Electrical – Thevenin Equivalent Network: How to treat Capacitor across 0v potential, with Dependent Sources

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.

It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for \$V_{ab}\$ in terms of \$i_{\Delta}\$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

Generally speaking, to find the Thevenin equivalent of a circuit with only dependent source(s), you must be sure to "activate" the dependent source(s) with a test source.

This is what was done above. We solved for the voltage across the port due to a test current source, \$I_{ab}\$ which, in this case, equals the controlling variable thus making this problem particularly easy to solve.