I am looking for the calculation that show what the amp draw will be on Each leg of a parellel feeder. Specifically if 1 leg is shorter then the other leg. I'm sure someone here has that calculation.

In general with everything else being equal the current will divide in inverse proportion to the percentage of length. Add the lengths of both (or all) runs, and divide the length of each run by the total. The most current will be on the shortest run. For example, with runs of 48' and 52', 52% of the total current will be on the 48' run and 48% on the 52' run. If you have 3 sets of 31, 33, and 36', you will have 36% of the current on the 31' run, 33% on the 33' one and 31% on the 36' run. However as reno said, the code requires that the runs of parallel conductors be of equal length. Don

Use ohm's law, I = E / R, first determine the resistance using Table 8 (or 9) in Chapter 9 of the NEC. In the longer runs, we determine that the conductors do not need to be exactly the same length, in the short runs you had better be very close.

If the cable has a reasonable length, more than 20 metres, 30 cm is not that much of an issue (1.5% of length). The shorter cable will take a bit more current, heats up a little more, resistance increases I²R and will equalise currents with the longer cable.

Ideally they should be the same length (as per code).

The product of rotation, excitation and flux produces electricty.

Earl, It works with any combination of lengths, but yes I chose easy numbers so I wouldn't need a calculator. It is always easier than ohms law because you never need to look up the resistance of the wire. Even to use ohms law, you need to know the total current to come up with the current on each of the parallel runs. Don