$\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{15}+\frac{1}{9}$ or $\quad \mathrm{R}_{\mathrm{eq}}=\frac{(15)(9)}{15+9}=\frac{135}{24}=5.62\, \Omega$

Current through circuit i $=\frac{10}{5.62}=1.78 \mathrm{\,A}$

If current through upper and lower branches of the network are $\mathrm{i}_{1}$ and $\mathrm{i}_{2}$, then according to Kirchoff's function law $-$

$\mathrm{i}_{1}+\mathrm{i}_{2}=1.78 \mathrm{\,A}$

and $\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}\right)+\left(\mathrm{V}_{\mathrm{b}}-\mathrm{V}_{\mathrm{c}}\right)=\left(\mathrm{V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{d}}\right)+\left(\mathrm{V}_{\mathrm{d}}-\mathrm{V}_{\mathrm{c}}\right)$

or $\quad \mathrm{i}_{1}(5+10)^{3 / 4}\, \mathrm{i}_{2}(3.0+6.0)$ or $(15) \mathrm{i}_{1}=(9.0) \mathrm{i}_{2}$

On solving, we get $-$ $\mathrm{i}_{1}=0.67 \mathrm{\,A}$ and $\mathrm{i}_{2}=1.1 \mathrm{\,A}$