0:00:00.920,0:00:05.595 >> In this video, we're going to[br]demonstrate how phasor diagrams 0:00:05.595,0:00:09.360 or showing the phasors[br]in the complex plane can 0:00:09.360,0:00:12.765 demonstrate the relative phases 0:00:12.765,0:00:17.370 and magnitudes of voltages and[br]currents within a circuit. 0:00:17.370,0:00:20.650 To do this, we're going[br]to remind ourselves that 0:00:20.650,0:00:27.125 phasor V is equal to the[br]phasor I times the impedance. 0:00:27.125,0:00:29.360 Now for example, in a resistor, 0:00:29.360,0:00:33.605 the impedance is R and V then is simply 0:00:33.605,0:00:37.980 equal to I times R. R is a real number. 0:00:37.980,0:00:41.975 So there is no phase term associated[br]with the impedance of a resistor, 0:00:41.975,0:00:45.680 and we say that the voltage is in 0:00:45.680,0:00:55.845 phase with the current in a resistor. 0:00:55.845,0:01:05.220 So if we have just arbitrarily choose[br]the current to have a zero phase angle, 0:01:06.860,0:01:10.520 then the phasor voltage 0:01:10.520,0:01:12.440 associated with the resistor or the voltage 0:01:12.440,0:01:14.885 across the resistor call it V sub r, 0:01:14.885,0:01:18.069 will have the same angle as I, 0:01:18.069,0:01:19.790 be in the same direction, 0:01:19.790,0:01:24.875 but just have a different magnitude than I. 0:01:24.875,0:01:30.990 In fact the magnitude of this[br]will be R times I. All right. 0:01:30.990,0:01:34.415 Unlike in a resistor, a capacitor[br]introduces a phase difference, 0:01:34.415,0:01:40.790 we know that V across the[br]capacitor is equal to I times Z. 0:01:40.790,0:01:44.620 Well, the impedance of[br]a capacitor is one over 0:01:44.620,0:01:49.400 j Omega c. We bring that[br]j up in the numerator 0:01:49.400,0:01:52.115 with a minus sign and we'll[br]have then this is equal to 0:01:52.115,0:01:58.320 I times a negative j over Omega c. Well, 0:01:58.320,0:02:06.230 that negative j in rectangular[br]coordinates is the same thing as 0:02:06.230,0:02:10.430 a minus 90 degree phase term[br]in polar coordinates 0:02:10.430,0:02:14.570 or we have then that V is equal 0:02:14.570,0:02:24.660 to I times one over Omega c[br]times e to the minus j90. 0:02:24.970,0:02:28.960 So whatever the phase of the current is, 0:02:28.960,0:02:31.460 when we multiply the current[br]times the impedance there is 0:02:31.460,0:02:34.790 a negative 90 degree phase term[br]that gets added to the current. 0:02:34.790,0:02:41.750 So once again if we let the current[br]have a zero reference angle, 0:02:41.750,0:02:47.480 then the voltage in a[br]capacitor is going to be 0:02:47.480,0:02:51.620 90 degrees less than 0:02:51.620,0:02:54.320 the phase of the current or we 0:02:54.320,0:02:59.300 say the voltage is 90 degrees[br]behind the current, 0:02:59.300,0:03:03.985 or we can say the current is[br]90 degrees ahead of the voltage. 0:03:03.985,0:03:07.625 Finally, we look at the[br]voltage across the inductor, 0:03:07.625,0:03:11.000 and we have phasor V. So that one, 0:03:11.000,0:03:13.280 of course these are all[br]phasor terms over here. 0:03:13.280,0:03:19.250 Phasor V sub l is equal to I times Z. 0:03:19.250,0:03:24.350 Well the Z the impedance of[br]an inductor is j Omega L, 0:03:24.350,0:03:31.620 which is equal to I times Omega L times j. 0:03:31.620,0:03:38.970 Well, j is the same thing[br]as e to the positive j90. 0:03:39.160,0:03:41.300 So in an inductor 0:03:41.300,0:03:45.430 the voltage will be 90 degrees[br]ahead of the phase of the current. 0:03:45.430,0:03:46.700 Whatever the phase of the current 0:03:46.700,0:03:49.910 is and it gets multiplied by[br]the impedance of the inductor, 0:03:49.910,0:03:52.510 which has this positive[br]90 phase term with it, 0:03:52.510,0:03:54.740 that positive 90 gets added to 0:03:54.740,0:03:57.020 the phase of the current[br]and the voltage will be 0:03:57.020,0:04:01.790 90 degrees ahead of the current. 0:04:01.790,0:04:06.320 So over here we had that being[br]I once again we'll reference I, 0:04:06.320,0:04:11.605 just arbitrarily choosing[br]it to have a zero phase. 0:04:11.605,0:04:18.709 I then V sub l will 0:04:18.709,0:04:25.220 be 90 degrees ahead of phasor I. 0:04:25.220,0:04:28.745 So we say that the current lags the voltage 0:04:28.745,0:04:33.590 in an inductor or the voltage[br]leads the current in an inductor. 0:04:33.590,0:04:37.610 Notice the voltages of the capacitor and 0:04:37.610,0:04:42.080 the inductor are 180 degrees[br]out of phase with each other. 0:04:42.080,0:04:44.915 The voltage and the current in the voltage 0:04:44.915,0:04:48.170 of the capacitor lags by 90 degrees, 0:04:48.170,0:04:52.645 the voltage in the inductor[br]leads the current by 90 degrees. 0:04:52.645,0:04:55.410 There's a mnemonic that[br]goes along with this, 0:04:55.410,0:04:59.870 today's clear back to the late 1800s[br]when much of electrical work that was 0:04:59.870,0:05:05.050 being done was associated with[br]refrigeration and creating ice, 0:05:05.050,0:05:12.730 and the mnemonic is ELI, the ICE man. 0:05:12.970,0:05:16.895 What this is suggesting is that E 0:05:16.895,0:05:20.620 represents voltage as in[br]physics we talk about the EMF. 0:05:20.620,0:05:22.105 So E is voltage, 0:05:22.105,0:05:25.015 I is current, L is an inductor. 0:05:25.015,0:05:27.145 So in an inductor, 0:05:27.145,0:05:33.905 the voltage leads[br]the current by 90 degrees. 0:05:33.905,0:05:40.010 In an inductor, the voltage[br]leads the current by 90 degrees. 0:05:40.010,0:05:47.720 In a capacitor, the current[br]leads the voltage by 90 degrees. 0:05:47.720,0:05:49.755 So in the capacitor c, 0:05:49.755,0:05:51.750 the current is 90 degrees ahead. 0:05:51.750,0:05:53.550 So as in the word ICE, 0:05:53.550,0:05:59.370 I comes ahead of E. So I is[br]leading the voltage over here. 0:05:59.370,0:06:03.840 In ELI, the voltage is leading the current. 0:06:05.860,0:06:09.860 Now let's demonstrate how[br]these phasor diagrams can be 0:06:09.860,0:06:13.100 used to demonstrate the various or 0:06:13.100,0:06:19.850 the relative phasors of different voltages[br]and currents within a circuit. 0:06:19.850,0:06:24.770 Here we have a single loop circuit[br]and writing one KVL, 0:06:24.770,0:06:32.130 we know then that phasor[br]V sub s is equal to. 0:06:33.470,0:06:39.650 Well, first of all V sub s sum of[br]the voltage is equal to V sub r, 0:06:39.650,0:06:41.480 plus V sub c, 0:06:41.480,0:06:48.830 plus V sub l and of course[br]these are all phasor voltages. 0:06:48.830,0:06:53.150 Now let's go ahead and put[br]in that V sub r is equal 0:06:53.150,0:06:58.295 to I times R. In this case, 0:06:58.295,0:07:00.900 R is eight ohms 0:07:02.030,0:07:07.340 plus the voltage across the capacitor[br]is just going to be I times Z sub c, 0:07:07.340,0:07:14.160 which is I times a negative j8 plus[br]the voltage across the inductor. 0:07:14.160,0:07:16.070 That's a J2 ohm inductor. 0:07:16.070,0:07:22.185 So the voltage across[br]that will be I times J2. 0:07:22.185,0:07:25.010 So we have three terms here. 0:07:25.010,0:07:27.800 Three different phasors. 0:07:27.800,0:07:30.830 These are phasor voltage is equal to 8I, 0:07:30.830,0:07:35.580 minus J8I, and positive J2I. 0:07:36.670,0:07:40.180 Showing this explicitly with the phases, 0:07:40.180,0:07:42.860 this will be I times eight. 0:07:42.860,0:07:45.140 There is no phase difference[br]between the voltage 0:07:45.140,0:07:48.260 across the resistor and the current. 0:07:48.260,0:07:53.600 On the other hand we have I times eight 0:07:53.600,0:07:58.770 times e to the minus j90 in the capacitor, 0:07:58.770,0:08:09.730 plus I times two times e to[br]the positive j90 for the inductor. 0:08:09.730,0:08:12.555 So again, what this is saying is that 0:08:12.555,0:08:20.540 the phase of the voltage 0:08:20.540,0:08:25.640 across the inductor is 90 degrees[br]greater than the phase of the current. 0:08:25.640,0:08:30.590 On the other hand, the phase[br]of the voltage across 0:08:30.590,0:08:36.049 the capacitor is 90 degrees[br]behind the phase of the current. 0:08:36.049,0:08:42.110 So let's just start out and[br]say that this is the current. 0:08:42.110,0:08:45.620 I will say that it has[br]a magnitude I naught. 0:08:45.620,0:08:51.574 The voltage across the resistor[br]is eight times that 0:08:51.574,0:08:58.670 or V sub r is equal to eight[br]times I naught in length, 0:08:58.670,0:09:03.110 and it has the same angle as I. 0:09:03.110,0:09:09.650 The capacitor on the other hand[br]is 90 degrees behind the current. 0:09:09.650,0:09:14.059 So the voltage V sub c is 90 degrees 0:09:14.059,0:09:20.775 behind the current V sub c, 0:09:20.775,0:09:29.570 and it's down here at[br]negative 8I naught, and the voltage. 0:09:29.570,0:09:31.835 Let's see that's the[br]voltage of the capacitor, 0:09:31.835,0:09:40.560 and the voltage of the inductor[br]is 90 degrees ahead of V sub l, 0:09:42.080,0:09:48.230 the length of it is equal to two times 0:09:48.230,0:09:53.930 I naught and it is at an angle of[br]90 degrees ahead of the current. 0:09:53.930,0:09:56.035 So our three voltages. 0:09:56.035,0:09:59.645 The voltage across the resistor[br]is in phase with the current, 0:09:59.645,0:10:02.990 capacitor is 90 degrees[br]behind the current and 0:10:02.990,0:10:06.380 the inductor voltage is 90 degrees ahead. 0:10:06.380,0:10:08.785 Now we have here that V sub s, 0:10:08.785,0:10:12.980 the source voltage is equal to[br]the sum of those three terms. 0:10:12.980,0:10:14.870 So when adding phasors, 0:10:14.870,0:10:17.660 you do it just like you do vectors[br]and that is tip to tell them. 0:10:17.660,0:10:22.280 Here's V sub r, add in V sub c, 0:10:22.280,0:10:29.590 that comes down here like that[br]and then add in V sub l. That 0:10:29.590,0:10:37.909 brings us back two units back that way[br]to this point here at negative 6I0, 0:10:38.080,0:10:43.555 and so we end up at that point[br]and that is V sub s then. 0:10:43.555,0:10:46.360 Let see how good of a job I[br]can do drawing a vector down, 0:10:46.360,0:10:48.160 they're not very good. 0:10:48.160,0:10:51.200 That's a straight line vector. 0:10:51.870,0:10:55.165 That's better I guess. 0:10:55.165,0:11:01.905 So V sub s then is equal to V sub r, 0:11:01.905,0:11:03.480 plus V sub c, plus 0:11:03.480,0:11:09.795 V sub l. This results in phasor[br]is V sub s. What does it tell us? 0:11:09.795,0:11:13.910 Well, it tells us that V sub s is something 0:11:13.910,0:11:18.920 less than 90 degrees[br]behind the current or that 0:11:18.920,0:11:22.310 the current in this circuit I is leading 0:11:22.310,0:11:28.790 the voltage source by some angle[br]less than 90 degrees. 0:11:28.790,0:11:37.050 It also tells us the relative length of[br]this voltage is longer than V sub r, 0:11:37.060,0:11:47.230 it's shorter than V sub c because[br]V sub c and V sub l are opposites. 0:11:49.610,0:11:54.570 I guess to be more accurate if[br]you to say that V sub c and 0:11:54.570,0:11:59.465 V sub l are in opposite directions[br]and they tend to cancel each other. 0:11:59.465,0:12:01.910 Let me say that one more[br]time just a little bit 0:12:01.910,0:12:04.295 differently and see if I can[br]make it a little more clear. 0:12:04.295,0:12:07.790 V sub s the source voltage is equal to 0:12:07.790,0:12:11.400 the phasor sum of V sub l, plus V sub r, 0:12:11.400,0:12:16.500 plus V sub c. By diagramming[br]or drawing in V sub l, 0:12:16.500,0:12:19.665 V sub r and V sub c and then[br]adding them tip to tail, 0:12:19.665,0:12:23.600 we can see the relative length and phase 0:12:23.600,0:12:27.740 of the source to the voltage[br]across the resistor, 0:12:27.740,0:12:29.705 which is in phase with the current. 0:12:29.705,0:12:33.340 The relative length and 0:12:33.340,0:12:36.980 the relative angle of the source[br]relative to the voltage across 0:12:36.980,0:12:40.370 the inductor and the relative length and 0:12:40.370,0:12:46.720 relative phase of the source relative[br]to the voltage across the capacitor. 0:12:46.720,0:12:48.980 Again, this doesn't give[br]us a real accurate, 0:12:48.980,0:12:52.640 but it does give us some feel[br]for the relative phases, 0:12:52.640,0:12:55.520 the phases of one of each of 0:12:55.520,0:12:59.880 these voltages relative to the current[br]and relative to the source.