recall and use R = ρL / A

Resistance and Resistivity

Learning Objective

Recall and use the relationship \$R = \frac{\rho L}{A}\$ where \$R\$ is resistance, \$\rho\$ is resistivity, \$L\$ is the length of the conductor and \$A\$ is its cross‑sectional area.

Key Concepts

• Resistance (\$R\$): opposition to the flow of electric current, measured in ohms (\$\Omega\$).
• Resistivity (\$\rho\$): intrinsic property of a material that quantifies how strongly it resists current, measured in \$\Omega\!\cdot\!m\$.
• Length (\$L\$): the distance the current travels through the material, measured in metres (m).
• Cross‑sectional area (\$A\$): the area perpendicular to the direction of current flow, measured in square metres (m²).

Derivation of \$R = \dfrac{\rho L}{A}\$

Starting from Ohm’s law for a uniform conductor:

\$V = IR\$

and the definition of electric field \$E\$ and current density \$J\$:

\$E = \rho J\$

Since \$E = V/L\$ and \$J = I/A\$, substituting gives:

\$\frac{V}{L} = \rho\frac{I}{A} \;\;\Longrightarrow\;\; V = I\frac{\rho L}{A}\$

Comparing with \$V = IR\$ yields the desired relationship.

Units and Typical \cdot alues

Factors Affecting Resistance

1. Material – determined by resistivity.
2. Length – resistance increases linearly with \$L\$.
3. Cross‑sectional area – resistance decreases with larger \$A\$.
4. Temperature – for most conductors, \$\rho\$ increases with temperature (approximately \$\rho = \rho0[1+\alpha(T-T0)]\$).

Example Calculation

Find the resistance of a copper wire 2.0 m long with a diameter of 1.0 mm. Use \$\rho_{\text{Cu}} = 1.68\times10^{-8}\,\Omega\!\cdot\!m\$.

1. Calculate the cross‑sectional area:

\$A = \pi\left(\frac{d}{2}\right)^2 = \pi\left(\frac{1.0\times10^{-3}}{2}\right)^2 = 7.85\times10^{-7}\,\text{m}^2\$

2. Apply the formula:

\$R = \frac{\rho L}{A} = \frac{1.68\times10^{-8}\times2.0}{7.85\times10^{-7}} \approx 0.043\,\Omega\$

Common Mistakes

• Confusing resistivity (\$\rho\$) with resistance (\$R\$). \$\rho\$ is a material property; \$R\$ depends on geometry.
• Using diameter instead of area directly in the formula. Always convert to \$A\$ (or use \$A = \pi r^2\$).
• Neglecting temperature effects when high currents cause significant heating.

Practice Questions

1. A nichrome wire ( \$\rho = 1.10\times10^{-6}\,\Omega\!\cdot\!m\$ ) is 5.0 cm long and has a cross‑sectional area of \$2.0\times10^{-8}\,\text{m}^2\$. Calculate its resistance.
2. Two wires of the same material have resistances \$R1 = 4.0\,\Omega\$ and \$R2 = 9.0\,\Omega\$. If \$R1\$ is twice as long as \$R2\$, find the ratio of their cross‑sectional areas \$A1/A2\$.
3. Explain how the resistance of a copper conductor changes when its temperature is increased from \$20^\circ\text{C}\$ to \$70^\circ\text{C}\$, given the temperature coefficient \$\alpha = 3.9\times10^{-3}\,\text{K}^{-1}\$.

Suggested Diagram