I am calculating the resistance of a 500 m aluminum wire with a 10 mm (AWG #2/0) diameter. I calculated the resistance according to this video, but I would like to double check with the community here to make sure the calculation is correct.

The calculation and result can be found here: WolframAlpha calculation

The result is 0.168704 Ω. If the community here could double check my calculations before I implement the wire into our community center project I would greatly appreciate it.

  • \$\begingroup\$ interfacebus.com/… \$\endgroup\$
    – jsotola
    Jan 2, 2023 at 19:46
  • \$\begingroup\$ this is an XY question ... you are asking about a solution to an unspecified problem, even if the solution may be misguided ... please ask about the actual problem you are trying to solve \$\endgroup\$
    – jsotola
    Jan 2, 2023 at 19:49
  • \$\begingroup\$ the question is a better fit at diy.stackexchange.com/questions \$\endgroup\$
    – jsotola
    Jan 2, 2023 at 19:49
  • \$\begingroup\$ Keep in mind that connectors, fuses, switches, everything else can also be lossy. Loss won't matter for 1 Ampere, but it could be critical for 100 Amps. \$\endgroup\$
    – rdtsc
    Jan 2, 2023 at 21:28
  • \$\begingroup\$ Your calculations are for a solid 10 mm wire/bar of aluminum. For an electrical distribution system you would likely be using aluminum wire that is stranded internally. The stranding helps the wire to be flexible. At that size the wire strands may be round or even rectangular in form. The resistance per meter will not match that of a solid piece of aluminum at the same apparent diameter. The best way to determine the wire resistance would be to inquire with the manufacturer of the wire. Also note that a size 2/0 AWG wire should have a diameter of 9.266 mm, (excluding the insulation). \$\endgroup\$
    – Nedd
    Jan 3, 2023 at 4:58

1 Answer 1


From this site, the resistivity of Al is \$\rho = 2.65\times 10^{-8} \ \mathrm{\Omega m}\$. The cross-section area of 5 mm radius wire is:

$$ \begin{aligned} A &= \pi \times (5\times 10^{-3})^2 \\ \\ &= 7.85 \times 10^{-5}\ \mathrm{mm^2} \\ \\ \end{aligned} $$

Resistance is:

$$ \begin{aligned} R &= \frac{\rho\times l}{A} \\ \\ &= \frac{2.65\times 10^{-8} \times 500}{7.85 \times 10^{-5}} \\ \\ &= 0.169\ \mathrm{\Omega} \end{aligned} $$


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