Wire Gauge Calculator

Wire size is decided by two separate questions, and the answer is whichever one demands the thicker wire.

The first question is heat. Every conductor has an ampacity — the current it can carry continuously without its insulation cooking. That comes straight from the NEC table, and it does not care how long your run is.

The second is voltage drop. Copper has resistance, so some of your voltage is lost as heat on the way out and on the way back. Over 10 feet nobody notices. Over 80 feet at 12 volts, a fridge browns out. This is why long runs need thick wire even when the current is modest.

1. Step 1 How many volts can you afford to lose? allowed_drop_V = system_voltage × drop_target% At 12V and 3%, that is only 0.36 V. At 240V and 3%, it is 7.2 V — which is why low-voltage runs need fat cable.
2. Step 2 Convert that into circular mils of copper required_cmil = (2 × K × amps × one_way_feet) ÷ allowed_drop_V K is 12.9 for copper and 21.2 for aluminum. The 2 is there because current travels out and back.
3. Step 3 Round up to a real wire size The next size in the AWG table whose circular mils meet or exceed the requirement
4. Step 4 Now check heat The wire must also meet the NEC 60°C ampacity for your current A 100 ft, 5-amp run might be drop-limited at 10 AWG; a 3 ft, 60-amp run is ampacity-limited at 4 AWG.
5. Step 5 Take the thicker of the two final_size = max(drop_result, ampacity_result)
6. Step 6 Split it across parallel conductors if you want to per_conductor_cmil = required_cmil ÷ runs per_conductor_amps = amps ÷ runs Two equal conductors have twice the cross-section and half the resistance, which is exactly three AWG sizes — two 12 AWG behave like one 9 AWG.

Example: 30 A at 12V over 20 ft one-way, 3% drop. Allowed drop is 0.36 V, so required circular mils = (2 × 12.9 × 30 × 20) ÷ 0.36 = 43,000 — that is 3 AWG. Ampacity only needs 10 AWG. Voltage drop wins, so you run 3 AWG. Split across two conductors, each needs 21,500 circular mils, which is 6 AWG — and two 6 AWG really is equivalent to one 3 AWG.

Cable size is decided by two separate questions, and the answer is whichever one demands the larger cross-section.

The first question is heat. Every conductor has a current-carrying capacity — the current it can carry continuously without cooking its insulation. That comes from the IEC 60364-5-52 tables (Table 4D2A in BS 7671), and it does not care how long your run is.

The second is voltage drop. Copper has resistance, so some of your voltage is lost as heat on the way out and on the way back. Over 5 metres nobody notices. Over 25 metres at 12 volts, a fridge browns out. This is why long runs need thicker cable even when the current is modest.

1. Step 1 How many volts can you afford to lose? allowed_drop_V = system_voltage × drop_target% At 12 V and 3% that is only 0.36 V. At 230 V and 3% it is 6.9 V — which is why low-voltage runs need much larger cable.
2. Step 2 Convert that into a cross-section required_mm² = (k × ρ × amps × length_m) ÷ allowed_drop_V ρ is 0.0214 Ω·mm²/m for copper and 0.0352 for aluminium. k is 2 for DC and single-phase, because current goes out and comes back, or √3 for a balanced three-phase circuit.
3. Step 3 Round up to a cable you can buy The next standard size: 1.0, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120 mm² …
4. Step 4 Now check heat The cable must also meet its tabulated capacity for your current A 40 m, 16 A run is drop-limited at 4 mm²; a 2 m, 45 A run is capacity-limited at 10 mm².
5. Step 5 Take the larger of the two final_size = max(drop_result, capacity_result)
6. Step 6 Split it across parallel conductors if you want to per_conductor_mm² = required_mm² ÷ runs per_conductor_amps = amps ÷ runs Two equal conductors have twice the cross-section and half the resistance, so two 4 mm² behave like one 8 mm² — rounded up, a 10 mm² cable.

Example: 16 A at 230 V over 40 m one-way, 3% drop. Allowed drop is 6.9 V, so required cross-section = (2 × 0.0214 × 16 × 40) ÷ 6.9 = 3.98 mm² — that rounds up to 4 mm². Capacity alone would only need 1.5 mm². Voltage drop wins, so you run 4 mm².

• Ampacity figures come from NEC Table 310.16, 60°C column. That is the conservative column and matches the terminal rating on most residential breakers and devices. Equipment rated for 75°C terminals may allow a smaller conductor — an electrician can confirm.
• No ambient-temperature or conduit-fill derating is applied. More than three current-carrying conductors in one conduit, or an attic above 86°F, both reduce ampacity — sometimes by 30% or more.
• The formula uses DC resistance. On AC circuits above roughly 1/0 AWG, inductive reactance adds a little more drop than shown, and the effect grows with conductor size.
• Resistivity constants of 12.9 (copper) and 21.2 (aluminum) ohm-cmil/ft assume a conductor at about 75°C. Cold wire has slightly less resistance and less drop.
• Distance is one-way. The calculator doubles it for the return path. Three-phase circuits use a different multiplier and are covered in the Europe & UK mode, not here.
• NEC 240.4(D) caps overcurrent protection at 15 A for 14 AWG, 20 A for 12 AWG and 30 A for 10 AWG copper regardless of what any other table says.
• Continuous loads — anything running more than three hours, like an EV charger — must be sized at 125% of the load current under NEC 210.19. Enter the already-adjusted figure if that applies to you.
• 18 and 16 AWG appear here for low-power DC accessory wiring only; they are not NEC branch-circuit sizes.
• Parallel conductors are assumed identical in material, size, length and termination, which is what makes the current divide evenly. NEC 310.10(H) only permits paralleling at 1/0 AWG and larger on AC circuits; below that it is a low-voltage DC technique used in vans, boats and battery banks.
• Resistance figures use the same 12.9 and 21.2 ohm-circular-mil-per-foot constants as the voltage-drop calculation, which correspond to a conductor at roughly 75°C. At room temperature copper is about 8% lower.
• The mm² equivalent shown is a direct area conversion, not a recommendation. The two systems use different ampacity tables, so a metric size that matches on cross-section may not be compliant for the same current.
• A 10% drop target reflects ABYC and SAE practice for non-critical automotive and marine loads only. It is not acceptable for building wiring under any code.

• Current-carrying capacities follow IEC 60364-5-52 and BS 7671 Table 4D2A, installation method C — cable clipped direct to a surface, 30 °C ambient, PVC insulation, two loaded conductors. This is the common domestic case.
• No correction factors are applied. Grouping several circuits together (Cg), ambient temperatures above 30 °C (Ca), thermal insulation and buried runs all reduce capacity, sometimes by 50% or more. Your electrician applies these; this calculator does not.
• Resistivity is 0.0214 Ω·mm²/m for copper and 0.0352 for aluminium, which correspond to the conventional 0.0225 and 0.036 figures quoted at 70 °C conductor temperature to within about 5%.
• Three-phase sizing assumes a balanced load and uses the √3 line-to-line factor. An unbalanced three-phase circuit needs the neutral considered separately.
• The formula uses DC resistance. On AC circuits above roughly 95 mm², inductive reactance adds a little more drop than shown.
• BS 7671 recommends a maximum drop of 3% for lighting and 5% for other uses, measured from the origin of the installation — so if part of the budget is already spent on a submain, tighten the target here.
• Circuit protection is not sized here. Cable capacity must exceed the protective device rating, and RCD/RCBO requirements for socket and outdoor circuits are separate obligations.
• 0.5 and 0.75 mm² are flexible-cable sizes for appliance flexes and low-power DC accessories, not fixed wiring. Aluminium is not used below 16 mm².
• Parallel conductors are assumed identical in material, cross-section, length and arrangement. IEC 60364 and BS 7671 permit paralleling but attach conditions on protection and arrangement that need designing rather than improvising.
• Resistance figures use 0.0214 and 0.0352 ohm-mm² per metre, the same constants as the voltage-drop calculation, corresponding to a conductor at roughly 70°C. At 20°C copper is about 8% lower.
• The AWG equivalent shown is a direct area conversion, not a recommendation. The two systems use different current-carrying tables, so an AWG size that matches on cross-section may not be compliant for the same current.
• A 10% drop target reflects ABYC and SAE practice for non-critical automotive and marine loads only. It is not acceptable for fixed building wiring under BS 7671 or IEC 60364.