## Fundamentals of Electrical Engineering

$I_{3}$ = 2 A ; $I_{2}$ = 6 A
Given, $I_{0}$ = -2 A $I_{1}$ = -4 A $I_{S}$ = 8 A $V_{S}$ = 12 V Applying Kirchhoff's Law at node 'a', we get $I_{0}$ + $I_{1}$ + $I_{2}$ = 0 or, -2 A + (- 4 A) + $I_{2}$ = 0 or, $I_{2}$ = 6 A Applying Kirchhoff's Law at node 'b', we get -$I_{3}$ + $I_{1}$ + $I_{S}$ + $I_{0}$ = 0 or,$I_{3}$ = $I_{1}$ + $I_{S}$ + $I_{0}$ or,$I_{3}$ = (-4 A) + 8 A + (-2 A) or,$I_{3}$ = 2 A