1. Problem statement: A power supply has an open-circuit terminal voltage set to $40$ V and when a load $R_L=500$ ohm is connected the terminal voltage falls to $38.5$ V. 2. What happened: The missing portion of the no-load voltage is the voltage dropped across the internal resistance of the source when current flows through it. 3. Compute the load current: The load current is $I_L=V_{load}/R_L=38.5/500=0.077$ A. 4. Compute the internal voltage drop: The internal drop is $V_{int}=V_{oc}-V_{load}=40-38.5=1.5$ V. 5. Compute the internal resistance: Using $r=V_{int}/I_L$ gives $r=1.5/0.077\approx19.48$ ohm. 6. Final answer: The remainder of the no-load voltage, $1.5$ V, is dropped across the internal resistance of the source, and the internal resistance is approximately $19.48$ ohm.