The topic of this problem is Mesh Analysis. And we're working with circuits with independent sources. The problem is to find the current I sub zero in the circuit shown below. We have a circuit where I sub zero is a current which is passing through a four kiloohm resistor at the top of the circuit. So the first thing we do when we're performing mesh analysis on a circuit is we assign the meshes or loops and we assign the mesh currents. So I'm going to assign my meshes in the clockwise direction and assign mesh currents I1. I2, and I3. So when we're going mesh analysis, we know that we're using Kirchhoff's Voltage Law as a way of determining the mesh current. So the result of mesh analysis is the mesh current. So all the mesh currents, in this case I1, I2, and I3. We're looking for the current I sub 0 in this problem. We know that I sub 0 is going to be equal to what we have now designated as I sub 2. So if we can find I2, then we know what the solution to the problem is. So the first thing we do is we start looking for I1, I2 and I3 and we notice that in our first loop that we can immediately determine I, because it is the only loop or mesh current this passing to our current source our 2 million source. So our first equation from our first loop is simply going to be I1 = -2 mA. So it's just flowing in the opposite direction of the way I1 has been designated as flowing, which is clockwise. If you look at loop 2, we can sum the voltages up around loop 2 and we can use Kirchhoff's voltage law to do that. So starting at the lower left hand corner of this loop or mesh we first encountered the 12 volts source and in fact the minus or negative polarity of the 12 volts source. We continue and we encounter the 2-kilohm resistor. We know the voltage drop across the 2-kilohm resistor is going to be 2 kiloohms times I2, which is flowing the same direction as we're summing the voltages, minus I1, which is flowing the opposite direction. So our second voltage drop as the 2-kilohm resistor I sub 2 minus I sub 1. Then at the top as we go through the 4 kilo ohm resistor we have another voltage drop which is simply 4 K I 2. Because I2 is the only loop where mesh current flowing through it. And the we get on the left hand side of this, sorry the right hand side of this. Loop and we encounter the 6 kilo ohm resistor and we know the voltage drop for that is going to be 6 K (I2- I3). And then we get back to the starting point of our loop and that sums up all of our voltages and that's equal to zero. So here we have an equation which is three unknowns in it. We have first equation wherein one of unknowns is found for us by using our circuit analysis skills, and so we're going to look at loop three next. And again, it's like loop 1. We notice that loop 3 gives us the loop current directly. Namely, I sub 3 = 4 milliamps. So we know I1, we know I3. We could put both of those into our second equation. And we can find I2. And so we find I2 which we said is equal to I sub 0, is equal to 2.67 mA.