using kVL in left loop
(120 + 74) i1 + 56 (i1 + i2) = 5.8 + 7
250 i1 + 56 i2 = 12.8... (i)
using kVL in right loop
(135 + 56) i2 + 56 i1 = - 3
56 i1 + 191 i2 = - 3...(ii)
solving (i) and (ii) (and considering the naming of current from the fig)
I1 = 0.02526 A
I2 = 0.05856 A
I3 = - 0.0329 A
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