Question

Exercise 25.11 A 1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 17.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.2 A , while at 92.0 ∘C it reads 17.4 A . You can ignore any thermal expansion of the rod.

Part B

Find the temperature coefficient of resistivity at 20 ∘C for the material of the rod.

Answer 1

Resistance = ρ * (L/A) and Rf = Ri * ([1 + α * (Tf – Ti)]

ρ = Resistivity
L = length in meters
A = cross sectional area in m^2
α = temperature coefficient of resistivity

L = 1.70 m
Area = π * r^2
r = d/2 = 0.275 cm = 0.00275 m
Area = π * (0.00275)^2

= 0.0000237 m^2


The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases.
Resistance = Voltage ÷ Current
At 20˚, R = 17 ÷ 18.2 = 0.935 ohm
At 92˚, R = 17 ÷ 17.4 = 0.978 ohm

Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures.
Resistance = ρ * (L/A)
ρ = Resistance * (A/L)

At 20˚, ρ = (0.935) * [ (0.0000237] ÷ 1.7 = 0.936
At 92˚, ρ = (0.978) * [ (0.0000237] ÷ 1.7 = 0979

Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod.

Rf = Ri * ([1 + α * (Tf – Ti)]
Rf = 0.978, Ri = 0.935, Tf = 92˚, Ti = 20˚

0.978 = 0.935 * [1 + α * (92 – 20)]
Multiply both sides by 1.08
1.08 * (0.978) = 1 + α * 72
Subtract 1 from both sides
(1.08) * (0.978) – 1 = α * 72
Divide both sides by 72
α = 1.08 * 10^-3