The topic of this problem is mesh analysis and we're working with circuits with dependent sources. The problem is to find the mesh currents in the circuit shown below. We have a circuit that has a 15A source, it's an independent source and it also has a voltage controlled current source, which is controlled by the voltage V sub x. And V sub x is defined as a voltage drop across the 3 ohm resistor on the right hand side of the circuit. So we have a dependent source and we have an independent source, and several resistors in the problem. What we're looking for again are the mesh currents. So if we're looking for the mesh currents and we're doing mesh analysis, then we know the first thing we do in the problem is we assign loops or meshes and we can assign mesh currents associated with them. So for instance we'll call this mesh one and have mesh current I sub 1. We'll call this mesh two and we have mesh current I sub 2. And we have mesh 3 at the bottom right hand side and mesh current I sub 3. So now we can write our mesh equations for each one of our loops or meshes. So we have three meshes and we could have three equations that are independent from one another from those three meshes. So let's start with our left most mesh. If we start in the lower left hand corner of this mesh and go around this mesh, what we find is that we have a voltage drop across the 15A source, another across a 1 ohm resistor, another across the dependent source and across the 2 ohm resistor. We know that when we're doing mesh analysis, what we're looking for are we're looking for the mesh currents. That's what results from our mesh analysis. So in this first loop, the left most loop, we know what I sub 1 is so we can write it directly. So our first equation is I sub 1 = 15A because it's the only mesh current which is associated with this 15A independent source on the left most side of our circuit. So let's continue on to mesh two and look at mesh two. If we want to sum up the voltages around mesh two, again we can start at the lower left hand corner and proceed around the mesh. So if we do that we first encounter a 1 ohm resistor with a voltage drop. And so the voltage drop for that 1 ohm resistor is 1 for the 1 ohm resistor times I2- I1. To continue around this mesh, we encounter the 2 ohm resistor next. And the voltage drop there is going to be 2 ohms times I sub 2. And if we continue, the last thing we encounter is the 3 ohm resistor. We know the voltage drop for the 3 ohm resistor is 3 I sub 2 because it's flowing in the same direction as we're summing the voltages, minus I sub 3 which is flowing in the opposite direction. And that's = 0. So that's our second equation we get from Kirchhoff's Voltage Law for loop two. And now we have a third loop that we want to look at. Our third loop is down here at the lower right hand corner of our circuit. And if we look at that, the voltage drop again starting in the lower left hand corner, we have a 2 ohm register that we can find the voltage drop for. Then we encounter the dependent source and we don't know what the voltage drop is for this dependent current source. Again, we could find the voltage drop for the 3 ohm resistor as we did in the last equation and for the 1 ohm resistor as well. But we have a problem in that we have this voltage source, or this current source in this loop. So we really can't write an equation that sums these voltages up, unless we want to introduce another unknown variable, the voltage drop across this dependent source. So in order to avoid that we use the concept of a super mesh and we can define a super mesh around this circuit that allows us to sum up the equation, sum up the voltages around another loop. And so if we try to do that we can try to do perhaps a loop around the outer portion, then coming down the center part of our circuit. But we can't do that because we again encounter this 15A source. And we don't know what the voltage drop is across it. So if we also look at another route, perhaps there's another route around this loop. And we don't see one, so what we need to do is we use our constraining equations that allow us to solve for this, to put this dependent source in terms of our loop or mesh currents. So we know that 1/9 V sub x, the current source, is equal to I3- I1. Okay, so that gives us another equation. So now we have how many unknowns? We have I sub 1, I sub 2, I sub 3 and V sub x. So we have four unknowns, so four. So we need one more equation. Our last equation can be for V sub x, where we know V sub x is going to be equal to 3 (I sub 3), because I sub 3 is flowing into the positive polarity of our voltage drop, V sub x, across the 3 ohm resistor. So passive sign convention tells us that it's the one flowing into the positive, that it is the positive current minus I sub 2 which is flowing in the opposite direction, through the 3 ohm resistor. Now we have our four equations and our four unknowns. So if we solve for these, first of all, we already know what I1 is, it's 15A. If we solve for I2, we get an I2 = 11A. We get I3 = 17A. And so we have each of our mesh or loop currents, I sub 1, I sub 2, and I sub 3.
