The topic of this problem is Mesh Analysis, and we're going to be working with circuits with independent sources. The circuit shown below has two independent voltage sources, a 7-volt source and a 6-volt source. It also has five resistors of varying values. The problem is to determine the mesh currents associated with this circuit from solving the mesh equations. We know that, when we're solving a problem using mesh analysis, the first thing we do is we use Kirchhoff's Voltage Law, the sum of the voltages around each of the independent loops. So, we're going to do that first. We're going to identify as we do the analysis problems, we identify the meshes and the mesh currents first. So, we're going to identify a mesh current for the loop on the left-hand side of the circuit in a clockwise fashion. I1, we also have a loop on the upper right hand side of the circuit loop two, so we're going to assign it loop current I2, and we also have a loop on the lower right hand side of the circuit and we're going to identify its current as I sub 3. So, we're going to add the voltage drops around each one of these loops to ultimately end up with three equations and three unknowns. We can then solve for our mesh currents I1, I2, and I3. Starting with mesh 1, lower left hand corner of mesh 1, and going around that loop, or that mesh, we first encounter a negative polarity of the 7 volt drop. So it's from the source, and the voltage drop associated with the source. We then can continue around the loop, and we have a voltage drop from the 1 ohm resistor at the top center of the leg of the circuit. The voltage drop across the 1 ohm resistor is going to be I1 minus I2, which is flowing the opposite direction through that 1 ohm resistor, Times the one ohm. So it's going to be plus one ohm times I1 minus I2. So now we have two voltage drops associated with this loop, and we have two more to go. We have the voltage drop across the two ohm resistor at the bottom of the circuit, And the 6v drop associated with the 6v drop associated with the 6v source as well. So continuing this loop, we've add up the voltages across the 7v source, across the 1 ohm resistor at the top center lane, and then now we're down at the 2 ohm resistor at the center lane. And so it's going to be + 2 (I1- I3) because I1's flowing positive fashion through the 2 ohm resistor. I3 is Going the other direction, the opposite direction, so it has a negative polarity associated with it. We then have the 6 volt source, so we have plus, since we had the positive polarity of the 6 volt source first, plus 6 volts and that's equal to zero. Because as we go around this loops, starting at the lower left hand corner we encounter those four voltage drops back to our original starting point. We then have loop 2 to look out. Loop two is the upper right hand loop as one voltage drop across one on resistor the center Another voltage drop across the 2 ohm resistor on the right-hand side of the circuit. And another voltage drop across the 3 ohm resistor in the center right-hand side of the circuit. So we're going to add those up. First of all, starting at the lower left-hand corner of that loop going up through the 1 ohm resistor. Is going to be one times i two minus i one for that voltage drop plus two ohms times i two Continuing through the three ohm resistor. It's going to be a plus 3 the positive currents I2 minus I3 which is flowing in the opposite direction through the 3 ohm resistor. And that's equal to 0 because it's back to our starting point. So that's our second equation for our second loop. We can look at those two equations and we see that we have three unknowns, we have I1, I2, and I3. Our three loop currents are the unknowns in our mesh equations. So whenever we're solving problems using mesh analysis and Kirchhoff's Voltage Law We're ultimately finding those loop currents. There were loop currents allow us to find the voltage drops across each of the elements, the power absorbed or supplied by each element and all other parameters of interest. So let's sum up our voltage drop around loop three. Again, starting at the lower left hand corner of loop 3. And going around that, we have the negative polarity of 6 volt source first, so it's -6 volts. We then encounter the 2-ohm resistor, it's going to be +2 for the resistor, 2 ohms. And then the current, the positive current through that is I3- I1. Then we have a 3 ohm resistor that we encounter, and it's going to be 3 times I3 minus Minus I2 And our last voltage drop is the 1 ohm resistor on the right hand side it's going to be 1 ohm times I3. That's equal to 0. So now we have three independent equations and three unknowns. If we solve for I1, I2, and I3, we get an I1 equals to 3 amps, And I2 = 2 amps. And then I3 = 3 amps.