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.