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I cannot find the Zth in the circuit i uploaded; i calculated the correct value of Vth with the following steps:

LKC @ N1: 5 + (0.2)Vo = -Vo/(8+4j)

LKV @ the outer loop: Vth + Vo - (4-2j)*0.2 - Vo = 0

And the Thevenin Voltage is exactly 7.35 L(72.9°).

At this point I usually connect the two terminals (a, b) and try to find the short circuit current (i put a visual reference in the picture) using the node method (or the loop method) and use the formula Vth/Isc = Zth, but nothing seems to work! Also adding the SC makes the circuit look really weird, as all the "block" on the right can be seen as a single node. Any ideas to find the Zth? The solutions are in the picture. Thank you :)

enter image description here

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  • \$\begingroup\$ What is Vo relative to? (a) or (b) or somewhere else? I can't tell. Normally, I'd just set b=0 by definition and consider Vo relative to that. But it might be from your N2 to N1, which is what I think you mean. \$\endgroup\$
    – jonk
    Aug 31, 2016 at 20:25
  • \$\begingroup\$ The Vo is the voltage to which the dependent source refers (0.2Vo); from the picture you can see it's the voltage across the (8 + 4j) impedance; the voltage between 'a' and 'b' is Vth in the relations I wrote. Yes if i could simply ignore that Vo it would be way easier ;; \$\endgroup\$
    – DSimow
    Aug 31, 2016 at 20:55
  • \$\begingroup\$ Think of a short-circuit as a voltage source of 0V. Does that help you with resolving the issue of everything on the right side becoming one node? \$\endgroup\$
    – rioraxe
    Aug 31, 2016 at 20:59
  • \$\begingroup\$ Yeah it's pretty helpful to perform a nodal analysis... but i get stuck anyway :(. I've tried every combination with nodal analysis and I think the correct method in this case is mesh analysis over the three loops, in fact: 1. I could consider [Isc] the current flowing in the right loop; 2. the voltage between 'a' and 'b' is 0. Applying simple LKV rules I get to a pretty manageable linear system... But the closest result I get to the book's anwser is: Zth = 4.47L(172.3°) Isc = (-0.27 - 1.62j) At this point the only thing that could help me is someone actually trying to solve it x) \$\endgroup\$
    – DSimow
    Aug 31, 2016 at 22:46
  • \$\begingroup\$ Current at N1: Vo/(8+4j) + 5 + Vo/(4-2j) = 0. Current through bottom leg (same as current at b): Vo/(4-2j) = 0.2Vo + Isc. My definition of Isc here is from b to a. \$\endgroup\$
    – rioraxe
    Aug 31, 2016 at 23:13

2 Answers 2

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With node a and b shorted together --

Current going into node N1:
$$ V_o/(8+4j) + 5 + V_o/(4-2j) = 0$$ $$ \Rightarrow V_o = -16.2 + 2.7j $$ Current going into node b (define \$I_{SC}\$ going from a to b): $$ V_o/(4-2j) - (0.2V_o - I_{SC}) = 0 $$ $$ \Rightarrow I_{SC} = 0.27 + 1.62j $$ Using \$V_{Th}\$ (which I can duplicate) as given: $$ Z_{Th} = \frac{V_{Th}}{I_{SC}} = 4.47\angle {-7.63}^\circ $$ I cannot duplicate \$Z_{Th}\$ as given.

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    \$\begingroup\$ the first should be node N2 KCL. \$\endgroup\$
    – matzeri
    Sep 1, 2016 at 8:40
  • \$\begingroup\$ @matzeri If you recognized that there are two straight forward ways to represent the current through the bottom leg (as in the second equation), KCL for N1 or N2 can be written down to be exactly the same directly. I chose the Vo way because it saved one step. \$\endgroup\$
    – rioraxe
    Sep 1, 2016 at 21:02
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You could add a test source at the output instead of trying to solve by means of the short circuit current.

For the test source method, you have to turn off the independent sources. So your circuit will look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Your Thevenin impedance is \$Z_{\mathrm{TH}}=\dfrac{V_t}{i_t}\$, where \$V_t\$ and \$i_t\$ are your test source and current source respectively.

Using KCL:

$$i_a+0.2V_o =i_t$$

$$\dfrac{0-V_t}{4\Omega-j2\Omega+8\Omega+j4\Omega}+0.2(8\Omega+j4\Omega)\bigg(\dfrac{0-V_t}{4\Omega-j2\Omega+8\Omega+j4\Omega}\bigg)=i_t $$

$$ \dfrac{-V_t}{12\Omega+j2\Omega}-0.2(8\Omega+j4\Omega)\bigg(\dfrac{V_t}{12\Omega+j2\Omega}\bigg)=i_t$$

$$ \dfrac{V_t}{12\Omega+j2\Omega}\bigg[-1-0.2(8\Omega+j4\Omega)\bigg]=i_t$$

$$ \dfrac{V_t}{12\Omega+j2\Omega}\bigg[-2.6\Omega-j0.8\Omega\bigg]=i_t$$ $$ \dfrac{V_t}{i_t}=\dfrac{12\Omega+j2\Omega}{-2.6\Omega-j0.8\Omega}$$

$$ \dfrac{V_t}{i_t}=Z_{\mathrm{TH}}=4.47\angle{172.36^{\mathrm{o}}}$$

Now, this answer is different from the one your are given. The only way I can see you getting the \$12.166\angle{136.6^\mathrm{o}}\$ answer is if you reversed the polarity of \$V_o\$, I am very confident that's where the sign affect their answer.

I see in one of your comments that you tried a circuit solver and got the following for \$i_{sc}\$

$$ i_{sc}=-0.27-j1.62=1.64\angle{-99.46}$$

This could be used to validate the answer I obtained:

$$Z_{\mathrm{TH}}=\dfrac{V_{\mathrm{TH}}}{i_{sc}}=\dfrac{7.35\angle{72.9}}{1.64\angle{-80.54}}=4.48\angle{172.36}$$

That should be the answer.

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