1 00:00:02,750 --> 00:00:05,900 Welcome back. again linear circuits today. 2 00:00:05,900 --> 00:00:08,490 It's going to be maximum power transform, transfer. 3 00:00:08,490 --> 00:00:12,970 This is actually something that you might be able to apply, and use in some of the 4 00:00:12,970 --> 00:00:16,590 things that you actually might be familiar with in electric, electronics. 5 00:00:16,590 --> 00:00:19,470 To do this we're going to be using the feminine equivalent that was presented 6 00:00:19,470 --> 00:00:23,380 before by Dr. Parrish to show the load that gives the 7 00:00:23,380 --> 00:00:26,855 maximum power transfer, we'll also talk about why this is important. 8 00:00:26,855 --> 00:00:33,360 From the previous lesson, there were systematic ways of getting circuit 9 00:00:33,360 --> 00:00:38,178 equations using node-voltage and mesh-current analysis and then Dr. 10 00:00:38,178 --> 00:00:42,220 Parrish presented Thevenin and Norton equivalent circuits where we can take a 11 00:00:42,220 --> 00:00:46,590 whole bunch of components, dependent sources, independent sources, resistors. 12 00:00:46,590 --> 00:00:52,370 Put them all together and get a singe independent source and a single resistor 13 00:00:52,370 --> 00:00:55,930 or a single current source and a single resistor to represent the whole thing. 14 00:00:58,520 --> 00:01:01,996 So today we're going to be doing the next power transfer the next time Wye-delta 15 00:01:01,996 --> 00:01:06,410 and Wheatstone bridge and then finally some demos and applications to conclude 16 00:01:06,410 --> 00:01:09,870 the module. So in the lesson we're going to find the 17 00:01:09,870 --> 00:01:13,760 load resistance that gives maximum power transfer and then calculate what the 18 00:01:13,760 --> 00:01:16,560 power is. How much power is consumed by the load as 19 00:01:16,560 --> 00:01:18,970 well as how much power is being consumed by the system. 20 00:01:21,450 --> 00:01:23,595 So here we have a two terminal linear circuit. 21 00:01:23,595 --> 00:01:26,900 And we know from what Doctor Parrish presented, will be containing the whole 22 00:01:26,900 --> 00:01:29,820 thing and replace it by a thevenin equivalent. 23 00:01:29,820 --> 00:01:33,344 Because we can do this, we're going to just do an analysis using a thevenin 24 00:01:33,344 --> 00:01:36,570 equivalent circuit. And then we'll know that the same results 25 00:01:36,570 --> 00:01:41,830 are going to then apply more generally to any maneur circuit that we might be 26 00:01:41,830 --> 00:01:45,890 working with. But before we get into that we need to 27 00:01:45,890 --> 00:01:50,510 understand how to power for resistance just to make our calculations a little 28 00:01:50,510 --> 00:01:53,040 bit more simpler. So, from what we know, we already know 29 00:01:53,040 --> 00:01:57,050 Ohm's law that v equals i times r. We also know that power is equal to the 30 00:01:57,050 --> 00:02:00,240 current times the voltage. By putting these two ideas together we 31 00:02:00,240 --> 00:02:04,300 discover that power can be calculated by i squared times r. 32 00:02:04,300 --> 00:02:07,180 Or, by taking v squared and dividing it by r. 33 00:02:11,230 --> 00:02:14,450 So, this is what we are talking about when we talk about load resistance. 34 00:02:14,450 --> 00:02:21,540 We have a system, represented by the themanin voltage source in the themanin 35 00:02:21,540 --> 00:02:25,630 resistance here. And then, we connect a load here. 36 00:02:27,040 --> 00:02:30,140 Now, suppose that this load resistance is very, very, very large. 37 00:02:31,220 --> 00:02:34,620 Well, what do we know about that? Well, we know that this resistance is 38 00:02:34,620 --> 00:02:38,120 very very large. First of all, from a voltage divider 39 00:02:38,120 --> 00:02:43,892 stand point, most of this v thevenin is going to be across rl. 40 00:02:43,892 --> 00:02:52,170 But, a very large resistance here is going to give a very, very small current. 41 00:02:53,230 --> 00:02:58,120 And so p is equal to v times i, the result is going to be a small power 42 00:02:58,120 --> 00:03:03,670 consumed by rl. If we look at our thevenin in the same 43 00:03:03,670 --> 00:03:06,760 configuration, our thevenin. The voltage across it is going to become 44 00:03:06,760 --> 00:03:09,560 very small. And this current is also going to become 45 00:03:09,560 --> 00:03:13,520 very, very small. So this resistance is also going to be 46 00:03:13,520 --> 00:03:18,120 consuming very little power. Now suppose that this resistance is very, 47 00:03:18,120 --> 00:03:23,600 very, very small. That this load voltage is going to become 48 00:03:23,600 --> 00:03:28,975 very, very small. Well this voltage here across the 49 00:03:28,975 --> 00:03:30,840 thevenin resistance is going to become fairly large. 50 00:03:31,890 --> 00:03:38,100 But, the current, the power that we are really interested in is this. 51 00:03:38,100 --> 00:03:43,230 So this current here is going to become larger since this is a small resistance. 52 00:03:43,230 --> 00:03:46,220 But the power consumed by the load is going to be fairly small. 53 00:03:46,220 --> 00:03:50,805 So most of the power is going to be consumed by this resistor. 54 00:03:50,805 --> 00:03:53,145 Which isn't really doing what we want either. 55 00:03:53,145 --> 00:03:56,210 So it turns out that there's this sweet spot. 56 00:03:56,210 --> 00:03:59,800 We don't want this resistance to be too large, making the current small and the 57 00:03:59,800 --> 00:04:02,660 power small. And we don't want it to be too small, 58 00:04:02,660 --> 00:04:05,650 making this voltage small. And so the, most of the power is being 59 00:04:05,650 --> 00:04:08,920 consumed here, in this resistor, where the power is being consumed in the 60 00:04:08,920 --> 00:04:12,890 system. So let's derive which resistance we want. 61 00:04:15,950 --> 00:04:21,880 We know that p equals i squared r. And so we can then do this voltage 62 00:04:21,880 --> 00:04:25,250 divider. Finding the voltage drop across this 63 00:04:25,250 --> 00:04:32,742 resistance, to get this equation, uth squared over rth plus rl quantity 64 00:04:32,742 --> 00:04:37,890 squared, and multiplied by rl. And when you do that it turns out that 65 00:04:37,890 --> 00:04:42,140 you can take a derivative and find where this is equal to 0 and do some analysis. 66 00:04:42,140 --> 00:04:45,640 But the, it turns out that the best value that you can pick that's going to 67 00:04:45,640 --> 00:04:51,054 maximize this p, which is the power that's consumed in the load, is equal to 68 00:04:51,054 --> 00:04:57,850 vth squared over 4 times rth. And this happens when rl equals rth. 69 00:04:57,850 --> 00:05:01,850 So the answers to the, the questions that were posed before, is, what load 70 00:05:01,850 --> 00:05:05,455 resistance is the optimal one to use if you want maximum power transfer? 71 00:05:05,455 --> 00:05:09,270 Well, you want it to match the thevenin resistance, the resistance here. 72 00:05:10,390 --> 00:05:13,945 And how much power is consumed by the load resistance in that configuration? 73 00:05:13,945 --> 00:05:20,090 Its calculated by this equation, vth squared over 4 times rth. 74 00:05:20,090 --> 00:05:27,285 It turns out that in that system because now this voltage is equal to this voltage 75 00:05:27,285 --> 00:05:30,970 and obviously being in series their currents are the same. 76 00:05:30,970 --> 00:05:34,680 The same amount of power is being consumed by the load resistance and by 77 00:05:34,680 --> 00:05:37,240 the thevenin resistance. So where can we use this? 78 00:05:37,240 --> 00:05:45,930 If this load resistance is a speaker, for example, and this system here is your 79 00:05:45,930 --> 00:05:49,940 amplifier. You want to choose a speaker that's going 80 00:05:49,940 --> 00:05:54,330 to give maximum power transfer to the speaker, because that equivalate, equates 81 00:05:54,330 --> 00:05:57,288 to more air being moved. More volume. 82 00:05:57,288 --> 00:05:59,700 More power. In your speaker. 83 00:05:59,700 --> 00:06:04,030 So you'd want this, to match this. So you might be familiar with speakers 84 00:06:04,030 --> 00:06:07,975 enough to know that, generally, speakers are labelled as eight ohms. 85 00:06:07,975 --> 00:06:12,060 And you want to make sure that your speakers match up with your amplifier so 86 00:06:12,060 --> 00:06:16,510 that your not wasting extra energy in your system. 87 00:06:16,510 --> 00:06:20,210 That all of the energy well at least as much as you can get is going to the flow 88 00:06:20,210 --> 00:06:23,850 of your system. Now if you pose this question slightly 89 00:06:23,850 --> 00:06:29,240 differently. If we want to know what r thevenin would 90 00:06:29,240 --> 00:06:33,852 get the the most power transferred, turns out that you want to minimize this. 91 00:06:33,852 --> 00:06:39,990 I'm not going to go through the derivation, but it's an excellent 92 00:06:39,990 --> 00:06:44,290 exercise to see why is it that we want to make r thevenin in the smallest possible 93 00:06:44,290 --> 00:06:49,067 to save power and make it, as much of it go to r lotus as we can? 94 00:06:49,067 --> 00:06:57,430 So to summarize, we specified power equations for resistors and we matched 95 00:06:57,430 --> 00:07:02,630 the load resistance for the system resistance to get maximum power transfer. 96 00:07:02,630 --> 00:07:06,750 So as long as you can get a Norton or thevenin equivalent, you can very easily 97 00:07:06,750 --> 00:07:08,950 find out what load resistance you want to use. 98 00:07:08,950 --> 00:07:12,055 We also specified the equation for calculating the maximum power transfer. 99 00:07:12,055 --> 00:07:17,490 In the next lesson we'll talk about a transform that allows us to to work with 100 00:07:17,490 --> 00:07:19,670 resistors that are not in parallel or in series. 101 00:07:19,670 --> 00:07:23,450 They're kind of in the middle ground. As well as analyze what is called the 102 00:07:23,450 --> 00:07:25,600 Wheatstone bridge. Which is a device that is used in 103 00:07:25,600 --> 00:07:28,372 practice for measuring a number of resistances. 104 00:07:28,372 --> 00:07:29,944 Until next time.