1. the conductivity of the material
2. temprature differences
3.thickness of the material
4. area of the material
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 = 1.50 m Area = π * r^2 r = d/2 = 0.25 cm = 2.5 * 10^-3 m Area = π * (2.5 * 10^-3)^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 = 15 ÷ 18.5 At 92˚, R = 15 ÷ 17.2
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 = (15 ÷ 18.5) * [π * (2.5 * 10^-3)^2] ÷ 1.5 = At 92˚, ρ = (15 ÷ 17.2) * [π * (2.5 * 10^-3)^2] ÷ 1.5 =
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 = 15 ÷ 17.2, Ri = 15 ÷ 18.5, Tf = 92˚, Ti = 20˚
15 ÷ 17.2 = 15 ÷ 18.5 * [1 + α * (92 – 20)] Multiply both sides by (18.5 ÷ 15) (18.5 ÷ 15) * (15 ÷ 17.2) = 1 + α * 72 Subtract 1 from both sides (18.5 ÷ 15) * (15 ÷ 17.2) – 1 = α * 72 Divide both sides by 72 α = 1.05 * 10^-3
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