The topic of this problem is Nodal Analysis. We want to work with circuits with independent sources. The problem is to find the current I1 in the circuit shown below. The circuit has two independent sources. It has a current source on the left hand side which is 2 milliamps. As a voltage source at the top of the circuit which is a three volt source. So, the first thing that we do when we're doing a nodal analysis problem is, we identify the nodes. So, let's do that first. We have a node one on the left hand side at the top and node two at the right hand side at the top. And we have our reference node which is our 0 volt per ground node, it's at the bottom of the circuit. So now, we would then write the node equation about those nodes using Kirchhoff's current law. And in this case we're going to sum the currents into the nodes. So if we do that, for node 1, we could see that node 1 has three currents flowing into it. The current from the 2mA source, the current through the 3k resistor in the center leg of the circuit, and the current through the 3V source on the top part of the circuit. So if we add those currents up, we would find that we would have an additional unknown current associated with a three volt source. We have no way of determining what that current's going to be through the three volt source as it stands. So whenever we have a voltage source which is tied between two nodes where one of the nodes is not a grand node or a reference node, we have to use the concept of super node analysis. So in this case what we're going to do is going to identify our super node as the node which encompasses node one and node two. And our voltage source are three voltage source, so this is our super node. We can then write our Kirchhoff's Current Law equation about that super node as one of our independent equations. So let's do that. So we have our super node. Prior to the equations, summing the currents into the super node. We have a current 2 milliamps flowing in. We have a current up through 3 kilo ohm resistor and we have a current up through the 6 kilo ohm resistor on the right hand side. So summing those currents, we have 2 milliamps flowing in. For the 3k resistor and the current through the 3k resistor is going to be the ground note- v1 divided by 3k + the current up through the 6 kiloohm resistor which's against the ground node minus V2 divided by 6K. And that's equal to 0. So, we look at our super node equation we see that we have two variables V1 and V2 the nodal voltages for node one and node two. So we need another equation which is independent of our supernode equation to solve the problem. So we can write an additional equation which relates V1 and V2 with the voltage drop between those. That's our constraining equation. 3V source is + to- as designated in the problem and so we have V1- V2 = 3V. So we have our two equations. Ultimately, what we're looking for in this problem is we're looking for the current I1 flowing down through the 3K ohm resistor. So really what were most interested in is finding V1 because looking for I1, we know that I1 is equal to V1 minus zero divided by 3k. V1 minus zero / 3K. So if we could find V1, we can find I1, which is what the problem is really about. So looking for V1 using our set of two equations and two end nodes, we got a V1 = 5 volts. If we find V1, plug that back into our equation for I1, then we have an I1 equal to 1.67 mA.