The topic of this problem is Maximum Power and Power Delivered to the Load. The problem is to find the maximum power delivered to a load from a general circuit reduced to a circuit's Thevenin's equivalent circuit. So we have some complex circuit which is on the left-hand side of our circuit. And then we've reduced it to its Thevenin's equivalent circuit which includes Voc and R Thevenin's. So we have V open circuit and R Thevenin's as our representation for our more complex circuit. And we have add the load back into the circuit on the right-hand side of our circuit. So in order to find the maximum power delivered to a load, we have to first of all understand what power is. And we're talking about power delivered to our load which is the resistor R sub L. So power to the load is equal to, we know we can have it equal to I square R, or we can also restate that as V sub L squared divided by R. And a current I would be this current which is around the loop and shared by all of the elements in the circuit. So there's our equations for power. So using our circuit that we have drawn, we can represent the power in terms of R Voc or also known as V Thevenin's. And by doing that, what we would do is we would take our second representation for power, and we would represent V sub L using a voltage division of our source, our Voc, across our R Thevenin's and our R sub L. So we know if we use that voltage division, that V sub L = Voc times R sub L divided by R sub L plus R Thevenin's. So we can plug that back into our power equation and we end up with Voc squared. R sub L divided by R sub L plus R Thevenin's squared divided by R sub L. Because if we're talking about power to the load, then our resistance is the resistance of the load and we can put our R sub L in there like that. So this is our equation for power to the load and it is a function of R Thevenin's equivalent the open circuit voltage and R resistors that we have in our circuits. So reducing that to its final form, we have power to the load equal to Voc squared times R sub L divided by R sub L plus R Thevenin's squared. Now we can start looking at the interest that we have in maximum power. So this is our general equation for power delivered to the load. If we're looking at the situation where we're trying to consider maximum power delivered to the load, whether or not we're trying to find the correct R sub L or if we're trying to find the value for that power, the maximum power is delivered to the load when the partial of the power of the load divided by with respect to R sub L is equal to 0. So when the partial of the power with respect to the load is equal to 0, that's when we have maximum power. So what we're going to do is we are going to take our equation for P sub L and we're going to perform this partial differential of that power. So if we do that, then what we end up with is partial power with respect to the load = Voc squared divided by R sub L+R sub Thevenin's squared. Minus 2 Voc squared. R sub L divided by R sub L+R Thevenin's cubed. So if we set that equal to 0 for maximum power, We see that we can then take one of these terms to the right-hand side of the equation. We could cancel out our Vocs and ultimately what we end up with is at R Thevenin's plus R sub L = 2R sub L. And so what that tells us is that R Thevenin's = R sub L for maximum power delivered to the load. So if we wanted to find the value for R sub L where maximum power were delivered from a generic circuit that was reduced to its Thevenin's equivalent circuit, we'd set our load resistance to R Thevenin's. The value that we found, the equivalent value of resistance that we found for our circuit looking back from the right to the left of the circuit. If you wanted to find the maximum power, then you would take this R sub L and you would find the current or the voltage V sub L or I and you would solve for power for that load.