Wire size calculator
Working backwards from the drop you can tolerate gives the cable you actually need, which is often bigger than the one that just carries the current.
A
V
%
Minimum conductor area
3.7478mm2
- Equivalent solid diameter
- 2.1845
- Nearest AWG Round down to the next smaller number, which is the thicker wire.
- 11.46
- Area in thousand circular mils
- 7.3964
- Resulting current density
- 8.005 A/mm2
- Voltage budget
- 6.900 V
Show the working
- Current = 30
- One-way run length = 25
- Supply voltage = 230
- Maximum acceptable drop = 3
- System = 2
- Conductor = 1.724e-8
- Voltage budget = 230 * 3 / 100 = 6.9
- area_m2 = 2 * 30 * 1.724e-8 * 25 / 6.9 = 0.000 003 747 83
- Minimum conductor area = 0.000 003 747 83 * 1e6 = 3.747 83
- Equivalent solid diameter = sqrt(4 * 3.747 83 / pi) = 2.184 46
- Nearest AWG = 36 - 39 * log(2.184 46 / 0.127, 92) = 11.4627
- Area in thousand circular mils = 3.747 83 / 0.5067074790974979 = 7.396 43
- Resulting current density = 3.747 83 > 0 ? 30 / 3.747 83 : 0 = 8.004 64
Values are shown to six significant figures here; the arithmetic itself runs on exact fractions and is only rounded where a number is printed.
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The formula
- A = k I rho L / allowed drop
- AWG from diameter: n = 36 - 39 log_92(d / 0.127 mm)
This sizes for voltage drop only. The conductor must also satisfy the ampacity and fault-current requirements of the wiring code, which can demand more.
Always round to the next larger conductor, meaning the next lower AWG number.