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.

Answer

(i)` 6.45 xx 10^(-3)^(@)C^(-1)`(ii)` 0.267 Omega `(ii)`1.948 xx 10^(-4) Omega m`

Explanation

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`.