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11_Chapter 17 HomeworkCH17 Current and Resistance

11_Chapter 17 HomeworkCH17 Current and Resistance - Current...

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Current and Resistance 79 17.19 The volume of material, ( ) 2 0 0 L r L π = = 0 V A , in the wire is constant. Thus, as the wire is stretched to decrease its radius, the length increases such that ( ) ( ) 2 0 f f r L r L π π = 2 0 giving ( ) 2 2 0 0 0 0 0 4.0 0.25 r r L r r = 2 0 16 f f L L L = = = 0 L The new resistance is then ( ) ( ) ( ) 2 0 0 2 2 2 0 0 16 16 4 256 4 f f f f f L L L L R R A r r r ρ ρ ρ ρ π π π = = = = = = 0 256 1.00 256 = 17.20 Solving ( ) 0 1 T T α = + 0 R R for the final temperature gives ( ) ( ) 3 0 0 1 -3 0 140 19 20 C 1.4 10 C 4.5 10 C 19 R R T T R α Ω − = + = ° + = × ° × ° 17.21 From Ohm’s law, i i f f V I R I R = = , so the current in Antarctica is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 0 1 3 1 3 C 58.0 C 20.0 C 1.00 1.98 A C 88.0 C 20.0 C i i f i f f T R I I R T + + ° ° ° = = ° ° ° 1 1 1 3.90 10 A 1 3.90 10 i R T I R T α α = =
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