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