Determine the current in each branch of the network shown in fig.

Explanation

Current flowing through various branches of the circuit is represented in the given figure

`I_(1)=` Current flowing through the outer circuit
`I_(2)`= Current flowing through branch AB
`I_(3)=` Current flowing through branch AD
`I_(2)-I_(4)=` Current flowing through branch BC
`I_(3)+I_(4)=` Current flowing through branch CD
`I_(4)=` Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
`10 I_(2)+5I_(4) - 5I_(3)=0`
`2I_(2)+I_(4) - I_(3) =0`
`I_(3)=2I_(2)+I_(4)`....(1)
For the closed circuit ABDA, potential is zero i.e.,
`10 I_(2)+5I_(4)-5I_(3) =0`
`2I_(2)+I_(4)-I_(3)=0`
`I_(3)=2I_(2)+I_(4)`....(1)
For the closed circuit BCDB, potential is zero i.e.,
`5(I_(2)-I_(4))-10(I_(3)+I_(4))-5I_(4)=0`
`5I_(2)+5I_(4)-10 I_(3) - 10 I_(4)-5I_(4)=0`
`5 I_(2)-10 I_(3)-20 I_(4) = 0`
`I_(2)=2I_(3)+4I_(4)`....(2)
For the closed circuit ABCFEA, potential is zero i.e.,
`- 10 + 10 (I_(1))+10 (I_(2))+5(I_(2)-I_(4))=0`
`10 = 15 I_(2)+10 I_(1)-5 I_(4)`
`3I_(2)+2I_(1) - I_(4) =2`....(3)
From equations (1) and (2), we obtain
`I_(3)=2(2I_(3)+4I_(4))+I_(4)`
`I_(3)=4I_(3) + 8I_(4) + I_(4)`
`-3I_(3) = 9I_(4)`
`- 3 I_(4) = + I_(3)`....(4)
putting equation (4) in equation (1), we obtain
`I_(3)=2I_(2)+I_(4)`
`-4I_(4)=2I_(2)`
`I_(2)=-2I_(4)`...(5)
It is evident from the given figure that,
`I_(1)=I_(3) +I_(2)`....(6)
putting equation (6) in equation (1), we obtain
`3I_(2)+2(I_(3)+I_(2))-I_(4)=2`
`5 I_(2)+2I_(3)-I_(4) = 2`....(7)
putting equations (4) and (5) in equation (7), we obtain
`5(-2 I_(4))+2(-3 I_(4))-I_(4)=2`
`-10 I_(4)-6 I_(4) - I_(4)=2`
`17 I_(4)=-2`
`I_(4)=(-2)/(17)A`
Equation (4) reduces to
`I_(3)=-3(I_(4))`
`= -3((-2)/(17))=(6)/(17)A`
`I_(2)=-2(I_(4))`
`=-2((-2)/(17))=(4)/(17)A`
`I_(2)-I_(4)=(4)/(17)-((-2)/(17))=(6)/(17)A`
`I_(1)=I_(3)+I_(2)`
`=(6)/(17)+(4)/(17)=(10)/(17)A`
Therefore, current in branch `AB = (4)/(17)A`
In branch `BC = (16)/(17)A`
In branch `CD = (-4)/(17)A`
In branch `AD = (6)/(17)A`
In branch `BD = ((-2)/(17))A`
Total current `= (4)/(17)+(6)/(17)+(-4)/(17)+(6)/(17)+(-2)/(17)=(10)/(17)A`.