[Script Info]
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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:03.06,Default,,0000,0000,0000,,>> In this video, we're going to\Ndemonstrate the actual process of
Dialogue: 0,0:00:03.06,0:00:05.94,Default,,0000,0000,0000,,analyzing circuits using\Nphasors and impedances,
Dialogue: 0,0:00:05.94,0:00:07.90,Default,,0000,0000,0000,,analyzing circuits in the phasor domain.
Dialogue: 0,0:00:07.90,0:00:09.45,Default,,0000,0000,0000,,We've already pointed out that in
Dialogue: 0,0:00:09.45,0:00:13.15,Default,,0000,0000,0000,,the phasor domain or rather\Nthat we can analyze a circuit,
Dialogue: 0,0:00:13.15,0:00:21.06,Default,,0000,0000,0000,,in the phasor domain by representing\Nthe voltages in terms of phasors,
Dialogue: 0,0:00:21.06,0:00:22.98,Default,,0000,0000,0000,,the currents in terms of phasors,
Dialogue: 0,0:00:22.98,0:00:27.86,Default,,0000,0000,0000,,and the impedances or representing\Nthe effects of these capacitors,
Dialogue: 0,0:00:27.86,0:00:30.88,Default,,0000,0000,0000,,resistors, inductors in\Nterms of their impedances.
Dialogue: 0,0:00:30.88,0:00:32.60,Default,,0000,0000,0000,,So first of all,
Dialogue: 0,0:00:32.60,0:00:35.65,Default,,0000,0000,0000,,let's just represent phasor V,
Dialogue: 0,0:00:35.65,0:00:42.32,Default,,0000,0000,0000,,the source as having an amplitude\NV sub m e to the j Theta.
Dialogue: 0,0:00:42.32,0:00:44.100,Default,,0000,0000,0000,,So, representing this time domain signal as
Dialogue: 0,0:00:44.100,0:00:48.91,Default,,0000,0000,0000,,its phasor with an amplitude of V\Nsub m and a phase shift of Theta.
Dialogue: 0,0:00:48.91,0:00:53.36,Default,,0000,0000,0000,,Now the impedance of the capacitor\Nis equal to one over
Dialogue: 0,0:00:53.36,0:00:58.72,Default,,0000,0000,0000,,j Omega C. The impedance of\Na resistor is just equal to R,
Dialogue: 0,0:00:58.72,0:01:01.27,Default,,0000,0000,0000,,and the impedance of\Nthe inductor is equal to j times
Dialogue: 0,0:01:01.27,0:01:04.81,Default,,0000,0000,0000,,Omega times L. So we'll\Nrepresent that or show that
Dialogue: 0,0:01:04.81,0:01:12.61,Default,,0000,0000,0000,,here is this impedance is one\Nover j Omega C. This is just R,
Dialogue: 0,0:01:12.61,0:01:15.84,Default,,0000,0000,0000,,and this impedance here is j Omega
Dialogue: 0,0:01:15.84,0:01:20.38,Default,,0000,0000,0000,,L. Now in the phasor domain
Dialogue: 0,0:01:20.38,0:01:25.34,Default,,0000,0000,0000,,there will be a current flowing\Nthat we will represent as phasor I,
Dialogue: 0,0:01:25.55,0:01:29.93,Default,,0000,0000,0000,,and with these representations\Nlet's go ahead now and write
Dialogue: 0,0:01:29.93,0:01:34.10,Default,,0000,0000,0000,,a Kirchhoff's voltage law equation\Nin terms of these phasers.
Dialogue: 0,0:01:34.10,0:01:36.17,Default,,0000,0000,0000,,Starting at this point and\Ngoing in this direction,
Dialogue: 0,0:01:36.17,0:01:38.00,Default,,0000,0000,0000,,I'm going to add up the voltage
Dialogue: 0,0:01:38.00,0:01:40.73,Default,,0000,0000,0000,,drops which means the voltage increase\Nwill be a negative.
Dialogue: 0,0:01:40.73,0:01:45.60,Default,,0000,0000,0000,,So we'll have negative V sub m e
Dialogue: 0,0:01:45.60,0:01:50.51,Default,,0000,0000,0000,,to the j Theta plus now going in\Nthe direction of current flow.
Dialogue: 0,0:01:50.51,0:01:54.48,Default,,0000,0000,0000,,So direct and current flow represents\Na voltage drop and it will be a plus.
Dialogue: 0,0:01:54.48,0:01:56.81,Default,,0000,0000,0000,,The voltage drop across the capacitor is
Dialogue: 0,0:01:56.81,0:01:59.52,Default,,0000,0000,0000,,just the impedance of\Nthe capacitor times the current
Dialogue: 0,0:01:59.52,0:02:05.47,Default,,0000,0000,0000,,or one over j Omega C times I,
Dialogue: 0,0:02:05.47,0:02:08.84,Default,,0000,0000,0000,,plus the voltage drop across the resistor\Nagain going in the direction of
Dialogue: 0,0:02:08.84,0:02:13.21,Default,,0000,0000,0000,,current flow will be just R times I.
Dialogue: 0,0:02:13.21,0:02:16.01,Default,,0000,0000,0000,,Plus the voltage drop across\Nthe inductor will be,
Dialogue: 0,0:02:16.01,0:02:21.06,Default,,0000,0000,0000,,the impedance of the inductor\Nwhich is j Omega L times I.
Dialogue: 0,0:02:21.06,0:02:24.38,Default,,0000,0000,0000,,The sum of those terms equals zero.
Dialogue: 0,0:02:24.38,0:02:27.86,Default,,0000,0000,0000,,Now, this is not a function of I.
Dialogue: 0,0:02:27.86,0:02:29.90,Default,,0000,0000,0000,,In fact, this is the constant\Nassociated with the source.
Dialogue: 0,0:02:29.90,0:02:31.91,Default,,0000,0000,0000,,So let's bring it to the other side
Dialogue: 0,0:02:31.91,0:02:34.82,Default,,0000,0000,0000,,and factor out the I that is\Ncommon in each of those terms,
Dialogue: 0,0:02:34.82,0:02:42.95,Default,,0000,0000,0000,,we have in that I not equal\Nto I times one over j Omega C
Dialogue: 0,0:02:42.95,0:02:53.47,Default,,0000,0000,0000,,plus R plus j omega L is equal\Nto V sub m e to the j Theta.
Dialogue: 0,0:02:53.47,0:02:56.12,Default,,0000,0000,0000,,We now solve for I by\Ndividing both sides of
Dialogue: 0,0:02:56.12,0:02:59.02,Default,,0000,0000,0000,,the equation by this term in\Nparentheses, and we have then I,
Dialogue: 0,0:02:59.02,0:03:04.13,Default,,0000,0000,0000,,the phasor representation of\NI is equal to V sub m e to
Dialogue: 0,0:03:04.13,0:03:10.00,Default,,0000,0000,0000,,the j Theta divided by one over j Omega C,
Dialogue: 0,0:03:10.00,0:03:15.24,Default,,0000,0000,0000,,plus R plus j Omega L.
Dialogue: 0,0:03:15.24,0:03:18.44,Default,,0000,0000,0000,,Phasor I the phasor representation is
Dialogue: 0,0:03:18.44,0:03:23.14,Default,,0000,0000,0000,,calculated by doing that complex division.