Let the current directions be as shown in Fig.
Apply Kirchhoff s Second law to the closed circuit ABDA, we get
5 - x - z + y = 0 or x - y + z =5 ... (i)
Similarly, circuit BCDB gives
- (x - z) + 5 + (y + z) + z= 0
or x - y - 3z = 5 ......(ii)
Lastly, from circuit ADCEA, we get
- y -(y + z) + 10 - (x + y) = 0
or x + 3y + z = 10 .....(iii)
From Eq. (i) and (ii), we get, z = 0
Substituting z = 0 either in Eq. (i) or (ii) and in Eq. (iii), we get
x - y = 5 .....(iv)
x + 3y = 10 .....(v)
Subtracting Eq. (v) from (iv), we get
- 4y = - 5 or y = 5/4 = 1.24 A
Eq. (iv) gives x = 25/4 A = 6.25A
Current in branch AB =current in branch BC = 6.25 A
Current in branch BD = 0; current in branch AD = current in branch DC =1.25 A; current in branch CEA = 6.25 + 1.25 = 7.5 A.
