I've seen this issue raised lots of times in lots of ways. However, none of the instances I can find address the significance (or lack thereof) of the issue. Let me attempt to frame this issue in a specific way:

Taking into account the following definitions:

[LIST][*]Vd = Voltage Drop[*]K = Resistance in ohms of one circular mil foot of conductor, or 12.9 for Copper at 75 degrees C, and 21.2 for Aluminum at 75 degrees C[*]L = One way length of circuit (ft)[*]I = Current (Amps)[*]Cm = Cross sectional area of conductor (cmils)[/LIST]

The equation to find voltage drop in a single phase circuit is:

Vd = (2K x L x I)/Cm

The equation to find voltage drop in a three phase circuit is:

Vd = (1.73K x L x I)/Cm

As seen in the two equations above, the only difference between the two is the constant, which changes from 1.73 to 2. In a perfectly balanced three phase delta system, the above three phase equation applies. For an unbalanced three phase delta system, this is no longer true. Rather than discuss phasor arithmetic or symmetrical systems analysis, I want to pose this question:

Would the resulting voltage drop calculation for an unbalanced three phase delta system fall somewhere in between the calculation for a single phase system and a balanced three phase system? Another way of putting it is, if you could boil down the unbalanced circuit to a constant, as the two formulae above contain, would the constant always reside between 1.73 and 2?

Specifically, this is for lighting, not motors, and assuming normal conditions, not short circuits or open circuits. Thank you for your time.

Taking into account the following definitions:

[LIST][*]Vd = Voltage Drop[*]K = Resistance in ohms of one circular mil foot of conductor, or 12.9 for Copper at 75 degrees C, and 21.2 for Aluminum at 75 degrees C[*]L = One way length of circuit (ft)[*]I = Current (Amps)[*]Cm = Cross sectional area of conductor (cmils)[/LIST]

The equation to find voltage drop in a single phase circuit is:

Vd = (2K x L x I)/Cm

The equation to find voltage drop in a three phase circuit is:

Vd = (1.73K x L x I)/Cm

As seen in the two equations above, the only difference between the two is the constant, which changes from 1.73 to 2. In a perfectly balanced three phase delta system, the above three phase equation applies. For an unbalanced three phase delta system, this is no longer true. Rather than discuss phasor arithmetic or symmetrical systems analysis, I want to pose this question:

Would the resulting voltage drop calculation for an unbalanced three phase delta system fall somewhere in between the calculation for a single phase system and a balanced three phase system? Another way of putting it is, if you could boil down the unbalanced circuit to a constant, as the two formulae above contain, would the constant always reside between 1.73 and 2?

Specifically, this is for lighting, not motors, and assuming normal conditions, not short circuits or open circuits. Thank you for your time.

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