${I_{1}^{2}=\frac{36}{9}=4} $             $\left( {As\,\,P = {I^2}R} \right)$

${I_{1}=2 \,\mathrm{A}}$

As the resistors $9\, \Omega$ and $6\, \Omega$ are connected in parallel, therefore potential difference across them is

same.

$\therefore \quad 9 I_{1}=6 I_{2} ; I_{2}=\frac{9 \times 2}{6}=3\, \mathrm{A}$

Current drawn from the battery is

$I=I_{1}+I_{2}=(2+3) \mathrm{A}=5\, \mathrm{A}$

The potential difference across the $2\, \Omega$ resistor is

$=(5 \,A)(2 \,\Omega)=10\, \mathrm{V}$