[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:03.15,Default,,0000,0000,0000,,>> We're now going to derive\Nthe Impedance of the Capacitor.
Dialogue: 0,0:00:03.15,0:00:06.15,Default,,0000,0000,0000,,Again impedance we have defined as
Dialogue: 0,0:00:06.15,0:00:09.66,Default,,0000,0000,0000,,being the ratio of the phasor\Nvoltage to the phasor current.
Dialogue: 0,0:00:09.66,0:00:14.12,Default,,0000,0000,0000,,So let's determine that relationship\Nnow for our capacitor.
Dialogue: 0,0:00:14.12,0:00:18.10,Default,,0000,0000,0000,,Once again, we're going to define\Na voltage or reference voltage v of
Dialogue: 0,0:00:18.10,0:00:23.09,Default,,0000,0000,0000,,t and a current flowing through\Nthe capacitor flowing or as reference,
Dialogue: 0,0:00:23.09,0:00:26.60,Default,,0000,0000,0000,,from the positive to\Nthe negative reference to terminal.
Dialogue: 0,0:00:26.60,0:00:29.48,Default,,0000,0000,0000,,Now this time around, let's\Nassume that the voltage and
Dialogue: 0,0:00:29.48,0:00:34.02,Default,,0000,0000,0000,,the capacitor is equal to\Nor is of the form V sub m,
Dialogue: 0,0:00:34.02,0:00:38.13,Default,,0000,0000,0000,,cosine of Omega t plus Theta sub V.
Dialogue: 0,0:00:38.13,0:00:44.09,Default,,0000,0000,0000,,We know that in a capacitor\Ni is equal to c times
Dialogue: 0,0:00:44.09,0:00:46.14,Default,,0000,0000,0000,,the derivative of the voltage,
Dialogue: 0,0:00:46.14,0:00:49.94,Default,,0000,0000,0000,,which is going to equal\Nthen c. Now once again,
Dialogue: 0,0:00:49.94,0:00:52.15,Default,,0000,0000,0000,,the derivative of cosine\Nis negative the sine,
Dialogue: 0,0:00:52.15,0:00:53.96,Default,,0000,0000,0000,,so we'll bring that\Nnegative out in front here,
Dialogue: 0,0:00:53.96,0:00:56.68,Default,,0000,0000,0000,,the chain rule is going to\Ngive us an Omega out in front.
Dialogue: 0,0:00:56.68,0:01:01.77,Default,,0000,0000,0000,,We have then the V sub m and\Nwe will have in the sine,
Dialogue: 0,0:01:01.77,0:01:03.95,Default,,0000,0000,0000,,the derivative of cosine is the sine of the
Dialogue: 0,0:01:03.95,0:01:06.74,Default,,0000,0000,0000,,negative already taken\Ncare of sine of Omega t
Dialogue: 0,0:01:06.74,0:01:12.08,Default,,0000,0000,0000,,plus Theta sub V. Once again,
Dialogue: 0,0:01:12.08,0:01:16.31,Default,,0000,0000,0000,,we want to write this in terms\Nof cosine rather than a sine.
Dialogue: 0,0:01:16.31,0:01:19.14,Default,,0000,0000,0000,,So this is going to equal negative C times
Dialogue: 0,0:01:19.14,0:01:22.14,Default,,0000,0000,0000,,Omega times V sub m times the cosine
Dialogue: 0,0:01:22.14,0:01:29.20,Default,,0000,0000,0000,,of Omega t plus Theta\Nsub V minus 90 degrees.
Dialogue: 0,0:01:29.27,0:01:32.90,Default,,0000,0000,0000,,Now, let's write both of\Nthese and their phasor forms.
Dialogue: 0,0:01:32.90,0:01:35.80,Default,,0000,0000,0000,,This time we have phasor V is equal to,
Dialogue: 0,0:01:35.80,0:01:43.50,Default,,0000,0000,0000,,V sub m e to the j Theta sub V.\NWe can write phasor I is equal to
Dialogue: 0,0:01:43.50,0:01:51.06,Default,,0000,0000,0000,,negative negative C Omega V sub m, e
Dialogue: 0,0:01:51.06,0:01:56.84,Default,,0000,0000,0000,,to the j Theta sub V minus 90 degrees.
Dialogue: 0,0:01:56.84,0:02:00.47,Default,,0000,0000,0000,,Again, using the product of
Dialogue: 0,0:02:00.47,0:02:05.57,Default,,0000,0000,0000,,two exponential terms is that you\Nadd the exponents and do that.
Dialogue: 0,0:02:05.57,0:02:12.14,Default,,0000,0000,0000,,So we have the answer is equal\Nto negative C Omega V sub m e to
Dialogue: 0,0:02:12.14,0:02:19.88,Default,,0000,0000,0000,,the j Theta sub V e to\Nthe minus j90 degrees.
Dialogue: 0,0:02:19.88,0:02:25.20,Default,,0000,0000,0000,,As before, this term right\Nthere is just minus j.
Dialogue: 0,0:02:25.20,0:02:29.08,Default,,0000,0000,0000,,So we take the minus times\Na minus that gives us a positive,
Dialogue: 0,0:02:29.08,0:02:35.84,Default,,0000,0000,0000,,and bringing the J into this makes\Nhere we have then J Omega C,
Dialogue: 0,0:02:35.84,0:02:41.42,Default,,0000,0000,0000,,V sub m e to the j Theta sub V. Again,
Dialogue: 0,0:02:41.42,0:02:43.40,Default,,0000,0000,0000,,we recognize that right there as being
Dialogue: 0,0:02:43.40,0:02:48.59,Default,,0000,0000,0000,,phasor V. So we can then write that phasor
Dialogue: 0,0:02:48.59,0:02:56.33,Default,,0000,0000,0000,,I is equal to j Omega\NC times phasor V. Now,
Dialogue: 0,0:02:56.33,0:02:58.91,Default,,0000,0000,0000,,let's go ahead and form the impedance by
Dialogue: 0,0:02:58.91,0:03:02.14,Default,,0000,0000,0000,,writing a ratio of phasor V to phasor I.
Dialogue: 0,0:03:02.14,0:03:08.73,Default,,0000,0000,0000,,We have Z then is equal to\Nphasor V divided by phasor I,
Dialogue: 0,0:03:08.73,0:03:16.72,Default,,0000,0000,0000,,but phasor I is just j Omega C times\Nphasor V. The phasor V is cancel.
Dialogue: 0,0:03:16.72,0:03:21.89,Default,,0000,0000,0000,,We get then that the impedance of\Na capacitor is equal to one over
Dialogue: 0,0:03:21.89,0:03:30.17,Default,,0000,0000,0000,,j Omega C. Sometimes we'll write that by\Nbringing the j up into the numerator,
Dialogue: 0,0:03:30.17,0:03:32.36,Default,,0000,0000,0000,,changing the sign on it which then gives us
Dialogue: 0,0:03:32.36,0:03:38.98,Default,,0000,0000,0000,,a negative J times 1\Nover Omega C. All right.
Dialogue: 0,0:03:38.98,0:03:40.64,Default,,0000,0000,0000,,We need to make some observations here.
Dialogue: 0,0:03:40.64,0:03:45.72,Default,,0000,0000,0000,,First of all, notice that the impedance\Nof the capacitor is negative.
Dialogue: 0,0:03:45.72,0:03:47.78,Default,,0000,0000,0000,,Second of all, notice
Dialogue: 0,0:03:47.78,0:03:55.24,Default,,0000,0000,0000,,the frequency dependency of\Nthe capacitor impedance is inversely.
Dialogue: 0,0:03:55.24,0:03:57.32,Default,,0000,0000,0000,,In other words, I'll say it better,
Dialogue: 0,0:03:57.32,0:04:01.84,Default,,0000,0000,0000,,the impedance of a capacitor\Nis inversely proportional,
Dialogue: 0,0:04:01.84,0:04:06.30,Default,,0000,0000,0000,,to the frequency of\Nthe circuit source that's driving it.
Dialogue: 0,0:04:06.30,0:04:08.60,Default,,0000,0000,0000,,Remind ourselves here.
Dialogue: 0,0:04:08.60,0:04:11.03,Default,,0000,0000,0000,,The impedance of an inductor\Nwas j times Omega
Dialogue: 0,0:04:11.03,0:04:16.55,Default,,0000,0000,0000,,L. The impedance of an inductor was\Ndirectly proportional to the frequency.
Dialogue: 0,0:04:16.55,0:04:21.32,Default,,0000,0000,0000,,The impedance of a capacitor is\Ninversely proportional to the frequency.
Dialogue: 0,0:04:21.32,0:04:23.72,Default,,0000,0000,0000,,When the frequency is small,
Dialogue: 0,0:04:23.72,0:04:27.32,Default,,0000,0000,0000,,the impedance of the capacitor\Nis large and in fact,
Dialogue: 0,0:04:27.32,0:04:29.70,Default,,0000,0000,0000,,that DC when Omega equals 0,
Dialogue: 0,0:04:29.70,0:04:32.22,Default,,0000,0000,0000,,the capacitor has an infinite impedance.
Dialogue: 0,0:04:32.22,0:04:35.16,Default,,0000,0000,0000,,It is effectively an open circuit.
Dialogue: 0,0:04:35.16,0:04:40.31,Default,,0000,0000,0000,,As the frequency gets larger and larger\Nas the frequency approaches infinity,
Dialogue: 0,0:04:40.31,0:04:44.51,Default,,0000,0000,0000,,a larger number in the denominator makes\Nthe impedance of the capacitor gets
Dialogue: 0,0:04:44.51,0:04:48.68,Default,,0000,0000,0000,,smaller and smaller and in the limit\Nas Omega goes to infinity,
Dialogue: 0,0:04:48.68,0:04:51.20,Default,,0000,0000,0000,,the impedance capacitor goes to zero.
Dialogue: 0,0:04:51.20,0:04:58.23,Default,,0000,0000,0000,,At high frequencies, the capacitor\Nis approximated by a short circuit.
Dialogue: 0,0:04:58.48,0:05:05.70,Default,,0000,0000,0000,,So for example, let's say that we've\Ngot a 10 microfarad capacitor.
Dialogue: 0,0:05:05.70,0:05:12.48,Default,,0000,0000,0000,,The source is driving this is some V sub
Dialogue: 0,0:05:12.48,0:05:19.59,Default,,0000,0000,0000,,s of t is equal to 5 cosine of 5,000t.
Dialogue: 0,0:05:19.59,0:05:25.64,Default,,0000,0000,0000,,In other words, Omega is equal\Nto 5,000 radians per second.
Dialogue: 0,0:05:25.64,0:05:28.38,Default,,0000,0000,0000,,Then the impedance of\Nthe capacitor in this circuit,
Dialogue: 0,0:05:28.38,0:05:35.22,Default,,0000,0000,0000,,this 10 microfarad capacitor would be\Nequal to 1 over j times Omega which is
Dialogue: 0,0:05:35.22,0:05:42.40,Default,,0000,0000,0000,,5,000 times c which is 10\Ntimes 10 to the minus 6.
Dialogue: 0,0:05:42.40,0:05:45.54,Default,,0000,0000,0000,,You do the math on that, again\Nbringing the j in the numerator.
Dialogue: 0,0:05:45.54,0:05:47.87,Default,,0000,0000,0000,,Here you'll get that the impedance\Nof this capacitor would
Dialogue: 0,0:05:47.87,0:05:51.38,Default,,0000,0000,0000,,be negative negative j20.
Dialogue: 0,0:05:51.38,0:05:54.65,Default,,0000,0000,0000,,As we mentioned, the impedance\Nof the capacitor is negative,
Dialogue: 0,0:05:54.65,0:05:57.14,Default,,0000,0000,0000,,the impedance of the inductor is positive.
Dialogue: 0,0:05:57.14,0:05:58.79,Default,,0000,0000,0000,,The impedance of the inductor,
Dialogue: 0,0:05:58.79,0:06:00.76,Default,,0000,0000,0000,,in fact, let's just make\Na little chart here.
Dialogue: 0,0:06:00.76,0:06:05.39,Default,,0000,0000,0000,,In the time domain R and the impedance for
Dialogue: 0,0:06:05.39,0:06:10.40,Default,,0000,0000,0000,,a resistor is just plain\Nold R. For a capacitor,
Dialogue: 0,0:06:10.40,0:06:15.65,Default,,0000,0000,0000,,the impedance of a capacitor\Nis 1 over j Omega C.
Dialogue: 0,0:06:15.65,0:06:22.61,Default,,0000,0000,0000,,For an inductor, the impedance\Nof the inductor is J Omega L.
Dialogue: 0,0:06:22.61,0:06:28.84,Default,,0000,0000,0000,,Finally, in each of\Nthese cases, Ohm's law applies.
Dialogue: 0,0:06:28.84,0:06:36.47,Default,,0000,0000,0000,,We have then that V is equal\Nto I times Z because we
Dialogue: 0,0:06:36.47,0:06:45.75,Default,,0000,0000,0000,,defined Z to be the ratio\Nof phasor V to phasor I.