Calculate the total electrical resistance of a wire from its length, cross-section or AWG gauge, and conductor material.
Calculate
Enter the wire properties below.
mm²
mm
Common Conductor Resistivities
Resistivity at 20 °C. Lower resistivity = better conductor.
Resistivity (ρ) at 20 °C
Silver (Ag)
1.59×10⁻⁸ Ω·m
Best conductor, expensive
Copper (Cu)
1.72×10⁻⁸ Ω·m
Standard wiring material
Gold (Au)
2.44×10⁻⁸ Ω·m
Connectors, PCB pads
Aluminum (Al)
2.65×10⁻⁸ Ω·m
Power cables, cheaper
Tungsten (W)
5.60×10⁻⁸ Ω·m
Light bulb filaments
Iron (Fe)
10.0×10⁻⁸ Ω·m
Structural, high loss
Worked Examples
110 m copper AWG 24
MaterialCopper ρ = 1.724×10⁻⁸
Length10 m
AWG 24 area0.205 mm²
R = ρ × L / A≈ 0.840 Ω
25 m aluminum 2.5 mm²
MaterialAluminum ρ = 2.65×10⁻⁸
Length5 m
Area2.5 mm²
R = ρ × L / A≈ 53 mΩ
3Voltage drop check
Wire resistance0.84 Ω
Current500 mA
V drop = I × R420 mV
Power loss = I²×R210 mW
4Round-trip resistance
One-way length5 m
Use 2× length10 m
Material / AWGCu, AWG 22
Round-trip R≈ 0.529 Ω
Frequently Asked Questions
How do I calculate wire resistance? ▾
Use R = ρ × L / A, where ρ is the material's resistivity (Ω·m), L is the wire length (m), and A is the cross-sectional area (m²). Copper has ρ = 1.724×10⁻⁸ Ω·m.
Why do I need to double the length for voltage drop? ▾
Current travels from the source to the load and back. Both conductors add resistance, so for voltage-drop calculations you should enter the round-trip length (2 × one-way distance).
Does wire resistance change with temperature? ▾
Yes. R(T) = R₀ × (1 + α × ΔT). Copper's temperature coefficient α ≈ 0.00393 /°C means a 50 °C rise increases resistance by ~20%. This calculator gives values at 20 °C.
What is the resistance of 1 m of AWG 22 copper wire? ▾
AWG 22 copper wire has an area of about 0.326 mm². Using R = 1.724×10⁻⁸ / (0.326×10⁻⁶) ≈ 52.9 mΩ per metre.
Why is copper used instead of silver for wiring? ▾
Silver has slightly lower resistivity (1.59 vs 1.72×10⁻⁸ Ω·m) but is roughly 80× more expensive than copper. The small conductivity advantage doesn't justify the cost for most applications.