Find I in the circuit using the superposition principle:
My work so far:
I\,=\,I_1\,+\,I_2\,+\,I_3
This reduces down to:
16\,V\,=\,I_1\,(8\Omega)
I_1\,=\,\frac{16\,V}{8\Omega}\,=\,2\,A
This reduces down to:
I_2\,=\,-\frac{12\,V}{8\Omega}\,=\,-\frac{3}{2}...