Resistivity in terms of temperature coefficient of resistance R = R 0 ∗ ( 1 + α ( T − T 0 ) )Where R 0 and T 0 are the reference resistance and temperature.Is it then safe to say that: ρ = ρ 0 ∗ ( 1 + α ( T − T 0 ) )(where ρ 0 is the reference resistivity)?

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Resistivity in terms of temperature coefficient of resistance  R  =      R    0    ∗  (  1  +  α  (  T  −      T    0    )  )Where       R    0   and       T    0   are the reference resistance and temperature.Is it then safe to say that:  ρ  =      ρ    0    ∗  (  1  +  α  (  T  −      T    0    )  )(where       ρ    0   is the reference resistivity)?

Jaylon Richmond · Accepted Answer

Yes; under the relationship   R  =  ρ  L      /    A, the two expressions are equivalent for a given geometry. Resistivity generally depends on temperature in a complex way, but for small temperature differences, this dependence is approximately linear, as captured by the coefficient   α. All resistivity values for materials (should) include a reference temperature because of this temperature sensitivity.

Resistivity in terms of temperature coefficient of resistance R = R 0 </msu

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