Answer
$I_{3}$ = 2 A ;
$I_{2}$ = 6 A
Work Step by Step
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
