ch27-p047 - 47. (a) The copper wire and the aluminum sheath...

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ba a CA 22 2 3 2 3 2 8 3 2 8 15 3 0 380 10 0 250 10 169 10 0 250 10 2 75 10 310 10 −+ = × −× ×⋅ −− c h c h c h c h c h c h ρρ .. . mm m m Thus, i C = ×× = 0 250 10 2 75 10 2 00 111 3 2 8 15 . . . A m A 3 ch c h b g . (b) Similarly, () ( ) 33 8 15 3 0.380 10 m 0.250 10 m 1.69 10 m 2.00A 0.893A. 3.10 10 m A i ⎡⎤ ×− × × ⎢⎥ ⎣⎦ == ×Ω (c) Consider the copper wire. If V is the potential difference, then the current is given by V = i C R C = i C ρ C L / π a 2 , so L aV i CC × = π π 2 3 2 8 0 250 10 12 0 111 169 10 126 b g b g b gc h mV Am m. 47. (a) The copper wire and the aluminum sheath are connected in parallel, so the potential difference is the same for them. Since the potential difference is the product of the current and the resistance, i C R C = i A R A , where 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
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