EECS: Find i_0 in a circuit with 4 resistors and 1 indep. current source

[VinnyCee]

(a) Using nodal analysis, determine $i_0$ in the circuit.

(b) Now use linearity to find $i_0$.

Work so far:

$$i_0\,=\,\frac{v_1\,-\,0}{6\,\ohm}\,=\,\frac{v_1}{6\,\ohm}$$

KCL@$v_1$: $$\frac{v_1\,-\,0}{3\,\ohm}\,+\,\frac{v_1\,-\,0}{6\,\ohm}\,+\,\frac{v_1\,-\,v_2}{2\,\ohm}\,=\,0$$

KCL@$v_2$: $$\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0$$

$$v_1\,-\,\frac{1}{2}\,v_2\,=\,0$$

$$-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,-9$$

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

$$v_1\,=\,-\,9V$$
$$v_2\,=\,-\,18V$$

Then I plug into the first equation:

$$i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(-\,9V)}{6\,\ohm}\,=\,-\,\frac{3}{2}A$$

Is this the correct value for $i_0$ or have I got the sign wrong?

 

[doodle]

KCL@$v_2$: $$\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0$$

You are repeating the same mistake here as you did in the previous exercise.

 

[VinnyCee]

What is that, can you explain?

 

[berkeman]

What is that, can you explain?

The sign on the 9A is wrong. The other two terms in that KCL are for currents *out* of the node. So since the 9A is into the node, it has to be -9A in that equation. Makes sense?

 

[VinnyCee]

OIC! There has to be a minus in there somewhere if all of the terms are on one side and it's equal to zero. Whoops!

Using your correction:

KCL @ $v_2$: $$\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,-\,9A\,=\,0$$

$$-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,9$$

Which switches the signs on the $v_1$ and $v_2$!

$$i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(9V)}{6\,\ohm}\,=\,\frac{3}{2}A$$

 

[doodle]

Yes, that's much better. And be careful when you formulate the KCLs next time.