1 00:00:02,280 --> 00:00:04,860 Welcome back to our continuing series on linear circuits work. 2 00:00:04,860 --> 00:00:10,000 Today, we talk about AC circuit analysis. Today, we're going to be identifying how 3 00:00:10,000 --> 00:00:13,900 past techniques that we applied to resistive circuits can be applied to 4 00:00:13,900 --> 00:00:18,320 circuits that are sinusoidal, and how we can use impedances and phasors, to be 5 00:00:18,320 --> 00:00:23,800 able to do analyses, for those types of systems. 6 00:00:23,800 --> 00:00:27,155 In the previous lesson we talked about impedance and its relationship to 7 00:00:27,155 --> 00:00:30,302 phasors. And then we looked at how impedance 8 00:00:30,302 --> 00:00:34,178 changes with frequency in reactive circuits like capacitors and in 9 00:00:34,178 --> 00:00:37,615 inductors. So now we're going to apply that to doing 10 00:00:37,615 --> 00:00:40,513 analyses to sinusoidal systems and that will lead us next time to talk about 11 00:00:40,513 --> 00:00:45,496 transfer functions. The objectives for today's lesson are to 12 00:00:45,496 --> 00:00:48,898 apply techniques from DC analysis to sinusoidal systems, find equivalent 13 00:00:48,898 --> 00:00:52,732 impedances for devices that are placed in series or parallel, use superposition for 14 00:00:52,732 --> 00:00:57,995 analysis, particularly for systems that have multiple frequencies. 15 00:00:57,995 --> 00:01:01,809 And then be able to analyze a system using all of these techniques. 16 00:01:04,150 --> 00:01:09,170 So first of all, we want to identify that impedance is a linear property. 17 00:01:09,170 --> 00:01:14,262 So if you remember back when we talked about linearity, we use, as an example, 18 00:01:14,262 --> 00:01:19,260 Ohm's law, where we have V is equal to I times R. 19 00:01:20,360 --> 00:01:23,805 We said that this is a linear property, because it's basically multiplying by 20 00:01:23,805 --> 00:01:27,100 constant value. But now, we're doing the exact same 21 00:01:27,100 --> 00:01:31,730 thing. In phasors, we have the property V is 22 00:01:31,730 --> 00:01:36,204 equal to I times Z. Now V and I are phasors, so they're 23 00:01:36,204 --> 00:01:41,860 complex valued things representative of. They assign certain functions. 24 00:01:41,860 --> 00:01:44,912 And Z is the impedance, or the relationship between the two. 25 00:01:44,912 --> 00:01:50,096 And because of this setup, this is also linear linear properties, just the same 26 00:01:50,096 --> 00:01:55,280 as Ohm's law, but now with complex numbers. 27 00:01:55,280 --> 00:02:00,130 And so consequently, some of the same implications will occur. 28 00:02:00,130 --> 00:02:04,766 So if I wanted to do an analysis of impedances in series, her I have voltage 29 00:02:04,766 --> 00:02:12,280 V and I believe a phaser I. And I have these three impedances. 30 00:02:12,280 --> 00:02:16,930 Using this relationship of impedance to voltage, if I wanted to find what I 31 00:02:16,930 --> 00:02:27,790 happens to be, I know that I. It's going to be equal to V over Z. 32 00:02:27,790 --> 00:02:35,963 And if I look at each of the individual voltages, here, here and here. 33 00:02:35,963 --> 00:02:42,043 They'd be one, this would be two and this would be three. 34 00:02:42,043 --> 00:02:51,254 The voltage here is going to be the current, I, times Z1, V2 is the current I 35 00:02:51,254 --> 00:03:00,368 times Z2, and V3 is the current I times Z3. 36 00:03:02,190 --> 00:03:05,886 They're all in series so all of these I's are the same so adding them all together 37 00:03:05,886 --> 00:03:11,020 because we know that using Kircoff's voltage that V is equal V1 + V2 + V3. 38 00:03:11,020 --> 00:03:18,445 So adding all these up and factoring out the I we see that. 39 00:03:18,445 --> 00:03:30,923 V is equal to i times z1 plus z2 plus z3. So we can basically come to find what z 40 00:03:30,923 --> 00:03:35,760 happens to be right looking at this format. 41 00:03:35,760 --> 00:03:39,595 We see that our equivalent z is just going to be equal to the sum of each of 42 00:03:39,595 --> 00:03:45,880 the individual z's, and here I'll just use the summation notation for gravity. 43 00:03:49,280 --> 00:03:52,976 You might recognize this as being the exact same analysis we used for resistors 44 00:03:52,976 --> 00:03:56,672 except now we have complex numbers in our calculation and we get the same result 45 00:03:56,672 --> 00:04:02,950 The equivalent of impedance devices in series is just the sum of the impedances. 46 00:04:02,950 --> 00:04:08,800 And the natural next step is to look at it when we have devices in parallel. 47 00:04:09,850 --> 00:04:13,267 And doing the exact same technique that we used before for resistors, we, again, 48 00:04:13,267 --> 00:04:19,050 basically come up with the same results. We'll see that this Zeq is going to be 49 00:04:19,050 --> 00:04:23,470 equal to the sum of the inverses of each of the individual impedances, and then 50 00:04:23,470 --> 00:04:28,570 you invert the whole thing. So again, for brevity, I put it in 51 00:04:28,570 --> 00:04:31,949 summation notation. So it's exactly what you would expect 52 00:04:31,949 --> 00:04:36,240 with resistors. Kirchhoff's laws also apply in phasors, 53 00:04:36,240 --> 00:04:41,460 and before well, all of the input currents and the output currents kind of 54 00:04:41,460 --> 00:04:46,634 had to match up. Now we have the additional property that 55 00:04:46,634 --> 00:04:50,599 they have to match up in the real and the imaginary parts, because the sum has to 56 00:04:50,599 --> 00:04:53,956 be zero. And you can add up all the real parts 57 00:04:53,956 --> 00:04:57,314 which you want, but you're never going to get an imaginary. 58 00:04:57,314 --> 00:05:00,434 So we see that, that gives us one extra little bit of help, while we're doing 59 00:05:00,434 --> 00:05:03,602 analysis with phasor systems, is we can look at the real parts and the imaginary 60 00:05:03,602 --> 00:05:09,905 parts, and they all have to match up. So just as a review, Kirchhoff''s current 61 00:05:09,905 --> 00:05:14,114 law, the input currents have to equal the output currents, even the phasor, and for 62 00:05:14,114 --> 00:05:18,495 Kirchhoff's voltage law, which we already used. 63 00:05:18,495 --> 00:05:25,253 The, if I do a sum around the circle, we see that Vs has to be equal to V1 plus V2 64 00:05:25,253 --> 00:05:30,532 plus V3. Source transformations also work, but now 65 00:05:30,532 --> 00:05:34,066 we're going to be having sources that are phasors, and, instead of having a 66 00:05:34,066 --> 00:05:39,100 Thevenin resistance, we have a Thevenin impedance. 67 00:05:39,100 --> 00:05:42,940 But the same basic relationship applies, that V Thevenin is equal to I Norton 68 00:05:42,940 --> 00:05:48,680 times Z Thevenin. Superposition also works. 69 00:05:48,680 --> 00:05:51,780 When we initially presented the superposition, you might have thought it 70 00:05:51,780 --> 00:05:55,080 was kind of a lot of work to do analysis for something because you'd have to break 71 00:05:55,080 --> 00:05:58,030 it up into each individual piece and then add up all the results from the 72 00:05:58,030 --> 00:06:04,258 individual analysis together. Sometimes it's useful if, by breaking it 73 00:06:04,258 --> 00:06:08,598 up into individual places and zeroing out sources, the system becomes trivial, but 74 00:06:08,598 --> 00:06:13,580 that's not always the case. However, one time where superposition is 75 00:06:13,580 --> 00:06:17,176 extraordinarily useful is if we have systems with different frequencies 76 00:06:17,176 --> 00:06:21,260 because its linear superposition still holds. 77 00:06:21,260 --> 00:06:25,100 But because of the frequencies are different, you can't immediately relate 78 00:06:25,100 --> 00:06:29,357 one system to the other. Because that, we know that as frequency 79 00:06:29,357 --> 00:06:32,900 changes, impedances change in reactive components. 80 00:06:32,900 --> 00:06:39,385 So let's look at this example. Again, I zero out sources the same way 81 00:06:39,385 --> 00:06:43,958 that was presented before. So these current sources are going to go 82 00:06:43,958 --> 00:06:48,376 to open circuits. So I will do them one at a time if I 83 00:06:48,376 --> 00:06:55,948 first get rid of i2 by zeroing it out and leaving i1 intact. 84 00:06:55,948 --> 00:07:02,644 We get i1, which is now a phasor, and going down to this, we see square root of 85 00:07:02,644 --> 00:07:12,162 two, with an angle of zero degrees. And that would come over, here we have a 86 00:07:12,162 --> 00:07:19,867 capacitor, a resistor, and the impedance of the capacitor is going to be equal to 87 00:07:19,867 --> 00:07:33,720 1 over j omega c. Here, our omega is 10 to the 6th. 88 00:07:33,720 --> 00:07:38,953 So this is equal to 1 over j times 10 to the 6th. 89 00:07:38,953 --> 00:07:46,261 The capacitance is 1 nanofarad, which is 10 to the negative 9th, put all that 90 00:07:46,261 --> 00:07:53,514 together, you get minus j 10 to the third ohms. 91 00:07:53,514 --> 00:07:58,018 And then here the impedance of our resistor is simply the resistance. 92 00:07:58,018 --> 00:08:07,550 So this is also equal to minus j times one kilo ohm. 93 00:08:12,920 --> 00:08:17,520 Now, what I'm trying to solve is the voltage v here. 94 00:08:17,520 --> 00:08:23,833 So here I label it as v1, because it's the voltage generated by our first 95 00:08:23,833 --> 00:08:30,723 source, and v1 is going to be equal to i times z. 96 00:08:30,723 --> 00:08:37,047 So i is squared of 2.0, z is going to be the sum of these 2, which it turns out to 97 00:08:37,047 --> 00:08:43,881 be the the square root of 2, arc minus pi fourths, so multiply this, taking two 98 00:08:43,881 --> 00:08:55,116 values together. That gives us 2 with the phase angle of 99 00:08:55,116 --> 00:09:05,202 minus pi fourths. In other words the voltage from the first 100 00:09:05,202 --> 00:09:12,114 element, it's going to be equal to 2 cosine 10 to the 6th t minus pi 4th 101 00:09:12,114 --> 00:09:20,650 volts. Now that we finish that part, we are 102 00:09:20,650 --> 00:09:26,782 going to now 0 out source 1. Probably should indicate that something 103 00:09:26,782 --> 00:09:33,020 was removed here. So when we remove source one we'll have 104 00:09:33,020 --> 00:09:36,350 something like this. Again we have our capacitance, our 105 00:09:36,350 --> 00:09:44,540 resistance. Now this, again, is, 1 over j omega c. 106 00:09:44,540 --> 00:09:51,415 This is r, which is just one kilo ohm, and the second source. 107 00:09:51,415 --> 00:09:56,608 Now here, notice that this is a sine function. 108 00:09:56,608 --> 00:10:05,968 So, this phaser is actually going to be three with a phase angle of minus, oops, 109 00:10:05,968 --> 00:10:14,207 minus pi halves. Now coming back to this one over j omega 110 00:10:14,207 --> 00:10:21,422 c, well this has a different frequency, so this time omega is equal to 1 over the 111 00:10:21,422 --> 00:10:32,780 square root of 3 times 10 to the 6th. So I take 1 over j times 1 over the 112 00:10:32,780 --> 00:10:47,066 square root of 3, times 10 to the 6th times c, again, 10 to the negative 9th. 113 00:10:47,066 --> 00:10:59,207 This equals minus j square root of 3 times 10 to the 3rd ohms, or minus j 114 00:10:59,207 --> 00:11:09,898 square root of 3 kilo ohms. So now that we can clearly see that 115 00:11:09,898 --> 00:11:13,332 capacitance is different for the system than it was in the system. 116 00:11:13,332 --> 00:11:18,290 Now remember the phases always have this implied frequency, and we can't ignore 117 00:11:18,290 --> 00:11:21,460 it. It's very important that we don't. 118 00:11:21,460 --> 00:11:28,628 So if I put all of this together, I discover that again we're finding this v 119 00:11:28,628 --> 00:11:35,531 here. So I'm calling, going to call it v2 is 120 00:11:35,531 --> 00:11:48,550 equal to i from 2 times this impedance. So that gives us 3, phase angle of minus 121 00:11:48,550 --> 00:11:56,223 pi halves, times. And here we if we look at this we see 122 00:11:56,223 --> 00:12:03,861 that, adding 1k and minus square root of 3k in this format that, the amplitude 123 00:12:03,861 --> 00:12:18,170 here amplitude here is 2. And this is minus pi thirds, the angle. 124 00:12:18,170 --> 00:12:24,450 So we're going to be multiplying this by 2, an angle of minus pi 3rds. 125 00:12:24,450 --> 00:12:30,402 Oh my mistake. This should be kilovolts, not volts, 126 00:12:30,402 --> 00:12:36,695 because remember there's these, these k's here. 127 00:12:36,695 --> 00:12:41,472 This should be 2k. This should be 2k. 128 00:12:41,472 --> 00:12:49,365 So, multiplying this together we get 6k, with an angle of minus 5 pi 6ths. 129 00:12:49,365 --> 00:12:57,366 And if we put this back into the other format, that means that v2 is going to be 130 00:12:57,366 --> 00:13:04,859 equal to 6 cosine of 1 over the square root of 3 times 10 to the 6th t minus 5 131 00:13:04,859 --> 00:13:16,821 pi 6ths kilovolts. Then my final v is going to be equal to 132 00:13:16,821 --> 00:13:21,720 v1 Plus v2. It turns out that in this particular 133 00:13:21,720 --> 00:13:24,620 example, we are not able to directly solve it in one step, we have to use 134 00:13:24,620 --> 00:13:29,690 superposition because the two sources are operating at different frequencies. 135 00:13:29,690 --> 00:13:33,375 So the consequently the capacitance leads to different impedances for each 136 00:13:33,375 --> 00:13:36,344 different piece. So, superposition is the only technique 137 00:13:36,344 --> 00:13:37,864 to use. But, you can see that it's not 138 00:13:37,864 --> 00:13:39,899 particularly difficult to do the solution. 139 00:13:43,230 --> 00:13:47,135 To review valid impedance techniques, we can use Kirchhoff's Laws, Superposition, 140 00:13:47,135 --> 00:13:50,930 Node-voltage analysis, Mesh-current analysis, Thevenin and Norton Equivalent 141 00:13:50,930 --> 00:13:56,016 Circuits and Source Transformations. And the only real difference from what we 142 00:13:56,016 --> 00:13:59,855 were doing with resistors is that now we are using complex-valued impedances. 143 00:13:59,855 --> 00:14:06,191 One more example that we will do, is this RLC circuit, and we're going to want, to 144 00:14:06,191 --> 00:14:11,554 find the value of vc. So to do this, we will, again put it into 145 00:14:11,554 --> 00:14:16,349 a phasor form. So, the input will be a V, and that is 146 00:14:16,349 --> 00:14:20,570 going to be equal to 1 with a base angle of zero. 147 00:14:20,570 --> 00:14:26,120 We're going to be operating at 780 hertz. So omega's two pi times 780. 148 00:14:26,120 --> 00:14:33,296 The impedance for our resistance is 20 kilo-ohms. 149 00:14:33,296 --> 00:14:39,056 The impedance for our inductance is equal to j times omega times l, and for the 150 00:14:39,056 --> 00:14:46,100 impedance with capacitance, it's 1 over j times omega times c. 151 00:14:47,810 --> 00:14:51,490 We can use a voltage divider here to find the value of VC. 152 00:14:51,490 --> 00:15:01,285 This is just going to be Zc over Zr + Zl + Zc. 153 00:15:02,830 --> 00:15:06,716 All times v, and obviously this is after the switch is closed. 154 00:15:06,716 --> 00:15:10,736 this is the same circuit that we analyzed when we were doing the differential 155 00:15:10,736 --> 00:15:15,263 equations for RLC circuits. But in this case we are, going to be 156 00:15:15,263 --> 00:15:19,479 looking at it and analyzing it with a sinusoidal input, as opposed to just a 157 00:15:19,479 --> 00:15:26,765 constant value that switches on. Now since v is 1 angle 0 this can 158 00:15:26,765 --> 00:15:34,396 basically be somewhat ignored. it's important because of the units but 159 00:15:34,396 --> 00:15:38,056 that just basically means that vc is going to be equal to zc over zr plus zl 160 00:15:38,056 --> 00:15:44,385 plus zc. Plugging in our values for both various 161 00:15:44,385 --> 00:15:52,193 things we get 1 over J omega C divided by R plus J I make it L plus 1 over J we'll 162 00:15:52,193 --> 00:16:01,062 make it C. Now, plugging in the values of the 163 00:16:01,062 --> 00:16:04,464 capacitance, the inductance, and the omegas here, and doing a little bit of 164 00:16:04,464 --> 00:16:12,370 calculation with complex numbers. We'll discover that this turns out to be 165 00:16:12,370 --> 00:16:21,970 about 0.51, minus j 0.5 zero, which is pretty close to 1 over the square root of 166 00:16:21,970 --> 00:16:37,242 2 with a phase of minus phi fourths. Putting this back into our time domain 167 00:16:37,242 --> 00:16:43,029 Would be c of t. This is going to be 1 over the square 168 00:16:43,029 --> 00:16:52,048 root of 2, times the cosine of 2 pi times 780 times t, minus pi 4ths volts. 169 00:16:52,048 --> 00:16:56,574 And so you an see that this is quite a bit simpler than doing the calculation 170 00:16:56,574 --> 00:17:01,644 using differential equations. It is entirely possible to go through and 171 00:17:01,644 --> 00:17:05,970 do a derivation to get the same results using differential equations. 172 00:17:05,970 --> 00:17:09,256 But using phasors gives us an excellent tool at being able to apply complex 173 00:17:09,256 --> 00:17:12,489 number number to, to give a solution to see how the system will respond to a 174 00:17:12,489 --> 00:17:17,942 particular sinusoidal input. In summary, we showed how DC analysis 175 00:17:17,942 --> 00:17:22,562 techniques can be directly applied to sinusoidal systems by using phasors and 176 00:17:22,562 --> 00:17:27,109 impedances. Then we use superposition to analyze the 177 00:17:27,109 --> 00:17:32,079 system that had multiple frequencies, and then solve the example system using these 178 00:17:32,079 --> 00:17:39,526 techniques. In the next lesson, there will be a demo 179 00:17:39,526 --> 00:17:44,450 showing sinusoidal response, and then a discussion of transfer functions. 180 00:17:44,450 --> 00:17:47,683 Transfer functions are essentially how systems react across different 181 00:17:47,683 --> 00:17:52,984 frequencies for various inputs. Until next time.