Current Sharing Between Parallel Conductors

Also known as parallel feeder current sharing · unequal length parallel conductors · paralleled cable current division · why parallel runs must be the same length

I1=ItL2L1+L2I_{1} = I_{t} \frac{L_{2}}{L_{1} + L_{2}}

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Learning zone

When a load is too large for a single conductor, the standard answer is to run two or more in parallel. The assumption everyone makes is that they share equally. They do not — they share in inverse proportion to their resistances, and for identical conductors of the same material and size, resistance is proportional to length. Four hundred amperes split between a 30 m run and a 36 m run puts 218 A in the short conductor and 182 A in the long one: a 20 % difference in length has produced a 20 % imbalance in current, and the short conductor is now carrying more than its ampacity was chosen for while its partner loafs.

This is exactly why the codes require paralleled conductors to be identical in length, size, insulation type, conductor material and termination method, and why they set a minimum size below which paralleling is not permitted at all. It is not fussiness. The imbalance is invisible without a clamp meter, it does not trip anything — the breaker sees only the total — and it shows up as an overheated termination years later. The rule that catches people out on site is the one about the path, not just the wire: run all the conductors of one phase in one raceway and all of another phase in a second, and the differing inductive reactance unbalances them even when the lengths match perfectly. Each raceway or cable group must carry one conductor of each phase.

Two refinements worth knowing. This page's length-ratio model is the DC and low-frequency picture; on large conductors and long runs, reactance rather than resistance dominates the impedance, and then it is spacing and configuration, not length, that decide the sharing. And for more than two conductors the tidy two-branch form here stops working — with three or more, work in conductances (the reciprocals of the resistances) and give each conductor the share its conductance represents. As always, confirm the paralleling rules that apply to your installation against your own edition of the code.

Current Sharing Between Parallel Conductors
I1=ItL2L1+L2I_{1} = I_{t} \frac{L_{2}}{L_{1} + L_{2}}
ItI1L1L2
Where
  • I1I_{1}= Current in conductor 1 (A)
  • ItI_{t}= Total current (A)
  • L1L_{1}= Length of conductor 1 (m)
  • L2L_{2}= Length of conductor 2 (m)
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