P28_035 - ( b 2 a 2 ) C + a 2 A = (0 . 380 10 3 m) 2 (0 ....

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35. (a) The copper wire and the aluminum sheath are connected in parallel, so the potential diference is the same For them. Since the potential diference is the product oF the current and the resistance, i C R C = i A R A ,whe re i C is the current in the copper, i A is the current in the aluminum, R C is the resistance oF the copper, and R A is the resistance oF the aluminum. The resistance oF either component is given by R = ρL/A ,whe re ρ is the resistivity, L is the length, and A is the cross- sectional area. The resistance oF the copper wire is R C = ρ C L/πa 2 , and the resistance oF the aluminum sheath is R A = ρ A L/π ( b 2 a 2 ). We substitute these expressions into i C R C = i A R A , and cancel the common Factors L and π to obtain i C ρ C a 2 = i A ρ A b 2 a 2 . We solve this equation simultaneously with i = i C + i A ,where i is the total current. We ±nd i C = r 2 C ρ C i ( r 2 A r 2 C ) ρ C + r 2 C ρ A and i A = ( r 2 A r 2 C ) ρ C i ( r 2 A r 2 C ) ρ C + r 2 C ρ A
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Unformatted text preview: ( b 2 a 2 ) C + a 2 A = (0 . 380 10 3 m) 2 (0 . 250 10 3 m) 2 (1 . 69 10 8 m) +(0 . 250 10 3 m) 2 (2 . 75 10 8 m) = 3 . 10 10 15 m 3 . Thus, i C = (0 . 250 10 3 m) 2 (2 . 75 10 8 m)(2 . 00 A) 3 . 10 10 15 m 3 = 1 . 11 A and i A = (0 . 380 10 3 m) 2 (0 . 250 10 3 m) 2 (1 . 69 10 8 m)(2 . 00 A) 3 . 10 10 15 m 3 = . 893 A . (b) Consider the copper wire. IF V is the potential diference, then the current is given by V = i C R C = i C C L/a 2 , so L = a 2 V i C C = ( )(0 . 250 10 3 m) 2 (12 . 0 V) (1 . 11 A)(1 . 69 10 8 m) = 126 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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