The network in Fig. 3.27 contains both a current-controlled voltage source and a voltage-controlled current source. Let us use MATLAB to determine the loop currents.

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Accepted Answer

The equations for the loop currents shown in the figure are

[latex]I_{1} = \frac{4}{k}[/latex]
[latex]I_{2} =\frac{V_{x}}{2k}[/latex]
-1k[latex]I_{x} + 2k(I_{3} - I_{1}) + 1k(I_{3} - I_{4})[/latex] = 0
1k[latex](I_{4} - I_{3}) + 1k(I_{4} - I_{2})[/latex] + 12 = 0

where

[latex]V_{x} = 2k(I_{3} - I_{1})[/latex]
[latex]I_{x} = I_{4} - I_{2}[/latex]

Combining these equations yields

[latex]I_{1} =\frac{4}{k}[/latex]
[latex]I_{1} + I_{2} - I_{3}[/latex] = 0
1k[latex]I_{2} + 3k I_{3} - 2k I_{4}[/latex] = 8
1k[latex]I_{2} + 1k I_{3} - 2k I_{4}[/latex] = 12

In matrix form the equations are

[latex]\begin{bmatrix} 1 & 0 & 0 & 0 \ 1 & 1 & - 1 & 0\ 0 & 1k & 3k & - 2k \ 0 & 1k & 1k &- 2k \end{bmatrix}\begin{bmatrix} I_{1} \ I_{2}\ I_{3}\I_{4}\end{bmatrix} = \begin{bmatrix} \frac{4}{k} \ 0 \ 8 \ 12 \end{bmatrix}[/latex]

The equations are in the form RI = V, and the solution to I = [latex]R^{-1}[/latex]V is

[latex]i_{1}[/latex] = 4.0 mA
[latex]i_{2}[/latex] = 6.0 mA
[latex]i_{3}[/latex] = -2.0 mA
[latex]i_{4}[/latex] = -1.0 mA

Question 3.19: The network in Fig. 3.27 contains both a current-controlled ...

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