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

Explanation

The current,s through the various amrs of the circuit have been shown in figure. According to Kirchhoffs second law,
`-10 +10 (i_(2) + i_(@)) + 10 i_(1) + 5 (i_(1) - i_(3)) = 0`
(or) `10 = 25 I + 10 I - 5i_(3)`
(or) `2 = 5i + 2i_(2) - i_(3)`
(or) `2 = 5i_(1) + 2i_(2) - i_(3)`........`to` (i)
In a closed circuit ABDA
`10 i_(1) + 5 i_(3) - 5 i_(2) = 0`
(or) `2 i_(1) + i_(3) - i_(2) = 0`
or `i_(2) = 2i_(1) + i_(3)` ......`to` (ii) In a closed circuit BCDB
`5(i_(1) - i_(3)) - 10 (i_(2) + i_(3)) - 5i_(3) = 0`
(or) `5i_(1) - 10 i_(2) = 20 i_(3) = 0`
`i_(1) = 2i_(1) + 4i_(3)`.... `to` (iii)
From (ii) and (iii)
`i_(1) = 2 (2i_(!) - i_(3)) + 4i_(3) = 4i_(1) + 6i_(3)`
(or) `3 i_(1) = 6i_(3)`
(or) `i_(1) = 2i_(3)`
Putting this value in (ii) : `i_(2) = 2 (_2i_(3)) i_(3) = - 3 i_(3)`
Putting values in (i)
`2 = 5 (- 2i_(3)) + 2 (-3i_(3)) - i_(3)` (or) `2 = - 17 i_(3)`
(or) `i_(3) = 2//17 A`
From (iv) `i_(1) = - (-2//17) = - 4//17 A`
`i_(1) + i_(2) = (4//17) + (6//17) = (10//17) A`
`i_(1) + i_(3) = (4//17) + (-2//17) = (6//17)A`
`i_(2) + i_(3) = (6//17) + (-2//17) = (4//17)A`.