The topic of this problem is nodal analysis. And we're going to work with circuits with dependent voltage sources. The problem is to find the current i sub 0 in the circuit. And so i sub 0 is the current which is flowing downward through through 12 kiloohm resistor on the left-hand side of the circuit. We've noticed that we have the voltage controlled voltage source in our circuit. Our voltage is 2v sub x where v sub x is the voltage drop across the 6 kiloohm resistor in the center leg of the circuit. We know that with node analysis, the first step in solving the problem is to identify the nodes in the circuit. So we'll do that first. We notice that we have a node one on the left-hand side of the circuit. We have a node two in the center of the circuit. We have a node three on the right-hand side of the circuit. And we have a reference node at zero volts at the bottom of the circuit. So now we've identified our nodes. So we have three nodes and we know with nodal analysis ultimately what we're looking for are the nodal voltages. Once we have the nodal voltages, we can find any of the other parameters that we would like to find in the circuit. So let's solve this circuit. The first thing we want to do is we want to sum the currents about each one of the nodes. So, if we look at node one, we see that we have three currents flowing into node one. We have the current flowing through the 6 kiloohm resistor right to left. We have the current flowing up through the 12 kiloohm resistor. We have the current flowing right to left through our dependent voltage source. So if we were to write those currents and add them up we would have a known current based on nodal voltages for the 12 kiloohm resistor and for the 6 kiloohm resistor. But we would have two entities, an additional unknown current for the dependent voltage source. So we'd have the current flowing up through the 12 kiloohm, the current flowing downward through the 6 kiloohm and the current flowing right to left through the voltage source. This current through the voltage source would introduce an additional unknown variable, i sub 2, v sub x. So we'd end up with four unknowns and three equations and we'd not be able to solve the problems using that approach. So what we have to do is we have to use the concept of supernodes to solve this problem. We use the concept of a supernode anytime that we have a voltage source which is tied between two nodes floating up in the circuit somewhere and one of those nodes is not a ground node. So we don't have our reference node on one side of that voltage source. So we have no way of determining the current through that voltage source. So what we're going to do is we going to use the concept of supernode. And we're going to first identify our supernode as a supernode which envelopes node one, node two, and that dependent source, as we see from the dashed lines. That's our supernode. So we're going to use Kirchhoff's current laws and we're going to sum the currents into the nodes starting with the supernode. And so for the supernode, our equation's going to be the sum of a current coming through this 6k resistor at the top, coming up through the 12k resistor on the left-hand side, coming up through this 6k at the center of the circuit and the current flowing right to left through the 12 kiloohm resistor in the center of the circuit. So there's going to be one, two, three, four currents as part of this equation. Starting with a 12 kiloohm on the left-hand side of the circuit, the current flowing up through that is going to be the ground node 0 volts minus v1 divided by 12k. We also have the current flowing through the 6 kiloohm resistor at the top of the circuit, right to left, v3 minus v1 divided by 6k. We have the current through the 12 kiloohm resistor on the right hand side of the circuit, v3 minus v2 divided by 12k. And we have a current which is flowing up through the 6 kiloohm resistor and the center of the circuit, 0 for the ground or reference node minus v2 divided by 6k. And that's equal to 0. That's our first independent equation for the supernode. We look at that equation, we see that we have three unknown nodal voltages in that equation. So we're going to need three equations to solve for these three unknowns. A second equation we're going to write is about node three. So I'm going to call that three for node three. And just as we've seen in previous problems, we notice that there is a 6 volt source between node three and the ground node. And again, noting that for nodal analysis what we're ultimately looking for are the nodal voltages, we can find for this problem that we can directly calculate or find the nodal voltage for node three from this relationship between the voltage source and our known ground node. So in this case, v3 which is tied to the positive polarity of the 6 volt source minus zero is equal to 6 volts. So we can find v3 directly is equal to 6 volts. That's one of our unknowns. Our last equation is our constrained equation for the supernode. And we know that for the supernode, v1, tied to the positive polarity of our independent source, minus V2, tied to the negative polarity, is equal to 2 v sub x. So now we have this additional unknown v sub x that we've introduced into our set of equations. Now we have four unknowns and so we use our second constraining equation for this problem which relates our controlling variable v sub x to our nodal voltages v1, v2, and v3. So that equation is going to be v2, which is tied to the positive polarity of v sub x minus 0 is equal to v sub x. So v sub x is equal to v sub 2. And so now we have four equations and four nodes that we can solve for. Remember for this problem ultimate, we're looking for i sub 0. I sub 0 can be found if we look at the equation for i of 0 using Ohm's law, it's going to be v1 minus 0 divide by 12 kiloohms. So what we really need to find is that nodal voltage for node one. That's really the only one of interest to us for this problem. Once we have v1, then we can solve for i0. So if we use our four equations that we have here and solve for v1, we end up with a v1 which is equal to nine halves of a volt. If v1 is equal to nine-halves of a volt, then we can find i0, and our i0 comes out to be three eighths milliamps, simply by plugging in our v1 into our equation for i0.
