Result

Geometry and material

R = ρ \frac{ℓ}{A}

Statement

Long and thin means high resistance; short and thick means low.

Symbols and units

$R$	resistance	Ω
$ρ$	resistivity of the material	Ω·m (Cu 1.7×10⁻⁸)
$ℓ$	length along the current	m
$A$	cross-sectional area	m²

When it applies

Uniform material and cross-section, steady current.

Don't use when

Assumes a uniform material and constant cross-section along the conductor's length — a tapered wire or a temperature gradient along its length requires integrating resistivity over the varying geometry instead.

Derivation

A uniform field along a wire of length ℓ gives:

Δ V = E ℓ

The current follows from the current density over the cross-section:

I = J A = σ E A

Form the ratio, and E cancels:

R = \frac{Δ V}{I} = \frac{E ℓ}{σ E A} = \frac{ℓ}{σ A}

Write it with resistivity:

R = ρ \frac{ℓ}{A}

Worked examples

Voltage drop in a cable run

10 m of 1.5 mm² copper:

R = \frac{( 1.7 \times 1 0 ^{- 8} ) ( 10 )}{1.5 \times 1 0 ^{- 6}} \approx 0.11 Ω

Carrying 10 A:

Δ V = I R = 1.1 V, P = I^{2} R = 11 W

Watch out for

Doubling the diameter cuts R by four, since A goes with the radius squared.

Connects to

Ohm's law Power Temperature coefficient