1. Statement of the problem: Before a load is applied the open-circuit terminal voltage is 40 V, and when a load of 500 \Omega is attached the terminal voltage falls to 38.5 V. What happened to the remainder of the no-load voltage, and what is the internal resistance of the source? 2. Known quantities. $V_{oc}=40\ \text{V}$ (no-load terminal voltage). $V_{term}=38.5\ \text{V}$ (terminal voltage with load attached). $R_L=500\ \Omega$ (load resistance). 3. Concept and approach. When current flows the internal resistance $r$ of the source causes a voltage drop $I r$ so the missing portion of the no-load voltage appears across $r$. We will compute the load current from $V_{term}$ and $R_L$, then compute $r$ from the voltage drop. 4. Compute the load current $I$. $$I=\frac{V_{term}}{R_L}=\frac{38.5}{500}=0.077\ \text{A}$$ 5. Voltage dropped across internal resistance. $$V_r=V_{oc}-V_{term}=40-38.5=1.5\ \text{V}$$ 6. Compute internal resistance $r$. $$r=\frac{V_r}{I}=\frac{1.5}{0.077}=\frac{1500}{77}\approx 19.480519\ \Omega\approx 19.48\ \Omega$$ 7. Interpretation. The remainder of the no-load voltage, which is 1.5 V, is dropped across the internal resistance and is dissipated inside the source as heat, so it is not available at the terminals. 8. Final answer. Internal resistance $r\approx 19.48\ \Omega$ and the missing 1.5 V appears across the internal resistance.