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Im not sure where to start with this question. I dont know what to do with the dependent source and im not sure what I can do with terminals A and B.

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3 Answers 3

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  • To find \$R_{th}\$, connect a fictitious voltage source of 1V between A and B, so that \$V_x = 1\$

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Now find I using KVL, $$-2 + 6 - 1- 2I = 0$$ which means, \$ I = 1.5 A \$

Therefore, the thevenins resistance \$R_{th}\$ seen from AB will be just \$ V_x/I = 1/1.5 = 0.667 ohms \$

  • To find \$ V_{th}\$, look at the original ckt. It is the open circuit voltage across AB = \$V_x\$

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\$V_x\$ should be equal to \$-2V_x + 6\$ because current through 2 ohms = 0.

$$ V_x = -2V_x + 6 $$ $$\implies V_x = 2V = V_{th} $$

  • Now you can draw your thev.eq.ckt using \$ R_{th} = 0.667A \$ and \$ V_{th} = 2V \$
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  • \$\begingroup\$ Why not choose a fictitious voltage source of 0V (short circuit across A and B)? That'd make it much easier. See my solution. \$\endgroup\$
    – Curd
    Feb 3, 2018 at 11:04
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schematic

simulate this circuit – Schematic created using CircuitLab

You should start as usually by finding \$V_{TH}\$ voltage using any technique you know.

You should start as usually by finding \$V_{THE}\$ voltage using any technique you know.

\$V_{TH} = V_X+V1 = V_X+6V\$

So all we need is ti find \$V_X\$ voltage. So we can write a KVL equation:

$$V_2-I1*R_2+2V_{AB} = 0 $$

$$V_2-I1*R_2+2(V_X+V_1) = 0 $$

$$V_2-I1*R_2+2((V_2 - I_1R2)+V_1) = 0 $$

$$I = \frac{2V_1 + 3V_2}{3R_2} $$

And the Thévenin voltage is:

$$V_{TH} = V_2 - I_1 R_2 + V_1 = V_2 - \frac{2V_1 + 3V_2}{3R_2} R_2 + V_1 = \frac{V_1}{3} = 2V $$

Next we need to find the \$R_{TH}\$ resistance using this circuit diagram.

schematic

simulate this circuit

As you can see I short A and B terminals. And I want to find this short circuit current \$I_{SC}\$

Additional notice that since we have a short across A-B terminal \$V_{TH} = 0V\$ hence \$V_X = 0A\$ So, the \$I_{SC}\$ current is:

$$I_{SC} = \frac{V_1}{R_1} = 3A$$

And the \$R_{TH}\$ resistance is

$$R_{TH} = \frac{V_{TH}}{I_{SC}} \approx 0.667 \Omega$$

And the equivalent circuit will look like this

schematic

simulate this circuit

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    \$\begingroup\$ It'd be nice if the downvoter would explain what he/she considers wrong with this answer? \$\endgroup\$
    – Curd
    Feb 3, 2018 at 11:58
  • \$\begingroup\$ Wonder why -1 ? O.o might be some personal vendetta going on. \$\endgroup\$
    – Mitu Raj
    Feb 3, 2018 at 16:27
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There are several ways to find the Thevenin equivalent.
Here is one that makes it particularyly easy to solve this problem:

Find short circuit current and find open circuit voltage:

Note: you can ignore completley the left subcircuit (4V voltage source and 6\$\Omega\$ resistor) as the voltage across its terminals is completely determined by the dependent voltage source and the dependent voltage source does not depend on any quantities of that left subcircuit.

open circuit voltage (i.e. \$i_X=0\$):

\$v_{oc} = -2v_X + 2\Omega \cdot i_X + 6V = -2v_{oc} + 6V\$ → \$v_{oc}=2V\$

short circuit current (i.e. \$v_X=0\$):

\$i_{sc} = \frac{6V - v_X}{2\Omega} = \frac{6V - 0V}{2\Omega} = 3A\$

So you get
\$v_{Th} = v_{oc}\ = 2V\$ and
\$R_{Th}=\frac{v_{oc}}{i_{sc}}=\frac{2}{3}\Omega\$

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