Recall and use the following relationships for a metallic electrical conductor:
All of the above can be combined into one handy equation:
\$R = \rho \frac{L}{A}\$
Where \$R\$ is resistance (Ω), \$L\$ is length (m), \$A\$ is cross‑sectional area (m²) and \$ρ\$ is the resistivity of the material (Ω·m).
Think of electricity as water flowing through a pipe.
Suppose you have a copper wire with resistivity \$ρ = 1.68 \times 10^{-8}\,\Omega\cdot m\$.
| Length (m) | Area (mm²) | Resistance (Ω) |
|---|---|---|
| 1 | 1 | 0.0168 |
| 2 | 1 | 0.0336 |
| 1 | 2 | 0.0084 |
Notice how doubling the length doubles the resistance, while doubling the area halves it.
🔌 If a 0.5 m long wire has a resistance of 0.1 Ω, what will be the resistance of a 2 m long wire of the same material and cross‑section?
Answer: 0.4 Ω (because resistance doubles with length).