${{R}_{5}}{{I}_{5}}={{R}_{6}}{{I}_{6}}\Rightarrow 6\times \frac{4}{3}=12{{I}_{6}}\Rightarrow {{I}_{6}}=\frac{2}{3}A$ ${{I}_{3}}={{I}_{5}}+{{I}_{6}}=\frac{2}{3}+\frac{4}{3}=2A$ $({{R}_{2}}+{{R}_{4}})\times {{I}_{2}}=({{R}_{3}}+{{R}_{5,6}}){{I}_{3}}\Rightarrow 20{{I}_{2}}=(6+\frac{6\times 12}{18}){{I}_{3}}\Rightarrow {{I}_{2}}=\frac{1}{2}{{I}_{3}}=1A$ ${{I}_{t}}={{I}_{3}}+{{I}_{2}}=2+1=3A$ توان مصرفی در مدار $=$ توان خروجی باتری : $P={{R}_{eq}}I_{t}^{2}=(5+\frac{20\times 10}{30})\times {{3}^{2}}=(5+\frac{20}{3})\times 9=45+60=105W$