Calculate resistivity from conductor dimensions
Use R = ρL/A to compare material resistivity with resistance set by wire length and area.
Section 9 · Electricity
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 9. Circuit calculations and component sketches are original study practice; full timed assessment, supervised practical competence and human subject-teacher review remain outstanding.
Before you begin
- Use charge, time, energy, power, area and consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity in R = ρL/A describes the material?
What you will learn
- Recall and use R = ρL/A
- Relate resistance to length and cross-sectional area
- Calculate resistivity with compatible SI units
Resistivity is a material property in the model
For a uniform conductor at a stated temperature, resistance R = ρL/A. Here ρ is resistivity in Ω m, L is conductor length in metres and A is cross-sectional area in square metres.
For the same material and temperature, increasing length increases resistance; increasing cross-sectional area decreases resistance. Resistivity distinguishes material properties from geometry. Do not substitute diameter for area: for a circular wire A = πd²/4.
Rearrange after converting geometry to SI
Find ρ = RA/L when resistance, length and area are known. If only diameter is supplied, calculate its cross-sectional area first, taking care to square the radius or diameter correctly.
Temperature must be comparable when treating ρ as fixed. Real materials can have temperature-dependent resistivity; this equation does not imply the same ρ at every temperature.
Pause and explain
For the same material and length, what happens to R if area doubles?
Worked example
A wire of length 2.0 m and area 1.0 × 10⁻⁶ m² has resistance 0.034 Ω. Find resistivity.
Show the worked solution
- ρ = RA/L.
- ρ = (0.034)(1.0 × 10⁻⁶)/2.0.
- State Ω m.
Answer ρ = 1.7 × 10⁻⁸ Ω m.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A material sample has ρ = 2.0 × 10⁻⁸ Ω m, L = 5.0 m and A = 2.0 × 10⁻⁶ m². Find resistance.
Give me a hint
Use R = ρL/A, keeping the area in square metres.
Compare my reasoning
- R = (2.0 × 10⁻⁸ × 5.0)/(2.0 × 10⁻⁶).
- Simplify the numerical factors and powers of ten.
- State Ω.
R = 0.050 Ω.
Look for these in your work
- Equation used
- Powers simplified
- Resistance unit stated
A wire has R = 0.20 Ω, L = 4.0 m and A = 1.0 × 10⁻⁶ m². Find resistivity.
Give me a hint
Rearrange to ρ = RA/L.
Compare my reasoning
- ρ = (0.20)(1.0 × 10⁻⁶)/4.0.
- Calculate the value.
- Give Ω m.
ρ = 5.0 × 10⁻⁸ Ω m.
Look for these in your work
- Equation rearranged
- Value calculated
- Resistivity unit stated
Two wires of the same material and length have diameters d and 2d. Compare their resistance, assuming the same temperature.
Give me a hint
Circular area is proportional to diameter squared and resistance is inversely proportional to area.
Compare my reasoning
- Doubling diameter makes cross-sectional area four times larger.
- With the same ρ and L, R = ρL/A.
- Therefore the thicker wire's resistance is one quarter.
The 2d wire has one quarter the resistance of the d wire.
Look for these in your work
- Area scaling squared
- Material and length controlled
- Resistance ratio stated
Common mistakes
- Using diameter instead of cross-sectional area
- Forgetting area must be in m²
- Confusing resistivity ρ with resistance R
What can you explain now?
Which quantity in R = ρL/A describes the material?
Compare with the explanation
Resistivity ρ
It is a material property at the stated temperature.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why does doubling wire diameter make its area four times larger, not twice as large?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.