Resistivity and factors affecting resistance

Three separate factors decide how much a uniform conductor resists current flow.

Length. A longer wire offers more resistance because a charge carrier collides with more atoms on its way from one end to the other. Resistance is directly proportional to length: $R \propto l$.

Cross-sectional area. A thicker wire offers less resistance because it gives current more parallel "lanes" to flow through — think of a wide pipe carrying water more easily than a narrow one. Resistance is inversely proportional to area: $R \propto 1/A$.

Nature of the material (resistivity). Even two wires of identical length and thickness resist current differently if made of different substances. This material-specific property is called resistivity, $\rho$ — it is what is left over once shape has been accounted for. Putting the three together gives the syllabus formula:

where: $R$ is resistance in ohms ($\Omega$), $l$ is the conductor's length in metres, $A$ is its cross-sectional area in square metres, and $\rho$ is the resistivity of the material in ohm-metres ($\Omega\cdot\text{m}$). Rearranged, $\rho = RA/l$ — this is how resistivity is actually measured in a lab: take a sample of known length and area, measure its resistance, and divide. Resistivity itself does not depend on how the wire is cut or drawn — a short fat copper rod and a long thin copper wire have wildly different resistances but exactly the same resistivity, because resistivity belongs to copper, not to any one sample of it. Good conductors such as copper and silver have very low resistivity (of order $10^{-8}\ \Omega\cdot\text{m}$); alloys such as nichrome, used in heating elements, have resistivity roughly a hundred times higher, combined with a high melting point that lets them glow red-hot without melting; insulators sit many orders of magnitude higher still. The figure below shows the same current squeezed through three geometries — a normal wire, a longer wire, and a thicker wire of the same material — to make the two proportionalities visual before you meet them in the derivation.

Temperature (qualitative only). For most metals, resistance rises as temperature rises — atoms vibrate more and scatter the current-carrying electrons more often. This is why a glowing filament bulb draws less current once hot than the simple $V=IR$ ratio at switch-on would suggest, and it is a purely qualitative fact at this level — no formula linking $\rho$ to temperature is required in the ICSE syllabus. This same formula quietly justifies two opposite design choices you meet in every circuit: household connecting wires are made thick and of low-resistivity copper, so their own resistance stays negligible and they don't waste power as heat; a heater's element or a bulb's filament is deliberately made thin and of high-resistivity material, so it does develop a substantial, controlled resistance and glows or heats up. One formula, two opposite engineering goals.

A nichrome wire has length $l = 2.0\ \text{m}$ and cross-sectional area $A = 4.0 \times 10^{-7}\ \text{m}^2$. The resistivity of nichrome is $\rho = 1.1 \times 10^{-6}\ \Omega\cdot\text{m}$. Find its resistance, and then find the resistance of a second nichrome wire, B, that is twice as long and has half the cross-sectional area of the first.

Step 1. Write the formula and substitute for the first wire. $R = \rho\,\dfrac{l}{A} = (1.1\times10^{-6}) \times \dfrac{2.0}{4.0\times10^{-7}}$.

Step 2. Compute. $\dfrac{2.0}{4.0\times10^{-7}} = 5\times10^{6}$, so $R = 1.1\times10^{-6} \times 5\times10^{6} = 5.5\ \Omega$.

Step 3. For wire B, use ratio reasoning instead of recomputing from scratch. Since both wires are nichrome, $\rho$ is the same for both, so $\dfrac{R_B}{R_A} = \dfrac{l_B/A_B}{l_A/A_A} = \dfrac{l_B}{l_A}\times\dfrac{A_A}{A_B}$. With $l_B = 2l_A$ and $A_B = A_A/2$, this gives $\dfrac{R_B}{R_A} = 2\times 2 = 4$.

Step 4. Combine. $R_B = 4 \times R_A = 4 \times 5.5 = 22\ \Omega$. The first wire has resistance $5.5\ \Omega$, and doubling its length while halving its area quadruples this to $22\ \Omega$ for wire B.

Competitive stretch

If instead wire B were made of copper ($\rho_{\text{Cu}} \approx 1.7\times10^{-8}\ \Omega\cdot\text{m}$) with the same doubled length and halved area as above, its resistance would be $R_B = \rho_{\text{Cu}}\,l_B/A_B = (1.7\times10^{-8})\times(4.0)/(2.0\times10^{-7}) = 0.34\ \Omega$ — over sixty times smaller than the nichrome value, purely because copper's resistivity is about sixty-five times lower, even though both wires share identical geometry.

• Confusing resistance and resistivity — $R$ changes when you cut the wire shorter or use a thicker sample; $\rho$ is intrinsic to the material and stays fixed at a given temperature regardless of the sample's shape or size.
• Forgetting the inverse relationship with area — students often correctly recall $R \propto l$ but then wrongly assume $R \propto A$ too, instead of $R \propto 1/A$.
• Mixing up units — resistivity is in $\Omega\cdot\text{m}$, never $\Omega/\text{m}$ or $\Omega\cdot\text{m}^2$; a common slip is forgetting to convert area from $\text{cm}^2$ or $\text{mm}^2$ into $\text{m}^2$ before substituting into $R = \rho l/A$.
• Treating resistance as temperature-independent — the formula $R=\rho l/A$ does not show temperature explicitly, but $\rho$ itself (and hence $R$) rises with temperature for metals, which is why heating elements and filament bulbs run hotter and draw progressively less current as they warm up.

Ask the tutor: Want to try more wire-comparison ratio problems, or see how this formula connects to power dissipation in a heater coil?