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

\({I_{1}=2 \,\mathrm{A}}\)

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

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