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

Explanation

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 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,
  `-8(I_(1) -I_(2))+4(I_(2) +I_(3) -I_(1))-2I_(1) +10`= 0
that is, `14I_(1) – 12_(2) –4I_(3)` = 10 ……..(1)
For the closed loop ABCA, we get `2I_(1) +12I_(2) +4I_(3)`= 10………….(2)
For the closed loop BCDEB, we get `4I_(1)-8I_(2)-8I_(3)` = -5…………(3)
Equations (1, 2, 3) are three simultaneous equations in three unknowns. These can be solved by the usual method to give `I_(1)` = 1.25A, `I_(2)`=`(5)/(16)A` ,`I_(3)`= `(15)/(16)A`