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Wire Size and Voltage Drop Calculator

Every wire has a little resistance, so a current through it loses some voltage on the way. Enter the current, the one-way length, the wire size and the supply voltage to see the drop for the round trip, the power lost as heat, and the smallest standard wire that keeps the drop under a target (3 % to start). This is about voltage drop only, not a safety rating.

Result

Voltage lost in the wire (there and back)

2.293 V

19.11 % of the 12 V supply

Your wire

Cross-section and thickness: 1.5 mm², 1.382 mm across
Resistance of the round trip11.47 mΩ for each metre of wire: 229.3 mΩ
Voltage drop: 2.293 V (19.11 %)
Voltage left at the load: 9.707 V
Power lost as heat in the wire: 22.93 W
Power reaching the load: 97.07 W

Smallest standard size for 3 % or less

Metricexactly 9.556 mm² would just do it: 10 mm² (2.87 % drop)
AWG: AWG 7 (2.72 % drop)

Careful: The drop is 19.11 % of the supply, above your 3 % target. Use at least 10 mm² (AWG 7).

This checks voltage drop only. It is not a safety rating: how much current a cable may carry safely depends on its insulation, how it is installed and the local wiring rules, so check those too.

The formula

Resistance of the wire there and back (convert mm² to m² first)

\htmlData{sym=R}{R} = \htmlData{sym=rho}{\rho}\,\dfrac{2\,\htmlData{sym=L_wire}{L_{\mathrm{wire}}}}{\htmlData{sym=A_wire}{A_{\mathrm{wire}}}}

The voltage lost in the wire

\htmlData{sym=U_drop:wire}{U_{\mathrm{drop}}} = \htmlData{sym=I}{I}\,\htmlData{sym=R}{R}

The power lost as heat in the wire

\htmlData{sym=P_loss}{P_{\mathrm{loss}}} = \htmlData{sym=I}{I}^2 \htmlData{sym=R}{R}

The diameter of an AWG wire

\htmlData{sym=D_wire}{D_{\mathrm{wire}}} = 0.127\ \htmlData{funit=mm}{\text{mm}} \times 92^{\,(36 - \htmlData{sym=N_AWG}{N_{\mathrm{AWG}}})/39}

The cross-section of a round wire

\htmlData{sym=A_wire}{A_{\mathrm{wire}}} = \dfrac{\htmlData{sym=pi}{\pi}\,\htmlData{sym=D_wire}{D_{\mathrm{wire}}}^2}{4}

What each letter means

$R$: Resistance (Ω · ohm) : How strongly something opposes current.
$\rho$: Resistivity ρ (rho) (Ω·m) : How well a material resists current, whatever its shape. Copper is very low; plastic very high.
$L_{\mathrm{wire}}$: One-way length (m · metre) : How long the cable run is from the supply to the load. The current goes there and back, so the resistance counts twice this length.
$A_{\mathrm{wire}}$: Wire cross-section (mm²) : The area of the metal if you cut the wire across: the bigger, the less it resists.
$U_{\mathrm{drop}}$: Voltage lost in the wire (V · volt) : The voltage that disappears along the cable, so the load gets that much less than the supply gives.
$I$: Current (A · ampere) : How much charge flows each second.
$P_{\mathrm{loss}}$: Power lost as heat (W · watt)
$D_{\mathrm{wire}}$: Wire diameter (mm) : How thick the bare copper (or aluminium) is, across.
$N_{\mathrm{AWG}}$: AWG number (no unit) : The American wire gauge number: a bigger number is a thinner wire. 0 is 1/0, −1 is 2/0, and so on.
$\pi$: π (pi) ≈ 3.1416 (no unit) : A fixed number. It shows up whenever something goes round in circles, such as a sine wave.

How it works

A cable is a go wire and a return wire, so the current passes through twice the one-way length. The resistance is the material’s resistivity times that length, divided by the cross-section area. At 20 °C copper is 1.72 × 10⁻⁸ Ω·m and aluminium 2.82 × 10⁻⁸ Ω·m, so aluminium needs about 1.6 times the area for the same drop. The drop is then the current times that resistance, and the heat in the wire is the current squared times the resistance.

The percentage drop matters more on low voltages. 2.3 V lost on a 230 V supply is 1 %, which no one notices, but the same 2.3 V lost on a 12 V supply is nearly 20 %, so the lamp at the end is dim and the wire is warm. That is why low-voltage wiring for car accessories, LED strips and solar panels needs far thicker wire for the same current. A drop of 3 % is a common limit for power circuits.

Sizes are in mm² in most of the world, and in AWG in the United States. A bigger AWG number is a thinner wire, and every 3 steps roughly halves the area. The diameter formula above is the standard one, and the calculator also lists the smallest standard size of each kind that meets your target. Resistance rises by about 0.4 % for every 10 K of heating, so a hot cable drops a little more than the 20 °C figure.

Voltage drop is not the same as how much current a wire may carry. The safe current depends on the insulation, on how many cables lie together, on the ambient temperature and on the local wiring rules. For mains wiring, follow your local electrical code and use a qualified electrician; this page does not replace either.

Worked example

A load takes 10 A through 10 m of 1.5 mm² copper cable. The round trip is 2 × 10 m = 20 m, so R = 1.72 × 10⁻⁸ Ω·m × 20 m ÷ 1.5 × 10⁻⁶ m² = 229.3 mΩ.
Drop = 10 A × 229.3 mΩ = 2.293 V. On a 230 V supply that is 1.00 %, which is fine. The wire turns I² × R = 100 × 229.3 mΩ = 22.93 W into heat over its whole length.
The same cable on a 12 V supply loses the same 2.293 V, which is 19.1 % of 12 V. Only 9.707 V reaches the load.
For 3 % of 12 V (0.36 V), the resistance must be at most 0.036 Ω, so the area must be at least 9.556 mm². The next standard size up is 10 mm², or AWG 7 if you use AWG.
AWG check: AWG 12 has a diameter of 0.127 mm × 92^(24 ÷ 39) = 2.053 mm, and a cross-section of π × d² ÷ 4 = 3.309 mm².

Common questions

How much voltage drop is acceptable?

For power wiring, 3 % is a common limit, and many rules allow up to 5 % in total. Sensitive loads can need less: a 12 V LED strip, or a motor that has to start, often wants under 2 %. The calculator lets you choose your own target.

Is this the right size for safety?

Not necessarily. This page only looks at voltage drop. A wire can have a small drop and still be too thin to carry its current safely for hours, or the other way round. Safe current depends on the insulation, the installation and local wiring rules, so check those, and for mains wiring let a qualified person decide.

Why does the calculator double the length?

Current flows out through one wire and back through the other, so both of them drop voltage. If your cable is 10 m long, the current travels 20 m of copper. If you only use one wire and the return is the metal body of a car, enter twice the cable length for the same answer only if that return is a similar wire.

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