Resistivity

In the previous lesson, you saw that the resistance of a wire depends on its length, cross-sectional area, and the material it is made from. Resistivity is the property that captures the material's inherent ability to resist the flow of current, independent of the wire's dimensions.

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Defining Resistivity

The resistance of a uniform conductor is given by:

R = ρL / A

where:

• R = resistance in ohms (Ω)
• ρ = resistivity of the material in ohm metres (Ω m)
• L = length of the conductor in metres (m)
• A = cross-sectional area in square metres (m²)

Rearranging for resistivity:

ρ = RA / L

Resistivity is a property of the material itself — it does not depend on the shape or size of the sample. A copper wire 1 m long has the same resistivity as a copper wire 10 m long, even though their resistances differ.

Comprehensive Resistivity Values

Material	Resistivity (Ω m)	Classification	Typical Use
Silver	1.6 × 10⁻⁸	Excellent conductor	High-end connectors, contacts
Copper	1.7 × 10⁻⁸	Excellent conductor	Wiring, cables, PCB traces
Aluminium	2.8 × 10⁻⁸	Good conductor	Overhead power lines
Tungsten	5.6 × 10⁻⁸	Conductor	Lamp filaments
Constantan	4.9 × 10⁻⁷	Alloy (moderate)	Resistance wire, strain gauges
Nichrome	1.1 × 10⁻⁶	Alloy (high)	Heating elements
Silicon	2.3 × 10³	Semiconductor	Transistors, diodes
Glass	~10¹⁰	Insulator	Insulation, optical fibre
Rubber	~10¹³	Insulator	Cable sheathing

Notice the enormous range — more than 20 orders of magnitude separate the best conductors from the best insulators.

Standard Wire Gauge Reference

When solving resistivity problems, you often need to convert wire diameters. Here are some common wire gauges:

SWG	Diameter (mm)	Area (mm²)	Area (m²)
18	1.22	1.17	1.17 × 10⁻⁶
22	0.711	0.397	3.97 × 10⁻⁷
26	0.457	0.164	1.64 × 10⁻⁷
30	0.315	0.0779	7.79 × 10⁻⁸
34	0.234	0.0430	4.30 × 10⁻⁸
36	0.193	0.0293	2.93 × 10⁻⁸

Unit conversion tip: To convert diameter in mm to area in m², use A = π(d × 10⁻³)²/4. A common error is forgetting to convert mm to m before squaring.

Worked Example 1

A nichrome wire has a length of 0.80 m and a diameter of 0.50 mm. The resistivity of nichrome is 1.1 × 10⁻⁶ Ω m. Calculate the resistance.

Solution:

• First find the cross-sectional area. The radius = 0.25 mm = 0.25 × 10⁻³ m
• A = πr² = π × (0.25 × 10⁻³)² = π × 6.25 × 10⁻⁸ = 1.96 × 10⁻⁷ m²
• R = ρL/A = (1.1 × 10⁻⁶ × 0.80) / (1.96 × 10⁻⁷)
• R = 8.8 × 10⁻⁷ / 1.96 × 10⁻⁷
• R = 4.5 Ω

Worked Example 2: Comparing Wires of Different Materials

An engineer needs a 1.0 m heating element with a resistance of at least 5.0 Ω. The wire diameter available is 0.40 mm. Which material should they use: copper (ρ = 1.7 × 10⁻⁸ Ω m) or nichrome (ρ = 1.1 × 10⁻⁶ Ω m)?

Solution:

• A = π(0.20 × 10⁻³)² = 1.257 × 10⁻⁷ m²
• Copper: R = (1.7 × 10⁻⁸ × 1.0) / (1.257 × 10⁻⁷) = 0.14 Ω (far too low)
• Nichrome: R = (1.1 × 10⁻⁶ × 1.0) / (1.257 × 10⁻⁷) = 8.75 Ω (meets the requirement)
• The engineer should use nichrome. This is exactly why nichrome is used in real heating elements — its high resistivity gives adequate resistance in a short length.

Worked Example 3: Finding Required Length

A student needs a 10.0 Ω constantan wire (ρ = 4.9 × 10⁻⁷ Ω m) of diameter 0.28 mm. What length should they cut?

Solution:

• A = π(0.14 × 10⁻³)² = 6.158 × 10⁻⁸ m²
• Rearranging R = ρL/A: L = RA/ρ = (10.0 × 6.158 × 10⁻⁸) / (4.9 × 10⁻⁷)
• L = 6.158 × 10⁻⁷ / 4.9 × 10⁻⁷ = 1.26 m

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Effect of Temperature on Resistivity

Metals

In metals, the charge carriers are free electrons. The number of free electrons is essentially fixed — it does not change significantly with temperature. However, as temperature rises, the lattice ions vibrate more vigorously, causing more frequent collisions with the electrons.

Result: the resistivity of a metal increases with temperature.

For most metals, the relationship is approximately linear over a moderate temperature range:

ρ = ρ₀(1 + αΔT)

where ρ₀ is the resistivity at a reference temperature, α is the temperature coefficient of resistivity, and ΔT is the temperature change.

Semiconductors

In semiconductors (such as silicon or germanium), increasing the temperature provides thermal energy that frees more electrons from their covalent bonds, creating additional charge carriers (electrons and holes). This increase in the number of charge carriers far outweighs the increased collision rate.

Result: the resistivity of a semiconductor decreases with temperature.

This is why NTC thermistors (made from semiconductor materials) have decreasing resistance at higher temperatures.

Comparison Table

Property	Metals	Semiconductors
Charge carriers	Free electrons (fixed number)	Electrons and holes (number increases with T)
Effect of ↑ temperature on collisions	More frequent → impedes flow	More frequent → impedes flow
Effect of ↑ temperature on carrier number	Negligible change	Large increase
Dominant effect	Collision rate increase	Carrier number increase
Net effect on ρ	Increases	Decreases

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Superconductors

A superconductor is a material whose resistivity drops to exactly zero below a critical temperature, known as the transition temperature or critical temperature (Tc).

When a material becomes superconducting, current can flow through it with absolutely no energy dissipation — no heating, no wasted energy. A current set flowing in a superconducting loop will flow indefinitely without a power supply.

Examples of Critical Temperatures

Material	Critical Temperature	Coolant Required
Mercury	4.2 K	Liquid helium
Lead	7.2 K	Liquid helium
Niobium	9.3 K	Liquid helium
Nb₃Sn	18 K	Liquid helium
MgB₂	39 K	Liquid helium/neon
YBCO (high-Tc)	93 K	Liquid nitrogen (77 K)

Most conventional superconductors have critical temperatures below 30 K, requiring liquid helium for cooling. High-temperature superconductors like YBCO can operate in liquid nitrogen (77 K), which is cheaper and more practical.

Applications of Superconductors

• MRI scanners: Superconducting magnets generate the strong, uniform magnetic fields needed for medical imaging.
• Particle accelerators: Superconducting magnets steer and focus beams of particles at facilities like CERN.
• Power transmission: Superconducting cables could transmit electricity with zero resistive losses (though cooling costs currently limit practical use).
• Maglev trains: Superconducting magnets can be used for magnetic levitation.

Limitations

The main limitation is that current superconductors require cooling to very low temperatures, which is expensive and energy-intensive. If a room-temperature superconductor were discovered, it would revolutionise electrical technology.

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Measuring Resistivity Experimentally

A common A-Level practical involves measuring the resistivity of a wire. The method is as follows:

1. Measure the diameter of the wire using a micrometer screw gauge at several points along the wire. Take at least three readings at different positions and different orientations, then calculate the mean. Calculate the cross-sectional area using A = πd²/4.

2. Set up the circuit with a battery, an ammeter in series, and a voltmeter in parallel with the section of wire being measured. Include a variable resistor to adjust the current.

3. Measure V and I for a known length of wire. Calculate R = V/I.

4. Repeat for different lengths — for example, 0.20 m, 0.40 m, 0.60 m, 0.80 m, 1.00 m.

5. Plot R against L. The graph should be a straight line through the origin. The gradient equals ρ/A, so:

ρ = gradient × A

flowchart TD
    A[Measure diameter with micrometer at 6+ points] --> B[Calculate mean d and A = πd²/4]
    B --> C[Set up circuit: battery, ammeter, voltmeter, variable resistor]
    C --> D[Measure V and I for length L₁]
    D --> E[Calculate R₁ = V/I]
    E --> F[Repeat for L₂, L₃, L₄, L₅]
    F --> G[Plot R against L]
    G --> H{Straight line through origin?}
    H -->|Yes| I[gradient = ρ/A, so ρ = gradient × A]
    H -->|No| J[Check for systematic error - contact resistance?]

Sources of Uncertainty

• The wire may heat up during the experiment, changing its resistivity. Keep currents low and take readings quickly.
• The micrometer reading for diameter has significant impact because area depends on d². A small error in d leads to a proportionally larger error in A.
• Contact resistance at the connections can add a small systematic error, causing the y-intercept to be slightly above zero.
• Ensure the wire is taut and straight — sagging changes the effective length.

Worked Example 4: Processing Experimental Data

A student measures the following data for a constantan wire of diameter 0.38 mm:

Length (m)	Resistance (Ω)
0.20	0.86
0.40	1.75
0.60	2.60
0.80	3.48
1.00	4.35

From the graph, the gradient is 4.35 Ω m⁻¹. Calculate the resistivity.

Solution:

• d = 0.38 mm = 0.38 × 10⁻³ m
• A = π(0.38 × 10⁻³)²/4 = 1.134 × 10⁻⁷ m²
• ρ = gradient × A = 4.35 × 1.134 × 10⁻⁷ = 4.93 × 10⁻⁷ Ω m
• This is very close to the accepted value of 4.9 × 10⁻⁷ Ω m for constantan.

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A-Level Deep Dive: Resistivity

Spec mapping

Edexcel 9PH0 specification, Topic 3 — Electric circuits, sub-topic 3.3 (resistivity) establishes $R = \rho L / A$R=ρL/A for a uniform conductor, the definition of resistivity $\rho$ρ as a material property with SI unit ohm metre ($\Omega\ \text{m}$Ω m), the qualitative effect of temperature on $\rho$ρ for metals and semiconductors, and superconductivity below a critical temperature $T_c$Tc​ (refer to the official Pearson Edexcel specification document for exact wording). Resistivity binds the whole of Topic 3 together: it converts the abstract $R$R of 3.2 (Ohm's law) into a quantity predictable from material data and wire dimensions, and it underpins CP6 — determination of resistivity from resistance vs length, examined on Paper 3 (9PH0-03, General and Practical Principles in Physics). The formula booklet lists $R = \rho L / A$R=ρL/A, but the interpretation — that $\rho$ρ is intrinsic to the material whereas $R$R depends on geometry — must be argued, not assumed.

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Worked example with full mark scheme

Question (8 marks):

A student is asked to design a heating element of resistance $R = 12.0\ \Omega$R=12.0 Ω using nichrome wire of diameter $d = 0.40\ \text{mm}$d=0.40 mm. The resistivity of nichrome at room temperature is $\rho = 1.10 \times 10^{-6}\ \Omega\ \text{m}$ρ=1.10×10−6 Ω m.

(a) State what is meant by the resistivity of a material. (1)

(b) Calculate the cross-sectional area of the wire in $\text{m}^2$m2. (2)

(c) Show that the required length of wire is approximately $1.4\ \text{m}$1.4 m. (3)

(d) The student then operates the element so that its temperature rises by $300\ \text{K}$300 K. The resistivity of nichrome increases by about $12\%$12% over this range. State and explain the effect on the actual resistance of the element if its length and diameter are unchanged. (2)

Solution with mark scheme:

(a) B1 — resistivity is the resistance of a sample of the material of unit length and unit cross-sectional area; equivalently, $\rho = RA/L$ρ=RA/L for a uniform conductor, with units $\Omega\ \text{m}$Ω m. A bare $\rho = RA/L$ρ=RA/L without naming the quantities is borderline; examiners reward the "property of the material, independent of shape and size" phrasing as the unambiguous statement.

(b) Step 1 — convert diameter to metres and find the radius.

$d = 0.40\ \text{mm} = 0.40 \times 10^{-3}\ \text{m}$d=0.40 mm=0.40×10−3 m, so $r = 0.20 \times 10^{-3}\ \text{m}$r=0.20×10−3 m.

Step 2 — apply $A = \pi r^2$A=πr2.

$A = \pi \times (0.20 \times 10^{-3})^2 = \pi \times 4.0 \times 10^{-8} \approx 1.26 \times 10^{-7}\ \text{m}^2$A=π×(0.20×10−3)2=π×4.0×10−8≈1.26×10−7 m2

M1 — converting mm to m before squaring (the most common slip is to square first and then convert, producing a value $10^6$106 times too large). A1 — $A \approx 1.26 \times 10^{-7}\ \text{m}^2$A≈1.26×10−7 m2 to 3 s.f.

(c) Step 1 — start from $R = \rho L / A$R=ρL/A and rearrange for $L$L.

$L = \dfrac{RA}{\rho}$L=ρRA​

M1 — correct rearrangement of $R = \rho L / A$R=ρL/A for length.

Step 2 — substitute consistent SI quantities.

$L = \dfrac{12.0 \times 1.26 \times 10^{-7}}{1.10 \times 10^{-6}} = \dfrac{1.51 \times 10^{-6}}{1.10 \times 10^{-6}}$L=1.10×10−612.0×1.26×10−7​=1.10×10−61.51×10−6​

M1 — substitution with $R$R in $\Omega$Ω, $A$A in $\text{m}^2$m2 and $\rho$ρ in $\Omega\ \text{m}$Ω m.

A1 — $L \approx 1.37\ \text{m}$L≈1.37 m, which rounds to $1.4\ \text{m}$1.4 m to 2 s.f. as required by the printed value. Because this is a "show that", the candidate must produce an answer with at least one extra significant figure beyond the printed value (so $1.37\ \text{m}$1.37 m, not $1.4\ \text{m}$1.4 m, secures the A1).

(d) B1 — the resistance increases (by about $12\%$12%, since $R = \rho L / A$R=ρL/A with $L$L and $A$A fixed implies $R \propto \rho$R∝ρ).

B1 — explanation: at higher temperature, lattice ions vibrate with greater amplitude, increasing the rate at which conduction electrons scatter; the mean free time between collisions falls, so $\rho$ρ rises and therefore $R$R rises in the same proportion. (Numerically, $R \approx 1.12 \times 12.0 = 13.4\ \Omega$R≈1.12×12.0=13.4 Ω, but the question asks only for direction and reasoning.)

Total: 8 marks (B1 + M1 A1 + M1 M1 A1 + B1 B1).

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Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A coil of copper wire has length $L = 4.50\ \text{m}$L=4.50 m and a circular cross-section of diameter $0.32\ \text{mm}$0.32 mm. The resistivity of copper is $1.70 \times 10^{-8}\ \Omega\ \text{m}$1.70×10−8 Ω m.

(a) Calculate the cross-sectional area of the wire. (1)

(b) Calculate the resistance of the coil. (3)

(c) The coil is replaced by an aluminium coil of the same length and resistance. The resistivity of aluminium is $2.80 \times 10^{-8}\ \Omega\ \text{m}$2.80×10−8 Ω m. By what factor must the cross-sectional area of the aluminium wire differ from that of the copper wire? (2)

Mark scheme decomposition by AO:

(a)

• B1 (AO2.2) — $A = \pi (0.16 \times 10^{-3})^2 = 8.04 \times 10^{-8}\ \text{m}^2$A=π(0.16×10−3)2=8.04×10−8 m2. The mm-to-m conversion before squaring is the AO2 step.

(b)

• M1 (AO1.1b) — quoting $R = \rho L / A$R=ρL/A from the formula booklet.
• M1 (AO1.1b) — substituting $R = (1.70 \times 10^{-8}) \times 4.50 / (8.04 \times 10^{-8})$R=(1.70×10−8)×4.50/(8.04×10−8).
• A1 (AO1.1b) — $R \approx 0.951\ \Omega$R≈0.951 Ω (numerator $7.65 \times 10^{-8}$7.65×10−8, divided by $8.04 \times 10^{-8}$8.04×10−8).

(c)

• M1 (AO2.1) — recognising that for fixed $R$R and $L$L, $A \propto \rho$A∝ρ, so $A_\text{Al}/A_\text{Cu} = \rho_\text{Al}/\rho_\text{Cu} = 2.80/1.70$AAl​/ACu​=ρAl​/ρCu​=2.80/1.70.
• A1 (AO2.5) — concluding that the aluminium wire must have a cross-section about $1.65$1.65 times that of the copper wire (or equivalently a diameter $\sqrt{1.65} \approx 1.28$1.65​≈1.28 times larger), with a brief comment that this is precisely why aluminium overhead lines are physically thicker than copper conductors of the same resistance.

Total: 6 marks split AO1 = 3, AO2 = 3. The pattern echoes the charge/current question earlier in the topic: AO1 marks for procedural $R = \rho L / A$R=ρL/A work, AO2 marks for proportional reasoning about which variable changes when. Edexcel routinely tests "by what factor" comparisons on resistivity precisely because the equation has three independent variables on the right-hand side.

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Synoptic links

Connects to:

1. Topic 3 sub-topic 3.1 — drift velocity: combining $V = IR$V=IR with $R = \rho L / A$R=ρL/A and $I = nAve$I=nAve gives $V = \rho L n e v$V=ρLnev, so for fixed geometry drift velocity is proportional to $V/\rho$V/ρ. A small $\rho$ρ corresponds to electrons that scatter rarely and therefore drift faster for the same applied field — the bridge between resistivity and the free-electron picture.