The topic of this problem is mesh analysis. And were working with circuits with independent sources. The problem is to determine V out in the circuits shown below. V out is a voltages drop across the two milliamps source in the right hand out of the circuits. We have another independent source it's a 12 volt source in our problem and we have four one kilohm resistors. So again we're looking for V out the voltage drop across the two milliamp source. So we're asked to use mesh analysis to solve this problem. So for using Mesh analysis we know the first step in solving the problem is to assign our meshes or loops and the associated mesh currents. So we are going to call this mesh one with mesh current one in the lower left hand side of our circuit. We're going to have mesh two. Is the mesh at the top of our circuit, and we're going to have mesh 3 as the mesh from a lower right hand side of our circuit. So now, we've assigned our meshes and we have assigned our mesh currents. The next thing we do is we sum up the voltages around each one of our meshes in order to create three independent equations where we can solve ultimately for our mesh currents. We know the result of mesh analysis is the solution to the diminished currents. So if we can find I 1, I 2 and I 3 then we can find any other quantity of interest for our circuit that is any other voltage dropped, any of the powers associated with the elements who our circuit shown in the problem. So we are going to find I1 and I2 and I3 and we're going to sum up the voltages around each one of this loops and solve these three equations simultaneously to find I1, I2 and I3. So if we start in the lower left end corner of this loop and we go around through this loop. We encounter first a voltage drop for the 1-kiloohm resistor. Namely, it's going to be 1k times I1. Continuing, we encounter the 12V source and the negative polarity of that 12V source, so it's minus 12 with this voltage drop across this element, And then we encounter a 1 kilohm resistor in the center lower part of our circuit and the voltage drop across this 1 kilohm resistor is going to be I sub 1- I sub 3 x 1k. Because I sub one is flowing in the same direction as we're summing the voltages, we're summing voltages clockwise. I sub one is flowing clockwise around our loop one. That's the last of our voltages before we get back to the starting point so Kirchhoff's voltage law tells us that the sum of those voltages is equal to zero. So we have an equation which has two unknown,s I sub one and I sub three in it. Off we go to our second loop and we start solving the voltages around this loop in the same fashion. Again starting at the lower left hand corner of our loop, we encounter the 1 kiloohm resistor first as a voltage drop of 1k times I sub two. And then we encounter another one kiloohm resistor. The voltage drop there is 1k, I sub two minus I sub three and then we encounter again the 12 volt source but this time we encounter the of that first plus 12 And that's equal to zero. It gets us back to our starting point. So we have a second independent equation and we have three unknowns. A third equation is for our third loop and we see that in our third loop. We could sum up the voltage drops across the two resistors but when we get to this current source we wouldn't know what the voltage drop is across this so we'd have to assign to get another variable for the voltage drop drop across is fi we're going to sum up the voltages around this loop. But we noticed that I sub three is given to us directly in this loop. They made that I sub three is equal to minus two miliamps. So we have I sub three and we have two equations that are independent of our third equation to find I1 and I2. Ultimately what we're looking for is Vout. And Vout again is this voltage drop across the 2 milliamp source. And we also similarly know it's a voltage drop from this node, output node, to the ground node. And that can be also determined by the voltage drop across the one kilohm resistor to the voltage drop across the Other 1k ohm resistor in the center bottom of our circuit. We add these two up, and we can get Vout. So, Vout = 1k, for this first voltage drop, I2- I3 And for the second voltage drop is going to be 1K, I1- I3 that's equal to V out. So if we have I1 and I2 and I3 then we can find V out So if we solve this then we end up with I1 equal to 5 milliamps and I2 equal to -7 milliamps, we have I3 already So, if we use these and solve for Vo, than we get a Vo=2V.