Resistance: geometry and temperature dependence

Geometric dependence. For a conductor of uniform cross-section, resistance is directly proportional to length $l$ and inversely proportional to cross-sectional area $A$:

where: $\rho$ is the resistivity of the material (SI unit $\Omega\cdot\text{m}$), a property of the substance itself, independent of shape or size at a given temperature. A longer wire forces electrons through more lattice collisions (more resistance); a thicker wire offers more parallel conducting channels (less resistance). Microscopically, $\rho = m/(ne^2\tau)$, where $n$ is the free-carrier density and $\tau$ is the mean time between collisions (relaxation time) — this depends only on the material's electron population and scattering rate, never on the conductor's dimensions.

Conductivity is the reciprocal, $\sigma = 1/\rho$, with SI unit siemens per metre ($\text{S/m}$), measuring how readily a material carries current. A good conductor like copper has small $\rho$ and large $\sigma$; an insulator has the reverse. Because $\rho$ is a fixed material constant at a given temperature, two samples of the same metal — a thin wire and a thick busbar — always share the same $\rho$ even though their measured resistances $R$ can differ by orders of magnitude once length and area are folded in through $R = \rho l/A$.

Temperature dependence. Heating a conductor changes $\rho$ (and hence $R$, for a fixed shape) approximately linearly over a moderate range:

where: $R_0$ is the resistance at reference temperature $T_0$ (often $0^\circ\text{C}$ or $20^\circ\text{C}$), and $\alpha$ is the temperature coefficient of resistance (unit $\text{K}^{-1}$ or $^\circ\text{C}^{-1}$). For metals, $\alpha$ is positive: hotter lattice ions vibrate more violently, cutting down $\tau$, so $\rho = m/(ne^2\tau)$ rises. For semiconductors, $\alpha$ is negative: heat frees far more valence electrons into the conduction band, and the resulting jump in carrier density $n$ overwhelms the modest drop in $\tau$, so overall resistivity falls — the working principle behind thermistors used as temperature sensors. Certain alloys such as manganin and constantan are engineered with $\alpha \approx 0$, so their resistance is nearly independent of ambient heating — ideal for standard resistance coils and precision instruments.

A wire of length $2.0\ \text{m}$ and uniform cross-sectional area $0.5\ \text{mm}^2$ has resistance $8.5\ \Omega$ at room temperature. Find (a) the resistivity of the material, and (b) the new resistance if the wire is stretched, without changing its volume, to twice its original length.

Step 1. Convert the area to SI units: $A = 0.5\ \text{mm}^2 = 0.5\times10^{-6}\ \text{m}^2$.

Step 2. Find resistivity from $R = \rho l/A$, rearranged as $\rho = RA/l$: $\rho = \dfrac{8.5 \times 0.5\times10^{-6}}{2.0} = 2.125\times10^{-6}\ \Omega\cdot\text{m}$.

Step 3. Apply the stretching result from the derivation, $R_{\text{new}}/R = (l_{\text{new}}/l)^2$, with $l_{\text{new}} = 2l$: $R_{\text{new}} = 4R = 4 \times 8.5 = 34\ \Omega$.

Answer: $\rho = 2.125\times10^{-6}\ \Omega\cdot\text{m}$, and the stretched wire has resistance $34\ \Omega$. Notice that the resistivity itself never changes during the stretch — it is a property of the copper (or whichever alloy the wire is made of), not of its shape; only the geometric factor $l/A$ has changed, and it changed by the square-law relation proved in the derivation above.

Competitive stretch

Suppose this same stretched wire (now $34\ \Omega$ at $20^\circ\text{C}$) is a copper coil with $\alpha = 4.0\times10^{-3}\ ^\circ\text{C}^{-1}$ (reference $T_0 = 20^\circ\text{C}$), and it heats up to $80^\circ\text{C}$ while carrying current. Using $R_T = R_0[1+\alpha(T-T_0)]$: $R_{80} = 34[1 + (4.0\times10^{-3})(60)] = 34 \times 1.24 = 42.16\ \Omega$. Stretching and heating combine multiplicatively — the geometric quadrupling from Step 3 and the roughly 24% thermal rise both apply on top of each other.

• Treating resistivity and resistance as the same quantity. Resistivity $\rho$ depends only on the material and temperature; resistance $R$ additionally depends on length and area — a short fat rod and a long thin wire of the same metal have very different $R$ but identical $\rho$.
• Assuming resistance scales linearly with stretched length. Because area shrinks as length grows at constant volume, $R \propto l^2$, not $R \propto l$ — the doubling case gives a factor of 4, not 2.
• Assuming every material gets more resistive when heated. True for metals (positive $\alpha$) but false for semiconductors (negative $\alpha$); always check the sign of $\alpha$ against the material type.
• Ignoring the reference temperature $T_0$. $R_T = R_0[1+\alpha(T-T_0)]$ only works with the $T_0$ that $\alpha$ was quoted against; mixing up $0^\circ\text{C}$ and $20^\circ\text{C}$ references gives numerically wrong answers.
• Applying the volume-conservation stretching rule to the wrong scenario. The $R \propto l^2$ result only holds when the wire is drawn out (its material redistributed, volume fixed); if instead a wire is simply cut shorter or a different-gauge wire of the same material is substituted, use $R = \rho l/A$ directly with the new $l$ and $A$, not the stretching shortcut.

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