A 1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply tha

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 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 18.3 A , while at 92.0 C it reads 17.1 A . You can ignore any thermal expansion of the rod.

A.Find the resistivity and for the material of the rod at 20 C. in(ohm-meters)

B. Find the temperature coefficient of resistivity at 20 C for the material of the rod. in(degrees celcius ^-1)

Explanation / Answer

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

= Resistivity
L = length in meters
A = cross sectional area in m^2
= temperature coefficient of resistivity

L = 1.70 m
Area = * r^2
r = d/2 = 0.275 cm = 2.75 * 10^-3 m
Area = * (2.75 * 10^-3)^2


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 = Voltage ÷ Current
At 20, R = 12 ÷ 18.3 = 0.656
At 92, R = 12 ÷ 17.1 = 0.702

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, = (12 ÷ 18.3) * [ * (2.75 * 10^-3)^2] ÷ 1.7 = 9.16*10^-6
At 92, = (12 ÷ 17.1) * [ * (2.75 * 10^-3)^2] ÷ 1.7 = 9.8*10^-6

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 = 0.702, Ri = 0.656, Tf = 92, Ti = 20

0.702 = 0.656 * [1 + * (92 – 20)]
1.07 = 1 + * 72
0.07 = * 72
= 9.72* 10^-4

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