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

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You'll recall that we have[br]defined the impedance as

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being the ratio of the phasor voltage[br]to the phasor current.

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So let us now derive that relationship,

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phasor voltage, phasor[br]current for inductor.

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Here we have an inductor,

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and we'll reference the voltage v of t

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

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i of t where the current is referenced from

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the high voltage reference in

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the direction of the flowing[br]from high to low voltage.

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

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

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So the current we're[br]assuming is oscillating at

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the same frequency as the source[br]is driving the circuit.

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

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Now, we know that in an[br]inductor the voltage across

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that inductor is proportional not to

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the current but rather proportional[br]to the derivative of the current,

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or v is equal to L times di dt.

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So, we've got then l. Now,

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

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and the derivative of[br]the cosine is negative sine.

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

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

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

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

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Ultimately we're going[br]to want to have this or

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represent v in terms of its phasor,

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

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So this then is equal to negative L,

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I sub m Omega times the cosine of

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Omega t plus Theta sub i minus 90 degrees.

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Shifting the sign degree.

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The sign with 90 degrees to the right,

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because it is a cosine wave.

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

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

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

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It's amplitude.

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So negative l Omega I sub m,

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and then it will be e to the j,

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Theta sub i minus 90 degrees.

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Now, using the property of[br]the product of exponents,

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let's rewrite this then as equaling.

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

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e to the j Theta sub i,

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e to the j.

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Let's say, negative j 90.

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

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Just drawing a little[br]phasor diagram over here,

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

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That's equivalent to[br]this e to the minus j 90

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

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

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Now, let's combine this minus sign[br]here with this minus sign here,

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bring the j here into this.

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Bring the j into the mix[br]here and we get then that

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phaser v is equal to minus times[br]minus is a positive,

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and we'll have a j Omega L,

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e to the j Theta sub i, oh,

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

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

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e to the j Theta sub i.

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Well, that's just phasor I.

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We have then that phaser v is equal to

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j Omega l times phasor I. Phasor I,

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phaser v, z is defined as

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

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

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The phasor I's cancel and we're left

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with the impedance of[br]the inductor is equal to

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

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In other words the impedance[br]of an inductor,

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is equal to j times Omega[br]times L. Note two things.

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

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It's also positive.

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

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Here, the frequency of[br]the source that's driving

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the circuit determines or impacts,

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or yes what is it?

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It is involved in the impedance,

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the impedance is related to the frequency.

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In other words, as the frequency[br]gets larger and larger,

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the impedance gets larger and larger.

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Or as the frequency gets smaller,[br]the impedance gets smaller.

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In fact and the limits,[br]when Omega equals zero,

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the impedance inductor is zero.

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The ratio of the voltage[br]to the current is zero.

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

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Though we knew that at DC,

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

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Now, as Omega goes to infinity,

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the impedance of the inductor[br]becomes infinite,

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or it would be equivalent[br]to an open circuit.

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In other words at high frequencies,

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the voltage across here,

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the derivative of the current goes to

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infinity and the voltage[br]across it goes to infinity.

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

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Let's just take an example.

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

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five cosine of 100 t. In other words,

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Omega is equal to 100.

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

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Let's let L equal 0.05 Henry.

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Then the impedance of[br]that inductor is z sub L equals

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j times Omega which is[br]100 times the inductance which is 0.05,

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and we get that the impedance of

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that inductor in a circuit[br]oscillating 100 radians per

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second is equal to j five Ohms.