[SOUND]. Welcome back to Linear Circuits, this is Dr. Ferri. This lesson is on the Norton Equivalent and Source Transformations. The objective is to introduce the Norton Equivalent Circuit which looks like this. And also to show how to simplify circuits using source transformation. The goal is take something fairly complicated, like this, and reduce it down to an equivalent circuit with one source and one resistor. And it'd be so much easier to analyze than something like this. This lesson builds upon the Thevenin equivalent circuit, which is where we take a circuit that looks like this. Which I show this as a box here, but it could be anything in here, something fairly complicated. And I replace it with this circuit here, much much simpler. And they're equivalent in terms of this voltage and current that we have across this terminal. Would be the same across this terminal if we put a resistor or some other circuit element in there. I'd have the same voltage and currents here as I would in this case. Now I just mentioned the idea of the voltage, the Thevenin equivalent circuit. And in our previous lesson, we showed how to find these equivalent values, the R Thevenin and the V Thevenin. Now the Norton equivalent circuit is equivalent to the Thevenin equivalent circuit. The only difference is we've got a parallel combination here between the current source and the Thevenin. Now this i sub sc, we defined it in our previous lesson. It is the short circuit current across these terminals right here, that's i sub sc, and that is the source right here. We also defined another way of finding it, and that was through Ohm's law. So V Thevenin, and I sub sc are related to R Thevenin. So, using all the techniques and methods that we found in our last lesson to find V Thevenin, or R Thevenin, or i short circuit, we use here. The only thing is we end up putting our circuits together looking like this, rather than like that. Now source transformation is the simplification process that I mentioned before to take a circuit and simplify it. Now in this particular case, what we do is we look at a part of the circuit. If I see anything that looks like this, a resistor in series with a voltage source. I could replace it with this combination right here. So, I interchange the Thevenin equivalent and the Norton equivalent circuits at will using this relationship in order to simplify the circuit. And the goal is to be able to see if I can combine resistors in series or resistors in parallel. Now it's going to make a lot more sense when I show you an example of it. All right, let's do an example. And I want to recognize first that this is a Thevenin equivalent circuit. And if I convert it to a Norton equivalent, that would put this resistor in parallel with that one and I could combine them. So let's go ahead and do that first. So I replace the Thevenin equivalent with a Norton equivalent. The 2 carries down and then this current is this voltage divided by that resistance. So that's one half and that's in parallel with the 4. Everything else stays the same. Now let's combine these two. In parallel it's the product over the sum, that's 8 over 6 or four thirds node. So let me go ahead and redraw that. I'm going to convert that. I've combined those and then I am going to convert that combination back to a Thevenin. So I would have the four-thirds times one-half would give me that voltage. So that would be two-thirds and the resistance here is at four-thirds. That is this whole combination right there, converted back to Thevenin. And since that's going to be in series, I want to convert this part here to a Thevenin as well. So my voltage is going to be the product of these two, 0.2 times 10. And notice the direction of the polarity, it's the same direction as that arrow. So that would be two volts, the resistance is the same, and then my output resistance stays the same. And I've got this voltage right there, which is 2 volts. Now everything is in series, I can combine these sources. And if I go around the loop, I find that this voltage source of 2 and this voltage source of 2 are exactly opposite. Because if I added them up, if I did a KCL I'd hit the minus first here, and then the plus first here. They would cancel each other out. So the only source I'm left with is the two-thirds. And this right here is 10 ohms carried from right there. So then this becomes 11, and 10 plus four thirds is 11 and one-third. And then I've got my R0. So it's a way of simplifying this circuit, almost graphically, from the original one just by applying the source transformations. So to summarize this lesson, we've gone over the key concept of transforming between sources in order to simplify the circuit by combining resistors in series and parallel. And the goal is to reduce some complicated circuit into something much, much simpler. That we might be able apply mesh law or node analysis in order to solve. But we want to simplify it first before we apply any other analysis methods. All right, thank you. See you on the forums. [MUSIC]