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Resistivity is a material property that quantifies how strongly a material opposes the flow of electric current. Unlike resistance, which depends on the shape and size of a conductor, resistivity is an intrinsic property of the material itself.
Spec mapping: This lesson maps to AQA 7408 specification section 3.5.1.4 — Resistivity, including the defining relationship R = ρL/A, the dependence of resistivity on temperature for metals and semiconductors, the phenomenon of superconductivity, and Required Practical 5 — determining the resistivity of a wire from gradient analysis of a plotted R against L graph. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- 3.4.1 — Materials: the cross-sectional area A used in the resistivity formula is the same geometric quantity used in stress σ = F/A calculations; problems that combine a stretched wire (which alters L and A simultaneously) are a standard A-Level synoptic move connecting resistivity to Young's modulus and elastic deformation.
- 3.6 — Thermal physics: the temperature dependence of resistivity is rooted in the same kinetic-theory description of lattice vibrations as thermal conductivity and specific-heat capacity; the Wiedemann–Franz relation (beyond spec) ties thermal and electrical conductivity together.
- 3.7 — Magnetic fields and electromagnetic induction: the very low resistivities of high-temperature superconductors enable the strong, stable magnetic fields used in MRI scanners and particle-accelerator dipole magnets — the link from this lesson into Year 13 electromagnetism.
ρ=LRA
Rearranged to give resistance:
R=AρL
where:
Key Point: The SI unit of resistivity is the ohm-metre (Ω m), NOT ohm per metre. Check the dimensions: R × A / L = Ω × m² / m = Ω m ✓
| Material | Resistivity (Ω m) at 20°C | Classification |
|---|---|---|
| Silver | 1.59 × 10⁻⁸ | Metal (conductor) |
| Copper | 1.68 × 10⁻⁸ | Metal (conductor) |
| Aluminium | 2.65 × 10⁻⁸ | Metal (conductor) |
| Nichrome | 1.10 × 10⁻⁶ | Alloy (higher resistivity) |
| Silicon (pure) | 6.40 × 10² | Semiconductor |
| Glass | 10¹⁰ – 10¹⁴ | Insulator |
| PTFE (Teflon) | 10²³ | Insulator |
Note the enormous range: from about 10⁻⁸ Ω m for good conductors to over 10²³ Ω m for excellent insulators — a factor of more than 10³⁰.
Calculate the resistance of a copper wire of length 2.0 m and diameter 0.50 mm. The resistivity of copper is 1.68 × 10⁻⁸ Ω m.
Solution:
L = 2.0 m
Diameter = 0.50 mm = 0.50 × 10⁻³ m, so radius r = 0.25 × 10⁻³ m
Cross-sectional area: A = πr² = π × (0.25 × 10⁻³)² = π × 6.25 × 10⁻⁸ = 1.963 × 10⁻⁷ m²
R = ρL/A = (1.68 × 10⁻⁸ × 2.0) / (1.963 × 10⁻⁷)
R = 3.36 × 10⁻⁸ / 1.963 × 10⁻⁷
R = 0.171 Ω
Exam Tip: Always convert diameter to radius, and convert mm to m, before calculating the area. A very common error is to forget to halve the diameter or to forget to square the radius. Also, remember to use the area formula for a circle: A = πr², not 2πr (which is the circumference).
A wire of length 1.50 m and cross-sectional area 3.0 × 10⁻⁷ m² has a resistance of 8.8 Ω. Calculate the resistivity of the material.
Solution:
ρ = RA/L = (8.8 × 3.0 × 10⁻⁷) / 1.50
ρ = 2.64 × 10⁻⁶ / 1.50
ρ = 1.76 × 10⁻⁶ Ω m
This value is close to the resistivity of nichrome, so the wire is likely made of nichrome (a nickel-chromium alloy used in heating elements).
Wire A is made of copper (ρ = 1.68 × 10⁻⁸ Ω m), is 3.0 m long and has a diameter of 1.0 mm. Wire B is made of the same material, is 1.5 m long and has a diameter of 0.50 mm. Calculate the resistance of each wire and determine which has the greater resistance.
Solution:
Wire A: r = 0.50 × 10⁻³ m, A = π(0.50 × 10⁻³)² = 7.854 × 10⁻⁷ m² R_A = ρL/A = (1.68 × 10⁻⁸ × 3.0) / 7.854 × 10⁻⁷ = 5.04 × 10⁻⁸ / 7.854 × 10⁻⁷ = 0.0642 Ω
Wire B: r = 0.25 × 10⁻³ m, A = π(0.25 × 10⁻³)² = 1.963 × 10⁻⁷ m² R_B = ρL/A = (1.68 × 10⁻⁸ × 1.5) / 1.963 × 10⁻⁷ = 2.52 × 10⁻⁸ / 1.963 × 10⁻⁷ = 0.128 Ω
Wire B has the greater resistance (0.128 Ω vs 0.064 Ω), even though it is shorter. This is because halving the diameter reduces the area by a factor of 4, which quadruples the resistance per unit length. Halving the length only halves the resistance. The net effect is that Wire B has double the resistance of Wire A.
For metals, resistivity increases with temperature.
Explanation at the microscopic level:
For many metals, resistivity increases approximately linearly with temperature over moderate ranges.
For semiconductors, resistivity decreases with temperature.
Explanation:
| Material type | Temperature ↑ | Resistivity | Reason |
|---|---|---|---|
| Metal | ↑ | ↑ Increases | More lattice scattering |
| Semiconductor | ↑ | ↓ Decreases | Many more charge carriers |
Some materials, when cooled below a critical temperature (T_c), exhibit zero resistivity. These materials are called superconductors.
Key properties of superconductors:
Critical temperatures of some superconductors:
| Material | Critical Temperature T_c (K) |
|---|---|
| Mercury | 4.2 |
| Lead | 7.2 |
| Niobium | 9.3 |
| YBa₂Cu₃O₇ (YBCO) | 93 |
| HgBa₂Ca₂Cu₃O₈ | 134 |
Applications of superconductors:
Limitations:
Exam Tip: You need to know that superconductors have zero resistivity below the critical temperature, and be able to describe at least two applications. You do not need to explain the mechanism of superconductivity (BCS theory) at A-Level.
From R = ρL/A:
A student measures the diameter of a constantan wire as 0.38 mm (mean of several measurements). They plot R vs L and obtain a straight line through the origin with gradient 5.8 Ω m⁻¹. Calculate the resistivity.
Solution:
d = 0.38 mm = 0.38 × 10⁻³ m, r = 0.19 × 10⁻³ m
A = πr² = π × (0.19 × 10⁻³)² = π × 3.61 × 10⁻⁸ = 1.134 × 10⁻⁷ m²
Gradient = ρ/A, so ρ = gradient × A
ρ = 5.8 × 1.134 × 10⁻⁷ = 6.6 × 10⁻⁷ Ω m
This is a reasonable value for constantan (expected: ~4.9 × 10⁻⁷ Ω m). The slight discrepancy could be due to systematic errors in measuring the diameter or slight temperature variations.
| Source | How to minimise |
|---|---|
| Wire diameter | Take multiple measurements at different positions and angles; use a micrometer |
| Zero error on micrometer | Check and record zero error; subtract from all readings |
| Temperature change | Use low currents to minimise heating; take readings quickly |
| Contact resistance | Use clean, tight connections; include crocodile clips in length measurement |
| Length measurement | Use a metre ruler; measure from the inside edges of the crocodile clips |
A nichrome heating wire has length L = 2.0 m, diameter d = 0.30 mm, and resistivity ρ_nichrome = 1.10 × 10⁻⁶ Ω m at 20 °C. (a) Calculate the resistance of this wire. (b) Compare the result with the resistance that an identical-geometry copper wire would have (ρ_copper = 1.68 × 10⁻⁸ Ω m). (c) Explain in one sentence why nichrome — not copper — is used for heating elements.
Solution — Part (a): Cross-sectional area and resistance.
Convert the diameter to SI metres: d = 0.30 × 10⁻³ m, giving radius r = 0.15 × 10⁻³ m.
A = πr² = π × (0.15 × 10⁻³)² = π × 2.25 × 10⁻⁸ = 7.07 × 10⁻⁸ m²
Apply R = ρL / A:
R_nichrome = (1.10 × 10⁻⁶ × 2.0) / (7.07 × 10⁻⁸) = (2.20 × 10⁻⁶) / (7.07 × 10⁻⁸) ≈ 31.1 Ω
Part (b): Copper comparison.
The geometric factor L/A is identical, so the ratio of resistances equals the ratio of resistivities:
R_copper / R_nichrome = ρ_copper / ρ_nichrome = (1.68 × 10⁻⁸) / (1.10 × 10⁻⁶) ≈ 0.0153
R_copper = 31.1 × 0.0153 ≈ 0.48 Ω
Copper has roughly 65 times less resistance than nichrome of the same geometry.
Part (c): Why nichrome for heating.
For a given supply voltage V, the power dissipated is P = V²/R. A higher resistance dissipates less power per unit length only if voltage is fixed and current is unrestricted — but in a domestic mains element, the wire's own resistance sets the current via I = V/R, and the dissipated power per metre of wire is what produces the heating. Nichrome is chosen because its high resistivity (around 65× copper's) lets you build a compact, controllable heating element from a manageable length of wire, and crucially because nichrome resists oxidation at the red-hot temperatures (800–1000 °C) at which heating elements operate. Copper would melt or oxidise rapidly long before reaching useful element temperatures.
Note: Always sanity-check resistivity calculations against textbook orders of magnitude. Pure metals: ρ ~ 10⁻⁸ Ω m. Alloys (nichrome, constantan): ρ ~ 10⁻⁶ Ω m. Semiconductors (silicon at room temp): ρ ~ 10² Ω m. Insulators: ρ > 10¹⁰ Ω m. An answer that falls outside the expected band by more than a factor of 10 is almost certainly a unit error — usually in converting mm² to m².
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