state Ohm’s law

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Resistance and Resistivity

Resistance and Resistivity

Learning Objective

State Ohm’s law and understand the relationship between resistance, resistivity, length and cross‑sectional area of a conductor.

Key Definitions

  • Resistance (R): The opposition offered by a material to the flow of electric current. Measured in ohms (Ω).
  • Resistivity (ρ): An intrinsic property of a material that quantifies how strongly it resists electric current. Measured in ohm‑metres (Ω·m).
  • Conductivity (σ): The reciprocal of resistivity, σ = 1/ρ, measured in siemens per metre (S·m⁻¹).

Ohm’s Law

Ohm’s law relates the potential difference (V) across a conductor to the current (I) flowing through it and its resistance (R):

\$ V = I R \$

In terms of current:

\$ I = \frac{V}{R} \$

Ohm’s law is valid for ohmic conductors where the resistance remains constant over a range of applied voltages.

Resistance of a Uniform Conductor

The resistance of a uniform cylindrical conductor can be expressed using its resistivity:

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

where:

  • \$\rho\$ = resistivity of the material (Ω·m)
  • \$L\$ = length of the conductor (m)
  • \$A\$ = cross‑sectional area (m²)

Factors Affecting Resistance

  1. Material: Different materials have different resistivities.
  2. Length (\$L\$): Resistance is directly proportional to length.
  3. Cross‑sectional area (\$A\$): Resistance is inversely proportional to area.
  4. Temperature: For most conductors, resistivity increases with temperature; for semiconductors it decreases.

Typical Resistivity \cdot alues

MaterialResistivity (Ω·m)
Copper1.68 × 10⁻⁸
Aluminium2.82 × 10⁻⁸
Silver1.59 × 10⁻⁸
Nickel–Chromium alloy (Nichrome)1.10 × 10⁻⁶
Glass (dry)10¹⁰ – 10¹⁴

Example Calculation

Calculate the resistance of a copper wire 2.0 m long with a cross‑sectional area of 0.5 mm².

Given:

\$ \rho_{\text{Cu}} = 1.68 \times 10^{-8}\ \text{Ω·m},\; L = 2.0\ \text{m},\; A = 0.5\ \text{mm}^2 = 0.5 \times 10^{-6}\ \text{m}^2 \$

Using \$R = \rho L / A\$:

\$ R = \frac{1.68 \times 10^{-8} \times 2.0}{0.5 \times 10^{-6}} = \frac{3.36 \times 10^{-8}}{5.0 \times 10^{-7}} = 0.0672\ \text{Ω} \$

Thus the resistance is approximately \$6.7 \times 10^{-2}\ \text{Ω}\$.

Suggested Diagram

Suggested diagram: A cylindrical conductor showing length \$L\$, cross‑sectional area \$A\$, and the direction of current \$I\$ with voltage \$V\$ applied across its ends.

Summary

  • Ohm’s law: \$V = IR\$ for ohmic materials.
  • Resistance depends on material (via resistivity), length, and cross‑sectional area: \$R = \rho L/A\$.
  • Resistivity is a fundamental property; lower resistivity means better conductivity.