In the adjoining circuit diagram each resistance is of $10$ $\Omega$. The current in the arm $AD$ will be
Diffcult
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Applying Kirchoff's law in mesh $ABCDA$
$ - 10(i - {i_1}) - 10{i_2} + 20{i_1} = 0$
$\Rightarrow $ $\,3{i_1} - {i_2} = i$ ....... $(i)$ and in mesh $BEFCB$
$ - \,20\,(i - {i_1} - {i_2}) + \,10\,({i_1} + {i_2}) + 10{i_2} = 0$
$\Rightarrow $ $3{i_1} + 4{i_2} = 2i$ ...... $(ii)$
From equation $(i)$ and $(ii)$
${i_1} = \frac{{2i}}{5},\,{i_2} = \frac{i}{5}$ $\Rightarrow $ ${i_{AD}} = \frac{{2i}}{5}$
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