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UY1: Resistance And Resistivity

Distinguish resistance from resistivity, derive R = rho L/A, and handle units, geometry, and temperature effects correctly.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

This page gives the UY1 working model/result for Resistance And Resistivity. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • Resistance R is for one specific component; resistivity ρ is a material property.
  • Key equations:
R = V/I, ρ = E/J, R = ρL/A
  • SI units: R in ohms (Ω), ρ in Ω m.
  • Modelling context: R = ρ L/A assumes uniform material, uniform cross-section, and ohmic behavior (linear V-I at fixed temperature).

Prerequisites: Current, Drift Velocity And Current Density
Next uses: Resistance Of A Cylindrical Resistor, Electromotive Force & Power In Circuits

2) Setup

Consider a uniform conductor of length L and cross-sectional area A carrying steady current I under potential difference V.

  • Ohmic region: V ∝ I, so R is constant.
  • Microscopic form: vector J = σ vector E, where conductivity σ = 1/ρ.
  • This lesson assumes uniform material and constant temperature unless stated otherwise.
Common traps (rho vs R, and geometry)
  • ρ is a material property; R depends on geometry. Don’t quote ρ in ohms or R in Ωm.
  • Area matters: doubling radius makes area 4 times larger, so resistance becomes 1/4 (for fixed L and ρ).
  • Temperature dependence: many metals have ρ(T) ≈ ρ₀[1 + α(T-T₀)]. If temperature changes, R changes even with fixed geometry.

3) Core derivation/explanation

From vector J = σ vector E:

J = (1/ρ)E.

For a uniform wire,

E ≈ V/L, J = I/A.

So

I/A = (1/ρ)V/L ⇒ V/I = ρL/A ⇒ R = ρL/A.

Interpretation:

  • Larger L gives larger resistance.
  • Larger A gives smaller resistance.
  • For fixed geometry, R changes if ρ changes (e.g., with temperature).

Near room temperature for many metals:

ρ(T) ≈ ρ₀[1 + α(T-T₀)].

Checks (sanity)

  • Units: ρ L/A must reduce to Ω.
  • Scaling: longer wires have larger R; thicker wires have smaller R.

4) Worked example(s)

A copper wire has ρ = 1.68 × 10⁻⁸ Ω m, length L = 2.0 m, and area A = 1.0 × 10⁻⁶ m².

R = ρL/A = ((1.68 × 10⁻⁸)(2.0))/(1.0 × 10⁻⁶) = 3.36 × 10⁻² Ω.

If I = 3.0 A flows, then

V = IR = (3.0)(0.0336) = 0.101 V.

5) Practice set (with hints + answers)

  1. A resistor has V = 12 V and I = 0.40 A. Find R.
  2. A wire length doubles while A and material stay unchanged. How does R change?
  3. For fixed L and material, radius is doubled. How does R change?

Hints

  • Use R = V/I.
  • Use R ∝ L.
  • Area of a circular wire is A = π r².

Answers

  1. R = 30 Ω.
  2. R doubles.
  3. A becomes four times larger, so R becomes one quarter.

6) Summary + next steps

  • R describes a component; ρ describes a material.
  • The geometry law R = ρ L/A is central for design and scaling.
  • Always check units: ρ L/A → Ω.

Next: Resistance Of A Cylindrical Resistor Previous: Current, Drift Velocity And Current Density Back To Electromagnetism