Video narration
How can you find the voltage across the 5 kilo-ohm load when the circuit around it looks complicated? Think of a battery driving a flashlight bulb: a source pushes current through resistance. Thevenin’s theorem lets us replace the entire network on the left with one simpler source and one resistance, while keeping the same behavior at terminals A and B. Here, the source is 48 volts. From the source, there is a 12 kilo-ohm resistor to an internal node. From that node to terminal B is 4 kilo-ohms, and from the node to terminal A is 8 kilo-ohms. The 5 kilo-ohm load is connected between A and B. First, remove the load mentally. To find the Thevenin voltage, or Vth, look at the open circuit across A and B. With the load gone, no current flows through the 8 kilo-ohm resistor, so there is no voltage drop across it. Therefore, A has the same voltage as the internal node. The 12 and 4 kilo-ohm resistors form a divider: Vth equals 48 times 4 over 12 plus 4, which equals 12 volts. Next, find the Thevenin resistance, Rth. Turn off the ideal 48-volt source by replacing it with a wire, or short circuit. Looking back into terminals A and B, the 12 kilo-ohm and 4 kilo-ohm resistors are now parallel: 12 parallel with 4 equals 3 kilo-ohms. That combination is in series with the 8 kilo-ohm resistor, so Rth equals 8 plus 3, or 11 kilo-ohms. Now reconnect the 5 kilo-ohm load. The simplified circuit is a 12-volt source in series with 11 kilo-ohms and 5 kilo-ohms. Using Ohm’s law, the load current is 12 divided by 11 plus 5, giving 0.75 milliamps. The load voltage is current times resistance: 0.75 milliamps times 5 kilo-ohms equals 3.75 volts. So, the voltage across the 5 kilo-ohm load is 3.75 volts, with terminal A positive relative to terminal B.