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9_Ch 17 College Physics ProblemCH17 Current and Resistance

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

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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 ×
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