Using kirchhoff's rules wire the experssion for the current `I_(1) ,I_(2) and I_(3)` in the circuit diagram

Explanation

The given circuit diagram is

According to Kirchhoff's second law , in close circuit ABCFA , we have
`-2-4 I_(1) + 3I_(2) + 1 = 0 " " [ because Sigma IR + Sigma E = 0]`
`-4 I_(1) + 3I_(2) - 1 = 0 " " .... (i)`
According to Kirchhoff's second law in close circuit FCEDF
`-1 -3 I_(2) - 2 I_(3) + 4 = 0`
`-3 I_(2) - 2 I_(3) + 3 = 0 " " ... (ii)`
According to Kirchhoff's first law , at junction F ,
`I_(1) + I_(2) = I_(3) " " ... (iii) `
On solving eq(s) (i) , (ii) and (iii) , we can get `I_(1) , I_(2)` and `I_(3)` respectively .