The topic of this problem is nodal analysis. And in this problem, we're going to solve a circuit with independent voltage sources. The problem itself is to find the current I sub 0, and I sub 0 is defined as a current flowing down through the 2 kilohm resistor in the center of the circuit. So what were doing now is the first thing we do is we identified the nodes in the circuits, so lets do that. The top node we going to call node one, we also have a node on the left-hand side of the circuit, and we're going to call that node two. We have a node in the center of the circuit, we're going to call that node three. We have a node at the right-hand side of the circuit, node four. And we have the reference node or ground node at the bottom of the circuit, which is at zero volts. So we're going to write the nodal equations about each of those nodes. And to do that, we use Kirchoff's current law, and in this case we're going to sum the current into the nodes. With Kirchoff's current law, we know that we can either sum the currents into or out of the nodes and either one will work. We just need be consistent through out the problem when we're solving it. So, the first thing I'm going to do is we're going to look at node one. And we look at node one and we see that the top node has three currents flowing into it. We have current flowing through the 2 kilohm resistor on the left hand side. We have current flowing through the 2 kilohm on the right hand side. And we have current flowing up through the 12 volt source. So if we were to write an equation about that node then we'd have to introduce an additional known current. Let's call it high 12 volt because we don't know the current that's flowing up to the 12 volt and we cannot identify it without introducing an additional variable. Since we have four nodes we expect to have four equations using our conventional Kirchhoff's current law solution process. But because we have this additional unknown for the high 12 volt, the current to the 12 volt source we would end up having four equations and five unknowns, and would not be able to solve the problem using that approach. So what we do in problems where we have independent voltage sources which aren't tied to ground, so that we know that the reference voltage on at least one side of that voltage source. We use the concept of a super-node analysis to solve the problems. So in this case, we're going to identify a super-node, which encompasses node one and node three, and the 12 volt source. So it will travel and connect and encompass node one, node 3 and the 12 volt source that is our super node. Once we've identified that we need a supernode in the problem, we can then go about summing the currents about the different nodes. So for the super node we see that we have a current on the left hand side coming in through the 2 kilohm resistor on left hand side. We have a current flowing into the super node through the 2 kilohm resistor on the right hand side. Similarly, we have currents flowing in through the 1 kilohm of the left hand side and in the 1 kilohm of the right hand side, and we have the current flowing up through the 2 kilohm resistor at the bottom of the circuit. So let's add those currents up, starting with the kilohm resistor on the left hand side and the current through it upwards into the super node. That current and this is the super node equation, Is V2- V1 divided by 2 kilohms for the current through the 2 kilohms resistor flowing into the super node on the left hand side of the circuit. Similarly for the 2 kilohm on the right hand side we have the current V4- V1 divided by 2 kilohms for the current flowing upward through the 2 kilohms resistor into the super node. We also have the current flowing left to right through the 1k resistor in the center of the circuit that's V2- V3. Divided by 1 kilohm the current through the other 1 kilohm resistor from right to left from the node 4 to the node 3 is V4- V3 divided by 1 kilohm. And then we have the current, which is flowing upward through the 2 kiloohm resistor in the bottom of the circuit, and that's going to be the reference voltage, which is zero, for the ground node minus V sub 3 divided by 2 kilohm and that's equal to 0. So we have our five currents flowing into the super node that constitute our super node equation. Now we're going to go to node 2 and we're going to send the currents into node 2. We notice that in node 2 that we have a voltage source tied to it which is the six volt source. We know that also in nodal analysis what we're ultimately looking for are the nodal voltages. So in this problem, since we have a voltage source between node 2 and the ground node which is at zero volts, we can determine the nodal voltage for node two very easily, and we can do that simply by writing the equation relating the nodal voltages with that source. The bottom node is at zero volts, it's tied to the positive polarity of the six volt source. So we have zero and we noticed that node 2 is tied to the negative polarity of the 6 volt source, so it's -V2 = 6 volts. We have equation for node 4. I'm going to go ahead and call that the node 4 equation. And in a similar fashion, we can find nodal voltage for node 4 directly from our circuit. And in this case, it's the V4 which is tied to the post polarity of the 12 volt source minus the ground. So it's V4, minus the reference voltage which is at ground, 0 volts, is equal to 12 volts. So we know V2 and V4 from these two equations. If we look at our super node equation, we have four unknowns. So right now we have three equations and we have four unknowns two of which we've found very easily. The last equation that we need is the equation which relates the node altages to our supernode. And so if we look at the supernode, we can easily see that for the supernode are constrained equation for our supernode. Is going to be V1- V3 = 12 volts. So now we have four equations and we have four unknowns V1, V2, V3, and V4. Ultimately what we're looking for in this problem is we're looking for I sub 0, I sub 0 is related V sub 3. If we were to write the equation for I sub 0, we see that I sub 0 = V3 minus the ground voltage, so that's coming down through, from V3 down to ground through the 2 kilohm resistor that's our I sub 0. So if we're looking for I sub 0, really what we're after is V sub 3. Because once we find V sub 3, we can solve for I sub 0. Right now we know of V2 is is minus 6 volts and we know of V4 is it's 12v but we don't know what V1 is and we don't know what V3 is. So we're going to solve our set of equations for V3. And if we do that, we end up with a V3 which is equal to minus six-sevenths of a volt. Once we have that, we can then solve for I0 and I0 is going to be, Using our equation that we have on the left-hand side of our board, we have an I0 which is equal to minus six-sevenths of a volt, divided by 2 kilohms and that comes out to be minus three-seventh milliamps.
