Question

A hollow cylindrical tube with a rectangular cross section has external dimensions of \(0.5 \mathrm{in}\). by 1 in. and a wall thickness of \(0.05 \mathrm{in}\). Assume that the material is brass, for which \(\sigma=1.5 \times 10^{7} \mathrm{~S} / \mathrm{m}\). A current of \(200 \mathrm{~A} \mathrm{dc}\) is flowing down the tube. \((a)\) What voltage drop is present across a \(1 \mathrm{~m}\) length of the tube? (b) Find the voltage drop if the interior of the tube is filled with conducting material for which \(\sigma=1.5 \times 10^{5} \mathrm{~S} / \mathrm{m}\).

Short answer

Question: Determine the voltage drop across a hollow brass cylindrical tube with a rectangular cross section carrying a direct current. Then, find the voltage drop when the interior of the tube is filled with a conducting material with a given conductivity. Answer: To find the voltage drop across the hollow tube and the filled tube, follow these steps: 1. Calculate the resistance of the hollow brass tube using its dimensions and conductivity. 2. Find the voltage drop across the tube using Ohm's law. 3. Calculate the resistance of the conducting material inside the tube when it's filled. 4. Find the total resistance when the tube is filled with conducting material. 5. Calculate the voltage drop across the filled tube using Ohm's law.

Step-by-step solution

Calculate the resistance of the hollow brass tube

First, we'll need to determine the tube's resistance. The formula for resistance R in a rectangular wire is given by: $$R = \frac{\rho L}{A}$$ where $$\rho$$ is the resistivity, L is the length of the wire, and A is the cross-sectional area. The conductivity $$\sigma$$ is the inverse of the resistivity, so $$\rho = \frac{1}{\sigma}$$. We are given the dimensions of the tube and its conductivity, so we can compute the resistivity and find the cross-sectional area. The cross-sectional area of the hollow tube is the difference between the outer and inner rectangular areas. Given the wall thickness $$t = 0.05\,\mathrm{in}$$, the inner rectangle has dimensions of $$0.4$$ and $$0.9$$ inches. Converting the dimensions to meters, and calculating the areas: $$A_\text{outer} = (0.5\,\mathrm{in} \times 1\,\mathrm{in}) \times (0.0254\,\mathrm{m}/\mathrm{in})^2$$ $$A_\text{inner} = (0.4\,\mathrm{in} \times 0.9 \mathrm{in}) \times (0.0254\,\mathrm{m}/\mathrm{in})^2$$ The cross-sectional area of the tube, $$A = A_\text{outer} - A_\text{inner}$$.

Find the voltage drop across the tube

Now we have the cross-sectional area, we can calculate the resistance of the tube using its resistivity and length. $$R_\text{tube} = \frac{\rho L}{A}$$ Given the current $$I=200\,\mathrm{A}$$, we can find the voltage drop across the tube, $$V_\text{tube}$$, using Ohm's law: $$V_\text{tube} = I \times R_\text{tube}$$ This will give us the answer to part (a).

Calculate the resistance of the conducting material inside the tube

Now, we need to calculate the resistance of the conducting material that fills the inner part of the tube. We can use the same formula for resistance as before, but this time, we'll use the inner rectangular cross-sectional area, $$A_\text{inner}$$, and the conductivity of the conducting material, $$\sigma_\text{material}$$. $$R_\text{material} = \frac{\rho_\text{material} L}{A_\text{inner}}$$

Find the total resistance when the tube is filled with conducting material

When the tube is filled with the conducting material, the two resistances - tube and material - are effectively in parallel. Therefore, the total resistance, $$R_\text{total}$$, can be calculated using the formula for resistances in parallel: $$\frac{1}{R_\text{total}} = \frac{1}{R_\text{tube}} + \frac{1}{R_\text{material}}$$

Calculate the voltage drop across the filled tube

Finally, with the total resistance, we can find the voltage drop across the tube when it is filled with the conducting material, $$V_\text{filled}$$: $$V_\text{filled} = I \times R_\text{total}$$ This will give us the answer to part (b).

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electromagnetism that links voltage, current, and resistance in electrical circuits. This law is expressed as \( V = I \times R \), where \( V \) represents the voltage across a conductor, \( I \) is the current flowing through it, and \( R \) is the resistance. This equation forms the basis for analyzing electric circuits, as it enables the calculation of voltage drop when the current and resistance are known.
Understanding Ohm's Law is crucial for solving problems involving electrical circuits, like the exercise presented, where the voltage drop across a tube is computed based on known current and resistance. This concept emphasizes the relationship between these three parameters, allowing one to predict how any of the components will behave when the others are changed.
Using Ohm's Law, you can determine the potential difference required to drive a certain amount of current through a resistive element. It helps in designing circuits and choosing the right materials to ensure efficient electrical conduction.

Resistance Calculation

Resistance is a key property of a material that impedes the flow of electric current. To calculate resistance, you can use the formula: \( R = \frac{\rho \times L}{A} \). Here, \( R \) is the resistance, \( \rho \) (rho) symbolizes resistivity, \( L \) is the length of the conducting material, and \( A \) is the cross-sectional area.
In the exercise, calculating the resistance of a hollow brass tube involves first determining its resistivity using its conductivity: \( \rho = \frac{1}{\sigma} \). With the resistivity known, you can find the resistance over a specified length by dividing this by the cross-sectional area.
This resistance value is essential to determine the voltage drop using Ohm's Law. A higher resistance leads to a larger voltage drop for a given current, highlighting why materials with low resistivity or high conductivity are preferred for electrical applications.
Remember, resistance can also be influenced by temperature and material properties, which are important considerations in practical applications.

Conductivity

Conductivity is a measure of a material's ability to permit the flow of electric current. It is represented by \( \sigma \) (sigma) and is the reciprocal of resistivity \( \rho \). Mathematically, this is expressed as \( \sigma = \frac{1}{\rho} \).
Good conductors, such as metals like copper and brass, have high conductivity because they allow electrons to flow easily through them. In the given exercise, brass with a conductivity of \( 1.5 \times 10^7 \, \mathrm{S/m} \) is used. This implies low resistance and high efficiency, making it suitable for electrical applications.
Conductivity is an intrinsic property that does not depend on the shape or size of the material. Instead, it depends on the material type and its temperature, where conductivity usually decreases with an increase in temperature for metals.
When selecting materials for wiring or other conductive purposes, understanding and comparing their conductivity is crucial for ensuring optimal performance and safety.

Cross-Sectional Area Calculation

Calculating the cross-sectional area of a conductor is vital for determining its resistance. The cross-sectional area \( A \) can be found by subtracting the inner area from the outer area for hollow shapes. For a rectangular tube with given dimensions, this involves converting measurements to consistent units and subtracting the area of the inner rectangle from the outer one.
In our exercise, with an outer dimension of \( 0.5 \text{ inch} \times 1 \text{ inch} \) and an inner section calculated using a wall thickness of \( 0.05 \text{ inch} \), we compute the areas in square meters.
The formula used is:

• \( A_\text{outer} = \text{outer dimension converted to SI units} \)
• \( A_\text{inner} = \text{inner dimension converted to SI units} \)
• \( A = A_\text{outer} - A_\text{inner} \)

The cross-sectional area then helps in calculating resistance, as resistance through a specific length of material is inversely proportional to the cross-section.
A larger cross-sectional area permits more current to flow, reducing resistance and thus minimizing potential voltage drops along the conductor's length.