c
This is a balanced wheatstone bridge circuit. Therefore, we can drop \(10 \,\Omega\) resistance across points band d for the further analysis of the bridge. In This case, combination of resistances in upper and lower branches are in series with equivalent resistances \(15 \,\Omega\) and \(9 \,\Omega\) respectively. These equivalent resistances are connected in parallel across junction \(\mathrm{a}\) and \(\mathrm{c}.\) Thus equivalent resistance of the circuit \(-\)

\(\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{15}+\frac{1}{9}\) 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 \(\mathrm{i}_{2}=1.1 \mathrm{\,A}\)