The topic of this problem is Norton's analysis, and we're working with circuits with independent sources. The problem is to use Norton's equivalent circuit to find the voltage V out across our load resistance in this circuit shown below. So we have a 6 kilo-ohm load resistance, we're measuring the voltage drop across it. The problem has a 6 volt source and a 2 milliamp independent source associated with the problem. So we know that when we're solving these problems using Norton's equivalent circuit, that we first have to find an equivalent representation for our circuit that looks like this. It has a current source in it for Norton's equivalent circuit, and it has a Thevenin's equivalent resistance as part of it as well. As current is known as I short circuit, And the resistance is R Thevenin's. We also know that with our Norton's equivalent circuit, that we retain our load resistance, R load, for the problem as a circuit B on the right-hand side of our circuit. So what we're doing is we're taking the load out of that original circuit, and we're looking at the remaining part of our circuit, and replacing it by an equivalent current source and resistance. Once we've done that, we found equivalent representation for it, the Norton's equivalent circuit, it's easy for us to find things like the voltage drop across our load resistance, which is what we're for in this problem. Because we can use current division to find how much of our short circuit current is travelling through the load resistance. And once we know that, we know V out is going to be the load resistance times whatever portion of I short circuit that's traveling through it, through Ohm's law. So if we want to find the Norton's equivalent circuit, we have to find I short circuit and R Thevenin's. So let's look at finding I short circuit first. To find I short circuit, the way that we do this is we take our original circuit, and we remove the load, and we replace it via short circuit. And then we find out what the current is through that short circuit that we've replaced the load with. So let's redraw the circuit so that we can see this condition. Everything else remains the same in the circuit, so we still have our 6 volt source, we still have our 2 milliamp source, and we still have the resistors as well. So we have the 4 kilo-ohm and 2 kilohm resistor as well as the other 2 kiloohm resistor. We've taken the load out, our load is circuit B separate from the rest of the circuit, they were finding equivalent representation for. And again, we replace that by a short circuit, so this is our short circuit current that we're looking for. If we can find our short circuit current, then that'll be the first variable that we need to find in our Norton's equivalent circuit. The next step would be to find the R Thevenin's, the equivalent resistance of our circuit, and we'll talk about that next. So how about we find I short circuit? There's a number of different ways that we can find it. But in this problem, let's use loop analysis to find, I short circuit. So the first thing we do in loop analysis or mesh analysis is we define our loops or meshes and the currents associated with them. So I'm going to call this loop one, I'm going to call the lower left-hand loop loop two, and loop three on the right-hand side of our circuit. I sub 3 is the current in loop three. We automatically see or we very quickly see upon inspection that I3 and I short circuit are the same currents. So when we're using mesh analysis to solve a circuit, we write our mesh equations or loop equations associated with each one of our independent loops. So let's start with the first loop. It's the upper left-hand loop. So we're going to sum the voltage drops around this loop, starting in the lower left-hand corner, going around it and returning to that lower left-hand corner. So first thing that we encounter as we go around this path is the 6 volt source, and we encounter the negative polarity of that 6 volt source. So we're going to put -6 volts for that voltage drop. We continue round, and we encounter the 4 kilo-ohm resistor, we know the voltage drop across the 4 kilo-ohm resistor is going to be 4k times the current flowing through it. And we're taking this as a positive voltage drop so the current is I1 minus I3, which is flowing through the 4 kilo-ohm resistor as well, but in the opposite direction. So the voltage drop across a 4 kilo-ohm resistor is 4k(I1- I3). We also have, in our path, a 2 kilo-ohm resistor with the same type of situation where it's a shared resistor as far as a voltage drop's concerned between a current that is in loop one and in loop two. Loop one in the positive direction of our voltage drops and current two in the opposite direction. So the voltage drop is going to be 2k(I1- I2), and we then get back to our starting point, that's our closed loop, and the sum of the voltage drops using Kirchhoff's voltage law is going to be equal to 0. So our mesh analysis that we're using relies on Kirchhoff's voltage law. Now let's look at loop two. Loop two is a lower left-hand loop, and if we start summing the voltages, starting at the lower left-hand corner of that loop and going around it, the first one that we encounter is 2 milliamp source. We really don't know what the voltage drop is across that, and we can't use Ohm's law to determine it. So we could, in our mesh equation, introduce another variable and maybe call it V 2 milliamps for this current source. But that would add another unknown to our set or unknowns. Right now, we have unknown I1, unknown I2 and unknown I3. If we can determine those three, then we can find any other parameter thst we want in our circuit. What if we introduce this additional variable? That's four variables that we have. But we're only going to have three loop currents or three mesh currents. So we would have more unknowns than we do equations, and we wouldn't be able to solve our problem. So we know that, when we're using mesh analysis to solve problems, what we're looking for are the mesh currents, I1, I2, and I3. So we, upon inspection, see that I2 is equal to 2 milliamps. So we can put that down directly without having to have an independent equation. It ist an equation, but it's not one containing unknowns I1 and I2. We have now unknown I2. Now if we look at our third equation for loop three, knowing that I sub 3 is equal to I short circuit. Starting in the lower left-hand corner as we travel around this loop, we have two voltage drops, one at the 2 kilo-ohm resistor and one at the 4 kilo-ohm resistor. So as we're going clockwise around this loop starting in the lower left-hand corner, the voltage drop for the 2 kilo-ohm resistor is going to be 2K(I3-, I2. And then up here, across a 4 kilo-ohm resistor, it's going to be 4k, I3- I1. And that's equal to 0. That's only two voltage drops around this right most loop. So now we have these three independent equations, and we can solve for I1, I2, and I3. What we really need for this problem is I short circuit. We're really not concerned about what I1 is, or in fact what I2 is, but I3 is equal to I short circuit, so that's the one we're really after. So if we solve for I sub 3, Which is equal to I short circuit, then we get sixteen-fifths milliamps, For I short circuit. So we have one of our variables, or what was an unknown for our Norton's equivalent circuit. The next one we need is we need to find R Thevenin's. So we again, go back to our original circuit, and we redraw it for the R Thevenin condition, so that we can find R Thevenin's. When we do that, when you're looking at the original circuit and you redraw it for R and in the R Thevenin case, you short-circuit all voltage sources, and you open circuit all current sources. So our voltage source looks like a short circuit. Our 2 milliamp current source would look like an open circuit. The resistors that are part of the problem or our circuit remain the same, except for we've taken the load resistance out, and we're looking for the Thevenin's equivalent circuit. And the Thevenin's equivalent resistance, Of our original circuit has been redrawn for the Thevenin's equivalent resistance condition. So we look at this, and we see that R Thevenin's is going to be a parallel and series combination of these resistors. First of all, we see that the 4 kilo-ohm at the top and the 2 kiloohm in the center of our redrawn circuit are in parallel with one another. They have common connections at one end and then also common connections at the other end as well. So it's a parallel combination. So we have a 4 kilo-ohm resistor in parallel with the 2 kilo-ohm resistor, and then we're going to add another 2 kilo-ohms to that to find the total R Thevenin's. Because this is the series with that initial parallel combination. So if we do this, then we end up with an R Thevenin's which is equal to ten-thirds kilo-ohm. So now we have R Thevenin's, and we have I short circuit. R Thevenin's is ten-thirds kilo-ohm, and I short circuit is equal to sixteen-fifths milliamp. So now we can redraw our circuit above, our Norton's equivalent circuit, with the actual values for I short circuit in this circuit and for R Thevenin's in this circuit, as well as our load as well. So R Thevenin's is ten-thirds kilo-ohm. I short circuit is sixteen-fifths milliamps. And we have to put our load resistance back in, in order to find the voltage drop across it. So there's our 6k load resistance, and we're looking for V out in this problem. So this representation that we have here, the Norton's equivalent circuit, is the same circuit as what we have over here, but we've replaced everything except for the load by equivalent current source and the equivalent resistance. So if we want to find V out, what is V out? V out is going to be the current through the six-column resistor. Let's call it I6k(6k) through Ohm's law. What is I6k? I6k is the amount of the current source which is flowing through the 6 kilo-ohm resistor. So we can use current division where we know that this current source is split between two pass containing the two different resistors. So let's do that to find I6k. We know that sixteen-fifths milliamps is split between these two pass. And if we don't find how much is going through the 6 kilo-ohm resistor, then we use the ratio of the opposite resistor and divide it by a sum of the two resistances. And if we do that then we end up with a V out, which is equal to forty-eight sevenths, Volts.