In the circuit shown in figure, find the current through the branch BD

The current in the circuit are assumed as shown in the fig.Applying KVL along the loop ABDA, we get – 6i1 – 3 i2 + 15 = 0 or  2i1 + i2 = 5       …..(i)Applying KVL along the loop BCDB, we get– 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 …..(ii)Solving equation (i) and (ii) for i2, we get i2 = 5 A