>> 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.