These can be solved by the usual method to give
I1 = 2 5A,   I2 = 5
8  A,   I3 = 7
18
  A
The currents in the various branches of the network are
AB : 5
8  A,   CA : 
1
2 2  A,   DEB : 
7
1 8   A
AD : 
7
18
 A,   CD : 0 A,   BC : 
1
2 2  A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network For example, the total voltage
drop over the closed loop BADEB
5
85
4
15
8
4
V
V
V
+
×


−
×



equal to zero, as required by Kirchhoff’s second rule 3