P27_050 - d of the copper wire: d = r 4 ρ c L πR = s 4(1...

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50. (a) We denote the copper wire with subscript c and the aluminum wire with subscript a . R = ρ a L A = (2 . 75 × 10 8 · m)(1 . 3m) (5 . 2 × 10 3 m) 2 =1 . 3 × 10 3 . (b) Let R = ρ c L/ ( πd 2 / 4) and solve for the diameter
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Unformatted text preview: d of the copper wire: d = r 4 ρ c L πR = s 4(1 . 69 × 10 − 8 Ω · m)(1 . 3 m) π (1 . 3 × 10 − 3 Ω) = 4 . 6 × 10 − 3 m ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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