Exercise 25.11 Part A A 1.40 m cylindrical rod of diameter 0.550 cm is connected
ID: 1774695 • Letter: E
Question
Exercise 25.11 Part A A 1.40 m cylindrical rod of diameter 0.550 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) te ammeter reads 18.7 A , while at 92.0 °C it reads 17.1 A.You can ignore any thermal expansion Find the resistivity and for the material of the rod at 20 °C. of the rod Submit My Answers Give Up Incorrect; Try Again; 18 attempts remaining Part B Find the temperature coefficient of resistivity at 20 °C for the material of the rod (C)-1 Submit My Answers Give UpExplanation / Answer
Resistance = p * (L/A) and Rf = Ri * ([1 + a * (Tf – Ti)]
where, p = Resistivity
L = length in meters
A = cross sectional area in m^2
a = temperature coefficient of resistivity
L = 1.40 m
Area = pi * r^2
r = d/2 = 0.55/2 = 0.275 cm = 2.75 * 10^-3 m
Area = pi * (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 deg C, R = 17 / 18.7
At 92 deg C, R = 17 / 17.1
Now, you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures.
Resistance = p * (L/A)
p = Resistance * (A/L)
At 20 deg C, p = (17/18.7) * [pi * (2.75 * 10^-3)^2] / 1.4 = 1.54 e-5 ohm m
At 92 deg C, p = (17/17.1) * [pi * (2.75 * 10^-3)^2] / 1.4 = 1.68 e-5 ohm 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 + a * (Tf – Ti)]
Rf = 17/17.1,
Ri = 17/18.7,
Tf = 92 deg C, Ti = 20 deg C
17/17.1 = 17 /18.7 * [1 + a * (92 – 20)]
18.7/17.1 = 1 + a*(92 - 20)
a = (1 - 18.7/17.1)/(92 - 20)
= 1.29*10^-3 deg C^-1 answer
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