Problem 27 - Fundamentals of Physics

Question

Two conductors are made of the same material and have the same length. Conductor \(A\) is a solid wire of diameter \(1.0 \mathrm{~mm} .\) Conductor \(B\) is a hollow tube of outside diameter \(2.0 \mathrm{~mm}\) and inside diameter \(1.0 \mathrm{~mm} .\) What is the resistance ratio \(R_{A} / R_{B}\), measured between their ends?

Step-by-Step Solution

Answer

The resistance ratio \( R_A / R_B \) is 3.

Step 1: Understanding the Problem

We need to find the resistance ratio between two conductors, one solid and the other hollow, made of the same material and length. We will use the formula for resistance, which is \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.

Step 2: Calculate Cross-Sectional Area of Conductor A

Conductor A is a solid wire with a diameter of 1.0 mm. The cross-sectional area \( A_A \) can be calculated using the formula for the area of a circle: \[ A_A = \pi \left( \frac{d_A}{2} \right)^2 = \pi \left( \frac{1.0}{2} \right)^2 = \pi \times 0.5^2 = 0.25\pi \text{ mm}^2. \]

Step 3: Calculate Cross-Sectional Area of Conductor B

Conductor B is a hollow cylinder with an outside diameter of 2.0 mm and inside diameter of 1.0 mm. The cross-sectional area \( A_B \) is the area of the outer circle minus the area of the inner circle:\[ A_B = \pi \left( \frac{d_{outer}}{2} \right)^2 - \pi \left( \frac{d_{inner}}{2} \right)^2 = \pi \times 1^2 - \pi \times 0.5^2 = \pi - 0.25\pi = 0.75\pi \text{ mm}^2. \]

Step 4: Use the Resistance Formula for Each Conductor

Using the resistance formula \( R = \frac{\rho L}{A} \), and noting that \( \rho \) and \( L \) are the same for both conductors, we express their resistances:- For conductor A: \( R_A = \frac{\rho L}{0.25\pi} \).- For conductor B: \( R_B = \frac{\rho L}{0.75\pi} \).

Step 5: Calculate the Resistance Ratio \( R_A / R_B \)

We calculate the resistance ratio by dividing \( R_A \) by \( R_B \):\[ \frac{R_A}{R_B} = \frac{\frac{\rho L}{0.25\pi}}{\frac{\rho L}{0.75\pi}} = \frac{0.75\pi}{0.25\pi} = 3. \] The resistance ratio is thus 3.

Key Concepts

Resistance Formula

Understanding the resistance of a conductor is key to solving electrical problems. One of the most essential formulas for this is the resistance formula: \( R = \frac{\rho L}{A} \). In this formula, \( R \) represents the resistance of the conductor, \( \rho \) is the resistivity of the material, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area of the conductor.
Resistivity (\( \rho \)) is a fundamental property that depends on the material itself. For instance, metals tend to have low resistivities, making them good conductors. Length (\( L \)) directly influences resistance: the longer the conductor, the higher the resistance, because electrons encounter more collisions as they travel.
Meanwhile, the cross-sectional area (\( A \)) inversely affects resistance. A larger area allows more room for electrons to flow through, reducing resistance. This means a thicker wire will have lower resistance compared to a thinner one of the same material and length.
When comparing conductors, like in this exercise, it's essential to assess how both length and area factor in when dealing with resistance. The formula simplifies the process by isolating these variables.

Cross-Sectional Area Calculation

The calculation of the cross-sectional area is crucial when determining a conductor's resistance. For round wires, this practice involves determining the area of a circle. The formula is used: \( A = \pi \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter of the wire.
Let's take Conductor A from our example, which is a solid wire with a diameter of 1.0 mm. Plugging it into the formula gives \( A_A = \pi \left( \frac{1.0}{2} \right)^2 = 0.25\pi \) mm². This represents the entire cross-section available for current to flow through.
Conductor B, being hollow, requires a slightly different approach. Here, we calculate the area of both the outer and inner circles to determine the material that's present for current passage. The cross-sectional area is determined by: \( A_B = \pi r_{outer}^2 - \pi r_{inner}^2 \). For this conductor, with outer diameter 2.0 mm and inner diameter 1.0 mm, it’s \( A_B = \pi (1)^2 - \pi (0.5)^2 = 0.75\pi \) mm². This calculation highlights the conductive area that is physically present.

Hollow versus Solid Conductor

When considering choosing between hollow and solid conductors, understanding their structural differences is essential. Solid conductors, like Conductor A, have homogenous material throughout their entire diameter, allowing a fixed cross-sectional area for electron movement. This type is particularly good for applications requiring firm mechanical strength.
Hollow conductors, on the other hand, have a hollow section that reduces the overall material while maintaining a specific outer dimension. Conductor B in our example exemplifies this. Despite having a larger outer diameter, its effective cross-sectional area is reduced by its hollow inner part.
The implications on resistance are significant. Given equal lengths and material, a hollow conductor tends to have higher resistance than a solid of the same outer diameter because there’s less conductive material for the same surface area. However, it can also offer benefits like reduced weight and better thermal management under certain circumstances.
Knowing when to use each type of conductor can optimize performance, efficiency, and practicality in electrical systems.