# Temperature Coefficient of Resistance   Whatsapp  #### 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: $${|T_i| + T_1 \over R_1} = {|T_i| + T_2 \over R_2} \tag{1}$$ By rearranging $${R_2 \over R_1} = {|T_i| + T_2 \over |T_i| + T_1}$$ here we are supposed to isolate $T_i$ from the equation, the rearrangement continuous as $$\begin{array} {rcl} {R_2 \over R_1} - 1 &=& {|T_i| + T_2 \over |T_i| + T_1}- 1 \\ {R_2 \over R_1} - 1&=& {|T_i| + T_2 - |T_i| - T_1 \over |T_i| + T_1}\\ {R_2 \over R_1} - 1&=&{T_2 - T_1 \over |T_i| + T_1}\\ {R_2 \over R_1}&=& 1 + {T_2 - T_1 \over |T_i| + T_1}\\ R_2 &=& R_1(1 + {T_2 - T_1 \over |T_i| + T_1})\\ R_2 &=& R_1(1 + \alpha(T_2 - T_1)) \tag{2} \end{array}$$
where $R_1$ is the current resistance at $T_1$ temperature, $R_2$ will be the resistance at certain temperature $T_2$. $\alpha$ is the temperature coefficient of resistance defining as: $$\alpha = {1 \over |T_i| + T_1}$$ 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 $$R_{al} = 40 (1 + 0.00391(50 - 20))$$ $$R_{al} = 44.692Ω$$

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