B.5.12—Resistance and resistivity
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Resistance of a uniform conductor
R=ρAL
where ρ is resistivity, L is length and A is cross-sectional area. Resistivity is a material property at the stated conditions.
Read the scaling
At fixed material, doubling length doubles R. Doubling diameter makes area four times larger and reduces R to one quarter. A longer, thinner wire has greater resistance.
Units
Resistivity has SI unit Ω m. Use A=πr2 for a circular wire and convert radius/diameter to metres before calculating.
Worked example from the mapped local textbook
Nichrome has ρ=1.1×10−6Ωm. A wire has L=1.96m and radius r=0.21mm=0.21×10−3m.
A=πr2=1.39×10−7m2
R=AρL=1.39×10−7(1.1×10−6)(1.96)=16Ω
Converting the radius before squaring prevents a factor-of-106 error.
Common trap
Do not treat resistivity as the resistance of every sample of a material. Geometry changes resistance even when ρ is unchanged.
The evidence asks for a wire radius from resistance data or for the new resistance after scaling length and diameter, so proportional reasoning and cross-sectional area are central.
Calculate / Determine
Use R=ρL/A and keep the geometry explicit. For a change in diameter, convert area using A∝d²; for a change in length, scale R directly with L. Give the final resistance or radius with units and explain which dimensions changed.
Scaling diameter as though it were area, rather than using A=πd²/4 so that area scales with the square of diameter.
Representative question
The total length of the metal wire is 5.0 m . Calculate the radius of the wire.
Resistivity of the high-resistance alloy =1.5×10−6Ω m
Use of ρ=IRA area =<Rρl=>7.8×10−6≪ m2>∨
radius <=7.8×10−6/π=>1.6×10−3< m>
Check for ECF from (a)(i)