The most common material used for sandpaper, silicon carbide, is also widely used in electrical applications. One common device is a tubular resistor made of a special grade of silicon carbide called carborundum. A particular carborundum resistor (see the figure) consists of a thick-walled cylindrical shell (a pipe) of inner radius \(a=\) \(1.50 \mathrm{~cm},\) outer radius \(b=2.50 \mathrm{~cm},\) and length \(L=60.0 \mathrm{~cm} .\) The resistance of this carborundum resistor at \(20 .{ }^{\circ} \mathrm{C}\) is \(1.00 \Omega\). a) Calculate the resistivity of carborundum at room temperature. Compare this to the resistivities of the most commonly used conductors (copper, aluminum, and silver). b) Carborundum has a high temperature coefficient of resistivity: \(\alpha=2.14 \cdot 10^{-3} \mathrm{~K}^{-1} .\) If, in a particular application, the carborundum resistor heats up to \(300 .{ }^{\circ} \mathrm{C},\) what is the percentage change in its resistance between room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and this operating temperature?

Step by step solution

01

Calculate the resistivity of carborundum at room temperature

To calculate the resistivity, we can use the formula: \(\text{Resistance} = \dfrac{\rho \cdot \text{Length}}{\text{Cross-sectional area}}\) The cross-sectional area of our cylindrical resistor can be expressed as the difference between the outer circle's area and the inner circle's area: \(\text{Cross-sectional area} = \pi(b^2 - a^2)\) Now we can rearrange the resistivity formula to find \(\rho\): \(\rho = \dfrac{\text{Resistance} \cdot \text{Cross-sectional area}}{\text{Length}}\) Plug in the values given in the problem statement: \(\rho = \dfrac{1.00 \Omega \cdot \pi(2.50^2 - 1.50^2)}{60.0}\) Calculate the resistivity: \(\rho \approx 7.04 \cdot 10^{-3} \text{ }\Omega\text{m}\)

02

Compare the resistivity of carborundum to common conductors

Let's compare the resistivity of carborundum (\(7.04 \cdot 10^{-3} \Omega\text{m}\)) to some common conductors: - Copper: \(\rho_{Cu} \approx 1.68 \cdot 10^{-8} \Omega\text{m}\) - Aluminum: \(\rho_{Al} \approx 2.82 \cdot 10^{-8} \Omega\text{m}\) - Silver: \(\rho_{Ag} \approx 1.59 \cdot 10^{-8} \Omega\text{m}\) Carborundum has a much higher resistivity compared to these common conductors, which is why it is used for resistive applications rather than conductive applications.

03

Calculate the percentage change in resistance

We can find the percentage change in resistance between the room temperature and the operating temperature using the formula: \(\Delta R (\%) = \alpha \cdot \Delta T \cdot 100\) Where \(\Delta R (\%)\) is the percentage change in resistance, \(\alpha = 2.14 \cdot 10^{-3} \mathrm{~K}^{-1}\) is the temperature coefficient of resistivity, and \(\Delta T\) is the temperature difference. The temperature difference is: \(\Delta T = 300^{\circ}C - 20^{\circ}C = 280^{\circ}C = 280\text{ K}\) Now find the percentage change in resistance: \(\Delta R(\%) = 2.14 \cdot 10^{-3} \mathrm{~K}^{-1} \cdot 280 \text{ K} \cdot 100\) \(\Delta R(\%) \approx 59.9 \%\) The resistance of the carborundum resistor increases by approximately \(59.9\%\) when its temperature increases from \(20^{\circ}C\) to \(300^{\circ}C\).

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