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UY1: Resistance Of A Cylindrical Resistor

Apply R = rho L/A to cylindrical conductors, including geometry scaling with radius and diameter.

  • 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 Of A Cylindrical Resistor. 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

  • For a solid cylinder of radius r:
R = ρL/(π r²)
  • Doubling L doubles R; doubling r makes R one quarter.
  • Keep SI units throughout: L,r in m and ρ in Ω m.
  • Modelling context: uniform material, uniform cross-section, and ohmic behavior at fixed temperature.

Prerequisites: Resistance And Resistivity
Next uses: Electromotive Force & Power In Circuits

2) Setup

Model a uniform cylindrical resistor (wire/rod):

  • Length: L
  • Radius: r (or diameter d = 2r)
  • Material resistivity: ρ
  • Current along the axis

Assume temperature is uniform and contact resistances are negligible.

Common traps (radius vs diameter)
  • If you are given diameter d, convert to radius: r = d/2. Then use A = π r² = π d²/4.
  • Convert mm to m before squaring: (2 mm)² = (2 × 10⁻³ m)².
  • Keep ρ in Ω m (not Ω/m).

3) Core derivation/explanation

From the geometry law,

R = ρL/A.

For a solid cylinder,

A = π r² = (π d²)/4.

Hence

R = ρL/(π r²) = (4ρ L)/(π d²).

This formula explains common design choices:

  • Long, thin resistors give large R.
  • Short, thick conductors give small R.

Unit check:

Ω m × m/m² = Ω.

Checks (sanity)

  • If r increases, resistance must decrease (area increases).
  • If L → 0, resistance tends to 0 (ideal contact-free model).

4) Worked example(s)

A carbon rod has ρ = 3.5 × 10⁻⁵ Ω m, length L = 0.12 m, and diameter d = 2.0 mm.

Convert diameter: d = 2.0 × 10⁻³ m, so

A = (π d²)/4 = (π(2.0 × 10⁻³)²)/4 = 3.14 × 10⁻⁶ m².

Then

R = ρL/A = ((3.5 × 10⁻⁵)(0.12))/(3.14 × 10⁻⁶) ≈ 1.34 Ω.

5) Practice set (with hints + answers)

  1. A wire has fixed material and radius. If length triples, what is the new resistance ratio R_new/R_old?
  2. A cylindrical resistor has ρ = 2.0 × 10⁻⁶ Ω m, L = 0.50 m, r = 0.50 mm. Find R.
  3. If diameter is halved (same ρ,L), what happens to R?

Hints

  • Use proportionality from R = ρ L/(π r²).
  • Convert mm to m first.
  • R ∝ 1/d².

Answers

  1. R_new/R_old = 3.
  2. r = 5.0 × 10⁻⁴ m, A = 7.85 × 10⁻⁷ m², so R ≈ 1.27 Ω.
  3. R becomes four times larger.

6) Summary + next steps

  • Cylindrical geometry turns R = ρ L/A into R = ρ L/(π r²).
  • Radius has a strong quadratic effect on resistance.
  • Always convert diameter/radius to metres before substitution.

Next: Electromotive Force & Power In Circuits Previous: Resistance And Resistivity Back To Electromagnetism