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Two wires of resistance `R_(1)` and `R_(2)` have temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effective temperature coefficient of resistance is |
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Answer» Let `R_(1),R_(2)` be the resistances of two wires at `0^(@)C`. `R_(t_(1)),R_(t_(2))` be the resistance of two wires at `t^(@)C`. Then `R_(t_(1))=R_(1)(1+alpha_(1)t)` and `R_(1_(2))=R_(2)(1+alpha_(2)t)` When wires are in series, then total resistance is `R_(t_(s))=R_(t_(1))+R_(t_(2))=R_(1)(1+alpha_(1)t)+R_(2)(1+alpha_(2)l)=(R_(1)+R_(2))+(R_(1)alpha_(1)+R_(2)alpha_(2))t` `=R_(s)+(R_(1)alpha_(1)+R_(2)alpha_(2))t` or `R_(t_(s))-T_(s)=(R_(1)alpha_(1)+R_(2)alpha_(2))t` where `R_(1)+R_(2)=R_(s)=` effective resistance of two wires in series at `0^(@)C` If `alpha_(s)` is the effective temperature coefficient of resistance of the two wires in series, then `alpha_(s)=(R_(t_(s))-T_(s))/(R_(s)xxt)=((R_(1)alpha_(1)+R_(2)alpha_(2))t)/((R_(1)+R_(2))t)=(R_(1)alpha_(1)+R_(2)alpha_(2))/(R_(1)+R_(2))` |
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