In this video we're going to take a look at a circuit that contains
the structure known as a super node.
A super node consist of two critical nodes separated only by a voltage source.
That voltage source can be either an independent source as it shown here or
a dependent source.
To understand what this leads to, let's go ahead and
analyse this circuit as we would any other circuit using the node voltage approach.
We start by identifying our critical nodes 1, 2, 3 critical nodes.
This one we have identified as our reference where V equals 0, and
we've assigned this node here the voltage V1 or the variable V1, and
this node here, the voltage of that node is, we have assigned the variable V2.
Now to understand what's happening let's just go ahead and
write the two node equations.
For this left hand node we have the current leaving going to the left
will be V1- 4, divided by 2, plus the current coming down
through the forum resistor would be V1 minus 0, divided by 4.
Now, here's where the challenge comes.
When we look at the current leaving this node and
going to the right, we see only a voltage source.
There is no other device in here,
like a resistor that we can use to calculate the current.
A voltage source and it's current are unrelated to each other.
There is no mathematical or formula that relates the current to the voltage.
In fact, an ideal voltage source produces
this same voltage regardless of what the current is.
So what do we do?
Well, let's just go ahead and identify a variable,
we'll call it i, representing the current leaving V1 and going to the right.
This is inconsistent with the philosophy of node voltage analysis,
where we wanted to write everything in terms of our node voltage variables.
We'll see here in just a second that this i is gonna cancel.
But for now, let us call that i, it's referenced leaving the node there so
it will be represented over here as a positive i, and
the sum of those three currents equals 0.
Now let's write the node equation over here at the right hand node.
Starting with the current leaving the node coming down to the 8 ohm resistor,
we have V2 divided by 8.
And then we have this 2 amp source coming in, so that's going to be -2.
And then we also have now i, the current in this branch.
Well, this time it's entering the right hand node, so
we'll represent it with a -i.
And the sum of those three currents must equal 0.
Note that we can eliminate this i term by adding these two equations.
Let's go ahead and do that.
We then get V1-4 divided by 2 + V1 over 4.
Again, the i's add to 0,
then we have the other two terms down here plus
V2 over 8, then- minus 2 equals 0.
So we've eliminated the i but in the process we've gone from having two
equations down to just one equation, and we still have the two unknowns V1 and V2.
We get the second equation by realizing that because there's
this ideal source between V1 and V2, V1 and V2 are not independent of each other.
In fact, we can say that V2 is equal to
the voltage of V1 going up 18 volts.
Or that V2 is equal to V1 plus 18 volts.
That then, we're gonna refer to that as the super node equation.
That becomes our second equation in terms of V1 and
V2 that will allow us to solve for the circuit.
Let's clean this up here just a little bit.
Combining like terms we have V1 times 1/2 + 1/4.
We have only one V2 term, as V2 times 1/8.
Now our constants, we have a negative 4 over t2, that gives us a negative 2, and
another minus 2, which we take to the other side as a positive 4.
Combining 1/2 plus 1/4, that gives us 3/4 V1,
Plus V2 times
1/8 equals 4.
Now that we've got this cleaned up a little bit, let's substitute
V2 here in this equation, with the relationship that we came up with here.
And we're left with then, V1 times three fourths,
plus replacing V2 with V1 plus 18,
times one eighth, is equal to 4.
Once again, let's combine like terms.
We have V1 here times the 3/4 and
then over here we have 1/8 v1 that we need to add in, so plus 1/8.
We've eliminated the V2 term, yet we have a constant
term now here of 18/8, which is the same as 9/4.
When we subtract 9/4 from both sides,
we're left with here on the right hand side, 7/4.
Solving this for V1 we get the V1 is equal to 2 volts.
And now that we know V1 we can determine
the V2 is 18 volts greater than V1, or
V2 is equal to 20 volts, 18 volts plus 2 is 20 volts.
Notice, now that we've done this, let's go back to the circuit and
look at this super node as one single node.
And add the currents that are leaving the super node.
Starting on this side of the super node,
we have a current leaving in this branch here, which we know to be simply V1- 4.
Minus 4 divided by 2.
Add to that the current leaving the super node coming down here through the 4
ohm resistor, which is V1 divided by 4.
Now, you'll notice that this current i doesn't leave the super node,
in fact it just is contained with it.
You can if you will, think of it as leaving one part of the super node and
ending up in the other part of the super node, but i does not leave the super node.
So we come over here and account for these other two or
the currents in these other two branches.
The current coming down here to the 8 atom resistor is V2 divided by 8, and
then of course we get the 2 ohms going in there which we account for
as a -2 equals 0.
Notice that writing this equation in the context of the super node,
gives us exactly the same equation that we came up with when we wrote
the individual node equations, and combined them to eliminate i.
What that means and this is always true,
we don't have to do this intermediate step of writing the equations and adding them.
We can simply account for the currents leaving the super node and
write the equation in 4 or summing those currents.
Now, of course,
the second equation comes just as it did over here relating to V1 and V2.
We've got two equations and two unknowns as we had over here, and from there on,
the calculations are exactly the same