Determine the current in each branch of the network shown in Fig.

Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff's rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns $I_1,I_2$ and $I_3$ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff's second rule for the closed loop ADCA gives, $10-4\left(I_1-I_2\right)+2\left(I_2+I_3-I_1\right)-I_1=0$$[(\mathrm{a})]$ that is, $7I_1-6I_2-2I_3=10$ For the closed loop ABCA, we get $10-4I_2-2\left(I_2+I_3\right)-I_1=0$ that is, $I_1+6I_2+2I_3=10$$[(\mathrm{b})]$ For the closed loop BCDEB, we get $5-2\left(I_2+I_3\right)-2\left(I_2+I_3-I_1\right)=0$ that is, $2I_1-4I_2-4I_3=-5$ Equations $(\mathrm{a},\mathrm{b},\mathrm{c})$ are three simultaneous equations in three unknowns. These can be solved by the usual method to give $I_1=2.5\mathrm{A},I_2=\frac{5}{8}\mathrm{A},I_3=1\frac{7}{8}\mathrm{A}$ The currents in the various branches of the network are $\mathrm{AB}:\frac{5}{8}\mathrm{A},\mathrm{CA}:2\frac{1}{2}\mathrm{A},\mathrm{DEB}:1\frac{7}{8}\mathrm{A}$$\mathrm{AD}:1\frac{7}{8}\mathrm{A},\mathrm{CD}:0\mathrm{A},\mathrm{BC}:2\frac{1}{2}\mathrm{A}$ It is easily verified that Kirchhoff's second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB $5\mathrm{V}+\left(\frac{5}{8}\times 4\right)\mathrm{V}-\left(\frac{15}{8}\times 4\right)\mathrm{V}$ equal to zero, as required by Kirchhoff's second rule.