Problem 21.33

21.33. a)
Let I₁ be the current through the $8 \Omega$ resistor. Let I₂ be flowing from the $5 \Omega$ resistor towards the 4V battery, and let I₃ flow from the 12V battery to the $3 \Omega$ resisitor. Then we have
I₃ = I₁ + I₂
Apply Kirchhoff's rule to the loop containing I₂ and I₃.
12 V - 4 I₃ - 6 I₂ - 4V = 0
or if we simplify,
8 = 4I₃ + 6I₂
Applying Kirchhoff's rule to the other loop, we have
-6 I₂ - 4V + 8 I₁ = 0
8I₁ = 4 + 6I₂
Solving the system of three equations and three unknowns, we find that
$I_{1} = \frac{11}{13} A$
$I_{2} = \frac{6}{13} A$
$I_{3} = \frac{17}{13} A$