0:03

The topic of this problem is nodal analysis and

Â we're going to work with circuits with independent sources, and in this case,

Â we have two independent current sources and we have a independent voltage source.

Â 0:17

And we'll use nodal analysis so that means that we first of all, we have to identify

Â our nodes and secondly, write the nodal equations about those nodes.

Â So first of all, finding our nodes and then identifying the nodes.

Â We have a node 1 at the left hand, top left hand side of the circuit,

Â node 2 to top right hand side of the circuit and

Â of course we have our ground node, our 0 volt node at the bottom of the circuit.

Â 0:59

So let's look at that and see if we can do that.

Â So first of all starting at node 1, what do we see for node 1?

Â Looking at the currents flowing into the node 1,

Â we have a 6 mA source flowing into node 1, so itâ€™s going to be 6 mA.

Â 1:17

We have a current flowing up through the 6 kilo ohm resistor and

Â thatâ€™s going to be the 0 volt reference node at the bottom,

Â current flowing up to the 6 kilo ohm to the node 1 with voltage V sub 1.

Â So it's 0 minus V sub 1 divided by 6k and then we also have

Â the current which is flowing through the 6 volts source at the top.

Â 1:47

So if we write this equation, we would then write an equation for

Â some unknown current through this 6 volt source.

Â So I'm going to call that I 6v and

Â I'm going to identify then the right to left direction, and that's equal to 0.

Â 2:09

So we know from using nodal analysis and

Â Kirchhoff's current laws that what we're looking for are the nodal voltages.

Â So typically with Kirchhoff's current law,

Â we would sum up the currents about each of our nodes, and we have

Â a simultaneous set of equations which have a certain number of unknowns.

Â In our case, this would have two unknowns, V sub 1 and V sub 2 for

Â the different nodes and we would then solve our two equations or

Â our two unknowns to find V sub 1 and V sub 2.

Â But in the approach that we've just taken, we've introduced another unknown.

Â We've introduced the current through the 6 volt source as an unknown as well

Â because we have no way of determining that current through the 6 volt source.

Â So this approach is not going to work for us in this problem and the reason for

Â it is because we have a voltage source, an independent voltage source which is

Â between two nodes, but neither of those nodes are the ground node.

Â So we don't have a reference voltage for

Â the 6 volt source from one node to another node.

Â So ultimately if we were to continue writing these equations,

Â we'd have two equations and we'd three unknowns and it would be unsolvable.

Â So the way we get around this is we introduce the concept of a super node.

Â And what this super node allows us to do is it allows us to kind of block off

Â that independent voltage source which is

Â floating up in the circuit between node 1 and node 2.

Â And if we do this, then we'll be able to then write our two equations about

Â our nodes and then we'll also be able to, actually we're going to write one

Â equation around a super-node and then we're going to have a second equation

Â which relates the nodal voltages to that 6 volt source.

Â So let's do this and see if we can figure out how to do this problem.

Â 4:09

So the first thing we do is we identify what we call a super node and

Â the super node in this problem is the node about the 6 volt source.

Â So we've taken this 6 volt source and

Â we've kind of isolated it from our analysis.

Â And what we're going to do is we're going to take this,

Â which we're going to call the super node, this is our super node and

Â we're going to sum the currents into the super node.

Â So this equation, we have equation one is not going to work for

Â us in solving this problem.

Â 4:56

We would sum the currents into that super node.

Â We have current flowing from the 6 mA source flowing into it,

Â we have current flowing up through the 6K resistor.

Â We have current flowing up through the 12K resistor into this super node and

Â the current flowing up through the right hand side of the circuit as well.

Â So let's add those currents up.

Â First of all, we have 6 mA on the left hand side,

Â we have a current flowing up through the 6 kilo ohm resistor, so

Â it's going to be 0 for the reference node minus V1 divided by 6K

Â 6:10

And that's all of them, so they're equals to 0.

Â And so if you look at this equation, we have two unknowns,

Â V1 and V sub 2, so that's our first equation.

Â 6:25

Our second equation relates that 6 volts source to the nodal voltages, V1 and V2.

Â And so that's our constraining equation for this problem and

Â if we want that equation, we going to have V1 minus V2 is equal to 6 volts.

Â 6:46

And so that's our second equation which is independent from the first from the super

Â node and we can solve this problem based on these two equations and two unknowns.

Â So if we take these two equations and we solve for V1 and V2,

Â V1 turns out to be 10 volts and

Â V2 is 4 volts.

Â 7:14

So this is the example of a super node problem and

Â we identified it as a super node problem because we had a voltage source

Â which was between two nodes and one of them wasn't a reference node.

Â It wasn't a known voltage, so we could not work this problem.

Â So we had to go back and identify a super node,

Â write the equation around the super node and also the additional

Â equation which relates to super node voltages to the nodal voltages.

Â