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Alright, hello my name is Edgar Garcia 
and today we're going to work a two part 

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problem on impedance. 
The results we get from the first part 

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are going to be used on our second part. 
Alright, so let's look at this problem 

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here. 
What we have here is a time varying 

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voltage source, a 100 Ohm resistor, a 1 
millifarad capacitor, a 2 millifarad 

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capacitor and a 5 millihenry inductor. 
So, what you want to do is, given a 

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frequency, we want to be able to find the 
equivalent impedance of these elements 

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such that it ends up being in parallel 
with our voltage source. 

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So, it would be connected to, let's call 
these A and B notes. 

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So, the first things you want to do is be 
able to find the individual impedance of 

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each element. 
So, the problem tells us that the 

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frequency is equal to 60 hertz. 
So, the first step would be to change it 

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into radians per second. 
So let's call that w. 

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And the formula for that is 2 times pi 
times f. 

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Which is approximate to 377 radians per 
second. 

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Now that we know our, our frequency in 
radians per second, we can calculate, 

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like I said the individual impedance of 
each element. 

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So, let's start with r. 
The impedance for r, let's call it, let's 

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call it ZR. 
It's just going to be equal to the same 

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value. 
because it's already in nodes. 

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Now lets go to our next element. 
Let's Z1. 

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Let's call it Zc1. 
So, our formula for this is going to be 1 

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over j. 
Times our frequency times our 

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capacitance. 
And so, what we can see here is that 

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since our j is in our denominator, 
remember that we can switch it up and 

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down from the numerator to the 
denominator by changing its its sign. 

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So, we can just say, negative j over 377, 
which is our frequency, times 10 to the 

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minus 3, which is our capacitance for 
this capacitor right here. 

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And we end up getting a value equal to 
negative j 2.65 ohms. 

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Alright, let's keep going. 
Now, lets find impedance for our second 

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capacitor let's call it Zc2. 
As we did for the previous capacitor, we 

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can have negative j in the denominator, 
the same frequency and it's capacitance. 

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So, let's just make a note here. 
This is for C1 this is for C2. 

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So, we end up having a negative j 377 
over 2 times 10 to the minus 3. 

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And we end up having a value of negative 
j 1.33 Ohms, alright. 

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And, finally, let's calculate the 
impedance of our inductor right here. 

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So let's call that ZL. 
For ZL, our formula for the impedance of 

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our inductors is going to, formula's 
going to be j times the frequency times 

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the inductance. 
So, j stays the same. 

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Frequency 377, and the inductor is 5 
times 10 to the minus 3. 

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We end up having a value of j 1.89 Ohms. 
Alright, so now that we have our values 

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for the independent individual impedances 
for each element, we can rewrite our, 

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redraw our 

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[UNKNOWN]. 

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Such that it has little boxes with these 
values. 

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So, let's redraw. 
We have our time varying voltage source. 

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We have our impedence for the resistor 
which is a 100 ohms. 

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We have the impedence for our capacitors, 
which let's call it ZC, just because we 

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can combine these two capacitors since 
there in series, we can add them up 

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together. 
So, the value of the impedance for both 

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capacitors, we can look at it here, is 
just the sum of both. 

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So, the impedance for the capacitor is 
going to be the impedance of capacitor 1 

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plus the impedance for capacitor 2. 
And this ends up being equal to negative 

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j 3.98 ohms. 
So, keep in mind that since they are in 

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series, we just add them up together. 
Since they're both imaginary values, it's 

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just adding them up. 
If they were in parallel we would've done 

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the parallel formula. 
But let's just plot our, new value into 

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this box here which is negative j 3.98 
ohms. 

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And finally, our impedance for the 
inductor, which is, j 1.89 ohms. 

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Now that we've simplified our circuit 
such that we only have the impedance 

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values for each element. 
We can keep combining them as we did for 

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the two capacitors. 
And what we can see here is that the 

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impedance for the capacitor, ZC, and the 
impedance for the conductor, ZL, are in 

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parallel together. 
So, we can combine these two. 

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As we've always done, ZC times ZL over ZC 
plus ZL. 

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And this is going to give us a value 
equal to, minus j 3.52 ohms. 

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So finally, what we have here is a new 
box with this value right here, with this 

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impedance. 
And by the end, we're going to have our 

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resistor, [UNKNOWN] for our resistor in 
series with this value. 

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So, as we were trying to search for, was 
the impedance between these two terminals 

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such that it's in parallel with our 
voltage source. 

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Let's call that Z equivalent. 
We end up having, let's write it down 

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here, that Z equivalent is going to be 
equal to the impedance of our resistor. 

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In series with the impedance of these two 
values in, in parallel which we already 

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calculated. 
So, minus j 3.52 ohms. 

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By substituting our, our the impedance 
for our resistor, we have a 100 minus j 

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3.52 ohms. 
And this is the impedance, again, that's 

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going to be in parallel with our voltage 
source. 

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So to go over the steps, when we have, 
when we're given a circuit with 

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individual elements. 
Such that were giving the value for 

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example resistance of resistors, the 
capacitance of capacitors, and the 

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inductance of the inductors. 
What we want to do first, is be able to 

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calculate the frequency such that it's 
given us in radians per second. 

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This value is going to be a, we're 
going to need this value as such that we 

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can substitute it in all of these 
formulas right here. 

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In order to calculate the impedances of 
each element. 

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Once we have all of these values, we can 
go back to our circuit and exchange the 

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individual elements for little boxes with 
their impedance values. 

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And after that we're just going to 
combine them as we've always done for 

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resistance, resistances sorry. 
If their in series we add them up. 

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If their in parallel we use our formula 
right here. 

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Thank you very much.