Current and Resistance7917.19The volume of material, ( )200LrLπ==0VA, in the wire is constant. Thus, as the wire is stretched to decrease its radius, the length increases such that ( )()20ff=20giving 2204.00.25rrL=2016ffLLL=0LThe new resistance is then 2222001616 42564fRRArrrρρρππ== ====0256 1.00 256=ΩΩ17.20Solving ( )01TTα=+−0for the final temperature gives ()( )3001-30140 19 20 C1.410 C4.519 R−−Ω−Ω=+=°+= × °×°Ω17.21From Ohm’s law, iiVIR IR∆==, so the current in Antarctica is 1313C58.0 C20.0 C1.001.98 AC88.0 C20.0 CiififfTRIIRT−−−+−°°−°° −°−°1113.9010A13.9010iRTI−+×17.22 The expression for the temperature variation of resistance,( )1with , gives the temperature coefficient of resistivity of this material as 020 CT=°( )
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