Using Kirchoff's laws in the electrical net work shown in figure, calculate the values of `I_(1),I_(2)` and `I_(3)`.

Explanation

According to Kirchhoff's first law at junction B
`I_(1) +I_(2)=I_(3)`
Using Kirchhoff's second law to loop ABEFA and BEDCB, we have
`2 I_(1) + 5 I_(1)=12`
`2 I_(3) + 3 I_(2) = 6`
From (ii) and (i), `2 (I_(1) +I_(2)) + 5 I_(1) = 12` or `7 I_(1) + 2 I_(2)=12`
From (iii) and (i), `2 (I_(1) +I_(2)) + 3 I_(2)=6`
or `2I_(1) + 5 I_(2)=6`
Solving equation (i), (ii) and (iv), we get
`I_(1)=48/31 A , I_(2) = 18/31 A , I_(3)=66/31A`