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).