>> In this video, we're going to demonstrate
the node analysis method of analyzing circuits in the Phasor Domain.
You'll recall from our previous discussions of
node analysis that the first step in analyzing circuits
using the node voltage approach
involves identifying the critical or essential nodes.
In this circuit we have got one,
two, three, four essential nodes.
Again, an essential node is a node where three or more branches are connected.
We choose one of these nodes to then serve as
our reference node where we by
definition we're going to say voltage is equal to zero.
In this case, I've selected this node here to be our reference node.
So we'll set that voltage equal to zero.
The other three nodes are
the three essential nodes we will then identify or assign variables to them.
Variable names. This one we're going to call V1.
The voltage at this node will refer to as
V2 and the voltage at this node here will refer to as V3.
Node voltage analysis then has this writing
Kirchhoff's Current Law or node equations at each of these essential nodes.
Summing the currents leaving each of those nodes in terms of
the node voltage variables and the circuit parameters in the circuit.
So let's go ahead and do this.
On this first node,
let's write KCL at node one.
The current leaving this node coming down in this direction here will be
V1 minus the voltage at this node minus the voltage at this point.
The voltage here by definition is zero,
so V1 minus 0. We won't even bother.
Well, start let's write it V1 minus 0 divided by
the impedance there which is 3 minus J4.
Plus the current leaving this node going in this direction,
which is V1 minus V3 divided by
the impedance here of two ohms plus
the current leaving this node going in this direction here.
We need to be a little bit careful here.
It's going to be the current leaving this node going in
that direction will be the current through that resistor.
The current through that resistor is equal to the voltage drop
across that resistor divided by the resistance of two ohms.
The voltage drop across that resistor is the voltage on this side which is V1
minus the voltage on
this side of the resistor and here's where we need to be careful.
The voltage here in terms of node voltages can be
thought of as being starting here at V2.
V2 we drop going from here to here crossing this voltage source of 12 volts.
We drop 12 volts..
So the voltage at this point then is going to be V2 minus 12 volts.
So then the current leaving this branch going in
that direction will be the voltage of
the left-hand side of that resistor which
is V1 minus the voltage of the right-hand side of
the resistor which is V2 minus 12.
We throw in the parentheses there to remind ourselves that we're
subtracting the voltage there which is V2 minus 12.
Finally, the current in that branch then will be that voltage drop divided by
two and that equals zero.
Now, KCL at node two.
The current leaving this node going to the left, again,
is going to be the current through this resistor only
this time going from right to
left and that voltage will be the voltage here,
which we've already established to be V2 minus 12 volts minus
the voltage over here which is V1 minus V1 divided by the two ohms.
The current coming down to the node or leaving this node coming
down is going to be then added in and that's
Simply V2 divided by the impedance of two plus one J or simply J.
Then the current leaving node to going to the right will be equal
to V2 minus V3 divided by two ohms.
The sum of those three currents equal zero.
Now, at the third node we've got KCL at node three.
The current leaving node three and come into the left is going to be
V3 minus V2 divided by two ohms.
Plus the current leaving this node going up and over to the left is going to
be V3 minus V1 divided by two ohms.
Finally, the current leaving node three coming down this
way is going to be the voltage drop across that 2-ohm resistor,
which will be V3 minus the voltage right there.
Now, the voltage at that point is equal to starting here
the reference node zero minus to plus
we go up a value of negative J6 volts.
So the voltage right here then is
zero minus J6 and then the current through here will be V3
minus a minus J6
divided by the two ohms and the sum of those three currents equal zero.
Those are the three note note equations.
All that's left now is to calculate the node voltages V1,
V2 and V3 is a little bit of algebra.
Let's start by combining like terms.
Let's do that on up here.
The V1 terms in the first equation.
I've got V1 over.
Let's see 3 minus J4.
I've got V1 over two and I've got another V1 over two.
So plus one over two plus one over two gives us the V1 terms.
Plus the V2 terms here we've
got only one right here and that's a minus V2 over two.
So if we pull the V2 out we'll have
a minus one half there and then our V3 terms there.
Again, is only one of them.
It's a minus V3 over two.
So we have V3 times a negative one half.
That then will equal.
Let's see, we've got a 12 volts here.
12 divided by two is six.
A minus times a minus is a plus.
I've got plus six on the left-hand side.
We subtract six from
both sides to bring our constant over there on the right-hand side.
Now, combining like terms in the second equation,
we have V1 times.
This one we've only got one term that's going to be a negative V1 over two.
So we pull the V out or the V1 out or left with
the negative one half plus V2 times.
I've got V2 over two,
V2 over two plus J and V2 over two.
So V2 or let's see V2 would be one half
plus one over two plus J plus another one half and then the V3 terms,
again, there's only one and that will be a negative one half.
Now, here we've got a minus 12 over two.
So I've got a negative six volts on the left-hand side.
We add six volts to plus to our both sides and that
gives us then equaling positive six volts on the right-hand side.
Now, equation three, we have V1 times a negative one half plus
V2 times got a negative one half again
plus V3 times I've got one half plus one half plus one half.
That's three halves, V3.
Now, here we've got a constant of negative times negative is
a positive J6 over two.
So on the left-hand side we have a positive J3,
which we subtract J3 from both sides gives us
then a negative J3 over here on this side.
All right. Moving ahead now just cleaning things up
so that we're ready to put it into a matrix solver.
Let's combine all those fractions here and when we do
that we've got V1 times 1.12
plus 0.16J plus V2 times a negative
0.5 plus V3 times a
negative 0.5 equals negative six volts.
The second equation we have V1 times a negative 0.5 plus V2 times
1.4 minus 0.2J plus
V3 times a negative 0.5 equals a positive six.
Finally, the third equation,
V1 times a negative 0.5 plus V2 times a negative
0.5 plus V3 times
1.5 is equal to a negative J3.
When you plug this into a matrix solver of some sort,
you come up with that V1 is equal to in rectangular coordinates,
a negative 4.72 minus 0.88J,
which in polar coordinates is 4.8 angle negative 169.5.
V2 is equal to 2.46 minus 0.87J,
which in polar coordinates is 2.6 angle negative 19.84 and V3 we
find to be equal to negative 0.76 minus
J2.59 in rectangular coordinates in polar
coordinates it's 2.7 angle negative 106.29.
Of course, are all degrees.
Now, that we have the voltages at each of the nodes we can
calculate any current or voltage that we might want to calculate.
For example, let's determine
the current leaving node two coming down this branch here.
Well, it's called in this circuit here which has been identified
as I5 or I sub L. So, I5 then,
is equal to V2 divided by two plus J,
which is equal to V2 is 2.6
angle negative 19.84 divided by two plus J.
We find then if that is equal to in polar coordinates 1.1.
six angle negative 46.4 degrees.