Why do I need ratios to choose the right wire?

A wire's resistance is its length divided by its cross-section, times a number for the metal. Double the length and resistance doubles; double the area and it halves. That ratio decides how much voltage and power a wire wastes, which is why long runs and big loads need thicker wire than the shortest-distance choice.

18 AWG16 AWG14 AWG12 AWG0.527 Ωcopper · out and back
Wire resistance
0.527 Ω
Total conductor
40 m
Cross-section
1.31×10^-6 m²

Challenge: Get the wire's resistance under 0.1 Ω.

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Play

Stretch the cable run and switch gauges. Thicker wire (smaller AWG number) keeps the line lower.

Challenge: Get the wire's resistance under 0.1 Ω. The box under the picture turns green when you get it.

Stuck? Pick one of the examples from the “Try an example” menu, or press “New example.”

Understand

R=ρLAR = \frac{\rho L}{A}

A wire resists like a long, narrow pipe resists water. Length makes it harder, cross-section makes it easier:

R=ρLAR = \frac{\rho L}{A}

That's a ratio: L/AL/A scaled by the metal's resistivity ρ\rho. On the graph, each line is one gauge. Resistance grows in a straight line with length, and thicker wire (smaller AWG number) gives a flatter line.

Use

Every input has a unit menu, so you can type values in the units you already have. Results follow your units.

Show the work

  1. Resistance of a wireR = \frac{\rho L}{A}
  2. Area of 16 AWGA = \frac{\pi d^2}{4} = \frac{\pi (1.291\,\mathrm{mm})^2}{4} = 1.309\,\mathrm{mm^2}
  3. With your numbersR = \frac{1.724\times 10^{-8}\,\Omega\cdot\mathrm{m} \times 40\,\mathrm{m}}{1.309\times 10^{-6}\,\mathrm{m^2}} = 0.5269\,\Omega

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Enter the one-way distance, the gauge, and the metal. Leave "count out and back" on for a normal two-wire circuit. Resistance is at 20 °C; hot wire resists a few percent more.

  • Link this resistance into the voltage-drop exhibit to see what a run loses.
  • Check local code for the minimum gauge allowed for your breaker.

For learning and estimation. Verify with applicable codes, standards, and a qualified professional before using in design, construction, or safety-critical work.

Cheat card

R=ρLAR = \frac{\rho L}{A}
A=πd24A = \frac{\pi d^2}{4}
dAWG=0.127×92(36−n)/39 mmd_{\text{AWG}} = 0.127 \times 92^{(36-n)/39}\ \text{mm}
SymbolMeaningUnit
RRresistanceΩ
ρ\rhoresistivity (copper 1.72×10⁻⁸, aluminum 2.82×10⁻⁸)Ω·m
LLtotal conductor lengthm
AAcross-section aream²
  • Every 3 AWG sizes smaller doubles the area and halves the resistance.
  • A circuit's current goes out and back, so count both wires.
  • Aluminum needs about two sizes bigger than copper for the same resistance.

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Where it’s used

  • Electrical & Electronics
    Electricians size branch circuits and feeders so long runs stay within voltage-drop limits.
  • Aerospace
    Aircraft wiring is sized for resistance and weight together, since every gram counts.
  • Home Projects
    It explains why a long, thin extension cord gets warm and makes tools run weakly.

Questions people ask

Why does a longer wire have more resistance?

The electricity has farther to push through the metal. Resistance is proportional to length, so twice the length is twice the resistance.

Why is a bigger AWG number a thinner wire?

AWG counts how many times the wire was drawn through dies to make it thinner. More steps, higher number, thinner wire.

Should I count the length once or twice?

Twice for a normal circuit. Current flows out on one wire and back on the other, so both add resistance.