The topic of this problem is Maximum Power, and we're working with circuits with dependent sources. The problem is to find the Thevenin's equivalent circuit and use it to determine the load resistance for maximum power and the value of the maximum power delivered to the load. So here's our initial circuit from our previous video problem. So this was the circuit that we started with before our previous problem, and then what we did was we worked through this problem finding Voc, finding I short circuit. So that we could ultimately write the circuit in the form of a Thevenin's equivalent circuit which contains the Voc, R Thevenin's or R Thevenin's is equal to Voc divided by I source circuit and the load resistance. So our initial circuit has everything we need. And we're going to determine what R sub L is for maximum power delivered to R sub L. And then we're going to determine the power associated with R sub L. So we know though the maximum power theorem that maximum power is delivered to the load when R sub L is equal to R Thevenins. We found, in this problem, from the previous video solution to it, that R Thevenin is equal to A kilo ohms. It's Voc, which is 8 volts, divided by I short circuit, which was 4 milliamps. And so that ended up giving us 2 kilo ohms for R Thevenin's. If we want four maximum power delivered to our load, the condition that R sub L is equal to R Thevenin's. Then our R sub L in this problem is going to be equal to 2 kilo ohms. That's our value for R sub L. Now if we know that and we want to find the maximum power delivered to that R sub L, then we can use our simple equation for power, which relates the power to the voltage. So we have V R sub L divided by V squared divided by R sub L. And we know what the voltage is going to be across R sub L using voltage division of our simplified Thevenin's equivalent circuit. We have an A volt source. We have a 8 kilo ohm resistor, we now have identified our load's resistance as 8 kilo ohms, so our voltage drop across this is going to be the voltage division of 8 volt source between these two resistors which are the same. So VRL is going to be 8 volts. And then R sub L which is 8k divided by 8k + 8k. That's the amount of the 8 volt source which Is across the load resistance of 8 kilo ohms. And we know that number is in fact going to be 4. So we have a 4 squared at the top, and we have a load resistance which is equal to 2 kilo ohms. So the power in this case is going to be 4 squared, which is 16 volts squared divided by 2 kilo ohms. So that gives us 8 milliwatts for the power. And that's a power max delivered to the load.
