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This is an exercise from Irwin's book, 9th edition.

schematic

simulate this circuit – Schematic created using CircuitLab

Although the book calls for the use of nodal analysis, I decided to do it by mesh analysis (this is what I am currently training).

The question asks for the value of VC. After performing Mesh Analysis, I arrived at the matrix equation:

$$\begin{pmatrix} 2 &-3 &-1 \\ -1 &2 &0 \\ -1 &0 &2 \end{pmatrix}\begin{pmatrix} i_A\\ i_B\\ i_C \end{pmatrix}=\begin{pmatrix} 0\\ -12m\\ 12m \end{pmatrix}$$

Whose determinant is zero.

When I found a solution book on the internet, performing a nodal analysis, I arrived at the value of VC = 9 V. However, when performing the Super Node Equation, it considers VC - VA = 2·k·I1, which does not seem correct to me.

Anyway, am I wrong in assuming that there is no solution for this circuit? In case there is no such solution, what explanation could I have for this?

Thinking about the idea of building a circuit, it bothers me a bit that they don't have a solution. Is this due to the topological restrictions imposed by Kirchhoff's Law and Ohm's Law? Or is it a flaw in the theory trying to describe a real physical system?

Is my assembly in the simulator correct? In particular, how to realize the current-controlled voltage source?

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  • \$\begingroup\$ Is that "R1.nA" really just I(R1)? And if so, which direction is considered positive? I don't understand the notation, I guess. Also, what's the point of nodes B and D? They are defined as 0 V. So I don't know why you placed them there. (And no, I don't have Irwin's book on hand.) \$\endgroup\$
    – jonk
    Aug 2 at 20:53
  • \$\begingroup\$ In fact, I agree that there is no need to indicate nodes B and D. As for the notation, I understand that R1.nA indicates the current going from node A towards resistor R1, in this case, the current that controls the voltage source is the one "going down" from node A towards GND. \$\endgroup\$ Aug 2 at 21:04
  • \$\begingroup\$ So is this image an accurate representation? \$\endgroup\$
    – jonk
    Aug 2 at 21:06
  • \$\begingroup\$ Exactly. I had just made the changes. I apologize for the confusion. \$\endgroup\$ Aug 2 at 21:10
  • \$\begingroup\$ Seems to solve out okay. Do you already have the answer from using a nodal approach? Okay. Well, never mind that! It solves out symbolically. But when the numbers are plugged in, then I think you may be right. I need to double check my work. \$\endgroup\$
    – jonk
    Aug 2 at 21:19

2 Answers 2

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Adapting your diagram, here's how I'd plan out the mesh analysis:

enter image description here

$$\begin{align*} 0\:\text{V} + V_1 -\left(I_1-I_3\right)R_2 - I_1 \,R_1 &= 0\:\text{V} \\\\ 0\:\text{V} - I_2 \,R_4 -\left(I_2-I_3\right)R_3-V_1&= 0\:\text{V} \\\\ 0\:\text{V} -\left(I_3-I_1\right)R_2 -\left(I_3-I_2\right)R_3+I_1\,R_{_\text{CCVS}}&=0\:\text{V} \end{align*}$$

So the matrix I get is:

$$\left[\begin{smallmatrix} -R_1-R_2&0&R_2 \\\\ 0 & - R_3 -R_4 &R_3 \\\\ R_2+R_{_\text{CCVS}}&R_3&-R_2 -R_3 \end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\\\\I_2\\\\I_3\end{smallmatrix}\right]=\left[\begin{smallmatrix}-V_1\vphantom{I_1}\\\\V_1\vphantom{I_1}\\\\0\vphantom{I_1}\end{smallmatrix}\right]$$

Or, dividing everything though by 1,000 and stuffing in values:

$$\left[\begin{smallmatrix} -2&0&1\vphantom{I_1} \\\\ 0 & -2 &1\vphantom{I_1} \\\\ 3&1&-2\vphantom{I_1} \end{smallmatrix}\right]\left[\begin{smallmatrix}I_1\\\\I_2\\\\I_3\end{smallmatrix}\right]=\left[\begin{smallmatrix}-12\:\text{m}\vphantom{I_1}\\\\12\:\text{m}\vphantom{I_1}\\\\0\vphantom{I_1}\end{smallmatrix}\right]$$

Signs different but the gist is the same.

For nodal analysis, I'd start here:

schematic

simulate this circuit – Schematic created using CircuitLab

But \$I_{R_1}=\frac{V_{_\text{A}}}{R_1}\$ so it follows that the voltage difference across the current controlled voltage source must be \$2\cdot V_{_\text{A}}\$. But this means that \$V_{_\text{C}}=-V_{_\text{A}}\$.

And, using Thevenin equivalents I find:

schematic

simulate this circuit

And there, the current must be the same throughout. But it cannot be the same throughout. So it cannot be resolved by nodal, either.

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Following the suggestion to do it generically, I took R1 = R.

Then the Mesh Analysis equations are:

\$2i_A-3i_B-i_C=0\$

\$-i_A+(R+1)i_B=-12\$

\$-i_A+2i_C=12\$

With \$i_A\$, \$i_B\$ and \$i_C\$ being the loop currents.

From the second and third equation, I put \$i_B\$ and \$i_C\$ as a function of \$i_A\$, substituting in the first equation and solving, I get:

\$i_A = (25+R)/(R-1)\$

Therefore, the system has no solution the way it is described in the book.

I would be happy if someone could comment on the other questions asked in the initial question.

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