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circuit

i have a problem with calculating the thevening resistance \$R_{th}\$ . Since the circuit has only dependent sources the thevening voltage \$V_{th}\$ = 0.
because of this you can't calculate \$R_{th}\$ like you would do with only independent sources.
I think the next step is adding a voltage source between point A and B and than do nodal analysis to find the current in point A. After that you can use Ohms law to calculate \$R_{th}\$
i have been trying this for a hour with no succes, Any help on how to solve a thevenin circuit like this is greatly appreciated.

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Since the circuit has only dependent sources the thevening voltage \$V_{th} = 0\$.

Correct.

because of this you can't calculate \$R_{th}\$ like you would do with only independent sources.

It's correct that in this case the short-circuit current doesn't give you an independent point on the I-V curve from which to determine \$R_{th}\$. But you can use any two points on the network's I-V curve to determine the slope, whether the sources involved are dependent or independent.

I think the next step is adding a voltage source between point A and B and than do nodal analysis to find the current in point A. After that you can use Ohms law to calculate \$R_{th}\$

This approach will work. You don't just need Ohm's law, you need the rules for combining series and parallel resistors (which you could in principle derive from Ohm's law). It would also be valid to use nodal or mesh analysis to determine the current flowing in to node A.

Any help on how to solve a Thevenin circuit like this is greatly appreciated.

You can solve this problem in general using mesh or nodal analysis.

For this particular circuit, you can take a shortcut. Notice that the dependent source is connected in parallel with the branch where its controlling current (\$I_x\$) is taken from. Compare these two circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

These circuits are equivalent if $$R_2 = \frac{R_1}{1+g}$$ You should work the math yourself to understand this.

With this substitution and the usual rules for combining series and parallel resistors, you can find \$R_{th}\$ in your problem.

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  • \$\begingroup\$ Thanks for the help, tomorrow i will give this a shot. (Going to sleep now) \$\endgroup\$ – Lethalbeast Sep 27 '15 at 0:49

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