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Considering the following circuit

schematic diagram

When calculating the Thevenin equivalent, I calculate the Thevenin resistance value Rth.

Why do I have to consider the controlled voltage generator? And furthermore, why can't I eliminate the current source I1 and why does it gets replaced by an 1A current source in this example? Is it in similar cases always true that the controlled generators are not to be eliminated and the other generators (current and voltage) are thus to be replaced by unitary generators?

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Why do I have to consider the controlled voltage generator?

Think of it this way, if you connect a test source across A and B terminals, the test source cannot affect the current through the independent source - that's why it's called an independent source - it's value does not depend on the attached circuit in any way.

However, the voltage across E1 will, in general, be affected by the test source and, thus, the equivalent resistance seen by the test source is modified by the presence of the dependent source.

And thereby why I cannot eliminate the current source I1 and in the example in question it gets replaced by an 1A current source?

If the 1A current source mentioned is, in fact, the test source, you should zero the 5A source to find the Thevenin resistance of the circuit.

With the 5A source activated, there will be an open circuit voltage, \$V_{AB_{(OC)}}\$. When you connect the test source, the voltage \$V_{AB}\$ will be different from the open circuit voltage. To find the Thevenin resistance, take the difference in the voltages and divide by the test source current.

But, you get the same result if you simply zero the 5A source which sets the open circuit voltage to zero. Then, you get the Thevenin resistance directly from the value of voltage across the test source.

In other cases it is true to assert that the controlled generators are not to be eliminated and so the other generators (current and voltage) are to be replaced by unitary generators?

I honestly don't know where this idea comes from. A unitary generator is typically used as a test source but I'm not aware of any reason to replace the other sources. Perhaps you should expand this question a bit. I suspect there's a misunderstanding here.

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  • \$\begingroup\$ What I didn't understand is why and when you consider using a test source, in my example the 5A current source gets replaced by a 1A current test source, is this always necessary when controlled generators are present? what about if instead of the current source I would have had an voltage source ? \$\endgroup\$ – corsibu Nov 19 '12 at 15:01
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    \$\begingroup\$ It's not correct to think of replacing the 5A source with the test source. I think this is where the confusion is. When finding the Thevenin resistance, all independent sources should be zeroed (current sources are opens, voltage sources are shorts). Then, the test source is connected and the voltage (in the case of a current test source) or current (in the case of a voltage test source) is calculated to find the Thevenin resistance. For this circuit, when the 5A source is zeroed and the test source connected, it only appears that the test source replaced the 5A source. \$\endgroup\$ – Alfred Centauri Nov 19 '12 at 16:37
  • \$\begingroup\$ Yes it was the example to create confusion, so independently from the dependent source I chose a voltage or current source depending on what my requisites are, for example if I must find Vth I chose a current source ? \$\endgroup\$ – corsibu Nov 19 '12 at 16:46
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    \$\begingroup\$ The test source, whether a current or voltage source, is used to find the Thevenin or Norton resistance. To find Vth, do not use a test source. Vth is just the open circuit voltage across the terminals of interest. \$\endgroup\$ – Alfred Centauri Nov 19 '12 at 16:49
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    \$\begingroup\$ Yes, I've mistyped that. Thanks a lot for your post, you were very exhaustive. \$\endgroup\$ – corsibu Nov 19 '12 at 16:58

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