Engineering 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.