Resistivity, thermistors and superconductivityAQA A-Level Physics: Revision notes

The resistance of a conductor depends on its dimensions and its material. Resistivity $\rho$ is the property of the material that does not depend on its size:

$\rho = \frac{R A}{L}$, so $R = \frac{\rho L}{A}$

The unit is the ohm metre (Ω m). Resistance is proportional to length and inversely proportional to cross-sectional area. Doubling the diameter multiplies the area by 4, so the resistance falls to a quarter.

Worked example: a wire of length 2.0 m and diameter 0.50 mm has $\rho = 4.9 \times 10^{-7}$ Ω m. $A = \pi \times (0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7}$ m², so $R = 4.9 \times 10^{-7} \times 2.0 / 1.96 \times 10^{-7} = 5.0$ Ω.

In a metal, the number of charge carriers (conduction electrons) is almost constant. As the temperature rises, the lattice ions vibrate with greater amplitude, so the electrons collide with them more often and the resistance increases.

This is why a filament lamp has a larger resistance when it is hot than when it is cold.

A negative temperature coefficient (ntc) thermistor is a semiconductor component whose resistance decreases as its temperature rises. As it gets hotter, more electrons are released from their atoms, so there are more charge carriers per unit volume. This effect outweighs the extra lattice vibrations, so the resistance falls.

Applications include temperature sensors (for example in incubators and thermostats), where the resistance is read from a resistance–temperature graph or used to change the pd in a circuit.

The resistance–temperature graph curves downwards from a high resistance at low temperature, falling more slowly at high temperature.

Some materials become superconductors: below a critical temperature their resistivity drops to zero. The critical temperature depends on the material.

With zero resistance, a current in a superconductor causes no energy transfer to thermal energy. Applications:

• Very strong magnetic fields produced by large currents in superconducting coils, for example in medical scanners and particle accelerators
• Reduced energy loss in the transmission of electric power

The material must be cooled below its critical temperature, for example with liquid nitrogen or liquid helium, which adds cost.

Method: use a long, uniform wire of constant diameter.

1. Measure the diameter with a micrometer at several points and in two perpendicular directions, and take the mean, then calculate $A = \pi d^2 / 4$
2. Connect the wire in series with a power supply and an ammeter, with a voltmeter across the length of wire being tested
3. Use a crocodile clip to change the length $L$, measured with a metre rule, and record $V$ and $I$ for each length
4. Calculate $R = V / I$ and plot $R$ against $L$
5. The gradient is $\rho / A$, so $\rho = \text{gradient} \times A$

Reducing errors: use a small current and switch off between readings so the wire does not heat up (heating increases R). The diameter gives the largest percentage uncertainty because it is small and squared in the area.