A 1.60 m cylindrical rod of diameter 0.500 cm is connected to a power supply tha
ID: 1521246 • Letter: A
Question
A 1.60 m cylindrical rod of diameter 0.500 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 18.8 A .while at 92.0 C it reads 17.3 A . You can ignore any thermal expansion of the rod. Part A Find the resistivity and for the material of the rod at 20 degree C. rho =______________Ohm middot m Part B Find the temperature coefficient of resistivity at 20 degree C for the material of the rod. alpha =___________(degree C)^-1Explanation / Answer
In my college physics book, I found the following equations.
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 = 1.60 m
Area = * r^2
r = d/2 = 0.25 cm = 2.5 * 10^-3 m
Area = * (2.5 * 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 = 16 ÷ 18.8
At 92, R = 16 ÷ 17.3
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, = (16 ÷ 18.8) * [ * (2.5 * 10^-3)^2] ÷ 1.6 = 1.044*10^-5
At 92, = (16 ÷ 17.3) * [ * (2.5 * 10^-3)^2] ÷ 1.6 = 1.135*10^-5
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 = 16 ÷ 17.3, Ri = 16 ÷ 18.8, Tf = 92, Ti = 20
16 ÷ 17.3 = 16 ÷ 18.8 * [1 + * (92 – 20)]
Multiply both sides by (18.8 ÷ 16)
(18.8 ÷ 16) * (16 ÷ 17.3) = 1 + * 72
Subtract 1 from both sides
(18.8 ÷ 16) * (16 ÷ 17.3) – 1 = * 72
Divide both sides by 72
= 1.204 * 10^-3
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