WEBVTT

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In this video we're going to take
a look at a circuit that contains

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the structure known as a super node.

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A super node consist of two critical
nodes separated only by a voltage source.

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That voltage source can be either
an independent source as it shown here or

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a dependent source.

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To understand what this leads to,
let's go ahead and

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analyse this circuit as we would any other
circuit using the node voltage approach.

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We start by identifying our critical
nodes 1, 2, 3 critical nodes.

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This one we have identified as our
reference where V equals 0, and

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we've assigned this node here
the voltage V1 or the variable V1, and

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this node here, the voltage of that node
is, we have assigned the variable V2.

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Now to understand what's
happening let's just go ahead and

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write the two node equations.

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For this left hand node we have
the current leaving going to the left

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will be V1- 4, divided by 2,
plus the current coming down

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through the forum resistor would
be V1 minus 0, divided by 4.

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Now, here's where the challenge comes.

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When we look at the current
leaving this node and

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going to the right,
we see only a voltage source.

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There is no other device in here,

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like a resistor that we can
use to calculate the current.

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A voltage source and
it's current are unrelated to each other.

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There is no mathematical or formula
that relates the current to the voltage.

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In fact, an ideal voltage source produces

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this same voltage regardless
of what the current is.

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So what do we do?

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Well, let's just go ahead and
identify a variable,

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we'll call it i, representing the current
leaving V1 and going to the right.

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This is inconsistent with the philosophy
of node voltage analysis,

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where we wanted to write everything in
terms of our node voltage variables.

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We'll see here in just a second
that this i is gonna cancel.

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But for now, let us call that i,
it's referenced leaving the node there so

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it will be represented over
here as a positive i, and

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the sum of those three currents equals 0.

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Now let's write the node equation
over here at the right hand node.

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Starting with the current leaving the node
coming down to the 8 ohm resistor,

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we have V2 divided by 8.

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And then we have this 2 amp source
coming in, so that's going to be -2.

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And then we also have now i,
the current in this branch.

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Well, this time it's entering
the right hand node, so

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we'll represent it with a -i.

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And the sum of those three
currents must equal 0.

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Note that we can eliminate this i
term by adding these two equations.

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Let's go ahead and do that.

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We then get V1-4 divided by 2 + V1 over 4.

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Again, the i's add to 0,

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then we have the other
two terms down here plus

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V2 over 8, then- minus 2 equals 0.

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So we've eliminated the i but
in the process we've gone from having two

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equations down to just one equation, and
we still have the two unknowns V1 and V2.

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We get the second equation by
realizing that because there's

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this ideal source between V1 and V2, V1
and V2 are not independent of each other.

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In fact, we can say that V2 is equal to

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the voltage of V1 going up 18 volts.

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Or that V2 is equal to V1 plus 18 volts.

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That then, we're gonna refer to
that as the super node equation.

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That becomes our second
equation in terms of V1 and

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V2 that will allow us to solve for
the circuit.

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Let's clean this up
here just a little bit.

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Combining like terms we
have V1 times 1/2 + 1/4.

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We have only one V2 term, as V2 times 1/8.

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Now our constants, we have a negative 4
over t2, that gives us a negative 2, and

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another minus 2, which we take to
the other side as a positive 4.

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Combining 1/2 plus 1/4,
that gives us 3/4 V1,

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Plus V2 times

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1/8 equals 4.

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Now that we've got this cleaned
up a little bit, let's substitute

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V2 here in this equation, with the
relationship that we came up with here.

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And we're left with then,
V1 times three fourths,

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plus replacing V2 with V1 plus 18,

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times one eighth, is equal to 4.

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Once again, let's combine like terms.

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We have V1 here times the 3/4 and

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then over here we have 1/8 v1 that
we need to add in, so plus 1/8.

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We've eliminated the V2 term,
yet we have a constant

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term now here of 18/8,
which is the same as 9/4.

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When we subtract 9/4 from both sides,

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we're left with here on
the right hand side, 7/4.

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Solving this for
V1 we get the V1 is equal to 2 volts.

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And now that we know V1 we can determine

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the V2 is 18 volts greater than V1, or

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V2 is equal to 20 volts,
18 volts plus 2 is 20 volts.

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Notice, now that we've done this,
let's go back to the circuit and

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look at this super node
as one single node.

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And add the currents that
are leaving the super node.

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Starting on this side of the super node,

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we have a current leaving in this branch
here, which we know to be simply V1- 4.

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Minus 4 divided by 2.

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Add to that the current leaving the super
node coming down here through the 4

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ohm resistor, which is V1 divided by 4.

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Now, you'll notice that this current
i doesn't leave the super node,

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in fact it just is contained with it.

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You can if you will, think of it as
leaving one part of the super node and

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ending up in the other part of the super
node, but i does not leave the super node.

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So we come over here and
account for these other two or

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the currents in these other two branches.

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The current coming down here to the 8
atom resistor is V2 divided by 8, and

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then of course we get the 2 ohms
going in there which we account for

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as a -2 equals 0.

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Notice that writing this equation
in the context of the super node,

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gives us exactly the same equation
that we came up with when we wrote

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the individual node equations,
and combined them to eliminate i.

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What that means and this is always true,

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we don't have to do this intermediate step
of writing the equations and adding them.

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We can simply account for
the currents leaving the super node and

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write the equation in 4 or
summing those currents.

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Now, of course,

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the second equation comes just as it
did over here relating to V1 and V2.

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We've got two equations and two unknowns
as we had over here, and from there on,

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the calculations are exactly the same