Alright, hello my name is Edgar Garcia 
and today we're going to work a two part 
problem on impedance. 
The results we get from the first part 
are going to be used on our second part. 
Alright, so let's look at this problem 
here. 
What we have here is a time varying 
voltage source, a 100 Ohm resistor, a 1 
millifarad capacitor, a 2 millifarad 
capacitor and a 5 millihenry inductor. 
So, what you want to do is, given a 
frequency, we want to be able to find the 
equivalent impedance of these elements 
such that it ends up being in parallel 
with our voltage source. 
So, it would be connected to, let's call 
these A and B notes. 
So, the first things you want to do is be 
able to find the individual impedance of 
each element. 
So, the problem tells us that the 
frequency is equal to 60 hertz. 
So, the first step would be to change it 
into radians per second. 
So let's call that w. 
And the formula for that is 2 times pi 
times f. 
Which is approximate to 377 radians per 
second. 
Now that we know our, our frequency in 
radians per second, we can calculate, 
like I said the individual impedance of 
each element. 
So, let's start with r. 
The impedance for r, let's call it, let's 
call it ZR. 
It's just going to be equal to the same 
value. 
because it's already in nodes. 
Now lets go to our next element. 
Let's Z1. 
Let's call it Zc1. 
So, our formula for this is going to be 1 
over j. 
Times our frequency times our 
capacitance. 
And so, what we can see here is that 
since our j is in our denominator, 
remember that we can switch it up and 
down from the numerator to the 
denominator by changing its its sign. 
So, we can just say, negative j over 377, 
which is our frequency, times 10 to the 
minus 3, which is our capacitance for 
this capacitor right here. 
And we end up getting a value equal to 
negative j 2.65 ohms. 
Alright, let's keep going. 
Now, lets find impedance for our second 
capacitor let's call it Zc2. 
As we did for the previous capacitor, we 
can have negative j in the denominator, 
the same frequency and it's capacitance. 
So, let's just make a note here. 
This is for C1 this is for C2. 
So, we end up having a negative j 377 
over 2 times 10 to the minus 3. 
And we end up having a value of negative 
j 1.33 Ohms, alright. 
And, finally, let's calculate the 
impedance of our inductor right here. 
So let's call that ZL. 
For ZL, our formula for the impedance of 
our inductors is going to, formula's 
going to be j times the frequency times 
the inductance. 
So, j stays the same. 
Frequency 377, and the inductor is 5 
times 10 to the minus 3. 
We end up having a value of j 1.89 Ohms. 
Alright, so now that we have our values 
for the independent individual impedances 
for each element, we can rewrite our, 
redraw our 
[UNKNOWN]. 
Such that it has little boxes with these 
values. 
So, let's redraw. 
We have our time varying voltage source. 
We have our impedence for the resistor 
which is a 100 ohms. 
We have the impedence for our capacitors, 
which let's call it ZC, just because we 
can combine these two capacitors since 
there in series, we can add them up 
together. 
So, the value of the impedance for both 
capacitors, we can look at it here, is 
just the sum of both. 
So, the impedance for the capacitor is 
going to be the impedance of capacitor 1 
plus the impedance for capacitor 2. 
And this ends up being equal to negative 
j 3.98 ohms. 
So, keep in mind that since they are in 
series, we just add them up together. 
Since they're both imaginary values, it's 
just adding them up. 
If they were in parallel we would've done 
the parallel formula. 
But let's just plot our, new value into 
this box here which is negative j 3.98 
ohms. 
And finally, our impedance for the 
inductor, which is, j 1.89 ohms. 
Now that we've simplified our circuit 
such that we only have the impedance 
values for each element. 
We can keep combining them as we did for 
the two capacitors. 
And what we can see here is that the 
impedance for the capacitor, ZC, and the 
impedance for the conductor, ZL, are in 
parallel together. 
So, we can combine these two. 
As we've always done, ZC times ZL over ZC 
plus ZL. 
And this is going to give us a value 
equal to, minus j 3.52 ohms. 
So finally, what we have here is a new 
box with this value right here, with this 
impedance. 
And by the end, we're going to have our 
resistor, [UNKNOWN] for our resistor in 
series with this value. 
So, as we were trying to search for, was 
the impedance between these two terminals 
such that it's in parallel with our 
voltage source. 
Let's call that Z equivalent. 
We end up having, let's write it down 
here, that Z equivalent is going to be 
equal to the impedance of our resistor. 
In series with the impedance of these two 
values in, in parallel which we already 
calculated. 
So, minus j 3.52 ohms. 
By substituting our, our the impedance 
for our resistor, we have a 100 minus j 
3.52 ohms. 
And this is the impedance, again, that's 
going to be in parallel with our voltage 
source. 
So to go over the steps, when we have, 
when we're given a circuit with 
individual elements. 
Such that were giving the value for 
example resistance of resistors, the 
capacitance of capacitors, and the 
inductance of the inductors. 
What we want to do first, is be able to 
calculate the frequency such that it's 
given us in radians per second. 
This value is going to be a, we're 
going to need this value as such that we 
can substitute it in all of these 
formulas right here. 
In order to calculate the impedances of 
each element. 
Once we have all of these values, we can 
go back to our circuit and exchange the 
individual elements for little boxes with 
their impedance values. 
And after that we're just going to 
combine them as we've always done for 
resistance, resistances sorry. 
If their in series we add them up. 
If their in parallel we use our formula 
right here. 
Thank you very much.