[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.85,0:00:04.49,Default,,0000,0000,0000,,All righty,\Nlet's work through some examples of
Dialogue: 0,0:00:04.49,0:00:07.02,Default,,0000,0000,0000,,finding Thevenin equivalent circuits.
Dialogue: 0,0:00:07.02,0:00:08.35,Default,,0000,0000,0000,,Take a look at this circuit right here.
Dialogue: 0,0:00:08.35,0:00:11.61,Default,,0000,0000,0000,,We want to determine\Nthe Thevenin equivalent circuit
Dialogue: 0,0:00:11.61,0:00:15.19,Default,,0000,0000,0000,,between the terminals a and b.
Dialogue: 0,0:00:15.19,0:00:17.69,Default,,0000,0000,0000,,That means we need to find\Nthe open circuit voltage and
Dialogue: 0,0:00:17.69,0:00:19.98,Default,,0000,0000,0000,,we need to determine the Thevenin\Nequivalent resistance.
Dialogue: 0,0:00:21.02,0:00:26.48,Default,,0000,0000,0000,,Now, you'll notice the open circuit\Nvoltage is the voltage across here, but
Dialogue: 0,0:00:26.48,0:00:32.70,Default,,0000,0000,0000,,because it's open circuit, that means that\Nthere's no current flowing through here.
Dialogue: 0,0:00:32.70,0:00:36.73,Default,,0000,0000,0000,,I equals zero, so if there's no\Ncurrent flowing through that resister,
Dialogue: 0,0:00:36.73,0:00:39.40,Default,,0000,0000,0000,,there's no voltage drop\Nacross that resister.
Dialogue: 0,0:00:39.40,0:00:45.15,Default,,0000,0000,0000,,And the voltage at the terminals,\Nthe open circuit voltage then,
Dialogue: 0,0:00:45.15,0:00:50.19,Default,,0000,0000,0000,,is, in fact,\Nthe voltage across this 30 ohm resistor.
Dialogue: 0,0:00:51.51,0:00:53.36,Default,,0000,0000,0000,,Again, because there's\Nno voltage drop here,
Dialogue: 0,0:00:53.36,0:00:57.31,Default,,0000,0000,0000,,the voltage that hit this point is\Nthe same as the voltage at this point.
Dialogue: 0,0:00:59.00,0:01:03.50,Default,,0000,0000,0000,,And again, because there's no current\Ngoing here, the current source here,
Dialogue: 0,0:01:03.50,0:01:09.22,Default,,0000,0000,0000,,0.5 amps flowing is driving these\Ntwo in series at this point.
Dialogue: 0,0:01:10.38,0:01:14.55,Default,,0000,0000,0000,,And we have then that V open\Ncircuit is simply equal
Dialogue: 0,0:01:14.55,0:01:19.35,Default,,0000,0000,0000,,to 0.5 times, 0.5 is the current,
Dialogue: 0,0:01:19.35,0:01:25.13,Default,,0000,0000,0000,,times the 30 ohms is equal to 15 volts.
Dialogue: 0,0:01:26.84,0:01:29.33,Default,,0000,0000,0000,,We now need to determine\Nthe Thevenin equivalent resistance,
Dialogue: 0,0:01:29.33,0:01:31.25,Default,,0000,0000,0000,,which we're gonna do in\Neach of the three methods.
Dialogue: 0,0:01:31.25,0:01:35.30,Default,,0000,0000,0000,,Or at least consider whether each of those\Nthree different methods will work for
Dialogue: 0,0:01:35.30,0:01:36.65,Default,,0000,0000,0000,,this circuit.
Dialogue: 0,0:01:36.65,0:01:41.78,Default,,0000,0000,0000,,Method one works when we have at least one\Nindependent source, which we have here.
Dialogue: 0,0:01:42.79,0:01:45.74,Default,,0000,0000,0000,,Method one involves finding\Nthe short circuit current.
Dialogue: 0,0:01:47.72,0:01:53.64,Default,,0000,0000,0000,,And then, forming the ratio, V,\Nopen circuit divided by I, short circuit.
Dialogue: 0,0:01:53.64,0:01:58.75,Default,,0000,0000,0000,,So if we short this out,\Nshort the terminals,
Dialogue: 0,0:01:58.75,0:02:05.59,Default,,0000,0000,0000,,the current flowing through\Nthat short is I, short circuit.
Dialogue: 0,0:02:05.59,0:02:10.25,Default,,0000,0000,0000,,Now, how do we find I,\Nshort circuit in this circuit?
Dialogue: 0,0:02:10.25,0:02:17.74,Default,,0000,0000,0000,,Well, by shorting that, out that brings\Nthis, the voltage here, down to ground.
Dialogue: 0,0:02:17.74,0:02:25.07,Default,,0000,0000,0000,,We have 0.5 amps flowing into now\Na node which consists of two resistors,
Dialogue: 0,0:02:25.07,0:02:29.78,Default,,0000,0000,0000,,this 30 ohm resistor and\Nthis 5 ohm resistor.
Dialogue: 0,0:02:29.78,0:02:33.30,Default,,0000,0000,0000,,Some of the current's gonna through the 5\Nohm resistor, that's our short circuit
Dialogue: 0,0:02:33.30,0:02:36.63,Default,,0000,0000,0000,,current, the rest of the current is\Ngonna come through this 30 ohm resistor.
Dialogue: 0,0:02:36.63,0:02:39.09,Default,,0000,0000,0000,,We have a current divider here.
Dialogue: 0,0:02:39.09,0:02:42.60,Default,,0000,0000,0000,,So using current divider,\Nwe can calculate short circuit directly.
Dialogue: 0,0:02:42.60,0:02:49.64,Default,,0000,0000,0000,,I short circuit is equal to 0.5 times,\Nand remember from the current divider,
Dialogue: 0,0:02:49.64,0:02:55.74,Default,,0000,0000,0000,,the term that goes in the numerator is\Nthe resistance in the other branch.
Dialogue: 0,0:02:55.74,0:03:01.34,Default,,0000,0000,0000,,So that would be 30 ohms divided\Nby the sum of those two,
Dialogue: 0,0:03:01.34,0:03:04.93,Default,,0000,0000,0000,,30 plus 5 is 35 ohms.
Dialogue: 0,0:03:04.93,0:03:10.74,Default,,0000,0000,0000,,And we get then the short circuit current
Dialogue: 0,0:03:10.74,0:03:16.97,Default,,0000,0000,0000,,is equal to 0.4286, 0.4286 amps.
Dialogue: 0,0:03:16.97,0:03:22.10,Default,,0000,0000,0000,,And then RTh is equal to Voc
Dialogue: 0,0:03:22.10,0:03:27.93,Default,,0000,0000,0000,,divided by Isc, which is 15
Dialogue: 0,0:03:27.93,0:03:34.10,Default,,0000,0000,0000,,divided by 0.4286.
Dialogue: 0,0:03:34.10,0:03:39.16,Default,,0000,0000,0000,,Which, when we get the calculator out on\Nthat, you'll find that equals 35 ohms.
Dialogue: 0,0:03:42.08,0:03:45.14,Default,,0000,0000,0000,,So, method 1.
Dialogue: 0,0:03:45.14,0:03:50.17,Default,,0000,0000,0000,,Method 2,\Nlet's clear this up just a little bit.
Dialogue: 0,0:03:50.17,0:03:50.84,Default,,0000,0000,0000,,All righty,
Dialogue: 0,0:03:50.84,0:03:54.77,Default,,0000,0000,0000,,now let's use method 2 to determine\Nthe Thevenin equivalent resistance.
Dialogue: 0,0:03:54.77,0:04:00.04,Default,,0000,0000,0000,,In method 2, which works if you have,
Dialogue: 0,0:04:00.04,0:04:04.20,Default,,0000,0000,0000,,only, independent sources.
Dialogue: 0,0:04:05.32,0:04:07.93,Default,,0000,0000,0000,,Method two, you deactivate the sources.
Dialogue: 0,0:04:07.93,0:04:10.39,Default,,0000,0000,0000,,In this case,\Nwe have an independent current source.
Dialogue: 0,0:04:10.39,0:04:12.77,Default,,0000,0000,0000,,We turn the independent\Ncurrent source to zero.
Dialogue: 0,0:04:12.77,0:04:17.46,Default,,0000,0000,0000,,And, that is effectively,\Nopen circuiting this branch,
Dialogue: 0,0:04:17.46,0:04:23.09,Default,,0000,0000,0000,,which means that there will be no\Ncurrent flowing through this branch.
Dialogue: 0,0:04:23.09,0:04:27.36,Default,,0000,0000,0000,,Such that as we look\Nback in to the circuit,
Dialogue: 0,0:04:27.36,0:04:33.66,Default,,0000,0000,0000,,this is just hanging unconnected and\Ndoes not enter into this.
Dialogue: 0,0:04:33.66,0:04:36.35,Default,,0000,0000,0000,,Effectively, we have,\Nbecause this is open circuit here,
Dialogue: 0,0:04:36.35,0:04:38.64,Default,,0000,0000,0000,,this is effectively\Nan infinite resistance here.
Dialogue: 0,0:04:39.84,0:04:46.15,Default,,0000,0000,0000,,And we see coming back into this then,\Nthe 5 ohms in series with the 35.
Dialogue: 0,0:04:46.15,0:04:47.84,Default,,0000,0000,0000,,So method 2,
Dialogue: 0,0:04:52.80,0:05:00.78,Default,,0000,0000,0000,,We get that R Thevenin is simply\Nthe 30 + 5 ohms = 35 ohms,
Dialogue: 0,0:05:00.78,0:05:04.86,Default,,0000,0000,0000,,same value we got for method 1.
Dialogue: 0,0:05:04.86,0:05:06.59,Default,,0000,0000,0000,,Now, let's take a look at method 3.
Dialogue: 0,0:05:06.59,0:05:12.06,Default,,0000,0000,0000,,Method 3 works regardless of\Nthe types of sources that we have,
Dialogue: 0,0:05:12.06,0:05:17.34,Default,,0000,0000,0000,,but in method 3,\Nwe do deactivate the independent sources.
Dialogue: 0,0:05:17.34,0:05:19.78,Default,,0000,0000,0000,,So once again,\Nwe have this independent current source.
Dialogue: 0,0:05:19.78,0:05:23.24,Default,,0000,0000,0000,,By deactivating it, we open the circuit.
Dialogue: 0,0:05:23.24,0:05:28.18,Default,,0000,0000,0000,,And then, method 3 says that\Nwe apply a test voltage,
Dialogue: 0,0:05:28.18,0:05:33.24,Default,,0000,0000,0000,,which then forces a test\Ncurrent into the AB terminals,
Dialogue: 0,0:05:33.24,0:05:39.50,Default,,0000,0000,0000,,and we then algebraically work to\Nmake the ratio of V test over I test.
Dialogue: 0,0:05:39.50,0:05:44.48,Default,,0000,0000,0000,,So method 3 then Is form the ratio V
Dialogue: 0,0:05:44.48,0:05:49.99,Default,,0000,0000,0000,,test over I test to give us R Thevenin.
Dialogue: 0,0:05:49.99,0:05:55.29,Default,,0000,0000,0000,,All righty, with that open circuit,\Nthis is a simple situation.
Dialogue: 0,0:05:55.29,0:06:00.99,Default,,0000,0000,0000,,I, going in here,\Nwe can write a KVL loop going
Dialogue: 0,0:06:00.99,0:06:05.97,Default,,0000,0000,0000,,around this loop in the direction of IT.
Dialogue: 0,0:06:05.97,0:06:12.45,Default,,0000,0000,0000,,So KVL, we have then, minus v test\Ngoing in the direction of current flow.
Dialogue: 0,0:06:12.45,0:06:19.35,Default,,0000,0000,0000,,We have then a voltage drop of plus five\NI sub t coming on down here plus 30.
Dialogue: 0,0:06:21.18,0:06:27.32,Default,,0000,0000,0000,,That's a plus sign, 30 I sub t,\Nthe sum of those terms equals zero.
Dialogue: 0,0:06:27.32,0:06:30.35,Default,,0000,0000,0000,,Let's bring in V test over to\Nthe other side of the positive V test,
Dialogue: 0,0:06:30.35,0:06:31.38,Default,,0000,0000,0000,,factor out the I test.
Dialogue: 0,0:06:31.38,0:06:38.50,Default,,0000,0000,0000,,We had I test x 5 + 30 is 35 = V test.
Dialogue: 0,0:06:38.50,0:06:42.97,Default,,0000,0000,0000,,Divide both sides by I test,\Nthat gives us our R Thevenin,
Dialogue: 0,0:06:42.97,0:06:47.56,Default,,0000,0000,0000,,which is equal to the ratio of\NV test over I test = 35 ohms.
Dialogue: 0,0:06:47.56,0:06:52.84,Default,,0000,0000,0000,,So we get the same results in\Nany one of the three methods,
Dialogue: 0,0:06:52.84,0:06:55.48,Default,,0000,0000,0000,,and we then, can or now can,
Dialogue: 0,0:06:58.75,0:07:03.97,Default,,0000,0000,0000,,Draw our Thevenin equivalent circuit,\Nwhere V Thevenin
Dialogue: 0,0:07:03.97,0:07:09.52,Default,,0000,0000,0000,,is equal to 15 volts, and\NR Thevenin is equal to 35 ohms.
Dialogue: 0,0:07:11.30,0:07:14.36,Default,,0000,0000,0000,,Notice that the original\Ncircuit had a current source.
Dialogue: 0,0:07:15.72,0:07:20.26,Default,,0000,0000,0000,,But the Thevenin Equivalent Circuit\Nreplaces that source and
Dialogue: 0,0:07:20.26,0:07:24.50,Default,,0000,0000,0000,,all of the resistance, the effects of\Nthe resistance within this circuit,
Dialogue: 0,0:07:25.60,0:07:30.84,Default,,0000,0000,0000,,are represented by a single voltage source\Nand the Thevenin Equivalent resistance.