For the given network calculate current I and `I_(2)`.

Explanation

KVl in loop (ABDA)
`-2I_(1)-I_(2)+3(I-I_(1))=0`
`rArr 5I_(1)+I_(2)=3I` ....(1)
VL in loop (BCDB)
`-3(I_(1)-I_(2))+2(I-I_(1)+I_(2))+I_(2)=0`
`rArr 5I_(1)-6I_(2)=2I` ...........(2)
By subtracting (2) from ( 1)
`7I_(2)=I rArrI_(2)=(I)/(7)` and also `I_(1)=(4I)/(7)`
KVL in loop (ABCA via battery)
`-2I_(1)-3(I_(1)-I_(2))-1+10=0`
`rArr(-8)/(7)I-(9)/(7)I-1+10=0rArr I=(35)/(12)A & I_(2)=(5)/(12)A`