1 00:00:02,100 --> 00:00:04,940 Welcome back. Let's do some more linear circuits. 2 00:00:04,940 --> 00:00:08,720 Last time we talked about this idea of resistance and looking at individual 3 00:00:08,720 --> 00:00:11,249 elements. Now we're going to start putting them 4 00:00:11,249 --> 00:00:13,910 together, and we're going to be discussing Kirchhoff's Laws. 5 00:00:13,910 --> 00:00:16,590 So we're going introduce Kirchhoff's Current Law and Kirchhoff's Voltage Law, 6 00:00:16,590 --> 00:00:19,279 and then we'll use it to solve a very simple circuit. 7 00:00:20,940 --> 00:00:25,234 So as I mentioned before in our previous lesson we introduced Ohm's law. 8 00:00:25,234 --> 00:00:27,935 We introduced this idea of resistance and then we calculated it. 9 00:00:27,935 --> 00:00:36,110 And then we found this way of relating the voltages and currents within devices. 10 00:00:36,110 --> 00:00:41,120 So now we're going to look at between devices in a system perspective. 11 00:00:41,120 --> 00:00:45,646 After we've covered this, were going to then be able to start talking about 12 00:00:45,646 --> 00:00:51,610 resistors as actual devices that are intended to exhibit Ohm's Law. 13 00:00:51,610 --> 00:00:54,619 So the objectives of this lesson are to first of all have you be able what 14 00:00:54,619 --> 00:00:57,628 Kirchoff's current law and Voltage law are, and describe this voltage 15 00:00:57,628 --> 00:01:02,980 relationship parallel elements. And current relationship of series 16 00:01:02,980 --> 00:01:06,514 elements as well as to use Kirhhoff's Law to find unknown values in a simple 17 00:01:06,514 --> 00:01:10,408 circut. Kirchoff's Voltage Law basically states 18 00:01:10,408 --> 00:01:14,698 it is, that you make your way around any closed path in a circuit summing up your 19 00:01:14,698 --> 00:01:18,715 voltages. Now the sum has to be zero. 20 00:01:18,715 --> 00:01:21,691 And this is because if you start in one place [INAUDIBLE] And you take a trip 21 00:01:21,691 --> 00:01:24,970 around and then you come back to where you started. 22 00:01:24,970 --> 00:01:29,140 You can't have gained or lost any, kind of, electrical elevation. 23 00:01:30,580 --> 00:01:36,061 So as an example, if I have this pathway here, I can sum up all of my various 24 00:01:36,061 --> 00:01:40,490 voltages. VA, VG, VH and VD. 25 00:01:40,490 --> 00:01:43,980 Because this makes a closed circuit and then when I get back to the beginning. 26 00:01:43,980 --> 00:01:46,070 The total sum of the voltage has to be zero. 27 00:01:46,070 --> 00:01:48,735 And you can go in any direction you want, this one happens to be clockwise, but I 28 00:01:48,735 --> 00:01:56,348 can also do it counterclockwise. Or in this case just flip the signs 29 00:01:56,348 --> 00:01:59,304 rather. And so now instead of having the arrows 30 00:01:59,304 --> 00:02:04,110 go from the pluses to the minuses; now they go from the minuses to the pluses. 31 00:02:04,110 --> 00:02:08,030 So if I sum up the a, the c, the f, the h, and the e, they still all have to 32 00:02:08,030 --> 00:02:12,260 equal 0. And this would be equivalent to, if I 33 00:02:12,260 --> 00:02:18,740 instead changed the direction. The E, the H, the F, the C, and the A. 34 00:02:18,740 --> 00:02:21,241 You see that these are 2 equivalent ways of going around and looking at the 35 00:02:21,241 --> 00:02:27,872 circle. So if I look at parallel circuts. 36 00:02:27,872 --> 00:02:34,160 And they make a nice little loop here. If I do my Kirchhoff's law, I see that VA 37 00:02:34,160 --> 00:02:40,255 minus VB has to be equal to 0. And this is because general trend is 38 00:02:40,255 --> 00:02:44,087 you're following the arrow. You can either choose to use the first 39 00:02:44,087 --> 00:02:46,850 sign that you see or the second sign that you see. 40 00:02:46,850 --> 00:02:55,820 So here I used the first sign. So I used plus vA and minus vB. 41 00:02:55,820 --> 00:02:58,875 When I do that, I could actually put the vB on the other side, and I discover that 42 00:02:58,875 --> 00:03:02,071 anytime you have two parallel elements, Which is to say two elements that connect 43 00:03:02,071 --> 00:03:04,985 the same two nodes that they have to have the same voltage across them due to 44 00:03:04,985 --> 00:03:09,550 Kirchhoff's current, or Kirchhoff's voltage law. 45 00:03:09,550 --> 00:03:14,550 For Kirchhoff's current law we're going to be looking at the currents. 46 00:03:14,550 --> 00:03:17,322 And Kirchhoff's current law states that the sum of all the currents going into a 47 00:03:17,322 --> 00:03:21,350 node must equal the sum of the currents going out of the node. 48 00:03:21,350 --> 00:03:23,438 And remember we've already kind of discussed nodes as being the set of 49 00:03:23,438 --> 00:03:26,144 interconnected wires. So if I color the blue node here and I 50 00:03:26,144 --> 00:03:29,072 give some ref instructions I get the equation that ia plus ib must equal id 51 00:03:29,072 --> 00:03:33,384 plus ie because ia and ib are going in id and e are going out. 52 00:03:33,384 --> 00:03:41,168 I'll look at another node this orange node. 53 00:03:41,168 --> 00:03:47,860 I see that IB plus IF, the two going out must equal IC, the one going in. 54 00:03:49,269 --> 00:03:52,990 Let's look at physical descriptions to why Kirchhoff's Current Law is so 55 00:03:52,990 --> 00:03:59,140 important. What if I had this system, where I have a 56 00:03:59,140 --> 00:04:03,773 one mili amp. Current source and I have two nodes 57 00:04:03,773 --> 00:04:07,927 separated by this one centimeter of distance and I have current that's now 58 00:04:07,927 --> 00:04:12,416 flowing through the system and Kirkov's current law does not hold but that means 59 00:04:12,416 --> 00:04:16,570 that I'm going to be building up charge here as I'm pushing You know pulling 60 00:04:16,570 --> 00:04:24,632 electrons, and pushing electrons through. So this is going to get a charge at a 61 00:04:24,632 --> 00:04:30,380 rate of ten to the negative third times T cololms, so T being in seconds. 62 00:04:30,380 --> 00:04:34,030 But on the other note, these, this charge has to come from somewhere. 63 00:04:34,030 --> 00:04:37,841 It's coming from Q 2. So we get negative ten to the third times 64 00:04:37,841 --> 00:04:43,572 T cololms. And then I can use coulombs law, 65 00:04:43,572 --> 00:04:50,100 remembering that KE is equal to this 8.987 times 10 to the ninth newton meters 66 00:04:50,100 --> 00:04:56,628 squared per coulombs squared, and plugging in these values, I discover that 67 00:04:56,628 --> 00:05:02,850 the strength of the attractive force between These two points is equal to 68 00:05:02,850 --> 00:05:12,521 89.87 times t squared meganewtons. And so let's see what's going on here. 69 00:05:12,521 --> 00:05:17,211 In the first millisecond I have 20 pounds of pressure, or attractive force, pulling 70 00:05:17,211 --> 00:05:21,444 these two together. After one second it's twenty million 71 00:05:21,444 --> 00:05:25,906 pounds, and then two seconds, eighty million pounds, or kilogram. 72 00:05:25,906 --> 00:05:28,538 They're all listed over here. So there's a tremendous amount of force 73 00:05:28,538 --> 00:05:31,290 that's going to be pulling these things together. 74 00:05:32,390 --> 00:05:36,599 And it's this electromagnetic force that motivates Kirchoff's law and why it's so 75 00:05:36,599 --> 00:05:42,158 important. If I place two devices in series in this 76 00:05:42,158 --> 00:05:45,830 configuration, it's possible for us to find another relation, similar to like we 77 00:05:45,830 --> 00:05:50,497 did with the parallel circuits. And when we say series, that means that 78 00:05:50,497 --> 00:05:53,506 there's two devices connected to each other, but there's nothing else 79 00:05:53,506 --> 00:05:59,180 connecting in that middle area. Because they're in series, the currents I 80 00:05:59,180 --> 00:06:02,750 a must equal i b because of Kirchhoff's Current Law. 81 00:06:02,750 --> 00:06:09,130 The current flowing in to that center node, here, has to be equal to the 82 00:06:09,130 --> 00:06:14,019 current flowing out. Now we're going to solve a simple 83 00:06:14,019 --> 00:06:17,577 problem. We have a few nodes and from the things 84 00:06:17,577 --> 00:06:22,360 we have learned so far, we know that P is equal to i times v. 85 00:06:25,160 --> 00:06:26,940 We also have Kirchhoff's Current Law and Kirchhoff's Voltage Law. 86 00:06:26,940 --> 00:06:29,194 So let's find the unknown values. First of all, at A, we have one amp of 87 00:06:29,194 --> 00:06:31,460 current flowing through it, and we have three watts. 88 00:06:39,280 --> 00:06:43,627 The arrow goes from the plus to the minus, which means that, because p equals 89 00:06:43,627 --> 00:06:52,644 i times v, this must be 3 volts. Now I can use Korkov's Voltage Law, and 90 00:06:52,644 --> 00:06:59,070 make this big loop around the outside here, so I get minus 3 volts plus vB plus 91 00:06:59,070 --> 00:07:09,217 5 is equal to 0. As long as the vB is minus 2 volts. 92 00:07:09,217 --> 00:07:13,771 Using Kirchoff's current law we see that, since we have 1 amp of current flowing 93 00:07:13,771 --> 00:07:17,704 this way, and A and B are in series but the arrows now point in opposite 94 00:07:17,704 --> 00:07:23,648 directions. This current must be minus one m. 95 00:07:23,648 --> 00:07:28,100 D and C are two devices that are in parallel with each other and so 96 00:07:28,100 --> 00:07:35,268 Kirchoff's voltage law states that they have to have the same voltage. 97 00:07:35,268 --> 00:07:41,592 That's 5 volts. We also know the power in c is five watts 98 00:07:41,592 --> 00:07:46,110 the arrow goes from the plus to the minus so p equals i times b. 99 00:07:47,440 --> 00:07:50,415 That means that the current flowing through here must be one ampere. 100 00:07:50,415 --> 00:07:55,485 Because we have minus one ampere coming here and one ampere going out here that 101 00:07:55,485 --> 00:08:01,810 would be the same as having. One ampere flowing this way. 102 00:08:01,810 --> 00:08:08,245 And Kurcoff's current law states that the sum of that, that, and that have to be 103 00:08:08,245 --> 00:08:15,249 zero, which means that this current is minus two amps. 104 00:08:15,249 --> 00:08:20,617 So finally the two unknown powers pb Is now equal two [UNKNOWN] plus to the 105 00:08:20,617 --> 00:08:26,337 minus, so minus two plus minus one...two watts...and Pd we have they are going 106 00:08:26,337 --> 00:08:31,969 from the plus to the minus so that equals I times V, 5 times -2 is -10 watts And if 107 00:08:31,969 --> 00:08:37,073 I add all of these up, 3 plus 5, plus 2, plus -10, is 0, which means we have 108 00:08:37,073 --> 00:08:42,617 conservation of power, and that gives us some confidence that our answer is 109 00:08:42,617 --> 00:08:52,913 correct. Finally remembering that if they're 110 00:08:52,913 --> 00:08:57,244 positive powers, we say that they're consuming power, and negative that 111 00:08:57,244 --> 00:09:02,144 they're Generating. This is a source, and the others are 112 00:09:02,144 --> 00:09:05,395 consuming power. And so we've been able to solve all of 113 00:09:05,395 --> 00:09:10,190 the voltages, currents, and powers, for this simple example. 114 00:09:12,690 --> 00:09:15,750 So to summarize, we've covered Kirkov's current law, voltage law. 115 00:09:15,750 --> 00:09:19,065 We applied Kirkov's voltage law, to show that parallel elements have the same 116 00:09:19,065 --> 00:09:21,851 voltage. And use Kirchhoff's Current Law to 117 00:09:21,851 --> 00:09:26,190 identify that series elements have the same currents. 118 00:09:26,190 --> 00:09:28,840 We gave an interesting justification for Kirchhoff's Current Law. 119 00:09:28,840 --> 00:09:31,895 And then solved a simple circuit by using Kirchhoff's Laws and, as well as, the 120 00:09:31,895 --> 00:09:37,622 Power Law that we learned before. In our next lesson we'll be introducing 121 00:09:37,622 --> 00:09:40,395 resistors, that are devices that are specifically intended. 122 00:09:40,395 --> 00:09:44,560 To exhibit Ohm's Law and inhibit current flow. 123 00:09:44,560 --> 00:09:47,011 And then we'll apply Ohm's Law and Kirchhoff's laws together to some 124 00:09:47,011 --> 00:09:50,070 resistor circuits. As always, if you've had any questions on 125 00:09:50,070 --> 00:09:53,423 this material, go to the forums. I look forward to seeing you there and 126 00:09:53,423 --> 00:09:57,319 seeing you in our next lesson. Until then.