Using kirchhoff's laws, find the currents `I_(1) I_(2)` and `I_(3)` of the network

Answer

`(6)/(13) A`, (22)/(13)A`, (28)/(13)A`

Explanation

According to kirchhoff's firse rule , at junction A
`l_(1) = l_(2) + l_(3) or l_(3) = (1) - l_(2)`
Appliying kirchhoff's second rule to the closed loop `BADCB`we have
`3l_(2) + 2l_(1) = - 10 + 12 = 2 or 2l_(1) + 3l_(2) = 2`…..(i)
to the loop `AFEDA`
`4l_(1) - 3l_(2) = 10 `
or `4(l_(1) - l_(2)) - 3l_(1) = 10 or 4l_(1) - 7l_(2) = 10`....(ii)
Multipiying (i) by (2) and subratercting (ii) from it we get
`6l_(1) + 7l_(2) = 4 - 10 = - or l_(2) = - (6)/(13) A`
From (i) `2l_(1) + 3 (-(6)/(13)) = 2`
or `2l_(1) = 2 + (18)/(13) = (44)/(13) or l_(1) = (22)/(13)A`
`l_(3) = l_(1) - l_(2) = (22)/(13) - (- (6)/(13)) = (28)/(13)A`