Find the Thevenin equivalent circuit of the circuit shown here : https://imgur.com/9DtrDF2

The first course of action is to find V_Thev. In order to do this, we remove the "Load" resistor, and calculate the voltage across the open connection points where the resistor used to be. In this particular case, a very useful thing to do is give a name to the node right above the 2A current source. Let's call it N, and assign it a voltage V_1. We then use Kirchhoff's Current Law (KCL) at node N to establish equations about the three currents flowing "in" and "out" of node N. (An important thing to notice is that if we were to name the node above the 20Ω resistor "M", then the voltage at M would also be V_1! Indeed, nodes N and M are actually the same node, although this is not apparent in the circuit diagram.) We use KCL at node N and Ohm's Law across the 20Ω resistor to finally solve for V_1, and since we notice that the 6Ω resistor draws no current because we replaced the "Load" resistor with an open circuit, we finally get V_1 = V_Thev = 16V.

To find R_Thev, and thus complete the exercise, we first need to replace the voltage and current sources in the diagram with their respective internal resistances. An ideal voltage source has no internal resistance, so we replace it with a short circuit. An ideal current source has infinite internal resistance, so we replace it with an open circuit. We also remove the "Load" resistor as we did when we calculated V_Thev. (This makes sense, since the whole point of Thevenin's Theorem is that it works for any value you assign to the load resistor.) From this point, finding R_Thev is pretty straightforward: we notice that the 5Ω and the 20Ω resistors are in parallel, so they simplify to a 4Ω equivalent resistor. Finally, we see that our "new" 4Ω resistor is in series with the 6Ω resistor; they add up to a single 10Ω equivalent resistor. This is R_Thev. To conclude, we draw the Thevenin equivalent circuit: a 16V voltage source, in series with a 10Ω resistor, in series with the load resistor.