The topic of this problem is mesh analysis. And we're working with circuits with independent voltage sources. The problem is to determine I sub 0 in the circuit shown below. I sub 0 is the output current through the 6 kiloohm resistor in the center leg of the circuit. We have two voltage sources in this problem. We have a 12 volts source on the left-hand slide, a 3 volt source on the right-hand side. So we're going to use mesh analysis, also know as loop analysis, to solve this problem. So, if we're going to use mesh analysis, we know that we're going to rely on Kirchhoff's voltage law to solve this problem. Kirchhoff's voltage law tells us that the sum of all the voltages around any closed loop is equal to zero. So if we sum up the voltages around the loops that we identify in the circuit, a sum of those voltages at any instant in time should be equal to zero. So we're going to first identify, as we do with any mesh analysis problem, we're going to identify our loop currents, or mesh currents, first. So in this problem, we have a mesh on the left-hand side, we have a mesh on the right-hand side. And if we wanted to, we can identify this mesh on the exterior, or the outer loop, of the circuit as well. But we need for this problem, two independent meshes. So we going to start with identifying mesh one on the left-hand side of the circuit. And we're going to identify mesh two on the right-hand side of the circuit. So we have a mesh current one and a mesh current two. And so, we're going to use Kirchhoff's voltage law, and we're going to sum up the voltages around both of those loops. The first one that we're going to work on is loop one. So if we start in the lower left-hand corner of loop one and we start traversing around it using the pass assign convention, we know our polarity drop is plus to minus across the 6 kiloohm in the top of the circuit. And also plus to minus across the 6 kiloohm resistor in the center of the circuit. So again, we're going to hand up the voltage drops as we go around loop one. And so, if we start at the lower left-hand corner and start up the left-hand side, we encounter the negative polarity of the 12v source first. So it's -12v for our first voltage drop. We come up and we meet the 6k resistor at the top, and we hit the positive polarity first. So it's going to be plus. And then the voltage drop there is going to be 6k(I1), because that's the current flowing through a 6k resistor at the top of the circuit. We then pass that resistor and come down through the resistor in the center part of the circuit. If you come down through the resistor in the center part of the circuit, we see that we have, again, a plus to minus voltage drop. But the current flowing through that 6k resistor is going to be I1 minus I2, because I2 is flowing the opposite direction. It's flowing through the 6k resistor, but it's a contribution from 2. So it's going to be + 6k(I1- I2). And if we go beyond that we get back to the beginning of the circuit, and that's end of our loop. And so that's equal to 0. So our first equation has two unknowns, loop current one and loop current two. Let's look at loop two. Ultimately, if we can find another independent equation, which has I1 and I2 in it, then we can solve these equations simultaneously we have two equations and two unknowns. So let's do that, let's start with the lower left hand corner of loop two come up through the 6K resistor, the 3k, and then the voltage source on the right-hand side of the circuit and back to the beginning. So, first thing we have is the voltage drop across the 6k in the center of the circuit. So starting there we have a negative polarity, it's negative. And it was 6k(I1- I2). Then we continue up, on the top of the circuit, we have the 3k resistor, which just has current I2 flowing through it. We know the polarity of the voltage drop is plus to minus using the passive sign convention, so it's 3k(I2). And then we continue, we hit the positive polarity of the 3v source, so it's + 3v. Continuing on back to where we started. And that's equal to 0. So if we look at equation two it has I1 and I2 in it. And so we have this set of equations where we have two equations and two unknowns, and we can solve for I1 and I2. So when doing when we're doing mesh analysis, that's kind of the first thing we do, is we solve for these mesh currents. So if we solve for I1 and I2 in this problem, we get an I1 = 1.25 mA. And we have an I2 which is equal to 0.5 mA. Now how do those relate to I0? because we're really looking for I0. And so, I0 is the current, again, through the 6k resistor in the center. Going downward through to that branch of the circuit. We know that I0 is going to be equal to I1, which is flowing the same direction as I0,- I2 which is flowing at the opposite direction up through the center leg of the circuit. So if that's the case, we could subtract I2 from I1, and we get 0.75 mA for I0. So we, again, summed up the voltages around the loops. So a point of interest that makes this a little easier, in some cases, to write your equations is looking at the 6k resistor in the center part of the circuit. We've identified that voltage drop as plus to minus 6k(I1- I2), okay? So it's 6k (I1- I2) for that voltage drop, as we're going around the circuit in the left-hand loop. As we go around the right-hand loop, we had to add up the negative of that as part of our loop. It's the very first term in our second equation. And it was a minus 6K(I1- I2). So I'm going to put a negative sign out here in front to reflect that. So that's our voltage drop as we go around the second loop associated with the 6k resistor. It's important to know that this -6k(I1- I2), it's the same thing as saying, 6k(I2- I1), okay? So as we go around this loop, let's look at it again. If we start here at the lower left-hand corner and we start up this loop and this is our first voltage drop we encounter in the second loop. This loop around the right-hand side of the circuit, it's equal to 6k(I2- I1). 6K(I2- I1), it's the same expression as the -6K(I1- I2). So in future problems, we're not going to add up the negative of this voltage drop like we did here for equation two. We're going to express it as we see down here in the equivalent form, just an easier way to remember how to solve these problems. It's taking the voltage drop as the resistor times the positive current in that loop minus the negative associated with the corresponding, or the adjacent loop. So the answer to this problem is I sub 0 = to 0.75 mA.