Physics - Current Electricity Question with Solution | TestHub

$P=\frac{V^2}{R}=\frac{{\left(200\right)}^2}{R_0[1+\alpha (T-T_0\left)\right]}$

T$\to$temperature at 't'

${\text{T}}_0=\mathrm{300K}\to$temperature at t = 0

$\text{T}-{\text{T}}_0=\frac{500-300}{30}\left(\text{t}\right)$

$\text{T}-{\text{T}}_0=\frac{200}{30}\text{t}$

$\text{T}-{\text{T}}_0=\frac{20\text{t}}{3}$

$\alpha =\frac{1}{R_0}\left(\frac{∆R}{∆T}\right)=\frac{1}{100}\left(\frac{20}{200}\right)=\frac{{10}^{-3}}{K}$

$H=\int Pdt=$

$$$\underset{0}{\overset{30}{\int }}\frac{{\left(200\right)}^2}{100\left(1+\alpha \frac{20\mathrm{t}}{3}\right)}\text{dt}=\frac{200\times 200}{100}\underset{0}{\overset{30}{\int }}\frac{\text{dt}}{1+\frac{20\alpha }{3}\text{t}}$

$=\frac{400\times 3}{20\alpha }\ell \text{n}\left(\frac{1+\frac{20\propto }{3}\times 30}{1}\right)$

$=6000\mathrm{0 }\ell \text{n }\left(\frac{6}{5}\right)$