>> We're now going to derive the Impedance of the Capacitor.
Again impedance we have defined as
being the ratio of the phasor voltage to the phasor current.
So let's determine that relationship now for our capacitor.
Once again, we're going to define a voltage or reference voltage v of
t and a current flowing through the capacitor flowing or as reference,
from the positive to the negative reference to terminal.
Now this time around, let's assume that the voltage and
the capacitor is equal to or is of the form V sub m,
cosine of Omega t plus Theta sub V.
We know that in a capacitor i is equal to c times
the derivative of the voltage,
which is going to equal then c. Now once again,
the derivative of cosine is negative the sine,
so we'll bring that negative out in front here,
the chain rule is going to give us an Omega out in front.
We have then the V sub m and we will have in the sine,
the derivative of cosine is the sine of the
negative already taken care of sine of Omega t
plus Theta sub V. Once again,
we want to write this in terms of cosine rather than a sine.
So this is going to equal negative C times
Omega times V sub m times the cosine
of Omega t plus Theta sub V minus 90 degrees.
Now, let's write both of these and their phasor forms.
This time we have phasor V is equal to,
V sub m e to the j Theta sub V. We can write phasor I is equal to
negative negative C Omega V sub m, e
to the j Theta sub V minus 90 degrees.
Again, using the product of
two exponential terms is that you add the exponents and do that.
So we have the answer is equal to negative C Omega V sub m e to
the j Theta sub V e to the minus j90 degrees.
As before, this term right there is just minus j.
So we take the minus times a minus that gives us a positive,
and bringing the J into this makes here we have then J Omega C,
V sub m e to the j Theta sub V. Again,
we recognize that right there as being
phasor V. So we can then write that phasor
I is equal to j Omega C times phasor V. Now,
let's go ahead and form the impedance by
writing a ratio of phasor V to phasor I.
We have Z then is equal to phasor V divided by phasor I,
but phasor I is just j Omega C times phasor V. The phasor V is cancel.
We get then that the impedance of a capacitor is equal to one over
j Omega C. Sometimes we'll write that by bringing the j up into the numerator,
changing the sign on it which then gives us
a negative J times 1 over Omega C. All right.
We need to make some observations here.
First of all, notice that the impedance of the capacitor is negative.
Second of all, notice
the frequency dependency of the capacitor impedance is inversely.
In other words, I'll say it better,
the impedance of a capacitor is inversely proportional,
to the frequency of the circuit source that's driving it.
Remind ourselves here.
The impedance of an inductor was j times Omega
L. The impedance of an inductor was directly proportional to the frequency.
The impedance of a capacitor is inversely proportional to the frequency.
When the frequency is small,
the impedance of the capacitor is large and in fact,
that DC when Omega equals 0,
the capacitor has an infinite impedance.
It is effectively an open circuit.
As the frequency gets larger and larger as the frequency approaches infinity,
a larger number in the denominator makes the impedance of the capacitor gets
smaller and smaller and in the limit as Omega goes to infinity,
the impedance capacitor goes to zero.
At high frequencies, the capacitor is approximated by a short circuit.
So for example, let's say that we've got a 10 microfarad capacitor.
The source is driving this is some V sub
s of t is equal to 5 cosine of 5,000t.
In other words, Omega is equal to 5,000 radians per second.
Then the impedance of the capacitor in this circuit,
this 10 microfarad capacitor would be equal to 1 over j times Omega which is
5,000 times c which is 10 times 10 to the minus 6.
You do the math on that, again bringing the j in the numerator.
Here you'll get that the impedance of this capacitor would
be negative negative j20.
As we mentioned, the impedance of the capacitor is negative,
the impedance of the inductor is positive.
The impedance of the inductor,
in fact, let's just make a little chart here.
In the time domain R and the impedance for
a resistor is just plain old R. For a capacitor,
the impedance of a capacitor is 1 over j Omega C.
For an inductor, the impedance of the inductor is J Omega L.
Finally, in each of these cases, Ohm's law applies.
We have then that V is equal to I times Z because we
defined Z to be the ratio of phasor V to phasor I.