Electrical – How was the transformer reduced in this Thevenin equivalent circuit

First start with the approximations L = L1 + L2 + L3 R = R1 + R3 C = C1 R2 is too large to do anything. The influence of C2 could possibly be somehow added in to the simple RLC model, but I would ignore it unless there is some specific reason to include it. Once there is a reason to include it, then there should be a way to calculate its influence on the outcome.

Ignoring C2, it looks like your resonant frequency is about 377kHz, and the Q is about 6.4. This is not a great match for a 1MHz source. I assume that your program is going to tell you how to tune things up to get your big voltage to light up the gas tube.

Take care with the high voltage and the bright lights! I know a guy who remembered not to touch the high voltage but forget about the sunburn.

Reading the question and the comments, I believe first of all some clarification is needed.

First of all you need to tell us where you intend to calculate the Thévenin equivalent: you can do that across any two terminals of that particular circuit since it is linear and the controlled voltage source control quantity is inside the circuit you want to reduce.

For example, if you want to calculate the Thévenin equivalent impedance of the net across the \$ V_{rms} \$ voltage source it would be zero since it is ideal.

To calculate the Thévenin equivalent impedance across any two other terminals you first need to turn off any independent voltage/current source, then connect a test source \$V_t\$ across the terminal you have chosen, solve the circuit, obtain the test current \$I_t\$ and finally compute \$Z_{th}=\frac{V_t}{I_p}\$.

I believe that what you want to do is calculate the Thévenin impedance seen from \$V_{rms}\$. To do that \$V_{rms}\$ must be removed, then usual circuit approach can be applied.

Let's call N the node where the resistor, the inductor and the capacitor are connected. From \$R_1\$ arrives a current \$I_t\$ that then splits in the two branches. Let's call \$I^*\$ the current that flows in \$L_1\$, and \$V_n\$ the voltage at node n. Let's finally tie the node \$V_{rms}\$, \$C_1\$, \$Vccs_1\$ to the ground. If \$Z_n\$ is the impedance seen across node n and ground removing \$R_1\$, you can write \$Z_{th} = R_1 + Z_n\$. Can we calculate \$Z_n\$ easily? Yes.

You can write: $$ I^* = I_t - \frac{V^*}{Z_{C_1}} $$ $$ I^* = \frac{V^*-\frac{I_t}{2}}{Z_{L_1}} $$ From these you can derive: $$ V^* = I_t\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}} $$ where \$Z_{L_1}//Z_{C_1} = \frac{Z_{L_1}Z_{C_1}}{Z_{L_1}+Z_{C_1}}\$

Now, \$Z_n = \frac{V_n}{I_t}=\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}}\$, and finally: $$ Z_{th} = R_1 + \frac{V_n}{I_t}=\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}}=(1.9375-0.24j)\Omega $$