Two methods
Approximate-K
The field formula. It uses the conductor's circular-mil area (Chapter 9, Table 8) and a resistivity constant K:
Vd = (F × K × I × L) ÷ CM
- F is 2 for single-phase and √3 (≈1.732) for three-phase.
- K is 12.9 for copper and 21.2 for aluminum.
- I is the load current in amps; L the one-way length in feet.
- CM is the conductor area in circular mils.
Exact AC (Chapter 9, Table 9)
Table 9 gives AC resistance (R) and reactance (X) per 1,000 ft by size, conductor material and raceway. The calculator combines them with the power factor into an effective impedance, Z = R·PF + X·sin θ, then Vd = F × I × Z × L ÷ 1,000.
Worked examples
- 20 A at 120 V, 100 ft, 12 AWG copper, single-phase. (2 × 12.9 × 20 × 100) ÷ 6,530 = 7.90 V, which is 6.58% of 120 V and well over 3%.
- Same circuit on 10 AWG. 4.97 V (4.14%), still over. The smallest copper that gets this run under 3% is 8 AWG, at 2.60%.
- 200 A at 480 V three-phase, 500 ft, 250 kcmil copper. (1.732 × 12.9 × 200 × 500) ÷ 250,000 = 8.94 V, or 1.86%. That's why long feeders run at 480 V.
Example 1 uses the full 20 A because that's how the question usually gets asked. Size for the load you expect: the same run carrying 16 A drops 5.27%. A 20 A circuit sized at 3% for its full rating is a good circuit; one sized at 3% for a 5 A load is a long run waiting for someone to plug in a heater.
What the NEC says, and doesn't
The 3% and 5% figures are informational notes in 210.19(A) (branch circuits) and 215.2(A) (feeders). Informational notes aren't enforceable (90.5(C)), so an inspector can't fail a job on them unless the jurisdiction has adopted a limit. Excessive drop shows up anyway: motors that run hot, lights that dip when a compressor starts, and electronics that reset.
Worked examples on this page are computed by the same engine as the Voltkit app. The page cites NEC 2023 sections but doesn't reproduce NFPA text or tables; use your copy of the code for the exact wording.