1 00:00:02,930 --> 00:00:04,890 Welcome back to our course on linear circuits. 2 00:00:04,890 --> 00:00:08,119 And today, we're going to be talking about, a couple of more advanced 3 00:00:08,119 --> 00:00:10,062 techniques. The y delta transform and the Wheatstone 4 00:00:10,062 --> 00:00:14,120 Bridge. And so, it's kind of two separate things 5 00:00:14,120 --> 00:00:16,810 that we're going to be looking at. But they kind of have some relation to 6 00:00:16,810 --> 00:00:18,930 each other. And we'll see where that is. 7 00:00:18,930 --> 00:00:22,220 From the previous lesson we talked about how we systematically obtain circuit 8 00:00:22,220 --> 00:00:25,200 equations and then the maximum power transfer. 9 00:00:25,200 --> 00:00:28,330 Finding the load resistance that gives the best power transfer. 10 00:00:28,330 --> 00:00:31,300 So this is only mildly related to some of this previous material. 11 00:00:33,420 --> 00:00:37,690 After we've completed this lesson there will be an application showing something 12 00:00:37,690 --> 00:00:40,630 about using Wheatstone bridges and using sensors. 13 00:00:40,630 --> 00:00:46,010 So there will be some more demos that we'll complete this module.The objectives 14 00:00:46,010 --> 00:00:50,220 of this lesson are to transform resistor circuits between wye configurations and 15 00:00:50,220 --> 00:00:54,680 delta configurations. To specify a test resistor which balances 16 00:00:54,680 --> 00:00:57,870 a Wheatstone bridge. And identify whether the resistor under 17 00:00:57,870 --> 00:01:00,940 test in a Wheatstone bridge is below or above the target resistance. 18 00:01:02,400 --> 00:01:03,999 So, first of all, the Wye-Delta Transform. 19 00:01:05,320 --> 00:01:11,640 If I have a circuit, a system where I have three resistors that kind of come 20 00:01:11,640 --> 00:01:18,460 into point, I can take this whole set and replace it with three resistors in kind 21 00:01:18,460 --> 00:01:23,360 of this triangle configuration. This is called a Wye-delta transformation 22 00:01:23,360 --> 00:01:28,870 or sometimes a Y-D transformation since the y, this kind of looks like a y, this 23 00:01:28,870 --> 00:01:32,340 kind of looks like Greek letter delta, which is just a triangle. 24 00:01:34,230 --> 00:01:40,572 In this system, this a corresponds to this a, this b to this b, this c to this 25 00:01:40,572 --> 00:01:44,920 c. And then we add an extra node d in the y 26 00:01:44,920 --> 00:01:49,520 configuration. You can use these equations to take you 27 00:01:49,520 --> 00:01:54,240 from a Wye into a delta and so the equations are all listed here. 28 00:01:54,240 --> 00:01:55,712 You can actually also go the other direction. 29 00:01:55,712 --> 00:02:00,850 If you have a delta and you want a wye you use this set of equations. 30 00:02:00,850 --> 00:02:02,520 Now I'm going to skip over the derivation. 31 00:02:02,520 --> 00:02:05,700 It takes a little while but if you're interested we can make those available as 32 00:02:05,700 --> 00:02:10,560 resources to you. But we will make use of the actual, of 33 00:02:10,560 --> 00:02:14,060 these actual equations and use it to solve and do a transformation. 34 00:02:15,240 --> 00:02:18,190 So suppose we have this circuit and we want to know the equivalent resistance 35 00:02:18,190 --> 00:02:22,840 from the top node to the bottom node. So we have several different things here 36 00:02:22,840 --> 00:02:28,890 and we'll recognize that nothing here is actually in parallel or in series. 37 00:02:29,960 --> 00:02:32,920 So we cant use the parallel and series combinations we were using before. 38 00:02:32,920 --> 00:02:39,350 However one method to use would be to stick a known source, so in this case I 39 00:02:39,350 --> 00:02:42,590 use a one amp current source and then measure the voltage that we see. 40 00:02:42,590 --> 00:02:45,810 And then you can use other analysis techniques that have already been 41 00:02:45,810 --> 00:02:48,690 presented. But since we now have the delta transform 42 00:02:48,690 --> 00:02:53,678 we can take this top triangle, which is a delta and we can turn it into a y. 43 00:02:53,678 --> 00:02:58,066 So we can call this node a, this node b, this node c. 44 00:02:58,066 --> 00:03:07,841 So transforming it, going to get a Oops. The resistor there. 45 00:03:07,841 --> 00:03:27,383 Now this is b and this is c. Now this 210 ohm resistor is part of the 46 00:03:27,383 --> 00:03:31,480 delta. So you see we no, don't have it there 47 00:03:31,480 --> 00:03:36,314 anymore. But these two resistors, the 10 and the 48 00:03:36,314 --> 00:03:41,164 30. So that's 10, that's 30. 49 00:03:41,164 --> 00:03:48,266 Now we need to figure out these three resistances. 50 00:03:48,266 --> 00:03:57,000 So for the top one, this resistor using the equations that were presented before 51 00:03:57,000 --> 00:04:19,630 that r is going to be equal to 168 times 63 divided by the sum of 168, 63 and 210. 52 00:04:19,630 --> 00:04:31,620 So calculating the out the solution is equal to 24. 53 00:04:31,620 --> 00:04:37,020 To find the resistor here you take 168 times 63 divided by the sum. 54 00:04:37,020 --> 00:04:39,590 That gives us 24. To find b we're going to basically do the 55 00:04:39,590 --> 00:04:44,760 same thing. Now it's 168 times 210 over the sum of 56 00:04:44,760 --> 00:04:50,170 all of these. So doing that calculation, we see that b 57 00:04:50,170 --> 00:04:59,654 is going to be 80 ohms. And with c, 63 times 210 over the sum 58 00:04:59,654 --> 00:05:04,630 giving us 30. One we've gotten it in this configuration 59 00:05:04,630 --> 00:05:09,289 we see that it becomes very simple. Now, this 80 ohm resistor and this 10 ohm 60 00:05:09,289 --> 00:05:14,430 resistor are in series so that becomes an equivalent 90 ohm's. 61 00:05:14,430 --> 00:05:19,200 The 30 and the 30 are in series becoming an equivalent 60 ohm's. 62 00:05:19,200 --> 00:05:24,740 Combining these two together and this 24 ohm here, we discover that the final 63 00:05:24,740 --> 00:05:33,940 resistance is 60 ohms. And so now instead of having to do one of 64 00:05:33,940 --> 00:05:38,330 these other analysis techniques, we were able to just combine resistors and by 65 00:05:38,330 --> 00:05:40,245 doing a transform and simplifying a system. 66 00:05:40,245 --> 00:05:43,490 Now it turns out that this does have some application. 67 00:05:43,490 --> 00:05:49,350 You might say, you might think this is a little bit a bit of work for what we're 68 00:05:49,350 --> 00:05:51,590 doing with it. But it turns out that in certain 69 00:05:51,590 --> 00:05:55,670 circumstances particularly with power, three-phase power, power generation, 70 00:05:55,670 --> 00:05:59,850 power consumption that these types of configurations are very common. 71 00:05:59,850 --> 00:06:04,190 And being able to convert from a delta to a y back and forth can make the analysis 72 00:06:04,190 --> 00:06:09,060 a lot easier. The other thing we're going to talk about 73 00:06:09,060 --> 00:06:12,480 is a Wheatstone Bridge. A Wheatstone Bridge is a device that is 74 00:06:12,480 --> 00:06:16,640 used to measure a resistance that's under test. 75 00:06:16,640 --> 00:06:21,660 So in this case, RX here is the resist, resistor we wish to test. 76 00:06:23,350 --> 00:06:29,184 So we have R1, R2 and R3. These little arrows, going through R2 and 77 00:06:29,184 --> 00:06:32,150 R3 indicate that these are resistors that we can change. 78 00:06:32,150 --> 00:06:35,205 Sometimes referred to as potenciometers. Where we, we can change what the 79 00:06:35,205 --> 00:06:37,470 resistances and we can measure what it is. 80 00:06:37,470 --> 00:06:42,100 We know what it is. There's some way of identifying it. 81 00:06:42,100 --> 00:06:47,276 The way that we do this type of measurement and test these Wheatstone 82 00:06:47,276 --> 00:06:51,830 bridges by taking a measurement here demonstrated by this little graphic. 83 00:06:53,340 --> 00:06:55,910 That measures the current that goes across what's called the bridge. 84 00:06:57,020 --> 00:07:00,740 This wire right here. We're going to say that this bridge is 85 00:07:00,740 --> 00:07:06,330 balanced when this current is equal to 0. No current is flowing across the gap. 86 00:07:06,330 --> 00:07:11,240 So what do we need to do to I get that current to be 0. 87 00:07:11,240 --> 00:07:19,300 Well firstly, what we need is for this voltage, which it will be 1. 88 00:07:19,300 --> 00:07:23,739 And this voltage, it will be 2. With a voltage across R3 and the voltage 89 00:07:23,739 --> 00:07:28,490 across RX have to be equal. Because when they're equal. 90 00:07:28,490 --> 00:07:31,880 Using Kirchhoff's voltage law we see that there's zero voltage drop between these 91 00:07:31,880 --> 00:07:35,350 two places. There's going to be no, no pressure for 92 00:07:35,350 --> 00:07:37,488 this to try and drive this current across the bridge. 93 00:07:37,488 --> 00:07:40,640 When I'm, discovering what type of scenario I want to get to see that 94 00:07:40,640 --> 00:07:47,806 through, you can look at this like a voltage divider. 95 00:07:47,806 --> 00:07:51,750 So ther's no vol, no currents going across this bridge at this point. 96 00:07:51,750 --> 00:07:56,710 You end up having two independent voltage divider circuits. 97 00:07:56,710 --> 00:08:01,160 Since I know that the voltage from this top node to this bottom node is specified 98 00:08:01,160 --> 00:08:06,180 by the s. So, consequently V1 using a voltage 99 00:08:06,180 --> 00:08:17,470 divider is equal to R3 divided by our 1 plus our 3 times the s and the 2 is going 100 00:08:17,470 --> 00:08:28,040 to be equal to Rx over R2 plus Rx. And then I know that they have to be 101 00:08:28,040 --> 00:08:30,957 equal. So by setting these two things equal and 102 00:08:30,957 --> 00:08:37,830 then, we can, factorise the VS's and doing a little bit of algebra. 103 00:08:37,830 --> 00:08:43,910 We get the result that R1 over R2 is equal to R3 over Rx. 104 00:08:43,910 --> 00:08:49,590 So when this happens, then this bridge said to be balanced, no currents going to 105 00:08:49,590 --> 00:08:53,530 flow across. And, because our x is an unknown value. 106 00:08:53,530 --> 00:08:59,440 All we do is we take the know values, R1, R2, and R3 and then solve for Rx because 107 00:08:59,440 --> 00:09:03,400 we've already seen in practice that this current is 0 we know what we're 108 00:09:03,400 --> 00:09:12,130 expecting. If this resistance is smaller than the 109 00:09:12,130 --> 00:09:20,000 balanced resistance, there's going to be some current flowing across this bridge, 110 00:09:20,000 --> 00:09:22,190 down this way. And so with this reference direction 111 00:09:22,190 --> 00:09:29,540 there'll be a positive current flowing. If it's larger, there's going to be some 112 00:09:29,540 --> 00:09:34,790 current flowing through here across the bridge this way. 113 00:09:34,790 --> 00:09:38,650 So with this reference direction, we would measure it in negative current 114 00:09:38,650 --> 00:09:46,760 flowing across the bridge. So in summary we showed the delta y 115 00:09:46,760 --> 00:09:50,260 transform or the y delta transform, to simplify circuits. 116 00:09:50,260 --> 00:09:53,900 And we balanced a Wheatstone bridge and identified as we change the test resistor 117 00:09:53,900 --> 00:09:59,138 if we see a positive current or negative current what that means for our resister 118 00:09:59,138 --> 00:10:04,250 under test. In the next lesson, we'll be able to look 119 00:10:04,250 --> 00:10:09,045 at an application of resisters in sensors as well do an experiment showing a 120 00:10:09,045 --> 00:10:11,220 Wheatstone bridge. And see exactly how they're used in 121 00:10:11,220 --> 00:10:15,710 practice which is a much better way of just looking at it from an analysis 122 00:10:15,710 --> 00:10:19,270 perspective because you can actually see how Wheatstone bridge can be used in 123 00:10:19,270 --> 00:10:23,920 practice. So after those two lab demos that will 124 00:10:23,920 --> 00:10:28,130 conclude module two and so I will see you again on the forums and in module three.