[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.92,0:00:05.60,Default,,0000,0000,0000,,>> In this video, we're going to\Ndemonstrate how phasor diagrams
Dialogue: 0,0:00:05.60,0:00:09.36,Default,,0000,0000,0000,,or showing the phasors\Nin the complex plane can
Dialogue: 0,0:00:09.36,0:00:12.76,Default,,0000,0000,0000,,demonstrate the relative phases
Dialogue: 0,0:00:12.76,0:00:17.37,Default,,0000,0000,0000,,and magnitudes of voltages and\Ncurrents within a circuit.
Dialogue: 0,0:00:17.37,0:00:20.65,Default,,0000,0000,0000,,To do this, we're going\Nto remind ourselves that
Dialogue: 0,0:00:20.65,0:00:27.12,Default,,0000,0000,0000,,phasor V is equal to the\Nphasor I times the impedance.
Dialogue: 0,0:00:27.12,0:00:29.36,Default,,0000,0000,0000,,Now for example, in a resistor,
Dialogue: 0,0:00:29.36,0:00:33.60,Default,,0000,0000,0000,,the impedance is R and V then is simply
Dialogue: 0,0:00:33.60,0:00:37.98,Default,,0000,0000,0000,,equal to I times R. R is a real number.
Dialogue: 0,0:00:37.98,0:00:41.98,Default,,0000,0000,0000,,So there is no phase term associated\Nwith the impedance of a resistor,
Dialogue: 0,0:00:41.98,0:00:45.68,Default,,0000,0000,0000,,and we say that the voltage is in
Dialogue: 0,0:00:45.68,0:00:55.84,Default,,0000,0000,0000,,phase with the current in a resistor.
Dialogue: 0,0:00:55.84,0:01:05.22,Default,,0000,0000,0000,,So if we have just arbitrarily choose\Nthe current to have a zero phase angle,
Dialogue: 0,0:01:06.86,0:01:10.52,Default,,0000,0000,0000,,then the phasor voltage
Dialogue: 0,0:01:10.52,0:01:12.44,Default,,0000,0000,0000,,associated with the resistor or the voltage
Dialogue: 0,0:01:12.44,0:01:14.88,Default,,0000,0000,0000,,across the resistor call it V sub r,
Dialogue: 0,0:01:14.88,0:01:18.07,Default,,0000,0000,0000,,will have the same angle as I,
Dialogue: 0,0:01:18.07,0:01:19.79,Default,,0000,0000,0000,,be in the same direction,
Dialogue: 0,0:01:19.79,0:01:24.88,Default,,0000,0000,0000,,but just have a different magnitude than I.
Dialogue: 0,0:01:24.88,0:01:30.99,Default,,0000,0000,0000,,In fact the magnitude of this\Nwill be R times I. All right.
Dialogue: 0,0:01:30.99,0:01:34.42,Default,,0000,0000,0000,,Unlike in a resistor, a capacitor\Nintroduces a phase difference,
Dialogue: 0,0:01:34.42,0:01:40.79,Default,,0000,0000,0000,,we know that V across the\Ncapacitor is equal to I times Z.
Dialogue: 0,0:01:40.79,0:01:44.62,Default,,0000,0000,0000,,Well, the impedance of\Na capacitor is one over
Dialogue: 0,0:01:44.62,0:01:49.40,Default,,0000,0000,0000,,j Omega c. We bring that\Nj up in the numerator
Dialogue: 0,0:01:49.40,0:01:52.12,Default,,0000,0000,0000,,with a minus sign and we'll\Nhave then this is equal to
Dialogue: 0,0:01:52.12,0:01:58.32,Default,,0000,0000,0000,,I times a negative j over Omega c. Well,
Dialogue: 0,0:01:58.32,0:02:06.23,Default,,0000,0000,0000,,that negative j in rectangular\Ncoordinates is the same thing as
Dialogue: 0,0:02:06.23,0:02:10.43,Default,,0000,0000,0000,,a minus 90 degree phase term\Nin polar coordinates
Dialogue: 0,0:02:10.43,0:02:14.57,Default,,0000,0000,0000,,or we have then that V is equal
Dialogue: 0,0:02:14.57,0:02:24.66,Default,,0000,0000,0000,,to I times one over Omega c\Ntimes e to the minus j90.
Dialogue: 0,0:02:24.97,0:02:28.96,Default,,0000,0000,0000,,So whatever the phase of the current is,
Dialogue: 0,0:02:28.96,0:02:31.46,Default,,0000,0000,0000,,when we multiply the current\Ntimes the impedance there is
Dialogue: 0,0:02:31.46,0:02:34.79,Default,,0000,0000,0000,,a negative 90 degree phase term\Nthat gets added to the current.
Dialogue: 0,0:02:34.79,0:02:41.75,Default,,0000,0000,0000,,So once again if we let the current\Nhave a zero reference angle,
Dialogue: 0,0:02:41.75,0:02:47.48,Default,,0000,0000,0000,,then the voltage in a\Ncapacitor is going to be
Dialogue: 0,0:02:47.48,0:02:51.62,Default,,0000,0000,0000,,90 degrees less than
Dialogue: 0,0:02:51.62,0:02:54.32,Default,,0000,0000,0000,,the phase of the current or we
Dialogue: 0,0:02:54.32,0:02:59.30,Default,,0000,0000,0000,,say the voltage is 90 degrees\Nbehind the current,
Dialogue: 0,0:02:59.30,0:03:03.98,Default,,0000,0000,0000,,or we can say the current is\N90 degrees ahead of the voltage.
Dialogue: 0,0:03:03.98,0:03:07.62,Default,,0000,0000,0000,,Finally, we look at the\Nvoltage across the inductor,
Dialogue: 0,0:03:07.62,0:03:11.00,Default,,0000,0000,0000,,and we have phasor V. So that one,
Dialogue: 0,0:03:11.00,0:03:13.28,Default,,0000,0000,0000,,of course these are all\Nphasor terms over here.
Dialogue: 0,0:03:13.28,0:03:19.25,Default,,0000,0000,0000,,Phasor V sub l is equal to I times Z.
Dialogue: 0,0:03:19.25,0:03:24.35,Default,,0000,0000,0000,,Well the Z the impedance of\Nan inductor is j Omega L,
Dialogue: 0,0:03:24.35,0:03:31.62,Default,,0000,0000,0000,,which is equal to I times Omega L times j.
Dialogue: 0,0:03:31.62,0:03:38.97,Default,,0000,0000,0000,,Well, j is the same thing\Nas e to the positive j90.
Dialogue: 0,0:03:39.16,0:03:41.30,Default,,0000,0000,0000,,So in an inductor
Dialogue: 0,0:03:41.30,0:03:45.43,Default,,0000,0000,0000,,the voltage will be 90 degrees\Nahead of the phase of the current.
Dialogue: 0,0:03:45.43,0:03:46.70,Default,,0000,0000,0000,,Whatever the phase of the current
Dialogue: 0,0:03:46.70,0:03:49.91,Default,,0000,0000,0000,,is and it gets multiplied by\Nthe impedance of the inductor,
Dialogue: 0,0:03:49.91,0:03:52.51,Default,,0000,0000,0000,,which has this positive\N90 phase term with it,
Dialogue: 0,0:03:52.51,0:03:54.74,Default,,0000,0000,0000,,that positive 90 gets added to
Dialogue: 0,0:03:54.74,0:03:57.02,Default,,0000,0000,0000,,the phase of the current\Nand the voltage will be
Dialogue: 0,0:03:57.02,0:04:01.79,Default,,0000,0000,0000,,90 degrees ahead of the current.
Dialogue: 0,0:04:01.79,0:04:06.32,Default,,0000,0000,0000,,So over here we had that being\NI once again we'll reference I,
Dialogue: 0,0:04:06.32,0:04:11.60,Default,,0000,0000,0000,,just arbitrarily choosing\Nit to have a zero phase.
Dialogue: 0,0:04:11.60,0:04:18.71,Default,,0000,0000,0000,,I then V sub l will
Dialogue: 0,0:04:18.71,0:04:25.22,Default,,0000,0000,0000,,be 90 degrees ahead of phasor I.
Dialogue: 0,0:04:25.22,0:04:28.74,Default,,0000,0000,0000,,So we say that the current lags the voltage
Dialogue: 0,0:04:28.74,0:04:33.59,Default,,0000,0000,0000,,in an inductor or the voltage\Nleads the current in an inductor.
Dialogue: 0,0:04:33.59,0:04:37.61,Default,,0000,0000,0000,,Notice the voltages of the capacitor and
Dialogue: 0,0:04:37.61,0:04:42.08,Default,,0000,0000,0000,,the inductor are 180 degrees\Nout of phase with each other.
Dialogue: 0,0:04:42.08,0:04:44.92,Default,,0000,0000,0000,,The voltage and the current in the voltage
Dialogue: 0,0:04:44.92,0:04:48.17,Default,,0000,0000,0000,,of the capacitor lags by 90 degrees,
Dialogue: 0,0:04:48.17,0:04:52.64,Default,,0000,0000,0000,,the voltage in the inductor\Nleads the current by 90 degrees.
Dialogue: 0,0:04:52.64,0:04:55.41,Default,,0000,0000,0000,,There's a mnemonic that\Ngoes along with this,
Dialogue: 0,0:04:55.41,0:04:59.87,Default,,0000,0000,0000,,today's clear back to the late 1800s\Nwhen much of electrical work that was
Dialogue: 0,0:04:59.87,0:05:05.05,Default,,0000,0000,0000,,being done was associated with\Nrefrigeration and creating ice,
Dialogue: 0,0:05:05.05,0:05:12.73,Default,,0000,0000,0000,,and the mnemonic is ELI, the ICE man.
Dialogue: 0,0:05:12.97,0:05:16.90,Default,,0000,0000,0000,,What this is suggesting is that E
Dialogue: 0,0:05:16.90,0:05:20.62,Default,,0000,0000,0000,,represents voltage as in\Nphysics we talk about the EMF.
Dialogue: 0,0:05:20.62,0:05:22.10,Default,,0000,0000,0000,,So E is voltage,
Dialogue: 0,0:05:22.10,0:05:25.02,Default,,0000,0000,0000,,I is current, L is an inductor.
Dialogue: 0,0:05:25.02,0:05:27.14,Default,,0000,0000,0000,,So in an inductor,
Dialogue: 0,0:05:27.14,0:05:33.90,Default,,0000,0000,0000,,the voltage leads\Nthe current by 90 degrees.
Dialogue: 0,0:05:33.90,0:05:40.01,Default,,0000,0000,0000,,In an inductor, the voltage\Nleads the current by 90 degrees.
Dialogue: 0,0:05:40.01,0:05:47.72,Default,,0000,0000,0000,,In a capacitor, the current\Nleads the voltage by 90 degrees.
Dialogue: 0,0:05:47.72,0:05:49.76,Default,,0000,0000,0000,,So in the capacitor c,
Dialogue: 0,0:05:49.76,0:05:51.75,Default,,0000,0000,0000,,the current is 90 degrees ahead.
Dialogue: 0,0:05:51.75,0:05:53.55,Default,,0000,0000,0000,,So as in the word ICE,
Dialogue: 0,0:05:53.55,0:05:59.37,Default,,0000,0000,0000,,I comes ahead of E. So I is\Nleading the voltage over here.
Dialogue: 0,0:05:59.37,0:06:03.84,Default,,0000,0000,0000,,In ELI, the voltage is leading the current.
Dialogue: 0,0:06:05.86,0:06:09.86,Default,,0000,0000,0000,,Now let's demonstrate how\Nthese phasor diagrams can be
Dialogue: 0,0:06:09.86,0:06:13.10,Default,,0000,0000,0000,,used to demonstrate the various or
Dialogue: 0,0:06:13.10,0:06:19.85,Default,,0000,0000,0000,,the relative phasors of different voltages\Nand currents within a circuit.
Dialogue: 0,0:06:19.85,0:06:24.77,Default,,0000,0000,0000,,Here we have a single loop circuit\Nand writing one KVL,
Dialogue: 0,0:06:24.77,0:06:32.13,Default,,0000,0000,0000,,we know then that phasor\NV sub s is equal to.
Dialogue: 0,0:06:33.47,0:06:39.65,Default,,0000,0000,0000,,Well, first of all V sub s sum of\Nthe voltage is equal to V sub r,
Dialogue: 0,0:06:39.65,0:06:41.48,Default,,0000,0000,0000,,plus V sub c,
Dialogue: 0,0:06:41.48,0:06:48.83,Default,,0000,0000,0000,,plus V sub l and of course\Nthese are all phasor voltages.
Dialogue: 0,0:06:48.83,0:06:53.15,Default,,0000,0000,0000,,Now let's go ahead and put\Nin that V sub r is equal
Dialogue: 0,0:06:53.15,0:06:58.30,Default,,0000,0000,0000,,to I times R. In this case,
Dialogue: 0,0:06:58.30,0:07:00.90,Default,,0000,0000,0000,,R is eight ohms
Dialogue: 0,0:07:02.03,0:07:07.34,Default,,0000,0000,0000,,plus the voltage across the capacitor\Nis just going to be I times Z sub c,
Dialogue: 0,0:07:07.34,0:07:14.16,Default,,0000,0000,0000,,which is I times a negative j8 plus\Nthe voltage across the inductor.
Dialogue: 0,0:07:14.16,0:07:16.07,Default,,0000,0000,0000,,That's a J2 ohm inductor.
Dialogue: 0,0:07:16.07,0:07:22.18,Default,,0000,0000,0000,,So the voltage across\Nthat will be I times J2.
Dialogue: 0,0:07:22.18,0:07:25.01,Default,,0000,0000,0000,,So we have three terms here.
Dialogue: 0,0:07:25.01,0:07:27.80,Default,,0000,0000,0000,,Three different phasors.
Dialogue: 0,0:07:27.80,0:07:30.83,Default,,0000,0000,0000,,These are phasor voltage is equal to 8I,
Dialogue: 0,0:07:30.83,0:07:35.58,Default,,0000,0000,0000,,minus J8I, and positive J2I.
Dialogue: 0,0:07:36.67,0:07:40.18,Default,,0000,0000,0000,,Showing this explicitly with the phases,
Dialogue: 0,0:07:40.18,0:07:42.86,Default,,0000,0000,0000,,this will be I times eight.
Dialogue: 0,0:07:42.86,0:07:45.14,Default,,0000,0000,0000,,There is no phase difference\Nbetween the voltage
Dialogue: 0,0:07:45.14,0:07:48.26,Default,,0000,0000,0000,,across the resistor and the current.
Dialogue: 0,0:07:48.26,0:07:53.60,Default,,0000,0000,0000,,On the other hand we have I times eight
Dialogue: 0,0:07:53.60,0:07:58.77,Default,,0000,0000,0000,,times e to the minus j90 in the capacitor,
Dialogue: 0,0:07:58.77,0:08:09.73,Default,,0000,0000,0000,,plus I times two times e to\Nthe positive j90 for the inductor.
Dialogue: 0,0:08:09.73,0:08:12.56,Default,,0000,0000,0000,,So again, what this is saying is that
Dialogue: 0,0:08:12.56,0:08:20.54,Default,,0000,0000,0000,,the phase of the voltage
Dialogue: 0,0:08:20.54,0:08:25.64,Default,,0000,0000,0000,,across the inductor is 90 degrees\Ngreater than the phase of the current.
Dialogue: 0,0:08:25.64,0:08:30.59,Default,,0000,0000,0000,,On the other hand, the phase\Nof the voltage across
Dialogue: 0,0:08:30.59,0:08:36.05,Default,,0000,0000,0000,,the capacitor is 90 degrees\Nbehind the phase of the current.
Dialogue: 0,0:08:36.05,0:08:42.11,Default,,0000,0000,0000,,So let's just start out and\Nsay that this is the current.
Dialogue: 0,0:08:42.11,0:08:45.62,Default,,0000,0000,0000,,I will say that it has\Na magnitude I naught.
Dialogue: 0,0:08:45.62,0:08:51.57,Default,,0000,0000,0000,,The voltage across the resistor\Nis eight times that
Dialogue: 0,0:08:51.57,0:08:58.67,Default,,0000,0000,0000,,or V sub r is equal to eight\Ntimes I naught in length,
Dialogue: 0,0:08:58.67,0:09:03.11,Default,,0000,0000,0000,,and it has the same angle as I.
Dialogue: 0,0:09:03.11,0:09:09.65,Default,,0000,0000,0000,,The capacitor on the other hand\Nis 90 degrees behind the current.
Dialogue: 0,0:09:09.65,0:09:14.06,Default,,0000,0000,0000,,So the voltage V sub c is 90 degrees
Dialogue: 0,0:09:14.06,0:09:20.78,Default,,0000,0000,0000,,behind the current V sub c,
Dialogue: 0,0:09:20.78,0:09:29.57,Default,,0000,0000,0000,,and it's down here at\Nnegative 8I naught, and the voltage.
Dialogue: 0,0:09:29.57,0:09:31.84,Default,,0000,0000,0000,,Let's see that's the\Nvoltage of the capacitor,
Dialogue: 0,0:09:31.84,0:09:40.56,Default,,0000,0000,0000,,and the voltage of the inductor\Nis 90 degrees ahead of V sub l,
Dialogue: 0,0:09:42.08,0:09:48.23,Default,,0000,0000,0000,,the length of it is equal to two times
Dialogue: 0,0:09:48.23,0:09:53.93,Default,,0000,0000,0000,,I naught and it is at an angle of\N90 degrees ahead of the current.
Dialogue: 0,0:09:53.93,0:09:56.04,Default,,0000,0000,0000,,So our three voltages.
Dialogue: 0,0:09:56.04,0:09:59.64,Default,,0000,0000,0000,,The voltage across the resistor\Nis in phase with the current,
Dialogue: 0,0:09:59.64,0:10:02.99,Default,,0000,0000,0000,,capacitor is 90 degrees\Nbehind the current and
Dialogue: 0,0:10:02.99,0:10:06.38,Default,,0000,0000,0000,,the inductor voltage is 90 degrees ahead.
Dialogue: 0,0:10:06.38,0:10:08.78,Default,,0000,0000,0000,,Now we have here that V sub s,
Dialogue: 0,0:10:08.78,0:10:12.98,Default,,0000,0000,0000,,the source voltage is equal to\Nthe sum of those three terms.
Dialogue: 0,0:10:12.98,0:10:14.87,Default,,0000,0000,0000,,So when adding phasors,
Dialogue: 0,0:10:14.87,0:10:17.66,Default,,0000,0000,0000,,you do it just like you do vectors\Nand that is tip to tell them.
Dialogue: 0,0:10:17.66,0:10:22.28,Default,,0000,0000,0000,,Here's V sub r, add in V sub c,
Dialogue: 0,0:10:22.28,0:10:29.59,Default,,0000,0000,0000,,that comes down here like that\Nand then add in V sub l. That
Dialogue: 0,0:10:29.59,0:10:37.91,Default,,0000,0000,0000,,brings us back two units back that way\Nto this point here at negative 6I0,
Dialogue: 0,0:10:38.08,0:10:43.56,Default,,0000,0000,0000,,and so we end up at that point\Nand that is V sub s then.
Dialogue: 0,0:10:43.56,0:10:46.36,Default,,0000,0000,0000,,Let see how good of a job I\Ncan do drawing a vector down,
Dialogue: 0,0:10:46.36,0:10:48.16,Default,,0000,0000,0000,,they're not very good.
Dialogue: 0,0:10:48.16,0:10:51.20,Default,,0000,0000,0000,,That's a straight line vector.
Dialogue: 0,0:10:51.87,0:10:55.16,Default,,0000,0000,0000,,That's better I guess.
Dialogue: 0,0:10:55.16,0:11:01.90,Default,,0000,0000,0000,,So V sub s then is equal to V sub r,
Dialogue: 0,0:11:01.90,0:11:03.48,Default,,0000,0000,0000,,plus V sub c, plus
Dialogue: 0,0:11:03.48,0:11:09.80,Default,,0000,0000,0000,,V sub l. This results in phasor\Nis V sub s. What does it tell us?
Dialogue: 0,0:11:09.80,0:11:13.91,Default,,0000,0000,0000,,Well, it tells us that V sub s is something
Dialogue: 0,0:11:13.91,0:11:18.92,Default,,0000,0000,0000,,less than 90 degrees\Nbehind the current or that
Dialogue: 0,0:11:18.92,0:11:22.31,Default,,0000,0000,0000,,the current in this circuit I is leading
Dialogue: 0,0:11:22.31,0:11:28.79,Default,,0000,0000,0000,,the voltage source by some angle\Nless than 90 degrees.
Dialogue: 0,0:11:28.79,0:11:37.05,Default,,0000,0000,0000,,It also tells us the relative length of\Nthis voltage is longer than V sub r,
Dialogue: 0,0:11:37.06,0:11:47.23,Default,,0000,0000,0000,,it's shorter than V sub c because\NV sub c and V sub l are opposites.
Dialogue: 0,0:11:49.61,0:11:54.57,Default,,0000,0000,0000,,I guess to be more accurate if\Nyou to say that V sub c and
Dialogue: 0,0:11:54.57,0:11:59.46,Default,,0000,0000,0000,,V sub l are in opposite directions\Nand they tend to cancel each other.
Dialogue: 0,0:11:59.46,0:12:01.91,Default,,0000,0000,0000,,Let me say that one more\Ntime just a little bit
Dialogue: 0,0:12:01.91,0:12:04.30,Default,,0000,0000,0000,,differently and see if I can\Nmake it a little more clear.
Dialogue: 0,0:12:04.30,0:12:07.79,Default,,0000,0000,0000,,V sub s the source voltage is equal to
Dialogue: 0,0:12:07.79,0:12:11.40,Default,,0000,0000,0000,,the phasor sum of V sub l, plus V sub r,
Dialogue: 0,0:12:11.40,0:12:16.50,Default,,0000,0000,0000,,plus V sub c. By diagramming\Nor drawing in V sub l,
Dialogue: 0,0:12:16.50,0:12:19.66,Default,,0000,0000,0000,,V sub r and V sub c and then\Nadding them tip to tail,
Dialogue: 0,0:12:19.66,0:12:23.60,Default,,0000,0000,0000,,we can see the relative length and phase
Dialogue: 0,0:12:23.60,0:12:27.74,Default,,0000,0000,0000,,of the source to the voltage\Nacross the resistor,
Dialogue: 0,0:12:27.74,0:12:29.70,Default,,0000,0000,0000,,which is in phase with the current.
Dialogue: 0,0:12:29.70,0:12:33.34,Default,,0000,0000,0000,,The relative length and
Dialogue: 0,0:12:33.34,0:12:36.98,Default,,0000,0000,0000,,the relative angle of the source\Nrelative to the voltage across
Dialogue: 0,0:12:36.98,0:12:40.37,Default,,0000,0000,0000,,the inductor and the relative length and
Dialogue: 0,0:12:40.37,0:12:46.72,Default,,0000,0000,0000,,relative phase of the source relative\Nto the voltage across the capacitor.
Dialogue: 0,0:12:46.72,0:12:48.98,Default,,0000,0000,0000,,Again, this doesn't give\Nus a real accurate,
Dialogue: 0,0:12:48.98,0:12:52.64,Default,,0000,0000,0000,,but it does give us some feel\Nfor the relative phases,
Dialogue: 0,0:12:52.64,0:12:55.52,Default,,0000,0000,0000,,the phases of one of each of
Dialogue: 0,0:12:55.52,0:12:59.88,Default,,0000,0000,0000,,these voltages relative to the current\Nand relative to the source.