A 1.70 m cylindrical rod of diameter 0.500 cm is connected to a power supply tha

Question

A 1.70 m cylindrical rod of diameter 0.500 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 Degree C) the ammeter reads 18.2 A, while at 92.0 Degree C it reads 17.0 A. You can ignore any thermal expansion of the rod. Find the resistivity and for the material of the rod at 20 Degree C. Find the temperature coefficient of resistivity at 20 Degree C for the material of the rod.

Explanation / Answer

Part A:

Apply the formula -

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

where -
= 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 = 0.25 cm = 2.5 * 10^-3 m
Area = * (2.5 * 10^-3)^2

Now, 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 = 17 / 18.2 = 0.934 ohm

And, at 92, R = 17 / 17 = 1.0 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)

So, at 20, = (0.934) * [ * (2.5 * 10^-3)^2] / 1.7 = 1.078 x 10^-5 ohm-m.

Part B:
Again, for determining the temperature coefficient of resistivity for the material of the rod, apply the formula -

Rf = Ri * ([1 + * (Tf – Ti)]

1.0 = 0.934 * [1 + * (92 – 20)]

=> 1 + * (92 – 20) = 1.07066

=> * 72 = 0.07066

=> = 9.8144 x 10^-4 (deg C)^-1

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