[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.02,0:00:04.09,Default,,0000,0000,0000,,>> We'll do that same example now only with
Dialogue: 0,0:00:04.09,0:00:08.37,Default,,0000,0000,0000,,the actual values for the circuit\Nelements and the source here.
Dialogue: 0,0:00:08.37,0:00:10.56,Default,,0000,0000,0000,,So again, the first thing we do is convert
Dialogue: 0,0:00:10.56,0:00:13.80,Default,,0000,0000,0000,,the time-domain source to\Nits phasor representation
Dialogue: 0,0:00:13.80,0:00:19.02,Default,,0000,0000,0000,,and then calculate the impedances\Nassociated with the capacitor,
Dialogue: 0,0:00:19.02,0:00:20.62,Default,,0000,0000,0000,,resistance and inductor here.
Dialogue: 0,0:00:20.62,0:00:24.80,Default,,0000,0000,0000,,So this the phasor\Nrepresentation V we'll call it
Dialogue: 0,0:00:24.80,0:00:31.11,Default,,0000,0000,0000,,cap V or phasor V is\Nequal to 10E to the J30.
Dialogue: 0,0:00:31.36,0:00:33.92,Default,,0000,0000,0000,,Now, let's be explicit here.
Dialogue: 0,0:00:33.92,0:00:37.71,Default,,0000,0000,0000,,Omega is 1,000 radians per second.
Dialogue: 0,0:00:38.20,0:00:41.34,Default,,0000,0000,0000,,Radians per second.
Dialogue: 0,0:00:41.34,0:00:46.79,Default,,0000,0000,0000,,Z_C is equal to one over J Omega C\Nwhich is equal to one over
Dialogue: 0,0:00:46.79,0:00:49.19,Default,,0000,0000,0000,,J times Omega which is
Dialogue: 0,0:00:49.19,0:00:53.92,Default,,0000,0000,0000,,1,000 times C which is 10 times\N10 to the minus six to six micro,
Dialogue: 0,0:00:53.92,0:01:02.55,Default,,0000,0000,0000,,10 microfarads and that then\Nequals negative J 100 Ohms.
Dialogue: 0,0:01:02.55,0:01:06.94,Default,,0000,0000,0000,,That's a negative J 100 Ohms.
Dialogue: 0,0:01:06.94,0:01:10.67,Default,,0000,0000,0000,,Z_R is just a value of\Nthe resistor which is
Dialogue: 0,0:01:10.67,0:01:16.61,Default,,0000,0000,0000,,50 Ohms and Z_L is equal to J\NOmega L which is equal to J times
Dialogue: 0,0:01:16.61,0:01:19.67,Default,,0000,0000,0000,,Omega which is 1,000 times L which is
Dialogue: 0,0:01:19.67,0:01:27.08,Default,,0000,0000,0000,,75 millihenries and that\Ngives us a positive J 75.
Dialogue: 0,0:01:27.08,0:01:30.52,Default,,0000,0000,0000,,So this is a J 75 there.
Dialogue: 0,0:01:30.52,0:01:36.46,Default,,0000,0000,0000,,This is R which is equal to the impedance\Nthat is equal to 50 ohms and this,
Dialogue: 0,0:01:36.46,0:01:42.40,Default,,0000,0000,0000,,the impedance of the capacitor\Nis a negative J 100.
Dialogue: 0,0:01:42.40,0:01:45.35,Default,,0000,0000,0000,,We'll go ahead and define the current
Dialogue: 0,0:01:45.35,0:01:47.87,Default,,0000,0000,0000,,flowing through here or the phasor\Nrepresentation of the current going
Dialogue: 0,0:01:47.87,0:01:51.32,Default,,0000,0000,0000,,through there call it phasor I and let's
Dialogue: 0,0:01:51.32,0:01:56.46,Default,,0000,0000,0000,,write a Kirchhoff's voltage law equation\Ngoing around that loop.
Dialogue: 0,0:01:56.46,0:01:59.70,Default,,0000,0000,0000,,Once again, adding up the voltage drops.
Dialogue: 0,0:01:59.70,0:02:02.15,Default,,0000,0000,0000,,So, a voltage increase will be a negative,
Dialogue: 0,0:02:02.15,0:02:10.25,Default,,0000,0000,0000,,that would be a negative 10 E to the J\N30 plus going across this capacitor,
Dialogue: 0,0:02:10.25,0:02:12.29,Default,,0000,0000,0000,,the voltage drop across that capacitor will
Dialogue: 0,0:02:12.29,0:02:14.78,Default,,0000,0000,0000,,be the current I times the impedance.
Dialogue: 0,0:02:14.78,0:02:17.66,Default,,0000,0000,0000,,So, that will be I times\Nthe impedance which is
Dialogue: 0,0:02:17.66,0:02:24.05,Default,,0000,0000,0000,,a negative J 100 plus the voltage drop\Nacross the resistor which is going
Dialogue: 0,0:02:24.05,0:02:30.58,Default,,0000,0000,0000,,to be I times 50 plus the voltage drop\Nacross that inductor is going to be
Dialogue: 0,0:02:30.58,0:02:33.32,Default,,0000,0000,0000,,I times the impedance of\Nthat inductor which is
Dialogue: 0,0:02:33.32,0:02:38.84,Default,,0000,0000,0000,,J 75 and the sum of those\Nterms then equals zero.
Dialogue: 0,0:02:38.84,0:02:42.08,Default,,0000,0000,0000,,We want to solve for I\Nthe current in this circuit.
Dialogue: 0,0:02:42.08,0:02:45.45,Default,,0000,0000,0000,,So we'll factor out the I\Nfrom these common terms
Dialogue: 0,0:02:45.45,0:02:50.30,Default,,0000,0000,0000,,here and bring this term here on\Nover other side and we get then
Dialogue: 0,0:02:50.30,0:02:57.14,Default,,0000,0000,0000,,that I times a negative J 100 plus 50
Dialogue: 0,0:02:57.14,0:03:06.87,Default,,0000,0000,0000,,plus J 75 is equal to\Na positive 10 E to the J 30.
Dialogue: 0,0:03:06.87,0:03:12.84,Default,,0000,0000,0000,,We have a negative J 100 plus J 75\Nthat gives us a negative J 25 plus 50.
Dialogue: 0,0:03:12.84,0:03:17.57,Default,,0000,0000,0000,,Dividing both sides by that term\Nthen gives us that I is equal
Dialogue: 0,0:03:17.57,0:03:23.83,Default,,0000,0000,0000,,to 10 E to the J 30 divided by,
Dialogue: 0,0:03:23.83,0:03:30.06,Default,,0000,0000,0000,,writing the real part first 50 minus J 25.
Dialogue: 0,0:03:30.06,0:03:32.27,Default,,0000,0000,0000,,You get out your calculator.
Dialogue: 0,0:03:32.27,0:03:37.85,Default,,0000,0000,0000,,First of all, let's convert\Njust this denominator here to polar form.
Dialogue: 0,0:03:37.85,0:03:40.72,Default,,0000,0000,0000,,It's a little bit easier to see\Nwhat's going on here if we do so.
Dialogue: 0,0:03:40.72,0:03:48.23,Default,,0000,0000,0000,,In polar form, that's equal to 55.9 E
Dialogue: 0,0:03:48.23,0:04:00.65,Default,,0000,0000,0000,,to negative J 26.57.
Dialogue: 0,0:04:00.65,0:04:06.17,Default,,0000,0000,0000,,Now, we can see there the division,
Dialogue: 0,0:04:06.17,0:04:13.55,Default,,0000,0000,0000,,it's 10 E to the J 30 divided\Nby 55.9 E to the minus J 26.57.
Dialogue: 0,0:04:13.55,0:04:18.40,Default,,0000,0000,0000,,So 10 divided by 55.9\Ngives us the magnitude.
Dialogue: 0,0:04:18.40,0:04:20.63,Default,,0000,0000,0000,,The magnitude of the numerator\Ndivided by the magnitude of
Dialogue: 0,0:04:20.63,0:04:25.82,Default,,0000,0000,0000,,the denominator gives us 0.179.
Dialogue: 0,0:04:25.82,0:04:28.14,Default,,0000,0000,0000,,Now, the phase E,
Dialogue: 0,0:04:28.14,0:04:31.24,Default,,0000,0000,0000,,it will be E to the J 30 in the numerator,
Dialogue: 0,0:04:31.24,0:04:35.52,Default,,0000,0000,0000,,E to the minus J 26.57 in the denominator.
Dialogue: 0,0:04:35.52,0:04:41.15,Default,,0000,0000,0000,,We've already demonstrated that when\Ndividing numbers involving exponents,
Dialogue: 0,0:04:41.15,0:04:45.83,Default,,0000,0000,0000,,it's the exponent of the numerator\Nminus the exponent in the denominator.
Dialogue: 0,0:04:45.83,0:04:51.32,Default,,0000,0000,0000,,So it's J 30 minus and minus\Nit becomes plus a 26.57 that
Dialogue: 0,0:04:51.32,0:05:01.82,Default,,0000,0000,0000,,gives us a positive J 56.57 degrees.
Dialogue: 0,0:05:01.82,0:05:08.46,Default,,0000,0000,0000,,This entire thing here is\Nthe phasor representation of the current.
Dialogue: 0,0:05:08.46,0:05:11.15,Default,,0000,0000,0000,,We can now take this back\Ninto the time domain
Dialogue: 0,0:05:11.15,0:05:13.40,Default,,0000,0000,0000,,by recognizing that the\Namplitude of the current
Dialogue: 0,0:05:13.40,0:05:17.87,Default,,0000,0000,0000,,is going to be the magnitude of\Nthis phasor or let's be explicit now.
Dialogue: 0,0:05:17.87,0:05:24.31,Default,,0000,0000,0000,,I of T then is equal to amplitude of 0.179.
Dialogue: 0,0:05:24.67,0:05:27.38,Default,,0000,0000,0000,,It will be a cosine term.
Dialogue: 0,0:05:27.38,0:05:32.08,Default,,0000,0000,0000,,We came in as a cosine will come back\Nto the time domain as a cosine term.
Dialogue: 0,0:05:32.08,0:05:35.48,Default,,0000,0000,0000,,Cosine, the frequency\Ndoesn't change that's still
Dialogue: 0,0:05:35.48,0:05:42.34,Default,,0000,0000,0000,,1,000 T. But now the phase of the\Ncurrent is equal to 56.57 degrees.
Dialogue: 0,0:05:42.34,0:05:49.56,Default,,0000,0000,0000,,So plus 56.57 degrees.
Dialogue: 0,0:05:49.56,0:05:56.32,Default,,0000,0000,0000,,That then is the time domain\Nexpression for the current.
Dialogue: 0,0:05:56.32,0:05:59.76,Default,,0000,0000,0000,,The voltage, the source driving\Nthis circuit is oscillating at
Dialogue: 0,0:05:59.76,0:06:03.98,Default,,0000,0000,0000,,1,000 radians per second and the current\Nis oscillating 1,000 times per second.
Dialogue: 0,0:06:03.98,0:06:05.63,Default,,0000,0000,0000,,It's going back and forth changing at
Dialogue: 0,0:06:05.63,0:06:08.90,Default,,0000,0000,0000,,the same frequency that\Nthe source is going into it.
Dialogue: 0,0:06:08.90,0:06:11.58,Default,,0000,0000,0000,,The source has an amplitude of 10 volts.
Dialogue: 0,0:06:11.58,0:06:17.90,Default,,0000,0000,0000,,The current flowing has\Nan amplitude of 0.179 Amps.
Dialogue: 0,0:06:17.90,0:06:20.63,Default,,0000,0000,0000,,So, the amplitude of the current\Nis smaller than the amplitude of
Dialogue: 0,0:06:20.63,0:06:24.44,Default,,0000,0000,0000,,the voltage and finally\Nlet's look at the phase.
Dialogue: 0,0:06:24.44,0:06:30.17,Default,,0000,0000,0000,,The phase angle of the voltage source\Nis 30 degrees, positive 30 degrees.
Dialogue: 0,0:06:30.17,0:06:35.92,Default,,0000,0000,0000,,The angle of the current is 56.57 degrees.
Dialogue: 0,0:06:35.92,0:06:40.91,Default,,0000,0000,0000,,So, we'd say then that the current\Nhas been shifted ahead
Dialogue: 0,0:06:40.91,0:06:47.64,Default,,0000,0000,0000,,of the source by the 26.57 degrees.
Dialogue: 0,0:06:49.55,0:06:51.99,Default,,0000,0000,0000,,Here are plots of those,
Dialogue: 0,0:06:51.99,0:06:53.90,Default,,0000,0000,0000,,the voltage in the red,
Dialogue: 0,0:06:53.90,0:07:04.05,Default,,0000,0000,0000,,V of T is equal to 10 cosine\Nof 1,000 T plus 30 degrees.
Dialogue: 0,0:07:04.05,0:07:09.72,Default,,0000,0000,0000,,I of T we have now found\Nto be equal to 0.179,
Dialogue: 0,0:07:09.72,0:07:20.26,Default,,0000,0000,0000,,cosine of 1,000 T plus 56.57 degrees.
Dialogue: 0,0:07:20.26,0:07:22.64,Default,,0000,0000,0000,,So here's the voltage.\NIt's got an amplitude of
Dialogue: 0,0:07:22.64,0:07:28.67,Default,,0000,0000,0000,,10 volts and it has been shifted\Nto the left by 30 degrees.
Dialogue: 0,0:07:28.67,0:07:31.94,Default,,0000,0000,0000,,The current has an amplitude of
Dialogue: 0,0:07:31.94,0:07:39.08,Default,,0000,0000,0000,,0.179 and it has been shifted\Nto the left 56.57 degrees.
Dialogue: 0,0:07:39.08,0:07:42.74,Default,,0000,0000,0000,,So the difference, I can\Ndraw this very well.
Dialogue: 0,0:07:42.74,0:07:44.77,Default,,0000,0000,0000,,I don't know if I can or not.
Dialogue: 0,0:07:44.77,0:07:47.87,Default,,0000,0000,0000,,That distance right there is
Dialogue: 0,0:07:47.87,0:07:52.43,Default,,0000,0000,0000,,a phase shift of the current\Nahead of the voltage.
Dialogue: 0,0:07:52.43,0:07:55.43,Default,,0000,0000,0000,,We say that the current\Nis leading the voltage,
Dialogue: 0,0:07:55.43,0:07:59.14,Default,,0000,0000,0000,,the current peaks out\Nbefore the voltage peaks.
Dialogue: 0,0:07:59.14,0:08:02.69,Default,,0000,0000,0000,,The current crosses the zero before\Nthe voltage crosses the zero.
Dialogue: 0,0:08:02.69,0:08:06.71,Default,,0000,0000,0000,,The current is leading the voltage by that
Dialogue: 0,0:08:06.71,0:08:11.28,Default,,0000,0000,0000,,much which corresponds to\Nagain that 26.57 degrees.
Dialogue: 0,0:08:11.28,0:08:17.60,Default,,0000,0000,0000,,Just one note about what appears\Nto be an inconsistency of units.
Dialogue: 0,0:08:17.60,0:08:19.90,Default,,0000,0000,0000,,Not only does it appear to be\Nan inconsistency of units,
Dialogue: 0,0:08:19.90,0:08:21.56,Default,,0000,0000,0000,,it isn't inconsistency of units.
Dialogue: 0,0:08:21.56,0:08:25.85,Default,,0000,0000,0000,,What are the units of omega,\Nthe radial frequency?
Dialogue: 0,0:08:25.85,0:08:31.22,Default,,0000,0000,0000,,That's 1,000 radians per second.
Dialogue: 0,0:08:31.22,0:08:36.20,Default,,0000,0000,0000,,Yet we're specifying the\Nphase shift of 30 degrees.
Dialogue: 0,0:08:36.20,0:08:38.48,Default,,0000,0000,0000,,We're specifying it in degrees.
Dialogue: 0,0:08:38.48,0:08:41.67,Default,,0000,0000,0000,,To be consistent and frankly,
Dialogue: 0,0:08:41.67,0:08:44.45,Default,,0000,0000,0000,,we're going to leave it in this form\NI think simply because it's
Dialogue: 0,0:08:44.45,0:08:47.27,Default,,0000,0000,0000,,a lot easier for us to visualize.
Dialogue: 0,0:08:47.27,0:08:49.84,Default,,0000,0000,0000,,We have a better intuitive feel for
Dialogue: 0,0:08:49.84,0:08:53.49,Default,,0000,0000,0000,,phase shifts in terms of degrees than\Nwe do radians but if wherever we're
Dialogue: 0,0:08:53.49,0:08:57.11,Default,,0000,0000,0000,,doing calculations we'll need\Nto convert E to the this into
Dialogue: 0,0:08:57.11,0:09:01.31,Default,,0000,0000,0000,,degrees per second or\Nchange this to radians.
Dialogue: 0,0:09:01.31,0:09:04.12,Default,,0000,0000,0000,,Each course the change of radius\Nis going to just be simply
Dialogue: 0,0:09:04.12,0:09:10.52,Default,,0000,0000,0000,,the 30 degrees times Pi divided by 180.
Dialogue: 0,0:09:10.52,0:09:12.56,Default,,0000,0000,0000,,So that phase shift there,
Dialogue: 0,0:09:12.56,0:09:14.87,Default,,0000,0000,0000,,the 30 degree phase shift represents
Dialogue: 0,0:09:14.87,0:09:18.68,Default,,0000,0000,0000,,Pi sixth phase there
Dialogue: 0,0:09:18.68,0:09:21.65,Default,,0000,0000,0000,,and somebody could do the same for\Nthe phase shift and the current.
Dialogue: 0,0:09:21.65,0:09:23.74,Default,,0000,0000,0000,,But the point is let's\Njust drive this home.
Dialogue: 0,0:09:23.74,0:09:26.66,Default,,0000,0000,0000,,This circuit here, the circuit that we just
Dialogue: 0,0:09:26.66,0:09:31.10,Default,,0000,0000,0000,,analyzed did not change the frequency.
Dialogue: 0,0:09:31.10,0:09:33.20,Default,,0000,0000,0000,,The frequency of the current flowing in
Dialogue: 0,0:09:33.20,0:09:35.72,Default,,0000,0000,0000,,here is the same as\Nthe frequency of the voltage.
Dialogue: 0,0:09:35.72,0:09:39.08,Default,,0000,0000,0000,,What did change was the\Namplitude and the phase.
Dialogue: 0,0:09:39.08,0:09:42.71,Default,,0000,0000,0000,,Using phasor analysis we were\Nable to determine the phase
Dialogue: 0,0:09:42.71,0:09:47.33,Default,,0000,0000,0000,,on the amplitude and phase of the current.