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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.20,0:00:03.69,Default,,0000,0000,0000,,>> In the last few videos, we've been\Ndeveloping the tools that we need
Dialogue: 0,0:00:03.69,0:00:07.10,Default,,0000,0000,0000,,to analyze these circuits that are\Nbeing driven by sinusoidal sources.
Dialogue: 0,0:00:07.10,0:00:10.68,Default,,0000,0000,0000,,Specifically, we've developed the construct
Dialogue: 0,0:00:10.68,0:00:12.66,Default,,0000,0000,0000,,of the mathematical tool of a phasor,
Dialogue: 0,0:00:12.66,0:00:15.33,Default,,0000,0000,0000,,we've introduced the concepts of impedance,
Dialogue: 0,0:00:15.33,0:00:17.58,Default,,0000,0000,0000,,and now we're ready to move on and show how
Dialogue: 0,0:00:17.58,0:00:21.75,Default,,0000,0000,0000,,Kirchhoff's and Ohm's Law is can be\Nused in terms of phasors impedances,
Dialogue: 0,0:00:21.75,0:00:23.55,Default,,0000,0000,0000,,to analyze these types of circuits.
Dialogue: 0,0:00:23.55,0:00:24.57,Default,,0000,0000,0000,,My name is Lee Brinton.
Dialogue: 0,0:00:24.57,0:00:25.83,Default,,0000,0000,0000,,I'm an Electrical Engineering Instructor
Dialogue: 0,0:00:25.83,0:00:28.09,Default,,0000,0000,0000,,in Salt Lake Community College.
Dialogue: 0,0:00:29.78,0:00:36.41,Default,,0000,0000,0000,,Specifically, we're going to derive or\Ndemonstrate how Kirchhoff's voltage and
Dialogue: 0,0:00:36.41,0:00:39.50,Default,,0000,0000,0000,,current laws apply in these\Ncircuits that are being
Dialogue: 0,0:00:39.50,0:00:43.22,Default,,0000,0000,0000,,driven by sinusoidal sources,
Dialogue: 0,0:00:43.22,0:00:46.07,Default,,0000,0000,0000,,and we'll be analyzing it\Nin this phasor domain,
Dialogue: 0,0:00:46.07,0:00:50.57,Default,,0000,0000,0000,,and then we'll give\Nan example of how these laws
Dialogue: 0,0:00:50.57,0:00:51.77,Default,,0000,0000,0000,,and these tools that we've developed at
Dialogue: 0,0:00:51.77,0:00:55.33,Default,,0000,0000,0000,,this point are used to analyze circuits.
Dialogue: 0,0:00:55.33,0:00:57.96,Default,,0000,0000,0000,,First of all, Kirchhoff's Voltage Law.
Dialogue: 0,0:00:57.96,0:01:00.38,Default,,0000,0000,0000,,We have here a circuit defined involving
Dialogue: 0,0:01:00.38,0:01:06.38,Default,,0000,0000,0000,,three different devices with a voltage\Nreference here plus to minus v_1 of t,
Dialogue: 0,0:01:06.38,0:01:08.31,Default,,0000,0000,0000,,plus to minus v_2 of t,
Dialogue: 0,0:01:08.31,0:01:11.04,Default,,0000,0000,0000,,and plus to minus v_3 of t. We know from
Dialogue: 0,0:01:11.04,0:01:12.86,Default,,0000,0000,0000,,Kirchhoff's Voltage Law\Nthat the sum of the voltage
Dialogue: 0,0:01:12.86,0:01:15.10,Default,,0000,0000,0000,,drops around that loop must equal 0,
Dialogue: 0,0:01:15.10,0:01:21.20,Default,,0000,0000,0000,,or v_1 of t plus v_2
Dialogue: 0,0:01:21.20,0:01:27.38,Default,,0000,0000,0000,,of t plus v_3 of t must all add to be 0.
Dialogue: 0,0:01:27.38,0:01:32.40,Default,,0000,0000,0000,,Now, if we assume that we're operating\Nin the sinusoidal steady-state,
Dialogue: 0,0:01:32.40,0:01:36.74,Default,,0000,0000,0000,,and we've asserted that in\Nthat type of a circuit,
Dialogue: 0,0:01:36.74,0:01:40.01,Default,,0000,0000,0000,,all of the voltages and\Ncurrents associated with
Dialogue: 0,0:01:40.01,0:01:43.68,Default,,0000,0000,0000,,that circuit will be oscillating\Nat the same frequency.
Dialogue: 0,0:01:43.68,0:01:46.58,Default,,0000,0000,0000,,Let's just say that\Nthe source is oscillating at
Dialogue: 0,0:01:46.58,0:01:49.100,Default,,0000,0000,0000,,some Omega radians per second,
Dialogue: 0,0:01:49.100,0:01:52.67,Default,,0000,0000,0000,,that means that v_1, v_2\Nand v_3 will also be
Dialogue: 0,0:01:52.67,0:01:55.13,Default,,0000,0000,0000,,oscillating at that same frequency,
Dialogue: 0,0:01:55.13,0:01:57.60,Default,,0000,0000,0000,,but they'll have different amplitudes\Nand different phases.
Dialogue: 0,0:01:57.60,0:01:59.40,Default,,0000,0000,0000,,Or more specifically then,
Dialogue: 0,0:01:59.40,0:02:05.53,Default,,0000,0000,0000,,let's write it as v sub m1,
Dialogue: 0,0:02:05.53,0:02:10.62,Default,,0000,0000,0000,,cosine of Omega t plus Theta 1,
Dialogue: 0,0:02:10.62,0:02:13.79,Default,,0000,0000,0000,,so the amplitude of v_1 is v sub m1,
Dialogue: 0,0:02:13.79,0:02:16.08,Default,,0000,0000,0000,,and it has a phase of Theta 1,
Dialogue: 0,0:02:16.08,0:02:23.82,Default,,0000,0000,0000,,then plus v sub m2 cosine\NOmega t, it's the same Omega.
Dialogue: 0,0:02:23.82,0:02:29.10,Default,,0000,0000,0000,,Omega t plus Theta 2 plus finally
Dialogue: 0,0:02:29.10,0:02:36.84,Default,,0000,0000,0000,,v sub m3 cosine Omega t plus Theta sub 3,
Dialogue: 0,0:02:36.84,0:02:40.12,Default,,0000,0000,0000,,and the sum of those three\Nterms has to equal 0.
Dialogue: 0,0:02:40.12,0:02:44.45,Default,,0000,0000,0000,,Now, let's represent these in\Nterms of phasors, and by that,
Dialogue: 0,0:02:44.45,0:02:49.83,Default,,0000,0000,0000,,we mean then that the real part\Nof whether this v sub
Dialogue: 0,0:02:49.83,0:02:57.34,Default,,0000,0000,0000,,m cosine Omega t plus Theta 1 can be\Nthought of as the real part of v sub m1,
Dialogue: 0,0:02:57.34,0:03:00.14,Default,,0000,0000,0000,,e to the j Theta 1,
Dialogue: 0,0:03:00.14,0:03:03.12,Default,,0000,0000,0000,,e to the j Omega t,
Dialogue: 0,0:03:03.12,0:03:06.20,Default,,0000,0000,0000,,plus the second term here\Ncan be thought of as being
Dialogue: 0,0:03:06.20,0:03:09.79,Default,,0000,0000,0000,,the real part of v sub m2,
Dialogue: 0,0:03:09.79,0:03:13.20,Default,,0000,0000,0000,,e to the j Theta 2, that's a j,
Dialogue: 0,0:03:13.20,0:03:18.07,Default,,0000,0000,0000,,Theta 2, e to the j Omega t,
Dialogue: 0,0:03:20.06,0:03:25.70,Default,,0000,0000,0000,,plus the real part of v sub m3,
Dialogue: 0,0:03:25.70,0:03:28.83,Default,,0000,0000,0000,,e to the j Theta 3,
Dialogue: 0,0:03:28.83,0:03:34.77,Default,,0000,0000,0000,,e to the j Omega t, must equal 0.
Dialogue: 0,0:03:34.77,0:03:38.06,Default,,0000,0000,0000,,Now, realizing that we're talking\Nabout the real part of all of
Dialogue: 0,0:03:38.06,0:03:41.77,Default,,0000,0000,0000,,these and that this e to the j Omega t\Nis common to all of them,
Dialogue: 0,0:03:41.77,0:03:46.44,Default,,0000,0000,0000,,we can rewrite this then\Nas the real part of,
Dialogue: 0,0:03:46.44,0:03:52.79,Default,,0000,0000,0000,,and also pointing out that that term\Nright there is just phasor v_1,
Dialogue: 0,0:03:52.79,0:03:57.38,Default,,0000,0000,0000,,and this right here is phasor v_2,
Dialogue: 0,0:03:57.38,0:04:03.33,Default,,0000,0000,0000,,and then of course right\Nthere is phasor v_3.
Dialogue: 0,0:04:03.33,0:04:06.85,Default,,0000,0000,0000,,We can now rewrite this as the real part of
Dialogue: 0,0:04:06.85,0:04:16.29,Default,,0000,0000,0000,,phasor v_1 plus phasor v_2 plus phasor v_3,
Dialogue: 0,0:04:16.29,0:04:20.38,Default,,0000,0000,0000,,e to the j Omega t,
Dialogue: 0,0:04:21.62,0:04:24.27,Default,,0000,0000,0000,,and that must equal 0.
Dialogue: 0,0:04:24.27,0:04:26.89,Default,,0000,0000,0000,,Well, we know that e to\Nthe j Omega t doesn't equal 0,
Dialogue: 0,0:04:26.89,0:04:29.11,Default,,0000,0000,0000,,therefore the sum of the phasors v_1,
Dialogue: 0,0:04:29.11,0:04:31.64,Default,,0000,0000,0000,,v_2, and v_3, must equal 0.
Dialogue: 0,0:04:31.64,0:04:36.40,Default,,0000,0000,0000,,Or specifically, phasor v_1 plus
Dialogue: 0,0:04:36.40,0:04:43.20,Default,,0000,0000,0000,,phasor v_2 plus phasor v_3 must equal 0.
Dialogue: 0,0:04:43.20,0:04:44.92,Default,,0000,0000,0000,,This is incredibly important.
Dialogue: 0,0:04:44.92,0:04:46.82,Default,,0000,0000,0000,,What that's saying is that up here,
Dialogue: 0,0:04:46.82,0:04:48.57,Default,,0000,0000,0000,,we have this circuit\Noperating in the time domain,
Dialogue: 0,0:04:48.57,0:04:51.80,Default,,0000,0000,0000,,we know that the sum of\Nthose three voltages has to equal zero.
Dialogue: 0,0:04:51.80,0:04:55.16,Default,,0000,0000,0000,,What this demonstrates is\Nthat not only must the sum
Dialogue: 0,0:04:55.16,0:04:58.24,Default,,0000,0000,0000,,of the time signals around\Nthat closed loop equals zero,
Dialogue: 0,0:04:58.24,0:05:01.55,Default,,0000,0000,0000,,but the sum of the phasor\Nrepresentations of
Dialogue: 0,0:05:01.55,0:05:05.50,Default,,0000,0000,0000,,each of these voltages\Nmust also equal zero.
Dialogue: 0,0:05:05.50,0:05:08.04,Default,,0000,0000,0000,,In other words, and this\Nis the final thing,
Dialogue: 0,0:05:08.04,0:05:12.78,Default,,0000,0000,0000,,Kirchhoff's Voltage Laws apply or Law,
Dialogue: 0,0:05:12.78,0:05:15.62,Default,,0000,0000,0000,,singular, the Kirchhoff's\NVoltage Law applies in
Dialogue: 0,0:05:15.62,0:05:21.47,Default,,0000,0000,0000,,this phasor domain in exactly the same way\Nthat it does in the time domain.
Dialogue: 0,0:05:21.47,0:05:24.74,Default,,0000,0000,0000,,Now, let's take a look at\NKirchhoff's Current Law.
Dialogue: 0,0:05:24.74,0:05:27.73,Default,,0000,0000,0000,,We have a similar set of circumstances here
Dialogue: 0,0:05:27.73,0:05:31.36,Default,,0000,0000,0000,,where the sum of the currents i_1 of t
Dialogue: 0,0:05:31.36,0:05:37.69,Default,,0000,0000,0000,,plus i_2 of t plus i_3 of t, must equal 0.
Dialogue: 0,0:05:37.69,0:05:39.59,Default,,0000,0000,0000,,I'm not going to take\Nthe time right now to do it,
Dialogue: 0,0:05:39.59,0:05:41.83,Default,,0000,0000,0000,,suffice it to say that\Nthe same type of analysis
Dialogue: 0,0:05:41.83,0:05:44.35,Default,,0000,0000,0000,,that we just did for\NKirchhoff's Voltage Law,
Dialogue: 0,0:05:44.35,0:05:47.23,Default,,0000,0000,0000,,results in the sum of
Dialogue: 0,0:05:47.23,0:05:52.12,Default,,0000,0000,0000,,the phasor currents associated with
Dialogue: 0,0:05:52.12,0:05:57.81,Default,,0000,0000,0000,,the node must also equal zero.
Dialogue: 0,0:05:57.81,0:06:01.48,Default,,0000,0000,0000,,So once again, in terms of\Nimpedances and phasors,
Dialogue: 0,0:06:01.48,0:06:05.12,Default,,0000,0000,0000,,the sum of the phasors or the sum of
Dialogue: 0,0:06:05.12,0:06:10.43,Default,,0000,0000,0000,,the currents represented in\Nthe phasor form must equal zero also.