Analyzing the Setup
Imagine you are heating up a piece of metal wire. As the temperature rises, the atoms inside the metal start vibrating more vigorously. These vibrating atoms scatter the flowing electrons more frequently, which means the resistance of the wire increases! For most standard conductors over a reasonable temperature range, this increase is beautifully linear.
The master equation that governs this physical reality is:
Rt=R0(1+αΔT)
Here, Rt is the resistance at some temperature t, R0 is the resistance at a reference temperature (usually 0∘C), α is the temperature coefficient of resistance, and ΔT is the change in temperature from the reference.
The Master Equations
In our problem, we are given two distinct states of the conductor.
First, at
15∘C, the resistance is
16 Ω. Let's plug this into our master equation, assuming our reference temperature is
0∘C:
16=R0(1+α×15)
Next, we are told that at
100∘C, the resistance climbs to
20 Ω. Setting up the second equation, we get:
20=R0(1+α×100)
We now have a system of two equations with two unknowns: R0 and α.
The Mathematical Execution
Since we are only interested in finding α, the most elegant way forward is to eliminate R0. How do we do that? By simply dividing the first equation by the second!
2016=R0(1+100α)R0(1+15α)
The
R0 terms cancel out perfectly. Simplifying the fraction on the left gives us:
54=1+100α1+15α
Now, it's just a matter of careful algebra. Let's cross-multiply to get rid of the fractions:
4(1+100α)=5(1+15α)
Expanding the brackets, we get:
4+400α=5+75α
Let's group all the
α terms on one side and the constants on the other:
400α−75α=5−4
325α=1
Final Calculation
Finally, isolating
α, we find:
α=3251
When you perform this division, you get:
α≈0.003∘C−1
And there you have it! The temperature coefficient of resistance for this conductor is 0.003∘C−1. This tells us exactly how much the resistance will fractionally increase for every single degree rise in temperature.