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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.83,0:00:04.52,Default,,0000,0000,0000,,Now let's talk about admittance and\Nhow to use it in current dividers.
Dialogue: 0,0:00:04.52,0:00:08.16,Default,,0000,0000,0000,,Remember that resistance is\Na value that's given in ohms.
Dialogue: 0,0:00:08.16,0:00:12.20,Default,,0000,0000,0000,,It's inverse,\N1 over R is called admittance.
Dialogue: 0,0:00:12.20,0:00:17.62,Default,,0000,0000,0000,,We usually use G to represent that and\Nit's given in inverse ohms or
Dialogue: 0,0:00:17.62,0:00:23.50,Default,,0000,0000,0000,,ohms to the -1, or 1 over ohms,\Nand sometimes we call that Mhos.
Dialogue: 0,0:00:23.50,0:00:26.04,Default,,0000,0000,0000,,Kinda see it's ohms backwards.
Dialogue: 0,0:00:26.04,0:00:31.23,Default,,0000,0000,0000,,So admittance,\Nright here is one over the resistance.
Dialogue: 0,0:00:31.23,0:00:32.79,Default,,0000,0000,0000,,We use that in a lot of good math.
Dialogue: 0,0:00:32.79,0:00:37.96,Default,,0000,0000,0000,,For example, remember that if we have\Nresistors in series, R1, R2, and
Dialogue: 0,0:00:37.96,0:00:43.42,Default,,0000,0000,0000,,R3, that if we want to\Nhave the total resistance
Dialogue: 0,0:00:43.42,0:00:48.51,Default,,0000,0000,0000,,over this region, R = R1 + R2 + R3.
Dialogue: 0,0:00:49.71,0:00:56.52,Default,,0000,0000,0000,,If we had values that were in parallel,\Nthen we had to handle that differently.
Dialogue: 0,0:00:56.52,0:01:00.81,Default,,0000,0000,0000,,One very easy way of handling that\Nis to put admittances in parallel.
Dialogue: 0,0:01:03.95,0:01:09.99,Default,,0000,0000,0000,,If we use values G1, G2,\Nand G3, then we can say
Dialogue: 0,0:01:09.99,0:01:16.26,Default,,0000,0000,0000,,that the entire admittance G,\Nfrom this point a to this point b,
Dialogue: 0,0:01:16.26,0:01:22.22,Default,,0000,0000,0000,,so G from a to b = G1 + G2 + G3.
Dialogue: 0,0:01:23.32,0:01:28.61,Default,,0000,0000,0000,,So it's very handy to add up resistors in\Nparallel by treating them as admittances
Dialogue: 0,0:01:28.61,0:01:29.91,Default,,0000,0000,0000,,instead of as resistors.
Dialogue: 0,0:01:31.52,0:01:34.58,Default,,0000,0000,0000,,Now let's do a current\Ndivider using resistance.
Dialogue: 0,0:01:34.58,0:01:37.18,Default,,0000,0000,0000,,Remember that when we've got\Na current divider like this,
Dialogue: 0,0:01:37.18,0:01:41.71,Default,,0000,0000,0000,,we first have to combine two\Nof the resistors in parallel.
Dialogue: 0,0:01:41.71,0:01:46.74,Default,,0000,0000,0000,,Let's say that what I want to find is I1,\Nso I would combine these in parallel.
Dialogue: 0,0:01:46.74,0:01:51.91,Default,,0000,0000,0000,,This would be equal to R2\Ntimes R3 over R2 + R3.
Dialogue: 0,0:01:51.91,0:01:54.15,Default,,0000,0000,0000,,Then if I wanted to find I1,
Dialogue: 0,0:01:54.15,0:02:00.62,Default,,0000,0000,0000,,that would be equal to I0 times\Nthe resistors that it's not going through.
Dialogue: 0,0:02:00.62,0:02:05.60,Default,,0000,0000,0000,,Which is this R2 times R3 over R2 + R3
Dialogue: 0,0:02:05.60,0:02:09.42,Default,,0000,0000,0000,,divided by all the resistors,
Dialogue: 0,0:02:09.42,0:02:14.12,Default,,0000,0000,0000,,which is R1 + R2R3 over R2 + R3.
Dialogue: 0,0:02:15.56,0:02:18.59,Default,,0000,0000,0000,,That's how we handle a current\Ndivider using resistance.
Dialogue: 0,0:02:18.59,0:02:21.18,Default,,0000,0000,0000,,But if we wanted to use\Nadmittance instead,
Dialogue: 0,0:02:21.18,0:02:23.37,Default,,0000,0000,0000,,here's the way that we can do that.
Dialogue: 0,0:02:23.37,0:02:28.79,Default,,0000,0000,0000,,We don't even need to combine\Nthis into series and parallel,
Dialogue: 0,0:02:28.79,0:02:34.74,Default,,0000,0000,0000,,we're just going to be able to say\Nthat I1 = Io times the admittance
Dialogue: 0,0:02:34.74,0:02:39.56,Default,,0000,0000,0000,,that it is going through,\Ndivided by G1 + G2 + G3.
Dialogue: 0,0:02:39.56,0:02:43.35,Default,,0000,0000,0000,,That's really a lot easier than what\Nwe did on the previous page, so
Dialogue: 0,0:02:43.35,0:02:47.80,Default,,0000,0000,0000,,I'm gonna say this is a really cool\Nway of doing current dividers.
Dialogue: 0,0:02:47.80,0:02:51.45,Default,,0000,0000,0000,,I'm going to show you how similar\Nthat is to doing voltage dividers.
Dialogue: 0,0:02:51.45,0:02:57.78,Default,,0000,0000,0000,,If we have, so this is a voltage divider,\Nif we have resistors in series,
Dialogue: 0,0:02:57.78,0:03:02.79,Default,,0000,0000,0000,,R1, R2, and R3,\Nlet's say that I wanted to find V1.
Dialogue: 0,0:03:02.79,0:03:09.76,Default,,0000,0000,0000,,Then V1 would be equal to,\Nthis is the total voltage across here,
Dialogue: 0,0:03:09.76,0:03:14.38,Default,,0000,0000,0000,,V0 times R1, divided by R1 + R2 + R3.
Dialogue: 0,0:03:14.38,0:03:19.97,Default,,0000,0000,0000,,You can see the form of these two\Nequations is equivalent except,
Dialogue: 0,0:03:19.97,0:03:26.48,Default,,0000,0000,0000,,that one is handling parallel admittances,\Nthe other is handling series resistances.
Dialogue: 0,0:03:26.48,0:03:32.07,Default,,0000,0000,0000,,Now let's do an example just with numbers,\Nlet's do a resistive current divider
Dialogue: 0,0:03:33.94,0:03:39.50,Default,,0000,0000,0000,,like this with 1 ohm, a 2 ohm,\Nand a 3 ohm resistor.
Dialogue: 0,0:03:39.50,0:03:44.63,Default,,0000,0000,0000,,And let's say that I want to find\NI1 given that I have 10 amps
Dialogue: 0,0:03:44.63,0:03:47.90,Default,,0000,0000,0000,,of current going in, that is I0.
Dialogue: 0,0:03:47.90,0:03:52.01,Default,,0000,0000,0000,,So we will first of all need to\Nconvert this two admittances and
Dialogue: 0,0:03:52.01,0:03:58.87,Default,,0000,0000,0000,,what we would have Is 1 divided by 3 mhos,
Dialogue: 0,0:03:58.87,0:04:05.28,Default,,0000,0000,0000,,1 divided by 2 mhos and\N1 divided by 1 mhos.
Dialogue: 0,0:04:05.28,0:04:10.49,Default,,0000,0000,0000,,And again, we're looking for\NI1 with an I0 of 10 amps going in.
Dialogue: 0,0:04:10.49,0:04:14.36,Default,,0000,0000,0000,,So what we would say is, I1 = 10 amps,
Dialogue: 0,0:04:14.36,0:04:20.78,Default,,0000,0000,0000,,that's the I0 value times the admittance\Nthat it is going through,
Dialogue: 0,0:04:20.78,0:04:26.34,Default,,0000,0000,0000,,1 mho, divided by the sum of\Nall of the other admittances.
Dialogue: 0,0:04:28.88,0:04:30.88,Default,,0000,0000,0000,,And that's our I1 value.
Dialogue: 0,0:04:30.88,0:04:35.52,Default,,0000,0000,0000,,That's really very simple, whole lot\Nsimpler than doing it with resistances.