The topic of this problem is mesh analysis, and we're working with circuits with dependent voltage sources, and also independent voltage sources in them. So we're going to work on a problem where we need to determine V sub out, which is the voltage drop across a 2 kiloohm load resistor on the right-hand side of the circuit. We see that we have a current controlled voltage source as our dependent source. And the current that controls this voltage source is the current that flows down through the 2 kiloohm resistor on the right-hand side of the circuit. So, that current controls the voltage level at the left-hand side of the circuit. So what we want to do is we'll use Mesh Analysis to solve this problem. So we immediately know that we're going to be using Kirchhoff's Voltage Laws, Kirchhoff's Voltage Law to solve the problem. We also know that when we're doing mesh analysis, the first thing we do is assign meshes, and then mesh currents associated with those meshes. So that we can then sum up the voltages around the meshes to create the equations that can be solved simultaneously to determine the mesh currents. So, we're going to start with assigning our meshes. Our first mesh we're going to assign in a clockwise fashion, on the left hand side, and we're going to call the current on it I-sub-1. And you might imagine that the next mesh is going to be on the right side of our circuit. It's going to have a mesh current I2. We immediately notice that I2 is equal to I sub x, because it's the current flowing down through the 2 kilometer resistor. So, we can write that down to help us out, because I sub x Introduces a third unknown variable. We have an unknown, current I sub 1, we have an unknown current, I sub 2, and we have an unknown current, I sub x. But this is our first equation, which is independent of our other equations, it's a constrain equation that relates I1, I2, and Ix. So, let's write the equation around loop one, as our first mesh equation. Starting at the lower left-hand corner, we encounter the negative polarity of the dependent voltage source. So, we're going to write that as a minus 2,000 Is of x voltage drop is our first voltage drop, as we travel around this loop1. The second one is the voltage drop across the 2 kilometer resistor, that's 2k, times I sub 1. The only current flowing through the 2 kilometer resistor at the top left is the I sub 1 current. Then we get to the 4 kilometer resistor, which is shared between Mesh 1 and Mesh 2. And so, the voltage drop here is going to be 4K and traveling clockwise around this loop, same direction as I1. The voltage drop is going to be 4K I1 minus I2, because I2 is going through the 4 kilometer in the opposite direction as I1. And then we continue back to our starting point, and that is our sum of the voltages around that close loop. Our next loop is loop two. And again, we have three unknowns. We have I sub x, we have I2, and we have I1. That's our three unknowns in our equations that we're going to come up. So if we're going to come up with the third equation, then we can solve for I1, I2 and I3 using various techniques for solving equations simultaneously. So let's add up our voltage drops around loop two, starting in the lower left-hand corner, first encountering the 4K resistor. We know the voltage drop there is 4K, we're going clockwise around loop two, I2- I1. Keep in mind that 4K, I2 minus I1 is equal to, we write it up here, 4K, I 2 minus I1 is equal to minus 4K, I1 minus I2, where this 4K I1 minus I2 was a voltage drop that we assigned from our first loop across that. And for coming up through the negative of this that some people might look at this problem and say well, we should take the negative of the voltage drop that we've already assigned in loop 1 as we sum up these voltages. And in fact, that's what we did. Here is the negative of that voltage, and is this, which is what we have in our equation 2. So, it is exactly what we thought it was. So we continue around this loop, and we run to the 6 volts source, the negative polarity first, so it's minus 6 volts. And continuing around, we have a two kilometer resistor with I2 dropped across it. So it's 4 2K. So, it's 2K I sub 2. And we continue around to the start of that loop, and so our sum is equal to 0. So there's our 3 equations and our 3 unknowns. What we're interested in this problem is V out, the output voltage across the 2 kilometer resistor. We know that V out is equal to 2k times I sub 2. We use a pass assign convention I2 is flowing clockwise around this loop, and we assign a voltage drop across the 2K resistor as so, and that is the same voltage as we are measuring, which is V out. So, V out is equal to 2KI2. And so, what we really need from this set of three equations and three unknowns as we need I2. And so, if we use those equations and we solve them simultaneously for I sub 2, then I sub 2 equal to 3 milliamps. Once we have I sub 2, we can come down and use our equation for V out, and we end up with the V out equals 6 volts.
