So for this example problem we are going to again find the seven and equivalent circuit. Um but we're going to have to apply a little bit of an analysis technique to this. Um, and. Again will be able to ignore this one kilo ohm resistor here. We can just drop that off here. So, um, let's go ahead and start working on the problem. So remember the first thing we typically find is the open circuit voltage. And so this resistor here is, uh, kind of what I call a hanging resistor, which we talked about in the previous video. And so I'm going to go ahead. Oops. I don't know what just happened there. I am going to go ahead and just erase that for now. And replace it with. Just a wire, because then this is the circuit we need to analyze to get the open circuit voltage. And so let's go ahead and again just think of this and think of what nodes we have here. Because I think sometimes people overcomplicate these problems. And they really are just um, what I would call um, simple. Um. Circuit analysis problems that we've learned how to do. Now I'm just going to go ahead and actually erase these things here for a second. Those bars that we're used to see. And if I solve this circuit right here. It's put the ground in. I could label the nodes. I have a node down here. Which is zero volts because it's ground. I have a node up here at the top left, which because of the power supply, I know that this is three volts. And then I have a node right here. And I don't know what that voltages. So I'm just going to label that voltage as v x. And then if I think about this, if I add these little. Arms with the. Node circles attached to them. All we're being asked to do is find the open circuit voltage. But what is that equivalent to? Well, if I look carefully at the circuit I had before. The open circuit is the same thing as that voltage v x, because this circle is the same node as v x. And this circle right here is the same node as ground. So the open circuit is equal to v x which also then of course is my v subnet. So whenever I see a current source attached to a node, I automatically want to do node analysis on this. So let's do a node analysis. In fact, it's kind of turned out we only need to have one node equation. So I'm going to draw here. Some currents here. I have a current coming down here. I'll call that I won a current going to the left. I'll call that i2. And then current that's going to the right. But then it will go down through the five kiloton resistor and I'll call that i3. And those are the only currents I have at that note. And so right away we know that I1 is 7,000,007 milliamps. Now, if you remember here, this is kind of a fun little thing here to remember. I'll kind of start off to the side here. Milliamp times killer. Ohm. Is equal to voltage. And then you can swap these around however you want. Voltage divided by q0 ohm equals milliamps. Or voltage divided by milli amps is equal to kill ohms. These can be useful because what it will allow me to do is drop the units from the equations that I write. As long as I keep in mind of things that are milliamps and km's. But since my resistors are kilo ohms, my current is milli amps, my voltage is volts. I can just write things down with no units. At this point I write my equation and know that my answers are going to be in this format. So let's go ahead and write then my node equation. So at the i1. Plus I two. Plus i3. Equals zero. So I want a seven. Now. Yes, it's really nice, but we know that. That's okay. We already got that taken care of. And then if I look at my circuit I have plus and then I two would be v x minus three. Divided by two km's and then I'd have plus. V x divided by five. Equals zero. And if you want to think about the VCs, it helps you to think of it this way. It's VCs -0 over five equals zero because it's the voltage at this top right here VCs minus the voltage down here which is zero divided by the five ohm resistor. And now I have a simple equation that has just one unknown. And so I'm just going to solve this. Now you could go ahead and use an online solver to solve this equation for you, and I'll show you that in a second. Let's go ahead and just do this by hand. Um, and then I'll show you how those online solvers can actually save you some time. And also avoid making algebra mistakes. So to solve this we are going to multiply both sides by ten. And that would get me. And why did I choose ten? What's two times five is ten? So that would get me ten times seven, which is 70. And then I'd have a two with the ten cancel and get me five. So I'd have plus five times v x -15. And then I'd have plus 2VX equals zero. And so ultimately I'd have. Seven. V x. Equals. And let me believe that would be -55. Let's just double check that on the calculator real quick. Yep. And then that gets made v x is equal to. Oops. 7.86 negative votes. Okay. So that's one way to solve this equation. And then that would be equal to V7 in. But there is a another way you can solve these equations. And this is something called symbol AB. Um. And if I do simple web equation solver. It's an online calculator. Um, like graphing calculators have the ability to do this as well. But we can just enter the problem. Now we need to use x as opposed to v x here. So let's go ahead. And I'm gonna have to kind of go back and forth here a little bit. Or actually what I'll do is I'm going to move this into my other screen, short for a brief amount of time so that we had. Seven. Plus. And then we'd have x minus three. Divided by two. Plus x divided by five equals zero. And then we just hit enter on it. That way it gets me. -55 over seven, which is the answer we got doing it by hand, which is negative. Rounding into two decimal spots -7.86V. And it's find that it's a negative voltage. Just has to do with how the voltage and current supply are connected in the circuit diagram. And so we now have my V7 in. And if I bring the original circuit here. I'm ready to find. Art definite. Now, instead of doing the I short circuit method, um, I'm going to again turn off just my power supplies. And so if I turn off the power supply here, I'd be turning off this. Three volt, which this becomes then. A short. And then the 7 million. That's a current source. So that becomes an open. And I'll just have it like that for now. But when it's an open circuit you can really just redraw it like that. Because you're literally just taking it completely out when you turn it off as an open circuit. And now this is the circuit that I'm left with that I have to find are equivalent for. And then without redrawing the circuit, I can see that the two k is in parallel with the five k. So I have two times five. Over to plus five. Which is. Ten. Over. Sudden. And now I'm going to again, because if I have to round this to two decimal spots, I'm going to definitely keep a few extra here just to be safe. So I'm going to have this out to, um, four decimal spots here. So I have this is 1.4 to. Eight six km's. Because if I'm going to round it to two, I want to carry some extra ones with me along while I'm doing my computations. The resulting circuit of that, though. Would be the following. And this would be the 1.4 286K. And this is the one kiloton. And those now are clearly in series. So we would just have that power equivalent. Which is the same thing as our seven and for us. Would be one plus 1.4286, which rounding that now to two decimal spots. 2.43kW. And so then my seven equivalent circuit. Would be my voltage, which we got -7.86V. And then 2.43kW. And so that would be my resulting feminine equivalent. All right. So there's a couple problems on the homework dealing with feminine equivalents but not too many. So this should give you a good starting place on how to do those problems.