Resistivity and Temperature Effects Revision Notes | A Level AQA Physics

1. Why Resistance Depends on More Than Just the Material

You already know that resistance opposes current. But here is a question worth sitting with: if you take two wires made of exactly the same metal, one short and fat and one long and thin, do they have the same resistance? They do not — and yet they are made of the same stuff. That tells you resistance is not purely a material property; it also depends on the geometry of the conductor.

To separate the material's intrinsic opposition to current from the effect of shape and size, physicists define a quantity called resistivity. Resistivity belongs to the material alone. Once you know it, you can predict the resistance of any sample of that material, whatever its dimensions.

2. Resistivity: The Equation and What It Means

Notice the structure of the equation. Resistance increases with length — a longer wire gives electrons more collisions to navigate. Resistance decreases with cross-sectional area — a wider wire offers more parallel paths for charge to travel. Resistivity captures the material's contribution after those geometric effects are factored out.

To put some numbers in context: copper has a resistivity of around $1.7 \times 10^{-8}$ Ω m, which is why it dominates electrical wiring. Nichrome (used in heating elements) sits near $1.1 \times 10^{-6}$ Ω m — roughly 65 times higher. Glass is around $10^{12}$ Ω m. The range across materials spans more than 20 orders of magnitude, which is one of the most striking facts in all of physics.

Worked example: Calculating Resistivity from Measured Values

A student measures the resistance of a nichrome wire to be $5.6$ Ω. The wire has a length of $0.80$ m and a circular cross-section with diameter $0.50$ mm. Calculate the resistivity of nichrome. [4 marks]

1. $d = 0.50 \text{ mm} = 0.50 \times 10^{-3} \text{ m} = 5.0 \times 10^{-4} \text{ m}$ $r = \frac{d}{2} = \frac{5.0 \times 10^{-4}}{2} = 2.5 \times 10^{-4} \text{ m}$ — The diameter must be halved to get the radius, and converted from mm to m before squaring. This is the most common place students lose marks — forgetting to halve the diameter. [M1]
2. $A = \pi r^2 = \pi \times (2.5 \times 10^{-4})^2$ $(2.5 \times 10^{-4})^2 = 2.5 \times 10^{-4} \times 2.5 \times 10^{-4} = 6.25 \times 10^{-8} \text{ m}^2$ $A = \pi \times 6.25 \times 10^{-8} = 1.963 \times 10^{-7} \text{ m}^2$ — Expand the square longhand: $2.5 \times 2.5 = 6.25$ and $10^{-4} \times 10^{-4} = 10^{-8}$. Then multiply by $\pi$. [M1]
3. $\rho = \frac{RA}{L} = \frac{5.6 \times 1.963 \times 10^{-7}}{0.80}$ — Write the formula first, then substitute with units. This earns the method mark even if arithmetic slips. [M1]
4. $\rho = \frac{5.6 \times 1.963 \times 10^{-7}}{0.80} = \frac{1.099 \times 10^{-6}}{0.80} = 1.37 \times 10^{-6} \text{ Ω m}$ — Numerator: $5.6 \times 1.963 \times 10^{-7} = 10.99 \times 10^{-7} = 1.099 \times 10^{-6}$. Dividing by 0.80 gives $1.37 \times 10^{-6}$ Ω m (3 s.f.). This is consistent with the known value for nichrome. [A1]

Answer: $\rho \approx 1.4 \times 10^{-6}$ Ω m (2 s.f.)

That retrieval question highlights a key analytical skill: the ability to reason about ratios rather than just plugging numbers in. Examiners frequently set questions where dimensions change by a factor, and the expected response is a ratio argument, not a full numerical calculation.

3. Required Practical — Determining the Resistivity of a Wire

The resistivity equation is only useful if you can measure $R$, $L$, and $A$ reliably. AQA's required practical asks you to do exactly that for a metal wire, and the examiner will probe your understanding of the method, the variables, and the sources of uncertainty.

The graph interpretation is worth dwelling on. Because $R = \frac{\rho}{A} L$, a plot of $R$ against $L$ is linear with gradient $\frac{\rho}{A}$ and should pass through the origin. A non-zero intercept would suggest a systematic error — perhaps contact resistance at the crocodile clips. Recognising and explaining that kind of anomaly is exactly what distinguishes a high-mark evaluation response.

The gradient of this graph is approximately $6.95 \div 1.00 = 6.95$ Ω m⁻¹. If the wire has a cross-sectional area of $5.0 \times 10^{-7}$ m², then $\rho = 6.95 \times 5.0 \times 10^{-7} \approx 3.5 \times 10^{-6}$ Ω m. The exact value depends on the wire material and its temperature during the experiment — which leads directly into the next section.

4. Temperature and Resistance in Metal Conductors

Resistivity is not fixed forever for a given material. For metals, it changes with temperature — and understanding why requires thinking at the level of free electrons and the metal lattice.

Picture the interior of a metal: a regular lattice of positive ions surrounded by a sea of delocalised (free) electrons. When a voltage is applied, those electrons drift through the lattice, constituting a current. Their progress is interrupted whenever they collide with a lattice ion. Each collision transfers energy and impedes the flow — that is resistance at the microscopic level.

Now raise the temperature. The lattice ions vibrate with greater amplitude. The electrons encounter a more disrupted lattice and collide more frequently. More collisions mean greater resistance. This is why the resistance of a metal conductor increases with temperature.

For most pure metals over moderate temperature ranges, resistance increases approximately linearly with temperature. This predictable behaviour is exploited in platinum resistance thermometers, where the resistance of a platinum wire is used to measure temperature precisely.

5. NTC Thermistors: The Opposite Behaviour

Metals are not the only conductors. Semiconducting materials behave very differently, and the NTC thermistor is the most important example at A Level. NTC stands for negative temperature coefficient — meaning resistance decreases as temperature increases. This is the opposite of a metal.

The mechanism is different from metals. In a semiconductor, the number of charge carriers available to conduct is not fixed — it increases with temperature. As the material warms, more electrons gain enough energy to break free and contribute to conduction. This surge in carrier number outweighs the increased collision rate, so overall resistance falls.

At low temperatures, an NTC thermistor has very high resistance — it is almost an insulator. At high temperatures, resistance drops dramatically. The relationship is steep and curved, not the gentle linear rise seen in metals.

6. Resistance–Temperature Graphs and Comparing the Two Behaviours

A resistance–temperature graph is one of the most frequently tested items in this topic. You need to be able to sketch and interpret both types, and to explain the shape of each curve in terms of the underlying physics.

The metal line is approximately straight with a positive gradient — resistance rises steadily with temperature. The thermistor curve is steep and concave, falling sharply at first and then flattening as temperature increases. Both shapes carry marks in an exam: the gradient direction and the curvature are both assessable features.

• Metal conductor: resistance increases with temperature; graph is approximately linear with positive gradient; explained by increased lattice ion vibration causing more frequent electron collisions.
• NTC thermistor: resistance decreases with temperature; graph is a steep, negatively-sloped curve (non-linear); explained by a large increase in the number of charge carriers as temperature rises.
• Both graphs should start from a positive resistance value at 0 °C — neither material has zero resistance at room temperature.
• The thermistor curve flattens at higher temperatures; the metal curve remains approximately linear over moderate ranges.

7. Applications of Thermistors: Temperature Sensors

The dramatic change in resistance with temperature makes the NTC thermistor extremely useful as a temperature sensor. The key circuit to understand is the potential divider arrangement, where the thermistor is placed in series with a fixed resistor and a supply voltage. The output voltage is taken across one of the two components.

Consider a thermistor in series with a fixed resistor $R_0$, connected across a supply $V_s$. The output voltage $V_{\text{out}}$ is measured across $R_0$. As temperature rises, the thermistor's resistance falls, so a greater fraction of $V_s$ appears across $R_0$ — meaning $V_{\text{out}}$ increases. This rising output voltage can trigger a switching circuit or be read by a microcontroller as a temperature signal.

If instead the output is taken across the thermistor, $V_{\text{out}}$ decreases as temperature rises. Choosing which component to take the output across determines whether the sensor gives a rising or falling output — a design decision that depends on the application.

Real-world applications include the temperature sensor in a car engine management system, the thermostat in a central heating controller, and the body-temperature sensor in a digital thermometer. In each case, the thermistor's resistance is converted to a voltage that can be processed electronically.

8. Exam Technique and Common Mistakes

This topic rewards precision in language and careful unit handling. The following guidance consolidates the most mark-sensitive points across all four specification requirements.

For extended-response questions that ask you to compare metals and thermistors, structure your answer around three elements: the direction of change, the shape of the graph, and the microscopic mechanism. A response that covers all three for both materials will consistently reach the top mark band.

Resistivity links the macroscopic measurement of resistance to the intrinsic properties of a material, while temperature effects reveal that those properties are not static — they respond to the thermal environment in ways that are both predictable and exploitable in real devices.

The resistivity equation required by AQA is ρ = RA/L. It shows that resistivity is a property of the material itself, while resistance also depends on the length L and cross-sectional area A of the sample used.