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InterviewSolution
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A copper coil has resistance of `20.0 Omega` at `0^@C` and a resistance of `26.4 Omega` at `80^@C`. Find the temperature coefficient of resistance of copper. |
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Answer» Correct Answer - (i)` 6.45 xx 10^(-3)^(@)C^(-1)`(ii)` 0.267 Omega `(ii)`1.948 xx 10^(-4) Omega m` Here,`r =1 mm = 10^(-3) , l = 50 cm = 0.50 m` , `R_(1)=0.31 Omega,t_(1)=25^(@)C,R_(2)=0.51 cm=0.50 m` , (i) Temperature coefficient of resistance `alpha=(R_(2)-R_(1))/(R_(1)(t_(2)-t_(1)))=(0.51-0.31)/(0.31(125-25))` `=(0.20)/(0.31xx100)=6.45xx10^(3).^(@)C^(-1)` (ii) Resistance at `0^(@)C` is `R_(0)=(R_(1))/(1+alphat_(1))=(0.31)/(1+(6.4xx10^(-3))xx25)=0.267^(@)Omega` (iii) Resistivity at `0^(@)C` `rho_(0)=(R_(0)A)/(l)=(R_(0)xxpir^(2))/(l)=(0.267xx3.142xx(10^(-3))^(2))/(0.50)` `=1.68 xx 10^(-6)m` Resistivity at `25^(@)C,rho_(25)=rho_(0)[1+alpha xx 25]` `=1.68xx10^(-6)[1+6.4xx10^(-3)xx25]` `=1.948xx10^(-6)Omegam`. |
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