Question

A copper wire has a resistance of 𝟏𝟓. 𝟎Ω at a temperature of 14℃. After a

current passes through the windings, the resistance rose to 𝟏𝟐. 𝟓Ω. To what

temperature was the wire heated?

Answer 1

We know that

$R=R_0(1+\alpha(T_2-T_1))$

$$15=12.5(1+3.93\times10^{-3}(14-T_0))$$

$1.2-1=3.93\times10^{-3}(14-T_0)$

$50.89=T_0-14\\T_0=64.89°C$