11_Chapter 16 HomeworkCH17 Current and Resistance

11_Chapter 16 HomeworkCH17 Current and Resistance - Current...

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Current and Resistance 79 17.19 The volume of material, ( ) 2 00 L rL π == 0 VA , in the wire is constant. Thus, as the wire is stretched to decrease its radius, the length increases such that ( ) () 2 0 ff = 2 0 giving 2 2 0 4.0 0.25 rr L  =   2 0 16 f f LL L = 0 L The new resistance is then 2 2 22 0 0 16 16 4 256 4 f RR Ar r r ρρ ρ ππ == = = = = 0 256 1.00 256 = 17.20 Solving ( ) 0 1 TT α =+− 0 for the final temperature gives ()( ) 3 0 0 1 -3 0 140 19 20 C 1.4 10 C 4.5 19 R −Ω =+ =°+ = × °  ×°  17.21 From Ohm’s law, ii VI R IR ∆= = , so the current in Antarctica is 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 fi f f T R II R T +− ° ° ° ° − ° ° 1 1 1 3.90 10 A 1 3.90 10 i RT I 17.22 The expression for the temperature variation of resistance, ( ) 1 with , gives the temperature coefficient of resistivity of this material as 0 20 C T ( )
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