Question

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?

Short answer

Answer: The resistivity of carborundum at room temperature is approximately \(7.04 \cdot 10^{-3} \Omega\text{m}\). It has a much higher resistivity compared to common conductors such as copper, aluminum, and silver. The resistance of the carborundum resistor increases by approximately 59.9% when its temperature increases from 20°C to 300°C.

Step-by-step solution

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: \(\rho \approx 7.04 \cdot 10^{-3} \text{ }\Omega\text{m}\)

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.

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 \(300^{\circ}C\).

Key concepts

Resistivity Calculation

Resistivity is a measure of how strongly a material opposes the flow of electric current. It's an intrinsic property that varies between different materials. In the case of a carborundum resistor, resistivity can be calculated using the formula: \( \text{Resistance} = \dfrac{\rho \cdot \text{Length}}{\text{Cross-sectional area}} \). The formula can be rearranged to find resistivity (\( \rho \)): \[ \rho = \dfrac{\text{Resistance} \cdot \text{Cross-sectional area}}{\text{Length}} \].For the carborundum resistor described, the cross-sectional area is determined by the difference in areas of two circles – the outer and inner parts of the cylindrical shell: \( \text{Cross-sectional area} = \pi(b^2 - a^2) \). Using the provided dimensions, the resistivity calculates to \( \rho \approx 7.04 \times 10^{-3} \, \Omega \text{m} \). This value is quite high compared to common conductive materials like copper (\(1.68 \times 10^{-8} \, \Omega\text{m}\)), aluminum (\(2.82 \times 10^{-8} \, \Omega\text{m}\)) and silver (\(1.59 \times 10^{-8} \, \Omega\text{m}\)), which makes carborundum excellent for resistive purposes.

Temperature Coefficient of Resistivity

The temperature coefficient of resistivity is a factor that indicates how much the resistivity of a material changes with temperature. For carborundum, this coefficient is given as \( \alpha = 2.14 \times 10^{-3} \, \text{K}^{-1} \).When temperature changes, the resistance of materials adjusts because their resistivity also changes. The formula to understand this relationship is: \( R(T) = R_0(1 + \alpha \cdot \Delta T) \), where:

• \( R(T) \) is the resistance at temperature \( T \).
• \( R_0 \) is the original resistance at room temperature.
• \( \alpha \) is the temperature coefficient of resistivity.
• \( \Delta T \) is the change in temperature.

A high temperature coefficient like in carborundum suggests significant sensitivity of resistivity and thus resistance with temperature variations, useful in applications requiring precise control over resistance.

Percentage Change in Resistance

The percentage change in resistance tells us how much the resistance of a material alters from its original value due to temperature changes. In this case, you use the formula \( \Delta R (\%) = \alpha \cdot \Delta T \cdot 100 \). Here, \( \Delta R (\%) \) represents the change we're interested in.For our specific carborundum resistor problem, the temperature increase from room temperature \( 20^{\circ}C \) to \( 300^{\circ}C \) is \( 280 \, \text{K} \). Substituting into our formula, we get \[ \Delta R(\%) = 2.14 \times 10^{-3} \, \text{K}^{-1} \times 280 \, \text{K} \times 100 \approx 59.9\% \].This calculation means the resistance increases by roughly 59.9% with that temperature change. Such significant change is crucial for designing circuits where resistance variability impacts performance or safety.