[Script Info]
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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.45,0:00:05.83,Default,,0000,0000,0000,,In this video,\Nwe introduce the concept of a super mesh.
Dialogue: 0,0:00:05.83,0:00:08.13,Default,,0000,0000,0000,,Here we have a circuit that\Ninvolves three meshes.
Dialogue: 0,0:00:08.13,0:00:12.98,Default,,0000,0000,0000,,The left-hand mesh, which has a mesh\Ncurrent associated with it of I1,
Dialogue: 0,0:00:12.98,0:00:19.40,Default,,0000,0000,0000,,a mesh here with a mesh current defined of\NI2, and a mesh current here defined of I3.
Dialogue: 0,0:00:19.40,0:00:24.30,Default,,0000,0000,0000,,The problem arises when\Nwe have a branch that
Dialogue: 0,0:00:24.30,0:00:29.08,Default,,0000,0000,0000,,contains an independent current source.
Dialogue: 0,0:00:29.08,0:00:34.04,Default,,0000,0000,0000,,By definition, an independent current\Nsource produces, in this case,
Dialogue: 0,0:00:34.04,0:00:38.20,Default,,0000,0000,0000,,I0 amps of current ,no matter\Nwhat the voltage is across it.
Dialogue: 0,0:00:38.20,0:00:42.48,Default,,0000,0000,0000,,In other words, there is no relationship\Nbetween a current source current and
Dialogue: 0,0:00:42.48,0:00:44.10,Default,,0000,0000,0000,,its voltage.
Dialogue: 0,0:00:44.10,0:00:47.66,Default,,0000,0000,0000,,Let's demonstrate the problem by\Nattempting to write the mesh equation
Dialogue: 0,0:00:47.66,0:00:52.53,Default,,0000,0000,0000,,around this loop right here starting\Nin the lower left-hand corner.
Dialogue: 0,0:00:52.53,0:00:57.27,Default,,0000,0000,0000,,We've got the voltage drop across R2,
Dialogue: 0,0:00:57.27,0:01:02.28,Default,,0000,0000,0000,,which would be R2 x the current through R2
Dialogue: 0,0:01:02.28,0:01:07.03,Default,,0000,0000,0000,,going in this direction, is I2- I1.
Dialogue: 0,0:01:07.03,0:01:12.90,Default,,0000,0000,0000,,Coming down the right-hand side, we have\Nthe voltage drop across R3 is R3 X I2.
Dialogue: 0,0:01:12.90,0:01:17.25,Default,,0000,0000,0000,,And now we get to this independent source.
Dialogue: 0,0:01:17.25,0:01:21.88,Default,,0000,0000,0000,,We're attempting to\Nwrite the voltage drops
Dialogue: 0,0:01:21.88,0:01:26.74,Default,,0000,0000,0000,,around this mesh, but\Nwe have no expression for
Dialogue: 0,0:01:26.74,0:01:31.98,Default,,0000,0000,0000,,the voltage drop across\NI0 in terms of I2 and I3.
Dialogue: 0,0:01:31.98,0:01:33.81,Default,,0000,0000,0000,,How do we resolve this?
Dialogue: 0,0:01:33.81,0:01:38.49,Default,,0000,0000,0000,,Well, to resolve it, we take\Nadvantage of the fact that the sum of
Dialogue: 0,0:01:38.49,0:01:42.66,Default,,0000,0000,0000,,the voltage drops around any\Nclosed loop must equal zero.
Dialogue: 0,0:01:43.73,0:01:48.67,Default,,0000,0000,0000,,So instead of going around each of these\Nmeshes separately, we're going to make
Dialogue: 0,0:01:48.67,0:01:53.81,Default,,0000,0000,0000,,a trip around the outside of\Nboth of these loops combined.
Dialogue: 0,0:01:55.13,0:01:59.55,Default,,0000,0000,0000,,Let's show what we mean here\Nby now analyzing this circuit.
Dialogue: 0,0:01:59.55,0:02:06.36,Default,,0000,0000,0000,,Let's start by writing the loop\Nequation around this left-hand mesh.
Dialogue: 0,0:02:06.36,0:02:11.71,Default,,0000,0000,0000,,We'll have -V0 + R1I1 + now
Dialogue: 0,0:02:11.71,0:02:16.69,Default,,0000,0000,0000,,coming down this branch here,
Dialogue: 0,0:02:16.69,0:02:24.35,Default,,0000,0000,0000,,voltage drop across R2\Nwill be R2 x I1- I2.
Dialogue: 0,0:02:24.35,0:02:29.92,Default,,0000,0000,0000,,Plus the voltage drop across R5,
Dialogue: 0,0:02:29.92,0:02:34.73,Default,,0000,0000,0000,,which will be R5 x I1- I3.
Dialogue: 0,0:02:34.73,0:02:39.59,Default,,0000,0000,0000,,The sum of those voltage drops\Nmust equal 0, nothing new there.
Dialogue: 0,0:02:39.59,0:02:45.36,Default,,0000,0000,0000,,Now let's write the KDL equation\Ngoing around our combined super mesh.
Dialogue: 0,0:02:45.36,0:02:50.66,Default,,0000,0000,0000,,Starting down here in the lower left-hand\Ncorner and going in a clockwise direction,
Dialogue: 0,0:02:50.66,0:02:55.53,Default,,0000,0000,0000,,we'll have voltage drop across R5 will\Nbe R5 times, let's be careful here.
Dialogue: 0,0:02:55.53,0:02:58.50,Default,,0000,0000,0000,,We're going up across R5.
Dialogue: 0,0:02:58.50,0:03:04.30,Default,,0000,0000,0000,,The current going up is I3- I1.
Dialogue: 0,0:03:04.30,0:03:07.48,Default,,0000,0000,0000,,So R5 x I3- I1,
Dialogue: 0,0:03:07.48,0:03:12.86,Default,,0000,0000,0000,,plus the voltage drop across R2
Dialogue: 0,0:03:12.86,0:03:18.04,Default,,0000,0000,0000,,going in this direction will be
Dialogue: 0,0:03:18.04,0:03:24.02,Default,,0000,0000,0000,,I2- I1 + the voltage drop across
Dialogue: 0,0:03:24.02,0:03:29.02,Default,,0000,0000,0000,,R3 is just gonna be R3 x I2.
Dialogue: 0,0:03:29.02,0:03:37.59,Default,,0000,0000,0000,,And finally, we've got the voltage\Ndrop across R6, which will be R6 x I3.
Dialogue: 0,0:03:37.59,0:03:42.09,Default,,0000,0000,0000,,And the sum of those\Nvoltage drops will equal 0.
Dialogue: 0,0:03:42.09,0:03:43.96,Default,,0000,0000,0000,,There, we've got the two equations.
Dialogue: 0,0:03:43.96,0:03:48.56,Default,,0000,0000,0000,,Again, we're gonna get our\Nthird equation by taking
Dialogue: 0,0:03:48.56,0:03:53.78,Default,,0000,0000,0000,,advantage of the fact that\Nthe current in this branch is I0.
Dialogue: 0,0:03:53.78,0:03:58.63,Default,,0000,0000,0000,,And that that current I0,\Nin terms of the two mesh currents,
Dialogue: 0,0:03:58.63,0:04:03.02,Default,,0000,0000,0000,,is equal to the mesh current\Nflowing in this direction,
Dialogue: 0,0:04:03.02,0:04:07.78,Default,,0000,0000,0000,,minus the mesh current\Nflowing in that direction.
Dialogue: 0,0:04:07.78,0:04:15.68,Default,,0000,0000,0000,,Or, I0 will equal I2- I3.
Dialogue: 0,0:04:15.68,0:04:20.63,Default,,0000,0000,0000,,Thus, we have the three equations that we\Nneed to cover the three variables that we
Dialogue: 0,0:04:20.63,0:04:21.28,Default,,0000,0000,0000,,defined.
Dialogue: 0,0:04:22.46,0:04:29.90,Default,,0000,0000,0000,,Again, our first equation came from going\Naround this mesh, as we have always done.
Dialogue: 0,0:04:31.84,0:04:36.82,Default,,0000,0000,0000,,The second equation came from\Ngoing around the super mesh,
Dialogue: 0,0:04:36.82,0:04:41.00,Default,,0000,0000,0000,,effectively avoiding the need\Nto go across the volt or
Dialogue: 0,0:04:41.00,0:04:44.14,Default,,0000,0000,0000,,to know the voltage across\Nthat current source.
Dialogue: 0,0:04:46.56,0:04:50.58,Default,,0000,0000,0000,,And then the third equation came from\Nthe fact that the branch current in this
Dialogue: 0,0:04:50.58,0:04:53.04,Default,,0000,0000,0000,,sources is in fact I0.
Dialogue: 0,0:04:53.04,0:04:59.37,Default,,0000,0000,0000,,And in terms of the mesh currents,\Nthe current in that branch is I2- I3.
Dialogue: 0,0:04:59.37,0:05:03.95,Default,,0000,0000,0000,,Three equations, three unknowns,\Nbe ready to solve it,
Dialogue: 0,0:05:03.95,0:05:09.67,Default,,0000,0000,0000,,using whatever method for solving\Nsystems equations you choose to use.
Dialogue: 0,0:05:10.80,0:05:14.09,Default,,0000,0000,0000,,In the next video, we'll do\Na numerical example of a super mesh.