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	Subject - How Important is Conductor Length? tony thrower A 1600 Ampere, 480 Volt, 3 phase, balanced load is connected to the power source using (4) parallel runs of 600 kcmil thwn copper. This maximum value of current is only reached periodically and then for only a short period of time, say 45 minutes. All the conductors are 25 feet in length with the exception of one "A" phase conductor. This conductor is 23 feet long. Using the resistance for uncoated copper conductors in table 8 what is the current through each "A" phase conductor at full load? This is just for fun! Scott Vickrey At the risk of learning something here, I get: 922.84 amps for the 3 - 25 foot conductors 922.91 amps for the single 23 conductor tony thrower 600 kcmil thwn copper has an allowable ampacity of 420 amperes 4 runs of 600 meeting the requirements of art. 310.4 should provide a conductor capable of carrying of 1680 amperes.In this case the current would divide evenly, 420 amps through each conductor.I believe this is the goal of art. 310.4. Since our conductors are not the same length the current will not evenly divide.More current will go through the shortest conductor because it's resistance is least of the (4) conductors in "A" phase. At the full load of 1600 amperes the current in the 25 ft. conductors would be 391.48 amperes each.The current in the 23 ft. conductor would be 425.53 aamperes. Since the allowable ampacity is 420 amperes the short conductor is overloaded. I think this would be correct. Ryan_J Tony: Would you mind showing the me the math? Thanks alot, Ryan Scott Vickrey Egads! I didn't think about checking the wire capacity. If I had this would have shown me how far off I was. I would also like to see how you worked this out. tony thrower The (4) 600 kcmil conductors are connected parallel each having a resistance based on individual conductor length. The resistance of 600 kcmil copper is .0000704 ohms per ft. @ 75 degrees(c). Conductor A(1) 23' x .0000704= .0016192 ohms Conductor A(2) 25' x .0000704= .00176 ohms Conductor A(3) 25' x .0000704= .00176 ohms Conductor A(4) 25' x .0000704= .00176 ohms Using the Reciprocal Method for calculating parallel resistance I determined the total resistance is .000430638 ohms. Since 1600 amperes will be flowing through this total resistance I determined the voltage drop across the parallel conductors, which must be equal seeing they are in parallel,using Ohm's Law. 1600 amperes x .000430638 ohms = .689021277 volts Next I'll use Ohm's Law to determine the current through each conductor.Current will take the path of least resistance. Conductor A(1) .689021277 volts / .0016192 ohms = 425.53 amperes Conductor A(2) .689021277 volts / .00176 ohms = 391.48 amperes Conductor A(3) .689021277 volts / .00176 ohms = 391.48 amperes Conductor A(4) .689021277 volts / .00176 ohms = 391.48 amperes JimmyDee I find this thread very interesting. As I was thinking about it, I read this, "600 kcmil copper is .0000704 ohms per ft. @ 75 degrees(c)". As the temperature of the short conductor rises, the resistance of it will rise also. Do you think that there would reach an equilibrium where all 4 conductors would carry equal amperage? I realize this is less than ideal but would it happen? Just wondering. Jim Ryan_J Well done Tony. Thanks :)