A 1.00-m cylindrical rod of diameter 0.460 cm is connected to a power supply tha

Question

A 1.00-m cylindrical rod of diameter 0.460 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 degree C) the ammeter reads 18.1 A, while at 92.0 degree C it reads 17.4 A. You can ignore any thermal expansion of the rod. (a) Find the resistivity at 20 degree C for the material of the rod. (b) Find the temperature coefficient of resistivity at 20 degree C for the material of the rod.

Explanation / Answer

part a )

R = rho*L/A

rho = restivity

L = length

A = area = pi*r^2

rho = RA/L

R at 2o degree = V/I =15/18.1

R at 92 degree = V/I = 15/17.4

rho = (15/18.1) x pi * (0.230 x 10^-2)^2 / 1

rho = 1.377 x 10^-5 ohm. m

part b )

Rf = Ri ( 1+ alpha (Tf-Ti)

alpha = thermal coffiecient of restivity

Rf = at 92 degree

Rf = 15/17.4

15/17.4 = 15/18.1[1+alpha(92-20)]

(15/17.4)/(15/18.1) = 1 + alpha(92-20)

18.1/17.4 = 1 + alpha(72)

alpha = 5.587 x 10^-4 C^-1

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