[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.78,0:00:05.33,Default,,0000,0000,0000,,In this video, we're gonna\Ndemonstrate the super mesh technique.
Dialogue: 0,0:00:05.33,0:00:06.76,Default,,0000,0000,0000,,As you look at this,
Dialogue: 0,0:00:06.76,0:00:11.45,Default,,0000,0000,0000,,you'll notice that there are three meshes,\Nthis one here to the left.
Dialogue: 0,0:00:11.45,0:00:15.40,Default,,0000,0000,0000,,And we'll label the current\Nin that mesh current I1.
Dialogue: 0,0:00:15.40,0:00:18.85,Default,,0000,0000,0000,,We have a mesh up here.
Dialogue: 0,0:00:18.85,0:00:21.79,Default,,0000,0000,0000,,And we'll call this mesh current I2, and
Dialogue: 0,0:00:21.79,0:00:26.37,Default,,0000,0000,0000,,we have a mesh here with a mesh\Ncurrent that we'll label I3.
Dialogue: 0,0:00:27.83,0:00:32.53,Default,,0000,0000,0000,,And further before we even get started\Nwe'll notice that separating mesh one from
Dialogue: 0,0:00:32.53,0:00:36.37,Default,,0000,0000,0000,,mesh three is a branch that\Ncontains a current source.
Dialogue: 0,0:00:36.37,0:00:39.18,Default,,0000,0000,0000,,And as we've mentioned in the past,\Nwe don't have a.
Dialogue: 0,0:00:39.18,0:00:44.25,Default,,0000,0000,0000,,A numerical or mathematical\Nrelationship between the voltage across
Dialogue: 0,0:00:44.25,0:00:46.24,Default,,0000,0000,0000,,a current source and the current itself.
Dialogue: 0,0:00:46.24,0:00:51.16,Default,,0000,0000,0000,,That will then require us to\Ndo a super mesh which we will
Dialogue: 0,0:00:52.71,0:00:55.56,Default,,0000,0000,0000,,Be writing around there.
Dialogue: 0,0:00:55.56,0:00:58.79,Default,,0000,0000,0000,,Let's get started first of all\Nthough by writing the mesh
Dialogue: 0,0:00:58.79,0:01:02.00,Default,,0000,0000,0000,,current equation around\Nthis mesh number one.
Dialogue: 0,0:01:02.00,0:01:06.41,Default,,0000,0000,0000,,Starting down here in the lower left hand\Ncorner, we have going minus to plus.
Dialogue: 0,0:01:06.41,0:01:08.17,Default,,0000,0000,0000,,It will be a negative 12 volts.
Dialogue: 0,0:01:09.41,0:01:15.92,Default,,0000,0000,0000,,And then across this resistor of a voltage\Ndrop of five resistance five times current
Dialogue: 0,0:01:15.92,0:01:22.59,Default,,0000,0000,0000,,rate through that which is I1-12
Dialogue: 0,0:01:22.59,0:01:27.69,Default,,0000,0000,0000,,plus the volts struck across this\Nfilm resistor which is going to be 3
Dialogue: 0,0:01:27.69,0:01:32.95,Default,,0000,0000,0000,,times I 1- I3 and
Dialogue: 0,0:01:32.95,0:01:37.81,Default,,0000,0000,0000,,the sum of those voltage\Ndrop must equal 0.
Dialogue: 0,0:01:37.81,0:01:43.22,Default,,0000,0000,0000,,Now, the Super Mesh starting here at this\Npoint going up and around the 15 ohm, and
Dialogue: 0,0:01:43.22,0:01:50.28,Default,,0000,0000,0000,,when we're across the 15 ohm resistor now\Nthe voltage drop there being 15 times I2.
Dialogue: 0,0:01:50.28,0:01:54.86,Default,,0000,0000,0000,,Now continuing on down here,
Dialogue: 0,0:01:54.86,0:01:59.84,Default,,0000,0000,0000,,not going around this branch here because\Nthat's where the current source is.
Dialogue: 0,0:01:59.84,0:02:04.82,Default,,0000,0000,0000,,Continuing on down here across the 10\Nohm resistor we have a voltage drop
Dialogue: 0,0:02:04.82,0:02:09.23,Default,,0000,0000,0000,,of +10 x (I3) Plus,
Dialogue: 0,0:02:09.23,0:02:13.72,Default,,0000,0000,0000,,now continuing around here,\Nacross the 3 ohm resistor, +3 times.
Dialogue: 0,0:02:13.72,0:02:19.90,Default,,0000,0000,0000,,Now the current going in\Nthis direction is I3-I1.
Dialogue: 0,0:02:19.90,0:02:24.98,Default,,0000,0000,0000,,I3-I1+ the voltage
Dialogue: 0,0:02:24.98,0:02:30.04,Default,,0000,0000,0000,,drop across that 5 ohm\Nresistor Is five times.
Dialogue: 0,0:02:30.04,0:02:32.17,Default,,0000,0000,0000,,And again,\Nbecause we're going from right to left,
Dialogue: 0,0:02:32.17,0:02:35.37,Default,,0000,0000,0000,,the current going in that\Ndirection will be I2 minus I1.
Dialogue: 0,0:02:35.37,0:02:42.11,Default,,0000,0000,0000,,I2 minus I1,\Nthe sum of those terms must equal 0.
Dialogue: 0,0:02:42.11,0:02:44.42,Default,,0000,0000,0000,,And finally the third equation we need for
Dialogue: 0,0:02:44.42,0:02:49.70,Default,,0000,0000,0000,,these three unknowns comes from\Nthe relationship between I2, I3.
Dialogue: 0,0:02:49.70,0:02:50.74,Default,,0000,0000,0000,,And this current source.
Dialogue: 0,0:02:51.94,0:02:56.06,Default,,0000,0000,0000,,In this case, I2 is going in\Nthe direction of the current source, so
Dialogue: 0,0:02:56.06,0:03:02.29,Default,,0000,0000,0000,,we're going to have I2 minus I3,
Dialogue: 0,0:03:02.29,0:03:08.53,Default,,0000,0000,0000,,which is the current going through that\Nbranch in terms of the mesh currents.
Dialogue: 0,0:03:08.53,0:03:11.04,Default,,0000,0000,0000,,It's going to equal 2 Amps.
Dialogue: 0,0:03:12.69,0:03:15.62,Default,,0000,0000,0000,,So we have our three equations and\Nthree unknowns.
Dialogue: 0,0:03:15.62,0:03:17.24,Default,,0000,0000,0000,,Let's go ahead and combine terms.
Dialogue: 0,0:03:17.24,0:03:24.28,Default,,0000,0000,0000,,Taking the first equation\Nwe have I1 times,
Dialogue: 0,0:03:24.28,0:03:29.09,Default,,0000,0000,0000,,we got 5 there plus 3 is 8 plus
Dialogue: 0,0:03:29.09,0:03:33.73,Default,,0000,0000,0000,,I2 times We have a negative 5
Dialogue: 0,0:03:36.35,0:03:41.64,Default,,0000,0000,0000,,Plus I3, we have a negative 3.
Dialogue: 0,0:03:44.38,0:03:48.69,Default,,0000,0000,0000,,And then we have this negative 12 volts on\Nthe left-hand side we bring to the other
Dialogue: 0,0:03:48.69,0:03:50.72,Default,,0000,0000,0000,,side as a positive 12 volts.
Dialogue: 0,0:03:50.72,0:03:51.100,Default,,0000,0000,0000,,And there is our first equation.
Dialogue: 0,0:03:54.12,0:03:57.26,Default,,0000,0000,0000,,Second equation combining like terms,\Nwe have I1.
Dialogue: 0,0:03:57.26,0:04:02.51,Default,,0000,0000,0000,,And in this we have -3 and a -5.
Dialogue: 0,0:04:02.51,0:04:07.29,Default,,0000,0000,0000,,That makes -8 times I1 + I2 times,
Dialogue: 0,0:04:07.29,0:04:11.96,Default,,0000,0000,0000,,we got 15
Dialogue: 0,0:04:13.66,0:04:20.03,Default,,0000,0000,0000,,plus five is 20 times i2 plus i3, we've
Dialogue: 0,0:04:21.26,0:04:26.78,Default,,0000,0000,0000,,got 10i3 plus 3 is 13i3 equals, and
Dialogue: 0,0:04:28.72,0:04:32.74,Default,,0000,0000,0000,,there are no constants there So
Dialogue: 0,0:04:32.74,0:04:36.72,Default,,0000,0000,0000,,the sum of those terms equals 0.
Dialogue: 0,0:04:36.72,0:04:40.99,Default,,0000,0000,0000,,Finally, our third equation is I2,
Dialogue: 0,0:04:43.90,0:04:52.42,Default,,0000,0000,0000,,Times 1 + I3(-1) = 2.
Dialogue: 0,0:04:54.06,0:04:56.17,Default,,0000,0000,0000,,And we have our three equations and\Nthree unknowns so
Dialogue: 0,0:04:56.17,0:05:00.88,Default,,0000,0000,0000,,get ready with your calculator or whatever\Nyou choose to solve through your systems
Dialogue: 0,0:05:01.94,0:05:07.70,Default,,0000,0000,0000,,through your equations and three unknowns.
Dialogue: 0,0:05:07.70,0:05:11.20,Default,,0000,0000,0000,,And when you do we get the following.
Dialogue: 0,0:05:11.20,0:05:12.23,Default,,0000,0000,0000,,I1=2.03 amps.
Dialogue: 0,0:05:13.73,0:05:20.17,Default,,0000,0000,0000,,I two is equal to one point two eight amps
Dialogue: 0,0:05:20.17,0:05:24.92,Default,,0000,0000,0000,,and I three is equal to negative
Dialogue: 0,0:05:24.92,0:05:29.43,Default,,0000,0000,0000,,point seven two amps.
Dialogue: 0,0:05:31.02,0:05:34.10,Default,,0000,0000,0000,,LEts do a couple of consistency checks\Nhere just to make sure we didn't make any
Dialogue: 0,0:05:34.10,0:05:39.21,Default,,0000,0000,0000,,mistakes not to prove that our work was\Ncorrect but at least to look at a couple
Dialogue: 0,0:05:39.21,0:05:43.58,Default,,0000,0000,0000,,of calculations that would be consistent\Nwith what we know to be true.
Dialogue: 0,0:05:43.58,0:05:48.56,Default,,0000,0000,0000,,For example we know that i two\Nminus i three equals two amps.
Dialogue: 0,0:05:48.56,0:05:53.59,Default,,0000,0000,0000,,So I2, that is 1.28
Dialogue: 0,0:05:53.59,0:05:57.69,Default,,0000,0000,0000,,minus I3, well I3 is a -0.72 amps.
Dialogue: 0,0:05:57.69,0:06:03.22,Default,,0000,0000,0000,,The sum of those,\Nthey're supposed to equal 2 amps,
Dialogue: 0,0:06:03.22,0:06:04.80,Default,,0000,0000,0000,,and sure enough that works there.
Dialogue: 0,0:06:06.80,0:06:10.52,Default,,0000,0000,0000,,And of course we could plug those values\Ninto any one of those equations And
Dialogue: 0,0:06:10.52,0:06:12.28,Default,,0000,0000,0000,,proves that our solutions are correct.
Dialogue: 0,0:06:13.57,0:06:16.06,Default,,0000,0000,0000,,But now that we know what I1,\NI2, and I3 are.
Dialogue: 0,0:06:16.06,0:06:18.84,Default,,0000,0000,0000,,Let's use them to say for
Dialogue: 0,0:06:18.84,0:06:24.48,Default,,0000,0000,0000,,example now determine what the voltages\Ndrop is across this current source.
Dialogue: 0,0:06:25.70,0:06:28.53,Default,,0000,0000,0000,,We said that we couldn't numerically\Ncalculate it based upon the current going
Dialogue: 0,0:06:28.53,0:06:33.26,Default,,0000,0000,0000,,through it, but now that we know\Nthe currents around it, we can write a KVL
Dialogue: 0,0:06:33.26,0:06:38.72,Default,,0000,0000,0000,,once again around that loop, and\Nsolve for the voltage across the current.
Dialogue: 0,0:06:38.72,0:06:44.13,Default,,0000,0000,0000,,Starting here and going clockwise,\Nacross that 3 ohm resistor,
Dialogue: 0,0:06:44.13,0:06:49.89,Default,,0000,0000,0000,,we're going to have Three\Ntimes I three minus I one.
Dialogue: 0,0:06:51.42,0:06:56.27,Default,,0000,0000,0000,,Now, going across this current source with\Nthe voltage as referenced plus to minus.
Dialogue: 0,0:06:56.27,0:07:00.08,Default,,0000,0000,0000,,That represents a voltage drop,\Nso it would be plus v.
Dialogue: 0,0:07:01.87,0:07:05.70,Default,,0000,0000,0000,,Now, coming down across here,\Nthat would be plus 10.
Dialogue: 0,0:07:06.88,0:07:12.68,Default,,0000,0000,0000,,Oops, plus 10 times I3
Dialogue: 0,0:07:14.74,0:07:19.73,Default,,0000,0000,0000,,equals that's where we started from, 0.
Dialogue: 0,0:07:19.73,0:07:25.11,Default,,0000,0000,0000,,Now I plug in our values for I1, I2,
Dialogue: 0,0:07:25.11,0:07:30.15,Default,,0000,0000,0000,,and I3 and we have 3 times I [SOUND] So
Dialogue: 0,0:07:30.15,0:07:36.88,Default,,0000,0000,0000,,3 times I3, well, I3 is a -0.72 amps,
Dialogue: 0,0:07:36.88,0:07:41.25,Default,,0000,0000,0000,,-0.72,- I1, which is
Dialogue: 0,0:07:41.25,0:07:46.80,Default,,0000,0000,0000,,2.03 amps, + V + 10 times I3,
Dialogue: 0,0:07:46.80,0:07:51.87,Default,,0000,0000,0000,,which is a negative 0.72 amps.
Dialogue: 0,0:07:51.87,0:07:54.97,Default,,0000,0000,0000,,The sum of those things has to equal 0.
Dialogue: 0,0:07:54.97,0:08:00.06,Default,,0000,0000,0000,,Solving this for V gives us
Dialogue: 0,0:08:00.06,0:08:04.70,Default,,0000,0000,0000,,V = 15.45 volts.
Dialogue: 0,0:08:07.59,0:08:12.36,Default,,0000,0000,0000,,All righty, now looking at this source,\Nwe see that the current is reference into
Dialogue: 0,0:08:12.36,0:08:16.66,Default,,0000,0000,0000,,the negative terminal of our voltage and\Nleaving the positive terminal.
Dialogue: 0,0:08:16.66,0:08:18.04,Default,,0000,0000,0000,,So it's acting as a source.
Dialogue: 0,0:08:18.04,0:08:24.22,Default,,0000,0000,0000,,And we can then say that the power being\Ngenerated by that source is equal to
Dialogue: 0,0:08:24.22,0:08:29.30,Default,,0000,0000,0000,,negaitive i times v Is equal to negative i
Dialogue: 0,0:08:29.30,0:08:34.20,Default,,0000,0000,0000,,which is two times v which is 15.45 and\Nthe power
Dialogue: 0,0:08:36.10,0:08:40.02,Default,,0000,0000,0000,,then would be equal to\Nan negative what is that
Dialogue: 0,0:08:40.02,0:08:45.93,Default,,0000,0000,0000,,30.9 watts the fact that
Dialogue: 0,0:08:45.93,0:08:50.23,Default,,0000,0000,0000,,it's negative means that it's energy being\Nput into the circuit from the source.
Dialogue: 0,0:08:51.54,0:08:54.62,Default,,0000,0000,0000,,So once again, once we know all\Nthree of the mass/f currents,
Dialogue: 0,0:08:54.62,0:08:57.83,Default,,0000,0000,0000,,we can determine any voltage or\Ncurrent or for that matter power or
Dialogue: 0,0:08:57.83,0:09:01.11,Default,,0000,0000,0000,,any other quantity that we might\Nwant to know within this circuit.