Answer
A and E are in series, C and D are in series.
$i_{c} + i_{d}$ = 0
$i_{b}$ = 4 A
$i_{d}$ = 2 A
Work Step by Step
a) Two elements are connected in series if they are connected end to end, and that point has no other element connected to it. According to this, A,E and C,D are two pairs of elements in series
b) Current through series connected elements is same in magnitude and direction. Therefore, $i_{c}$ = - $i_{d}$
c) $i_{d}$ = -(-2 A) = 2 A. KCL at node connecting A,B,C gives $i_{a}$ + $i_{c}$ - $i_{b}$ = 0
$i_{b}$ = 6 + (-2) = 4 A
