Video narration
What voltage and current does the 5 kΩ load receive? Think breadboard: replace everything before the load with one source and resistance. Remove the load: Vth = 48 × 4/(12 + 4) = 12 V. The 8 kΩ carries no current, so it drops nothing. Deactivate the source by shorting the 48 V source: Rth = 8 kΩ + (12 kΩ || 4 kΩ) = 11 kΩ. Your breadboard is now 12 V in series with 11 kΩ. Reconnect: IL = 12/(11 + 5) kΩ = 0.75 mA, and VL = 3.75 V.