9_Ch 17 College Physics ProblemCH17 Current and Resistance

9_Ch 17 College Physics ProblemCH17 Current and Resistance...

Info iconThis 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 LL R  ×Ω ==    ( L Inserting this expression for A into equation (1) gives ) 82 7 3 3.40 10 m 1.12 10 m L −− ×= × , which yields 1.82 m L = (b) From equation (1), 27 1.12 10 m 4 d A L π × 3 , or 73 4 4 1.12 10 m 4 1.12 10 m 1.82 m 2.80 10 m 0.280 mm d L ππ ×× = 17.13 From L R A = , we obtain 2 4 dL A R , or ( ) 4 4 5.6 10 m 2.0 10 m 4 1.7 10 m 0.17 mm 0.050 L d R × = 17.14 ( )( ) 8 2 2 3 41 .7 10 m 15
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online