The output current value cited is the average current of a sine wave, which mathemagically is two over pi times the peak value. This can be proven by integrating I * sin(x) from zero to pi, multiplying by 2 to mimic the rectification (both peaks in the same direction) and dividing by 2 * pi since we're dealing with a whole AC cycle, and we get our 2 over pi factor.

EDIT: A derivation (calculus = rusty, apologizes for glaring errors)

\$ I_o = 2\dfrac{\int\limits_{0}^{\pi}{I_{ac}}^{pk}\cdot \textrm{sin}(x) \textrm{d}x}{2\pi} \$

\$ I_o = 2\dfrac{{I_{ac}}^{pk}\cdot \Big(-\textrm{cos}(\pi) - \big(-\textrm{cos}(0)\big)\Big)}{2\pi}\$

\$ I_o = 2\dfrac{{I_{ac}}^{pk}\cdot \Big(-(-1) + (1)\Big)}{2\pi}\$

\$ I_o = 2\dfrac{2{I_{ac}}^{pk}}{2\pi}\$

\$ I_o = \dfrac{2{I_{ac}}^{pk}}{\pi}\$

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.

## Best Answer

This is a recursive infinite series problem. You can start by considering only one strip with two rows of horizontal resistors and one row of vertical between them. Pick one of the vertical resistor and imagine the current that flows when 1V is put accross it. Note that the infinite strips on either side will contribute to the same current and are independent, so you solve for only one side and double the result. Now write the infinite series with a new term each new vertical resistor. With that you can solve the original infinite strip. Similar technique, althoug a bit more complicated, will let you add a infinite number of strips.

Numerically, you could make this a sufficiently large grid so that current in the edge resistors are below so small as to not matter for the desired accuracy. Now put a fixed voltage accross one resistor and perform iterative relaxation on the voltage and currents in the grid. After this settles sufficiently, you have the total current drawn when 1V is applied accross a single resistor, from which you can compute its the effective resistance.