Chapter 1 - 1.4 - Problems - Kirchhoff's Current Laws - Page 38: P1.32

$i_a=-2A$ $i_c=1A$ $i_d=4A$

Work Step by Step

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

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