A 1.50 m cylindrical rod of cross-sectional area 1.963 times 10^-5 m^2 (radius 0
ID: 1544454 • Letter: A
Question
A 1.50 m cylindrical rod of cross-sectional area 1.963 times 10^-5 m^2 (radius 0.500 cm or 0.00500 m) is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 degree C) the ammeter reads 18.5 A while at 90.0 degree C it reads 17.2 A. At 90.0 degree C this gives a resistance of 0.872 ohm. Find the resistance and resistivity at 20.0 degree C and the temperature coefficient of resistivity at 20.0 degree C for the material of the rod. It turns out that the current is not uniform across this conductor. The magnitude J(r) of the current density in this wire (at 20.0 degree C) as a function of radial distance from the center of the wire's cross section is J(r) = Br. Find the coefficient B.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 = 1.50 m
Area = * r^2
r = 5 * 10^-3 m
Area = * (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.
a) Resistance = Voltage ÷ Current
At 20, R = 15 ÷ 18.5 = 0.81 ohm
At 90, R = 15 ÷ 17.2 = 0.87 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, = (15 ÷ 18.5) * 1.963*10^-5 ÷ 1.5 = 1.06*10^-5 ohm m
At 92, = (15 ÷ 17.2) * 1.963*10^-5 ÷ 1.5 = 1.06*10^-5 ohm m
b) 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 = 15 ÷ 17.2, Ri = 15 ÷ 18.5, Tf = 90, Ti = 20
15 ÷ 17.2 = 15 ÷ 18.5 * [1 + * (90 – 20)]
Multiply both sides by (18.5 ÷ 15)
(18.5 ÷ 15) * (15 ÷ 17.2) = 1 + * 70
Subtract 1 from both sides
(18.5 ÷ 15) * (15 ÷ 17.2) – 1 = * 70
Divide both sides by 70
= 1.079 * 10^-3
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