35 mm2 cable size, resistance and voltage drop calculator
Metric cable is sized by its nominal copper cross-section, so 35 mm2 is the area itself rather than a gauge number. That makes the arithmetic direct: resistance is resistivity times length divided by area, and everything else follows from it.
- Drop as a percentage
- 0.257 % Within the usual 3% final-circuit limit
- Nominal cross-section
- 35 mm2
- Equivalent solid diameter
- 6.675 58
- Equivalent AWG Fractional. Round down to the next AWG number for the next thicker wire.
- 1.83
- Area in circular mils
- 69.0734
- Resistance per kilometre
- 0.492 571 ohm/km
- Resistance of the one-way run
- 0.014 777 1 ohm
- Power lost as heat
- 11.822 W
- Conductor weight, one core
- 9.408
- Longest run within 3% drop
- 350.20
Show the working
- One-way run length = 30
- Load current = 20
- Supply voltage = 230
- Circuit = 2
- Conductor = 1.724e-8
- Conductor temperature = 20
- Conductor density = 8960
- Nominal cross-section = 35 = 35
- Equivalent solid diameter = sqrt(4 * 35 / pi) = 6.675 58
- Area in circular mils = (6.675 58 / 25.4)^2 * 1000 = 69.0734
- Equivalent AWG = 36 - 39 * ln(6.675 58 / 0.127) / ln(92) = 1.827 91
- r_per_m_20 = 1.724e-8 / (35 * 1e-6) = 0.000 492 571
- r_per_m = 0.000 492 571 * (1 + 0.00393 * (20 - 20)) = 0.000 492 571
- Resistance per kilometre = 0.000 492 571 * 1000 = 0.492 571
- Resistance of the one-way run = 0.000 492 571 * 30 = 0.014 777 1
- Voltage drop over the run = 2 * 20 * 0.014 777 1 = 0.591 086
- Drop as a percentage = 0.591 086 / 230 * 100 = 0.256 994
- Power lost as heat = 2 * 20^2 * 0.014 777 1 = 11.8217
- Conductor weight, one core = 35 * 1e-6 * 30 * 8960 = 9.408
- Longest run within 3% drop = 230 * 0.03 / (2 * 20 * 0.000 492 571) = 350.203
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
- resistance = rho x length / area
- single phase drop = 2 x I x R (out and back)
- three phase drop = sqrt(3) x I x R (line to line)
- temperature correction: R(T) = R(20) x (1 + 0.00393 (T - 20)) for copper
The cross-section is nominal. IEC 60228 fixes the maximum resistance per kilometre rather than an exact area, so a real stranded conductor is slightly larger than the nominal figure and its resistance is a little above the ideal solid value computed here.
The equivalent diameter is what a solid conductor of this area would measure. A stranded conductor of the same area is larger overall because of the gaps between strands.
This is a voltage drop calculation, not a current rating. Ampacity depends on the insulation, the installation method, grouping and ambient temperature.
Frequently asked
What is 35 mm2 in AWG?
Inverting the gauge formula gives n = 36 - 39 log_92(d / 0.127 mm), and the result is shown above. It is almost never a whole number, so pick the next lower AWG number to stay at or above the metric area.
How long a run can 35 mm2 carry before the drop matters?
The calculator shows the length at which the drop reaches 3% of the supply voltage for the current you entered. Halve the current and that length doubles.