The topic of this problem is mesh analysis. And we're working with circuits with independent sources. This problem has three independent sources. It has a current source, 2 milliamps, and two voltage sources, a 3 volt and a 6 volt source. The problem is to determine I sub 0. That's the current through the 6 coulomb resistor. It's an output current through the load resistor, which, again, is the 6 coulomb resistor. So if we're going to use mesh analysis to solve this problem, the first thing that we recognize is that we are going to use Kirchhoff's voltage law to solve the problem. And Kirchhoff's voltage law states that the sum of all the voltage drops across any closed loop is equal to 0 at any instant in time. So ultimately, we're looking for I sub 0, which is the current, through the 6 coulomb resistor at the top of the circuit. So to solve this problem, we can identity a variety of meshes. We can then find the mesh equations, and we can solve for I sub 0. But just look at this problem a little closer and see if we can find a shortcut to solving the problem. So we, first of all, notice that we can assign loop currents, let's go ahead and do that, here's our loop currents. It's the first thing when we do, when we are solving mesh analysis problems. Loop 1, Loop 2, Loop 3, And loop 4. Let's look at loop 3. And let's sum the voltages around loop 3. So if we start here at the lower left hand corner of loop 3, and we travel all the way around loop 3 back to the beginning, the first thing we encounter as we take this path is the posipolarity of the 6 volt source. So we have 6 volts, this is for loop 3, 6 volts for our first voltage drop. We continue around this loop, and we have the voltage drop across the 6k resistor at the top. So it's going to be 6k(I3). Continuing around this loop, we encounter the positive polarity of the 3 volt source. It's plus 3 volts, and then we get back to the beginning of our loop, so it's a closed loop. The sum of those voltage drops is equal to 0. So we see that we can instantly solve for I3, using just one single loop equation. And the reason we're able to do this is because that particular loop had two independent voltage sources associated with it. So I3, which is equal to I0, is equal to -1.5 mA.