Sample Problem: Node Analysis (Independ Sources) 5

The topic of this problem is nodal analysis and

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

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

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

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

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

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

node 2 to top right hand side of the circuit and

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

So we're going to use Kirchhoff's current law to solve this,

Kirchhoff's current laws and we're going to sum the currents into the nodes.

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

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

Looking at the currents flowing into the node 1,

we have a 6 mA source flowing into node 1, so it’s going to be 6 mA.

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

that’s going to be the 0 volt reference node at the bottom,

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

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

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

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

some unknown current through this 6 volt source.

So I'm going to call that I 6v and

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

So we know from using nodal analysis and

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

So typically with Kirchhoff's current law,

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

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

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

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

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

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

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

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

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

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

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

So we don't have a reference voltage for

the 6 volt source from one node to another node.

So ultimately if we were to continue writing these equations,

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

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

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

that independent voltage source which is

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

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

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

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

which relates the nodal voltages to that 6 volt source.

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

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

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

So we've taken this 6 volt source and

we've kind of isolated it from our analysis.

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

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

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

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

us in solving this problem.

So if we write our equation about the super node

We would sum the currents into that super node.

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

we have current flowing up through the 6K resistor.

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

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

So let's add those currents up.

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

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

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

We have the current flowing up to the 12 kilo ohm in a similar fashion.

So ground voltage of 0 volts divided by node 2 voltage, V sub 2 divided by 12K.

So it's going to be 0 minus V sub 2 divided by 12K.

We also have the fourth current which is flowing up

through the right hand side of the circuit and we know that's going to be minus 4 mA.

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

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

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

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

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

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

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

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

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

V1 turns out to be 10 volts and

V2 is 4 volts.

So this is the example of a super node problem and

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

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

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

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

write the equation around the super node and also the additional

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

Linear Circuits 1: DC Analysis

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Taught By

Dr. Bonnie H. Ferri

Dr. Joyelle Harris

Dr Mary Ann Weitnauer

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