Chapter 3, Problem 3.1P

Formulate node-voltage equations for the circuit in Figure P3-1. Arrange the results in matrix form $\text{Ax}=\text{b}\text{.}$

To determine

The node voltage equations in matrix form $\text{Ax}=\text{b}$ .

Answer to Problem 3.1P

The required node voltage equation in the form of matrix is $\left[\begin{array}{ccc}\frac{1}{R_5}+\frac{1}{R_1}& -\frac{1}{R_1}& -\frac{1}{R_5}\\ -\frac{1}{R_1}& \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}& -\frac{1}{R_3}\\ -\frac{1}{R_5}& -\frac{1}{R_3}& \frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\end{array}\right]+\left[\begin{array}{c}{v}_{\text{A}}\\ v_{\text{B}}\\ v_{\text{C}}\end{array}\right]=\left[\begin{array}{c}{i}_{\text{S}}\\ 0\\ 0\end{array}\right]$ .

Explanation of Solution

Calculation:

The given circuit diagram is shown in Figure 1

  

Mark the branch current in the given diagram.

The required diagram is shown in Figure 2

  

Apply the KCL at the node $v_{\text{A}}$ .

  $\begin{array}{l}{i}_{\text{S}}-\frac{v_{\text{A}}-v_{\text{C}}}{R_5}-\frac{v_{\text{A}}-v_{\text{B}}}{R_1}=0\\ \frac{v_{\text{A}}-v_{\text{C}}}{R_5}+\frac{v_{\text{A}}-v_{\text{B}}}{R_1}=i_{\text{S}}\\ v_{\text{A}}\left(\frac{1}{R_5}+\frac{1}{R_1}\right)-v_{\text{B}}\left(\frac{1}{R_1}\right)-v_{\text{C}}\left(\frac{1}{R_5}\right)=i_{\text{S}}\end{array}$ ....... (1)

Apply the KCL at the node $v_{\text{B}}$ .

  $\begin{array}{l}\frac{v_{\text{A}}-v_{\text{B}}}{R_1}-\frac{v_{\text{B}}-v_{\text{C}}}{R_3}-\frac{v_{\text{B}}}{R_2}=0\\ v_{\text{B}}\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)-v_{\text{A}}\left(\frac{1}{R_1}\right)-v_{\text{C}}\left(\frac{1}{R_3}\right)=0\\ -v_{\text{A}}\left(\frac{1}{R_1}\right)+v_{\text{B}}\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)-v_{\text{C}}\left(\frac{1}{R_3}\right)=0\end{array}$ ....... (2)

Apply the KCL at the node $v_{\text{C}}$ .

  $\begin{array}{l}\frac{v_{\text{B}}-v_{\text{C}}}{R_3}+\frac{v_{\text{A}}-v_{\text{C}}}{R_5}-\frac{v_{\text{C}}}{R_4}=0\\ -v_{\text{C}}\left(\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\right)+v_{\text{A}}\left(\frac{1}{R_5}\right)+v_{\text{B}}\left(\frac{1}{R_3}\right)=0\\ -v_{\text{A}}\left(\frac{1}{R_5}\right)-v_{\text{B}}\left(\frac{1}{R_3}\right)+v_{\text{C}}\left(\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\right)=0\end{array}$ ....... (3)

Arrange the Equation (1), (2) and (3) in the matrix form.

  $\left[\begin{array}{ccc}\frac{1}{R_5}+\frac{1}{R_1}& -\frac{1}{R_1}& -\frac{1}{R_5}\\ -\frac{1}{R_1}& \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}& -\frac{1}{R_3}\\ -\frac{1}{R_5}& -\frac{1}{R_3}& \frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\end{array}\right]+\left[\begin{array}{c}{v}_{\text{A}}\\ v_{\text{B}}\\ v_{\text{C}}\end{array}\right]=\left[\begin{array}{c}{i}_{\text{S}}\\ 0\\ 0\end{array}\right]$

Conclusion:

Therefore, the required node voltage equation in the form of matrix is $\left[\begin{array}{ccc}\frac{1}{R_5}+\frac{1}{R_1}& -\frac{1}{R_1}& -\frac{1}{R_5}\\ -\frac{1}{R_1}& \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}& -\frac{1}{R_3}\\ -\frac{1}{R_5}& -\frac{1}{R_3}& \frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\end{array}\right]+\left[\begin{array}{c}{v}_{\text{A}}\\ v_{\text{B}}\\ v_{\text{C}}\end{array}\right]=\left[\begin{array}{c}{i}_{\text{S}}\\ 0\\ 0\end{array}\right]$ .