{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

9_Chapter 17 HomeworkCH17 Current and Resistance

9_Chapter 17 HomeworkCH17 Current and Resistance - Current...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Current and Resistance 77 (a) From L R A ρ = , we find that ( ) 8 8 1.70 10 m 3.40 10 m 0.500 A L L R ρ × Ω⋅ = = = × ( L Inserting this expression for A into equation (1) gives ) 8 2 7 3 3.40 10 m 1.12 10 m L × = × , which yields 1.82 m L = (b) From equation (1), 2 7 1.12 10 m 4 d A L π × = = ( ) 3 , or ( ) ( ) 7 3 7 3 4 4 1.12 10 m 4 1.12 10 m 1.82 m 2.80 10 m 0.280 mm d L π π × × = = = × = 17.13 From L R A ρ = , we obtain 2 4 d L A R π ρ = = , or ( )( ) ( ) 8 2 4 4 5.6 10 m 2.0 10 m 4 1.7 10 m 0.17 mm 0.050 L d R ρ π π ×
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern