L7 2 5 Impedance of Capacitor

>> 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.