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

Quantity	Symbol	SI Unit	Typical Range (selected materials)
Resistance	$R$	$\Omega$	0.01 Ω (copper wire) – 10⁶ Ω (insulators)
Resistivity	$\rho$	$\Omega\!\cdot\!m$	1.7 × 10⁻⁸ Ω·m (copper) – 10¹⁴ Ω·m (glass)
Length	$L$	m	0.001 m – 10 m (typical lab wires)
Area	$A$	m²	10⁻⁸ m² – 10⁻⁴ m² (common wire gauges)

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 $\rho = \rho_0[1+\alpha(T-T_0)]$).

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 $R_1 = 4.0\,\Omega$ and $R_2 = 9.0\,\Omega$. If $R_1$ is twice as long as $R_2$, find the ratio of their cross‑sectional areas $A_1/A_2$.
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