Wire Size Calculator (Current and Voltage Drop)

Wire Size Calculator (Current and Voltage Drop)

The smallest standard cable size that keeps the voltage drop within your limit, for DC, single-phase or three-phase circuits in copper or aluminium — with the actual drop, the voltage at the load, the power lost in the cable and a check against the ampacity from your code’s table.

Cable size for voltage drop

Current + length + drop → mm² and AWG
From the supply to the load. The page counts the return conductor itself.
For three-phase, the line-to-line voltage (e.g. 400 V).
Resistance rises with temperature. 70 °C is the limit for PVC insulation at full load; 20 °C gives the cold figure.
The NEC’s informational notes suggest 3% for a branch circuit and 5% in total; for 12 V systems 3% is common.
Look up the size this page recommends in your electrical code’s table or the cable datasheet, for your installation method and insulation.
From your code’s table; 1 at its reference ambient.
From your code’s table; 1 for a single circuit.
The cable drawn as two resistors, one for the conductor out and one for the return, each with the voltage it drops; the load gets what is left. For three-phase the page's drop is between lines (√3 × I × R). The dots show the load current (for AC, its RMS size). The conductors turn red if the current is over the derated ampacity you entered.
2.5mm²Example

16 A single-phase at 230 V over 25 m of copper, 70 °C, 3% allowed

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Voltage drop and cable size

Amin = k × L × I × ρT ÷ ΔVmax;   ΔV = k × I × ρT × L ÷ A;   ρT = ρ20 × (1 + α (T − 20));   k = 2 (DC, single-phase) or √3 (three-phase)
L
one-way cable length in metres
ρ20, α
copper 0.017241 Ω·mm²/m and 0.00393 /°C; aluminium 0.028264 and 0.00403
ΔVmax
allowed drop = supply voltage × percentage ÷ 100
A
the conductor area chosen from the IEC 60228 series, mm²

Worked example

16 A single-phase at 230 V over 25 m of copper, 70 °C, 3% allowed
ρ at 70 °C = 0.017241 × (1 + 0.00393 × 50) = 0.020629 Ω·mm²/m; allowed drop = 230 × 3% = 6.9 V
Amin = 2 × 25 × 16 × 0.020629 ÷ 6.9 = 2.39 mm² → next IEC size 2.5 mm²
R per conductor = 0.020629 × 25 ÷ 2.5 = 206.3 mΩ; drop = 2 × 16 × 0.2063 = 6.60 V (2.87%)
Load voltage 223.4 V; 105.6 W lost in the cable
AWG: the smallest gauge at least 2.39 mm² is 13 AWG (2.62 mm²); 12 AWG is the usual stocked size

Size for a 3% drop, 230 V single-phase, copper at 70 °C

One-way length10 A16 A20 A32 A
5 m1 mm²1 mm²1 mm²1 mm²
10 m1 mm²1 mm²1.5 mm²2.5 mm²
20 m1.5 mm²2.5 mm²2.5 mm²4 mm²
30 m2.5 mm²4 mm²4 mm²6 mm²
50 m4 mm²6 mm²6 mm²10 mm²
75 m6 mm²10 mm²10 mm²16 mm²
100 m6 mm²10 mm²16 mm²25 mm²
Voltage drop only, from this page’s formula; 1 mm² is the smallest size the page offers. Check every size against the ampacity your code allows for the installation.

Low-voltage DC runs: size for a 3% drop, copper at 20 °C

One-way length12 V, 5 A12 V, 20 A24 V, 20 A48 V, 20 A
1 m1 mm² (1.4%)2.5 mm² (2.3%)1 mm² (2.9%)1 mm² (1.4%)
2 m1 mm² (2.9%)4 mm² (2.9%)2.5 mm² (2.3%)1 mm² (2.9%)
5 m2.5 mm² (2.9%)10 mm² (2.9%)6 mm² (2.4%)2.5 mm² (2.9%)
10 m6 mm² (2.4%)25 mm² (2.3%)10 mm² (2.9%)6 mm² (2.4%)
Actual drop with the chosen size in brackets. Doubling the voltage quarters the area needed for the same power.

Sizing a cable for voltage drop

Every conductor has resistance, so current flowing through it loses voltage on the way. The load sees the supply voltage minus the drop in both the outgoing and the return conductor. The drop is current × resistance, and resistance is resistivity × length ÷ area. So for a given current and length, the only thing you choose is the area: this page finds the smallest area that keeps the drop within your limit, then rounds up to the next standard size from IEC 60228 (1.5, 2.5, 4, 6, 10 mm² and so on).

The example. A 16 A load at the end of 25 m of single-phase cable at 230 V, with 3% allowed, needs 2.39 mm² of copper at 70 °C, so 2.5 mm², which drops 6.60 V (2.87%) and wastes 106 W as heat. Cold copper at 20 °C would need only 2.00 mm², but a cable carrying its full load runs warm, so sizing at the conductor’s working temperature is the safe choice. Aluminium, with about 64% more resistivity, needs 3.94 mm², so 4 mm².

Low-voltage DC is where voltage drop bites. The same percentage is a far smaller number of volts: 10 A over 5 m at 12 V needs 4.79 mm² (6 mm²) for 3%. That is why solar, automotive and marine systems use thick cable, and why a higher system voltage (24 or 48 V) makes long runs so much cheaper: for the same power and the same percentage drop, doubling the voltage halves the current and doubles the volts you can afford to lose, so the area needed falls to a quarter.

Three-phase. In a balanced three-phase circuit the neutral carries no current, and the line-to-line drop is √3 × I × R rather than 2 × I × R. Enter the line-to-line voltage (400 V in Europe, 480 V or 208 V in North America). 32 A over 60 m at 400 V needs 3.43 mm², so 4 mm² for a 4.29% drop with a 5% limit.

Voltage drop is only half the job. A cable must also carry its current without overheating, and that limit, the ampacity, depends on the insulation’s temperature rating, how the cable is installed (in conduit, clipped to a wall, buried, in free air), the ambient temperature and how many circuits are grouped together. Those figures come from your electrical code (the NEC in the US, IEC 60364-5-52 or BS 7671 elsewhere) or the cable maker’s datasheet, and they are not reproduced here. Look up the size this page suggests, enter its tabulated ampacity and your code’s correction factors, and the page checks the load against them; if it fails, go up a size and look again. The overcurrent device must protect the cable too. For the physical size and resistance of an AWG gauge, see the AWG wire size calculator.

How much drop to allow. The NEC’s informational notes suggest no more than 3% on a branch circuit and 5% from the service to the farthest outlet; they are advice, not rules. IEC 60364-5-52 gives its own recommended maxima in an informative annex. Motors, LED drivers and electronics each have their own tolerance; long runs to pumps and outbuildings are where it most often matters.

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Frequently asked questions

What size cable do I need for 16 A over 25 m?

For a 3% drop at 230 V single-phase in copper at 70 °C, 2.39 mm² minimum, so 2.5 mm², giving a 2.87% drop. Then check 2.5 mm² against the ampacity your code gives for the way it is installed.

How do I calculate voltage drop?

For DC or single-phase: ΔV = 2 × length × current × resistivity ÷ area (length one way in metres, area in mm², copper about 0.0172 Ω·mm²/m cold). For three-phase, replace the 2 with √3 and compare with the line-to-line voltage.

Why does the length count twice?

Current flows out along one conductor and back along the other, so the resistance in the loop is two conductors’ worth. In a balanced three-phase circuit the return currents cancel and the factor is √3 instead.

Does this page tell me the ampacity?

No. Ampacity tables belong to your electrical code and depend on the installation. Enter the figure from your table and its correction factors, and the page checks the current against it.

Is aluminium cable worse than copper?

For the same area it has about 64% more resistance, so it needs roughly 1.6 times the area for the same drop. It is lighter and cheaper, and it needs terminals rated for aluminium.

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References

  1. IEC 60228:2004. Conductors of insulated cables: nominal cross-sectional areas.
  2. IEC 60028. International standard of resistance for copper; IEC 60889 Hard-drawn aluminium wire (resistivities at 20 °C).
  3. NFPA 70, National Electrical Code: informational notes to 210.19(A) and 215.2(A) (3% branch, 5% total voltage drop, advisory).
  4. IEC 60364-5-52. Low-voltage electrical installations — Selection and erection of electrical equipment — Wiring systems (current-carrying capacities, correction factors and voltage drop).