Ohm's law, $V = IR$, says how much voltage it takes to push a current through a resistance $R$. But what decides $R$? For a wire, three things: what it is made of, how long it is, and how thick it is.
Why it works
Electrons drifting through a metal keep bumping into its vibrating atoms. A longer wire means more bumps along the way, like more resistors in a row: $R \propto L$. A thicker wire gives the electrons more paths side by side, like more resistors in parallel: $R \propto 1/A$. What is left depends on the material, its resistivity $\rho$. In most metals the atoms vibrate more when hot, so $\rho$ rises with temperature.
Key idea
$$R = \rho\frac{L}{A}$$ with $\rho$ in $\Omega\cdot\text{m}$, $L$ in meters and $A$ in square meters. For a round wire, $A = \pi r^2 = \pi d^2/4$.
| Material | $\rho$ near room temperature ($\Omega\cdot\text{m}$) |
|---|---|
| Silver | $1.59 \times 10^{-8}$ |
| Copper | $1.68 \times 10^{-8}$ |
| Aluminum | $2.65 \times 10^{-8}$ |
| Tungsten | $5.6 \times 10^{-8}$ |
| Nichrome (heater wire) | about $1.1 \times 10^{-6}$ |
Worked examples
Worked example: A copper wire
Find the resistance of $20\,\text{m}$ of copper wire $1.0\,\text{mm}$ in diameter.
- Area: $A = \pi(0.50 \times 10^{-3}\,\text{m})^2 = 7.85 \times 10^{-7}\,\text{m}^2$.
- $R = \rho\dfrac{L}{A} = (1.68 \times 10^{-8})\dfrac{20}{7.85 \times 10^{-7}} = 0.428\,\Omega$.
Answer: $0.428\,\Omega$
Worked example: Scaling a wire
A wire has $R = 6.0\,\Omega$. Another of the same metal is twice as long with twice the diameter. Find its resistance.
- Twice the length doubles $R$. Twice the diameter gives four times the area, which divides $R$ by 4.
- $R = 6.0 \times \dfrac{2}{4} = 3.0\,\Omega$.
Answer: $3.0\,\Omega$
Worked example: Sizing a heater element
How long must a nichrome wire ($\rho = 1.10 \times 10^{-6}\,\Omega\cdot\text{m}$) $0.40\,\text{mm}$ in diameter be to have $20\,\Omega$?
- $A = \pi(0.20 \times 10^{-3})^2 = 1.26 \times 10^{-7}\,\text{m}^2$.
- $L = \dfrac{RA}{\rho} = \dfrac{(20)(1.26 \times 10^{-7})}{1.10 \times 10^{-6}} = 2.29\,\text{m}$.
Answer: $2.29\,\text{m}$
Common mistake
Thinking a thicker wire has more resistance because there is more metal in the way. The extra metal is extra path, side by side, not extra obstacle in a row. That is why thick cables are used for big currents. Check: $A$ is in the denominator of $\rho L/A$, so a bigger area means a smaller resistance.