The given circuit is redrawn with assumed distribution of currents in Fig. (b).
Applying KVL to different closed loops, we get
Circuit EF ADE
- 2y + 10z + (x - y - 6) = 0 or x - 3y + 10z = 6 ......(i)
Circuit ABCDA
2(y + z + 6) - 10 + 3(x - y - z - 6) - 10z = 0
or 3x - 5y - 14z = 40 .....(ii)
Circuit EDCGE
- (x - y - 6) - 3(x - y - z - 6) - 4x + 24 = 0
or 8x - 4y - 3z = 48 ......(iii)
From above equations we get x = 4.1A
