Okay, now we need to talk about real voltage and current sources. And that's going to bring us to the point where we can then start talking about impedance matching. So now, up to now, we've been talking about only ideal voltage sources. So, an ideal voltage source. The definition of that is that, it has a fixed voltage no matter how much current is drawn from the source. And we just represented that with, with this sort of symbol. Now, a real voltage source can only provide a limited amount of current. There is a you know, if you take, for example, a 1 1 half volt battery, it's 1 1 half volts. but if I, completely short-circuit that battery, it cannot provide an infinite amount of current. A very large current, but it's limited. And the way to represent that is to add a small internal resistance inside the battery. There actually is a small amount of resistance inside the battery. And that limits the maximum amount of current available from that source. So then, if we take this model of a real voltage source here, and then I connect a short circuit across it. I'm going to have some current flowing through this short. And the amount of current that flows through that short is going to be limited by the internal resistance of the battery. So, let's say this is a 1 and 1 half volt battery. And it has an internal resistance of 0.1 ohms, for example. Then, the total amount of current flowing through this short will be limited at 15 amps. So, that's a lot of current, but it's not infinite. Now the, in a practical sense then we're getting closer to the idea of impedance matching. So, in a practical circuit. I want to let's say I'm trying to build a heater circuit. And I want to get the maximum of power transfer from the battery into this load resistor to make the maximum amount of heat. And now, in practice, the load resistance and the internal resistance form a voltage divider. So remember, this is inside the battery and this is the external load. So, the voltage developed across RL, the load, is just going to be familiar voltage divider equation. So, it's the total voltage times this factor, which is the ratio RL over the total series resistance RL plus R internal. Now, we're ready to talk about the subject of impedance matching. So, this is the same circuit that we just looked at, you have a volt, voltage source, a real voltage source with some internal impedance. And we attached some lower resistor and there's a current flowing around this circuit. So the question is, what is the value of the load resistance rl that will give us maximum power transfer from this source to the load? Now, to calculate that we have to go back and remember the formula for the power. Dissipated in this resistor is just the current going through the resistor times the voltage across the resistor, VL. Then, the current going through the resistor is just the voltage divided by the series combination of the two resistances. And the voltage across this resistor is just the current going through it times RL. So, I can use this expression for I and plug it in here, and so this becomes V over R internal plus RL times RL. So, now I can take both of these expressions. The one for VL and plug it in here. The one for I and plug it in there. And so, you get the power dissipated in the low resistor. It goes as V squared. So, there's a V times a V. And then it, it's proportional to RL divided by this factor R internal plus R L squared. So, that's the power dissipated in that resistor, which transferred to that resistor. And so now what we want to do is find the value of RL that makes this a maximum, as big as it can be. Now, we can do this using a little bit of calculus. And we'll do that in a minute. But first, let's just take a look at a graph of this. So I just use Microsoft Excel and plotted this factor as a function of RL over R internal. And so you, you see that there's a peak here when RL equals R internal. This curve reaches its maximum, and the value of that maximum is .25. And so, here's the answer to, the question that RL equals R internal gives us the maximum of this curve. And at that point, this factor has a value of one fourth. Now we, as I said, we could also use a little bit of calculus to find this maximum. And so, what we have to do is take the factor that we're trying to maximize with respect to RL. So, you take the derivative with respect to RL, set the whole thing equal to zero. And then, solve for RL. If you do that, you find that the solution to that is going to be RL equals R internal. That's where the peak is going to be. And then, if I plug that back in, I make RL equal R internal. The maximum power transfer then is just v squared over 4R internal. So at that value of rl I get maximum power transfer to the load resistor