A 1.30m cylindrical rod of diameter 0.500cm is connected to a power supply that

Question

A 1.30m cylindrical rod of diameter 0.500cm is connected to a power supply that maintains a constant potential difference of 14.0V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ?C) the ammeter reads 18.8A , while at 92.0?C it reads 17.3A . You can ignore any thermal expansion of the rod.

Part A

Find the resistivity and for the material of the rod at 20 ?C.

Part B

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

Explanation / Answer

A)

Here, using ohm's law

V = IR

R = V/ I

R = 14/18.8

R = 0.744 Ohm

0.744 = p*1.30 /(pi*.0025^2)

solving for p

p = 1.124 * 10-5 Ohm . m

the resistivity and for the material of the rod is 1.124 * 10-5 Ohm . m

B)

at 92 C , using Ohm's law

Rf = 14/17.3

Rf = 0.809 Ohm

temperature coefficient of resistivity = change in Resistance/(R*change in temperature)

temperature coefficient of resistivity=   (0.809 -0.744 )/(0.744*(92 -20))

temperature coefficient of resistivity= 1.22 *10^-3 oC-1

temperature coefficient of resistivity is 1.22 *10^-3 oC-1

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