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

Explanation

As per Kirchhoff.s first rule the distribution of current in various branches is shown in Fig. Now applying Kirchhoff.s second law to mesh ABDA, we have
` - 10.I_1 - 5.I_2 + 5.(I - I_1) = 0`
or ` + 5I - 15I_1 - 5I_2 = ` ...(i)
Again for mesh BDCB , we have
` - 5.I_2 - 10.(I - I_1 + I_2) + 5.(I_1 - I_2) = 0`
or `-10 I + 15I_1 - 20I_2 = 0` ...(ii)
and for mesh ABCEA , we have
` - 10 . I_1 - 5.(I_1 - I_2) - 10.I + 10 = 0`
or `10I + 15I_1 - 5I_2 = 10` ....(iii)
Solving these three equations , we find that
`I =10/17 , I_1 = 4/17 A and I_2 = -2/17A`
Hence , electric current in branch `AB , I_1 = 4/17`A, in branch `AD, (I - I_1) = 6/17 A` , in branch BD , `I_2 = -2/17A` , in branch BC , `(I_1 -I_2) = 6/17 A,` in branch `DC, (I - I_1 + I_2) = 4/17A` and the total current drawn from battery `I = 10/17 A`