In the circuit shown in figure, find the current through the branch $BD$ ............. $A$
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The current in the circuit are assumed as shown in the fig.
Applying $KVL$ along the loop $ABDA$, we get
$-6i_1 -3 i_2 + 15 = 0$ or $2i_1 + i_2 = 5$ ….. $(i)$
Applying $KVL$ along the loop $BCDB$, we get
$-3(i_1 -i_2) -30 + 3i_2 = 0$ or $-i_1 + 2i_2 = 10$ ….. $(ii)$
Solving equation $(i)$ and $(ii)$ for $i_2$, we get $i_2 = 5\, A$.
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