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I think I'm overlooking something very basic in this problem but I'm having trouble determining what that is! I'm pretty sure the issue is stemming from my equation for mesh 1, but the assumption I made about what \$I_x\$ is might also be incorrect.

I got the following equations:

$$ 4I_x + I_1(1-j) - I_2 + I_3 = 0 $$ $$ -I_1 + 2I_2 = -4 $$ $$ I_1(j) + I_3(1-j) = 4 $$

I also assumed that: \$I_x = I_2\$

I should get \$V_0 = 6.32 V\$, angle -18.43 degrees as my final answer, but I keep getting values that are way off!

enter image description here

The circuit diagram shows an AC steady-state circuit with three principal nodes and several components. It contains a current-dependent current source on the left with a value of \$4I_x\$​, where \$I_x\$​ is the current flowing down through a 1 Ω resistor on the bottom left. The circuit also includes three resistors (1 Ω, 1 Ω, and 1 Ω), a capacitor with an impedance of −j1 Ω, and an AC voltage source with a value of 4∠0∘ V placed in the lower-middle section. The voltage to be found, \$V_o\$​, is defined across the rightmost 1 Ω resistor, which is connected between the top and bottom rails of the right-hand side of the circuit.

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  • \$\begingroup\$ I see that you want understanding; however, it is not possible unless you show your working. \$\endgroup\$ Commented Oct 1 at 11:17

4 Answers 4

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Here's the circuit and mesh conventions I chose (with a hope that it was close enough to your choices):

enter image description here

(In the above, I've included your equations as I've tried to assign them to my loops. I don't agree with your equations if I got the loop assignments assigned correctly. But I kept them in the above image for clarity, not accuracy.)

The three mesh equations, and the dependent source equation, are then:

$$\begin{align*} 0\:\text{V}+v_{\small{4\, ix}}-\left(-1\,j\:\Omega\right)\cdot\left(i_1-i_3\right)-1\:\Omega\cdot\left(i_1-i_2\right)&=0\:\text{V}\tag{$i_1$} \\\\ 0\:\text{V}-1\:\Omega\cdot i_2-1\:\Omega\cdot\left(i_2-i_1\right)-4\:\text{V}&=0\:\text{V}\tag{$i_2$} \\\\ 0\:\text{V}+4\:\text{V}-\left(-1\,j\:\Omega\right)\cdot\left(i_3-i_1\right)-1\:\Omega\cdot i_3&=0\:\text{V}\tag{$i_3$} \\\\ i_1&=4\cdot i_2\tag{$4\,i_x$} \end{align*}$$

This solves out as: \$i_1= 8\:\text{A}\$, \$i_2= 2\:\text{A}\$, \$i_3= \left(6 - 2\,j\right)\:\text{A}\$, and \$v_{\small{4\, ix}}= \left(8 - 2\,j\right)\:\text{V}\$.

Assuming \$v_{o_-}=0\:\text{V}\$ then it follows that \$v_{o_+}=1\:\Omega\cdot\left(6 - 2\,j\right)\:\text{A}=\left(6 - 2\,j\right)\:\text{V}\$.

Which works out to:

abs(6 - 2*j).n()
6.32455532033676
(arg(6 - 2*j) * (360/(2*pi))).n()
-18.4349488229220

or \$\approx 6.325\:\text{V}\:\angle -18.435^\circ\$.

I'll let you examine my equations and mesh current assignments to spot differences.

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Answer is ok ... Equations (?).

Here is a simulation with microcap v12 (Dynamic Analysis AC)

enter image description here

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I present to you my network analysis based on network topology:

enter image description here

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analysis 1

analysis2

here is the analysis. I trust this will help.

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