Chapter 3, Problem 3.21HP

In the circuit in Figure P3.21, assume the source voltage and source current and all resistances are known.
a. Write the node equations required to determine the node voltages.
b. Write the matrix solution for each node voltage in terms of the known parameters.

To determine

(a)

Thenode equations required to obtain the node voltages.

Answer to Problem 3.21HP

The required node equations are $V_1\left[\frac{1}{R_2}+\frac{1}{R_1}\right]-\frac{V_2}{R_1}=\left(I_S+\frac{V_S}{R_2}\right)$ and $-V_1\left[\frac{1}{R_1}\right]+V_2\left[\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right]=\frac{V_S}{R_3}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Apply KVL at the node $V_1$ .

  $\begin{array}{l}{I}_S=\frac{V_1-V_S}{R_2}+\frac{V_1-V_2}{R_1}\\ I_S+\frac{V_S}{R_2}=V_1\left[\frac{1}{R_2}+\frac{1}{R_1}\right]-\frac{V_2}{R_1}\\ V_1\left[\frac{1}{R_2}+\frac{1}{R_1}\right]-\frac{V_2}{R_1}=\left(I_S+\frac{V_S}{R_2}\right)\end{array}$ ..... (1)

Apply KVL at the node $V_2$ .

  $\begin{array}{l}\frac{V_2-V_1}{R_1}+\frac{V_2-V_S}{R_3}+\frac{V_2}{R_4}=0\\ -V_1\left[\frac{1}{R_1}\right]+V_2\left[\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right]-\frac{V_S}{R_3}=0\\ -V_1\left[\frac{1}{R_1}\right]+V_2\left[\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right]=\frac{V_S}{R_3}\end{array}$   ....... (2)

Conclusion:

Therefore, the required node equations are $V_1\left[\frac{1}{R_2}+\frac{1}{R_1}\right]-\frac{V_2}{R_1}=\left(I_S+\frac{V_S}{R_2}\right)$ and $-V_1\left[\frac{1}{R_1}\right]+V_2\left[\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right]=\frac{V_S}{R_3}$ .

To determine

(b)

The matrix solutions for each of the node voltages in terms of the known parameters.

Answer to Problem 3.21HP

The required node voltage equations are $V_1=\frac{\left(I_S+\frac{V_S}{R_2}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)+\frac{V_S}{R_3{R}_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$ and $V_2=\frac{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\frac{V_S}{R_3}+\left(I_S+\frac{V_S}{R_2}\right)\frac{1}{R_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$ .

Explanation of Solution

Calculation:

The equation (1) equation (2) in the matrix form is given by,

  $\left[\begin{array}{cc}\frac{1}{R_2}+\frac{1}{R_1}& -\frac{1}{R_1}\\ -\frac{1}{R_1}& \frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\end{array}\right]\left[\begin{array}{c}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{c}\left(I_S+\frac{V_S}{R_2}\right)\\ \frac{V_S}{R_3}\end{array}\right]$

The determinant of the square matrix is given by,

  $\begin{array}{c}\Delta =\left[\begin{array}{cc}\frac{1}{R_2}+\frac{1}{R_1}& -\frac{1}{R_1}\\ -\frac{1}{R_1}& \frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\end{array}\right]\\ =\left[\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}\right]\end{array}$

The first determinant form of the matrix form is given by,

  $\begin{array}{c}{\Delta }_1=\left[\begin{array}{cc}{I}_S+\frac{V_S}{R_2}& -\frac{1}{R_1}\\ \frac{V_S}{R_3}& \frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\end{array}\right]\\ =\left[\left(I_S+\frac{V_S}{R_2}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)+\frac{V_S}{R_3{R}_1}\right]\end{array}$

The second determinant is given by,

  $\begin{array}{c}{\Delta }_2=\left[\begin{array}{cc}\frac{1}{R_2}+\frac{1}{R_1}& I_S+\frac{V_S}{R_2}\\ -\frac{1}{R_1}& \frac{V_S}{R_3}\end{array}\right]\\ =\left[\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\frac{V_S}{R_3}+\left(I_S+\frac{V_S}{R_2}\right)\frac{1}{R_1}\right]\end{array}$

The expression to determine the node voltage $V_1$ is given by,

  $V_1=\frac{{\Delta }_1}{\Delta }$

Substitute $\left(I_S+\frac{V_S}{R_2}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)+\frac{V_S}{R_3{R}_1}$ for ${\Delta }_1$ and $\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}$ for $\Delta$ in the above equation.

  $V_1=\frac{\left(I_S+\frac{V_S}{R_2}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)+\frac{V_S}{R_3{R}_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$

The expression for the node voltage $V_2$ is given by,

  $V_2=\frac{{\Delta }_2}{\Delta }$

Substitute $\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\frac{V_S}{R_3}+\left(I_S+\frac{V_S}{R_2}\right)\frac{1}{R_1}$ for ${\Delta }_2$ and $\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}$ for $\Delta$ in the above equation.

  $V_2=\frac{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\frac{V_S}{R_3}+\left(I_S+\frac{V_S}{R_2}\right)\frac{1}{R_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$

Conclusion:

Therefore, the required node voltage equations are $V_1=\frac{\left(I_S+\frac{V_S}{R_2}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)+\frac{V_S}{R_3{R}_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$ and $V_2=\frac{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\frac{V_S}{R_3}+\left(I_S+\frac{V_S}{R_2}\right)\frac{1}{R_1}}{\left(\frac{1}{R_2}+\frac{1}{R_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_4}\right)-\frac{1}{R_1^2}}$ .