Voltage Drop Calculator — Percent Drop & End Voltage

Calculate how many volts a wire run loses over distance via circular mils, checked against the NEC's 3% guideline — always verify sizing with an electrician.

Your run
A
ft
V
Results update as you type — no submit
ResultsLIVE
Voltage drop
6.58%
7.9 V lost over 100 ft (30.5 m)
Volts lost7.9 V
Voltage at load112.1 V
NEC 3% targetOver 3% — size up
Formula
VD = 2 × 12.9 × 20 A × 100 ft ÷ CM
= 7.9 V = 6.58%
This run exceeds the NEC-recommended 3% branch/feeder target. Stepping up one wire size lowers the drop — confirm the final conductor with a licensed electrician.

Voltage drop is the voltage a conductor loses to its own resistance on the way to the load. Over a long run it can leave motors humming, lights dim, and electronics unhappy, so it is worth checking before you pull wire. This calculator uses the circular-mils method, the same one in most electrical references: VD = 2 × K × I × L ÷ CM. Here K is the conductor constant — 12.9 for copper and 21.2 for aluminum — I is the load current, L is the one-way run length in feet, and CM is the wire's circular-mil area from NEC Chapter 9, Table 8. The results are planning estimates; confirm any final design with a licensed electrician.

The leading 2 accounts for the round trip: current flows out on one conductor and back on the other, so the electrons travel twice the run length. For a three-phase circuit the calculator swaps that 2 for √3 (about 1.732), the standard line-to-line factor. Enter your amperage, one-way length, gauge, material, and system voltage, and it returns the volts lost, the percent of source voltage, and the voltage that actually reaches the load.

The number to watch is the percentage. The NEC does not mandate a limit, but Informational Note No. 4 to 210.19(A) and Informational Note No. 2 to 215.2(A)(1) recommend keeping a branch circuit or feeder at or below 3%, and total drop at or below 5%. If your run comes in over 3%, the usual fix is to step up one wire size, which increases CM and cuts the drop.

Worked example
Load 20 A · one-way length 100 ft · 12 AWG copper (6530 cmil) · 120 V · single-phase
VD = 2 × 12.9 × 20 A × 100 ft ÷ 6530 = 7.90 V
percent = 7.90 V ÷ 120 V = 6.59% (over the 3% target)
end voltage = 120 V − 7.90 V = 112.1 V — step up to 10 AWG to fix it

Use VD = 2 × K × I × L ÷ CM: two times the conductor constant (12.9 copper, 21.2 aluminum), times the current, times the one-way length in feet, divided by the wire's circular mils. Divide the result by the source voltage for the percentage. A 20 A, 100 ft, 12 AWG copper run at 120 V drops about 7.9 V, or 6.6%.

SOURCES NFPA 70, National Electrical Code, 2023 edition, Chapter 9, Table 8 (Conductor Properties) (the "Area, Circular Mils" column gives 14 AWG as 4,110 cmil, 12 AWG as 6,530 and 10 AWG as 10,380, the CM values this calculator divides by. The same table's DC resistance column is where K comes from rather than from any published constant: stranded uncoated copper is 1.98 Ω/kFT at 12 AWG and 1.24 at 10 AWG, and resistance per thousand feet times circular mils divided by 1,000 gives 12.93 and 12.87 Ω·cmil/ft — the 12.9 used here. Aluminum's 3.25 and 2.04 Ω/kFT give 21.22 and 21.18, the 21.2. NFPA sells the Code; free read-only access requires registration, so no stable public URL is cited.) · NFPA 70, National Electrical Code, 2023 edition, 210.19(A) Informational Note No. 4 (branch circuits) and 215.2(A)(1) Informational Note No. 2 (feeders) (both recommend sizing conductors so voltage drop does not exceed 3 percent at the farthest outlet, with combined drop on feeders and branch circuits not exceeding 5 percent. NFPA 70 §90.5 states that informational notes are explanatory and not enforceable as Code requirements, which is why this page presents the figures as a target rather than a limit; individual jurisdictions sometimes adopt them as mandatory local amendments. Cited without a URL for the same reason as above.) · NIST Guide to the SI (Special Publication 811), Appendix B.9: Factors for units listed by kind of quantity or field of science (inch to metre 2.54 E-02, printed in boldface, which the table's caution note defines as exact. A circular mil is the area of a circle one thousandth of an inch in diameter, so 1 cmil = (π/4)(0.0254 mm)² = 5.067075×10⁻⁴ mm²; 6,530 cmil is therefore 3.3088 mm², which matches the 3.31 mm² printed in NEC Table 8's own metric column — an independent check that the circular-mil areas above and their SI equivalents agree.)
NEC-referenced estimate for planning only — electrical work must meet local code and be done or inspected by a licensed electrician; local code governs.