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.