The topic of this problem is Nodal Analysis. And we're going to do it work on a circuit with Independent Voltage Sources associated with it. And the problem is to find the Nodal Voltages associated with the circuit. So we can see from our circuit that we have two Independent Voltage Sources, a 12 volt source on the left hand side, a 6 volt source on the right hand side. We also see that there's four resistors in series and parallel combinations and with the other elements in the circuit. So as we deal with all Nodal Analysis problems, our first tip is to identify the nodes in this circuit. And so, we have a node on the left hand side, we'll call that node 1. We have a node in the center of the circuit, node 2. Node on the right hand side which is node 3. And we have the ground node at the bottom. And the ground node at the bottom is our reference node and it set zero volts. So I'm going to write the Nodal equations around each one of these nodes. And again, what we're looking for when we do Nodal Analysis is the nodal voltages. So if we can find the nodal voltages for node 1, node 2 and node 3. Then we've accomplished what we've set out to do using Nodal Analysis. So first of all, we can look at node 1. And if we look at node 1 we see that there are three currents flowing into node 1. We have a current flowing up through the 12 volt source. We have a current through the 12 kiloohm resistor. And we have a current through the 9 kiloohm resistor that all feed into node one. But again, when we're dong Nodal Analysis, we have to remember that we're looking for the node voltages. So when you have Circuits with Independent Voltage Sources which are tied to ground, like what we see in this example. Then we know what the voltage is at node 1 because we know what the reference voltage is at the bottom of the circuit. We have a 0 volts at the bottom of the circuit, therefore the voltage at node 1 has to be 12 volts. We have V sub 1 using Nodal Analysis. We have V sub 1 -0 for the bottom node is = 12 volts. So we know that V sub 1 is 12 volts. So that solves for our first unknown V sub 1. The next node, if we look at node 2. We have three currents flowing into node two, which are coming through the three resistors, the two 12 K resistors and the 6 K resistor. So we're going to sum up those currents. And we're going to use Kirchhoff's current law to do this. And we're going to sum the currents into. Node 2. So if we do that, first of all, for the 12 kiloohm resistor on the left hand side of the circuit. Current flowing from left to right through the 12k resistor is going to be V1- V2 over 12 kiloohms. Similarly, for the current flowing through the other 12 kiloohm resistor on the right hand side, from right to left, is V3- V2 divided by 12. So it's going to be V3- V2 divided by 12 kiloohms. And we have the third current which is flowing through out the 6 kiloohm resistor. And if we add that current is going to be zero for the bottom node- V2 divided by 6K. That's the current flowing up through the 6 kiloohm, from bottom to top as shown. And that's all the currents associated with node 2 so that's equal to 0. Now, we have a third equation for node 3. And similarly to node 1, we have a voltage source between node 3 and ground. Ground has 0 volts, it's our reference node, so we can then determine what node 3 voltage is. And we know that 0- V3, 0 down here, positive polarity of the source, -V3 is = 6 volts. Therefore, V3 = minus 6 volts. We know that V1, from the first equation, is equal to 12 volts. And if we know those two volts just then we can use our second equation and solve for V2. And if we do that, we end up with a V2 equal to three halves volts.