Question

A 1.60 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 16.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.8 A , while at 92.0 ∘C it reads 17.0 A . You can ignore any thermal expansion of the rod.

Find the resistivity and for the material of the rod at 20 ∘

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

Answer 1

Resistance = p * (L/A)

R_f = R_i * ([1 + alpha * (T_f – T_i)]

L = 1.60 m, Area = pi * r²

r = d / 2 = 0.25 cm = 2.5 * 10^-3 m

Area = pi * (2.5 * 10^-3)²

Resistance = Voltage / Current

At 20˚, R = 16 / 18.8

At 92˚, R = 16 / 17

p = Resistance * (A/L)

At 20˚, p = (16 / 18.8) * [pi * (2.5 * 10^-3)² ] / 1.6

At 92˚, p = (16 / 17) * [pi * (2.5 * 10^-3)² ] / 1.6

The temperature coefficient of resistivity for the material of the rod R_f = R_i * ([1 + alpha * (T_f – T_i)]

[16 / 17] = [16 / 18.8] * [1 + alpha * (92 – 20)]

(18.8 / 16) * (16 / 17) = 1 + alpha * 72

alpha = 1.47 * 10^-3