Power dissipated across the $8\Omega$ resistor in the circuit shown here is 2 W. The power dissipated in watt units across the $3\Omega$ resistor is

Resistances $1\Omega$ and $3\Omega$ are connected in series, so effective resistance

${\mathrm{R}}^{\text{'}}=1+3=4\mathrm{\Omega }$

Now, $R^{\text{'}}$ and $8\Omega$ are in parallel. We know that potential difference across resistances in parallel order is the same.

Hence, ${\mathrm{R}}^{\text{'}}\times {\mathrm{i}}_1=8{\mathrm{i}}_2$

$\mathrm{or}4\times {\mathrm{i}}_1=8{\mathrm{i}}_2\mathrm{or}{\mathrm{i}}_1=\frac{8}{4}{\mathrm{i}}_2=2{\mathrm{i}}_2$

or  i₁ = 2i₂                           …..(i)

Power dissipated across $8\Omega$ resistance is 

$\begin{array}{l}{\mathrm{i}}_2^2\left(8\right)\mathrm{t}=2\mathrm{W}\\ \mathrm{or} {\mathrm{i}}_2^2\mathrm{t}=\frac{2}{8}=0.25\mathrm{W}\end{array}$          ….(ii)

Power dissipated across $3\Omega$ resistance is 

$\begin{array}{l}\mathrm{H}={\mathrm{i}}_1^2\left(3\right)\mathrm{t}={\left(2{\mathrm{i}}_2\right)}^2\left(3\right)\mathrm{t}=12{\mathrm{i}}_2^2\mathrm{t}\\ \mathrm{But}{\mathrm{i}}_2^2\mathrm{t}=0.25\mathrm{W}\\ \mathrm{H}=12\times 0.25=3\mathrm{W}\end{array}$