## Electrical Engineering: Principles & Applications (6th Edition)

$i_a=-2A$ $i_c=1A$ $i_d=4A$
Since $i_a+i_b=0, i_a=-i_b=-2A$ $i_c+i_b=3A$, so $i_c=3-i_b\ A$ $3A+i_e=i_d=3A+1A$