Week 5: Resistance 175
is probably upon temperature – we expect thermal speed to increase with temperature, which in
turn should decrease the time between collisions and hence decrease the average speed accumulated
during this time. Buthowshould it depend on temperature?
Recall that in the Drude model,〈v〉=√
3 kT /m, henceτtherm=d/〈v〉 ∝ 1 /√
T. From this we
expect the conductivity to vary like:
σ=nq^2 τtherm
m∝ 1 /
√
T or ρ∝√
T (318)
We thus expect from the Drude model that the resistance of metals should increase with temperature
like the square root of the temperature (and incidentally, this alters the scaling of conductivity with
the mass of the charge carriers as well).
This is incorrect – the Drude model, recall, is classical but in real materials conductivity is a
quantum phenomenon. In fact, the resistivities of most materials increase approximatelylinearly
with temperature, at least in the range of temperatures near room temperature. Close to room
temperature the resistivity is described accurately enough by the(linearized) Taylor series expansion:
ρ(T) =ρ 0 {1 +α(T−T 0 )} (319)whereρ 0 is the resistivity at temperatureT 0 andα= ρ^10 ∂T∂ρ evaluated atT =T 0. αis called
thetemperature coefficient of resistivity. This equation allow resistivity to be accurately computed
across a moderate, relevant, range of temperatures by means of two tabulated quantities.
However, it is too complicated for most of our purposes here^54. In this introductoryclassical
textbook we we will generally assume thatα ≈0 so thatρ= ρ 0 for any given material and
concentrate instead below on the simple scaling of resistance with length and area of the resistor.
Obviously, one cannot do this if one is designing circuits that heat up significantly as they operate
or that have to function correctly across a wide range of temperatures, and this whole approach
fails for things like semiconductors or superconductors that can be understood only with a correct
treatment in quantum theory.
L
J
E
A
ρ
Figure 54: A simple resistor with resistivityρ, lengthL, and cross sectional areaA.Now, consider an archetypical “resistor”: a uniform conductor with resistivityρ, lengthL, and
cross-sectional areaA(where the ends are at right angles to the sides), as pictured in figure 54. We
can rearrange the current density equation as:
E~=ρJ~ (320)The electric field and current density inside of this volume are both uniform (in steady state, all of
the charges must move through the volume at the same speed or charge would build up somewhere
in the volume). The electrical current is the flux of the current through either end, so:
∫
E~·ˆndA=EA=ρ
∫
J~·nˆdA=ρIthrough the resistor (321)(^54) Unless, of course, you are a physics major or are interested in electrical engineering, in which case you would do
well to at the very least earmark this discussion for future reference in more advanced courses.