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Thread: Unbalanced three phase loads return current path?

  1. #1
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    Unbalanced three phase loads return current path?

    Dear All,

    I would like to have your kind help regarding this dilemma i am facing about the unbalanced return current (in neutral in attached figure shown to be 50 Amp) for a three phases transformer and three phases loads. the question is: Is the path going to be path A or path B, and what i mean is the unbalanced current returning in neutral will go back to the transformer star point and get down into the ground connected at the star point of the transformer? or keep in circling between the transformer and phase A and transformer wont throw it in the ground? i made a small sketch just to make my self clear it shows what path A and B exactly are!!! Thank you!!!


    I would like your help please any clues?


    Best Regards all Sirs,
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  2. #2
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    The unbalanced current in a 3-phase four wire system (A, B, C, N) would normally flow in the neutral.
    The loads are connected between phase and neutral.

  3. #3
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    The short answer to your question is that “Path A” is not involved at all. Current does not flow into the ground (i.e., planet Earth) unless there is a fault somewhere in the system.

    However, there are several things wrong with your sketch. First and foremost, you are incorrectly showing a connection between the current-carrying portion of the load and the ground. The ground wire at the load end is attached to the external metal portions of the load (e.g., the metal frame of a motor), not to any of the wires inside the component. If you disconnect the green wire from the load to the ground on the right side of your sketch, you will see that “Path A” does not exist anymore, as should be the case.

    Secondly, when current leaves the transformer along one of the phase conductors, it will return to the transformer on one or both of the other phase conductors. At any moment in time, the total amount of current flowing in the four wires (A, B, C, and N) is zero. But the way you are showing current flows, you have 350 amps flowing away from the transformer on A, B, and C, and you have 50 amps flowing back to the transformer on N. That is not how current flows.

    Finally, you can’t add current in the three phases as though you were adding up the number of apples in one box to the number of apples in another box to come up with the total number of apples. The three currents do not reach their respective peak values at the same time, so you can’t add them up as though they were all the same type of quantity. To get the numerical value of the neutral current, therefore, you don’t add up A plus B plus C and assign whatever is leftover to N. This is the formula (with the symbology of A*2 meaning A squared):

    N*2 = A*2 + B*2 + C*2 –AB – AC – BC.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  4. #4
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    Quote Originally Posted by charlie b View Post

    Finally, you can’t add current in the three phases as though you were adding up the number of apples in one box to the number of apples in another box to come up with the total number of apples. The three currents do not reach their respective peak values at the same time, so you can’t add them up as though they were all the same type of quantity. To get the numerical value of the neutral current, therefore, you don’t add up A plus B plus C and assign whatever is leftover to N. This is the formula (with the symbology of A*2 meaning A squared):

    N*2 = A*2 + B*2 + C*2 –AB – AC – BC.
    We've been here before. That formula works only if the power factor is identical for all three phases which is not the general case.
    For example, if all three currents are equal it yields In = 0A

    Here is what can happen if you have equal currents but differing power factors:



    Clearly In ≠ 0A even with balanced currents.

    I've seen that formula posted a few times. I don't know why it is taught.
    Last edited by Besoeker; 09-18-12 at 01:21 PM.

  5. #5
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    what if the green grounding wire is disconnected

    Dear Charlie,

    First i would like to thank you for this outstanding elaboration in explaining how it works,i really appreciate it,there might a small additional question i would like to know your answer for,well suppose we removed this green connection to ground on the load side and we made it only as a connection to the metal enclosure or cabinet of this load to the ground i.e:normal grounding to body of a fridge lets say, still the path would be path B right?

  6. #6
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    Still the path would be B. However, when we connect a ground wire (green insulation, or bare copper) to the metal case of the motor (or the fridge), we do not connect the other end of the wire to the ground (i.e., into dirt, using a ground rod). Rather, the other end of the green wire is run alongside the black and white wires all the way back to the panel. The purpose of the green wire is to provide a low-resistance path for current to flow back to the source, in the event there is a failure inside the component (e.g., wire breaks free inside the motor, and touches the metal frame). That will cause a large current to flow (very briefly) in the green wire, a current so large that it will force the breaker to trip. When the breaker trips, the failure event is over, and it happens so fast that a person touching the motor frame when the failure occurred would not feel a shock.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  7. #7
    T.M.Haja Sahib Guest
    hisham1986:
    The path-A does exist! Remember parallel circuits: current in each is inversely proportional to the resistance (impedance) in each. Here the path-A is in parallel with the system neutral to load neutral path . The division of current between them is inversely proportional to the resistance or impedance in each of them. So the lower the earth resistance of neutral at system and load ends,greater is the flow of current via ground.

  8. #8
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    Clarification please

    Dear All gents,

    It seems a bit confusing what i have received as info but to make my self more clear what i want to mainly know is the following:

    DOES THE CURRENT RETURNING FROM THE LOAD( UNBALANCED CURRENT IN THE GREY COLORED LINE) ONCE IT REACHES THE STAR POINT OF THE TRANSFORMER GOES DOWN TO THE GROUND LOOP CONNECTED TO THE STAR POINT OF THE TRANSFORMER(PATH A) OR MAKES A CLOSED LOOP AND GOES BACK TO THE LOAD AGAIN (PATH B) AND FOMRS A CLOSED CIRCUIT CURRENT BETWEEN TRANSFORMER AND LOAD??


    HOPE IT GETS MORE CLEAR AM SORRY IF ANYONE MISUNDERSTOOD ME THANK YOU ALL SIRS IN ADVANCE FOR YOUR HELP

  9. #9
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    N.B

    AM SORRY JUST FOR CLARIFICATION: THE GROUNDING CONNECTION AT THE LOAD (GREEN COLOR) IS INTENDED TO BE CONNECTED TO THE BODY OF THE LOAD AND NOT TO THE NEUTRAL POINT OF THE LOAD ITSELF!!

    THANKS AGAIN

  10. #10
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    Quote Originally Posted by T.M.Haja Sahib View Post
    The path-A does exist!
    No it doesn't. It would, as shown on the sketch. But the OP has clarified that the sketch has an error, in that the green wire at the load is intended to be connected to the external case, not to the current-carrying wires.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

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