1 00:00:00,025 --> 00:00:05,830 Okay, so now, we're going to be talking about capacitors, the actual physical 2 00:00:05,830 --> 00:00:08,850 devices. So, we'll start by presenting how these 3 00:00:08,850 --> 00:00:12,834 capacitors work in a system, identify the behavior in DC circuits. 4 00:00:12,834 --> 00:00:18,010 And then graphically represent the relationships between current, voltage, 5 00:00:18,010 --> 00:00:22,330 power, and energy within capacitors. To get some feel for how these devices 6 00:00:22,330 --> 00:00:24,090 are actually going to be used and how they work. 7 00:00:25,240 --> 00:00:29,260 From out previous class we talked about capacitance, we talked about finding the 8 00:00:29,260 --> 00:00:32,660 capacitance of things. And got some idea of the relationship 9 00:00:32,660 --> 00:00:36,530 between current and voltage with capacitors and it was similar to what we 10 00:00:36,530 --> 00:00:40,430 did with resistance. And we now have some idea about what 11 00:00:40,430 --> 00:00:45,270 capacitance means, basically, it's how well this device holds holds these 12 00:00:45,270 --> 00:00:48,550 electric fields. So, how much charge is going to be stored 13 00:00:48,550 --> 00:00:51,820 by that capacitor for any amount of voltage that we're putting across it. 14 00:00:53,270 --> 00:00:55,765 After we finish talking about capacitors, we're going to be talking about 15 00:00:55,765 --> 00:01:00,130 inductors, which are different devices. But you'll notice a lot of similar 16 00:01:00,130 --> 00:01:02,780 behavior, a lot of parallelism between the two. 17 00:01:02,780 --> 00:01:06,800 And so, it can be good to go back and forth and look at them together to see 18 00:01:06,800 --> 00:01:09,165 how what things are similar and what things are different. 19 00:01:09,165 --> 00:01:13,660 The objectives for this particular lesson are to allow you to analyze the 20 00:01:13,660 --> 00:01:16,958 capacitors when they are placed in parallel and series configurations. 21 00:01:16,958 --> 00:01:20,165 To analyze DC circuits that have capacitors in them. 22 00:01:20,165 --> 00:01:24,755 DC meaning that there's no changes in the system, everything is a constant value. 23 00:01:24,755 --> 00:01:27,340 And then calculating the energy of a capacitor. 24 00:01:27,340 --> 00:01:29,590 We'll do a calculation and a derivation for that. 25 00:01:29,590 --> 00:01:35,690 And then we'll also sketch these current and voltage and power energy curves to 26 00:01:35,690 --> 00:01:41,160 see the relationships between them. So, here we've placed capacitors into a 27 00:01:41,160 --> 00:01:44,890 parallel configuration. Due to Kirchhoff's current law, we know 28 00:01:44,890 --> 00:01:49,170 that i, the current coming in, is equal to i1 plus i2, the two currents that are 29 00:01:49,170 --> 00:01:52,110 flowing. We also know that because these devices 30 00:01:52,110 --> 00:01:56,242 are in parallel, the voltage across them is equivalent, thanks to Kirchoff's 31 00:01:56,242 --> 00:02:00,930 voltage law. That allows us to do this derivation. 32 00:02:00,930 --> 00:02:05,840 Since i is equal to i1 plus i2, and these currents can then be related by voltages 33 00:02:05,840 --> 00:02:12,872 C1 dv dt plus C2 dv dt. If you factor out the Cs, put the dv dt 34 00:02:12,872 --> 00:02:18,940 together, then you see that C1 plus C2 is our capacitance. 35 00:02:18,940 --> 00:02:26,386 Since capacitance was basically defined to be the C that gives us this behavior C 36 00:02:26,386 --> 00:02:35,360 dv dt so, there's our C. If we place them in series instead, now 37 00:02:35,360 --> 00:02:39,809 we're going to have to use the, the other equation but we're still making use of 38 00:02:39,809 --> 00:02:42,230 Kirchhoff's laws. The current is equivalent between the 39 00:02:42,230 --> 00:02:44,690 two. Now the voltages are different. 40 00:02:46,040 --> 00:02:52,960 And my voltage ab when we're using this notation the first letter is the one with 41 00:02:52,960 --> 00:02:55,410 the plus the second letter is the one with the minus. 42 00:02:55,410 --> 00:02:58,520 That voltage is going to be equal to v1 plus v2. 43 00:02:58,520 --> 00:03:02,330 So let's start form there, Vab is equal to v1 plus v2. 44 00:03:02,330 --> 00:03:08,850 Using our integration properties, finding voltage as a function of current, we get 45 00:03:08,850 --> 00:03:11,010 this. And after moving things around, doing 46 00:03:11,010 --> 00:03:14,940 some manipulation, here we make use of the fundamental theorem of calculus and 47 00:03:14,940 --> 00:03:20,510 take a derivative. We eventually come to this, where i is 48 00:03:20,510 --> 00:03:25,030 equal to C dv dt, where that is our C. What we're doing is we're inverting each 49 00:03:25,030 --> 00:03:29,758 of the Cs, adding them together, and then inverting the result. 50 00:03:29,758 --> 00:03:34,620 So, you might think back to the resistors, capacitors in series behave 51 00:03:34,620 --> 00:03:40,640 very similarly to resistors in parallel. Capacitors in parallel behave a lot like 52 00:03:40,640 --> 00:03:45,810 resistors in series in the sense that to find equivalence, here we're doing this 53 00:03:45,810 --> 00:03:50,770 inverse sum and then inverting. And for the parallel, we just added up 54 00:03:50,770 --> 00:03:53,770 the individual capacitances, just like with the resistors in series adding up 55 00:03:53,770 --> 00:03:56,889 individual resistances. So, its good to notice that, that 56 00:03:56,889 --> 00:03:59,380 combination so you can remember these equations. 57 00:04:00,400 --> 00:04:03,700 When we put these things into a dc circuit what will happen is, since 58 00:04:03,700 --> 00:04:06,700 everything is constant nothings changing in time. 59 00:04:06,700 --> 00:04:09,210 The capacitors going to find it's way into an equilibrium. 60 00:04:10,510 --> 00:04:13,030 Where it finds its way into that equilibrium is because charge is then 61 00:04:13,030 --> 00:04:17,142 pushed onto the plates, to a point where it's kind of at this natural voltage. 62 00:04:17,142 --> 00:04:23,160 And the, essentially, for an analysis point of view, is equivalent to taking 63 00:04:23,160 --> 00:04:27,440 this capacitor, and just removing it. And if I want to know, for example, the 64 00:04:27,440 --> 00:04:32,680 voltage across this capacitor, what I do in this DC system, start off by removing 65 00:04:32,680 --> 00:04:36,900 the capacitor. Now, this gap here, the voltage for this 66 00:04:36,900 --> 00:04:41,170 gap is the voltage across the capacitor. Now, since I don't have any current 67 00:04:41,170 --> 00:04:45,202 flowing through this resistor here, I can't have any voltage across this 68 00:04:45,202 --> 00:04:47,180 resistor because it's just dangling off the end. 69 00:04:47,180 --> 00:04:51,590 So this is really unimportant for our analysis, we can pretend like it doesn't 70 00:04:51,590 --> 00:04:54,379 even exist at this point, for the purposes of what we're doing. 71 00:04:56,920 --> 00:05:00,700 At this point the system reduces to essentially a voltage divider, we have 72 00:05:00,700 --> 00:05:03,680 one volt here one ohm here and three ohms here. 73 00:05:03,680 --> 00:05:06,373 Which means that the drop across this resistor is three fourths of a volt and 74 00:05:06,373 --> 00:05:14,959 since this voltage is measuring the difference from this node to this node. 75 00:05:16,740 --> 00:05:18,860 It's essentially in parallel with this resistor. 76 00:05:18,860 --> 00:05:23,440 So, this voltage here is also going to be equal to three fourths of a volt. 77 00:05:29,280 --> 00:05:32,890 To find the stored energy in a capacitor we're going to start by doing an energy 78 00:05:32,890 --> 00:05:35,270 calculation and we'll start from what we know. 79 00:05:35,270 --> 00:05:39,570 First we know that power is equal to current times voltage. 80 00:05:39,570 --> 00:05:44,410 Secondly we know we can calculate energy, by integrating that power and then adding 81 00:05:44,410 --> 00:05:48,310 our initial energy. Finally we know that the relationship 82 00:05:48,310 --> 00:05:53,300 between current and voltage in the capacitor is i is equal to C dv vt. 83 00:05:53,300 --> 00:06:00,090 So we will plug in for p i times v in our energy calculation and then replace i 84 00:06:00,090 --> 00:06:06,810 with C dv dt to give us this equation. After doing a little bit of manipulation 85 00:06:06,810 --> 00:06:11,500 this is doing a change of variables. If you kind of get lost in this step go 86 00:06:11,500 --> 00:06:14,240 back and review changes of variables from calculus. 87 00:06:17,000 --> 00:06:20,710 That leads us to this equation, and finally we get this result. 88 00:06:20,710 --> 00:06:25,600 The energy stored in a capacitor is equal to one half Cv squared. 89 00:06:25,600 --> 00:06:29,230 So the bigger the voltage, the bigger the energy is stored by our capacitor and 90 00:06:29,230 --> 00:06:32,954 also it's scaled by the capacitance of our capacitor as well. 91 00:06:32,954 --> 00:06:40,050 And we'll take that and apply it to a graphically seeing how all of these 92 00:06:40,050 --> 00:06:42,990 things relate. So, first of all we're going to take a 93 00:06:42,990 --> 00:06:45,260 curve that represents the voltage through the capacitor. 94 00:06:46,860 --> 00:06:50,260 And here notice that our capacitance is going to be one farad, it's a really big 95 00:06:50,260 --> 00:06:53,080 capacitance. But it makes it easy for doing this 96 00:06:53,080 --> 00:06:55,310 example, it makes the math very simple and clear. 97 00:06:57,580 --> 00:07:02,110 If I want to find the current from the voltage, i is C dv dt. 98 00:07:02,110 --> 00:07:09,630 Since C is 1, this basically just amounts to finding the derivative of the voltage, 99 00:07:09,630 --> 00:07:13,220 or the slope of this line. So, the slope of this line it goes up 100 00:07:13,220 --> 00:07:16,440 two, in two seconds. So the slope is one. 101 00:07:16,440 --> 00:07:19,888 So our current, in this graph, is one amp. 102 00:07:19,888 --> 00:07:25,520 So this flattens out here, it has zero slope. 103 00:07:25,520 --> 00:07:29,580 Which means our current goes to zero. And then since it drops down here, at 104 00:07:29,580 --> 00:07:36,750 twice the rate, our slope is 2, our current here is going to be minus 2 amps. 105 00:07:36,750 --> 00:07:41,270 To find our power, which is in this graph, we're going to be combining these 106 00:07:41,270 --> 00:07:46,035 two by multiplication. So, if I multiply these two functions, I 107 00:07:46,035 --> 00:07:52,419 start at zero going up to 2 times 1, so it's going to increase like this. 108 00:07:54,429 --> 00:07:57,980 Then at this point where this flattens out the current goes to zero so, so does 109 00:07:57,980 --> 00:08:02,170 our power. And this goes down but this is now a 110 00:08:02,170 --> 00:08:07,180 negative current, so we take minus 2 times 2 volts which puts us at this 111 00:08:07,180 --> 00:08:09,460 point. And then we're just basically multiplying 112 00:08:09,460 --> 00:08:15,240 a constant value by this line so we get a curve linearly increasing here. 113 00:08:17,600 --> 00:08:20,650 So this is what our power's going to be. Now remember this is not a problem, power 114 00:08:20,650 --> 00:08:23,490 can jump as much as it likes. But something that's interesting to 115 00:08:23,490 --> 00:08:29,330 notice is that because of this behavior, the C dv dt behavior, our voltage in the 116 00:08:29,330 --> 00:08:34,760 capacitor will always be continuous. If it wasn't continuous, that would mean 117 00:08:34,760 --> 00:08:39,690 that we had an unbounded current. And since that's not possible, if you 118 00:08:39,690 --> 00:08:42,640 ever get a result where your voltages across your capacitors are jumping 119 00:08:42,640 --> 00:08:46,030 around, you've made a mistake. But it's fine if current does, current 120 00:08:46,030 --> 00:08:47,990 can jump around as much as it likes in a capacitor. 121 00:08:49,660 --> 00:08:55,190 To calculate our energy here, we're basically going to be integrating this 122 00:08:55,190 --> 00:09:01,450 line, and so we know it starts at zero. We're going to assume that, well, maybe 123 00:09:01,450 --> 00:09:06,960 not, that's going to be zero. And as we integrate over this triangle, 124 00:09:06,960 --> 00:09:11,360 this is going to then increase. And since this is a line and we're 125 00:09:11,360 --> 00:09:14,110 integrating it, basically integrating by x. 126 00:09:14,110 --> 00:09:16,490 So, we're going to get something that's and X square, so it's something that's 127 00:09:16,490 --> 00:09:18,540 parabolic. So it's going to parabolically increase 128 00:09:18,540 --> 00:09:23,000 and we can find this final point by integrating the area under that curve. 129 00:09:23,000 --> 00:09:27,696 So, 2 times 2 times one half, which brings us up to 2 joules. 130 00:09:27,696 --> 00:09:31,410 Now one of the problems that my students often made is that as this power drops 131 00:09:31,410 --> 00:09:35,755 down to zero here, they would say, okay great, that means my energy comes down to 132 00:09:35,755 --> 00:09:40,280 zero. It does not, remember, energy is 133 00:09:40,280 --> 00:09:43,380 something that has to be continuous. You can't instantaneously change energy, 134 00:09:43,380 --> 00:09:46,130 because to do so would require unbounded power. 135 00:09:47,300 --> 00:09:51,090 It's just going to go flat. This power being zero just in just means 136 00:09:51,090 --> 00:09:53,830 that no power's being generated or consumed. 137 00:09:53,830 --> 00:09:56,120 So it goes flat here, and then this is basically backwards. 138 00:09:56,120 --> 00:10:00,930 And this is now going to parabolically decrease until it reaches back at zero. 139 00:10:00,930 --> 00:10:04,240 Since the area under this curve is the same as the area under this one. 140 00:10:07,040 --> 00:10:13,130 In order to check our work, we remember that w is equal to one half, Cd squared. 141 00:10:13,130 --> 00:10:18,400 [INAUDIBLE] C here is 1, and we do one half times all these values squared. 142 00:10:18,400 --> 00:10:24,168 We'll see that this, if you square it and divide it by 2, gives us this as well. 143 00:10:24,168 --> 00:10:28,239 And so we come at it from two different directions, we get the same answer. 144 00:10:28,239 --> 00:10:32,060 It's a great way to check your work. And if they match, you probably did it 145 00:10:32,060 --> 00:10:38,130 correctly. because, so to summarize, we calculated 146 00:10:38,130 --> 00:10:41,600 capacitance for capacitors in both parallel and series configurations, and 147 00:10:41,600 --> 00:10:44,250 did the derivations. Now, the derivations were kind of gone 148 00:10:44,250 --> 00:10:47,560 over very quickly, I put all of the, the steps there. 149 00:10:47,560 --> 00:10:51,460 But if you got lost by any of those derivations, it might be a good thing for 150 00:10:51,460 --> 00:10:53,260 you to go to the forums. And ask questions there, to make sure 151 00:10:53,260 --> 00:10:54,645 that you really understand what we were doing. 152 00:10:54,645 --> 00:11:00,220 We also identified how capacitors in DC circuits behave like open circuits. 153 00:11:00,220 --> 00:11:04,700 We derived an equation for energy stored in a capacitor as an electric field. 154 00:11:04,700 --> 00:11:08,140 And then we showed graphically the relationships between our voltage, our 155 00:11:08,140 --> 00:11:11,820 current, our power and energy in capacitors. 156 00:11:14,060 --> 00:11:16,992 Now that we've talked about capacitors, the next topic of discussion will be 157 00:11:16,992 --> 00:11:20,160 inductors. Where capacitors hold their energies as 158 00:11:20,160 --> 00:11:23,170 electric fields, we're now going to be looking at magnetic fields. 159 00:11:23,170 --> 00:11:27,590 And so we're going to be talking a little bit about how electricity and magnetism 160 00:11:27,590 --> 00:11:29,890 interreact. and that allow us to start talking about 161 00:11:29,890 --> 00:11:32,849 something called inductance, so until then