Electronic – How to calculate thevenin’s voltage for this circuit

Dependent sources typically will change the Thevenin resistance* so, if you zero the dependent source and combine the resistors into an equivalent, you typically will get the wrong answer.

Since your first schematic has two ports, it's not clear which one you're finding the Thevenin equivalent of but, for example, let's assume it's the 2nd port.

Now that you've found the open circuit voltage, \$U_i\$, there are two approaches to finding the Thevenin resistance \$R_i\$.

One approach is to place a wire across the 2nd port and calculate the current through this wire, the short-circuit current \$I_{SC}\$. Then

$$R_i = \dfrac{U_i}{I_{SC}} $$

A second approach is to find the Thevenin resistance directly by zeroing the independent source \$u_1\$, connecting a 1A current source (the test source) to port 2, and calculating the voltage across the test source \$U_S\$. Then

$$R_i = \dfrac{U_S}{1A}$$

The test source will "activate" the dependent source and so its effect on the Thevenin resistance will become apparent.

*For example, consider the equivalent resistance of the following dependent source in parallel with a resistor:

^{simulate this circuit – Schematic created using CircuitLab}

The dependent current source produces a current that is k times the resistor current.

If I place a 1V test source across this circuit, the source current will be

$$I_S = (1 + k)I_{R1} = (1 + k)\dfrac{1V}{100}$$

The equivalent resistance seen by the test source is just

$$R_{EQ} = \dfrac{V_S}{I_S} = \dfrac{100}{1 + k}\Omega $$

So, the effect of the dependent source has been to make the resistor appear smaller by factor of \$ \dfrac{1}{1 + k}\$.

Since the bottom node is your reference node, Vx = Vb. Now you've got three equations and three unknowns, and that's do-able.

Check your equations, as well. You mix up current flowing into the node as positive sometimes (e.g. Node B), negative other times (e.g. Node A). You need to be consistent. Almost everyone uses the convention that current flowing out of the node is positive.