The current through the branch BD in the give | vedclass

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(B) Let $i_1$ and $i_2$ be the currents drawn from the $15 \,V$ and $30 \,V$ batteries,respectively,as shown in the circuit diagram.By Kirchhoff's Cur

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$5.0$

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(B) Let $i_1$ and $i_2$ be the currents drawn from the $15 \,V$ and $30 \,V$ batteries,respectively,as shown in the circuit diagram.By Kirchhoff's Current Law $(KCL)$ at junction $B$,the net current passing through branch $BD$ is:$i_3 = i_1 + i_2$ ...$(i)$Applying Kirchhoff's Voltage Law $(KVL)$ in loop $ABDA$:$15 - 6 i_1 - 3(i_1 + i_2) = 0$$15 - 6 i_1 - 3 i_1 - 3 i_2 = 0$$9 i_1 + 3 i_2 = 15$$3 i_1 + i_2 = 5$ ...(ii)Applying $KVL$ in loop $CBDC$:$30 - 3 i_2 - 3(i_1 + i_2) = 0$$30 - 3 i_2 - 3 i_1 - 3 i_2 = 0$$3 i_1 + 6 i_2 = 30$$i_1 + 2 i_2 = 10$ ...(iii)Solving equations (ii) and (iii):From (ii),$i_2 = 5 - 3 i_1$. Substituting this into (iii):$i_1 + 2(5 - 3 i_1) = 10$$i_1 + 10 - 6 i_1 = 10$$-5 i_1 = 0 \Rightarrow i_1 = 0 \,A$Substituting $i_1 = 0$ into (ii):$3(0) + i_2 = 5 \Rightarrow i_2 = 5 \,A$Therefore,the current through branch $BD$ is:$i_3 = i_1 + i_2 = 0 + 5 = 5 \,A$

The current through the branch $BD$ in the given circuit is: (in $\,A$)

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