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I'm trying to solve this problem, and i seem to be unable to solve it, part of the reason being that the there is a dependent current source in here, so I can't find the Thevenin equivalent resistance and voltage, I did find out the Ix current value of 5 Amps tho.

schematic

simulate this circuit – Schematic created using CircuitLab

So I would like to ask, if anyone knows the steps to simplifying a circuit with dependent current sources, independent voltage sources and resistors, into a Thevenin equivalent circuit.

I tried to use Kirchoff's current law, to find voltage at nodes and then using that I calculated the value of the current flow, but I am not able to find out how to find the Thevenin equivalent resistance of the circuit, which is the very last step I need to take to get the Thevenin equivalent of the circuit I'm trying to simplify.

Got the circuit now! The Question was: Find the Thevenin equivalent of the circuit to the left of terminals A and B (do not include the load resistor RL)

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  • \$\begingroup\$ You can draw your circuit by editing your question and hitting Ctrl-M \$\endgroup\$ – jippie Jun 7 '13 at 21:11
  • \$\begingroup\$ finally got the circuit on the post, i managed to calculate using nodal analysis and got the Ix value of 5Amps, not sure if im right but i dont know how to continue from here \$\endgroup\$ – Liquified Jun 8 '13 at 1:01
  • \$\begingroup\$ The solution depends on which direction you're measuring Ix. You need to draw an arrow showing which direction is considered "positive current". \$\endgroup\$ – Dave Tweed Jun 8 '13 at 2:08
  • \$\begingroup\$ @DaveTweed My Ix current is flowing downwards from the 1 ohm resistor above the above \$\endgroup\$ – Liquified Jun 8 '13 at 13:40
  • \$\begingroup\$ I put in the direction for Ix using a Unicode arrow character, and labeled the components for easier discussion. \$\endgroup\$ – Kaz Jun 8 '13 at 16:51
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I think Thevenin (or Norton) equivalent circuits do not consider variable sources. The same refers to non-linear resistors (and other elements in AC scope). But I understand what you mean: you would like to have something like these.

In your case you should first select all the elements that are not dependent on other and do not alter other elements, and simplify them. The next step is to find all independent voltage/current sources.

Now combine non-linear static elements, like resistors. The combination of a linear object and a non-linear object is also non-linear object (but there is a theoretical possibility that two non-linear functions make a linear one).

At this moment you get: combined resistances that are (generally speaking) non-linear and do not alter anything and independent and dependent sources, and the elements that alter sources. If possible, combine independent sources.

That's the hardest task now: to combine independent sources with dependent. The Kirchoff's laws might be necessary here.

UPDATE

According to your circuit, this is not that difficult as it seems on the first sight. Please forgive me there are no exact calculations as I did them last time almost 20 years ago...

First of all, take a look at the non-ideal current source I1. Because it has R1 in parallel you can convert it to a non-ideal voltage source, which has resistance in series. This voltage source would have internal resistance 1 Ohm too and voltage R1 * 4Ix that is 4*Ix volts as R1 = 1 Ohm. I will name this new source as V2.

At the moment on the left side of the circuit you have non-ideal voltage source V2 (equivalent to I1 current source), its internal resistance (equivalent to R1), than voltage source V1. The R1 resistance is gone as it became internal load of voltage source. More reading about source transformation.

Because in the same branch there are two voltage sources you can combine them. So it is E = V1 + V2 which leads to (4 Ix - 10) V (- because V1 is in opposition to V2).

Now we have the first part of our task, the source. Now we're going to find equivalent resistance, and, moreover, we need to drive out Ix from source equation, because after combining resistances to one there will be no Ix.

As we know from Mr. Kirchoff, the load current (the one in R3), say I, divides in two: Ix and IL (IL flows through R3). The Ix is U2 / R2 and IL is U2 / (R3 + RL). You can write down proper equations yourself :).

Now you can find relation between Ix and IL (you need IL in equation of voltage source) and make E function of IL. If this source is no more function of Ix, you can combine other resistances to one equivalent. Do not forget source E internal resistance (the one driven from R1).

Please note that this method will lead you to have voltage source that is a function of load current (so in fact load resistance RL). This is normal as U2 depends on this load (that's why I've written at the very beginning it is not true Thevenin method).

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  • \$\begingroup\$ thanks for the advice, i am still a little confused though, do you mind giving me a walkthrough on the circuit i just added onto my post? i have manage to calculate a value of Ix of 5Amps using nodal analysis @Voitcus \$\endgroup\$ – Liquified Jun 8 '13 at 13:47
  • \$\begingroup\$ @Liquified I will help you on the late evening (I live in the CET timezone) \$\endgroup\$ – Voitcus Jun 8 '13 at 14:52
  • \$\begingroup\$ @Liquified I updated answer. I hope this guide will help you somehow. \$\endgroup\$ – Voitcus Jun 8 '13 at 20:58
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Try solving it for a Norton equivalent first (current source with resistance across it). This can be easily converted to the Thevenin equivalent (voltage source, series resistor). Open circuit voltage, Short circuit current, calculate resistance. Simples!

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  • \$\begingroup\$ Mathematically there is no difference between Thevenin and Norton like there is no difference between impedance and admittance. The solving method are exactly the same and are driven from the same basis. \$\endgroup\$ – Voitcus Jun 7 '13 at 21:10
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Imagine that RL is removed: the load is open.

The voltage source has an impedance of 0 ohms, and so R1 and R2 are in parallel with current source I1. They are equal and so the 4Ix current of I1 splits equally between R1 and R2, which means that the R2 current must be 2Ix. But, oops, the schematic asserts that the current is in fact Ix.

What this means is that we cannot remove RL from the circuit.

And what does that tell us? If we have a circuit which blows up when we remove the load, its equivalent circuit must be this:

schematic

simulate this circuit – Schematic created using CircuitLab

Circuits that abhor having their their loads removed are current sources. We cannot model those as Thévenin, because a current source is a very large VTH voltage with a very large RTH. For instance a 5A current source is basically \$\displaystyle\lim_{x\to\infty}\frac{5x V}{x \Omega}\$: a voltage that is as large as possible, with a resistor of 1/5th that voltage in series.

If the device terminals are a current source, it is probably best to just leave it as a current source and just document the amperage value.

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  • \$\begingroup\$ I'm at the moment not sure, but I think that Norton method, which you use instead of Thevenin, requires the current source to be non-ideal, having some non-zero admittance (or conductivity in this case). If there were any, a current source can be transformed to a voltage source and vice versa. Here you have none. I also had the same thinking that this RL branch could not be removed as voltage source cannot be depending on itself; in my answer I've managed to do this in general way, without any calculations; is my way of thinking wrong? \$\endgroup\$ – Voitcus Jun 8 '13 at 21:33

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