The Thévenin Equivalent
Sweep the load: the terminals trace one line
Raise and lower R_L and watch the terminal voltage. Find the load where the terminal voltage is exactly half the open-circuit voltage, and that R_L is the box’s internal resistance R_th.
Thévenin voltage = open-circuit voltage
Remove the load entirely and the current falls to zero. No voltage is dropped across the internal resistance, so the voltage appearing at the terminals is the most the box can push. This is the Thévenin voltage V_th — the open-circuit voltage.
Thévenin resistance = slope of the line
Lower the load and the current rises while the terminal voltage drops. How many volts the terminal loses per added ampere is the internal resistance R_th. Drive the load to zero (a short) and the terminal voltage reaches zero while the current climbs to the short-circuit current I_sc = V_th/R_th.
Half voltage points to R_th
The instant the load equals the internal resistance, the two split the voltage evenly and the terminal voltage is exactly half the open-circuit value. So the load that halves the terminal voltage is R_th itself — you have measured the internal resistance without ever opening the box.
Folding the box into one line
Knowing only the open-circuit voltage V_th and the internal resistance R_th, you cannot tell from outside whether the box holds ten resistors or several sources. The single V-I line those two values fix stands in for the whole box. When you analyze one part of a large circuit, folding the rest into this one line makes the problem simple.
Back to the first screen
As you swept R_L, the operating point always slid along the same straight line. Removing the load, the line met the horizontal axis at the open-circuit voltage V_th; shorting it, the line met the vertical axis at the short-circuit current I_sc. Where the terminal voltage was halved, R_L was exactly R_th. You never looked inside the box, yet one line stood in for everything within it.