Problem 11 - University Physics with Modern Physics

Question

A 1.50 -m cylindrical rod of diameter 0.500 \(\mathrm{cm}\) is connected to a power supply that maintains a constant potential difference of 15.0 \(\mathrm{V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads \(18.5 \mathrm{A},\) while at \(92.0^{\circ} \mathrm{C}\) it reads 17.2 \(\mathrm{A} .\) You can ignore any thermal expansion of the rod. Find (a) the resistivity at \(20.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

Step-by-Step Solution

Answer

(a) \(1.061 \times 10^{-5} \ \mathrm{\Omega \cdot m}\); (b) \(0.00105 \ \mathrm{C}^{-1}\).

Step 1: Calculate the Resistance at 20°C

Using Ohm's Law, we know that \( V = IR \). Thus, the resistance \( R \) can be calculated as \( R = \frac{V}{I} \). For \( 20^{\circ} \mathrm{C} \), the potential difference \( V = 15.0 \ \mathrm{V} \) and the current \( I = 18.5 \ \mathrm{A} \), so:\[ R = \frac{15.0}{18.5} = 0.811 \ \Omega \]

Step 2: Calculate Resistivity at 20°C

The resistance \( R \) of a cylindrical conductor is given by:\[ R = \rho \frac{L}{A} \]where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. The area for a cylinder \( A = \pi r^2 \), with radius \( r = 0.250 \ \mathrm{cm} = 0.00250 \ \mathrm{m} \), so:\[ A = \pi (0.00250)^2 = 1.9635 \times 10^{-5} \ \mathrm{m}^2 \]Given \( L = 1.50 \ \mathrm{m} \) and \( R = 0.811 \ \Omega \), rearranging the resistance formula gives:\[ \rho = R \frac{A}{L} = 0.811 \frac{1.9635 \times 10^{-5}}{1.50} = 1.061 \times 10^{-5} \ \Omega \cdot \mathrm{m} \]

Step 3: Calculate the Resistance at 92°C

Using Ohm's Law again at \( 92^{\circ} \mathrm{C} \), with \( V = 15.0 \ \mathrm{V} \) and current at \( 92^{\circ} \mathrm{C} \) as \( 17.2 \ \mathrm{A} \):\[ R_{92} = \frac{15.0}{17.2} = 0.872 \ \Omega \]

Step 4: Determine the Temperature Coefficient of Resistivity

The temperature dependence of resistance is given by:\[ R_T = R_0 (1 + \alpha (T - T_0)) \]Rearranging to solve for \( \alpha \), the temperature coefficient of resistivity, we have:\[ \alpha = \frac{R_T - R_0}{R_0 \times (T - T_0)} \]Substituting \( R_{92} = 0.872 \ \Omega \), \( R_{20} = 0.811 \ \Omega \), \( T = 92^{\circ} \mathrm{C} \), and \( T_0 = 20^{\circ} \mathrm{C} \), we find:\[ \alpha = \frac{0.872 - 0.811}{0.811 \times (92 - 20)} = \frac{0.061}{0.811 \times 72} = 0.00105 \ \mathrm{C}^{-1} \]

Key Concepts

Ohm's Law

Understanding Ohm's Law is crucial when working with electrical circuits. Ohm's Law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points.
This relationship is mathematically represented as:\( V = IR \)
where:

• \( V \) is the voltage (potential difference) in volts,
• \( I \) is the current in amperes,
• \( R \) is the resistance in ohms (\( \Omega \)).

In our case, the cylindrical rod is measured at 20°C with an ammeter reading 18.5 A while a 15.0 V potential difference is applied. Using Ohm's Law, we calculate the resistance at this temperature by rearranging the formula to \( R = \frac{V}{I} \). This helps us understand how changes in temperature or material affect electrical resistance.

Temperature Coefficient of Resistivity

The temperature coefficient of resistivity is a measure of how a material's resistivity changes with temperature. Different materials react differently to temperature changes. Most conductors increase in resistivity (and hence resistance) with an increase in temperature.
Mathematically, this relationship is given by the formula:
\[ R_T = R_0 (1 + \alpha (T - T_0)) \]
where:

• \( R_T \) is the resistance at temperature \( T \),
• \( R_0 \) is the original resistance at temperature \( T_0 \),
• \( \alpha \) is the temperature coefficient of resistivity,
• \( T \text{ and } T_0 \) are the final and initial temperatures respectively.

By calculating \( \alpha \) using the resistances at two temperatures (92°C and 20°C in the exercise), we can assess how significant the temperature's impact is on the resistance of the cylindrical rod's material.

Cylindrical Rod Resistance

The resistance of a cylindrical rod depends on its resistivity, length, and cross-sectional area. It's important to understand that a cylinder's dimensions play a crucial role in its electrical resistance.
The formula describing this relationship is:
\[ R = \rho \frac{L}{A} \]
where:

• \( \rho \) is the resistivity,
• \( L \) is the length of the cylinder,
• \( A \) is the cross-sectional area, calculated as \( \pi r^2 \) for a cylinder.

Resistivity \( \rho \) itself is an intrinsic property of the material, unaffected by its shape or size, although the resistance \( R \) derived from it certainly is. In this example, given the length (1.5 m) and the radius (0.0025 m) of the rod, we calculated resistivity by rearranging the resistance formula, illustrating how these physical characteristics factor into electrical properties.