Question

<module 3a
electrical resistance and temperature conceptual question
the electrical resistance of most materials is dependent on temperature. for many materials, the resistance decreases as the temperature of the material decreases. the data below show the resistance of a disk of yba₂cu₃o₇, a hard, brittle ceramic, at temperatures between 200 k and 240 k (figure 1)
figure
< 1 of 1

10
8
6
4
2
0
40 80 120 160 200 240
temperature (k)
r(10⁻³ω)
part a
based on the trend of these data, predict the expected resistance r of the sample at 160 k. express your answer in milliohms (10⁻³ ω).

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r =
×10⁻³ ω
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Explanation:

Step1: Assume linear - like relationship

Since no other information about the relationship between resistance and temperature is given, assume a linear - like relationship between resistance $R$ and temperature $T$ in the given range. Let $R=aT + b$. We can take two points from the graph (e.g., if we assume we can read approximate values at two points in the range $200K - 240K$). But since the graph details are not fully clear, we can also use the general idea of the trend. If we assume the resistance $R$ changes linearly with $T$, and we know that resistance decreases as temperature decreases.

Step2: Estimate from the trend

Looking at the graph, as the temperature decreases from $200K$ to $160K$ (a decrease of $\Delta T=40K$), and assuming a roughly linear decrease in resistance. If we assume at $200K$ the resistance is around $8\times10^{-3}\Omega$ (a rough estimate from the graph) and at $240K$ it is around $10\times10^{-3}\Omega$. The slope of the line $m=\frac{\Delta R}{\Delta T}=\frac{(8 - 10)\times10^{-3}}{200 - 240}=\frac{- 2\times10^{-3}}{-40}=5\times10^{-5}\Omega/K$.
Using the point - slope form $R - R_1=m(T - T_1)$, taking $(T_1,R_1)=(200,8\times10^{-3})$, when $T = 160K$, we have $R-8\times10^{-3}=5\times10^{-5}(160 - 200)$.
$R-8\times10^{-3}=5\times10^{-5}\times(- 40)=-2\times10^{-3}$.
$R=(8\times10^{-3}-2\times10^{-3}) = 6\times10^{-3}\Omega$.

Answer:

$6$