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

Explanation

KVL in loop (ABDA)
` -21 _1 -I_2+3( I-I_1) =0`
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` rArr 5I_1 +I_2=3I`
KVL in loop (BCDB)
`" " - 3(I_1-I_2) +2(I-I_1+I_2) +I_2=0`
`rArr " "5I_1-6I_2=2I`
By subtracting (2) from (1)
` 7I_2=I rArr I_2 =(I)/(7) ` and also ` I_1 =(4I)/( 7) `
KVL in loop (ABCA via battery )
` " " -2I_1- 3( I_1-I_2) -I+10 =0`
`rArr " "(-8)/( 7) I-( 9)/(7) I-I+10 =0rArr I=(35)/( 12)A &I_2=(5)/(12) A `