Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 68.9 A. The resistance per unit length is to be 0.132 0/km. The densities of copper and aluminum are 8960 and 2600 kg/m³, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length À for a copper cable and (c) J and (d) À for an aluminum cable. (a) Number i (b) Number i IN (c) Number (d) Number i Units Units Units Units
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 68.9 A. The resistance per unit length is to be 0.132 0/km. The densities of copper and aluminum are 8960 and 2600 kg/m³, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length À for a copper cable and (c) J and (d) À for an aluminum cable. (a) Number i (b) Number i IN (c) Number (d) Number i Units Units Units Units
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Statement:**
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 68.9 A. The resistance per unit length is to be 0.132 Ω/km. The densities of copper and aluminum are 8960 and 2600 kg/m³, respectively. Compute the following:
(a) The magnitude \( J \) of the current density for a copper cable.
(b) The mass per unit length \( \lambda \) for a copper cable.
(c) The magnitude \( J \) of the current density for an aluminum cable.
(d) The mass per unit length \( \lambda \) for an aluminum cable.
**Inputs for Your Calculations:**
(a) \[\text{Number}\]
Units [ _select units_ ]
(b) \[\text{Number}\]
Units [ _select units_ ]
(c) \[\text{Number}\]
Units [ _select units_ ]
(d) \[\text{Number}\]
Units [ _select units_ ]
**Explanation:**
In mathematical terms, the current density \( J \) is given by the formula:
\[ J = \frac{I}{A} \]
where:
- \( I \) is the current (68.9 A in this case).
- \( A \) is the cross-sectional area of the cable.
Furthermore, the mass per unit length \( \lambda \) is determined by:
\[ \lambda = \rho \cdot A \]
where:
- \( \rho \) is the density of the material.
- \( A \) is the cross-sectional area of the cable, which can be derived from the resistance and the electrical properties of the materials, using:
\[ R = \frac{\rho_{\text{electrical}} \cdot l}{A} \]
where:
- \( \rho_{\text{electrical}} \) is the resistivity of the material.
- \( l \) is the length of the cable.
- \( R \) is the resistance per unit length (0.132 Ω/km).
Make sure to use appropriate units and conversion factors in your calculations.
**Note:**
There are no graphs or diagrams in this image. The problem requires performing a series of calculations with the information provided.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2582877-f8f8-44db-83ab-009b8ab7497d%2F03d7f6ec-cc4a-4aa7-8e08-69e02ae1d9e3%2Fcsx05c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 68.9 A. The resistance per unit length is to be 0.132 Ω/km. The densities of copper and aluminum are 8960 and 2600 kg/m³, respectively. Compute the following:
(a) The magnitude \( J \) of the current density for a copper cable.
(b) The mass per unit length \( \lambda \) for a copper cable.
(c) The magnitude \( J \) of the current density for an aluminum cable.
(d) The mass per unit length \( \lambda \) for an aluminum cable.
**Inputs for Your Calculations:**
(a) \[\text{Number}\]
Units [ _select units_ ]
(b) \[\text{Number}\]
Units [ _select units_ ]
(c) \[\text{Number}\]
Units [ _select units_ ]
(d) \[\text{Number}\]
Units [ _select units_ ]
**Explanation:**
In mathematical terms, the current density \( J \) is given by the formula:
\[ J = \frac{I}{A} \]
where:
- \( I \) is the current (68.9 A in this case).
- \( A \) is the cross-sectional area of the cable.
Furthermore, the mass per unit length \( \lambda \) is determined by:
\[ \lambda = \rho \cdot A \]
where:
- \( \rho \) is the density of the material.
- \( A \) is the cross-sectional area of the cable, which can be derived from the resistance and the electrical properties of the materials, using:
\[ R = \frac{\rho_{\text{electrical}} \cdot l}{A} \]
where:
- \( \rho_{\text{electrical}} \) is the resistivity of the material.
- \( l \) is the length of the cable.
- \( R \) is the resistance per unit length (0.132 Ω/km).
Make sure to use appropriate units and conversion factors in your calculations.
**Note:**
There are no graphs or diagrams in this image. The problem requires performing a series of calculations with the information provided.
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