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

Question

A 1.30 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 12.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 C ) the ammeter reads 18.8 A , while at 92.0 C it reads 17.1 A . You can ignore any thermal expansion of the rod.

Part A

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

Answer: p= 9.64*10^-6 m     

Part B (Find part B and show work please)

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

Answer: = ? (C)1    

Explanation / Answer

Resistance = * (L/A) and Rf = Ri * ([1 + * (Tf – Ti)]

= Resistivity
L = length in meters
A = cross sectional area in m2
= temperature coefficient of resistivity

L = 1.30 m
Area = * r2= * (d/2)2= * (2.5 * 10-3)2=1.9625*10-5 m2

The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases.
Resistance = V/I
At 20, R = 12/18.8=0.638 ohm
At 92, R = 12/17.1=0.701 ohm


Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures.
Resistance = * (L/A)
= Resistance * (A/L)

At 20, = (0.638) * (1.9625*10-5) /1.3 =9.63* 10-6 m


Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod.

Rf = Ri * ([1 + * (Tf – Ti)]
Rf =0.701 ohm , Ri = 0.638 ohm, Tf = 92, Ti = 20

0.701=0.638*(1 + * (92– 20))

1.099=1 + *72

= 1.38 * 10-3 (C)-1

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