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Thread: Calculating Vd of 120/208 V circuit

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    Calculating Vd of 120/208 V circuit

    when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?
    Last edited by unsaint33; 02-12-19 at 04:24 PM. Reason: Claritying

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    Quote Originally Posted by unsaint33 View Post
    when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?
    if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
    Bob

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    Do you have a reduced neutral?
    Tom
    TBLO

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    Quote Originally Posted by petersonra View Post
    if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
    So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)

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    Quote Originally Posted by unsaint33 View Post
    So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)
    If you have 3% VD on the L-N side, that would be 116.4 V. That would make the L-L voltage 201.6 V, or a 3% VD.


    Quote Originally Posted by petersonra View Post
    if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
    I should have said the VD in % is the same. Obviously the voltage drop in volts is not the same.
    Bob

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    Quote Originally Posted by unsaint33 View Post
    So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)
    They are all telling you straight.
    3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V

    It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
    120V X 1.732 = 207.8V

    Works the same for the Voltage Drops
    3% X 120V = 3.6V
    3.6V X 1.732 = 6.24V

    .03 X 207.8V = 6.23V

    The only difference we are seeing is the round off error.

    The attached sketches show the angles.
    Attached Files Attached Files
    Without data you’re just another person with an opinion – Edwards Deming

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    Quote Originally Posted by iceworm View Post
    They are all telling you straight.
    3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V

    It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
    120V X 1.732 = 207.8V

    Works the same for the Voltage Drops
    3% X 120V = 3.6V
    3.6V X 1.732 = 6.24V

    .03 X 207.8V = 6.23V

    The only difference we are seeing is the round off error.

    The attached sketches show the angles.
    I deal with balanced lines (3 phase PV inverters), so for me the answer is simple; it's just the one way drop on an individual conductor @120V. There's no current on the neutral, so no voltage drop.

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