Voltage Drop Explained: The K Formula, 3% and 5%

Where voltage drop comes from

Every conductor has resistance, the ratio of the voltage across it to the current through it, which is Ohm's law (HyperPhysics). On a single-phase circuit the current flows out to the load and back, so both conductors drop voltage, and the equipment sees the source voltage minus that drop.

Why it matters beyond efficiency: IAEI points out that manufacturers' installation instructions, which NEC 110.3(B) requires you to follow, often call for a minimum voltage for the equipment to work as intended (IAEI, 2017). A run that is fine on ampacity can still fall short of a nameplate requirement.

The K-factor formula

• Single-phase: VD = 2 × K × I × L ÷ CM
• Three-phase: VD = √3 × K × I × L ÷ CM
• Percent drop: VD ÷ source voltage × 100

VD is volts dropped, I is the load current in amps, L is the one-way length of the run in feet, and CM is the conductor area in circular mils. K is the resistance of a conductor one circular mil in area and one foot long: 12.9 ohm-circular mils per foot for copper and 21.2 for aluminum.

IAEI derives those K values from the dc resistances at 75 °C in NEC Chapter 9, Table 8, and calls the formulas generally acceptable for voltage drop work. The 75 °C basis reflects that most new lugs are rated 75 °C. IAEI's 2019 follow-up adds that the dc values are used for simplicity and consistency and are very close to the AC values in Table 9 (IAEI, 2019). We don't reproduce either table.

You can check copper's K from first principles. NBS Handbook 100 gives annealed copper a resistivity of 10.371 ohm-circular mils per foot at 20 °C, rising 0.0409 per degree, and allows 2% extra for stranding: (10.371 + 55 × 0.0409) × 1.02 = 12.87 at 75 °C, within 0.2% of 12.9. We found no matching primary source for aluminum conductors, so 21.2 rests on IAEI's figure alone.

The same formula rearranges two ways (IAEI):

• Area needed for a target: CM = 2 × K × I × L ÷ allowed VD
• Longest run for a size: L = CM × allowed VD ÷ (2 × K × I)

For three-phase, swap the 2 for √3. IAEI writes 1.732; our calculator uses the exact square root, which differs in the fifth significant figure.

Circular mils from the AWG definition

You don't need a printed table for conductor area. The American Wire Gauge is defined by two diameters, No. 0000 at 0.4600 in and No. 36 at 0.0050 in, with 38 sizes between them in geometric progression. Area in circular mils is the square of the diameter in mils (NBS Handbook 100). So for gauge n:

• diameter (mils) = 5 × 92^((36 − n) ÷ 39), with 1/0 = 0, 2/0 = −1, 3/0 = −2, 4/0 = −3
• area (cmil) = diameter squared

These are computed from the definition. Printed tables round the diameter first, so the fourth significant figure can differ: IAEI uses 83,690 cmil for 1 AWG, where the definition gives 83,693. Sizes above 4/0 are named by area (250 kcmil is 250,000 cmil). NBS's shortcut holds too: every three gauge sizes, the area doubles (the exact ratio is 2.005).

Worked examples

All numbers below are computed from the formulas above with unrounded areas.

1. 120 V branch circuit, 12 AWG copper, 16 A, 100 ft one way. 2 × 12.9 × 16 × 100 ÷ 6,529.95 = 6.32 V, a 5.27% drop, leaving about 113.68 V at the load. On 10 AWG the same run drops 3.98 V, or 3.31%.
2. The same 1,920 W load at 240 V. Current falls to 8 A. On 12 AWG over 100 ft the drop is 3.16 V, or 1.32%. Doubling the voltage halved the volts dropped and cut the percentage to a quarter.
3. 240 V condenser circuit, 10 AWG copper, 24 A, 75 ft. 4.47 V, or 1.86%.
4. Three-phase feeder, 480 V, 4/0 copper, 150 A, 200 ft. √3 × 12.9 × 150 × 200 ÷ 211,600 = 3.17 V, or 0.66%. The single-phase formula with the same inputs gives 3.66 V; three-phase is always √3 ÷ 2, or 0.866, of the single-phase result.
5. Copper vs aluminum, 240 V single-phase, 100 A, 150 ft. 1/0 copper drops 3.67 V (1.53%); 1/0 aluminum drops 6.03 V (2.51%). Aluminum's K is 1.64 times copper's, so it needs about 64% more area for the same drop. Going two gauge sizes larger only adds 59% more area, which is why 3/0 aluminum (3.79 V, 1.58%) still drops slightly more than 1/0 copper.

What the NEC says about 3% and 5%

The familiar numbers come from informational notes. In the wording EC&M quoted in 2015 (Informational Note No. 4 to 210.19(A)), branch-circuit conductors sized to keep the drop under 3% at the farthest outlet, with feeders and branch circuits together under 5%, "provide reasonable efficiency of operation" (EC&M). A matching note covers feeders.

Informational notes are explanatory material, not mandatory code requirements; IAEI cites NEC 90.5(C) for that. EC&M makes the practical point: these are recommendations, so a design sized by the ampacity tables can pass inspection without meeting them (EC&M). EC&M also notes a place where the NEC does set a mandatory limit: Article 647, for sensitive electronic equipment, at half those values (1.5% and 2.5%). It isn't the only one: IAEI notes that some sections require a voltage-drop calculation, and its list includes 695.7, for fire pumps.

Note numbers move between editions. The 3% note was No. 4 in the 2015 and 2017 articles cited here, and NFPA says the NEC is being reorganized, starting with the 2023 edition and aiming to finish in 2029 (NFPA). Read the edition your jurisdiction enforces through NFPA's free access, which is read-only and requires a sign-in. This is general information, not code advice; the authority having jurisdiction decides what applies on your job.

When to upsize, and when the formula isn't the whole answer

Take a 240 V, 30 A load at 150 ft on copper with a 3% target. The allowed drop is 7.2 V, so the area needed is 2 × 12.9 × 30 × 150 ÷ 7.2 = 16,125 cmil. 10 AWG (10,383 cmil) falls short at a 4.66% drop; 8 AWG (16,510 cmil) gives 7.03 V, or 2.93%.

Before you write that up:

• Voltage drop is not ampacity. EC&M notes that the NEC ampacity tables don't take voltage drop into account. The reverse is just as true: the K formula knows nothing about ampacity, insulation rating or terminal temperature. Size for each and install the larger conductor. Our calculator does not check ampacity.
• Upsize where it pays. EC&M suggests making the long home runs larger rather than every conductor, and distributing at the higher voltage for as long as you can; example 2 shows why.
• Know the method's limits. The K formula uses dc resistance at 75 °C. AC methods that add conductor reactance and the load's power factor can return a somewhat different number for the same run; IAEI calls the dc method generally acceptable for voltage drop calculations.
• Read the nameplate. If the equipment instructions set a minimum voltage, that, not 3%, is the number your run has to meet.

Next step

Run your circuit through the voltage drop calculator. It gives the drop for a chosen size, the smallest size for a target percentage, or the longest run for a size, for copper or aluminum, single- or three-phase. If you are starting from kilowatts or kVA rather than amps, convert first with the Ohm's law calculator, using the nameplate power factor; the kW to amps guide shows the single- and three-phase arithmetic.