Welcome back, and today we're going to be talking about a very important property
of Linear Circuits and actually use it for analysis.
It's called Superposition and so we're going to describe what it is and how we
can use it to analyze our circuits. In the previous lesson we talked about
what it meant for something to be linear. And why this class is called Linear
Circuits. And in order for superposition the thing
we're using today to work, we need to have everything to be linear.
If it's not things start to break down. But we'll talk a little more about that.
After we've finished discussing Superposition, Doctor Fairly will be
presenting some systematic ways of going about finding systems of equations.
To solve for these types of circuits. A lot of them actually make use of
superposition. But superposition itself is a tool you
can use to solve some of these circuits and do some analysis.
The executives for this particular lesson are, given a complicated system, generate
a set of simpler systems. Each with a single, independent source.
And then, use those individual solutions and combine them together to come up with
a complete solution for a system. In order to use superposition, the first
step is you need to be able to zero out your sources.
And what do we mean by zeroing them out? Well first of all, if I have a voltage
source, and I want to replace it by some type of device that will have a zero
voltage. What's the device that I want to choose
to replace it with? It turns out that the wire works best
because if I measure the voltage between here and here, because there's this wire.
We know that the wire is believed to be equipotential.
There's no voltage drop across it. So, that means that if I want to zero out
the voltage, I turn it into a wire. For a current source, if I want zero
current to be flowing across a pathway, the best circuit element to use is going
to be an open circuit. Since this is basically defined as
something that's disconnected, so no current is able to flow.
The Steps For Superposition. So first of all, begin by following this,
zeroing out all of the independent sources.
After they've all been zeroed out, you return each of the sources, one at a
time, and then you do the analysis of the circuit.
And solve the quantity that you're searching for.
After you've completed that, you take the arithmetic sum of these values for the
final quantity. Now you might wonder, why would we use
superposition, if we took a problem that we had to solve once and now we have to
solve it multiple times. It turns out that sometimes it can be
useful if by using Superposition, you're able to make the individual simple
circuits very easy to analyze. so it does, sometimes, find practical
applications in those circumstances. But superposition is often an underlying
method, of some of the more systematic methods, that are, generally, used more
frequently. That will be covered next time.
Okay, so let's do an example of solving a system using superposition.
In this example we have three sources and we have four resistors, so it's a fairly
complicated system. What we're going to do is start by
removing each of the sources and then putting them in one at a time, solving
for v. Which is the value that we're looking to,
to find in the circuit. And after we've done this three times,
then we take all of the individual solutions, add them together to get our
final solution. Right, so in this example, we have a 3
volt source here, and we're going to be zeroing out this voltage source, and this
current source. So doing that, I get something like this,
3 volt source, 4 ohm resistor. the current source are being removed, so
we can just take this 36 ohm resistor and move it over here.
There's a 45 ohm resistor over here. The 30 ohm resistor stays.
And now this is a wire, because we've removed that other source.
Within this we see that the 30 ohm resistor and the 45 ohm resistor are in
parallel with each other. So you can combine them, that then allows
us to take this combination and this 36 ohm resistors and this just because a
voltage divider. So, we have 3 volts, split across this,
where this is the voltage we're interested in finding.
When you do that, you discover that the voltage is equal to 18 over 18 plus 36
times 3 is 1 volt. So that means that from part a, we've
discovered the voltage drop contributed by this 3 volt source is 1 volt.
Now let's do it again, but with a different source.
Here I've noted that our first result was 1 volt.
Now we're going to remove this voltage source and this voltage source and leave
the current source. So again we will draw a circuit.
This is 30 ohms, this is 45 ohms and this is the voltage that we're trying to find.
Notice though, that because we've zeroed out this voltage source, there's a wire
connecting the two sides of this 54 ohm resistor.
Consequently this resistor isn't going to be interesting.
Any current that flows is going to circumvent it.
And so we'll see no voltage drop across it, no current going through it.
So we can remove it from our circuit. So that the current source remains at one
fourth amp. And then, finally, a 36 ohm resistor.
At this point, we see that we have 3 resistors.
Which are all in parallel, which we can combine, because they connect the same
two nodes, as a single resistor. And then by multiplying the current that
is going through that resistor, which is one quarter of an amp, we can discover
what the voltage, v happens to be. In doing that, we see that the v is equal
to 3 volts. Simply making use of combining resistors
and then Ohm's Law. Still one more to go.
Now, we put the five volt source back and we remove the current source again.
Still we see that this, voltage source, because we've zeroed it out, gives us a
wire across this 54 ohm resistor. So again, the 54 ohm resistor's not going
to be interesting to us. We're drawing the circuit again.
The five volts is going to then go across the 30 ohm resistor, and then we have our
45 ohm resistor here. Cross for measuring a voltage.
That will also be going across a 36 ohm resistor.
So to do the analysis for this problem. We can take these two resistors and
combine them, because they're in parallel, into a single resistor.
And at that point, the whole problem simplifies down to a voltage divider.
The equivalent resistance here is 20 ohms, which means that we have 20 over 20
plus 30 or 50, so two fifths times 5 volts is equal to two volts.
So to get our final solution of v, I don't know if this is vc.
We simply take v from part a. Add a v from part b, add a v from part c.
Getting a result of one plus two plus three, Six volts.
Now, it might be easier to use a different technique, something not like
superposition, for a circuit like this. But, this is a good way of seeing how
superposition can be applied to a circuit.
And sometimes, being able to zero out the sources can simplify your system enough
that it's worth using. Additionally, a lot of the systematic
methods that will be presented for solving these types of circuits.
Have underlying them the requirement that the system follows superposition or that
everything be linear. But because this is a linear circuit
class, you're going to pretty much be able to use the techniques that we're
presenting here, in all of the circuits that you're analyzing.
When you're working with dependent sources, it becomes a little trickier.
You can't just zero out your dependent sources because the dependent sources are
functions of something else that's going on in the system.
In order for them to work, there's a couple of caveats.
First of all you have to leave them in in every single implementation.
Every time you remove your independent sources you leave in your dependent
sources when you put back in Individual independent sources.
And analyze the whole thing with your dependent sources.
And in order for this to work, the dependent sources must be linear.
And typically, what that means is that your dependent source must be some
constant times some measured value in the system.
So if I look at this example, I have K1 times v1, where K1 is a constant 2 volts
per volt. That's just fine.
And that will work with no problem. But in the other example here I do K2, 2
milliamps per volt squared and I'm multiplying it by this V i squared.
And because we're multiplying by something that's kind of a derived
quantity, and not just something that was measured.
This is no longer linear. Which means that if you have something
like this in your circuit, super position will stop, stop working.
And you won't be able to use it anymore. So that one's out.
We will make available, some examples. To show you how you can analyze a circuit
using dependent sources. But typically we're not going to use
superposition to analyze a circuit when there are dependent sources present.
In summary, we used superposition to solve circuits, and we presented how to
do that with independent sources only. And the technique for doing it with
dependent sources as well. In the next lesson we will talk about
systematic application of using Ohm's Law, Kirchhoff's Current Law, and
Kirchhoff's Voltage Law to generate a system of equations.
As well as, what's called, the Thevenin and Norton equivalent circuit, that
allows you take a very complicated system.
And replace it by a very simple equivalent system.
Which is going to be used in some of our later analysis.