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.