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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.08,0:00:03.95,Default,,0000,0000,0000,,In this video we're going to take\Na look at a circuit that contains
Dialogue: 0,0:00:03.95,0:00:06.13,Default,,0000,0000,0000,,the structure known as a super node.
Dialogue: 0,0:00:06.13,0:00:13.95,Default,,0000,0000,0000,,A super node consist of two critical\Nnodes separated only by a voltage source.
Dialogue: 0,0:00:13.95,0:00:17.43,Default,,0000,0000,0000,,That voltage source can be either\Nan independent source as it shown here or
Dialogue: 0,0:00:17.43,0:00:19.04,Default,,0000,0000,0000,,a dependent source.
Dialogue: 0,0:00:19.04,0:00:23.50,Default,,0000,0000,0000,,To understand what this leads to,\Nlet's go ahead and
Dialogue: 0,0:00:23.50,0:00:27.66,Default,,0000,0000,0000,,analyse this circuit as we would any other\Ncircuit using the node voltage approach.
Dialogue: 0,0:00:27.66,0:00:32.27,Default,,0000,0000,0000,,We start by identifying our critical\Nnodes 1, 2, 3 critical nodes.
Dialogue: 0,0:00:32.27,0:00:37.06,Default,,0000,0000,0000,,This one we have identified as our\Nreference where V equals 0, and
Dialogue: 0,0:00:37.06,0:00:42.19,Default,,0000,0000,0000,,we've assigned this node here\Nthe voltage V1 or the variable V1, and
Dialogue: 0,0:00:42.19,0:00:48.03,Default,,0000,0000,0000,,this node here, the voltage of that node\Nis, we have assigned the variable V2.
Dialogue: 0,0:00:48.03,0:00:51.03,Default,,0000,0000,0000,,Now to understand what's\Nhappening let's just go ahead and
Dialogue: 0,0:00:51.03,0:00:52.56,Default,,0000,0000,0000,,write the two node equations.
Dialogue: 0,0:00:52.56,0:00:57.100,Default,,0000,0000,0000,,For this left hand node we have\Nthe current leaving going to the left
Dialogue: 0,0:00:57.100,0:01:03.04,Default,,0000,0000,0000,,will be V1- 4, divided by 2,\Nplus the current coming down
Dialogue: 0,0:01:03.04,0:01:08.39,Default,,0000,0000,0000,,through the forum resistor would\Nbe V1 minus 0, divided by 4.
Dialogue: 0,0:01:08.39,0:01:09.80,Default,,0000,0000,0000,,Now, here's where the challenge comes.
Dialogue: 0,0:01:11.10,0:01:13.89,Default,,0000,0000,0000,,When we look at the current\Nleaving this node and
Dialogue: 0,0:01:13.89,0:01:16.95,Default,,0000,0000,0000,,going to the right,\Nwe see only a voltage source.
Dialogue: 0,0:01:16.95,0:01:18.35,Default,,0000,0000,0000,,There is no other device in here,
Dialogue: 0,0:01:18.35,0:01:20.73,Default,,0000,0000,0000,,like a resistor that we can\Nuse to calculate the current.
Dialogue: 0,0:01:22.31,0:01:25.84,Default,,0000,0000,0000,,A voltage source and\Nit's current are unrelated to each other.
Dialogue: 0,0:01:25.84,0:01:32.90,Default,,0000,0000,0000,,There is no mathematical or formula\Nthat relates the current to the voltage.
Dialogue: 0,0:01:32.90,0:01:35.77,Default,,0000,0000,0000,,In fact, an ideal voltage source produces
Dialogue: 0,0:01:35.77,0:01:39.29,Default,,0000,0000,0000,,this same voltage regardless\Nof what the current is.
Dialogue: 0,0:01:40.78,0:01:41.67,Default,,0000,0000,0000,,So what do we do?
Dialogue: 0,0:01:41.67,0:01:45.74,Default,,0000,0000,0000,,Well, let's just go ahead and\Nidentify a variable,
Dialogue: 0,0:01:45.74,0:01:52.13,Default,,0000,0000,0000,,we'll call it i, representing the current\Nleaving V1 and going to the right.
Dialogue: 0,0:01:52.13,0:01:56.06,Default,,0000,0000,0000,,This is inconsistent with the philosophy\Nof node voltage analysis,
Dialogue: 0,0:01:56.06,0:02:00.41,Default,,0000,0000,0000,,where we wanted to write everything in\Nterms of our node voltage variables.
Dialogue: 0,0:02:00.41,0:02:02.84,Default,,0000,0000,0000,,We'll see here in just a second\Nthat this i is gonna cancel.
Dialogue: 0,0:02:02.84,0:02:07.79,Default,,0000,0000,0000,,But for now, let us call that i,\Nit's referenced leaving the node there so
Dialogue: 0,0:02:07.79,0:02:11.25,Default,,0000,0000,0000,,it will be represented over\Nhere as a positive i, and
Dialogue: 0,0:02:11.25,0:02:14.12,Default,,0000,0000,0000,,the sum of those three currents equals 0.
Dialogue: 0,0:02:14.12,0:02:18.28,Default,,0000,0000,0000,,Now let's write the node equation\Nover here at the right hand node.
Dialogue: 0,0:02:18.28,0:02:23.87,Default,,0000,0000,0000,,Starting with the current leaving the node\Ncoming down to the 8 ohm resistor,
Dialogue: 0,0:02:23.87,0:02:26.29,Default,,0000,0000,0000,,we have V2 divided by 8.
Dialogue: 0,0:02:26.29,0:02:31.12,Default,,0000,0000,0000,,And then we have this 2 amp source\Ncoming in, so that's going to be -2.
Dialogue: 0,0:02:31.12,0:02:35.72,Default,,0000,0000,0000,,And then we also have now i,\Nthe current in this branch.
Dialogue: 0,0:02:35.72,0:02:39.78,Default,,0000,0000,0000,,Well, this time it's entering\Nthe right hand node, so
Dialogue: 0,0:02:39.78,0:02:42.13,Default,,0000,0000,0000,,we'll represent it with a -i.
Dialogue: 0,0:02:42.13,0:02:45.59,Default,,0000,0000,0000,,And the sum of those three\Ncurrents must equal 0.
Dialogue: 0,0:02:45.59,0:02:52.38,Default,,0000,0000,0000,,Note that we can eliminate this i\Nterm by adding these two equations.
Dialogue: 0,0:02:52.38,0:02:53.12,Default,,0000,0000,0000,,Let's go ahead and do that.
Dialogue: 0,0:02:53.12,0:03:00.28,Default,,0000,0000,0000,,We then get V1-4 divided by 2 + V1 over 4.
Dialogue: 0,0:03:00.28,0:03:03.62,Default,,0000,0000,0000,,Again, the i's add to 0,
Dialogue: 0,0:03:03.62,0:03:09.14,Default,,0000,0000,0000,,then we have the other\Ntwo terms down here plus
Dialogue: 0,0:03:09.14,0:03:14.09,Default,,0000,0000,0000,,V2 over 8, then- minus 2 equals 0.
Dialogue: 0,0:03:14.09,0:03:17.19,Default,,0000,0000,0000,,So we've eliminated the i but\Nin the process we've gone from having two
Dialogue: 0,0:03:17.19,0:03:20.86,Default,,0000,0000,0000,,equations down to just one equation, and\Nwe still have the two unknowns V1 and V2.
Dialogue: 0,0:03:22.03,0:03:25.60,Default,,0000,0000,0000,,We get the second equation by\Nrealizing that because there's
Dialogue: 0,0:03:25.60,0:03:30.55,Default,,0000,0000,0000,,this ideal source between V1 and V2, V1\Nand V2 are not independent of each other.
Dialogue: 0,0:03:30.55,0:03:35.34,Default,,0000,0000,0000,,In fact, we can say that V2 is equal to
Dialogue: 0,0:03:35.34,0:03:40.29,Default,,0000,0000,0000,,the voltage of V1 going up 18 volts.
Dialogue: 0,0:03:40.29,0:03:44.98,Default,,0000,0000,0000,,Or that V2 is equal to V1 plus 18 volts.
Dialogue: 0,0:03:49.45,0:03:53.43,Default,,0000,0000,0000,,That then, we're gonna refer to\Nthat as the super node equation.
Dialogue: 0,0:03:53.43,0:03:56.61,Default,,0000,0000,0000,,That becomes our second\Nequation in terms of V1 and
Dialogue: 0,0:03:56.61,0:03:59.94,Default,,0000,0000,0000,,V2 that will allow us to solve for\Nthe circuit.
Dialogue: 0,0:03:59.94,0:04:01.76,Default,,0000,0000,0000,,Let's clean this up\Nhere just a little bit.
Dialogue: 0,0:04:01.76,0:04:07.38,Default,,0000,0000,0000,,Combining like terms we\Nhave V1 times 1/2 + 1/4.
Dialogue: 0,0:04:07.38,0:04:11.32,Default,,0000,0000,0000,,We have only one V2 term, as V2 times 1/8.
Dialogue: 0,0:04:11.32,0:04:15.95,Default,,0000,0000,0000,,Now our constants, we have a negative 4\Nover t2, that gives us a negative 2, and
Dialogue: 0,0:04:15.95,0:04:19.67,Default,,0000,0000,0000,,another minus 2, which we take to\Nthe other side as a positive 4.
Dialogue: 0,0:04:19.67,0:04:23.48,Default,,0000,0000,0000,,Combining 1/2 plus 1/4,\Nthat gives us 3/4 V1,
Dialogue: 0,0:04:25.93,0:04:29.68,Default,,0000,0000,0000,,Plus V2 times
Dialogue: 0,0:04:29.68,0:04:34.69,Default,,0000,0000,0000,,1/8 equals 4.
Dialogue: 0,0:04:34.69,0:04:39.76,Default,,0000,0000,0000,,Now that we've got this cleaned\Nup a little bit, let's substitute
Dialogue: 0,0:04:39.76,0:04:46.20,Default,,0000,0000,0000,,V2 here in this equation, with the\Nrelationship that we came up with here.
Dialogue: 0,0:04:46.20,0:04:50.65,Default,,0000,0000,0000,,And we're left with then,\NV1 times three fourths,
Dialogue: 0,0:04:50.65,0:04:56.66,Default,,0000,0000,0000,,plus replacing V2 with V1 plus 18,
Dialogue: 0,0:04:56.66,0:05:01.80,Default,,0000,0000,0000,,times one eighth, is equal to 4.
Dialogue: 0,0:05:01.80,0:05:03.94,Default,,0000,0000,0000,,Once again, let's combine like terms.
Dialogue: 0,0:05:03.94,0:05:09.28,Default,,0000,0000,0000,,We have V1 here times the 3/4 and
Dialogue: 0,0:05:09.28,0:05:15.37,Default,,0000,0000,0000,,then over here we have 1/8 v1 that\Nwe need to add in, so plus 1/8.
Dialogue: 0,0:05:15.37,0:05:19.98,Default,,0000,0000,0000,,We've eliminated the V2 term,\Nyet we have a constant
Dialogue: 0,0:05:19.98,0:05:24.69,Default,,0000,0000,0000,,term now here of 18/8,\Nwhich is the same as 9/4.
Dialogue: 0,0:05:24.69,0:05:28.69,Default,,0000,0000,0000,,When we subtract 9/4 from both sides,
Dialogue: 0,0:05:28.69,0:05:33.97,Default,,0000,0000,0000,,we're left with here on\Nthe right hand side, 7/4.
Dialogue: 0,0:05:35.42,0:05:41.45,Default,,0000,0000,0000,,Solving this for\NV1 we get the V1 is equal to 2 volts.
Dialogue: 0,0:05:41.45,0:05:46.01,Default,,0000,0000,0000,,And now that we know V1 we can determine
Dialogue: 0,0:05:46.01,0:05:50.84,Default,,0000,0000,0000,,the V2 is 18 volts greater than V1, or
Dialogue: 0,0:05:50.84,0:05:57.91,Default,,0000,0000,0000,,V2 is equal to 20 volts,\N18 volts plus 2 is 20 volts.
Dialogue: 0,0:06:00.12,0:06:06.31,Default,,0000,0000,0000,,Notice, now that we've done this,\Nlet's go back to the circuit and
Dialogue: 0,0:06:06.31,0:06:10.16,Default,,0000,0000,0000,,look at this super node\Nas one single node.
Dialogue: 0,0:06:12.64,0:06:17.31,Default,,0000,0000,0000,,And add the currents that\Nare leaving the super node.
Dialogue: 0,0:06:17.31,0:06:19.60,Default,,0000,0000,0000,,Starting on this side of the super node,
Dialogue: 0,0:06:19.60,0:06:24.28,Default,,0000,0000,0000,,we have a current leaving in this branch\Nhere, which we know to be simply V1- 4.
Dialogue: 0,0:06:26.14,0:06:29.43,Default,,0000,0000,0000,,Minus 4 divided by 2.
Dialogue: 0,0:06:29.43,0:06:33.08,Default,,0000,0000,0000,,Add to that the current leaving the super\Nnode coming down here through the 4
Dialogue: 0,0:06:33.08,0:06:37.40,Default,,0000,0000,0000,,ohm resistor, which is V1 divided by 4.
Dialogue: 0,0:06:39.02,0:06:43.29,Default,,0000,0000,0000,,Now, you'll notice that this current\Ni doesn't leave the super node,
Dialogue: 0,0:06:43.29,0:06:45.90,Default,,0000,0000,0000,,in fact it just is contained with it.
Dialogue: 0,0:06:45.90,0:06:48.98,Default,,0000,0000,0000,,You can if you will, think of it as\Nleaving one part of the super node and
Dialogue: 0,0:06:48.98,0:06:54.29,Default,,0000,0000,0000,,ending up in the other part of the super\Nnode, but i does not leave the super node.
Dialogue: 0,0:06:54.29,0:06:57.41,Default,,0000,0000,0000,,So we come over here and\Naccount for these other two or
Dialogue: 0,0:06:57.41,0:06:59.97,Default,,0000,0000,0000,,the currents in these other two branches.
Dialogue: 0,0:06:59.97,0:07:06.40,Default,,0000,0000,0000,,The current coming down here to the 8\Natom resistor is V2 divided by 8, and
Dialogue: 0,0:07:06.40,0:07:11.93,Default,,0000,0000,0000,,then of course we get the 2 ohms\Ngoing in there which we account for
Dialogue: 0,0:07:11.93,0:07:13.65,Default,,0000,0000,0000,,as a -2 equals 0.
Dialogue: 0,0:07:13.65,0:07:17.100,Default,,0000,0000,0000,,Notice that writing this equation\Nin the context of the super node,
Dialogue: 0,0:07:17.100,0:07:22.27,Default,,0000,0000,0000,,gives us exactly the same equation\Nthat we came up with when we wrote
Dialogue: 0,0:07:22.27,0:07:26.64,Default,,0000,0000,0000,,the individual node equations,\Nand combined them to eliminate i.
Dialogue: 0,0:07:26.64,0:07:29.13,Default,,0000,0000,0000,,What that means and this is always true,
Dialogue: 0,0:07:29.13,0:07:34.42,Default,,0000,0000,0000,,we don't have to do this intermediate step\Nof writing the equations and adding them.
Dialogue: 0,0:07:34.42,0:07:40.65,Default,,0000,0000,0000,,We can simply account for\Nthe currents leaving the super node and
Dialogue: 0,0:07:40.65,0:07:45.74,Default,,0000,0000,0000,,write the equation in 4 or\Nsumming those currents.
Dialogue: 0,0:07:45.74,0:07:47.11,Default,,0000,0000,0000,,Now, of course,
Dialogue: 0,0:07:47.11,0:07:52.82,Default,,0000,0000,0000,,the second equation comes just as it\Ndid over here relating to V1 and V2.
Dialogue: 0,0:07:52.82,0:07:57.28,Default,,0000,0000,0000,,We've got two equations and two unknowns\Nas we had over here, and from there on,
Dialogue: 0,0:07:57.28,0:07:59.41,Default,,0000,0000,0000,,the calculations are exactly the same