A 1.40 m cylindrical rod of diameter 0.550 cm is connected to a power supply tha

Question

A 1.40 m cylindrical rod of diameter 0.550 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 18.7 A , while at 92.0 C it reads 17.2 A . You can ignore any thermal expansion of the rod.

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

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

Explanation / Answer

from the given data

L = 1.4 m

cross sectional area of the rod, A = pi*d^2/4

= pi*(0.55*10^-2)^2/4

= 2.376*10^-5 m^2

a)
Let R1 is the resistance at 20 C

Apply Ohm's law

V = I*R1

R1 = V/I

= 16/18.7

= 0.8556 ohms

we know, R1 = rho1*L/A

==> rho1 = R1*A/L

= 0.8556*2.376*10^-5/1.4

= 1.452*10^-5 ohm.m

b) Let R2 is the resistance at 92 C

Apply Ohm's law

V = I*R2

R2 = V/I

= 16/17.2

= 0.93 ohms

we know, R2 = rho2*L/A

==> rho2 = R2*A/L

= 0.93*2.376*10^-5/1.4

= 1.578*10^-5 ohm.m

now the temperature coefficient of resistivity,

alfa = (rho2 - rho1)/(rho1*(T2-T1))

= (1.578 - 1.452)*10^-5/(1.452*10^-5*(92 - 20))

= 1.205*10^-3 C^-1

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