Temperature Coefficient of Resistance

What is Temperature Coefficient of Resistance?

There is a second popular equation for calculating the resistance of a conductor at different temperatures. This equation is derived from the inferred absolute temperature given as: $${|T_i| + T1 \over R1} = {|T_i| + T2 \over R2} \tag{1}$$ By rearranging $${R2 \over R1} = {|T_i| + T2 \over |T_i| + T1} $$ here we are supposed to isolate $T_i$ from the equation, the rearrangement continuous as $$\begin{array} {rcl} {R2 \over R1} - 1 &=& {|T_i| + T2 \over |T_i| + T1}- 1 \\ {R2 \over R1} - 1&=& {|T_i| + T2 - |T_i| - T1 \over |T_i| + T1}\\ {R2 \over R1} - 1&=&{T2 - T1 \over |T_i| + T1}\\ {R2 \over R1}&=& 1 + {T2 - T1 \over |T_i| + T1}\\ R2 &=& R1(1 + {T2 - T1 \over |T_i| + T1})\\ R2 &=& R1(1 + \alpha(T2 - T1)) \tag{2} \end{array}$$
where R1 is the current resistance at T1 temperature, R2 will be the resistance at certain temperature T2. $\alpha$ is the temperature coefficient of resistance defining as: $$\alpha = {1 \over |T_i| + T1}$$ Temperature coefficient of resistance at a temperature of 20℃ is $$\alpha _{20} = {1 \over |T_i| + 20℃}$$ and temperature coefficient of resistance of copper (cu) at a temperature of 20℃ $$\alpha _{20} = {1 \over |234.5℃| + 20℃}$$ $$\alpha _{20} = 0.00393$$ The values of $\alpha_{20}$ for different materials have been evaluated, and a few are listed in Table 1.
Table 1: Temperature coefficient of resistance for various conductors at 20℃.
since the values of alpha of the conductors are calculated at 20℃(at room temperature). we can also write the equation(2) in terms of change of temperature from 20℃ $$\bbox[10px,border:1px solid grey]{R = R_{20}(1 + \alpha_{20}(T - 20℃)) } \tag{3}$$
Example 1: What is the resistance of copper and aluminum conductors at 50℃. If the resistance of the same copper conductor is 30Ω and that of aluminum is 40Ω
Solution: Resistance of copper conductor:
$$R_{20} = 30Ω$$ $$\alpha_{20} = 0.00393$$ Putting the values in equation 3 we get $$R_{cu} = 30(1 + 0.00939{20}(50 - 20))$$ $$R_{cu} = 33.537Ω$$ Resistance of aluminum conductor: $$R_{20} = 40Ω$$ $$\alpha_{20} = 0.00391$$ Putting the values in equation 3 we get $$R_{al} = 40 (1 + 0.00391(50 - 20))$$ $$R_{al} = 44.692Ω$$