Resistivity & Superconductivity

Material properties, the resistivity equation, temperature dependence, and what happens at absolute zero.

Spec Points Covered
  • Define resistivityA material property that quantifies how strongly it resists currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohm-metres (Ω m). and use the equation $\rho = RA/L$.
  • Explain why doubling length doubles resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω). and doubling diameter quarters resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω)..
  • Describe and evaluate the required practical for measuring resistivityA material property that quantifies how strongly it resists current. Measured in ohm-metres (Ω m). of a wire.
  • Explain the mechanism of resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω). change in metals and NTC thermistors.
  • Define superconductivityThe property of certain materials to have zero resistivityA material property that quantifies how strongly it resists current. Measured in ohm-metres (Ω m). below a critical temperature (the transition temperature). and critical temperature, and state applications of superconductors.
  • Sketch and interpret resistance-temperature graphs for metals, thermistors, and superconductors.
Σ Key Equations Full Reference →
On Data Sheet
Not on Data Sheet
Resistivity equation
$$\rho = \frac{RA}{L}$$
  • Where:
    • $ρ$ = resistivity (Ω m)
    • $R$ = resistance (Ω)
    • $A$ = cross-sectional area (m²)
    • $L$ = length (m)
Material property. Rearranges to R = ρL/A for calculations.
Cross-sectional area of a wire
$$A = \frac{\pi d^{2}}{4}$$
  • Where:
    • $A$ = cross-sectional area (m²)
    • $d$ = diameter (m)
Essential for the resistivity practical. Convert d from mm to m first.
Resistance from resistivity
$$R = \frac{\rho L}{A}$$
  • Where:
    • $R$ = resistance (Ω)
    • $ρ$ = resistivity (Ω m)
    • $L$ = length (m)
    • $A$ = cross-sectional area (m²)
The form used in most exam calculations. Gradient of R vs L graph = ρ/A.
Q Retrieval Practice All 12 Questions →
Q1. Define resistivity and state its unit.
  • Resistivity is a material property: the resistance of a 1 m length of material with a cross-sectional area of 1 m². ρ = RA/L.
  • Unit: Ω m.
Q2. A wire's diameter doubles. What happens to its resistance?
  • Resistance drops to one quarter.
  • Area = πd²/4, so doubling d quadruples A.
  • Since R = ρL/A, R falls by a factor of 4.
Q3. In the resistivity practical, what graph is plotted and what does the gradient give?
  • Plot R (y-axis) against L (x-axis).
  • The gradient equals ρ/A.
  • Multiply the gradient by the cross-sectional area to find resistivity.
Q4. Why must the wire be switched off between readings in the resistivity practical?
  • To prevent the wire from heating up.
  • Higher temperature changes the resistivity, which would make the results unreliable.
Q5. Explain why the resistance of a metal increases with temperature.
  • Higher temperature causes metal ions to vibrate with greater amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m)..
  • Conduction electronsDelocalised electrons in a metal that are free to move through the lattice structure and carry electric current. collide more frequently with the ions, increasing opposition to current flow.