Use Kricchoff's rules to find out the values of the current `I_(1) and I_(2)` in the electrical network as shown.

Explanation

From loop FEDAF: `-20I_(2) + 80 + 40-40 I_(3) = 0` …(i)
At junction D, applying Kirchhoff.s junctions law, we get `I_(3) = I_(1) + I_(2)`
Substituting this value in (i), we get `60 I_(2) + 40 I_(1) = +120`
`3I_(2) + 2I_(1)= 6` …(ii)
From loop ABCDA: `-20 I_(1) + 40-40 I_(1) - 40I_(2) = 0`
`+60 I_(1) + 40I_(2) = 40`
`I_(1) + 2I_(2) = 2` ...(iii)
Solving (ii) and (iii), we get the value of `I_(1) and I_(2)`
`I_(2) = (14)/(5)A`
`I_(1) = - (6)/(5)A`