parallel feeder lengths

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The code gives no tolerance and specifies that the conductors be of equal length. That being said, the current divides based on the inverse of the length. That is the longest length will have the least current and the shortest length will have the most. A simple way to check is to add of the total length of all of the conductors for a single phase and then divide the individual conductor length by the total. You use this number to find the current on a phase conductor. An example. 4 1/0 in parallel for a 600 amp circuit. The lengths are 10, 10.5, 11 and 11.5'. The total length is 43'. Use the percentage of the length of the longest times the current to find the current in the shortest.
10' =23.36% of length
10.5'=24.42% of length
11'=25.58% of length
11.5'=26.74% of length
10'=26.74% of total current or 160.44A
10.5'=25.58% of total current or 153.48A
11' = 24.42% of total current or 146.52A
11.5' = 23.36% of total current or 140.16A
The maximum permitted current for a 1/0 is 150 amps. You can see that the two shorter conductors will be operating at above their rated ampacity. In this case the overload will not likely cause any harm, but on higher capacity circuits and larger wire with short runs, the shortest conductor can be overloaded enough to cause problems.
Yes, there will be some change in the current division as the higher loaded conductors heat up and their resistance goes up reducing the amount of current on that conductor.
Don
 
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