1 00:00:01,630 --> 00:00:03,510 Welcome back to our linear cirucuits class. 2 00:00:03,510 --> 00:00:07,630 And today we're going ot be covering the liner model for transformers. 3 00:00:07,630 --> 00:00:10,309 We'll use the Linear Model of Transformers to analyze a transformer 4 00:00:10,309 --> 00:00:12,938 circuit. In the previous lesson we introduced 5 00:00:12,938 --> 00:00:16,688 tranformers and how they work. And then we identified some analysis 6 00:00:16,688 --> 00:00:20,969 models that exist for transformers. Mainly the linear and ideal models. 7 00:00:22,220 --> 00:00:26,000 And the objectives for this lesson are to identify the linear model of transformers 8 00:00:26,000 --> 00:00:29,456 so to be able to express that linear model, to use circuit analysis to analyze 9 00:00:29,456 --> 00:00:33,020 the behavior of a transformer system, and then apply this analysis to solving a 10 00:00:33,020 --> 00:00:40,788 transformer circuit problem. The linear model of transformers treats 11 00:00:40,788 --> 00:00:45,408 each of the two inductors as inductance's, and because we're operating 12 00:00:45,408 --> 00:00:50,930 in AC, you can put this as J omega L1 and J omega L2. 13 00:00:50,930 --> 00:00:55,332 There's also this mutual inductance, j omega M, that relates the one to the 14 00:00:55,332 --> 00:01:00,515 other, and so the way that we do the analysis here is we. 15 00:01:00,515 --> 00:01:05,123 Can make a current here, I1 and a current here, I2, to express the two currents 16 00:01:05,123 --> 00:01:09,443 that are going through each of the portions of the system, and then we can 17 00:01:09,443 --> 00:01:13,475 use Kirchoff's Laws to come up with systems of equations to find the 18 00:01:13,475 --> 00:01:21,450 relationship between these. So here using 19 00:01:21,450 --> 00:01:23,748 [INAUDIBLE] 20 00:01:23,748 --> 00:01:37,157 is going to be equal to I1 MC7 plus I1 times jambic L1. 21 00:01:40,230 --> 00:01:42,586 But the thing is, there's still this mutual inductance term that we need to 22 00:01:42,586 --> 00:01:45,373 account for. Now, remember that these dots are 23 00:01:45,373 --> 00:01:48,380 representative of the reference directions. 24 00:01:48,380 --> 00:01:52,430 Here, my current is going in the dot and out the not dot. 25 00:01:52,430 --> 00:01:55,745 Here I'm going in the not dot and out the dot, so they're referenced in opposite 26 00:01:55,745 --> 00:01:59,740 directions. So, this will be minus I, 2, times, J, 27 00:01:59,740 --> 00:02:04,130 Omega, L, 2. Because it's mutual. 28 00:02:04,130 --> 00:02:13,215 It's based on the current that's moving through, this, conductor. 29 00:02:13,215 --> 00:02:19,150 This should not be L, 2, it should be M. Those based on this [UNKNOWN] M the 30 00:02:19,150 --> 00:02:26,326 mutual invectance based on this current. The other equation for this side, again 31 00:02:26,326 --> 00:02:36,397 using Kurkot's voltage lot is zero is equal to I2 times J omega L2 plus ZL. 32 00:02:37,780 --> 00:02:42,988 But again we have to account for the mutual inductions term minus i1 j omega 33 00:02:42,988 --> 00:02:46,876 n. Again because of the opposite reference 34 00:02:46,876 --> 00:02:50,458 directions of the dots. If this dot were down at the bottom 35 00:02:50,458 --> 00:02:54,550 instead of at the top these then would become pluses, so using this and some 36 00:02:54,550 --> 00:03:01,030 linear algebra we can actually find and solve for the values of i1 and i2. 37 00:03:01,030 --> 00:03:04,120 So these are the type set equations that we were using. 38 00:03:04,120 --> 00:03:08,626 And then here's an expression of i1. It's a fairly lengthy and convoluted 39 00:03:08,626 --> 00:03:11,892 equation. But it turns out that it might sometimes 40 00:03:11,892 --> 00:03:16,262 be a little bit easier to work with an equivalent impedence. 41 00:03:16,262 --> 00:03:21,150 So if I want to know the impedence. That this source sees in all of this. 42 00:03:22,970 --> 00:03:27,726 That is going to be equal to the [UNKNOWN] voltage divided by i 1. 43 00:03:27,726 --> 00:03:32,742 Since the impedence is defined to be that ratio between the voltage and the, the 44 00:03:32,742 --> 00:03:37,636 current. And putting all of that together, we can 45 00:03:37,636 --> 00:03:42,310 find that z is zq is going to be equivalent to Zth7 plus j omega L1, plus 46 00:03:42,310 --> 00:03:49,540 this term the omega squared M squared over ZL plus j omega L2. 47 00:03:49,540 --> 00:03:54,652 We notice that this zth7 and j omega L1 correspond to this and this piece. 48 00:03:54,652 --> 00:04:00,640 And then this final portion Is sometimes referred to as the reflected impedance. 49 00:04:00,640 --> 00:04:04,727 This is the impedance that results by the impedance on this side being intracted 50 00:04:04,727 --> 00:04:08,814 through the mutual inductance twice, and so this side does actually impact this 51 00:04:08,814 --> 00:04:12,535 side even though it's not directly connected, and so that gives us an idea 52 00:04:12,535 --> 00:04:19,726 of how this type of circuit is analyzed. Now let's actually use these calculations 53 00:04:19,726 --> 00:04:25,475 to solve an example problem. So, consider this example. 54 00:04:25,475 --> 00:04:29,699 If we want to calculate the values we are going to be calculating a few different 55 00:04:29,699 --> 00:04:34,969 things. First of all, we're going to calculate 56 00:04:34,969 --> 00:04:41,103 I1. We're going to calculate V1, calculate 57 00:04:41,103 --> 00:04:48,654 V2, and then we'll calculate I2. Now, to make it a little bit easier, 58 00:04:48,654 --> 00:04:51,519 we're not going to do the whole derivation. 59 00:04:51,519 --> 00:04:54,419 We're just going to make use of the derived values of I1 and I2, so we can 60 00:04:54,419 --> 00:04:58,972 just plug in some values. And so, they're here listed. 61 00:04:58,972 --> 00:05:06,532 And you can do this just by the same analysis that you've already used for 62 00:05:06,532 --> 00:05:12,257 other phasors. So, in doing this we just need to plug in 63 00:05:12,257 --> 00:05:17,806 values for I1. So, on the top we have ZL. 64 00:05:17,806 --> 00:05:29,579 Just 50 minus j100 plus j omega l2 plus j100 divided by z7 plus j omega l1 is 4 65 00:05:29,579 --> 00:05:38,503 plus j2 plus j1. Multiply by zl plus j omega l2, which is 66 00:05:38,503 --> 00:05:42,403 the same thing that's in the numerator here, so 50 minus j 100 plus j 100 is 67 00:05:42,403 --> 00:05:47,221 simply plus 50. And then there's this plus omega square, 68 00:05:47,221 --> 00:05:51,243 n square. We know that this is j omega m is equal 69 00:05:51,243 --> 00:05:54,957 to j 10. So, that means that omega m is 10, so 70 00:05:54,957 --> 00:05:59,539 that should be plus 100. Then all of that gets multiplied by 71 00:05:59,539 --> 00:06:04,007 [UNKNOWN] which is 150, 150 arc zero or just 150. 72 00:06:04,007 --> 00:06:12,071 Putting all of that together, we get a, an I 1 value of 20 minus j10 in amps, 73 00:06:12,071 --> 00:06:22,307 which is equal to 22.4, with an angle of minus 26.6 degrees. 74 00:06:22,307 --> 00:06:24,756 To find I two, we plug into this equation. 75 00:06:24,756 --> 00:06:29,970 So on the top we have j10, that's divided by and it turns out the denominator here 76 00:06:29,970 --> 00:06:42,472 is the same as the denominator for I1. 4 plus j2 plus j1 times 50 plus 100. 77 00:06:42,472 --> 00:06:52,659 All times 150 is going to be 7, giving a value of 4 minus j2 amps or 4.5, an angle 78 00:06:52,659 --> 00:07:02,003 of minus 26.6 degrees. If I want to find my voltage 1. 79 00:07:02,003 --> 00:07:05,720 We have not only this inductance to consider, if I multiply my current times 80 00:07:05,720 --> 00:07:10,683 my impedence, but we also have the mutual inductance to consider. 81 00:07:10,683 --> 00:07:18,363 So it's actually going to be easier for us to calcualte V1 as V7 minus what I'll 82 00:07:18,363 --> 00:07:25,607 call V zed. Which corresponds to the voltage across 83 00:07:25,607 --> 00:07:26,182 the z 84 00:07:26,182 --> 00:07:28,284 [INAUDIBLE]. 85 00:07:28,284 --> 00:07:41,254 So it's going to be equal to 150 minus 4 plus j2 times 20 minus j10. 86 00:07:41,254 --> 00:07:45,408 And typically what you're going to do is convert these each into polar, multiply 87 00:07:45,408 --> 00:07:49,500 them together, convert them back into rectangular to do the addition, but I'll 88 00:07:49,500 --> 00:07:55,311 save the algebra there. What you get is 50 with an angle of zero 89 00:07:55,311 --> 00:07:59,096 degrees false. The 2 can simply be calculated as i2 90 00:07:59,096 --> 00:08:02,326 times zl. If I did i2 times this j omega l, I could 91 00:08:02,326 --> 00:08:07,619 end up getting the wrong value. Because there's also the mutual 92 00:08:07,619 --> 00:08:13,029 inductance to consider. But because these two are in parallel, we 93 00:08:13,029 --> 00:08:17,673 can simply say, b2 is equal to i2 times set l. 94 00:08:17,673 --> 00:08:29,019 This is going to be equal to 50 minus j100 times 4 minus j2, which is equal to 95 00:08:29,019 --> 00:08:40,039 500 [INAUDIBLE] angle of minus 90 degrees. 96 00:08:40,039 --> 00:08:46,018 And so now we know the voltage here v1. the voltage here v2, and then the two 97 00:08:46,018 --> 00:08:49,198 currents as well. So it turns out that you can then 98 00:08:49,198 --> 00:08:53,258 calculate the power by multiplying the the rms values of the, the various things 99 00:08:53,258 --> 00:08:58,081 that use all of the same techniques that we've already used. 100 00:08:58,081 --> 00:09:02,537 to find out how the system works. But one thing that's worth noting is that 101 00:09:02,537 --> 00:09:07,323 v2 has a high voltage with a low current on this side. 102 00:09:07,323 --> 00:09:09,722 Here we have a higher current with a lower voltage. 103 00:09:09,722 --> 00:09:12,680 And that's something that will be common through transformers. 104 00:09:12,680 --> 00:09:16,136 We don't get any more power by sending something through a transformer But it 105 00:09:16,136 --> 00:09:20,665 converts more of the power from being in voltage to being in energy. 106 00:09:20,665 --> 00:09:24,261 To summarize we presented the linear model, and derived the phenomenon of 107 00:09:24,261 --> 00:09:27,857 reflected impedance and then used circuit analysis to analyze an example 108 00:09:27,857 --> 00:09:32,160 transformer circuit. In the next lesson, we will look at the 109 00:09:32,160 --> 00:09:35,900 ideal transformer model and contrast it with the linear transformer model until 110 00:09:35,900 --> 00:09:37,440 then.