Current through the branch $B D$, in the given circuit is

Question

Aicharya · Accepted Answer

Let $i_1$ and $i_2$ be current drawn from battery 15 $\mathrm{V}$ and $30 \mathrm{~V}$, respectively as shown below

Hence, by KCL, Net current passing through branch $B D$ $i_3=i_1+i_2$ ...(i) By KVL, in loop $A B D A$, $15-6 i_1-3\left(i_1+i_2\right)=0$ $\Rightarrow \quad-6 i_1-3 i_1-3 i_2=-15$ $\Rightarrow \quad 3 i_1+i_2=5$ ...(ii) In loop $C B D C, 30-3 i_2-3\left(i_1+i_2\right)=0$ $\Rightarrow \quad i_1+2 i_2=10$ ...(iii) By solving Eqs. (ii) and (iii), we get $i_1=0$ and $i_2=5 \mathrm{~A}$ Substituting these values in Eq. (i), we get Current through branch $B D$, $i_3=i_1+i_2=0+5=5 \mathrm{~A}$

Current through the branch $B D$, in the given circuit is

Solution