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 1.

Step 1: Calculate Cross-sectional Area of Conductor A

Conductor A is a solid wire with a diameter of \(1.0 \ ext{mm}\). The cross-sectional area \(A_A\) can be calculated using the formula for the area of a circle: \[ A_A = \pi \left( \frac{d}{2} \right)^2 \]Substituting \(d = 1.0 \ ext{mm} = 0.001 \ ext{m}\):\[ A_A = \pi \left( \frac{0.001}{2} \right)^2 = \pi \times 0.0005^2 \approx 7.85 \times 10^{-7} \ ext{m}^2 \]

Step 2: Calculate Cross-sectional Area of Conductor B

Conductor B is a hollow tube with an outside diameter of \(2.0 \ ext{mm}\) and an inside diameter of \(1.0 \ ext{mm}\). The cross-sectional area \(A_B\) is calculated by subtracting the area of the inner circle from the area of the outer circle:\[ A_B = \pi \left( \frac{D_o}{2} \right)^2 - \pi \left( \frac{D_i}{2} \right)^2 \]Substitute \(D_o = 2.0 \ ext{mm} = 0.002 \text{m}\) and \(D_i = 1.0 \text{mm} = 0.001 \text{m}\):\[ A_B = \pi \left( 0.001 \right)^2 - \pi \left( 0.0005 \right)^2 = \pi (0.0001 - 0.000025) = 7.85 \times 10^{-7} \ ext{m}^2 \]

Step 3: Understand the Resistance Formula

The resistance \( R \) of a conductor is given by:\[ R = \frac{\rho L}{A} \]where \( \rho \) is the resistivity, \( L \) is the length of the conductor, and \( A \) is the cross-sectional area.

Step 4: Derive the Resistance Ratio

Since both conductors are made of the same material and have the same length, they have the same resistivity \( \rho \) and length \( L \). Therefore, the resistances are\[ R_A = \frac{\rho L}{A_A} \quad \text{and} \quad R_B = \frac{\rho L}{A_B} \]Thus, the resistance ratio is:\[ \frac{R_A}{R_B} = \frac{A_B}{A_A} \]Substitute the calculated areas:\[ \frac{R_A}{R_B} = \frac{7.85 \times 10^{-7}}{7.85 \times 10^{-7}} = 1 \]

Step 5: Conclude the Solution

The resistance ratio \( \frac{R_A}{R_B} \) is equal to 1, meaning that the resistances of both conductors, A and B, are the same.

Key Concepts

Conductors

Conductors are materials that allow electric current to flow through them easily. This ease of flow is due to the movement of electrons. Metals, such as copper and aluminum, are excellent conductors because they have a high density of free electrons that can move when an electric field is applied.
When discussing conductors in the context of resistance, it is important to consider their physical properties such as length, thickness, and material composition. Conductors play a vital role in the design and functioning of circuits and electrical devices.

In this exercise, we look at two conductors made of the same material: one is a solid wire and the other is a hollow tube. Both conductors share the same length, which impacts their electrical resistance.

Cross-sectional Area

The cross-sectional area is crucial in determining the resistance of a conductor. It is the area of the cut surface that you see when the conductor is sliced perpendicularly to its length. Remember, the larger this area, the lower the resistance, because more electrons can pass through at once.
For conductor A, which is a solid wire, the cross-sectional area is calculated using the formula for the area of a circle: \[ A = \pi \left( \frac{d}{2} \right)^2 \]where \(d\) is the diameter of the wire.

• With a diameter of \(1.0\, \mathrm{mm}\), this becomes \(7.85 \times 10^{-7} \mathrm{m}^2\).

For conductor B, which is a hollow tube, we calculate the area by considering both the outer and inner diameters:\[ A = \pi \left( \frac{D_o}{2} \right)^2 - \pi \left( \frac{D_i}{2} \right)^2 \]where \(D_o\) is the outer diameter and \(D_i\) is the inner diameter.

• This results in the same cross-sectional area of \(7.85 \times 10^{-7} \mathrm{m}^2\). This means both conductors have the same ability to carry current.

Resistivity

Resistivity is a material property that quantifies how strongly a material opposes the flow of electric current. It is denoted by \(\rho\) and remains constant for a material regardless of its shape or size. Materials with low resistivity are good conductors, while those with high resistivity are insulators.
In the formula for resistance \(R = \frac{\rho L}{A}\), resistivity \(\rho\) plays a key role alongside length \(L\) and cross-sectional area \(A\) to determine the actual resistance of the material.
For this exercise, both conductors have the same resistivity since they are made of the same material, thus affecting them equally in their resistance calculations. It helps simplify the evaluation of their resistance ratio, as resistivity and length cancel each other out when comparing resistance in identical materials.

Resistance Ratio

The resistance ratio, \(\frac{R_A}{R_B}\), represents the comparative measure of resistance between two conductors. Given a scenario where two conductors have the same resistivity and length, as seen in our exercise, the resistance is inversely proportional to their cross-sectional areas.
Thus, for the resistance ratio:\[ \frac{R_A}{R_B} = \frac{A_B}{A_A} \] Here:

• Both conductor A and conductor B were found to have the same cross-sectional area, meaning their resistance ratio becomes \(1\).

This result implies that the resistances of both conductors are identical, even if one is a hollow tube and the other is a solid wire. Such insights aid engineers in designing systems where equivalent resistances are needed without altering material properties or changing physical layout dramatically.