Resistance and resistivity

• Resistance

  • (R) is the opposition a conductor offers to the flow of current.
  • Unit: ohm (Ω).
  • One ohm is the resistance of a conductor in which a current of one ampere flows when a potential difference of one volt is applied across its ends.

• At constant temperature the resistance of a uniform wire depends on:

  1. its length L (R ∝ L): a longer wire means the electrons collide with more atoms on the way,
  2. its cross-sectional area A (R ∝ 1/A): a thicker wire offers more parallel paths,
  3. the material it is made of (through the resistivity ρ).

  Combining the three factors:

  R = ρ L / A

  L in metre, A in m², R in Ω, so ρ is in Ω·m.

• Resistivity

  • (ρ): the resistance of a wire of that material of unit length and unit cross-sectional area (a 1 m cube between opposite faces).
  • It depends only on the type of material and the temperature, not on the shape or size of the sample.
  • It is why we can compare materials: silver, copper and aluminium have very small ρ; nichrome and constantan have large ρ; glass and rubber have enormous ρ.

• Conductivity (σ) is the reciprocal of resistivity:

  σ = 1 / ρ with unit Ω⁻¹·m⁻¹ (siemens per metre).

  Good conductors have small ρ and large σ; insulators the opposite.

• For a round wire A = π r² = π d² / 4, so doubling the diameter divides the resistance by 4 (same length and material). Reading the formula the other way: a wire with half the diameter has four times the resistance.

• • Resistance is not a property of the material alone (that is ρ).
  • Two copper wires of different length and thickness have different resistances but identical resistivities.

• If a wire is stretched (or drawn) to a new length, its volume stays constant: L₁A₁ = L₂A₂. So when the length is multiplied by k, the area is divided by k and the resistance is multiplied by k².

Worked example 1.3 (resistivity). A nichrome wire has resistivity 1.1 × 10⁻⁶ Ω·m, length 5 m and diameter 0.40 mm. Find its resistance. Step 1: radius = 0.20 mm = 2.0 × 10⁻⁴ m. Step 2: A = π r² = 3.14 × (2.0 × 10⁻⁴)² = 1.257 × 10⁻⁷ m². Step 3: R = ρ L / A = 1.1 × 10⁻⁶ × 5 / 1.257 × 10⁻⁷ = 43.8 Ω.

Worked example 1.4 (ratio problem). Wire 1 has length L and diameter d. Wire 2 is made of the same material, but its length is 3L and its diameter is d / 2. Find R₂ : R₁. Same material, so ρ is the same. Area ∝ d², so A₂ = A₁ / 4. Hence R₂ / R₁ = (L₂ / L₁) × (A₁ / A₂) = 3 × 4 = 12. So R₂ : R₁ = 12 : 1.

Worked example 1.5 (stretching). A wire of resistance 10 Ω is drawn uniformly until its length is doubled. The volume is constant, so the area halves and R becomes 10 × 2² = 40 Ω.