Power is given as, $\mathrm{P}=50 \mathrm{J} {\mathrm{s}}^{-1}$

$\mathrm{P}=\mathrm{Vi}\mathrm{ } \Rightarrow \mathrm{P}={\mathrm{i}}^2\mathrm{R}$

$\Rightarrow$ ${\mathrm{i}}_2^2=\frac{\mathrm{P}}{\mathrm{R}}=\frac{50}{5}=10\mathrm{ }{\mathrm{A}}^2$

$\mathrm{V}={\mathrm{i}}^2{\mathrm{R}}_{\left(5\mathrm{ }\mathrm{\Omega }\right)}$

$=\mathrm{ }\sqrt{10}\times 5=\mathrm{ }\sqrt{250}\mathrm{ }\mathrm{V}$

and $2 \mathrm{\Omega }$ and $8 \mathrm{\Omega }$ are in series.

So, the required resistance $=2 \mathrm{\Omega }+8 \mathrm{\Omega }=10 \mathrm{\Omega }$

${\mathrm{i}}_1=\frac{\mathrm{V}}{{\mathrm{R}}_{\left(10\mathrm{ }\mathrm{\Omega }\right)}}$

${\mathrm{i}}_1=\frac{\sqrt{250}}{10}\mathrm{ }\mathrm{A}$

The heat generated per second in $2 \mathrm{\Omega }$ is given by,

$\mathrm{P}={\mathrm{Vi}}_1$

$={\mathrm{i}}_1^2\times \mathrm{R}$

$={\left(\frac{\sqrt{250}}{10}\right)}^2\times 2$

$=\frac{250}{100}\times 2=\frac{25}{10}\times 2=\frac{25}{5}=5 \mathrm{J} {\mathrm{s}}^{-1}$