Skip to content
Measure the Build logoMeasure the Build

Electrical

Voltage Drop Formula

Wire has resistance, so some of the voltage you send never arrives. On a short run that loss is irrelevant. On a long one it decides your conductor size, and it will happily overrule the ampacity table you just used.

The formula

VD = (2 × R × I × L) ÷ 1000

  • VD — voltage drop, in volts
  • R — conductor resistance, ohms per 1,000 ft
  • I — load current, amps
  • L — one-way run length, feet

The 2 is the round trip. Current leaves the panel, travels the length of the run, and comes back, so it crosses that distance twice and loses voltage in both directions. Dividing by 1,000 converts the published per-1,000-foot resistance to your actual length. Percentage drop is then simply VD divided by the nominal system voltage.

Three-phase uses √3 (about 1.732) in place of the 2. That factor is trigonometry rather than a code value, and everything else in the calculation is unchanged.

Worked example: the detached garage

A 20 A, 120 V circuit running 100 feet out to a detached garage, in 12 AWG copper. Ampacity is not remotely a problem — 12 AWG copper is rated 20 A in the 60°C column against a 20 A load. Now the other check:

  1. Resistance of 12 AWG copper: 1.93 Ω per 1,000 ft.
  2. VD = (2 × 1.93 × 20 × 100) ÷ 1000 = 7.72 V.
  3. As a percentage: 7.72 ÷ 120 = 6.43%.
  4. Voltage arriving at the garage: 112.3 V.

That is more than double the 3% recommendation, on a conductor that passed its ampacity check with room to spare. A motor starting on 112 volts draws more current to make the same power, heats up, and lives a shorter life. Incandescent light output falls off sharply. Electronics mostly cope until they do not.

Walking up the gauges

Same circuit, same 100 feet, larger conductors:

CopperΩ/1000 ftDropPercentAt 3%
12 AWG1.937.72 V6.43%fails
10 AWG1.214.84 V4.03%fails
8 AWG0.7643.06 V2.55%passes
6 AWG0.4911.96 V1.64%passes

You need 8 AWG — two sizes above what ampacity called for — to get a 20 amp load 100 feet away on 120 volts. This is the reason a wire size chart alone cannot answer "what size wire do I need" for anything but a short run.

How far each gauge will actually go

Rearranging the formula for length gives the run at which each conductor hits 3%. One-way feet, copper:

Copper15 A @ 120 V20 A @ 120 V30 A @ 240 V
14 AWG39 ft29 ft39 ft
12 AWG62 ft47 ft62 ft
10 AWG99 ft74 ft99 ft
8 AWG157 ft118 ft157 ft
6 AWG244 ft183 ft244 ft
4 AWG390 ft292 ft390 ft

The 240 V column is roughly four times the 120 V distance at the same power, not twice. Doubling the voltage halves the current for a given load and doubles the percentage headroom, and the two effects compound. That is the whole argument for running a subpanel at 240 V and splitting it locally rather than pulling long 120 V branch circuits.

Aluminum drops more at the same gauge

Aluminum's higher resistivity shows up directly in this calculation:

GaugeCopper ΩAluminum ΩRatio
6 AWG0.4910.8081.65×
4 AWG0.3080.5081.65×
2 AWG0.1940.3191.64×
1/0 AWG0.1220.2011.65×

An aluminum conductor of the same gauge drops roughly 1.6 times as much voltage over the same distance. Aluminum is still frequently the right economic choice on long service and feeder runs, but the size you land on has to satisfy both ampacity and drop, and drop is often what sets it.

Recommendation, not requirement — mostly

The 3% and 5% figures come from an Informational Note attached to NEC 210.19(A). Under NEC 90.1, Informational Notes are explanatory material and are not enforceable as code requirements. An inspector cannot fail a residential branch circuit purely for exceeding 3% in a jurisdiction that has adopted the NEC unmodified.

That said, a number of jurisdictions adopt the figures as mandatory amendments, and commercial energy codes such as ASHRAE 90.1 impose their own limits. More to the point, the recommendation is good engineering regardless of enforceability: nobody has ever regretted a conductor that delivered full voltage. Treat 3% as a design target rather than a rule you are permitted to break.

Where this fits

Voltage drop is the third of three checks, and the one most often skipped. Ampacity from NEC Table 310.16 says whether the conductor overheats. Voltage drop says whether the load gets usable voltage. Only after both are settled does conduit fill tell you what pipe holds the result.

The wire gauge calculator runs the first two together and reports which constraint is binding, which is the part worth knowing — if voltage drop is what set your conductor size, a shorter route or a higher system voltage may be cheaper than the copper.

Sources

Grouped by manufacturer or publisher, then by document. Expand a group for the cited document, its verified-on date, and every value taken from it — all of it stays in the HTML either way.

NFPA(14 documents, 102 values)

Frequently asked questions

What is the voltage drop formula?

For single-phase or DC: VD = 2 × R × I × L ÷ 1000, where R is conductor resistance in ohms per 1,000 feet, I is the load in amps and L is the one-way run in feet. The 2 accounts for the round trip — current flows out and back, so it crosses the full length twice.

How do I calculate voltage drop percentage?

Divide the drop in volts by the nominal system voltage and multiply by 100. In the worked example on this page, 7.72 V lost on a 120 V circuit is 6.43%. Percentage matters more than volts, because the same drop is twice as damaging at 120 V as at 240 V.

What is the maximum allowable voltage drop?

NEC 210.19(A) Informational Note recommends 3% for a branch circuit alone and 5% for feeder and branch circuit combined. Informational Notes are explanatory and not enforceable under NEC 90.1, though a number of jurisdictions and commercial energy codes adopt the figures as mandatory.

Why is voltage drop worse on 120 volt circuits?

Because the percentage is measured against the system voltage. Losing 3.6 V is 3% of a 120 V circuit and only 1.5% of a 240 V one. Doubling the voltage also halves the current for the same power, and the drop falls with it, so a 240 V circuit gains twice over.

Does voltage drop use one-way or round-trip length?

Enter the one-way distance and let the formula's factor of 2 handle the return. Measuring the round trip and then also multiplying by 2 is the most common arithmetic error here, and it produces a drop twice as large as reality — which sends you up two gauge sizes you do not need.

How do I fix a circuit that fails voltage drop?

Increase the conductor size, shorten the run, or raise the system voltage. Increasing wire size is usually the only one available. In the example here, going from 12 AWG to 8 AWG brings a 6.4% drop down to 2.55% — two gauge sizes for the same length.