[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.02,0:00:03.17,Default,,0000,0000,0000,,Now, we want to solve this circuit for vx.
Dialogue: 0,0:00:03.17,0:00:07.79,Default,,0000,0000,0000,,Let's do it this particular way where\Nwe're going to use a current divider and
Dialogue: 0,0:00:07.79,0:00:11.11,Default,,0000,0000,0000,,let's find the current right here,\Nlet's call that I1 and
Dialogue: 0,0:00:11.11,0:00:14.71,Default,,0000,0000,0000,,let's find this current right here,\Nlet's call that Ix.
Dialogue: 0,0:00:14.71,0:00:17.92,Default,,0000,0000,0000,,We can say that we can find the voltage\Nbecause we know the resistance.
Dialogue: 0,0:00:17.92,0:00:21.71,Default,,0000,0000,0000,,So if we do this to the current, we'd be\Nable to find the voltage using Ohm's law.
Dialogue: 0,0:00:21.71,0:00:23.38,Default,,0000,0000,0000,,This is the current divider.
Dialogue: 0,0:00:23.38,0:00:26.97,Default,,0000,0000,0000,,When we have a current divider,\Nthe easiest way to handle that is to take
Dialogue: 0,0:00:26.97,0:00:31.52,Default,,0000,0000,0000,,our source and divide it through,\Nlet's say, R1 and R2.
Dialogue: 0,0:00:32.70,0:00:36.71,Default,,0000,0000,0000,,The current I1 and\NI2 is going to be found this way.
Dialogue: 0,0:00:36.71,0:00:41.15,Default,,0000,0000,0000,,Let's take our R and convert it to G,\Nwhich is 1 over R1, and
Dialogue: 0,0:00:41.15,0:00:45.01,Default,,0000,0000,0000,,let's convert this one to G2,\Nwhich is 1 over R2.
Dialogue: 0,0:00:45.01,0:00:50.16,Default,,0000,0000,0000,,Then we know that I1 is\Nequal to G1 over G1 + G2,
Dialogue: 0,0:00:50.16,0:00:53.88,Default,,0000,0000,0000,,times our original source current.
Dialogue: 0,0:00:53.88,0:01:00.52,Default,,0000,0000,0000,,And I2 is equal to G2 over G1 + G2,\Nalso times our source.
Dialogue: 0,0:01:00.52,0:01:02.36,Default,,0000,0000,0000,,So let's do that over here.
Dialogue: 0,0:01:02.36,0:01:05.81,Default,,0000,0000,0000,,I can see that if I combine these two\Nin series, that would be 12 Ohms,
Dialogue: 0,0:01:05.81,0:01:07.77,Default,,0000,0000,0000,,that's what I'll be using in that arm.
Dialogue: 0,0:01:07.77,0:01:11.76,Default,,0000,0000,0000,,So if I wanted to have G1,\Nthat would be one-third and
Dialogue: 0,0:01:11.76,0:01:17.09,Default,,0000,0000,0000,,then converting this to G2,\Nthat would be 1 divided by one-twelfth.
Dialogue: 0,0:01:17.09,0:01:24.87,Default,,0000,0000,0000,,So I1 is going to be 10\NAmps times G1 over G1 + G2.
Dialogue: 0,0:01:24.87,0:01:30.88,Default,,0000,0000,0000,,I2 is going to be 10 Amps\Ntimes G2 over G1 + G2.
Dialogue: 0,0:01:30.88,0:01:33.26,Default,,0000,0000,0000,,Do the math to work out\Nwhat the current will be.
Dialogue: 0,0:01:33.26,0:01:39.09,Default,,0000,0000,0000,,Then because I now know I2, oops,\NI guess I called it x2, but
Dialogue: 0,0:01:39.09,0:01:47.02,Default,,0000,0000,0000,,there's I2, we can say that vx is going to\Nbe = I2 times the resistance which is 4.
Dialogue: 0,0:01:47.02,0:01:49.94,Default,,0000,0000,0000,,Mm-hm, let's work out\Nthe polarity on this.
Dialogue: 0,0:01:49.94,0:01:52.62,Default,,0000,0000,0000,,We can see that vx is\Ndefined from plus to minus.
Dialogue: 0,0:01:52.62,0:01:54.21,Default,,0000,0000,0000,,Let's make sure that's right.
Dialogue: 0,0:01:54.21,0:01:57.89,Default,,0000,0000,0000,,My current is defined in this\Ndirection from plus to minus, and so
Dialogue: 0,0:01:57.89,0:02:02.92,Default,,0000,0000,0000,,because these two are matching,\NI have the correct polarity.
Dialogue: 0,0:02:02.92,0:02:05.73,Default,,0000,0000,0000,,If I had defined my current\Nthe other direction,
Dialogue: 0,0:02:05.73,0:02:09.27,Default,,0000,0000,0000,,then I would have had a minus\Ninstead of a plus here.
Dialogue: 0,0:02:09.27,0:02:11.59,Default,,0000,0000,0000,,Okay, great,\Nthat's one way of doing this problem.
Dialogue: 0,0:02:11.59,0:02:13.58,Default,,0000,0000,0000,,Let's go do it another way.
Dialogue: 0,0:02:13.58,0:02:19.60,Default,,0000,0000,0000,,Let's say that I have the current,\NSo here's 3, 4, and 8.
Dialogue: 0,0:02:19.60,0:02:22.36,Default,,0000,0000,0000,,And this is the voltage\Nthat I want to find.
Dialogue: 0,0:02:22.36,0:02:23.39,Default,,0000,0000,0000,,This is 10 amps.
Dialogue: 0,0:02:25.78,0:02:30.58,Default,,0000,0000,0000,,One thing that we can do is convert\Nthis thing into a voltage source and
Dialogue: 0,0:02:30.58,0:02:33.90,Default,,0000,0000,0000,,then be able to do a voltage divider.
Dialogue: 0,0:02:33.90,0:02:38.64,Default,,0000,0000,0000,,So to convert this to a voltage source, I\Nsimply say, what is the voltage across it?
Dialogue: 0,0:02:38.64,0:02:40.64,Default,,0000,0000,0000,,It's going to be 10 amps times 3.
Dialogue: 0,0:02:40.64,0:02:45.62,Default,,0000,0000,0000,,So I'm going to have a voltage\Nsource plus to minus vs of 10 times
Dialogue: 0,0:02:45.62,0:02:47.44,Default,,0000,0000,0000,,3 which is 30 volts.
Dialogue: 0,0:02:48.71,0:02:52.36,Default,,0000,0000,0000,,And then my original current this 3,\Nthat's right here,
Dialogue: 0,0:02:52.36,0:02:56.39,Default,,0000,0000,0000,,is going to end up being in\Nseries instead of in parallel.
Dialogue: 0,0:02:56.39,0:03:01.27,Default,,0000,0000,0000,,So just 3 in parallel, sorry,\N3 in parallel became 3 in series.
Dialogue: 0,0:03:02.28,0:03:07.15,Default,,0000,0000,0000,,So these two things are equal,\Nthen I have my 4 Ohm and
Dialogue: 0,0:03:07.15,0:03:09.91,Default,,0000,0000,0000,,my 8 Ohm resistance, 4 and 8.
Dialogue: 0,0:03:09.91,0:03:12.98,Default,,0000,0000,0000,,And what I want to find is vx here.
Dialogue: 0,0:03:12.98,0:03:17.55,Default,,0000,0000,0000,,Remember when I have\Na voltage divider like so and
Dialogue: 0,0:03:17.55,0:03:23.13,Default,,0000,0000,0000,,I have a series of three resistances,\NR1, R2, and R3.
Dialogue: 0,0:03:23.13,0:03:28.96,Default,,0000,0000,0000,,Vs =, sorry V1 like this = vs times
Dialogue: 0,0:03:28.96,0:03:34.05,Default,,0000,0000,0000,,R1 divided by R1 + R2 + R3,
Dialogue: 0,0:03:34.05,0:03:39.83,Default,,0000,0000,0000,,similar for V2, V3, and V4.
Dialogue: 0,0:03:39.83,0:03:47.91,Default,,0000,0000,0000,,So finding xx now, I can just say that is\Nequal to vs times 4, divided by 3 + 4 + 8.
Dialogue: 0,0:03:47.91,0:03:52.16,Default,,0000,0000,0000,,And by vs is 30 volts.
Dialogue: 0,0:03:52.16,0:03:56.87,Default,,0000,0000,0000,,So again, 30 times 4 times 3 + 4 + 8.
Dialogue: 0,0:03:56.87,0:03:59.70,Default,,0000,0000,0000,,So that's the second way of\Nbeing able to solve the circuit.