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[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.44,0:00:02.67,Default,,0000,0000,0000,,>> In these next few videos,
Dialogue: 0,0:00:02.67,0:00:07.05,Default,,0000,0000,0000,,we're going to demonstrate the application\Nof Thevenin's equivalent circuits
Dialogue: 0,0:00:07.05,0:00:10.26,Default,,0000,0000,0000,,in this phasor domain\Nusing complex impedances.
Dialogue: 0,0:00:10.26,0:00:14.31,Default,,0000,0000,0000,,You'll recall from our previous\Ndiscussions on Thevenin Equivalency,
Dialogue: 0,0:00:14.31,0:00:18.15,Default,,0000,0000,0000,,that the idea was that\Nif we had some circuit,
Dialogue: 0,0:00:18.15,0:00:20.22,Default,,0000,0000,0000,,some relatively complicated circuit,
Dialogue: 0,0:00:20.22,0:00:25.78,Default,,0000,0000,0000,,and we were only interested in the\Nterminal characteristics of that circuit,
Dialogue: 0,0:00:25.78,0:00:29.13,Default,,0000,0000,0000,,we could replace that circuit\Nwith its Thevenin equivalent
Dialogue: 0,0:00:29.13,0:00:32.58,Default,,0000,0000,0000,,which consisted of a single voltage source
Dialogue: 0,0:00:32.58,0:00:35.19,Default,,0000,0000,0000,,and back then when we are talking about
Dialogue: 0,0:00:35.19,0:00:38.15,Default,,0000,0000,0000,,a single resistance called\Nthe Thevenin resistance,
Dialogue: 0,0:00:38.15,0:00:41.90,Default,,0000,0000,0000,,but now since we are talking about\Nimpedances in this complex domain,
Dialogue: 0,0:00:41.90,0:00:43.88,Default,,0000,0000,0000,,the Thevenin equivalent circuit involves
Dialogue: 0,0:00:43.88,0:00:47.18,Default,,0000,0000,0000,,a voltage source - a phasor voltage -
Dialogue: 0,0:00:47.18,0:00:49.40,Default,,0000,0000,0000,,and a complex impedance.
Dialogue: 0,0:00:49.40,0:00:58.33,Default,,0000,0000,0000,,By equivalent, we mean that if you\Nattach a load to the actual circuit,
Dialogue: 0,0:00:58.33,0:01:00.11,Default,,0000,0000,0000,,a certain current draws,
Dialogue: 0,0:01:00.11,0:01:04.04,Default,,0000,0000,0000,,and as you start drawing current,\Nthe terminal voltage starts to drop.
Dialogue: 0,0:01:04.04,0:01:10.97,Default,,0000,0000,0000,,This Thevenin equivalent circuit\Nwill model that current flowing
Dialogue: 0,0:01:10.97,0:01:14.57,Default,,0000,0000,0000,,and the voltage drop that will\Nhappen as current starts to flow
Dialogue: 0,0:01:14.57,0:01:21.26,Default,,0000,0000,0000,,and be equivalent to or be\Nthe same current and voltage drop
Dialogue: 0,0:01:21.26,0:01:24.34,Default,,0000,0000,0000,,that you would experience in\Nthe more complicated circuit.
Dialogue: 0,0:01:24.34,0:01:27.04,Default,,0000,0000,0000,,Again, it's simply a method of modeling
Dialogue: 0,0:01:27.04,0:01:30.13,Default,,0000,0000,0000,,a more complicated circuit by\Na less complicated circuit
Dialogue: 0,0:01:30.13,0:01:33.29,Default,,0000,0000,0000,,that has the same terminal characteristics.
Dialogue: 0,0:01:33.29,0:01:34.83,Default,,0000,0000,0000,,You'll recall that to do so,
Dialogue: 0,0:01:34.83,0:01:37.39,Default,,0000,0000,0000,,we had to determine what\Nthe open circuit voltage was,
Dialogue: 0,0:01:37.39,0:01:40.45,Default,,0000,0000,0000,,the open-circuit voltage being simply\Nthe voltage across the terminals AB
Dialogue: 0,0:01:40.45,0:01:45.01,Default,,0000,0000,0000,,when no load is connected.
Dialogue: 0,0:01:45.01,0:01:48.10,Default,,0000,0000,0000,,That's the Thevenin voltage.
Dialogue: 0,0:01:48.10,0:01:52.06,Default,,0000,0000,0000,,The Thevenin impedance, we can find\Nin one of three different ways.
Dialogue: 0,0:01:52.06,0:01:54.64,Default,,0000,0000,0000,,We can determine the equivalent\Nimpedance seen looking
Dialogue: 0,0:01:54.64,0:02:00.04,Default,,0000,0000,0000,,back in to the circuit with\Nall independent sources deactivated;
Dialogue: 0,0:02:00.04,0:02:03.42,Default,,0000,0000,0000,,for one method.
Dialogue: 0,0:02:03.42,0:02:06.58,Default,,0000,0000,0000,,Another method involves\Nputting a short circuit,
Dialogue: 0,0:02:06.58,0:02:11.90,Default,,0000,0000,0000,,actually grounding or shorting\Nout the AB terminals,
Dialogue: 0,0:02:11.90,0:02:15.05,Default,,0000,0000,0000,,and calculating that short-circuit current,
Dialogue: 0,0:02:15.05,0:02:19.25,Default,,0000,0000,0000,,and then this Thevenin\Nimpedance is equal to
Dialogue: 0,0:02:19.25,0:02:23.90,Default,,0000,0000,0000,,the ratio of the open-circuit voltage\Nto the short circuit current.
Dialogue: 0,0:02:23.90,0:02:25.78,Default,,0000,0000,0000,,We'll do an example using that.
Dialogue: 0,0:02:25.78,0:02:30.57,Default,,0000,0000,0000,,And the third method involves\Ndeactivating the source,
Dialogue: 0,0:02:30.57,0:02:32.68,Default,,0000,0000,0000,,any independent sources,
Dialogue: 0,0:02:32.68,0:02:37.43,Default,,0000,0000,0000,,applying an external source\Nto the circuit terminals,
Dialogue: 0,0:02:37.43,0:02:42.02,Default,,0000,0000,0000,,and calculating the ratio of\Nthe voltage, that external voltage,
Dialogue: 0,0:02:42.02,0:02:46.16,Default,,0000,0000,0000,,to the current that then flows.
Dialogue: 0,0:02:46.61,0:02:49.28,Default,,0000,0000,0000,,Okay, let's go ahead and do this.
Dialogue: 0,0:02:49.28,0:02:50.87,Default,,0000,0000,0000,,Here we've got a time domain circuit.
Dialogue: 0,0:02:50.87,0:02:54.20,Default,,0000,0000,0000,,Relatively complicated, not terribly\Nbut a good one for our example.
Dialogue: 0,0:02:54.20,0:02:56.18,Default,,0000,0000,0000,,The first thing we need to do is convert it
Dialogue: 0,0:02:56.18,0:02:58.19,Default,,0000,0000,0000,,to its phasor domain representation.
Dialogue: 0,0:02:58.19,0:03:02.45,Default,,0000,0000,0000,,So, this 10cosine(500t) has been converted
Dialogue: 0,0:03:02.45,0:03:07.22,Default,,0000,0000,0000,,to a complex impedance voltage\Nof 10e to the j0
Dialogue: 0,0:03:07.22,0:03:10.49,Default,,0000,0000,0000,,and similarly each of the impedances\Nare represented over here,
Dialogue: 0,0:03:10.49,0:03:14.84,Default,,0000,0000,0000,,and our task then is to determine\Nthe Thevenin equivalent circuit
Dialogue: 0,0:03:14.84,0:03:17.42,Default,,0000,0000,0000,,for this circuit.
Dialogue: 0,0:03:17.42,0:03:20.50,Default,,0000,0000,0000,,To do so, we need the open-circuit voltage.
Dialogue: 0,0:03:20.50,0:03:23.15,Default,,0000,0000,0000,,The open circuit voltage is\Nthe voltage that we would
Dialogue: 0,0:03:23.15,0:03:27.86,Default,,0000,0000,0000,,measure across the terminals\Nwith no load connected.
Dialogue: 0,0:03:27.86,0:03:30.10,Default,,0000,0000,0000,,Because there is no load connected,
Dialogue: 0,0:03:30.10,0:03:34.23,Default,,0000,0000,0000,,there is no current flowing\Nthrough that inductor
Dialogue: 0,0:03:34.23,0:03:37.40,Default,,0000,0000,0000,,and any current that is\Nproduced by this voltage source
Dialogue: 0,0:03:37.40,0:03:43.55,Default,,0000,0000,0000,,will be going through a series combination\Nof the capacitor and the resistor.
Dialogue: 0,0:03:43.55,0:03:47.00,Default,,0000,0000,0000,,So the open-circuit voltage\Nis simply going to be
Dialogue: 0,0:03:47.00,0:03:49.32,Default,,0000,0000,0000,,the voltage across this 20-Ohm resistor.
Dialogue: 0,0:03:49.32,0:03:51.50,Default,,0000,0000,0000,,Again, because there's no current\Nflowing through that inductor,
Dialogue: 0,0:03:51.50,0:03:53.48,Default,,0000,0000,0000,,there'll be no voltage drop\Nacross that inductor
Dialogue: 0,0:03:53.48,0:03:56.03,Default,,0000,0000,0000,,and the open-circuit voltage would just be
Dialogue: 0,0:03:56.03,0:03:57.56,Default,,0000,0000,0000,,the voltage across that 20-Ohm resistor.
Dialogue: 0,0:03:57.56,0:03:59.51,Default,,0000,0000,0000,,Easy way to get that, it appears to me,
Dialogue: 0,0:03:59.51,0:04:00.98,Default,,0000,0000,0000,,would be to use a voltage divider,
Dialogue: 0,0:04:00.98,0:04:04.34,Default,,0000,0000,0000,,and say then that V\Nopen circuit is equal to
Dialogue: 0,0:04:04.34,0:04:14.46,Default,,0000,0000,0000,,10 times 20 divided by 20 minus j25,
Dialogue: 0,0:04:14.46,0:04:21.56,Default,,0000,0000,0000,,and when you do that, you get\Nthat the open-circuit voltage
Dialogue: 0,0:04:21.56,0:04:29.64,Default,,0000,0000,0000,,is equal to 6.25e to the j51.34.
Dialogue: 0,0:04:29.64,0:04:34.67,Default,,0000,0000,0000,,Because we are going to be\Nwriting so many of these phasors,
Dialogue: 0,0:04:34.67,0:04:36.95,Default,,0000,0000,0000,,rather than writing in\Nthe exponential form,
Dialogue: 0,0:04:36.95,0:04:45.95,Default,,0000,0000,0000,,I'm going to start writing them as\Na magnitude and an angle of 51.34.
Dialogue: 0,0:04:45.95,0:04:51.62,Default,,0000,0000,0000,,So this form and this form\Nare both polar forms.
Dialogue: 0,0:04:51.62,0:04:53.48,Default,,0000,0000,0000,,This just represents\Nthe magnitude and the phase
Dialogue: 0,0:04:53.48,0:04:55.28,Default,,0000,0000,0000,,and here we've got the complex exponential
Dialogue: 0,0:04:55.28,0:04:57.70,Default,,0000,0000,0000,,that again demonstrates\Nthe mathematics behind it,
Dialogue: 0,0:04:57.70,0:04:58.84,Default,,0000,0000,0000,,what is actually happening
Dialogue: 0,0:04:58.84,0:05:01.86,Default,,0000,0000,0000,,but whatever form we choose,
Dialogue: 0,0:05:01.86,0:05:04.49,Default,,0000,0000,0000,,we now have our Thevenin equivalent voltage
Dialogue: 0,0:05:04.49,0:05:06.50,Default,,0000,0000,0000,,because the Thevenin equivalent voltage is
Dialogue: 0,0:05:06.50,0:05:08.42,Default,,0000,0000,0000,,simply equal to the open circuit voltage
Dialogue: 0,0:05:08.42,0:05:15.82,Default,,0000,0000,0000,,which in this case is the\N6.25 angle 51.34 degrees.
Dialogue: 0,0:05:15.82,0:05:18.74,Default,,0000,0000,0000,,In the next video, we'll demonstrate\Nthe three different methods
Dialogue: 0,0:05:18.74,0:05:22.08,Default,,0000,0000,0000,,of determining the Thevenin\Nequivalent impedance