecting to the midpoint of the source you are assuring half the source voltage is what is applied to each load.

    It is higher potential from end to end of the source so that is where the current is going to want to flow. If you have equal loads the "neutral" current theoretically is zero, realistically may be some very small current as the loads probably are not absolutely equal in resistance.
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  4. #24
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    Quote Originally Posted by GoldDigger View Post
    The sine and cosine functions are 90 degrees out of phase, not 180.
    The two lines in 120/240 3-wire single phase are +sin(omega*t) and -sin(omega*t) respectively.

    Sent from my XT1585 using Tapatalk
    Yeah, I caught that right after I typed it.

  5. #25
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    See if this works for anybody

    The loads are in parallel with the phase & neutral.
    We know that, in a parallel circuit the voltage across each load is the same voltage as the source it is in parallel with.

    The voltage that is driving the current through the load is from the phase & neutral. The charge can only flow between those two conductors.
    Advise is a dangerous gift, even from the wise to the wise.

  6. #26
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    Quote Originally Posted by SG-1 View Post
    The loads are in parallel with the phase & neutral.
    We know that, in a parallel circuit the voltage across each load is the same voltage as the source it is in parallel with.

    The voltage that is driving the current through the load is from the phase & neutral. The charge can only flow between those two conductors.
    Um, no, not really. If two loads on different legs were perfectly balanced with no neutral current, then current flow in that moment would resemble a series circuit instead of a parallel circuit.

  7. #27
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    Quote Originally Posted by jaggedben View Post
    Um, no, not really. If two loads on different legs were perfectly balanced with no neutral current, then current flow in that moment would resemble a series circuit instead of a parallel circuit.
    There would still be current on the neutral, they would just cancel out. If both ungrounded conductors were from the same phase, the neutral would carry twice the current, and if you lifted the neutral conductor nothing would work as there would be zero potential across the loads.

    ~~~~~~~~~~~~~~„Äč~~~~~~~~~~~

    If one were to wire up two 60 watt light bulbs, one on each leg of a multiwire branch circuit, and lifted the neutral, virtually nothing would happen to the lamps... If they had exactly the same resistance they would continue glowing at the same intensity. if one had a 120W and a 24 W bulbs wired on a multiwire branch , from a split phase source, they will see 120 volts each.. if you lifted the neutral in that configuration, the 24 watt bulb, having five times the resistance of the hundred 20 watt bulb, would see 200 volts to the 40 volts of the 120 watt bulb... It would glow brightly for a moment, burnout, open the now series circuit, and then the 120 watt bulb would no longer have a complete path and go out.

    Edited to add... The reason why people burn up Electronics when they lose a service neutral is because most of them are lower wattage, higher resistance, therefore they have a higher voltage drop across them then something like a coffee pot, toaster, microwave, things that have a higher wattage
    Last edited by JFletcher; 04-10-18 at 07:06 PM.
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  8. #28
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    You can think of it either way; as two groups of 120v loads in series supplied by a 240v source, sharing a neutral, or two separate groups of 120v loads supplied by two separate 120v sources, and those two independent circuits happen to share a conductor. The latter should be easier to understand, and you can simultaneously supply line-to-line 240v loads. That's your home service.

    You could theoretically have two different 2-wire voltage systems, say one of 120v and one of 277v, share a neutral conductor, and the relative phase angle doesn't even matter. As long as only 120v loads receive their power from the 120v line conductor to the neutral, and only 277v loads receive their power from the 277v conductor to the neutral, they could share the neutral.
    Code references based on 2005 NEC
    Larry B. Fine
    Master Electrician
    Electrical Contractor
    Richmond, VA

  9. #29
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    Quote Originally Posted by jaggedben View Post
    Um, no, not really. If two loads on different legs were perfectly balanced with no neutral current, then current flow in that moment would resemble a series circuit instead of a parallel circuit.
    Resembling & behaving that way is two different things. If the polarity were swapped on one of the transformer windings you would see the neutral current, because it would add in the neutral conductor.

    If the loads are always perfectly balanced no neutral would be required. It's when people start throwing switches into the mix...

    I can say a MWBC is two parallel circuits with a common return. It's the grounded conductor that forces the voltage across each load to be L-N, not L-L.
    Advise is a dangerous gift, even from the wise to the wise.

  10. #30
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    Quote Originally Posted by JFletcher View Post
    There would still be current on the neutral, they would just cancel out.
    I like that and it makes sense. On single phase three wire they are 180 degrees from each other and would cancel out.

    On three phase with just two phases and a neutral they are 120 degrees from each other, neutral carries about same as the phase conductors when both have same load but add the third phase to the mix and they all cancel one another out on the neutral if balanced.
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