L7 8 1 Node Voltage in Phasor Domain

Analyzing circuits in the Phasor Domain using Node Voltage techniques

>> 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.