A-Level — physicsMedium3 min read

Resistance and Resistivity

R = ρL/A; factors affecting resistance; I-V characteristics of components

Tutor AI Team
Tutor AI Team

# Resistance and Resistivity — A-Level Physics

Resistance describes how much a component opposes current flow. Resistivity is a material property that allows us to predict resistance from dimensions.

1. Ohm's Law

V=IRV = IR

For ohmic conductors at constant temperature: V∝IV \propto I (straight-line I-V graph through origin). Resistance is constant.

2. Resistivity

R=ρLA\boxed{R = \frac{\rho L}{A}}

Where:

  • RR = resistance (Ω)
  • ρ\rho = resistivity (Ω·m)
  • LL = length (m)
  • AA = cross-sectional area (m²)

Typical Resistivities

Material ρ\rho (Ω·m)
Copper 1.7×10−81.7 \times 10^{-8}
Aluminium 2.8×10−82.8 \times 10^{-8}
Nichrome 1.1×10−61.1 \times 10^{-6}
Silicon 0.1–60
Glass 101010^{10}

Effect of Temperature on Resistance

Metals: Resistance increases with temperature. Higher T → more lattice vibration → more collisions → higher resistivity.

Semiconductors/Thermistors (NTC): Resistance decreases with temperature. Higher T → more charge carriers released → lower resistivity.

Superconductors: Below critical temperature, resistance drops to exactly zero.

3. I-V Characteristics

Ohmic resistor: Straight line through origin. RR constant.

Filament lamp: Curved — RR increases with current (temperature increases).

Thermistor (NTC): RR decreases as temperature increases.

LDR: RR decreases as light intensity increases.

Diode: Very high RR in reverse; very low RR in forward bias above ~0.7 V.

Semiconductor diode: Exponential I-V curve in forward bias.

4. Required Practical: Measuring Resistivity

  1. Measure the diameter of the wire at several points using a micrometer
  2. Calculate mean diameter → area A=πd2/4A = \pi d^2/4
  3. Set up circuit with ammeter in series, voltmeter in parallel
  4. Measure VV and II for different lengths LL
  5. Calculate R=V/IR = V/I for each length
  6. Plot RR vs LL — gradient = ρ/A\rho/A
  7. ρ=gradient×A\rho = \text{gradient} \times A

Worked Example: Example 1

Problem

A copper wire (ρ=1.7×10−8\rho = 1.7 \times 10^{-8} Ω·m) is 3 m long with diameter 0.5 mm. Find R.

A=π(0.25×10−3)2=1.96×10−7A = \pi(0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7} m² R=ρL/A=1.7×10−8×3/1.96×10−7=0.26R = \rho L/A = 1.7 \times 10^{-8} \times 3/1.96 \times 10^{-7} = 0.26 Ω

Solution

Worked Example: Example 2

Problem

A wire of resistance 4 Ω, length 2 m, and diameter 1 mm. Find the resistivity.

A=π(0.5×10−3)2=7.85×10−7A = \pi(0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} m² ρ=RA/L=4×7.85×10−7/2=1.57×10−6\rho = RA/L = 4 \times 7.85 \times 10^{-7}/2 = 1.57 \times 10^{-6} Ω·m (nichrome-like)

Solution

6. Practice Questions

    1. A nichrome wire (ρ=1.1×10−6\rho = 1.1 \times 10^{-6} Ω·m) is 50 cm long with diameter 0.4 mm. Calculate R. (3 marks)
    1. Explain why the resistance of a filament lamp increases when it is switched on. (3 marks)
    1. What is a superconductor? State one application. (2 marks)

    Answers

    1. A=π(0.2×10−3)2=1.26×10−7A = \pi(0.2 \times 10^{-3})^2 = 1.26 \times 10^{-7} m². R=1.1×10−6×0.5/1.26×10−7=4.37R = 1.1 \times 10^{-6} \times 0.5/1.26 \times 10^{-7} = 4.37 Ω.
    1. When switched on, current flows and the filament heats up. At higher temperatures, metal ions vibrate more, causing more frequent collisions with electrons. This increases the resistivity and hence the resistance.

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Summary

  • R=ρL/AR = \rho L/A; resistivity is a material property
  • Metals: ρ\rho increases with temperature; semiconductors: ρ\rho decreases
  • Superconductors: R=0R = 0 below critical temperature
  • Required practical: plot RR vs LL to find ρ\rho
Tags:
#a-level#physics#electricity#resistance#resistivity#ohms-law

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