A 1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 14.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 degrees C) the ammeter reads 18.5 A, while at 92.0 degrees C it reads 17.2 A. You can ignore any thermal expansion of the rod.

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

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

Question

A 1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 14.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 degrees C) the ammeter reads 18.5 A, while at 92.0 degrees C it reads 17.2 A. You can ignore any thermal expansion of the rod. 

Find the resistivity and for the material of the rod at 20 degrees C. 

Find the temperature coefficient of resistivity at 20 degrees C for the material of the rod.

anonymous · Answer

find resistance at 20C by V/I
find cross-sectional area
find resistivity

anonymous · Answer

$$R = \rho \frac{ l }{ A }$$

anonymous · Answer

Still confused on how to do this

anonymous · Answer

which part?

anonymous · Answer

dividing V by I?
or multiplying the square of d/2 by pi?
or plugging those values into the equation and solving for rho?

anonymous · Answer

how do you find the cross sectional area i mean?

anonymous · Answer

multiply the square of d/2 by pi

anonymous · Answer

ok, so for the resistance I got 14/1.7 which is 8.24. The the cross sectional area, I got pi(.0055/2)^2 which is 2.38*10^-5. Am I doing it right so far?

masumanwar · Answer

thats good

masumanwar · Answer

me integral

anonymous · Answer

alright, so after i find the resistance and cross sectional area, where do I go from here?