1 00:00:01,450 --> 00:00:05,829 In this video, we introduce the concept of a super mesh. 2 00:00:05,829 --> 00:00:08,134 Here we have a circuit that involves three meshes. 3 00:00:08,134 --> 00:00:12,982 The left-hand mesh, which has a mesh current associated with it of I1, 4 00:00:12,982 --> 00:00:19,400 a mesh here with a mesh current defined of I2, and a mesh current here defined of I3. 5 00:00:19,400 --> 00:00:24,302 The problem arises when we have a branch that 6 00:00:24,302 --> 00:00:29,081 contains an independent current source. 7 00:00:29,081 --> 00:00:34,041 By definition, an independent current source produces, in this case, 8 00:00:34,041 --> 00:00:38,202 I0 amps of current ,no matter what the voltage is across it. 9 00:00:38,202 --> 00:00:42,478 In other words, there is no relationship between a current source current and 10 00:00:42,478 --> 00:00:44,100 its voltage. 11 00:00:44,100 --> 00:00:47,660 Let's demonstrate the problem by attempting to write the mesh equation 12 00:00:47,660 --> 00:00:52,530 around this loop right here starting in the lower left-hand corner. 13 00:00:52,530 --> 00:00:57,266 We've got the voltage drop across R2, 14 00:00:57,266 --> 00:01:02,281 which would be R2 x the current through R2 15 00:01:02,281 --> 00:01:07,028 going in this direction, is I2- I1. 16 00:01:07,028 --> 00:01:12,896 Coming down the right-hand side, we have the voltage drop across R3 is R3 X I2. 17 00:01:12,896 --> 00:01:17,254 And now we get to this independent source. 18 00:01:17,254 --> 00:01:21,875 We're attempting to write the voltage drops 19 00:01:21,875 --> 00:01:26,738 around this mesh, but we have no expression for 20 00:01:26,738 --> 00:01:31,979 the voltage drop across I0 in terms of I2 and I3. 21 00:01:31,979 --> 00:01:33,813 How do we resolve this? 22 00:01:33,813 --> 00:01:38,488 Well, to resolve it, we take advantage of the fact that the sum of 23 00:01:38,488 --> 00:01:42,660 the voltage drops around any closed loop must equal zero. 24 00:01:43,730 --> 00:01:48,670 So instead of going around each of these meshes separately, we're going to make 25 00:01:48,670 --> 00:01:53,810 a trip around the outside of both of these loops combined. 26 00:01:55,130 --> 00:01:59,550 Let's show what we mean here by now analyzing this circuit. 27 00:01:59,550 --> 00:02:06,358 Let's start by writing the loop equation around this left-hand mesh. 28 00:02:06,358 --> 00:02:11,713 We'll have -V0 + R1I1 + now 29 00:02:11,713 --> 00:02:16,686 coming down this branch here, 30 00:02:16,686 --> 00:02:24,346 voltage drop across R2 will be R2 x I1- I2. 31 00:02:24,346 --> 00:02:29,919 Plus the voltage drop across R5, 32 00:02:29,919 --> 00:02:34,731 which will be R5 x I1- I3. 33 00:02:34,731 --> 00:02:39,591 The sum of those voltage drops must equal 0, nothing new there. 34 00:02:39,591 --> 00:02:45,360 Now let's write the KDL equation going around our combined super mesh. 35 00:02:45,360 --> 00:02:50,655 Starting down here in the lower left-hand corner and going in a clockwise direction, 36 00:02:50,655 --> 00:02:55,529 we'll have voltage drop across R5 will be R5 times, let's be careful here. 37 00:02:55,529 --> 00:02:58,495 We're going up across R5. 38 00:02:58,495 --> 00:03:04,301 The current going up is I3- I1. 39 00:03:04,301 --> 00:03:07,485 So R5 x I3- I1, 40 00:03:07,485 --> 00:03:12,863 plus the voltage drop across R2 41 00:03:12,863 --> 00:03:18,042 going in this direction will be 42 00:03:18,042 --> 00:03:24,016 I2- I1 + the voltage drop across 43 00:03:24,016 --> 00:03:29,017 R3 is just gonna be R3 x I2. 44 00:03:29,017 --> 00:03:37,587 And finally, we've got the voltage drop across R6, which will be R6 x I3. 45 00:03:37,587 --> 00:03:42,086 And the sum of those voltage drops will equal 0. 46 00:03:42,086 --> 00:03:43,960 There, we've got the two equations. 47 00:03:43,960 --> 00:03:48,557 Again, we're gonna get our third equation by taking 48 00:03:48,557 --> 00:03:53,783 advantage of the fact that the current in this branch is I0. 49 00:03:53,783 --> 00:03:58,634 And that that current I0, in terms of the two mesh currents, 50 00:03:58,634 --> 00:04:03,019 is equal to the mesh current flowing in this direction, 51 00:04:03,019 --> 00:04:07,780 minus the mesh current flowing in that direction. 52 00:04:07,780 --> 00:04:15,681 Or, I0 will equal I2- I3. 53 00:04:15,681 --> 00:04:20,627 Thus, we have the three equations that we need to cover the three variables that we 54 00:04:20,627 --> 00:04:21,279 defined. 55 00:04:22,460 --> 00:04:29,900 Again, our first equation came from going around this mesh, as we have always done. 56 00:04:31,840 --> 00:04:36,820 The second equation came from going around the super mesh, 57 00:04:36,820 --> 00:04:41,000 effectively avoiding the need to go across the volt or 58 00:04:41,000 --> 00:04:44,140 to know the voltage across that current source. 59 00:04:46,560 --> 00:04:50,580 And then the third equation came from the fact that the branch current in this 60 00:04:50,580 --> 00:04:53,040 sources is in fact I0. 61 00:04:53,040 --> 00:04:59,368 And in terms of the mesh currents, the current in that branch is I2- I3. 62 00:04:59,368 --> 00:05:03,953 Three equations, three unknowns, be ready to solve it, 63 00:05:03,953 --> 00:05:09,670 using whatever method for solving systems equations you choose to use. 64 00:05:10,800 --> 00:05:14,090 In the next video, we'll do a numerical example of a super mesh.