The topic of this problem is mesh and nodal analysis. And we're working with circuits with independent sources. The problem is to determine V sub 0 and the circuit shown below. It's a circuit that has two independent sources, a 6 volt source and a 2 milliamp source as well as four resistors. Vout is the voltage that's measured across the 6 kilo ohm load resistance on the right hand side of the circuit. So we're going to use both mesh analysis and nodal analysis to find Vout. In fact what we'll find is that whether we use or the other to get the same Vout for this problem. So let's start with mesh analysis. We know when we're doing the mesh analysis, and using this solution, approach for our problem. We are finding the mesh currents. So the first thing that we do for mesh analysis is we assign the meshes and the mesh currents. Ultimately, what we get from our mesh analysis is the value for each of the mesh currents, I1, I2 and I sub 3. Once we know those three currents, we can find any other value of interest for any of the elements in the circuit. We can find the currents, we can find the voltages, we can find the power that's absorb nor supplied by each of those elements. In this case, what we're looking for is V out. So with regard to mesh analysis, V out is simply going to be 6 kilo ohms times the mesh current I sub 2. If we can find I sub 2, then we can find V out because we know that V out for mesh analysis is going to be I sub 2 times 6K. So let's start with mesh one and let's add our voltages up around mesh one or loop one to find our first independent equation. So starting in the lower left hand corner, we first encounter the -6V source and the negative polarity of that 6V source. We continue, we can count to the 4 kilo ohm resistor, the voltage for the 4 kilo ohm resistor is 4k and it's I1 which is flowing in the same direction as we're summing our voltages minus I2 which is flowing the opposite direction. And in coming down to the 2 kilo ohm resistor, the voltage drop there is going to be 2k I1 because is again flowing in the direction as we're summing our voltages minus I3. And they get spare to the starting point and the sum is equal to zero. So if you look at our first independent equation, we have three unknowns, I1, I2, and I3. So we need at least three equations, which are independent, to solve for these three unknowns. So our second equation comes from our mesh 2, or loop 2. Again, starting at the lower left hand corner, summing up voltages as we go around this loop, we encounter the 2 kilo-ohm resistor first. The voltage drop there is 2k (I2- I3). And then we encounter the 4K resistor. Voltage drop there is 4K(I2- I sub 1). And then the voltage drop for our 6 kilo ohm resistor which is 6K I sub 2 and that's equal to 0. That's all the voltage drops around this loop before we get back to the beginning. Our third equations for third mesh of third loop, and we see that we can find our mesh current I sub 3 directly from this loop namely I sub 3 is equal to 2 milliamps. Because ultimately, what we're getting from mesh analysis is we're getting the mesh current and we're able to solve for I sub 3 directly from inspection of our circuit. So now we have our three equations and three unknowns, and we can solve for either the two unknowns. We can solve for I1 and I2. What we're really interested in is finding I2. If we could find I2, then we can find Vout. So I sub 2, I'm solving this equation simultaneously, it comes out to 1.14 milliamps. And if we want to find V out from that, the Vout is going to be 6K times I sub 2 and that results in a voltage of 6.86 volts. So Vout which is what we're looking for in the circuit through Mesh analysis comes out to 6.86 volts. Now I'm going to solve the problem using nodal analysis. And when we're using nodal analysis, we're summing the currents into each one of the nodes. And if we do that, then we can find the nodal voltages. Once we know the nodal voltage is, we can then solve for any other current or any other power associated with the elements in the circuit. So the first thing we do when we're performing nodal analysis on a circuit is we assign the nodes. So we're going to start with this leftmost node and we're going to call it Node 1. When it call the way in the center of the circuit node 2, we're going to call the one at the top of the circuit node 3. And we're going to sum the currents into each one of those nodes. So we're going to use node analysis somewhere to sum the currents into the nodes. And so starting with node 1, so many currents into node 1. We have the current 2 mA and then we also have this current which is associated with the 6 volt source. And we don't know what the current is associated with this 6 volt source. So, we have to use another approach to solve the problem. Let's look at another node, let's look at node 2. If we go to node 2 and we sum the currents in the node 2, then the current through this 2 kilo ohm resistor is going to be V1 minus V2 divided by 2k. We see the current through the 2 kilo ohm resistor in the bottom center of the circuit is going to be 0 minus V2 Divided by 2K for the voltage drop 0 minus V2 because this is our ground node at the bottom divided by 2K. And our last current in the node 2 is the current that we have here namely V3 minus V2 over 4k. And that's equal to 0. So if we look at this equation, we have three unknowns, V1, V2, and V3. And if we look at node 3, we have a similar situation we had with node 1. We have a current which is flowing through this 6V source and we don't know what that current is. We know it's I1 but that's all we know. And we don't know how to otherwise find the current flowing in to that direction, but we do have a constraint equation from this voltage source that helps us. That is V3- V1 Is equal to 6 volts. So now we have two equations for our nodal analysis. We need a third equation which is independent of our first two because we have three unknowns, we have V sub 1, V sub 2, and V sub 3. Our third equation comes from the concept of a super node, where we take this 6 volt source which is in our circuit. And we isolate it and create a super node where we're going to sum the currents In to this super node, namely will sum the 2 milliamps with what's flowing to the 2 kilo ohm, which is what's flowing to the 4 kilo ohm, which what's flowing to the 6 kilo ohm. So our third equation is 2 milliamps, which is the current flowing into this node plus V2 minus V1. Over 2K + V2- V3 Over 4 K plus 0 minus V3 divided by 6 K and that's all four currents that flowing in to our super node. And the sum of those is equal to 0. So now we have three equations and we have three unknowns. So we can find V1, V2, and V3. Really what we're interested in is finding V3 because we know that V3 Is equal to V out. So if we can find V3 then we can find Vout and so if we solve these equations simultaneously, then we find that V sub 3 is equal to 6.86 volts which is the same solution for Vout as we obtain using mesh analysis early in the problem.