Question

A 1.40 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 17.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.6 A , while at 92.0 ∘C it reads 17.3 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.

Answer 1

I = V/R ( I is current, V is potential difference and R is resistance )
R is also equal to ' (resistivity) L/A ) ' where L is lenght of conductor and A is it's area of cross section
Hence (resistivity) L/A = V/I
resistivity = V A / L I = V pi d² / 4 L I d is diameter of conductor
hence resistivity of rod at 20^oC = 17x3.14x0.45²x10^-4 / 4x1.4x18.6 = 1.04x10^-5 ohm-m

b) R_T2 = R_T1 ( 1 + alpha ( T2 - T1) ) (assuming no change in length of rod with tempreture)
where R_T is resistance at tempreture T and 'alpha' is coeffiecient of resistivity at tempreture T1
taking T1 = 20 and T2 = 92 and using R = V/I we have
17/17.3 = 17/18.6 ( 1 + alpha ( 92 - 20 ) )
alpha = 1.04x10^-3 per ^oC