1 00:00:02,080 --> 00:00:03,390 In this next set of videos, 2 00:00:03,390 --> 00:00:08,410 we're going to introduce a circuit analysis method known as mesh analysis. 3 00:00:08,410 --> 00:00:09,640 My name is Lee Brinton, 4 00:00:09,640 --> 00:00:12,430 I'm an Electrical Engineering Instructor at Salt Lake Community College. 5 00:00:13,940 --> 00:00:17,050 In this videos we'll make the distinction between a loop and 6 00:00:17,050 --> 00:00:22,610 a mesh, we'll make distinctions between branch currents and mesh currents 7 00:00:22,610 --> 00:00:27,120 We'll demonstrate the process of analyzing circuits using mesh analysis. 8 00:00:27,120 --> 00:00:31,070 Look at how dependent sources impact our analysis approach. 9 00:00:31,070 --> 00:00:34,430 And we'll also introduce concept of supermeshes. 10 00:00:34,430 --> 00:00:38,050 We're gonna find that a supermesh is analagous to a supernode that we ran into 11 00:00:38,050 --> 00:00:43,620 doing node analysis First of all the difference between a loop and a mesh. 12 00:00:43,620 --> 00:00:46,150 A loop is any closed path. 13 00:00:46,150 --> 00:00:52,550 In this circuit we have a closed path here, a closed path there, 14 00:00:52,550 --> 00:00:55,150 and yet one more closed path. 15 00:00:55,150 --> 00:00:57,600 So this circuit has three different loops. 16 00:00:58,870 --> 00:01:01,120 Let us formally define a mesh now. 17 00:01:01,120 --> 00:01:07,610 A mesh is a closed Path that contains no other closed paths within it. 18 00:01:08,750 --> 00:01:11,588 So in this case, we have a mesh here, 19 00:01:11,588 --> 00:01:17,360 a mesh here, but this outer loop, 20 00:01:17,360 --> 00:01:22,350 because it contains two loops within it, does not qualify as a mesh. 21 00:01:24,660 --> 00:01:28,960 Lets also make the distinction between branch currents and mesh currents. 22 00:01:28,960 --> 00:01:33,320 As we know, a branch current is the current flowing in 23 00:01:33,320 --> 00:01:36,330 any single branch of a circuit. 24 00:01:36,330 --> 00:01:39,030 This circuit has three different branches, there is this branch here, 25 00:01:39,030 --> 00:01:42,880 I sub a, this branch here that has the current to find I sub b, and 26 00:01:42,880 --> 00:01:44,810 this branch here that has the current I sub c. 27 00:01:46,270 --> 00:01:48,300 A mesh current is somewhat different. 28 00:01:49,690 --> 00:01:58,090 A mesh current is something of a contrived or a tool that we had define to 29 00:01:58,090 --> 00:02:03,140 assist in this analysis and to understand it think simply about or 30 00:02:03,140 --> 00:02:08,949 imagine that there was a current That flowed around this loop. 31 00:02:10,449 --> 00:02:12,300 That is a mesh current. 32 00:02:12,300 --> 00:02:18,240 Similarly imagine that there’s a current flowing around this loop. 33 00:02:19,590 --> 00:02:21,320 Those two mesh currents. 34 00:02:21,320 --> 00:02:26,060 We're now gonna see that those two miss current are all that we need to represent 35 00:02:26,060 --> 00:02:30,310 the minimum number of variables that we need to analyze the circuit. 36 00:02:30,310 --> 00:02:34,000 So what the difference between i1 and i2? 37 00:02:34,000 --> 00:02:36,520 Don't they both impact this branch? 38 00:02:36,520 --> 00:02:37,530 And in fact they do. 39 00:02:38,730 --> 00:02:43,840 Is defined as the current going down r3, it's a branch current. 40 00:02:45,020 --> 00:02:47,710 In terms of our mesh currents, I sub b 41 00:02:50,780 --> 00:02:56,500 would equal our mesh current I1 flowing down in the same direction as I sub b 42 00:02:57,660 --> 00:03:03,590 minus the I2 mesh current that's flowing or opposing that. 43 00:03:03,590 --> 00:03:08,560 So I sub b then would equal I1 minus I2. 44 00:03:08,560 --> 00:03:13,580 On the other hand, 45 00:03:13,580 --> 00:03:17,010 I1 is this current flowing through here. 46 00:03:20,150 --> 00:03:23,310 And that current is in fact the branch current Ia. 47 00:03:23,310 --> 00:03:30,010 So we can say, then, that Ia = I1. 48 00:03:30,010 --> 00:03:33,210 Or I guess more appropriately, more intuitively, 49 00:03:33,210 --> 00:03:37,480 we'd say that I1 is simply the branch current Ia. 50 00:03:39,180 --> 00:03:44,540 And similarly over here, this mesh current I2 Is 51 00:03:44,540 --> 00:03:48,070 the same as the branch current I sub c. 52 00:03:48,070 --> 00:03:52,460 So we can then see that I sub 2 is equal to I sub c. 53 00:03:53,880 --> 00:03:57,730 Thus, we can define any branch current in this circuit in terms of 54 00:03:57,730 --> 00:03:58,550 the mesh currents I1 and I2. 55 00:03:58,550 --> 00:04:03,000 Now let's see how we 56 00:04:03,000 --> 00:04:07,970 would go about writing The equations to solve this, 57 00:04:07,970 --> 00:04:12,840 or to analyze this kind of circuit.In node analysis we write KCL equations. 58 00:04:12,840 --> 00:04:15,430 One KCL for each node. 59 00:04:15,430 --> 00:04:18,910 In mesh analysis, we're going to write mesh equations or 60 00:04:20,070 --> 00:04:26,000 KVL equations, summing the voltage drops around the closed loops. 61 00:04:26,000 --> 00:04:27,760 One for each mesh. 62 00:04:27,760 --> 00:04:30,760 So for example, let's start with this left hand mesh, and 63 00:04:30,760 --> 00:04:34,100 write a KVL equation starting right here. 64 00:04:34,100 --> 00:04:39,450 a we go across the voltage source, we're increasing V0 volt, and we're 65 00:04:39,450 --> 00:04:43,009 gonna sum the voltage drops, therefore the voltage increase would be negative. 66 00:04:44,160 --> 00:04:51,880 And we have then -Vo continuing on around here the full stop across that resistor. 67 00:04:51,880 --> 00:04:56,825 In terms of I1 and I2, our mesh currency is going to be, 68 00:04:56,825 --> 00:05:01,154 and because we're going in the direction of I1, 69 00:05:01,154 --> 00:05:05,380 it'll be a plus R1I1 Continuing on down here, 70 00:05:05,380 --> 00:05:10,842 the voltage drop across R3 in terms of mesh currency is going to 71 00:05:10,842 --> 00:05:17,470 be R3 times the current flowing in that direction, which is I1 minus I2. 72 00:05:21,399 --> 00:05:25,500 And the sum of those three terms must equal zero. 73 00:05:25,500 --> 00:05:27,800 Now look at the right-hand mesh. 74 00:05:27,800 --> 00:05:33,030 Starting right here, and let's go in the direction that I2 has referenced. 75 00:05:33,030 --> 00:05:38,220 We have, then, the voltage drop going plus to minus, 76 00:05:38,220 --> 00:05:44,100 going in the voltage, direction of current flow, we will have R3 times 77 00:05:44,100 --> 00:05:51,865 the current going up that branch Which is (I2- I1). 78 00:05:51,865 --> 00:05:56,850 Continuing on around this loop we have the voltage 79 00:05:56,850 --> 00:06:01,786 drop across R2 is going to be + R2 times I2 and 80 00:06:01,786 --> 00:06:05,634 the sum of those two terms then must equal 0 81 00:06:06,980 --> 00:06:10,990 We have two equations with two unknowns, we've got everything we need to solve so 82 00:06:10,990 --> 00:06:13,750 the next step with then simply be to combine like terms. 83 00:06:13,750 --> 00:06:18,180 The first equation, lets factor or combine the I1 terms and factor I1 out. 84 00:06:18,180 --> 00:06:25,200 In doing so we're left with an R1 + R3. 85 00:06:27,440 --> 00:06:30,710 Similarly, for I2, there's only one I2 term. 86 00:06:30,710 --> 00:06:33,664 It's got a -R3 times that. 87 00:06:33,664 --> 00:06:35,415 So -R3 there. 88 00:06:35,415 --> 00:06:40,650 Bringing this -V0 to the other side of the positive, 89 00:06:40,650 --> 00:06:42,861 we have on the left hand side = V0. 90 00:06:42,861 --> 00:06:46,780 The second equation will similarly factor out I1. 91 00:06:46,780 --> 00:06:50,990 When we do, we got a negative R3 92 00:06:52,870 --> 00:06:57,560 now working with I2 we have, what is that? 93 00:06:57,560 --> 00:06:59,640 R2 plus R3. 94 00:06:59,640 --> 00:07:04,050 And the sum of those terms has to equal 0. 95 00:07:04,050 --> 00:07:07,970 So, we have two equations, two unknowns if we knew what R1, R2, and R3 and 96 00:07:07,970 --> 00:07:13,230 V0 are we could plug in those values and we would be able t solve for I1. 97 00:07:13,230 --> 00:07:13,980 And I2. 98 00:07:15,070 --> 00:07:19,190 Given those two values, we could then go back and calculate any branch voltage or 99 00:07:19,190 --> 00:07:22,880 branch currents that we were interested in in this circuit.