L3 1 4Super Node Example

Example analyzing a circuit containing a supernode (two nodes separated by only a voltage source

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