$i_b=-1A$ $i_f=7A$ $i_d=5A$ $i_e=-2A$
Work Step by Step
Given $i_a=2A, i_c=-3A,i_g=6A,$ and $i_h=1A$. Once again apply KCL at nodes to find: $i_b=i_c+i_a=-1A$ $i_f=i_g+i_h=7A$ $i_d=i_f-i_a=5A$ $i_e=i_c+i_h=-2A$
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