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A and B are equipotential so for the time being treat them as one single node (lets call it F).
Current flowing from F to C = (Vfc)/3
So current from F to D = [14-(Vfc)/3]
Now apply KCL at node D.
Therefore the current from D to C = i1 + [14-(Vfc)/3] = [16-(Vfc)/3]
Now apply KCL at C.
Current flowing from C to D = (Vfc)/3 + i2.
[16-(Vfc)/3]=-[(Vfc)/3 + i2]
i2=-16.
Now find Vfc applying KVL.
Vfc=Vac.
To find i take the original circuit.
Current flowing from A to D = [14-(Vfc)/3]
So i = 10 - [14-(Vfc)/3].
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