Temperature 80513

A copper wire coil winding has a resistance of 10 Ω at a temperature of 14°C. The passing current by the coil heats up, and its resistance increases to 12.2 Ω. To what temperature did the coil winding heat up? α = 3.92 * 10^-3 1/K.

Correct answer:

t₂ =  70.1224 °C

Step-by-step explanation:

$$\begin{gathered} t_1=14\text{°C} \\ {R}_1=10\Omega \\ {R}_2=12.2\Omega \\ \alpha =3.92\cdot 10^{-3}\doteq 0.00391\text{/K} \\ {R}_2=R_1\cdot (1+\alpha \cdot \Delta ) \\ \Delta =(R_2/R_1-1)/\alpha =(12.2/10-1)/0.0039=\frac{2750}{49}\doteq 56.1224\text{°C} \\ \Delta =t_2-t_1 \\ {t}_2=t_1+\Delta =14+56.1224=70.1224\text{°C} \end{gathered}$$

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