[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:02.71,0:00:07.72,Default,,0000,0000,0000,,Let's use this as an example\Nto see how mesh-current
Dialogue: 0,0:00:07.72,0:00:13.06,Default,,0000,0000,0000,,technique can be used to identify or\Nto calculate, say,
Dialogue: 0,0:00:13.06,0:00:19.19,Default,,0000,0000,0000,,the voltage across that pair\Nof combination resistors.
Dialogue: 0,0:00:19.19,0:00:22.29,Default,,0000,0000,0000,,In order to do that we are going to need\Nto know current flowing through the either
Dialogue: 0,0:00:22.29,0:00:23.62,Default,,0000,0000,0000,,the 15 ohm, the 12 ohm and
Dialogue: 0,0:00:23.62,0:00:28.62,Default,,0000,0000,0000,,then the voltage across that will be,\Nthat current times the resistance.
Dialogue: 0,0:00:28.62,0:00:30.80,Default,,0000,0000,0000,,So lets get started here.
Dialogue: 0,0:00:30.80,0:00:33.62,Default,,0000,0000,0000,,Our first step is to identify the meshes.
Dialogue: 0,0:00:33.62,0:00:39.01,Default,,0000,0000,0000,,As you can see we have three meshes,\None here, this one in the center.
Dialogue: 0,0:00:39.01,0:00:40.60,Default,,0000,0000,0000,,And then one there also.
Dialogue: 0,0:00:41.87,0:00:45.51,Default,,0000,0000,0000,,Now let's define mesh-currents\Nin each of those meshes.
Dialogue: 0,0:00:45.51,0:00:50.73,Default,,0000,0000,0000,,We'll call this I1,\Nwe'll call this one here
Dialogue: 0,0:00:52.34,0:00:57.92,Default,,0000,0000,0000,,I2, and\Ncall this mesh current right there I3.
Dialogue: 0,0:00:59.48,0:01:02.27,Default,,0000,0000,0000,,Now, let's look first of\Nall at that left hand mesh.
Dialogue: 0,0:01:02.27,0:01:03.68,Default,,0000,0000,0000,,It's interesting here.
Dialogue: 0,0:01:03.68,0:01:09.14,Default,,0000,0000,0000,,We have an independent current source of\N100 milliamps flowing in that branch.
Dialogue: 0,0:01:09.14,0:01:14.06,Default,,0000,0000,0000,,Well, that branch current happens to be\Nthe same as this mesh current we have
Dialogue: 0,0:01:14.06,0:01:14.84,Default,,0000,0000,0000,,identified I1.
Dialogue: 0,0:01:15.93,0:01:17.88,Default,,0000,0000,0000,,This we know right now.
Dialogue: 0,0:01:17.88,0:01:24.38,Default,,0000,0000,0000,,Without writing anything else,\NI one is equal to 100 million or .1 amps.
Dialogue: 0,0:01:25.53,0:01:27.19,Default,,0000,0000,0000,,That gives us one of our equations, also.
Dialogue: 0,0:01:30.01,0:01:31.14,Default,,0000,0000,0000,,All right.
Dialogue: 0,0:01:31.14,0:01:34.15,Default,,0000,0000,0000,,Let's look at the center mesh, here.
Dialogue: 0,0:01:34.15,0:01:41.02,Default,,0000,0000,0000,,Starting at that point,\NWe've got 60 Ohms times I2- I1 and
Dialogue: 0,0:01:41.02,0:01:47.84,Default,,0000,0000,0000,,we can certainly say -0.1 because we know\Nwhat I1 is but let's just go ahead and
Dialogue: 0,0:01:47.84,0:01:51.53,Default,,0000,0000,0000,,write two additional equations in terms of\NI1, I2 and I3 so we'll have a system of
Dialogue: 0,0:01:51.53,0:01:55.61,Default,,0000,0000,0000,,three equations and three unknowns and\Nwe can solve it that way.
Dialogue: 0,0:01:55.61,0:02:02.02,Default,,0000,0000,0000,,So we have 60(I2-I1) +,
Dialogue: 0,0:02:02.02,0:02:09.62,Default,,0000,0000,0000,,coming across here, 30( I2).
Dialogue: 0,0:02:09.62,0:02:13.30,Default,,0000,0000,0000,,Now we have a voltage source here.
Dialogue: 0,0:02:13.30,0:02:14.70,Default,,0000,0000,0000,,We're going from minus to plus.
Dialogue: 0,0:02:14.70,0:02:16.18,Default,,0000,0000,0000,,That represents a voltage increase.
Dialogue: 0,0:02:16.18,0:02:20.69,Default,,0000,0000,0000,,Therefore, it will be a -10 volts.
Dialogue: 0,0:02:20.69,0:02:24.64,Default,,0000,0000,0000,,Coming on down this 15 ohm resistor,\Nwe'll have plus 15,
Dialogue: 0,0:02:24.64,0:02:29.72,Default,,0000,0000,0000,,times the current in that 15 ohm resistor\Nwhere the current is referenced going
Dialogue: 0,0:02:29.72,0:02:34.81,Default,,0000,0000,0000,,down cuz that's the direction we're\Ngoing The current going down is I2-I3.
Dialogue: 0,0:02:38.59,0:02:40.85,Default,,0000,0000,0000,,And that brings us back\Nto where we started, so
Dialogue: 0,0:02:40.85,0:02:42.48,Default,,0000,0000,0000,,the sum of those terms equals 0.
Dialogue: 0,0:02:42.48,0:02:48.37,Default,,0000,0000,0000,,And then finally, the going around\Nthe right-hand mesh, we have 15
Dialogue: 0,0:02:48.37,0:02:53.50,Default,,0000,0000,0000,,times I3- I2 + 2
Dialogue: 0,0:02:53.50,0:02:59.67,Default,,0000,0000,0000,,times just I3 = 0.
Dialogue: 0,0:02:59.67,0:03:05.18,Default,,0000,0000,0000,,That gives us three equations\Nwith three unknowns.
Dialogue: 0,0:03:05.18,0:03:08.41,Default,,0000,0000,0000,,I've already pointed out that we know\Nwhat I1 is but let's just go ahead and
Dialogue: 0,0:03:08.41,0:03:14.55,Default,,0000,0000,0000,,write them Combing common terms and
Dialogue: 0,0:03:14.55,0:03:17.77,Default,,0000,0000,0000,,factoring out our mesh current variables.
Dialogue: 0,0:03:17.77,0:03:23.58,Default,,0000,0000,0000,,So starting with the top\Nequation we have I1, equals .1.
Dialogue: 0,0:03:23.58,0:03:32.16,Default,,0000,0000,0000,,The second equation factoring out I1 we're\Nleft with We've only got one i1 term.
Dialogue: 0,0:03:32.16,0:03:36.22,Default,,0000,0000,0000,,It's a negative in front of it with a 60,
Dialogue: 0,0:03:36.22,0:03:40.11,Default,,0000,0000,0000,,it will be a -60 + i2.
Dialogue: 0,0:03:40.11,0:03:46.61,Default,,0000,0000,0000,,We have three i2 terms.
Dialogue: 0,0:03:46.61,0:03:48.04,Default,,0000,0000,0000,,we got 60 +30 +15.
Dialogue: 0,0:03:48.04,0:03:49.63,Default,,0000,0000,0000,,60 +30 +15.
Dialogue: 0,0:03:49.63,0:03:57.99,Default,,0000,0000,0000,,Then for I3 we have just one I3\Nterm it's got a minus sign and
Dialogue: 0,0:03:57.99,0:04:03.03,Default,,0000,0000,0000,,being multiplied by 15 so minus 15 and
Dialogue: 0,0:04:03.03,0:04:08.96,Default,,0000,0000,0000,,then we've got this constant voltage here\Nof a minus 10 we bring to the other side.
Dialogue: 0,0:04:08.96,0:04:11.29,Default,,0000,0000,0000,,As a plus ten.
Dialogue: 0,0:04:11.29,0:04:14.57,Default,,0000,0000,0000,,So there's our second equation\Nwith the terms combined.
Dialogue: 0,0:04:15.87,0:04:19.73,Default,,0000,0000,0000,,Now the third equation,\Nthe third equation has no I1 terms.
Dialogue: 0,0:04:19.73,0:04:23.07,Default,,0000,0000,0000,,So lets put a I1 with a zero\Nthere to hold the place.
Dialogue: 0,0:04:23.07,0:04:28.58,Default,,0000,0000,0000,,Plus I2 times the negative 15
Dialogue: 0,0:04:28.58,0:04:34.08,Default,,0000,0000,0000,,Plus I3 times 15 plus 12 and
Dialogue: 0,0:04:34.08,0:04:40.40,Default,,0000,0000,0000,,the sum of those terms must equal 0.
Dialogue: 0,0:04:43.10,0:04:45.90,Default,,0000,0000,0000,,There again are three equations and\Nthree unknowns.
Dialogue: 0,0:04:45.90,0:04:47.76,Default,,0000,0000,0000,,Now let's just go ahead and\Ncalculate these.
Dialogue: 0,0:04:47.76,0:04:52.03,Default,,0000,0000,0000,,Let's see this is gonna be 60\Nplus 30 is 90 plus 15 is 105.
Dialogue: 0,0:04:52.03,0:04:58.18,Default,,0000,0000,0000,,And then this here is 15 plus 12 is 27.
Dialogue: 0,0:04:58.18,0:05:04.19,Default,,0000,0000,0000,,We go ahead and plug that into our\Ncalculator either using the solve button
Dialogue: 0,0:05:04.19,0:05:09.79,Default,,0000,0000,0000,,or the solve facility or
Dialogue: 0,0:05:09.79,0:05:14.39,Default,,0000,0000,0000,,matrix Algebra, and\Nwhen we do that, we find that I1
Dialogue: 0,0:05:17.00,0:05:23.71,Default,,0000,0000,0000,,is equal to, we already knew what I1 was,\N100 milliamps, or 0.1 amps.
Dialogue: 0,0:05:23.71,0:05:27.45,Default,,0000,0000,0000,,I2 = 0.1655 amps.
Dialogue: 0,0:05:30.02,0:05:34.84,Default,,0000,0000,0000,,And I3 is equal to .092 amps.
Dialogue: 0,0:05:38.92,0:05:42.52,Default,,0000,0000,0000,,But we weren't asked to determine\Nwhat the currents were.
Dialogue: 0,0:05:42.52,0:05:44.96,Default,,0000,0000,0000,,We were asked to find what Vout is.
Dialogue: 0,0:05:44.96,0:05:50.11,Default,,0000,0000,0000,,Vout is equal to 12 times The current
Dialogue: 0,0:05:50.11,0:05:54.84,Default,,0000,0000,0000,,flow through that which is\Nour mesh current I three.
Dialogue: 0,0:05:58.52,0:06:01.64,Default,,0000,0000,0000,,So 12 I three.
Dialogue: 0,0:06:01.64,0:06:04.99,Default,,0000,0000,0000,,12 times .092 and
Dialogue: 0,0:06:04.99,0:06:10.81,Default,,0000,0000,0000,,that equals 1.10 4 volts.
Dialogue: 0,0:06:14.14,0:06:16.96,Default,,0000,0000,0000,,We can determine any other volt for\Nthe current that we want to.
Dialogue: 0,0:06:16.96,0:06:24.85,Default,,0000,0000,0000,,For example, what is the current flowing\Ndown through that 60 ohms resistor?
Dialogue: 0,0:06:26.22,0:06:27.63,Default,,0000,0000,0000,,Let's call it i sub 60.
Dialogue: 0,0:06:29.93,0:06:34.83,Default,,0000,0000,0000,,Well, I sub 60 Is
Dialogue: 0,0:06:34.83,0:06:39.38,Default,,0000,0000,0000,,just going to be I1 minus I2, or
Dialogue: 0,0:06:39.38,0:06:43.93,Default,,0000,0000,0000,,.1 minus 0.1655 Amps and
Dialogue: 0,0:06:43.93,0:06:51.74,Default,,0000,0000,0000,,of course that then equals\Nnegative 0.0655 Amps.