Let us apply what we have just learned to write the equations forth e network in Fig. 3.7 by inspection. Then given the following parameters, we will determine the node voltages using MATLAB: R1 = R2 = 2 kΩ, R3 = R4 = 4 kΩ, R5 = 1 kΩ, iA = 4 mA, and iB = 2 mA.

Question

Let us apply what we have just learned to write the equations forth e network in Fig. 3.7 by inspection. Then given the following parameters, we will determine the node voltages using MATLAB: [latex]R_1 = R_2[/latex] = 2 kΩ, [latex]R_3 = R_4[/latex] = 4 kΩ, [latex]R_5[/latex] = 1 kΩ, [latex]i_A[/latex] = 4 mA, and [latex]i_B[/latex] = 2 mA.

Accepted Answer

The equations are

[latex]v_{1}\left(\frac{1}{R_{1}}+\frac{1}{R_{2}} ight)-v_{2}(0)-v_{3}\left(\frac{1}{R_{1}} ight) =-i_{A} [/latex]

[latex]-v_{1}(0)+v_{2}\left(\frac{1}{R_{3}}+\frac{1}{R_{4}} ight)-v_{3}\left(\frac{1}{R_{4}} ight) =i_{A}-i_{B} [/latex]

[latex]-v_{1}\left(\frac{1}{R_{1}} ight)-v_{2}\left(\frac{1}{R_{4}} ight)+v_{3}\left(\frac{1}{R_{1}}+\frac{1}{R_{4}}+\frac{1}{R_{5}} ight) =0[/latex]

which can also be written directly in matrix form as

[latex]\left[\begin{array}{ccc}\frac{1}{R_{1}}+\frac{1}{R_{2}} & 0 & -\frac{1}{R_{1}} \\quad\0 & \frac{1}{R_{3}}+\frac{1}{R_{4}} & -\frac{1}{R_{4}} \\quad\-\frac{1}{R_{1}} & -\frac{1}{R_{4}} & \frac{1}{R_{1}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}\end{array} ight]\left[\begin{array}{l}v_{1} \\quad\v_{2} \\quad\v_{3}\end{array} ight]=\left[\begin{array}{c}-i_{A} \\quad\i_{A}-i_{B} \\quad\0\end{array} ight][/latex]

Both the equations and the G matrix exhibit the symmetry that will always be present in circuits that contain only resistors and current sources.
If the component values are now used, the matrix equation becomes

[latex]\left[\begin{array}{ccc}\frac{1}{2 \mathrm{k}}+\frac{1}{2 \mathrm{k}} & 0 & -\frac{1}{2 \mathrm{k}} \\quad\0 & \frac{1}{4 \mathrm{k}}+\frac{1}{4 \mathrm{k}} & -\frac{1}{4 \mathrm{k}} \\quad\-\frac{1}{2 \mathrm{k}} & -\frac{1}{4 \mathrm{k}} & \frac{1}{2 \mathrm{k}}+\frac{1}{4 \mathrm{k}}+\frac{1}{1 \mathrm{k}}\end{array} ight]\left[\begin{array}{c}v_{1} \\quad\v_{2} \\quad\v_{3}\end{array} ight]=\left[\begin{array}{c}-0.004 \\quad\0.002 \\quad\0\end{array} ight][/latex]

or

[latex]\left[\begin{array}{ccc}0.001 & 0 & -0.0005 \\quad\0 & 0.0005 & -0.00025 \\quad\-0.0005 & -0.00025 & 0.00175\end{array} ight]\left[\begin{array}{l}v_{1} \\quad\v_{2} \\quad\v_{3}\end{array} ight]=\left[\begin{array}{c}-0.004 \\quad\0.002 \\quad\0\end{array} ight][/latex]

If we now employ these data with the MATLAB software, the computer screen containing the data and the results of the MATLAB analysis is as shown next.

Question 3.2: Let us apply what we have just learned to write the equation...

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