# Node Voltage Problem, possible to simplify?

I have a circuit here which has to be solved using only direct input into a matrix using the node voltage method.

Here is my attempt to solve it, the blue lines separate each matrix entry. I am wondering if there is a better way to solve it, or if I made a mistake or anything. I don't believe I did it correctly largely because of the $3V_1$ in term $b_{33}$ of the matrix.

• the original circuit shown labels $V_1$ as the voltage across $C_1$, but you also labeled $V_1 = V_S$. Sep 19, 2016 at 23:49
• It wouldn't be equal to that?
– ANZ
Sep 20, 2016 at 1:26
• It's not clear to me why it should be; R2 separates the two nodes. Unless this is a steady-state DC problem? In that case, a lot of simplifications can be made to the system. Sep 20, 2016 at 1:53
• Then it appears I have no idea what I am doing. The problem says nothing about assuming DC steady-state. It just says solve using node to datum voltages variables in a matrix.
– ANZ
Sep 20, 2016 at 1:58
• The problem statement has already use $v_1$ to designate the voltage across C1. It leads to big confusion when you designate the node above $v_s$ as $v_1$ also. Sep 20, 2016 at 5:59

I think you need $-3\cdot V_1$ in the first column, third row; removing it from the third column, third row. (It's incoming, so negative, if I take your other signs to be correct.) Since the first column implies $V_1$, you'd just put $-3$ there, not $-3\cdot V_1$, in keeping with your notation. I think you need to associate the columns with the nodes and you didn't do that, there. The rest of that row looks good to me.

I looked at the first and second rows and they look good to me, as well. I didn't look over the fourth row, but since you didn't ask about it and you've done really well elsewhere, I'll leave it unchecked for now.

The second row I got was:

\begin{align*} \left[-\frac{V_1}{R_2}\right] + \left[\frac{V_2}{R_2} + C_1\frac{\textrm{d}V_2}{\textrm{d}t} + C_2\frac{\textrm{d}V_2}{\textrm{d}t}\right] +\left[- C_2\frac{\textrm{d}V_3}{\textrm{d}t}\right] +\left[0\cdot V_4\right]&= 0 \end{align*}

The third row I got was:

\begin{align*} \left[-3\cdot V_1\right] +\left[- C_2\frac{\textrm{d}V_2}{\textrm{d}t}\right] + \left[\frac{V_3}{R_3} + C_2\frac{\textrm{d}V_3}{\textrm{d}t} + \frac{1}{L_1}\int V_3\; \textrm{d}t\right] - \left[\frac{1}{L_1}\int V_4\; \textrm{d}t\right] = 0 \end{align*}

Resistor R1 is shunted by a voltage source so it can be safely removed without interfering with the voltage.

The next step would be to use the Norton equivalent for V1 and R2. This way you can effectively reduce the matrix to a 3x3 matrix.

For the dependent current source, you could consider putting $+3\cdot v_1$ in the right-hand side vector (which is what you always do with an independent current source). Then you can bring it over to the left side, which makes it $-3\cdot v_1$. This has to be in the same current equation, so same row of the matrix. And it is multiplied with $v_1$ so it $-3$ to be inserted into the first column ($v_1$'s column).

simulate this circuit – Schematic created using CircuitLab

$\begin{bmatrix} \frac{1}{R_2}+(C_1+C_2)\frac{d}{dt}&&-C_2\frac{d}{dt}&&0\\ -3-C_2\frac{d}{dt}&&\frac{1}{R_3}+C_2\frac{d}{dt}+\frac{1}{L_1}\int\ dt&& -\frac{1}{L_1}\int\ dt\\ 0 && -\frac{1}{L_1}\int\ dt&& \frac{1}{R_4} + \frac{1}{L_1}\int\ dt \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} \frac{v_s}{R_2}\\ 0\\ 0 \end{bmatrix}$

[EDIT]

I also noticed some further improvements in your work. For one, the currents $i_2, i_3, i_4$ should not be there. They will have to be zero to satisfy Kirchoff's law. Only independent sources end up at the right-hand side.

Secondly, if you want to use an ideal voltage source in nodal analysis without reducing complexity, you have to go to so-called "Modified Nodal Analysis" (MNA). In MNA, you generally add equations that that are not necessarily a KCL law equation. However, if you add an equation and still want to keep the circuit solvable, you have to add an unknown as well (typically the current). So you will then get a 5x5 matrix, rather than a 4x4. For a voltage source, MNA will give

$v_1=v_s$, the equation that fixes the voltage.

$i_{v_s}$, the current through $v_s$ is added as an unknown. This current is added into v1. If no unknown current is added, then the circuit would become unsolvable.

$\begin{bmatrix} \frac{1}{R_2} && -\frac{1}{R_2} && 0 && 0 && 1\\ -\frac{1}{R_2} && \frac{1}{R_2}+(C_1+C_2)\frac{d}{dt} && -C_2\frac{d}{dt} && 0 && 0\\ 0 && -3-C_2\frac{d}{dt} && \frac{1}{R_3}+C_2\frac{d}{dt}+\frac{1}{L_1}\int\ &&-\frac{1}{L_1}\int\ dt&& 0\\ 0 && 0 && -\frac{1}{L_2}\int\ dt && \frac{1}{R_4}+\frac{1}{L_1}\int\ dt && 0\\ 1 && 0 && 0 && 0 && 0\\ \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4\\ i_{v_s} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ v_s \end{bmatrix}$

As you can see, the matrix becomes larger rather quickly. Reducing the circuit complexity is therefore always preferred. This method is usually used in circuit simulators. I guess a computer doesn't care that much about one extra equation.