A topic of this problem is Thevenin's Analysis, and we're working with circuits with dependent sources. The problem is defined that Thevenin's equivalent circuit for the circuit shown below. The circuit has two sources in it, it has an independent 4 miliamp source, and it has a dependent source in it which is a current controlled voltage source. The current controlled voltage source is 2000I sub x where I sub x is the current that's flowing through the 2k ohm resistor that's part of our circuit is flowing down through the 2k ohm resistor. So that current controls the level of the voltage and another part of the circuit namely at the current control voltage source. So when we're doing these problems and trying to find the Thevenin's equivalent circuit, the first thing that we want to realize is that Thevenin's equivalent circuit in general form looks like this. We have a voltage source, we have a resistance, and then we have separated our load to another side, which we call circuit B. So this is our load, R sub L. This resistance is the Thevenin's equivalent resistance which we'll have to determine. And the voltage is the open circuit voltage which we'll also have to determine. So, we'll take our initial circuit and we'll redraw our initial circuit with the load removed. And we'll redraw it for an equivalent circuit where we've taken this part of our overall circuit and we've replaced it by a voc, and a r Thevenin's. And then we'll put our load back in as we show here, and we'll solve for whatever we're interesting in solving for. It could be the voltage across the load, it could be the current through the load. So, we're going to work this problem, and solve for Voc and R Thevanin's. Let's start with Voc. So, to find Voc, we have to redraw our circuit for the open circuit condition. What we'll do is we'll take our load and we'll remove it and replace it by an open circuit. So we'll essentially take it out and not put anything back in it's place. So we want to redraw the circuit for that condition. It has a 3 kilo ohm resistor, and a 1 kilo-ohm resistor at the top. We still have our dependent source, our current control voltage source, and we still have the 4 kilo-ohm resistor. But we've taken the load out the far right hand side of our circuit. So I'm redrawing it with everything else still intact. All the sources are there and the resistances are there except for the load resistance. Since this circuit is a little bit different, it's not the same circuit we started with, our I sub x I'm going to call I sub x prime. Continuing to label, the resistances, this is what we have. And what we're going to be measuring if we take our load out is we're going to be measuring Voc. So if we can find the voltage from that top to bottom position, in our circuit, then we can find Voc. One thing we realize in this redrawn circuit is there's no current flow through the 4 kilowatts resistor, it's an open circuit. If there's no current flow through the 4 kilo-ohm resistor then there's no voltage drop there. So Voc can just as easily be measured from this point to ground as well. So this is Voc from this node, at the top of the circuit to ground. Okay, so now we need to find what that voltage is, and we can see that that voltage from this node at the top, to ground, is a voltage drop that's across the two kilo-ohm resistor. So Voc, in this case, is going to be equal to 2 kilo-ohms times the current that flows through that 2 kilo ohm resistor, and it's I sub x prime. That's going to be Voc. So if we can find I sub x prime in our redrawn circuit, then we can find Voc. So let's see how we might do that. There's a number of different ways we could do it. But perhaps we might use Kirchhoff's voltage law and mesh analysis to solve this problem. So if we're going to use measure analysis, again it's Kirchhoff's law, that we would implement for mesh analysis. The first thing we do is we assign meshes. And so this is our first mesh and it's going to have mesh current I sub 1. We'll assign another mesh over here, and this [INAUDIBLE] here is a 4 miliamp, and we're going to call this I sub 2. And we immediately notice that if we're looking at the loop 2, that I sub 2 is equal to I sub x prime. So if we can find I sub 2 then we can find I sub x prime. If we can find I sub x prime, then we can find Voc. So let's start with this right most loop and see what would happen and when we sum the voltages around this loop. What we'd find is that we can find the voltage drop across our three kilo-ohms resistor is going to be 3k times I1. Similarly for the 1k, it's going to be 1k times I1. So looking at loop 1, we know the voltage drop across a 3 kilo-ohm resistor, we know the voltage drop across a 1 kilo-ohm resistor. The last one in this loop is the voltage drop across this 4 milliamp source. We don't know what the voltage drop is across the 4 milliamp source. So we'd have to introduce another variable, perhaps V 4 milliamps for this voltage drop. Then we'd have three unknowns, I1, I2, and V 4 milliamps. Yet, we only have two equations. We have one equation for loop one, or mesh one, we have one equation for mesh two, and we're going to have this initial equation for our controlling parameter I sub x prime. So, we'd end up with four unknowns and three equations. And we wouldn't have enough equations to solve for all of the unknowns. So what we rely on when we're doing mesh analysis is this concept of super mesh where, if we have a current source, which is shared between two meshes like this one is, it's shared between mesh one or loop one and loop two or mesh two. We use this concept of a super mesh in order to solve the problems. So if we do that for this problem, we assign a super mesh around the circuit and avoid the 4 miliamp source. Now we can sum up voltages around this loop very easily. And we still have our second loop here that we can sum the voltages around if we wanted to, but we've run into the same problem with this side of the circuit as well. So we need to come up with that second equation relating the 4 milliamp to the I1 and I2 currents and we'll do that in just a minute. So the first thing we want to do when we solve this problem is we want to use the concept of super mesh and use Kirchhoff's Voltage Law to sum the voltages around our super mesh. So looking at the super mesh, In the lower left hand corner and going around it in a clockwise fashion back to the beginning, we first encounter a three kilo-ohm resistor. And the voltage drop for that is going to be 3k times I sub one, since that's the only current flowing through the three kilo-ohm resistor. Similarly at the top, we have a one kilo-ohm resistor. The only current flowing through it is I sub 1, so the voltage drop is 1K times I sub 1. Continuing around our loop, we encounter our dependent source, the negative polarity of that first. So it's going to be a -2,000 I sub x prime for that voltage drop and then the voltage drop across the 2K resistor. And in that case, it's going to be 2k times I sub x because we know that I sub x prime is equal to I2. And so that's our last voltage drop as we return to our starting point. So that's our first equation and if we look at that equation, we have two unknowns. We have I sub 1 and I sub x prime. So we need another equation, another independent equation that would allow us to solve this problem. And so, we can use our constraining equation since we can't sum voltages around loop two, can't sum voltages around loop one. We had to use a concept of a super mesh to get our first Kirchhoff's Voltage Law equation. Our second equation has to relate I1 and I2 as well and if we look at this circuit we see that the 4 milliamp source that we were avoiding when we were doing our Kirchhoff Voltage Law gives us that relationship between I1 and I2. Namely 4 milliamps is equal to I2 minus I1. And we know I2 is equal to I sub x prime, so we're going to put that into our equation. So 4 milliamps is equal to I sub x prime, minus I1, which is flowing the opposite direction of the four milliamps. So, now we have these two equations and two unknowns and so we can solve for I sub x prime. And ultimately we can solve for V open circuit. So if we continue with this and we solve for I sub x prime in this problem and then plug that into the equation for Voc above, we get a Voc which is equal to 8 volts. Now, we need to find our Thevenin's. because we have Voc in our Thevenin's equivalent circuit but we need to find our Thevenin's. When we have these circuits, our initial circuit, that has an independent source in, a dependent source in it, like 2000 I sub x that we have at the top. When we're finding our Thevenin's for these types of circuits, our Thevenin's is going to be Voc divided by I short circuit. So to find our Thevenin's we already have Voc, but we have to find this I short circuit condition. So we do similarly what we have done for Voc to find I short circuit. We're going to take our initial circuit, we're going to remove the load and we're going to replace it this time with a short circuit. And then the current through that short circuit is in fact I short circuit. So the first thing we do is we redraw the circuit for the condition where we've replaced the load by a short circuit Everything else remains in the circuit. The 4 milliamp of source is still in there, the resistors are still in there, a dependent source is still in there. And we're going to label all of those so that the 2k and 4k resistors on the right hand side, our dependent source which is has a current flowing through it. And we've redrawn a circuit to yet another configuration, so I'm going to call this I sub x double prime. We have a 4 miliamp source, and our resistors still intact. Our short circuit as we had said is the current through the short circuit that we've put in place of the load. So once we find I short circuit, we can find R Thevenin's. So to find I short circuit, we can use a number of different techniques to do that. Perhaps an easy one for us to use is again mesh analysis on our circuit that we have shown. So let's use that as a way of solving for the problem. So we're looking for I short circuit and I'm going to use the concept of super mesh again to do that. So starting at the lower left hand corner, going around, we know that we can't go through that source that we have the 4 milliamp source because we don't know the voltage drop across it as we explained previously in the problem. That's going to be our super mesh that we're going to sum the voltages around. So starting in the lower left hand corner and assigning the same. Meshes as we did previously. We have mesh one, mesh two and now we have this third mesh, mesh three. We can write our equation for the super mesh so the first voltage drop is 3 kilo-ohm times I sub 1 is the only current flowing through this 3 kilo-ohm resistor. Similarly for the 1 kilo-ohm resistor at the top, we have the same current flowing through it. We have our voltage drop. We encounter the negative polarity of that voltage drop from the dependent source first. So it's minus 2,000, I sub x double prime. Continuing we have a 4 kilo-ohm resistor which has I short circuit flowing through it so it's plus 4k I short circuit and that's equal to 0. So if we look at this equation we have three unknowns. We have an unknown I sub x, we have an unknown I short circuit, and we have the unknown I sub 1. We still have the same relationship between I1 and I sub x double prime through the 4 milliamp source. Namely that constrained equation is 4 milliamps is equal to Ix double prime since it's flowing the same direction as the 4 milliamps source, minus I1. So that gives us two equations which are independent of one another. We still need one more in order to be able to solve the problem. And is equation that could be obtained from this right most loop or mesh in our circuit. And so let's write that, this is for our short circuit mesh. It's mesh three, if you will, on the right hand side of the circuit. Summing up the voltages, we start at our lower left hand corner and going around this loop at a clockwise fashion, is going to be 2 kilo-ohm for the first resistor that we encounter times I short circuit minus I sub x double prime. We then continue around this loop and the 4 kilo-ohms resistor is the next resistor we encounter. We know the voltage drop across is 4k times I short circuit. Continuing on, we don't have any other resistances or any other voltage drops that we encounter so that's equal to 0. So if we continue with this and we solve for I short circuit in this problem using our three equations and three unknowns, we end up with I short circuit. Which is equal to 4 milliamps. So five short circuit is equal to 4 milliamps. We know that V open circuit is equal to 8 volts. We get that from our first circuit analysis. Then we can find R Thevenin's. R Thevenin's is equal to Voc which is 8 volts divided by I short circuit which is 4 milliamps and so in the end we get 2 kilo-ohms for R Thevenin's. So we'll take our Thevenin's equivalent circuit above and we'll redraw it. And we're going to redraw it with our values for V open circuit in the circuit and the values for R Thevenin's in the circuit. And you'll see in this case the value for Voc was 8 volts. So that's what our Thevenin's equivalent circuit looks like. If we had a value for the load resistance and we can find the voltage drop across that load resistance using voltage division of our 8 volt source between ak resistor and whatever the load resistance is, we could also find the current through the load resistor if we wanted, the power associated with the load resistance. All those can be determined from this equivalent circuit that we've drawn for our initial circuit that we had on the left-hand side of our board. So what we'll do is we're going to use this problem or this solution that we have so far, we're going to use it in the next problem that we look at in this series of videos.