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خطا

${{V}_{1}}={{R}_{1}}{{I}_{1}}=2\times 3=6V\,\,,\,\,{{V}_{2}}={{V}_{1}}$ ${{V}_{2}}={{R}_{2}}{{I}_{2}}\Rightarrow 6=3{{I}_{2}}\Rightarrow {{I}_{2}}=2A$ ${{I}_{3}}={{I}_{1}}+{{I}_{2}}\Rightarrow {{I}_{3}}=3+2=5A$ $V={{V}_{1}}+{{V}_{3}}=6+5=11V$