[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:03.56,Default,,0000,0000,0000,,>> We're going to now derive\Nthe impedance of an inductor.
Dialogue: 0,0:00:03.56,0:00:07.20,Default,,0000,0000,0000,,You'll recall that we have\Ndefined the impedance as
Dialogue: 0,0:00:07.20,0:00:11.36,Default,,0000,0000,0000,,being the ratio of the phasor voltage\Nto the phasor current.
Dialogue: 0,0:00:11.36,0:00:13.41,Default,,0000,0000,0000,,So let us now derive that relationship,
Dialogue: 0,0:00:13.41,0:00:16.80,Default,,0000,0000,0000,,phasor voltage, phasor\Ncurrent for inductor.
Dialogue: 0,0:00:16.80,0:00:19.17,Default,,0000,0000,0000,,Here we have an inductor,
Dialogue: 0,0:00:19.17,0:00:22.17,Default,,0000,0000,0000,,and we'll reference the voltage v of t
Dialogue: 0,0:00:22.17,0:00:25.78,Default,,0000,0000,0000,,plus the minus like that and also\Nreference the current through it,
Dialogue: 0,0:00:25.78,0:00:28.34,Default,,0000,0000,0000,,i of t where the current is referenced from
Dialogue: 0,0:00:28.34,0:00:30.53,Default,,0000,0000,0000,,the high voltage reference in
Dialogue: 0,0:00:30.53,0:00:33.76,Default,,0000,0000,0000,,the direction of the flowing\Nfrom high to low voltage.
Dialogue: 0,0:00:33.76,0:00:38.96,Default,,0000,0000,0000,,Now, let's just assume\Nthat i of t is of the form
Dialogue: 0,0:00:38.96,0:00:45.58,Default,,0000,0000,0000,,I sub m cosine of omega t plus Theta sub i.
Dialogue: 0,0:00:45.58,0:00:47.69,Default,,0000,0000,0000,,So the current we're\Nassuming is oscillating at
Dialogue: 0,0:00:47.69,0:00:50.22,Default,,0000,0000,0000,,the same frequency as the source\Nis driving the circuit.
Dialogue: 0,0:00:50.22,0:00:54.40,Default,,0000,0000,0000,,It has an amplitude I sub m and the phase\Nangle associated with the Thetas Y.
Dialogue: 0,0:00:54.40,0:00:58.10,Default,,0000,0000,0000,,Now, we know that in an\Ninductor the voltage across
Dialogue: 0,0:00:58.10,0:01:00.29,Default,,0000,0000,0000,,that inductor is proportional not to
Dialogue: 0,0:01:00.29,0:01:03.00,Default,,0000,0000,0000,,the current but rather proportional\Nto the derivative of the current,
Dialogue: 0,0:01:03.00,0:01:09.01,Default,,0000,0000,0000,,or v is equal to L times di dt.
Dialogue: 0,0:01:09.01,0:01:12.64,Default,,0000,0000,0000,,So, we've got then l. Now,
Dialogue: 0,0:01:12.64,0:01:17.18,Default,,0000,0000,0000,,di dt we're going to have I sub m,
Dialogue: 0,0:01:17.18,0:01:20.27,Default,,0000,0000,0000,,and the derivative of\Nthe cosine is negative sine.
Dialogue: 0,0:01:20.27,0:01:22.39,Default,,0000,0000,0000,,So let's bring the minus\Nsign out here in front,
Dialogue: 0,0:01:22.39,0:01:26.18,Default,,0000,0000,0000,,and let's also go ahead and we're going\Nto have a chain rule invoked here.
Dialogue: 0,0:01:26.18,0:01:30.90,Default,,0000,0000,0000,,So let's bring the Omega out in front\Nassociated with the chain rule,
Dialogue: 0,0:01:30.90,0:01:38.36,Default,,0000,0000,0000,,and then we will have the sine\Nof Omega t plus Theta sub i.
Dialogue: 0,0:01:38.36,0:01:40.97,Default,,0000,0000,0000,,Ultimately we're going\Nto want to have this or
Dialogue: 0,0:01:40.97,0:01:43.00,Default,,0000,0000,0000,,represent v in terms of its phasor,
Dialogue: 0,0:01:43.00,0:01:46.40,Default,,0000,0000,0000,,and so we need to convert\Nthe sine term to a cosine term.
Dialogue: 0,0:01:46.40,0:01:50.21,Default,,0000,0000,0000,,So this then is equal to negative L,
Dialogue: 0,0:01:50.21,0:01:55.11,Default,,0000,0000,0000,,I sub m Omega times the cosine of
Dialogue: 0,0:01:55.11,0:02:01.18,Default,,0000,0000,0000,,Omega t plus Theta sub i minus 90 degrees.
Dialogue: 0,0:02:01.18,0:02:02.34,Default,,0000,0000,0000,,Shifting the sign degree.
Dialogue: 0,0:02:02.34,0:02:04.24,Default,,0000,0000,0000,,The sign with 90 degrees to the right,
Dialogue: 0,0:02:04.24,0:02:06.64,Default,,0000,0000,0000,,because it is a cosine wave.
Dialogue: 0,0:02:06.64,0:02:10.13,Default,,0000,0000,0000,,Now, we can write I sub m\Nin terms of this phasor,
Dialogue: 0,0:02:10.13,0:02:17.81,Default,,0000,0000,0000,,or a phasor I is equal to I\Nsub m e to the j Theta sub i,
Dialogue: 0,0:02:17.81,0:02:27.53,Default,,0000,0000,0000,,and phaser v, will be equal to I've\Ngot the negative sign the L I Omega.
Dialogue: 0,0:02:27.53,0:02:28.97,Default,,0000,0000,0000,,It's amplitude.
Dialogue: 0,0:02:28.97,0:02:34.54,Default,,0000,0000,0000,,So negative l Omega I sub m,
Dialogue: 0,0:02:34.54,0:02:37.82,Default,,0000,0000,0000,,and then it will be e to the j,
Dialogue: 0,0:02:37.82,0:02:43.35,Default,,0000,0000,0000,,Theta sub i minus 90 degrees.
Dialogue: 0,0:02:43.35,0:02:47.30,Default,,0000,0000,0000,,Now, using the property of\Nthe product of exponents,
Dialogue: 0,0:02:47.30,0:02:51.10,Default,,0000,0000,0000,,let's rewrite this then as equaling.
Dialogue: 0,0:02:51.10,0:02:57.45,Default,,0000,0000,0000,,Let's see, negative L Omega i sub m,
Dialogue: 0,0:02:57.45,0:03:02.19,Default,,0000,0000,0000,,e to the j Theta sub i,
Dialogue: 0,0:03:02.19,0:03:04.88,Default,,0000,0000,0000,,e to the j.
Dialogue: 0,0:03:04.88,0:03:09.56,Default,,0000,0000,0000,,Let's say, negative j 90.
Dialogue: 0,0:03:09.56,0:03:14.78,Default,,0000,0000,0000,,Now, let's look at this term\Ne to the minus j 90.
Dialogue: 0,0:03:14.78,0:03:18.70,Default,,0000,0000,0000,,Just drawing a little\Nphasor diagram over here,
Dialogue: 0,0:03:18.70,0:03:26.92,Default,,0000,0000,0000,,e to the minus j 90 has a magnitude of\None and it has an angle of minus 90.
Dialogue: 0,0:03:26.92,0:03:30.49,Default,,0000,0000,0000,,That's equivalent to\Nthis e to the minus j 90
Dialogue: 0,0:03:30.49,0:03:34.16,Default,,0000,0000,0000,,then is in the negative j\Ndirection with a length of one,
Dialogue: 0,0:03:34.16,0:03:40.37,Default,,0000,0000,0000,,or we can rewrite this term\Nright here as a minus j.
Dialogue: 0,0:03:40.37,0:03:45.88,Default,,0000,0000,0000,,Now, let's combine this minus sign\Nhere with this minus sign here,
Dialogue: 0,0:03:45.88,0:03:50.27,Default,,0000,0000,0000,,bring the j here into this.
Dialogue: 0,0:03:52.26,0:03:56.02,Default,,0000,0000,0000,,Bring the j into the mix\Nhere and we get then that
Dialogue: 0,0:03:56.02,0:04:00.23,Default,,0000,0000,0000,,phaser v is equal to minus times\Nminus is a positive,
Dialogue: 0,0:04:00.23,0:04:04.67,Default,,0000,0000,0000,,and we'll have a j Omega L,
Dialogue: 0,0:04:04.67,0:04:09.48,Default,,0000,0000,0000,,e to the j Theta sub i, oh,
Dialogue: 0,0:04:09.48,0:04:14.52,Default,,0000,0000,0000,,and I left an I sub m term\Nmultiplying all of that also.
Dialogue: 0,0:04:14.52,0:04:18.95,Default,,0000,0000,0000,,Now, we notice that right there is I sub m,
Dialogue: 0,0:04:18.95,0:04:20.99,Default,,0000,0000,0000,,e to the j Theta sub i.
Dialogue: 0,0:04:20.99,0:04:24.38,Default,,0000,0000,0000,,Well, that's just phasor I.
Dialogue: 0,0:04:24.38,0:04:29.82,Default,,0000,0000,0000,,We have then that phaser v is equal to
Dialogue: 0,0:04:29.82,0:04:37.72,Default,,0000,0000,0000,,j Omega l times phasor I. Phasor I,
Dialogue: 0,0:04:37.72,0:04:41.66,Default,,0000,0000,0000,,phaser v, z is defined as
Dialogue: 0,0:04:41.66,0:04:46.10,Default,,0000,0000,0000,,the ratio phaser V to phasor\NI which is equal to then,
Dialogue: 0,0:04:46.10,0:04:55.47,Default,,0000,0000,0000,,phaser v is j Omega l times phasor\NI divided by phasor I.
Dialogue: 0,0:04:55.47,0:04:58.04,Default,,0000,0000,0000,,The phasor I's cancel and we're left
Dialogue: 0,0:04:58.04,0:05:00.62,Default,,0000,0000,0000,,with the impedance of\Nthe inductor is equal to
Dialogue: 0,0:05:00.62,0:05:05.76,Default,,0000,0000,0000,,j Omega L. Let's just look\Nat this for a second.
Dialogue: 0,0:05:05.76,0:05:08.54,Default,,0000,0000,0000,,In other words the impedance\Nof an inductor,
Dialogue: 0,0:05:08.54,0:05:12.14,Default,,0000,0000,0000,,is equal to j times Omega\Ntimes L. Note two things.
Dialogue: 0,0:05:12.14,0:05:15.52,Default,,0000,0000,0000,,First of all, the impedance\Nis imaginary, it's j.
Dialogue: 0,0:05:15.52,0:05:17.98,Default,,0000,0000,0000,,It's also positive.
Dialogue: 0,0:05:17.98,0:05:22.32,Default,,0000,0000,0000,,The second thing we'd like to point out\Nis that it is a function of frequency.
Dialogue: 0,0:05:22.32,0:05:25.67,Default,,0000,0000,0000,,Here, the frequency of\Nthe source that's driving
Dialogue: 0,0:05:25.67,0:05:30.83,Default,,0000,0000,0000,,the circuit determines or impacts,
Dialogue: 0,0:05:30.83,0:05:33.46,Default,,0000,0000,0000,,or yes what is it?
Dialogue: 0,0:05:33.46,0:05:35.87,Default,,0000,0000,0000,,It is involved in the impedance,
Dialogue: 0,0:05:35.87,0:05:38.28,Default,,0000,0000,0000,,the impedance is related to the frequency.
Dialogue: 0,0:05:38.28,0:05:42.22,Default,,0000,0000,0000,,In other words, as the frequency\Ngets larger and larger,
Dialogue: 0,0:05:42.22,0:05:44.40,Default,,0000,0000,0000,,the impedance gets larger and larger.
Dialogue: 0,0:05:44.40,0:05:46.91,Default,,0000,0000,0000,,Or as the frequency gets smaller,\Nthe impedance gets smaller.
Dialogue: 0,0:05:46.91,0:05:50.67,Default,,0000,0000,0000,,In fact and the limits,\Nwhen Omega equals zero,
Dialogue: 0,0:05:50.67,0:05:53.75,Default,,0000,0000,0000,,the impedance inductor is zero.
Dialogue: 0,0:05:53.75,0:05:56.75,Default,,0000,0000,0000,,The ratio of the voltage\Nto the current is zero.
Dialogue: 0,0:05:56.75,0:06:01.92,Default,,0000,0000,0000,,In other words, at DC there is\Nno voltage drop across the inductor.
Dialogue: 0,0:06:01.92,0:06:04.19,Default,,0000,0000,0000,,Though we knew that at DC,
Dialogue: 0,0:06:04.19,0:06:08.08,Default,,0000,0000,0000,,di dt is equal to zero and the voltage\Nacross the inductor is zero.
Dialogue: 0,0:06:08.08,0:06:10.39,Default,,0000,0000,0000,,Now, as Omega goes to infinity,
Dialogue: 0,0:06:10.39,0:06:13.37,Default,,0000,0000,0000,,the impedance of the inductor\Nbecomes infinite,
Dialogue: 0,0:06:13.37,0:06:16.52,Default,,0000,0000,0000,,or it would be equivalent\Nto an open circuit.
Dialogue: 0,0:06:16.52,0:06:19.90,Default,,0000,0000,0000,,In other words at high frequencies,
Dialogue: 0,0:06:19.90,0:06:22.62,Default,,0000,0000,0000,,the voltage across here,
Dialogue: 0,0:06:22.62,0:06:25.49,Default,,0000,0000,0000,,the derivative of the current goes to
Dialogue: 0,0:06:25.49,0:06:29.22,Default,,0000,0000,0000,,infinity and the voltage\Nacross it goes to infinity.
Dialogue: 0,0:06:29.22,0:06:38.16,Default,,0000,0000,0000,,It's an open circuit at that point.\NNothing gets through it.
Dialogue: 0,0:06:38.16,0:06:41.40,Default,,0000,0000,0000,,Let's just take an example.
Dialogue: 0,0:06:41.40,0:06:47.04,Default,,0000,0000,0000,,Let us assume that we have\Na source v of t is equal to
Dialogue: 0,0:06:47.04,0:06:53.81,Default,,0000,0000,0000,,five cosine of 100 t. In other words,
Dialogue: 0,0:06:53.81,0:06:56.70,Default,,0000,0000,0000,,Omega is equal to 100.
Dialogue: 0,0:06:56.70,0:07:02.69,Default,,0000,0000,0000,,Let's say we've got\Na 50 millihenry inductor.
Dialogue: 0,0:07:02.69,0:07:08.20,Default,,0000,0000,0000,,Let's let L equal 0.05 Henry.
Dialogue: 0,0:07:08.20,0:07:12.12,Default,,0000,0000,0000,,Then the impedance of\Nthat inductor is z sub L equals
Dialogue: 0,0:07:12.12,0:07:19.56,Default,,0000,0000,0000,,j times Omega which is\N100 times the inductance which is 0.05,
Dialogue: 0,0:07:19.56,0:07:21.14,Default,,0000,0000,0000,,and we get that the impedance of
Dialogue: 0,0:07:21.14,0:07:23.93,Default,,0000,0000,0000,,that inductor in a circuit\Noscillating 100 radians per
Dialogue: 0,0:07:23.93,0:07:29.94,Default,,0000,0000,0000,,second is equal to j five Ohms.