LECT9_176182 - Temperature coefficient of resistivity e=...

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Temperature coefficient of resistivity )] T T ( 1 [ o o α + ρ = ρ )] T T ( 1 [ R R o o α + = ρ T 0 = reference temperature α = temperature coefficient of resistivity, units of (ºC) -1 For Ag, Cu, Au, Al, W, Fe, Pt, Pb: values of α are ~ 3-5 × 10 -3 (ºC) -1 T slope = α
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Typical tungsten filament: ~1 m long, but 0.05mm in radius. Calculate typical R. A = π (5x10 -5 m) 2 = 7.9x10 -9 m 2 ρ = 5.6x10 -8 Ω m (Table 17.1) R = ρ L/A = (5.6x10 -8 Ω m) (1m)/ 7.9x10 -9 m 2 = 7.1 Ω Note: As per section 17.6, the resistivity value used above is valid only at a temperature of 20°C, so this derived value of R holds only for T=20°C. Please note -- I edited some values here compared to the slide I presented in lecture!
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Calculate ρ at T=4000°C, assuming a linear ρ -T relation: For tungsten, α = 4.5x10 -3 /°C ρ = ρ 0 [1+ α (T-T 0 )] = 8.3x10 -7 Ω m R = ρ L/A = 106 Ω . (note-- this is still less than the estimate of >200 Ω we’ll derive in class in a few minutes… I suspect the ρ -T relation in reality may not be strictly linear over such a wide range of temperature; my guess would be that the above value of α may only be valid for temperatures of tens to hundreds of °C)
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Superconductors For some materials, as temperature drops, resistance suddenly plummets to 0 below some T c . Once a current is set up, it can persist without any applied voltage because R 0!
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Superconductors Applications: •Energy storage at power plants •Superconducting magnets with much stronger magnetic fields than normal electromagnets •Super conducting distribution power lines could eliminate resistive losses
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More recently: As the field has advanced, materials with higher values of T c get discovered
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Electrical Energy and Power
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