Let us write the mesh equations by inspection for the network in Fig. 3.21. Then we will use MATLAB to solve for the mesh currents.

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Accepted Answer

The three linearly independent simultaneous equations are

(4k + 6k)[latex]I_{1} - (0)I_{2} - (6k)I_{3}[/latex] = –6
-(0)[latex]I_{1} + (9k + 3k)I_{2} - (3k)I_{3}[/latex] = 6
-(6k)[latex]I_{1} - (3k)I_{2} + (3k + 6k + 12k)I_{3}[/latex] = 0

or in matrix form

[latex]\begin{bmatrix} 10k & 0 & -6k \ 0 & 12k & -3k \ -6k & -3k & 21k \end{bmatrix}\begin{bmatrix} I_{1} \ I_{2} \ I_{3} \end{bmatrix} = \begin{bmatrix} –6 \ 6 \0 \end{bmatrix}[/latex]

Note the symmetrical form of the equations. The general form of the matrix equation is

RI = V

and the solution of this matrix equation is

I = [latex]R^{-1}[/latex]V

Performing the indicated operation yields the following loop currents:

[latex]i_{1}[/latex] = -0.6757 mA
[latex]i_{2}[/latex] = 0.4685 mA
[latex]i_{3}[/latex] = -0.1261 mA

Question 3.13: Let us write the mesh equations by inspection for the networ...

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