How to Formulate Circuit Analysis Equations into Matrix Form?

[VinnyCee]

Using superposition, find v_0 in the following circuit.

My work so far:

v_1\,=\,2\,i_2\,+\,v_{0\,1}

v_1\,=\,4\,i_1

v_{0\,1}\,=\,3\,i_3

v_{0\,1}\,=\,6\,i_4

KCL @ v1:
1\,A\,=\,i_1\,+\,i_2

i_2\,=\,i_3\,+\,i_4


Using these six equations, with 6 variables:

v_{0\,1}\,=\,\frac{12}{7}\,V

Is that correct?

My real question is how to put the six equations above into matrix form to enable solving using RREF.

Thanks

EDIT: I have fixed this part with Paallikko's guidance (thanks!)

i_1\,=\,\frac{1}{4}\,v_1

i_2\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}

i_3\,=\,\frac{1}{3}\,v_{01}

i_4\,=\,\frac{1}{6}\,v_{01}

i_1\,+\,i_2\,=\,1\,A\,\,\,\Rightarrow\,\,\,\frac{3}{4}\,v_1\,-\,\frac{1}{2}\,v_{01}\,=\,1

i_3\,+\,i_4\,=\,i_2\,\,\,\Rightarrow\,\,\,\frac{1}{3}\,v_{01}\,+\,\frac{1}{6}\,v_{01}\,=\,\frac{1}{2}\,v_1\,-\,\frac{1}{2}\,v_{01}

v_{01}\,=\,1\,V

 

[VinnyCee]

Apply superposition two:

i_3\,=\,i_2\,-\,2\,A

i_1\,=\,i_0\,+\,i_2

v_{0\,2}\,=\,6\,i_1\,=\,6\,i_3\,=\,3\,i_o

From these five equations, I put into a matrix and solve.

v_{0\,2}\,=\,-6\,V


Next, I do the last superposition:

The circuit above can be simplified:

v_{0\,3}\,=\,3\,I

v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right)\,=\,20\,V

Total them up:

v_0\,=\,v_{0\,1}\,+\,v_{0\,2}\,+\,v_{0\,3}

v_0\,=\,\left(\frac{12}{7}\,V\right)\,+\,(-6\,V)\,+\,(20\,V)

v_0\,=\,15.7\,V

Does this look right? If not, please explain the errors and how to correct them. Thanks

 

[Päällikkö]

Well, you've understood how to use superposition (setting the sources to 0), you even seem to have most of the equations right*. However, all your calculations are wrong.

The first one in matrix form would be (solve for i_x from the first equations and substitute into the KCL equations):

<br /> \left[ {\begin{array}{*{20}c}<br /> {3/4} &amp; { - 1/2} \\<br /> { - 1/2} &amp; 1 \\<br /> <br /> \end{array} } \right]\left[ {\begin{array}{*{20}c}<br /> {V_1} \\<br /> {V_2} \\<br /> <br /> \end{array} } \right] = \left[ {\begin{array}{*{20}c}<br /> 1 \\<br /> 0 \\<br /> <br /> \end{array} } \right]<br />

* the two equations below are wrong
i_1=i_0 + i_2
v_{0\,3}\,=\,3\,\left(\frac{20\,V}{3\,\ohm}\right) \,=\,20\,VPS. If you're not familiar with PSpice, there's a free version available for download eg. here: http://www.electronics-lab.com/downloads/schematic/013/ you can quickly check if your answers are correct with it.
PPS. the topic's more engineering or introductory physics than advanced physics.

 

[VinnyCee]

For the first incorrect equation (in the 2nd part of problem), I see that all the currents are going out of the node. Does this mean that the actual equation is i_0\,+\,i_1\,+\,i_2\,=\,0?

In the third part of the problem, is the error becuase I didn't include both resistors in the I = equation?

v_{03}\,=\,3\,I

v_{03}\,=\,3\,\left(\frac{20\,V}{6\,\ohm}\right)\,=\,10\,V

Is that right now?

 

[Päällikkö]

Yep, they're right now.

 

[VinnyCee]

How do I solve the second part? I keep getting 0V for v_{02}.

I also used Pspice and got an 8V differential at the resistor for v_0.

v_0\,=\,v_{01}\,+\,v_{02}\,+\,v_{03}

v_0\,=\,1V\,+\,0V\,+\,10V\,=\,11V

This is 3V higher than Pspice reports though.

 

[Päällikkö]

You've written:
v₀₂ = 3i₀ = 6i₁ = 6i₃ (1)
i₂ - i₃ = 2 (2)
i₀ + i₁ + i₂ = 0 (3)

Substituting (2) into (3) (Eliminating i₂):
i₀ + i₁ + i₃ = -2

Substituting (1) into the above:
v₀₂(1/3 + 1/6 + 1/6) = -2

=> v₀₂ = -3 (V)