5
\$\begingroup\$

I am currently studying circuit analysing techniques using Engineering Circuit Analysis 8th Ed. by Hayt, Kemmerly, and Durbin. In Section 4.2, the authors describe a "difficult approach" to handling voltage sources between two non-reference nodes (the red highlighted box in the image below). Instead of using a supernode, the method involves:

  • Assigning an unknown current to the branch containing the voltage source.
  • Applying KCL (Kirchhoff's Current Law) at each of the three non-reference nodes.
  • Applying one KVL (Kirchhoff's Voltage Law) equation.

enter image description here

I have attempted to set up the system of four equations with four unknowns. I've defined \$i_x\$ (unknown branch current) as flowing from node 2 to node 3.

KCL 1: \$ -8 \text{ A}-\frac{v_1-v_2}{3\ \Omega} -3 \text{ A} -\frac{v_1-v_3}{4\ \Omega} = 0 \text{ A} \$

KCL 2: \$ 3 \text{ A} + \frac{v_1-v_2}{3\ \Omega} - \frac{v_2}{1\ \Omega} -i_x =0 \text{ A}\$

KCL 3: \$ \frac{v_1-v_3}{4\ \Omega} +i_x -\frac{v_3}{5\ \Omega} + 25 \text{ A}= 0 \text{ A}\$

KVL: \$ -v_2 -22 \text{ V} + v_3 = 0 \text{ V} \$

Resulting Matrix Equation:

$$ \begin{bmatrix} -7 & 7 & 0 & 0\ \Omega \\ 1 & -4 & 0 & -3\ \Omega \\ 5 & 0 & -9 & 20\ \Omega \\ 0 & -1 & 1 & 0\ \Omega \end{bmatrix} \cdot \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ i_x \end{pmatrix} = \begin{pmatrix} 132 \text{ V} \\ -9 \text{ V} \\ -500 \text{ V} \\ 22 \text{ V} \end{pmatrix} $$

Could you please verify the coefficients and constants in this matrix equation?

\$\endgroup\$
1
  • \$\begingroup\$ The other form of matrix equations is [V][Y] = [I] where [V] is nodal voltage matrix, [Y] is admittance matrix . Admittance is reciprocal of impedance (resistance) . [I] is current matrix. \$\endgroup\$ Commented 17 hours ago

2 Answers 2

Reset to default
4
\$\begingroup\$

Your matrix is incorrect for the formulated equations. The simplified equations are:

\$-7V_1+4V_2+3V_3=132\$

\$V_1-V_2-3V_2-3i_x=-9\$

\$5V_1-9V_3+20i_x=-500\$

\$V_3-V_2=22\$

Putting these equations into matrix form yields:

$$ \begin{bmatrix} -7 & 4 & 3 & 0\ \Omega \\ 1 & -4 & 0 & -3\ \Omega \\ 5 & 0 & -9 & 20\ \Omega \\ 0 & -1 & 1 & 0\ \Omega \end{bmatrix} \cdot \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ i_x \end{pmatrix} = \begin{pmatrix} 132 \text{ V} \\ -9 \text{ V} \\ -500 \text{ V} \\ 22 \text{ V} \end{pmatrix} $$

Solving this in Matlab gives the following X matrix (for AX=B):

X = $$\begin{bmatrix} 1.0714\\ 10.5000\\ 32.5000\\ -10.6429\end{bmatrix}$$

To verify, let's subtract \$V_2\$ from \$V_3\$. Here, it is: 32.5 - (10.5) which is 22 V! This validates our solution. You can try this for the other voltages as well and see if the equations are satisfied.

\$\endgroup\$
6
  • \$\begingroup\$ Are you sure that this is the correct answer? \$\endgroup\$ Commented 18 hours ago
  • \$\begingroup\$ @G36 Pretty sure it is a simplification issue. \$\endgroup\$ Commented 18 hours ago
  • 1
    \$\begingroup\$ @G36 :) You are right. I messed up. The equations are however correct. Thanks. \$\endgroup\$ Commented 18 hours ago
  • 1
    \$\begingroup\$ +1 for showing us the correct answer. \$\endgroup\$ Commented 18 hours ago
  • 1
    \$\begingroup\$ @ :) Thank you. \$\endgroup\$ Commented 18 hours ago
2
\$\begingroup\$

The first thing I would do to eliminate any kind of confusion is to convert the negative current sources into their equivalent positive ones. Therefore, we have \$8A\$ coming out of node \$v_1\$, \$3A\$ entering node \$v_2\$, and \$25A\$ entering node \$v_3\$.

Summation of current entering a node = Summation of current leaving a node

Also, let's assume \$i_x\$ flows from node \$v_2\$ to node \$v_3\$.

Now applying K.C.L at nodes \$v_1\$, \$v_2\$, and \$v_3\$ respectively we get:

\$\frac{v_2 - v_1}{3}\$ \$+\$ \$\frac{v_3 - v_1}{4}\$ \$=\$ \$8\$ \$+\$ \$3\$ \$=\$ \$11\$

\$3\$ = \$\frac{v_2 - v_1}{3}\$ \$+\$ \$\frac{v_2}{1}\$ \$+\$ \$i_x\$

\$i_x\$ \$+\$ \$25\$ \$=\$ \$\frac{v_3}{5}\$ \$+\$ \$\frac{v_3 - v_1}{4}\$

We get one more equation because of the voltage source present between \${v_2}\$, and \${v_3}\$, which is:

\${v_3}\$ \$-\$ \${v_2}\$ \$=\$ \$22\$

Rearranging the 4 equations we get:

\$-7v_1\$ \$+\$ \$4v_2\$ \$+\$ \$3v_3\$ \$+\$ \$0i_x\$ \$=\$ \$132\$

\$-v_1\$ \$+\$ \$4v_2\$ \$+\$ \$0v_3\$ \$+\$ \$3i_x\$ \$=\$ \$9\$

\$-5v_1\$ \$+\$ \$0v_2\$ \$+\$ \$9v_3\$ \$-\$ \$20i_x\$ \$=\$ \$500\$

\$0v_1\$ \$-\$ \$v_2\$ \$+\$ \$v_3\$ \$+\$ \$0i_x\$ \$=\$ \$22\$

Converting them into matrix form, we get:

\$\begin{bmatrix}-7&4&3&0\\-1&4&0&3\\-5&0&9&-20\\0&-1&1&0\end{bmatrix}\$\$\begin{bmatrix}v_1\\v_2\\v_3\\i_x\end{bmatrix}\$ \$=\$ \$\begin{bmatrix}132\\9\\500\\22\end{bmatrix}\$

Now that you have your equation in matrix format, you may solve them easily using any kind of computational tool such as wxMaxima (open source) or any other tool of your preference.

\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.