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This cheat sheet covers how a material’s resistance depends on its size, resistivity, and temperature. Students need these ideas to analyze wires, circuits, sensors, and real electrical materials. It connects microscopic material properties to the measurable resistance of a conductor. The reference is organized around the formulas most often used in high school physics problems. The key relationship is R=ρLAR = \rho \frac{L}{A}, where resistance increases with length and decreases with cross-sectional area. Resistivity ρ\rho describes how strongly a material opposes electric current, while conductivity σ\sigma measures how easily charge flows. For many metals over a moderate temperature range, resistivity changes approximately as ρ=ρ0[1+α(TT0)]\rho = \rho_0[1 + \alpha(T - T_0)]. The same linear temperature model can often be used for resistance when the wire’s dimensions do not change much.

Key Facts

  • Resistance is related to resistivity by R=ρLAR = \rho \frac{L}{A}, where LL is length and AA is cross-sectional area.
  • Resistivity can be found from a measured resistance using ρ=RAL\rho = \frac{RA}{L}.
  • Conductivity is the reciprocal of resistivity, so σ=1ρ\sigma = \frac{1}{\rho}.
  • For many metals, resistivity changes with temperature according to ρ=ρ0[1+α(TT0)]\rho = \rho_0[1 + \alpha(T - T_0)].
  • If the conductor’s dimensions stay nearly constant, resistance changes with temperature as R=R0[1+α(TT0)]R = R_0[1 + \alpha(T - T_0)].
  • A positive temperature coefficient α\alpha means resistance usually increases as temperature increases.
  • A negative temperature coefficient α\alpha means resistance usually decreases as temperature increases, which is common in many semiconductors.
  • The cross-sectional area of a round wire is A=πr2=π(d2)2A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2.

Vocabulary

Resistance
Resistance is the opposition to electric current in a specific object, measured in ohms Ω\Omega.
Resistivity
Resistivity is a material property that describes how strongly a substance opposes current, measured in Ωm\Omega\cdot\text{m}.
Conductivity
Conductivity is a material property that describes how easily charge flows, given by σ=1ρ\sigma = \frac{1}{\rho}.
Temperature Coefficient
The temperature coefficient α\alpha tells how much resistivity or resistance changes per degree of temperature change.
Cross-Sectional Area
Cross-sectional area is the area of the cut face of a conductor, such as A=πr2A = \pi r^2 for a round wire.
Reference Temperature
The reference temperature T0T_0 is the starting temperature at which ρ0\rho_0 or R0R_0 is known.

Common Mistakes to Avoid

  • Confusing resistance with resistivity is wrong because RR depends on the object’s length and area, while ρ\rho depends mainly on the material and temperature.
  • Forgetting to convert diameter to radius is wrong because the area formula uses rr, so a wire with diameter dd has A=π(d2)2A = \pi \left(\frac{d}{2}\right)^2.
  • Using Celsius change incorrectly is wrong because the formula needs the temperature difference TT0T - T_0, not just the final temperature.
  • Assuming all materials have positive α\alpha is wrong because many semiconductors and some special materials can have negative temperature coefficients.
  • Treating the linear model as exact at all temperatures is wrong because ρ=ρ0[1+α(TT0)]\rho = \rho_0[1 + \alpha(T - T_0)] is an approximation over a limited temperature range.

Practice Questions

  1. 1 A copper wire has ρ=1.68×108 Ωm\rho = 1.68 \times 10^{-8}\ \Omega\cdot\text{m}, L=2.0 mL = 2.0\ \text{m}, and A=1.0×106 m2A = 1.0 \times 10^{-6}\ \text{m}^2. Find its resistance using R=ρLAR = \rho \frac{L}{A}.
  2. 2 A wire has resistance R=4.0 ΩR = 4.0\ \Omega, length L=5.0 mL = 5.0\ \text{m}, and cross-sectional area A=2.0×106 m2A = 2.0 \times 10^{-6}\ \text{m}^2. Find its resistivity using ρ=RAL\rho = \frac{RA}{L}.
  3. 3 A metal resistor has R0=10.0 ΩR_0 = 10.0\ \Omega at T0=20CT_0 = 20^\circ\text{C} and α=0.0040 C1\alpha = 0.0040\ ^\circ\text{C}^{-1}. Find RR at T=70CT = 70^\circ\text{C} using R=R0[1+α(TT0)]R = R_0[1 + \alpha(T - T_0)].
  4. 4 Explain why a long, thin wire made of the same material has a larger resistance than a short, thick wire at the same temperature.