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**(a) Using nodal analysis, determine [itex]i_0[/itex] in the circuit.**

(b) Now use linearity to find [itex]i_0[/itex].

(b) Now use linearity to find [itex]i_0[/itex].

__Work so far__:

[tex]i_0\,=\,\frac{v_1\,-\,0}{6\,\ohm}\,=\,\frac{v_1}{6\,\ohm}[/tex]

KCL@[itex]v_1[/itex]: [tex]\frac{v_1\,-\,0}{3\,\ohm}\,+\,\frac{v_1\,-\,0}{6\,\ohm}\,+\,\frac{v_1\,-\,v_2}{2\,\ohm}\,=\,0[/tex]

KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]

[tex]v_1\,-\,\frac{1}{2}\,v_2\,=\,0[/tex]

[tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,-9[/tex]

Whe I put these two equaitons into a 3 X 2 coefficient matrix, I get

[tex]v_1\,=\,-\,9V[/tex]

[tex]v_2\,=\,-\,18V[/tex]

Then I plug into the first equation:

[tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(-\,9V)}{6\,\ohm}\,=\,-\,\frac{3}{2}A[/tex]

Is this the correct value for [itex]i_0[/itex] or have I got the sign wrong?