Did I wire this correctly? Trying to understand Thevenin and Loading of circuits
Assume the following schematic and that you are measuring your voltages relative to the ground symbol shown:







simulate this circuit &ndash; Schematic created using CircuitLab

To use Thevenin, you first reduce the three items, \$R_1\$ and \$R_2\$ and \$V_{CC}\$, into two equivalents: \$V_{TH}\$ and \$R_{TH}\$. Ignore \$R_L\$ during this process. (As shown above.) To do that, given you've said that \$R_1=R_2=100\:\Omega\$ and \$V_{CC}=12.8\:\textrm{V}\$, simply find that \$R_{TH}=R_!\vert\vert R_2 =\frac{R_1\cdot R_2}{R_1+R_2}=500\:\Omega\$ and that \$V_{TH}=V_{CC}\cdot\frac{R_2}{R_1+R_2}=6.4\:\textrm{V}\$. That's your new circuit.

With \$R_L=100\:\textrm{k}\Omega\$, you now find that \$V_X=V_{TH}\cdot\frac{R_L}{R_L+T_{TH}}\approx 6.37\:\textrm{V}\$. Not much change. This should be obvious to you because \$R_L\$ is close to an open circuit, compared to \$R_{TH}\$. So it won't have much impact. Very little current will flow through it compared to what \$R_{TH}\$ can dish out.

But if you make \$R_L\$ closer in value, then you will see an effect. Suppose you use \$R_L=4.7\:\textrm{k}\Omega\$. Then you should find that \$V_X\approx 5.78\:\textrm{V}\$, which is enough to notice. Dropping \$R_L=1\:\textrm{k}\Omega\$, you should find \$V_X\approx 4.27\:\textrm{V}\$. Even more noticeable. Etc.