The topic of this problem is nodal analysis, and we're going to work with circuits with dependent sources. The problem is to write the nodal equations for the circuit shown. As you can see, the circuit has a dependent source on the left hand side the dependent source is beta I0. So we have a current controlled current source. The I0, which is controlling the current controlled current source, is the current down through resistor 3 in the center-right side of the circuit. So we're going to write the nodal equations, and what we want to remember when we have these circuits with dependent sources is we're going to need an additional equation relating the controlling parameter for the dependent source with our nodal voltages. So we're going to find an equation which relates I sub 0 with our nodal voltages as our third equation. So as we have done in all our nodal analysis circuits the first thing we do is we identify the nodes, that's the first step in solving the problem. So, we see that we have a node at the left hand side of the circuit at the top and I'll call that node one and we have a node at the right hand top part of the circuit and we're going to call that node two. We also have a ground node it's at the bottom of the circuit. That we know is at 0 volts reference so that we have a reference point for our calculations. So we're going to go through it and write our nodal equations at this point after we have identified our nodes. The first one is node 1 and we're going to use Kirchoff's current laws to write our nodal equations. That is, the sum of the currents into any node at any instant in time is equal to zero. So we're going to use Kirchoff's current law, and we're going to sum the currents into the nodes. Again, with kirchoff's current law we can choose to sum the current into the nodes. Or we can sum the currents out of the node, we just need to be consistent throughout the problem. So we're going to Kirchhoff's current law, and we're going to sum the currents into the node, starting with node number one. So for node number one, we have beta I sub 0 flowing out of that node. So we're going to put in a minus Beta I0 for that current we also have current flowing up through R sub 1 and if we add that current is going to be the current flowing up through R1 so it's going to be the bottom node. Minus the top node voltage divided by the resistance, R1. So that's going to be the bottom node, which is 0 volts- V sub 1, which is our nodal voltage at the top for node one, divided by R1. And then we have one more current flowing into that node. These are current flowing through the resistor R2 at the top of the circuit. That resistance is between node 1 and node 2. So if we're looking at the current flowing into node 1 through resistor R2. It's V2 minus V1 divided by R2, V2- V1 divided by R2. And that's equal to 0, that's our last current flowing into node 1. So that's our first independent equation. We have a second equation that we can write around node 2. And again, summing the currents into node 2. We know that we have I sub A flowing into node 2, it's on the right hand side of the circuit. And then we have the currents which are flowing through R3 and R2 into node 2 as well. So looking at R2 first, we have the current flowing left to right. So it's going to be V1 minus V2 divided by R2. V1 minus V2 divided by R2 and we have one other current the current flowing up through R3 flowing into node 2. So that's going to be our ground nodes, 0 volts- the nodal voltage at that second V2, divided by R sub 3. And that's all of our currents flowing into node 2. So that's sum is going to be equal to 0. So now we have two independent equation. And as we can see, we have the two equations that we have three unknowns, we have V1 and V2 for the nodal voltages and we also have I sub 0 which is in there as well and I sub 0 comes from our dependent source, it's beta I sub 0. So again, we need that third equation which relates our controlling parameter I sub 0 to our nodal voltages and so we have to think about that third equation for our set of three equations and our three unknowns. So how do we make that relationship? We see that I sub 0 is flowing down through R3 and then if we wanted to find I0 we can simply use Ohm's Law to find that. Where we know that I0. Is equal to V2. It's a node voltage at the top, minus the node voltage at the bottom which is 0 volts divided by the resistance R3. So, now we have this third independent equation. So, we have three equations, three unknowns, and we could solve for I0, V1, and V2.