L3 1 3Node Example Dependent Source

Node Voltage with dependent sources

In this video we're gonna take a look at
using node analysis techniques to analyze

a circuit that's got a dependent source.

You'll notice right off the bat that
we got a dependent current source

here in this branch.

This dependent current source,
the amount of current that this current is

producing depends upon the current i
flowing through the 6 ohm resistor and

this current source, then,
is two times that I.

We're gonna take that into account as we
go through and define our node voltages

and we'll come up with an expression for
I in terms of our node voltages.

So let's go ahead and get started.

We've got three critical nodes again.

One here, one here, and
one along the bottom.

Once again, we'll go ahead and
call this our reference node,

where we've got v equals 0.

And once again, we've got a voltage
source tied to the reference,

so we know the voltage here at this point
of our circuit is equal to 5.3 volts.

We'll define this node here to be V1,
this node here to be V2.

And with those two defined
voltages now we can express i,

the current i In terms of V1 and V2.

In fact I is equal to V1
minus V2 divided by 6 ohms,

it's just the voltage
here on the left minus

the voltage on the right
divided by 6 ohms.

When we're all done with this analysis,

we're going to use that expression
to calculate I let's say

to calculate the power being
produced by this dependent source.

With those definitions, let's go ahead and

write the node equations
at these two nodes.

First one here on the left
the current leaving.

The first node here going
to the left is going to be,

V1-5.3 divided by 4 ohms
+ the current coming down

here through the 3 ohm resistor
is going to be V1 divided by

3 + the current leaving this
node going to the right.

Well notice that that's that
current I has been defined for us.

We don't want to use I we want to use or
we want an expression for

that current in terms of
the node voltages V1 and V2, and

we've already expressed that I over
here as V1 minus V2 divided by 6.

So plus V1-V2 divided by 6.

The sum of those three
currents has to equal 0.

Now let's write the node
equation of the V2 node.

Current going to the left is going to

be V2-V1 divided by 6
+ V2-0 divided by 12.

Minus, okay, now let's be careful.

Minus two times I, it's minus because
it's referenced into the node.

Two times I is this dependent source,
but I don't want to put I in here.

We want to write everything in
terms of our node variables or

node voltage variables V1 and V2.

So I've got i is equal to two,
or v1 minus v2 divided by 6.

This current here is two times I so
it's going

to be -2 times v1 minus v2 divided by 6,

the sum of those three currents equals 0.

I suggest that you stop the video right
now and just take a look at that,

there's a couple of things going on here.

The fact that it's going into
the node gives us the minus sign.

It's gonna be minus
the current flowing in.

Well, the current flowing in is 2 times or

has a value of 2 times whatever I is.

We defined I as V1- V2 divided by 6.

So we have, then,
-2 times V1- V2 divided by 6.

Alrighty, now that you've taken a moment
to make sure you understand what we did

there with the signs, let's go ahead and

solve this system of two equations with
two unknowns by combining like terms.

For the first equation we've got

V1 times one-fourth + one-third

+ one- sixth + V2 times.

This is just one V2 term that's got
a negative one- sixth with it is

equal to negative 5 divided by 4 take it
to the other side as I'm sorry negative.

Let's see, it's -5.3 divided by 4,

take it to the other side
is a +5.3 divided by 4

will write it like that for
now 5.3 divided by 4.

Alrighty, the second equation and

here we're going to need to be careful
because we've got this stuff going on

there with the last term but
once again factoring out the V1 terms.

Here I've got a V1 term and I've got
a V1 term here, let's be careful there.

The first one is negative one- sixth.

But what do I actually have over here?

I've got -2 times V1 divided by 6.

So that's negative -2 over 6,
that's negative one third V1.

Factor out the V1, and that leaves me

with a negative one-third there.

Alrighty, now for
the V2 terms, + V2 times,

I've got a V2 over 6, factoring
out the V2 leaves me a one-sixth.

I've got a V2 over 12, so + one-twelfth.

Now once again, slow down here.

You've got signs and
signs like red flags waving.

That means slow down, you've got
an opportunity to make a sign error and

we don't like sign errors.

We've got -2 times a -V2 over 6.

Negative times a negative
is a positive to V2 over 6,

so that's gonna be positive
factor out the V2,

I've got 6 that's one-third, and
the sum of those three currents equals 0.

Once again, I'd encourage you to
stop the video at this point and

just make sure that you understand
what we did with the signs.

Alrighty, let's just Be careful here
that we keep our equations separate,

and now let's go ahead and
combine the fractions.

We've got here then for
the first equation we have v1 times

one-fourth plus one-third plus
one-sixth is three-forth.

Plus V2 time negative one-sixth

= 5.3 divided by 4 is 1.325.

And on the second equation
we have V1 times negative

one-sixth minus one-third
is a negative one-half.

Plus V2 times one-sixth + one-twelfths

+ one-third is seven-twelfths and
it's positive equals 0.

So we've got our two equations there.

Once again,
plug that into your matrix solver,

use the solve button on your calculator or
you can use MATLAB for that matter.

And when you do, you'll get

that V1 equals 2.1824,

2.1824 and V2 is equal to 1.87.

Now what can we do with V1 and
V2 while we can do anything we need to do.

Just for example in this one, why do
we calculate the power of this being

generated by this dependency source and

let's be careful here because we've
got issues with reference direction.

There is a dependence
source it has a value

of 2 times I, reference going up.

And what is the voltage across that power
supply, well is the reference plus to

minus V2 that power supply that dependent
current source has V2 on the top and

it's tied on the bottom to our reference
so the voltage across that is just V2.

Now we know that power
is equal to I times V.

And we also know that we got to be
careful about the sign in front of this.

Our passive sign convention says that
if current is referenced going from

the positive to the negative terminal,
it will be a positive power.

That's not what we have here, in this
case, we have the current going up,

referenced into the negative terminal and
out the positive terminal.

So that means that we
need a minus sign there.

Now, what is I?

Well, we've got I up here,
I is V1- V2 divided by 6.

So let's go ahead and calculate that.

V1 was 2.1824 minus V2.

Which was 1.87-1.87,

and the whole thing divided by 6.

That gives us a value for

I of 0.052 amps, or

52 milliamps, so 0.052 amps.

Now this current source right
here is equal to if I sub x,

the current that it generates
is equal to 2 times I,

we now know I is 0.052.

So that current source is producing

2 times .052 or .104 amps.

That's the current that we need
down here to calculate the power.

This current source is
producing 0.104 amps going up.

So the power down here is going to
be equal to negative I is 0.104.

Times V, which V did we say it was?

It was V2.

So times V2, which is 1.87 volts,

and when you multiple those two together,

you get a negative 0.194 watts.

Negative, what does the negative
sign mean on power?

It means that it's producing energy,

it's the source is putting
energy into the circuit.