What is the resistance of a 20-mile aluminum wire transmission line?

  • Thread starter kriegera
  • Start date
In summary, to find the resistance of 20 miles of a transmission line made of 1 inch diameter aluminum wire, we can use the formula R=pl/A where l=32,186.88 meters and the resistivity of aluminum is 2.8 x 10-8. After converting the diameter to meters and finding the cross-sectional area, we can plug in the values and get a resistance of 1.79 ohm-metres.
  • #1
kriegera
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Homework Statement



A transmission line is made of aluminum wire 1 inch in diameter. What is the resistance of 20 miles of this wire?

Homework Equations



Resistance varies directly with the length of a wire and inversely with its cross-sectional area.
Resistivity of aluminum = 2.8 x 10-8
R=pl/A
Where l=32,186.88 meters
Find cross-sectional area.

R= 2.8 x 10-8 (31,186.88)/A =


The Attempt at a Solution



I'm stuck trying to find the cross-sectional area, A.

Also, the answer has to be in the same units as the problem, so does that mean I have to convert to inches and miles? The answer should be in ohms correct?
 
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  • #2
The wire is like a long shaft or a long cylinder, so the cross-sectional area is?

Also, you need to convert everything to same set of units. What is the resistivity measured in?
 
  • #3
rock.freak667 said:
The wire is like a long shaft or a long cylinder, so the cross-sectional area is?

Also, you need to convert everything to same set of units. What is the resistivity measured in?

The cross-sectional area would be the diamter then? Converted to meters, we get


R= 2.8 x 10-8 (32,186.88)/0.0254 = .0354816 ohm-metres

correct?
 
  • #4
kriegera said:
The cross-sectional area would be the diamter then?


No, the cross-section of the wire is just a circle with diameter 'd'. So the area of a circle is ?
 
  • #5
rock.freak667 said:
No, the cross-section of the wire is just a circle with diameter 'd'. So the area of a circle is ?

SO - Cross-Sectional Area= = Area: (pi)(.0127) ² = 5.067E-4

R= 2.8 x 10-8 (32,186.88)/ 5.067E-4 = 1.79 ohm-metres
 
  • #6
Yes that should be correct now.
 

FAQ: What is the resistance of a 20-mile aluminum wire transmission line?

1. What is the "Stuck-Resistivity problem"?

The "Stuck-Resistivity problem" refers to a phenomenon in which the resistivity of a material becomes stuck or "frozen" at a certain value, preventing further changes or adjustments.

2. What causes the "Stuck-Resistivity problem"?

The exact cause of the "Stuck-Resistivity problem" is not fully understood, but it is believed to be linked to defects or imperfections in the material's structure, such as impurities or dislocations.

3. How does the "Stuck-Resistivity problem" affect scientific research?

The "Stuck-Resistivity problem" can significantly impact research in various fields, especially in materials science and engineering. It can make it challenging to accurately measure and predict the properties of materials, which can hinder the development of new technologies and materials.

4. Is there a solution to the "Stuck-Resistivity problem"?

There is no one-size-fits-all solution to the "Stuck-Resistivity problem" as it can vary depending on the specific material and its properties. However, some approaches, such as annealing or doping, have shown some success in reducing the effects of the problem.

5. How can scientists prevent or mitigate the "Stuck-Resistivity problem" in their research?

To prevent or mitigate the "Stuck-Resistivity problem," scientists can carefully choose the materials they work with, conduct thorough characterizations of the material's structure and properties, and use techniques like doping or annealing to modify the material's properties.

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