Cable runs · mm² · metres
Set the cable size, the run length and the load, and see how much of the supply the cable itself consumes. Opens on metric sizes and metres; the AWG and feet switches are still there if that is what you are working in.
3.89% is above the 3% limit you set. About 3.89% of the 230 V supply is lost in the conductors under these conditions, leaving 221.05 V at the load. That is 8.95 V dropped across the run, out and back.
Your voltage drop appears here
Enter a supply voltage, a load current and a cable length. The result updates as you type.
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3.89%
8.95 V lost against a3% limit
3.89% is above the 3% limit you set. About 3.89% of the 230 V supply is lost in the conductors under these conditions, leaving 221.05 V at the load. That is 8.95 V dropped across the run, out and back.
To bring it down: increase the conductor size, shorten the run, split the load across parallel conductors, or supply the load at a higher voltage. Doubling the conductor area roughly halves the drop; halving the run length halves it exactly.
One-way run length
L = entered lengthL = 40 m × 3.2808Result: 131.23 ft
The formulas below take the one-way distance from source to load and apply the system multiplier to it.
Conductor resistance per 1000 ft
R = table value for size and material6 mm² copper, at 75 °CResult: 1.065 Ω per 1000 ft
Calculated from IACS resistivity for the nominal cross-sectional area.
Voltage drop
V_drop = 2 × I × R × LV_drop = 2 × 32 A × 1.065 Ω/1000 ft × 0.1312 (1000 ft)Result: 8.95 V
A two-wire circuit carries the full load current out and back, so the current travels twice the one-way run length.
Voltage drop percentage
%V_drop = (V_drop ÷ V_source) × 100%V_drop = (8.95 V ÷ 230 V) × 100Result: 3.89%
Voltage at the load
V_load = V_source − V_dropV_load = 230 V − 8.95 VResult: 221.05 V
Power lost in the conductors
P_loss = I² × R_path × number of current-carrying pathsP_loss = 32² × 0.1398 Ω × 2Result: 286.3 W
Only resistance dissipates power. Reactance stores and returns energy each cycle rather than turning it into heat.
Resistance: calculated from IACS resistivity for the nominal area
Formula
The engine behind this page works in ohms per 1000 ft, because that is the unit the published conductor tables use, and converts on the way in and out. The two forms are the same calculation: 1 Ω per 1000 ft is 3.2808 Ω per km.
Reference
Multiply a figure below by your load current and by the run in hundreds of metres to get an approximate single-phase drop, before you reach for the calculator at all.
| Cable size | Copper Ω/km | Copper V per A per 100 m | Aluminium V per A per 100 m |
|---|---|---|---|
| 1.5 mm² | 13.978 | 2.796 | 4.604 |
| 2.5 mm² | 8.387 | 1.677 | 2.762 |
| 4 mm² | 5.242 | 1.048 | 1.726 |
| 6 mm² | 3.495 | 0.699 | 1.151 |
| 10 mm² | 2.097 | 0.419 | 0.691 |
| 16 mm² | 1.310 | 0.262 | 0.432 |
| 25 mm² | 0.839 | 0.168 | 0.276 |
| 35 mm² | 0.599 | 0.120 | 0.197 |
Computed with the site calculator from conductor resistance at 75 °C. Copper resistance per km is shown for reference. Resistance only, no reactance term, so the figures are exact at unity power factor.
Read the pattern rather than the individual numbers: doubling the conductor area roughly halves the drop, and aluminium sits about 1.6 times higher than copper at the same size. The full table for every size, in both AWG and mm², is on the voltage drop chart.
Questions
Multiply the load current by the resistance of the cable run and by the system multiplier. In metric terms: voltage drop = 2 × I × (R per km ÷ 1000) × L, with I in amperes, R per km the conductor resistance in ohms per kilometre and L the one-way cable run in metres. Use √3 in place of the 2 for a balanced three-phase cable. Divide the result by the supply voltage and multiply by 100 to get the percentage.
Enter the distance from the supply to the load and leave the calculator set to one-way, or enter the total length of conductor in the circuit and switch it to round-trip. The 2 in the formula is already accounting for the return conductor, so entering a there-and-back figure as a one-way length doubles the answer. For a two-core cable, the cable length is the one-way distance: the second core is the return path the multiplier covers.
It depends on the current, the run length and the supply voltage together, which is why there is no single answer. Set your limit in the calculator and it flags whether the size you picked clears it; the wire size calculator solves the same problem the other way round and returns the smallest size that fits. Remember that the answer also has to satisfy the cable’s current-carrying capacity, which is a separate check.
The resistance of the conductors is the same either way, so at unity power factor the drop is the same. The difference shows up in the reactance, which depends on how far apart the conductors sit: cores bundled inside one sheath are closer together than single conductors spaced in a raceway, so the inductive reactance is lower. The reference data on this site is for single conductors in conduit, which makes a multi-core result slightly conservative on AC below unity power factor.
It should be, and this calculator lets you set it. Conductor resistance rises with temperature, so a cable running warm under load drops more than the same cable cold. The published data here is referenced to 75 °C, which is a loaded conductor rather than an ambient one. Calculating at 20 °C because that is what the cable data sheet quotes will understate the drop of a cable that is actually working.
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