The formula used by the calculator
For DC and simplified single-phase circuits, required CM = 2 × K × I × L ÷ allowed volts lost. For a balanced three-phase circuit, replace 2 with √3. In both equations, K represents the conductor resistance constant (12.9 for copper, 21.2 for aluminum at 75°C), I is load current in amps, and L is one-way length in feet.
Multiply system voltage by your percentage target to find allowed volts lost: a 3% target on 240 V gives 7.2 V, whereas 3% on a 24 V DC run leaves only 0.72 V.
How do you calculate wire size?
The size of an electrical conductor refers to its cross-sectional area (measured in circular mils (CM) or AWG in the US). The larger the cross-sectional area, the more current the wire can transmit with lower resistance—very similar to water flowing through a wider pipe with less pressure loss. We derive the required conductor area from Ohm's law and conductor resistivity:
where:
- V_drop – Voltage drop calculated between the supply source and the farthest end of the wire, measured in volts (V_drop = V_source × (% drop ÷ 100));
- I – Maximum current running through the conductor, measured in amps (convert power using the watts to amps calculator);
- K – Specific resistance constant of the conductor material at 75°C (12.9 Ω·CM/ft for copper and 21.2 Ω·CM/ft for aluminum);
- L – One-way length of the wire run from panel to load, measured in feet; and
- CM – Cross-sectional area of the conductor, measured in circular mils (where 1,000 CM = 1 kcmil).
Calculating wire sizes for three-phase AC
For a balanced three-phase AC circuit, three current-carrying phase conductors are used instead of a two-wire loop. The calculator accepts line-to-line voltage and line current, modifying the required area of a single conductor to:
The factor of √3 ≈ 1.732 converts between phase and line quantities in a balanced three-phase system, replacing the factor of 2. The calculator automatically selects the next AWG size up from its listed 14 AWG – 4/0 table so the voltage-drop limit is never exceeded. To check apparent power feeding that circuit, use the kVA to amps calculator.
Worked example: 240 V, 30 A, 100 ft copper
Using K = 12.9 and a 3% target: CM = 2 × 12.9 × 30 × 100 ÷ 7.2 = 10,750 circular mils. Because 10 AWG offers 10,380 circular mils, it sits below the minimum area, so the calculator steps up to 8 AWG at 16,510 circular mils.
Rechecking 8 AWG copper over that same 100 ft run gives 4.69 V lost, or 1.95%. You must still check NEC Table 310.16 ampacity, 125% continuous-load sizing, lug temperature ratings, and breaker size before pulling wire.
| Check step | Value | Outcome |
|---|---|---|
| Allowed voltage drop (3% of 240 V) | 7.20 V | Target limit |
| Calculated minimum area | 10,750 CM | Requires ≥ 10,750 CM |
| 10 AWG area | 10,380 CM | Too small by 370 CM |
| 8 AWG area | 16,510 CM | Selected candidate (1.95% drop) |
Input checklist
- 01
Use design current
Do not automatically substitute breaker rating for the calculated load.
- 02
Measure one-way distance
Enter source to load once; the formula applies its circuit multiplier.
- 03
Select material and phase
Copper, aluminum and balanced three-phase runs use different inputs.
- 04
Round up by area
Choose the first available conductor whose circular-mil area meets or exceeds the result.
- 05
Complete the safety checks
Verify ampacity, terminals, correction, adjustment, protection and installation method.
What the formula cannot decide
- Whether the load calculation is correct
- Which ampacity temperature column applies
- Whether adjustment or correction factors reduce allowable ampacity
- Whether the conductor fits and is listed for every terminal
- Whether local rules or equipment instructions require a different design