The topic of this problem is Mesh Analysis. And we're going to work with Circuits with Independent Sources. The Problem is to Determine the mesh currents in this problem. And, we're going to introduce the concept of super mesh. In order do this. We see that we have two independent sources in this problem. We have a 7 volt source on the left hand side of the circuit, and a 7 amp source in the center bottom leg of the circuit. We have also have four resistors, that are associated with the circuit as well. What we're going to to do, is we're going to use mesh analysis to solve the circuit. So we know that we're going to be using Kirchhoff's Voltage Law to solve the circuit. We know that Kirchhoff's Voltage Law tells us that, the sum of the voltages around any closed loop at any estimate time is equal to zero. So we use that concept and that law to find three equations for our three loops that we have in this problem we identify the three independent loops. We'll ultimately have three loop currents, that we solve for also known as, the mesh currents. So let's first identify the loops for the meshes, that's the first step when you're doing mesh analysis. So our first loop, we're going to identify as loop one, on the left hand side of the circuit there has a current of I1 associated with, we have a mesh two at the top, right hand side, the current I2 associated with it. And on mesh three, lower right-hand side with the current I sub 3 associated with. So those are our three meshes, they're independent. And the three currents associated with those meshes. So let's look at this, and let's try and solve these problems. We're going to first try and solve it, and see what we come up with, without using the concept of a super mesh. So, let's look at the first loop and see what happens when we try to sum the voltages around that loop. Loop one. Starting at the lower left hand corner of this loop and travelling around in the clockwise version, we first encounter the 7 volt source and the negative polarity of the 7 volt source so it's -7 volts. We then continue around this loop and we encounter the 1 ohm resistor. It has the voltage drop of 1 ohm I1 which has a positive current flowing through it, minus I2 which is flowing the opposite direction through the 1 ohm resistor. So it's +1 (I1- I2). Let's continue. So we go down through the 1 ohm resistor and the voltage drop across it. The next one we want to add is the voltage drop across 2 ohm resistor in the middle of the circle. It's going to be +2 (I1-I3), which is the one in opposite direction of I1. We then go down, and then we encounter the 7A source, and the voltage are up across the 7A source. We don't know what that voltage drop is, so we'll have to add a term, maybe call it V7 amp for that voltage source and that's equal to 0. That's our first voltage blue for our first mesh that we've have had the voltages around, and gets us form the beginning around each element and back to their starting point. If you look at that equation, we'll see that we have a leak current, I 1, I 2 and I 3 which are unknowns, but we've introduced another unknown. And then as the voltage drop across a 7 amp source. Unfortunately, we can't come up with four equations. We have three independent equations for loops 1, 2, and 3. So if we try to solve this problem like this, we're going to have more unknowns than we do equations. We're going to have three equations from our three loops, and we’d have four unknowns from the way we've worked the problem. So that's not going to work, we can't solve this problem with four unknowns and three equations. So this brings up the concept of a super mesh. And what a super mesh allows us to do, is it allows us to write are three unknown, are three independent equations without encountering this voltage drop associating with the 7 amp source. So when we're doing Mesh Analysis, and we have a current source which is shared between two loops, we have to use the concept of super mesh in order to solve the problem, because if we don't, we're always going to have this problem of this, this two voltage drop across the current source that we can't solve for. So let's use the concept of super mesh. And we're going to acknowledge that this first equation, while it's written correctly, isn't going to get us a solution because it has four unknowns in it. We only have three equations. So we're going to use the concept of a super mesh. So the super mesh, is a mesh that allows us to, again, write an equation which is independent of our other two mesh equations, but doesn't travel through our 7 amp current source. So let's start over here in the lower left hand corner and let's identify what our mesh might look like. So it may come up here, come around, come down. And we don't want to go through the 7 amp source, so we're go around it. Like this back to the beginning. So if we travel clockwise around this loop. It will be a loop which is independent of loop 2 and loop 3, if we will. But, it gives us this equation that we're looking for. So let's do this. Let's call it loop 1, our super mesh loop. And let's write the equations. So the equation for the super mesh loop. Starting at the lower left hand corner, we have a -7 volts. Continuing on, we have the voltage drop across one on the resistor. Which is 1 (I1-I2). Continuing around this loop, we have voltage drop across the three ohm resistor, which is three, and and then a positive current through this, and that direction's going to be I3-I2, so it's going to be +3 (I3- I2). Continuing onward, we go down through the resistor on the right hand side of the circuits, at 1 ohm resistor, and it's going to be 1 (I3), which is the current flowing through the 1 ohm resistor. Continuing, back to the start and that's = 0. So now we have a super mesh equation that has a three unknowns, I1, I2, and I3 so we have this first equation which is our first independent equation. Now let's go to loop 2, and do the same thing. The sum up the currents or sum up the voltages around loop two. So we start again at the lower left hand corner of loop two and we encounter the 1 ohm resistor first, it's 1 (I2-I1) since I 1's flowing the, or the opposite direction. We come over and we encounter the 2 ohm resistor on the right hand side of the circuit. It's going to be + 2 I2, and then we encounter the 3 ohm resistor, so it's going to be 3 times the positive current flowing through this is I2, so it's + 3 (I2-I3), because the three caught in the third loop is flowing the opposite direction. And that's going to be equal to 0, that gets us back to the beginning point. So we have a second independent equation which has three unknowns. I1, I2, and I3. So we need one more equation. If we look at loop 3, we're going to have the same problem with loop three as we had with loop one. If this is our loop three, the first thing we encounter is the 7 amp source. We don't know what the voltage drop is across it, we'd have to introduce another variable, v 7 amps, but we're not going to do that. What we're going to do is we're going to look at this current source, and we're going to recognize that, that current source, 7A, is equal to what? It's equal to I1, which is flowing in the same direction as the current source, minus I3. So that gives us three independent equations. The one for the super mesh, the one for loop 2, and a constraining equation which I called, equation 3 is really not loop 3, it's really, kind of a constraining equation which relates that current source with the loop currents. So we have three equations and three unknowns. If we solve for I1, I2, and I3, we get I1= to 9A, I2 = 2.5 A. And I3 = 2 A.
