Let us find Vo in the circuit in Fig. 3.25, which contains a voltage-controlled voltage source.

Question

Let us find [latex]V_{o}[/latex] in the circuit in Fig. 3.25, which contains a voltage-controlled voltage source.

Accepted Answer

The equations for the loop currents shown in the figure are

- 2[latex]V_{x} + 2k(I_{1} + I_{2}) + 4k I_{1}[/latex] = 0
- 2[latex]V_{x} + 2k(I_{1} + I_{2}) - 3 + 6k I_{2}[/latex] = 0

where

[latex]V_{x} = 4k I_{1}[/latex]

These equations can be combined to produce

- 2k [latex]I_{1} + 2k I_{2}[/latex] = 0
- 6k [latex]I_{1} + 8k I_{2}[/latex] = 3

These equations can be placed in the form RI = V, where

[latex]R = \begin{matrix} - 2000 & 2000 \ - 6000 & 8000 \end{matrix}\qquad I = \begin{matrix} i_{1}\ i_{2} \end{matrix} \qquad V = \begin{matrix} 0 \ 3 \end{matrix}[/latex]

The solution is I = [latex]R^{-1}[/latex]V, and this operation yields

[latex]i_{1}[/latex] = 1.5 mA
[latex]i_{2}[/latex] = 1.5 mA

and therefore,

[latex]V_{o} = 6k I_{2}[/latex] = 9 V

For comparison, we will also solve the problem using nodal analysis. The presence of the voltage sources indicates that this method could be simpler. Treating the 3-V source and its connecting nodes as a supernode and writing the KCL equation for this supernode yields

[latex]\frac{V_{x} - 2V_{x}}{2k}+\frac{V_{x}}{4k}+\frac{V_{x}+ 3}{6k}[/latex] = 0

where

[latex]V_{o} = V_{x}[/latex] + 3

These equations also yield [latex]V_{o}[/latex] = 9 V.

Question 3.17: Let us find Vo in the circuit in Fig. 3.25, which contains a...

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