WEBVTT 00:00:02.711 --> 00:00:07.719 Let's use this as an example to see how mesh-current 00:00:07.719 --> 00:00:13.059 technique can be used to identify or to calculate, say, 00:00:13.059 --> 00:00:19.190 the voltage across that pair of combination resistors. 00:00:19.190 --> 00:00:22.290 In order to do that we are going to need to know current flowing through the either 00:00:22.290 --> 00:00:23.620 the 15 ohm, the 12 ohm and 00:00:23.620 --> 00:00:28.620 then the voltage across that will be, that current times the resistance. 00:00:28.620 --> 00:00:30.800 So lets get started here. 00:00:30.800 --> 00:00:33.620 Our first step is to identify the meshes. 00:00:33.620 --> 00:00:39.010 As you can see we have three meshes, one here, this one in the center. 00:00:39.010 --> 00:00:40.600 And then one there also. 00:00:41.870 --> 00:00:45.510 Now let's define mesh-currents in each of those meshes. 00:00:45.510 --> 00:00:50.730 We'll call this I1, we'll call this one here 00:00:52.340 --> 00:00:57.920 I2, and call this mesh current right there I3. 00:00:59.480 --> 00:01:02.270 Now, let's look first of all at that left hand mesh. 00:01:02.270 --> 00:01:03.680 It's interesting here. 00:01:03.680 --> 00:01:09.140 We have an independent current source of 100 milliamps flowing in that branch. 00:01:09.140 --> 00:01:14.060 Well, that branch current happens to be the same as this mesh current we have 00:01:14.060 --> 00:01:14.840 identified I1. 00:01:15.930 --> 00:01:17.880 This we know right now. 00:01:17.880 --> 00:01:24.380 Without writing anything else, I one is equal to 100 million or .1 amps. 00:01:25.530 --> 00:01:27.190 That gives us one of our equations, also. 00:01:30.009 --> 00:01:31.140 All right. 00:01:31.140 --> 00:01:34.150 Let's look at the center mesh, here. 00:01:34.150 --> 00:01:41.020 Starting at that point, We've got 60 Ohms times I2- I1 and 00:01:41.020 --> 00:01:47.840 we can certainly say -0.1 because we know what I1 is but let's just go ahead and 00:01:47.840 --> 00:01:51.530 write two additional equations in terms of I1, I2 and I3 so we'll have a system of 00:01:51.530 --> 00:01:55.610 three equations and three unknowns and we can solve it that way. 00:01:55.610 --> 00:02:02.024 So we have 60(I2-I1) +, 00:02:02.024 --> 00:02:09.621 coming across here, 30( I2). 00:02:09.621 --> 00:02:13.300 Now we have a voltage source here. 00:02:13.300 --> 00:02:14.700 We're going from minus to plus. 00:02:14.700 --> 00:02:16.180 That represents a voltage increase. 00:02:16.180 --> 00:02:20.690 Therefore, it will be a -10 volts. 00:02:20.690 --> 00:02:24.642 Coming on down this 15 ohm resistor, we'll have plus 15, 00:02:24.642 --> 00:02:29.715 times the current in that 15 ohm resistor where the current is referenced going 00:02:29.715 --> 00:02:34.809 down cuz that's the direction we're going The current going down is I2-I3. 00:02:38.588 --> 00:02:40.850 And that brings us back to where we started, so 00:02:40.850 --> 00:02:42.480 the sum of those terms equals 0. 00:02:42.480 --> 00:02:48.370 And then finally, the going around the right-hand mesh, we have 15 00:02:48.370 --> 00:02:53.500 times I3- I2 + 2 00:02:53.500 --> 00:02:59.672 times just I3 = 0. 00:02:59.672 --> 00:03:05.180 That gives us three equations with three unknowns. 00:03:05.180 --> 00:03:08.410 I've already pointed out that we know what I1 is but let's just go ahead and 00:03:08.410 --> 00:03:14.550 write them Combing common terms and 00:03:14.550 --> 00:03:17.770 factoring out our mesh current variables. 00:03:17.770 --> 00:03:23.584 So starting with the top equation we have I1, equals .1. 00:03:23.584 --> 00:03:32.160 The second equation factoring out I1 we're left with We've only got one i1 term. 00:03:32.160 --> 00:03:36.220 It's a negative in front of it with a 60, 00:03:36.220 --> 00:03:40.110 it will be a -60 + i2. 00:03:40.110 --> 00:03:46.610 We have three i2 terms. 00:03:46.610 --> 00:03:48.035 we got 60 +30 +15. 00:03:48.035 --> 00:03:49.630 60 +30 +15. 00:03:49.630 --> 00:03:57.990 Then for I3 we have just one I3 term it's got a minus sign and 00:03:57.990 --> 00:04:03.030 being multiplied by 15 so minus 15 and 00:04:03.030 --> 00:04:08.960 then we've got this constant voltage here of a minus 10 we bring to the other side. 00:04:08.960 --> 00:04:11.290 As a plus ten. 00:04:11.290 --> 00:04:14.570 So there's our second equation with the terms combined. 00:04:15.870 --> 00:04:19.730 Now the third equation, the third equation has no I1 terms. 00:04:19.730 --> 00:04:23.074 So lets put a I1 with a zero there to hold the place. 00:04:23.074 --> 00:04:28.576 Plus I2 times the negative 15 00:04:28.576 --> 00:04:34.077 Plus I3 times 15 plus 12 and 00:04:34.077 --> 00:04:40.404 the sum of those terms must equal 0. 00:04:43.098 --> 00:04:45.900 There again are three equations and three unknowns. 00:04:45.900 --> 00:04:47.760 Now let's just go ahead and calculate these. 00:04:47.760 --> 00:04:52.030 Let's see this is gonna be 60 plus 30 is 90 plus 15 is 105. 00:04:52.030 --> 00:04:58.180 And then this here is 15 plus 12 is 27. 00:04:58.180 --> 00:05:04.188 We go ahead and plug that into our calculator either using the solve button 00:05:04.188 --> 00:05:09.790 or the solve facility or 00:05:09.790 --> 00:05:14.390 matrix Algebra, and when we do that, we find that I1 00:05:17.000 --> 00:05:23.713 is equal to, we already knew what I1 was, 100 milliamps, or 0.1 amps. 00:05:23.713 --> 00:05:27.452 I2 = 0.1655 amps. 00:05:30.018 --> 00:05:34.843 And I3 is equal to .092 amps. 00:05:38.921 --> 00:05:42.520 But we weren't asked to determine what the currents were. 00:05:42.520 --> 00:05:44.960 We were asked to find what Vout is. 00:05:44.960 --> 00:05:50.110 Vout is equal to 12 times The current 00:05:50.110 --> 00:05:54.840 flow through that which is our mesh current I three. 00:05:58.520 --> 00:06:01.640 So 12 I three. 00:06:01.640 --> 00:06:04.991 12 times .092 and 00:06:04.991 --> 00:06:10.810 that equals 1.10 4 volts. 00:06:14.136 --> 00:06:16.960 We can determine any other volt for the current that we want to. 00:06:16.960 --> 00:06:24.850 For example, what is the current flowing down through that 60 ohms resistor? 00:06:26.220 --> 00:06:27.626 Let's call it i sub 60. 00:06:29.931 --> 00:06:34.834 Well, I sub 60 Is 00:06:34.834 --> 00:06:39.382 just going to be I1 minus I2, or 00:06:39.382 --> 00:06:43.931 .1 minus 0.1655 Amps and 00:06:43.931 --> 00:06:51.735 of course that then equals negative 0.0655 Amps.