Physics - Current Electricity Question with Solution | TestHub

Here,$R_{20}=20, R_{500}=60\mathrm{\Omega },{\mathrm{R}}_{\mathrm{t}}=25\mathrm{\Omega },\mathrm{ }$

$\because \mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{R}}_{\mathrm{t}}={\mathrm{R}}_0\left(1+\alpha \mathrm{t}\right)$(where$R_0=$Resistance at$0^{\text{o}}\text{C}$)

Where$\alpha$is the temperature coefficient of resistance.

$\therefore R_{20}=R_0\left(1+\alpha \times 20\right)$

$or 20=R_0\left(1+\alpha \times 20\right) ..\left(\mathrm{i}\right)$

$R_{500}=R_0\left(1+\alpha \times 500\right)$

$or 60=R_0\left(1+\alpha \times 500\right) \dots \left(\mathrm{i}\mathrm{i}\right)$

Dividing Eq. (ii) by Eq. (i), we get

$\frac{60}{20}=\frac{1+\alpha \times 500}{1+\alpha \times 20}$

$\mathrm{o}\mathrm{r} 3+60\alpha =1+500\alpha$

$\mathrm{o}\mathrm{r} \alpha =\frac{2}{440}=\frac{1}{220}{.}^{\text{o}}{\text{C}}^{-1}$

Again,$R_{20}=R_0\left(1+\alpha \times 20\right)$

$\mathrm{o}\mathrm{r} 20=R_0\left(1+\frac{1}{220}\times 20\right)\dots \left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$

$R_t=R_0\left(1+\alpha t\right)$

$25=R_0\left(1+\frac{1}{220}\times t\right) ..\left(iv\right)$

Dividing Eq. (iv) by Eq (iii), we get

$\frac{25}{20}=\frac{\left(1+\frac{1}{220}\times t\right)}{\left(1+\frac{1}{220}\times 20\right)}⟹4+\frac{4t}{220}=5+\frac{100}{220}$

$\mathrm{o}\mathrm{r}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }t=80°\mathrm{C}$