Wire Size Calculator

A conductor has to pass two separate tests, and this calculator runs both and hands you whichever answer is larger. The tests measure different things and neither implies the other, which is the whole reason a page like this exists.

Test one is ampacity. The conductor has to carry the current without its insulation cooking. That figure comes from NEC Table 310.16, read at the temperature column of the insulation you are installing. Distance does not enter into it at all — a 10 ft run and a 500 ft run of the same conductor have exactly the same ampacity. Two rules ride along with the table: 240.4(D) caps the overcurrent device on 14, 12 and 10 AWG regardless of what the column says, and 110.14(C) holds the circuit to its lowest-rated termination, which is why the 75 °C column is usually the honest one to work from.

Test two is voltage drop. The conductor also has to deliver usable voltage at the far end, and here distance is the entire problem. The calculation solves for cross-sectional area directly, in circular mils:

CM = (2 × K × I × D) ÷ Vdrop for single phase, with 1.732 in place of the 2 for a balanced three-phase load. D is the one-way distance in feet, Vdrop is your percentage target converted to volts. K comes from a single site-wide definition — 12.9 for copper, with aluminium 64% higher because it conducts that much worse — and thevoltage drop calculator carries the full note on where that figure comes from and how far it moves between references. What matters here is which way the error points: the value used runs slightly pessimistic, so a size chosen against it has margin rather than needing some.

Both tests are run at the design current rather than the raw nameplate figure: a continuous load — three hours or more at maximum — is multiplied by 1.25 first, and that inflated figure is carried into the drop calculation as well as the ampacity lookup. That is the conservative reading, and the result panel shows the drop at the real operating current alongside it so you can see the difference the choice makes.

The larger of the two answers wins, and the tool names which one it was. That label matters more than it looks: if drop won, a shorter route or a higher system voltage buys you a size back, and more copper is the expensive way to solve it.

A 40 A continuous load — a piece of shop equipment running all day — at 240 V single phase, 150 ft from the panel, copper conductors with 75 °C insulation, and a 3% drop target. Both paths, end to end.

The design current. The load is continuous, so 40 × 1.25 =50 A. Everything below runs on that figure.

The ampacity path. Reading down the 75 °C copper column of Table 310.16: 10 AWG is 35 A, which is short. 8 AWG is 50 A — exactly the design current, and 8 AWG is past the 240.4(D) sizes so nothing caps it. The ampacity-constrained answer is 8 AWG, with zero margin.

The voltage-drop path. The budget is 3% of 240 V =7.2 V. So the area needed is (2 × 12.9 × 50 × 150) ÷ 7.2 = 193,500 ÷ 7.2 = 26,875 circular mils. Looking up the ladder: 6 AWG is 26,240 cmil — 635 short. 4 AWG is 41,740 cmil, which clears it. The drop-constrained answer is 4 AWG.

The answer. 4 AWG is two sizes larger than 8 AWG, so4 AWG is the recommendation and voltage drop is what drove it. At that size the actual drop is 193,500 ÷ 41,740 = 4.64 V, or1.93% — well inside target, and 1.55% at the real 40 A operating current.

The interesting number is the one that lost. 6 AWG comes to 193,500 ÷ 26,240 = 7.37 V, or 3.07%. It misses a 3% target by seven hundredths of a percentage point, and jumping to 4 AWG for that margin is a real cost across a long pull. This is exactly where the 3% figure being a recommendation rather than enforceable code becomes a design conversation rather than a compliance one — and exactly why the calculator reports what the next size down would have done.

The calculator is worth running for any circuit you are actually going to install. This table is for the other situation — standing in front of a panel deciding whether a proposed run is going to be a problem at all, and wanting a feel for where the boundary sits rather than a specific answer.

Every row is 240 V single phase, copper, 75 °C insulation, against a 3% target, and the load column is taken as the design current already: multiply a continuous load by 1.25 before you look it up. Read the last column first. The pattern it shows is that the crossover distance falls as the load grows in the small sizes and then rises again — 20 A is already drop-bound at 100 ft, 40 A is still balanced there, and 100 A is back to being ampacity-bound. That is not a quirk of the numbers; it is the ampacity ladder stepping in coarser jumps than the circular-mil requirement does, so how much slack a size has depends on where on the ladder you land.

What the table cannot tell you is what to do about a row you do not like. A load that comes out drop-bound by two sizes is usually cheaper to fix by moving the supply than by upsizing the conductor — and at 480 V the same load in the same conductor drops a quarter as much.