Feeders, motor circuits and distribution at 208, 480 and 600 V — line-to-line drop with the √3 factor, power factor correction and parallel sets built in.
In a balanced three-phase circuit the return current of each phase flows through the other two phases rather than a dedicated return conductor, so the effective round-trip factor drops from 2 to √3 (≈1.732). That 13% saving — plus the higher voltages — is a big part of why industrial power is three-phase.
A motor's starting inrush is typically 6× its running current, so a feeder that drops 3% at full load can sag 18% during starting — enough to drop out contactors or stall the start. For motor feeders, many engineers size for 2% running drop or verify the starting sag separately.
| I | line current, amps |
| L | one-way length, feet |
| R | Ω/kft, Chapter 9 Table 8 |
| PF | power factor of the load |
√3 (≈1.732) replaces the single-phase ×2 — three-phase has no dedicated return conductor. Percent uses line-to-line voltage.
This calculator computes line-to-line drop for a balanced load, which is the standard feeder calculation. For single-phase loads tapped off one phase, treat them as a 1-phase circuit at the phase voltage.
Each added parallel set divides the effective conductor resistance — two sets halve the drop. Per NEC 310.10(G), paralleled conductors must be 1/0 AWG or larger and matched in length, material and termination.