>> We're going to now derive
the impedance of an inductor.

You'll recall that we have
defined the impedance as

being the ratio of the phasor voltage
to the phasor current.

So let us now derive that relationship,

phasor voltage, phasor
current for inductor.

Here we have an inductor,

and we'll reference the voltage v of t

plus the minus like that and also
reference the current through it,

i of t where the current is referenced from

the high voltage reference in

the direction of the flowing
from high to low voltage.

Now, let's just assume
that i of t is of the form

I sub m cosine of omega t plus Theta sub i.

So the current we're
assuming is oscillating at

the same frequency as the source
is driving the circuit.

It has an amplitude I sub m and the phase
angle associated with the Thetas Y.

Now, we know that in an
inductor the voltage across

that inductor is proportional not to

the current but rather proportional
to the derivative of the current,

or v is equal to L times di dt.

So, we've got then l. Now,

di dt we're going to have I sub m,

and the derivative of
the cosine is negative sine.

So let's bring the minus
sign out here in front,

and let's also go ahead and we're going
to have a chain rule invoked here.

So let's bring the Omega out in front
associated with the chain rule,

and then we will have the sine
of Omega t plus Theta sub i.

Ultimately we're going
to want to have this or

represent v in terms of its phasor,

and so we need to convert
the sine term to a cosine term.

So this then is equal to negative L,

I sub m Omega times the cosine of

Omega t plus Theta sub i minus 90 degrees.

Shifting the sign degree.

The sign with 90 degrees to the right,

because it is a cosine wave.

Now, we can write I sub m
in terms of this phasor,

or a phasor I is equal to I
sub m e to the j Theta sub i,

and phaser v, will be equal to I've
got the negative sign the L I Omega.

It's amplitude.

So negative l Omega I sub m,

and then it will be e to the j,

Theta sub i minus 90 degrees.

Now, using the property of
the product of exponents,

let's rewrite this then as equaling.

Let's see, negative L Omega i sub m,

e to the j Theta sub i,

e to the j.

Let's say, negative j 90.

Now, let's look at this term
e to the minus j 90.

Just drawing a little
phasor diagram over here,

e to the minus j 90 has a magnitude of
one and it has an angle of minus 90.

That's equivalent to
this e to the minus j 90

then is in the negative j
direction with a length of one,

or we can rewrite this term
right here as a minus j.

Now, let's combine this minus sign
here with this minus sign here,

bring the j here into this.

Bring the j into the mix
here and we get then that

phaser v is equal to minus times
minus is a positive,

and we'll have a j Omega L,

e to the j Theta sub i, oh,

and I left an I sub m term
multiplying all of that also.

Now, we notice that right there is I sub m,

e to the j Theta sub i.

Well, that's just phasor I.

We have then that phaser v is equal to

j Omega l times phasor I. Phasor I,

phaser v, z is defined as

the ratio phaser V to phasor
I which is equal to then,

phaser v is j Omega l times phasor
I divided by phasor I.

The phasor I's cancel and we're left

with the impedance of
the inductor is equal to

j Omega L. Let's just look
at this for a second.

In other words the impedance
of an inductor,

is equal to j times Omega
times L. Note two things.

First of all, the impedance
is imaginary, it's j.

It's also positive.

The second thing we'd like to point out
is that it is a function of frequency.

Here, the frequency of
the source that's driving

the circuit determines or impacts,

or yes what is it?

It is involved in the impedance,

the impedance is related to the frequency.

In other words, as the frequency
gets larger and larger,

the impedance gets larger and larger.

Or as the frequency gets smaller,
the impedance gets smaller.

In fact and the limits,
when Omega equals zero,

the impedance inductor is zero.

The ratio of the voltage
to the current is zero.

In other words, at DC there is
no voltage drop across the inductor.

Though we knew that at DC,

di dt is equal to zero and the voltage
across the inductor is zero.

Now, as Omega goes to infinity,

the impedance of the inductor
becomes infinite,

or it would be equivalent
to an open circuit.

In other words at high frequencies,

the voltage across here,

the derivative of the current goes to

infinity and the voltage
across it goes to infinity.

It's an open circuit at that point.
Nothing gets through it.

Let's just take an example.

Let us assume that we have
a source v of t is equal to

five cosine of 100 t. In other words,

Omega is equal to 100.

Let's say we've got
a 50 millihenry inductor.

Let's let L equal 0.05 Henry.

Then the impedance of
that inductor is z sub L equals

j times Omega which is
100 times the inductance which is 0.05,

and we get that the impedance of

that inductor in a circuit
oscillating 100 radians per

second is equal to j five Ohms.