Question

The resistance of a coil of aluminum wire at ${18}^{0}C$ is 200 ohm. The temperature of the wire is increased and the resistance rises to 240 ohm. If the temperature coefficient of resistance of aluminum is $0.0039{/}^{0}C$at ${18}^{0}C$, determine the temperature to which the coil has risen.

Answer 1

Given:
At t${}_1$ = 18${}^{\circ }$C, R${}_1$ = 200$\Omega$
At t${}_2$${}^{\circ }$C, R${}_2$ = 240$\Omega$
And $\propto$ = 0.0039/${}^{\circ }$C or 3.9$\times$10${}^{-3}$/${}^{\circ }$C
We know that the resistance of a resistor varies with temperature as, R = R${}_0$ (1 + $\propto$ t) $\to$~(1)
Where R${}_0$$\to$ resistance at 0${}^{\circ }$C
and t $\to$ temperatre t${}^{\circ }$C
Taking (1) for two different temperatures :
R${}_1$ = R${}_0$ (1 + $\propto$ t${}_1$) and R${}_2$ = R${}_0$ (1 + $\propto$ t${}_2$)
Taking the ratio and substituting the values,
$\frac{R_1}{R_2}$= $\frac{R_0}{R_0}\frac{(1+\propto t_1)}{(1+\propto t_2)}$
$\Rightarrow$$\frac{200}{240}$= $\frac{(1+3.9\times {10}^{-3}\times 18)}{(1+3.9\times {10}^{-3}\times t)}$
Solving, we get t = 72.88${}^{\circ }$C