In the closed circuit ABCA,
`(I_(1) - I_(3) 4 + (I_(1) + I_(2)-I_(3))2 + I_(1) xx 1 = 10`
or ` 7 I_(1) + 2 I_(2) - 6 I_(3)=10`
In the closed circuit ADCA,
`I_(3) xx 4 + (I_(3) - I_(2)) 2 +I_(1) xx 1 = 10`
or `I_(1) - 2 I_(2) + 6 I_(3)=10`
In the closed current ABEDA
`(I_(1) - I_(3)) 4 - I_(3) xx 4 = - 5`
or `4 I_(1) - 8 I_(3) = - 5`
On solving (i), (ii) and (iii), we get
`I_(1) = 2.5A`
`I_(2) = 1.875 A = I_(3)`
