Voltage dividers were brought up. They are a nice way to lower voltage in a circuit. Coming from a power source (a battery of 9V for example), you go into a resistor to drop the voltage. you then take that dropped voltage and pipe it out to your smaller circuit. After the branch, you add more resisters to bring the voltage back to zero. To figure out the divided voltage:
Vout = Vin * (R2/(R1+R2))
This equation assumes you know the resistances of R1 and R2.
Thevenin's equivalent circuit:
When faced with a complex, linear circuit, you can just treat like a black box in series by calculating Thevenin's voltage and resistance. to find the voltage, remove the the load element and measure the circuit by any means usable. to find the resistance, remove the power source and short it, then measure the resistance by any means usable. Now treat the circuit isolated as a black box in series with the load.
To find Norton's Current, calculate the Thevenin resistance of the circuit. Then short the load and then calculate the voltage source divided but the Thevenin resistance. Between Thevenin's resistance, his voltage, and Norton's current, you can black-box any complex circuit.
There was brief mention about AC and power mains. I also started the section about capacitors.
