Audited —·Last updated 31 Jul 2026·5 citations·Tier 1·0 uses

Voltage Drop Calculator — Volts, Percent and the 3 % Length Limit

Work out voltage drop from wire size, current and run length using NEC Chapter 9 Table 8 resistances, with percent drop and the 3 % length limit.

Voltage Drop Calculator

System
Conductor material
Conductor size
The current the circuit actually carries, not the breaker rating. Voltage drop is proportional to it, so a circuit loaded to 80 % of its breaker drops 80 % as much as one loaded to the trip point.
A
The distance from the source to the load, measured one way along the route the cable actually takes. Do not double it — the calculator already accounts for the return conductor.
ft
The nominal system voltage at the source — 120, 240, 208 or 480 for common systems, line-to-line for three-phase. It sets the percentage, not the volts lost: the same drop is a bigger problem on a lower-voltage system.
V
Voltage drop
5.28
The drop as a share of the source voltage — the number the NEC's informational notes are written in terms of, and the one that transfers between systems. 3 % is the branch-circuit suggestion, 5 % the feeder-plus-branch suggestion, and neither is an enforceable requirement.
Volts lost in the conductors
6.336 V
Voltage at the load
113.664 V
One-way length at 3 %
56.8182 ft
Conductor resistance
1.98 Ω/1000 ft
Conductor area
6,530 cmil
One-way conductor resistance
0.198 Ω
Power wasted as heat
101.376 W
Reading the result
12 AWG copper on a single-phase or DC circuit, 16 A over 100 ft one way: 6.336 V lost in the conductors, which is 5.28 % of a 120 V supply, leaving 113.664 V at the load. The run also turns 101.4 W into heat in the wire. That is past both figures the informational notes suggest — 3 % for a branch circuit and 5 % for feeder and branch combined. The 3 % point for this conductor and current is only 56.8 ft one way. Step up a conductor size, shorten the run, raise the system voltage, or split the load. Those percentages come from Informational Notes — NEC 210.19(A) Informational Note No. 4 and 215.2(A)(1) Informational Note No. 2 in the 2023 edition — and NFPA 70 §90.5 says informational notes "are informational only and are not enforceable as requirements". They are design guidance, not a pass/fail test. The adopted NEC edition and local amendments vary by jurisdiction. Scope: this page sizes nothing. Ampacity governs conductor selection first, with derating for temperature, conduit fill and continuous load, and voltage drop can only ever push a conductor larger — never smaller. The resistances are the DC values from NEC Chapter 9 Table 8 at 75 °C; AC impedance including inductive reactance (Table 9) is higher on large conductors in steel conduit, and unity power factor is assumed. A licensed electrician or engineer must sign off before anything is installed.

Background.

Every conductor is a resistor, and every resistor carrying current loses voltage. Run 16 A down 100 feet of 12 AWG copper and 6.34 V never arrives — a 120 V circuit delivers 113.66 V at the far end, and just over a hundred watts is turned into heat inside the wall. That is a 5.28 % loss, and it is why long runs get fatter wire even when the smaller wire could carry the current safely all day.

This calculator takes the conductor size, the metal, the current, the one-way run length and the system voltage, and returns the drop in volts, the drop as a percentage, the voltage left at the load, the heat wasted in the conductors, and — the output that usually answers the real question — the length at which this conductor at this current would reach 3 %.

The resistances come from NEC Chapter 9, Table 8, 'Conductor Properties', which lists DC resistance at 75 °C along with each size's circular-mil area. The stranded columns are used throughout, including for the sizes where the table also lists solid, because stranded has slightly higher resistance and therefore gives the more conservative answer. Two independently published copies of that table were compared before any value was used, and they agree on every size this page carries.

The 3 % and 5 % figures need care, because they are widely quoted as if they were rules and they are not. NEC 210.19(A) Informational Note No. 4 describes branch-circuit conductors sized to prevent a voltage drop exceeding 3 percent at the farthest outlet, with the total on feeders and branch circuits together not exceeding 5 percent, as providing reasonable efficiency of operation. NEC 215.2(A)(1) Informational Note No. 2 carries the parallel language for feeders. But NFPA 70 §90.5 states plainly that informational notes are informational only and are not enforceable as requirements. They are design guidance that the whole trade has adopted as a target, not a pass/fail test that an inspector applies — and the adopted NEC edition and any local amendments vary by jurisdiction.

What this page does not do matters as much as what it does. It does not select a conductor. Ampacity governs that choice first — with derating for ambient temperature, for the number of current-carrying conductors in a raceway, and for continuous load — and voltage drop can only ever push a conductor larger, never smaller. It does not size overcurrent protection. It does not apply a power factor: the arithmetic assumes unity, which is optimistic for motor loads. And the resistances are DC values. Real AC circuits also carry inductive reactance, which NEC Chapter 9 Table 9 covers and this page does not; on large conductors in steel conduit the AC impedance is materially higher than the DC resistance, so the drop computed here is understated in exactly the installations where it matters most.

Use the result the way a designer would: as the number that tells you whether the run is comfortable, marginal or hopeless, and as the length limit that tells you how much room is left. Then have a licensed electrician or engineer confirm the design against the code edition their jurisdiction has actually adopted. A voltage-drop figure is an input to that conversation, not a substitute for it.

What is voltage drop calculator?

Voltage drop is the voltage consumed by the conductors themselves rather than delivered to the load. It follows straight from Ohm's law: the conductor has resistance, the load current flows through it, and the product of the two is voltage that appears across the wire instead of across the equipment. On a single-phase or DC circuit the current travels out on one conductor and back on another, so both drop voltage and the total is 2 × R × I, where R is the resistance of one conductor over the run. On a balanced three-phase circuit there is no neutral carrying the full current, and the line-to-line drop works out to √3 × R × I — about 87 % of what the same conductors would drop single-phase, which is one of several reasons three-phase distribution is used for large loads. Expressed as a percentage of the source voltage, the drop is what the code's informational notes and most equipment specifications are written in terms of, because it transfers between systems: 3 % is 3.6 V on a 120 V circuit and 14.4 V on a 480 V one, and the effect on the equipment is comparable. The consequences of excessive drop are practical rather than dramatic. Incandescent and resistive loads simply run cooler and dimmer. Motors are the real problem: torque falls with the square of the voltage, so a 10 % drop costs about 19 % of the starting torque, and the motor draws more current to compensate, which increases the drop further and heats both the motor and the conductors.

How to use this calculator.

  1. Choose single-phase (or DC) or three-phase. The multiplier is 2 for the first and √3 for the second, and using the wrong one is a 15 % error.
  2. Choose the conductor material and size. Aluminium of the same size drops about 60 % more than copper.
  3. Enter the current the circuit will actually carry, not the breaker rating — drop is proportional to current.
  4. Enter the one-way length, measured along the route the cable takes. Do not double it; the calculator accounts for the return conductor already.
  5. Enter the nominal source voltage, line-to-line for three-phase.
  6. Read the percentage first, then the length limit beside it: that is how far this conductor could run at this current before reaching 3 %.
  7. If the drop is too high, the options in order of effectiveness are: raise the system voltage, shorten the run, increase the conductor size, or split the load across circuits.
  8. Confirm the conductor against ampacity requirements separately — this page does not do that — and have a licensed professional review the design.

The formula.

V_drop = 2 · R · I · L (single-phase or DC) V_drop = √3 · R · I · L (three-phase) % = V_drop ⁄ V_source × 100
R · conductor resistance per foot = (NEC Table 8 Ω/1000 ft) ⁄ 1000I · load current in amps — the actual load, not the breaker sizeL · one-way circuit length in feet2 · single-phase and DC: current goes out on one conductor and back on another√3 · three-phase line-to-line, balanced loadV_source · nominal system voltage, line-to-line for three-phaselength at 3 % · 0.03 · V_source · 1000 ⁄ (multiplier · Ω/1000 ft · I)heat · I²R × 2 conductors single-phase, × 3 conductors three-phase

Start with the conductor. NEC Chapter 9 Table 8 gives stranded 12 AWG uncoated copper a DC resistance of 1.98 Ω per 1000 ft at 75 °C, and a cross-sectional area of 6,530 circular mils. Over a 100 ft run that is 0.198 Ω in one conductor.

On a single-phase or DC circuit the current has to come back, so it passes through two conductors of that length. The drop is 2 × 0.198 Ω × 16 A = 6.336 V. Against a 120 V source that is 5.28 %, leaving 113.664 V at the load. On a balanced three-phase circuit the same conductors and current would drop √3 × 0.198 × 16 = 5.487 V instead — a factor of √3/2, about 87 %, because there is no single return conductor carrying the whole current.

The heat is a separate calculation and it is easy to get wrong. It is I²R summed over the conductors that actually carry current: two of them single-phase, giving 16² × 0.198 × 2 = 101.376 W. For a single-phase circuit that happens to equal the drop times the current, which is why the shortcut usually goes unnoticed — but on a three-phase circuit it does not, because the drop carries a √3 and the heat carries a 3. This page computes each from its own definition.

The length limit inverts the whole thing. Setting the drop to 3 % of the source voltage and solving for length gives 0.03 × 120 × 1000 ÷ (2 × 1.98 × 16) = 56.82 ft. That is the number most people actually want: this conductor, at this current, can run 56.82 ft one way before the drop reaches 3 %, and the 100 ft run in the example is nearly twice that.

Rounding happens once, at the return boundary, to ten decimal places, and both classification bands read the unrounded percentage. That matters directly here: the notes say 'not exceeding 3 percent', so a drop of exactly 3.000 % is inside the recommendation, and a drop of 3.0000001 % is not. The bands are tested at exactly those edges using 10 AWG aluminium, which Table 8 gives as exactly 2.00 Ω/1000 ft — at 15 A on 120 V that is precisely 0.06 V per foot, so 3 % falls at exactly 60 ft and 5 % at exactly 100 ft, with no rounding anywhere in the chain.

One cross-check worth knowing about. The resistances used here are code-table values, but they can be reproduced from first principles: the ASTM geometric definition of the AWG scale puts 12 AWG at 80.808 mils diameter and therefore 6,529.95 circular mils, matching the table's 6,530; and the International Annealed Copper Standard resistivity of 1.7241 µΩ·cm at 20 °C, corrected to 75 °C, gives 1.932 Ω per 1000 ft for solid 12 AWG copper against the table's published 1.93. The stranded figure this page uses, 1.98, is about 2.5 % higher than that, which is the lay of the strands — a real physical difference, not a discrepancy. The agreement of an entirely independent derivation with the published table is the strongest evidence available that the numbers were transcribed correctly.

What the arithmetic assumes: unity power factor, a balanced three-phase load, DC resistance rather than AC impedance, a 75 °C conductor, and no parallel conductor sets. Each of those makes the real drop equal to or larger than the computed one, so the answer should be read as a floor rather than a worst case.

A worked example.

Example

A 20 A branch circuit in 12 AWG copper, loaded to 16 A — the 80 % continuous-load figure — running 100 ft one way from a 120 V panel. NEC Chapter 9 Table 8 gives stranded 12 AWG uncoated copper 1.98 Ω per 1000 ft at 75 °C and an area of 6,530 circular mils, so one conductor over this run is 0.198 Ω. Current out and back through two of them drops 2 × 0.198 × 16 = 6.336 V, which is 5.28 % of 120 V and leaves 113.664 V at the outlet under load. The two conductors together turn 101.376 W into heat inside the wall. That 5.28 % is past both figures the informational notes suggest — 3 % for a branch circuit and 5 % for feeder and branch together — and the length output says why: at 16 A this conductor reaches 3 % after 56.8181818182 ft one way, so the run is nearly twice as long as the guidance contemplates. The fixes, in order of effectiveness: run it at 240 V instead, which quarters the percentage; shorten the run; or step up to 10 AWG, whose 1.24 Ω per 1000 ft would bring the same circuit to 3.31 %. What this result does not settle is whether 12 AWG is the right conductor at all — that is an ampacity question, decided first and separately, and voltage drop can only ever make the answer larger.

conductor Materialcopper
conductor Size12
one Way Length Ft100
source Voltage120
load Current16
system Typesingle-phase

Frequently asked questions.

Is 3 % voltage drop a code requirement?
No. The 3 % branch-circuit figure and the 5 % feeder-plus-branch figure appear in NEC 210.19(A) Informational Note No. 4 and 215.2(A)(1) Informational Note No. 2, and NFPA 70 §90.5 states that informational notes 'are informational only and are not enforceable as requirements'. They describe conductors that provide reasonable efficiency of operation. In practice the whole trade designs to them and many specifications and local amendments make them contractual, so treat them as a design target you should have a reason to miss — but an inspector is not failing an installation on the note itself. Check what your jurisdiction has actually adopted.
Do I use the one-way length or the round-trip length?
One way. Enter the distance from the source to the load along the route the cable actually takes, and the calculator handles the return path itself — that is what the factor of 2 in the single-phase formula is. Entering the round-trip distance doubles the answer. The other half of this mistake is measuring in a straight line: cable follows walls, joists and conduit runs, so the real length is usually a good deal more than the map distance, and it is worth pacing the route rather than estimating.
Why is the three-phase multiplier √3 instead of 2?
Because there is no single conductor carrying the whole current back. In a balanced three-phase circuit the three line currents sum to zero at every instant, so each conductor carries current out and the others carry it back, and working the line-to-line voltage difference through the phasor arithmetic leaves √3 × R × I rather than 2 × R × I. The practical result is that the same conductors, current and length drop about 87 % as much on three-phase as on single-phase. Note that the conductor heating does not follow the same factor: that is I²R in three conductors, not two, so a three-phase run wastes more heat while dropping less voltage.
How much worse is aluminium than copper?
About 60 % more voltage drop for the same size. Table 8 gives stranded 12 AWG copper 1.98 Ω per 1000 ft and aluminium 3.18, a ratio of 1.61, and that ratio holds within a few percent across every size on this page. The usual response is to go up two AWG sizes in aluminium to land near the copper figure — 2 AWG aluminium at 0.319 Ω per 1000 ft is close to 4 AWG copper at 0.308. This page reads both columns from the table directly rather than applying a multiplier, because the ratio is not exactly constant.
What actually goes wrong if the drop is too high?
Less than people fear on resistive loads and more than they expect on motors. Heaters and incandescent lamps simply deliver less power — a 5 % voltage drop costs about 10 % of the heat output, since power goes as voltage squared. Electronics with switching power supplies mostly do not care until the input falls below their range, at which point they draw more current, making the drop worse. Motors are the real problem: torque falls with the square of the voltage, so a 10 % drop costs roughly 19 % of starting torque, and a motor that struggles to start draws locked-rotor current for longer, heating both itself and the conductors.
Does this calculator tell me what size wire to use?
No, and it should not. Conductor selection starts with ampacity — the current the conductor can carry without exceeding its insulation temperature — and that involves the ampacity tables plus derating for ambient temperature, for the number of current-carrying conductors bundled together, and for continuous load. Voltage drop is applied afterwards and can only ever push the conductor larger, never smaller. This page tells you whether the conductor you have chosen will deliver acceptable voltage over the distance, which is the second question, not the first. A licensed electrician or engineer must do both.
Why does my answer differ slightly from another calculator?
Usually one of four reasons, all of which this page states openly. First, solid versus stranded: for 14 through 8 AWG the table lists both, and stranded is a couple of percent higher — this page uses stranded throughout, giving the more conservative answer. Second, DC resistance versus AC impedance: this page uses Table 8's DC values, while some calculators use Chapter 9 Table 9's AC impedance, which is higher on large conductors in steel conduit. Third, power factor: this page assumes unity, while some tools apply a load power factor and get a larger drop. Fourth, temperature: Table 8's values are at 75 °C, and a calculator working at 20 °C or 90 °C will differ by several percent in either direction.
How long can my circuit be?
That is exactly what the length output on this page answers. It solves the same equation backwards: at your current, conductor and system voltage, it reports the one-way distance at which the drop would reach 3 %. In the worked example — 12 AWG copper at 16 A on 120 V — the answer is 56.82 ft. Two things make that number move a long way. Doubling the system voltage quadruples the allowable length, because the volts available at 3 % double while the drop per foot stays the same. And halving the current doubles it. Raising the voltage is almost always the cheapest fix on a long run.

References& sources.

  1. [1]NFPA 70, National Electrical Code, Chapter 9, Table 8 'Conductor Properties' — the source of every resistance and area used on this page. The table gives DC resistance at 75 °C (167 °F) in ohms per 1000 ft for uncoated copper (solid and stranded) and for stranded aluminium, together with each size's circular-mil area. The values were taken from this published copy of the 2014 edition of the table: 14 AWG 4,110 cmil / 3.14 Ω-kFT Cu stranded / 5.06 Al; 12 AWG 6,530 / 1.98 / 3.18; 10 AWG 10,380 / 1.24 / 2.00; 8 AWG 16,510 / 0.778 / 1.26; 6 AWG 26,240 / 0.491 / 0.808; 4 AWG 41,740 / 0.308 / 0.508; 2 AWG 66,360 / 0.194 / 0.319; 1/0 105,600 / 0.122 / 0.201; 2/0 133,100 / 0.0967 / 0.159; 4/0 211,600 / 0.0608 / 0.100; 250 kcmil 0.0515 / 0.0847; 350 kcmil 0.0367 / 0.0605; 500 kcmil 0.0258 / 0.0424. The edition retrieved is stated because NFPA 70 itself is copyrighted; confirm against the edition your jurisdiction has adopted. Retrieved 2026-07-29.
  2. [2]Voltage Lab, 'NEC Chapter 9 Table 8 Explained: Conductor Properties Guide' — an independent second reproduction of the same table, using 'NEC 2023 as the reference baseline' and advising readers to 'verify the final conductor values against the official NEC edition adopted in your jurisdiction'. Consulted specifically to cross-check the numbers taken from the first source rather than to supply them. It agrees on every value shared between the two: 14 AWG 4,110 cmil / 3.14 Ω-kFT, 12 AWG 6,530 / 1.98, 10 AWG 10,380 / 1.24, 8 AWG 16,510 / 0.778, 6 AWG 26,240 / 0.491, 4 AWG 41,740 / 0.308, 2 AWG 66,360 / 0.194, 1/0 105,600 / 0.122, 4/0 211,600 / 0.0608, 250 kcmil 250,000 / 0.0515, 500 kcmil 500,000 / 0.0258 — eleven sizes, from a copy referenced to a different NEC edition, with no disagreement. Retrieved 2026-07-29.
  3. [3]OrbitalJump, 'Voltage Drop Limits in NEC 210.19: Recommendation vs. Requirement and When It Controls Wire Size' — source of the informational-note wording and section numbering used on this page, quoting NEC 210.19(A) Informational Note No. 4 (2023 edition) on conductors 'sized to prevent a voltage drop exceeding 3 percent at the farthest outlet … and where the maximum total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent', the parallel language in 215.2(A)(1) Informational Note No. 2 for feeders, and NFPA 70 §90.5: 'Informational notes are informational only and are not enforceable as requirements.' Used as a secondary source because the primary text is behind NFPA's registration wall. Retrieved 2026-07-29.
  4. [4]NFPA 70, National Electrical Code — the authoritative text for Chapter 9 Table 8, 210.19(A), 215.2(A)(1) and §90.5. NFPA publishes free read-only access to its codes behind account registration. This document was NOT opened during the preparation of this page: every code figure quoted here is attributed to the two published reproductions cited above and to the secondary source for the informational notes, and the edition each of those referenced is stated rather than implied. Readers designing a real installation should read the adopted edition directly, together with any local amendments. Access: gated (registration required). Reference date 2026-07-29.
  5. [5]OpenStax (Rice University), University Physics Volume 2, §15.4 'Power in an AC Circuit'. Source of the conductor-heating relation used for the wasted-power output: Equation 15.13 gives the average power in a resistance as 'P_ave = ½I_0V_0 = I_rms·V_rms = I_rms²R'. This page applies the I²R form to each current-carrying conductor separately — two on a single-phase run, three on a three-phase one — rather than multiplying the line-to-line drop by the current, which would be wrong for three-phase by a factor of √3. Retrieved 2026-07-29.
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