Recall and use the following relationship for a metallic electrical conductor: (a) resistance is directly proportional to length (b) resistance is inversely proportional to cross-sectional area

4.2.4 Resistance

Learning Objective

Recall and use the relationships for a metallic electrical conductor:

• Resistance is directly proportional to length.
• Resistance is inversely proportional to cross‑sectional area.

Key Concepts

For a uniform metallic conductor the resistance \$R\$ depends on three factors:

1. Material (expressed by its resistivity \$\rho\$).
2. Length \$L\$ of the conductor.
3. Cross‑sectional area \$A\$ of the conductor.

The quantitative relationship is

\$R = \rho \frac{L}{A}\$

where \$\rho\$ is a constant for a given material (measured in \$\Omega\!\cdot\!m\$).

Derivation of the Proportionalities

Starting from the definition \$R = \rho L/A\$:

• If \$A\$ is kept constant, \$R \propto L\$. Doubling the length doubles the resistance.
• If \$L\$ is kept constant, \$R \propto 1/A\$. Doubling the area halves the resistance.

Practical Implications

• Long, thin wires have high resistance – useful for heating elements.
• Short, thick wires have low resistance – ideal for power transmission.
• Choosing a material with low resistivity (e.g., copper) reduces resistance for a given \$L\$ and \$A\$.

Sample Calculations

Example 1 – Changing Length

A copper wire (\$\rho = 1.68\times10^{-8}\ \Omega\!\cdot\!m\$) has \$L = 2.0\ \text{m}\$, \$A = 1.0\times10^{-6}\ \text{m}^2\$. Find its resistance.

\$R = (1.68\times10^{-8})\frac{2.0}{1.0\times10^{-6}} = 0.0336\ \Omega\$

If the length is increased to \$4.0\ \text{m}\$ (area unchanged), the new resistance is

\$R' = (1.68\times10^{-8})\frac{4.0}{1.0\times10^{-6}} = 0.0672\ \Omega\$

Resistance has doubled, confirming \$R \propto L\$.

Example 2 – Changing Area

Using the same wire, keep \$L = 2.0\ \text{m}\$ but double the area to \$2.0\times10^{-6}\ \text{m}^2\$.

\$R'' = (1.68\times10^{-8})\frac{2.0}{2.0\times10^{-6}} = 0.0168\ \Omega\$

Resistance is halved, confirming \$R \propto 1/A\$.

Common Misconceptions

• “Longer wires always have more voltage drop.” – The drop depends on both resistance and the current flowing (Ohm’s law \$V = IR\$).
• “All metals have the same resistance for a given size.” – Resistivity \$\rho\$ varies between metals; copper, aluminium, and nichrome differ markedly.
• “Cross‑sectional area is the same as the diameter.” – For a circular wire \$A = \pi r^2 = \pi (d/2)^2\$; a small change in diameter produces a large change in area.

Summary Table

Practice Questions

1. A 3.0 m length of aluminium wire (\$\rho = 2.82\times10^{-8}\ \Omega\!\cdot\!m\$) has a cross‑sectional area of \$0.5\ \text{mm}^2\$. Calculate its resistance.
2. If the same aluminium wire is stretched so that its length becomes 4.5 m and its area reduces to \$0.35\ \text{mm}^2\$, what is the new resistance? Compare with the original value and comment on the effect of stretching.
3. Two wires of the same material have the same resistance of \$0.10\ \Omega\$. Wire A is 1.0 m long, while Wire B is 2.0 m long. Which wire has the larger cross‑sectional area? By what factor?
4. Explain why power transmission lines use aluminium conductors that are many square centimeters in cross‑section, even though aluminium’s resistivity is higher than copper’s.