[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:05.86,Default,,0000,0000,0000,,Hello, this is Doctor Cynthia, first at the University of Utah, and today we're going to talk about
Dialogue: 0,0:00:05.86,0:00:09.66,Default,,0000,0000,0000,,Thevenin and Norton equivalent circuits. Thevenin and Norton are used when
Dialogue: 0,0:00:09.66,0:00:15.39,Default,,0000,0000,0000,,we analyze systems to allow us to simplify a complex system into simpler block diagrams that we can
Dialogue: 0,0:00:15.39,0:00:19.65,Default,,0000,0000,0000,,analyze or design individually. The concept of input and output resistance
Dialogue: 0,0:00:19.65,0:00:24.20,Default,,0000,0000,0000,,becomes very important in this case and can or not lead to voltage loading.
Dialogue: 0,0:00:24.20,0:00:27.66,Default,,0000,0000,0000,,We always consider designing our circuit as simple feminine equivalent
Dialogue: 0,0:00:27.66,0:00:32.74,Default,,0000,0000,0000,,to a Norton equivalent shown here, and they can be converted back and forth between each other using
Dialogue: 0,0:00:32.74,0:00:38.08,Default,,0000,0000,0000,,source transformation. We'll also talk about maximum power transfer and the updated ECE toolbox
Dialogue: 0,0:00:38.08,0:00:42.46,Default,,0000,0000,0000,,using these two powerful methods. We actually do this all the time we if
Dialogue: 0,0:00:42.46,0:00:47.26,Default,,0000,0000,0000,,we want to design a simple lamp or a fan, we don't actually consider the entire
Dialogue: 0,0:00:47.26,0:00:52.57,Default,,0000,0000,0000,,Power Distribution shown in this box. We convert it instead to a feminine equivalent circuit.
Dialogue: 0,0:00:52.57,0:00:58.67,Default,,0000,0000,0000,,Here is the 110 Volt source in your house and an equivalent source resistance, say that's 10 ohms.
Dialogue: 0,0:00:58.67,0:01:03.94,Default,,0000,0000,0000,,What we're saying is that the voltage and the current seen at these two
Dialogue: 0,0:01:03.94,0:01:08.33,Default,,0000,0000,0000,,points are the same as the voltage and current that they're seen at these two points.
Dialogue: 0,0:01:08.33,0:01:12.05,Default,,0000,0000,0000,,So we've done an equivalent circuit. We can then analyze this circuit
Dialogue: 0,0:01:12.05,0:01:18.02,Default,,0000,0000,0000,,as a set of blocks, as shown here. Let's consider the total resistance and find the current,
Dialogue: 0,0:01:18.02,0:01:21.94,Default,,0000,0000,0000,,and let's see what voltage we're measuring across our fan in our lamp.
Dialogue: 0,0:01:21.94,0:01:26.70,Default,,0000,0000,0000,,We can see that these two are in parallel, so 100 in parallel with 100 is 50.
Dialogue: 0,0:01:26.70,0:01:32.70,Default,,0000,0000,0000,,Plus the 10 in series means that our total resistance is 60 ohms. If we consider the current here,
Dialogue: 0,0:01:32.70,0:01:38.62,Default,,0000,0000,0000,,that's 110 volts divided by the total resistance, or 1.83 lamps, 3 amps.
Dialogue: 0,0:01:38.62,0:01:43.27,Default,,0000,0000,0000,,Now let's consider what's the voltage across the fan in the lamp using a voltage divider.
Dialogue: 0,0:01:43.27,0:01:50.13,Default,,0000,0000,0000,,That's going to be the 110 volts divided by their fifty ohms divided by sorry times their fifty ohms
Dialogue: 0,0:01:50.13,0:01:55.49,Default,,0000,0000,0000,,divided by 10 + 50 or 60 ohms, which gives us 92 volts. Now wait a minute.
Dialogue: 0,0:01:55.49,0:01:59.93,Default,,0000,0000,0000,,That is significantly less than the voltage that we started with. That's called voltage loading.
Dialogue: 0,0:01:59.93,0:02:06.08,Default,,0000,0000,0000,,Where did the voltage go? It went across our S right here. What could we have done to reduce
Dialogue: 0,0:02:06.08,0:02:10.56,Default,,0000,0000,0000,,this voltage loading? We could have reduced our source resistance to a much smaller value
Dialogue: 0,0:02:10.56,0:02:15.17,Default,,0000,0000,0000,,so that we got a higher value here. Take a minute, stop the video and see if you
Dialogue: 0,0:02:15.17,0:02:18.35,Default,,0000,0000,0000,,can calculate what the voltage is across the fan in the lamp.
Dialogue: 0,0:02:18.35,0:02:24.79,Default,,0000,0000,0000,,Is it closer to 110 volts? Now let's talk about the input and output resistance.
Dialogue: 0,0:02:24.79,0:02:30.61,Default,,0000,0000,0000,,The input resistance is the resistance looking into the circuit, and the output resistance is the is
Dialogue: 0,0:02:30.61,0:02:33.70,Default,,0000,0000,0000,,the resistance that's looking into the output part of the circuit.
Dialogue: 0,0:02:33.70,0:02:39.34,Default,,0000,0000,0000,,So here is an output resistance and here is another input resistance. Let's number these.
Dialogue: 0,0:02:39.34,0:02:42.34,Default,,0000,0000,0000,,This is R out for the first block, R in and R out for
Dialogue: 0,0:02:42.34,0:02:45.35,Default,,0000,0000,0000,,the second block, and R in for the third block.
Dialogue: 0,0:02:45.35,0:02:48.95,Default,,0000,0000,0000,,Here's how you can decide if you need to consider voltage loading or not.
Dialogue: 0,0:02:48.95,0:02:54.67,Default,,0000,0000,0000,,If R out from our first block is much less than, not less than,
Dialogue: 0,0:02:54.67,0:03:01.12,Default,,0000,0000,0000,,or equal to much less than R in for our second block, then no voltage loading occurs and the
Dialogue: 0,0:03:01.12,0:03:04.75,Default,,0000,0000,0000,,circuits can be considered decoupled. So in this case,
Dialogue: 0,0:03:04.75,0:03:09.77,Default,,0000,0000,0000,,is 10 ohms much less than 100 ohms? No, In fact, probably not.
Dialogue: 0,0:03:09.77,0:03:16.65,Default,,0000,0000,0000,,How about one ohms? Yes. So in this case they would still be considered coupled.
Dialogue: 0,0:03:16.65,0:03:22.77,Default,,0000,0000,0000,,In this case they would be considered decoupled. Let's consider the next set.
Dialogue: 0,0:03:22.77,0:03:26.33,Default,,0000,0000,0000,,Let's say R out two.
Dialogue: 0,0:03:26.33,0:03:31.49,Default,,0000,0000,0000,,Is that much less than R in 2R in three? Absolutely not.
Dialogue: 0,0:03:31.49,0:03:38.33,Default,,0000,0000,0000,,In this case we have 100 ohms. Is that much less than the 100 ohms input to R3? No.
Dialogue: 0,0:03:38.33,0:03:45.23,Default,,0000,0000,0000,,So these circuits must always be considered coupled because they are not much less than.
Dialogue: 0,0:03:45.23,0:03:49.45,Default,,0000,0000,0000,,This tells us that we need to analyze these two together, and in fact we did because we
Dialogue: 0,0:03:49.45,0:03:53.31,Default,,0000,0000,0000,,considered them in parallel. But if we were using our one ohm source,
Dialogue: 0,0:03:53.31,0:03:58.06,Default,,0000,0000,0000,,if this was a simple one ohm source, we could at least consider the source
Dialogue: 0,0:03:58.06,0:04:04.17,Default,,0000,0000,0000,,block independent from the load blocks. OK, now let's go back to look at
Dialogue: 0,0:04:04.17,0:04:08.42,Default,,0000,0000,0000,,our Thevenin equivalent circuit. What we have done is considered two points,
Dialogue: 0,0:04:08.42,0:04:15.36,Default,,0000,0000,0000,,A and B in our circuit, and we have analyzed those as a very We've taken our actual circuit and
Dialogue: 0,0:04:15.36,0:04:19.86,Default,,0000,0000,0000,,converted it into a much simpler Febinine equivalent circuit. In order to do this,
Dialogue: 0,0:04:19.86,0:04:25.90,Default,,0000,0000,0000,,we need to find two things, the V Thevenin and the R Thevenin. Let's see how to do that.
Dialogue: 0,0:04:25.90,0:04:30.10,Default,,0000,0000,0000,,In order to find the V thevan and the first thing we do is what's called the open circuit method.
Dialogue: 0,0:04:30.10,0:04:35.20,Default,,0000,0000,0000,,We remove the loads and we open circuit the actual circuit and we measure, calculate or simulate the
Dialogue: 0,0:04:35.20,0:04:41.41,Default,,0000,0000,0000,,voltage between points A and B. That's called the open circuit voltage. So similarly,
Dialogue: 0,0:04:41.41,0:04:48.14,Default,,0000,0000,0000,,as soon as we measure that open circuit voltage, we can see that Vth is equal to Voc.
Dialogue: 0,0:04:48.14,0:04:54.00,Default,,0000,0000,0000,,Again, we can measure, calculate, or simulate this. There are three ways to
Dialogue: 0,0:04:54.00,0:04:58.40,Default,,0000,0000,0000,,find the Fevodin resistance, and I'm going to show you all three. The first one can only be used if
Dialogue: 0,0:04:58.40,0:05:01.91,Default,,0000,0000,0000,,there are no dependent sources, and it's probably the easiest of the methods.
Dialogue: 0,0:05:01.91,0:05:07.15,Default,,0000,0000,0000,,What you do is you deactivate all the independent sources, you convert a voltage to a short circuit,
Dialogue: 0,0:05:07.15,0:05:11.39,Default,,0000,0000,0000,,a current to an open circuit, and of course you can't have dependent sources.
Dialogue: 0,0:05:11.39,0:05:16.59,Default,,0000,0000,0000,,Then you simply look into points A and B and you see what the equivalent resistance is there.
Dialogue: 0,0:05:16.59,0:05:22.10,Default,,0000,0000,0000,,And that's equal to the Fevodin resistance. Let's do this. Here's an example.
Dialogue: 0,0:05:22.10,0:05:28.30,Default,,0000,0000,0000,,We want to do a feminine equivalent circuit of this more complicated circuit between points A and B.
Dialogue: 0,0:05:28.30,0:05:31.70,Default,,0000,0000,0000,,The first thing that we do is deactivate the voltage source by shorting it,
Dialogue: 0,0:05:31.70,0:05:34.75,Default,,0000,0000,0000,,and deactivate the current source by opening it
Dialogue: 0,0:05:34.75,0:05:38.08,Default,,0000,0000,0000,,up as we have here. Then we look to see what does the
Dialogue: 0,0:05:38.08,0:05:42.81,Default,,0000,0000,0000,,resistance look like when we look inside. Well, we can see that we have 250
Dialogue: 0,0:05:42.81,0:05:49.27,Default,,0000,0000,0000,,ohm resistors in parallel, so that gives us a 25 ohm resistance. That block of two would be
Dialogue: 0,0:05:49.27,0:05:55.23,Default,,0000,0000,0000,,in series with this 35 ohm. So combine the 25 and 35 and that's going to give you 60 ohms,
Dialogue: 0,0:05:55.23,0:06:01.43,Default,,0000,0000,0000,,which when you put it in parallel with this 30 ohm is going to be the Rth is equal to 20 ohms.
Dialogue: 0,0:06:01.43,0:06:07.39,Default,,0000,0000,0000,,We could have measured, simulated or calculated this. In this case we calculated it.
Dialogue: 0,0:06:07.39,0:06:12.44,Default,,0000,0000,0000,,Now here's the 2nd way to do this. Let's use the open circuit short circuit method.
Dialogue: 0,0:06:12.44,0:06:18.65,Default,,0000,0000,0000,,We have already opened the circuit in order to find Voc. We can and so that's our. That's the Voc.
Dialogue: 0,0:06:18.65,0:06:23.82,Default,,0000,0000,0000,,Then we can short circuit the load and measure the current. In that case,
Dialogue: 0,0:06:23.82,0:06:29.28,Default,,0000,0000,0000,,R Thevenin is equal to Voc divided by ISC. Again, we can measure,
Dialogue: 0,0:06:29.28,0:06:34.13,Default,,0000,0000,0000,,calculate or simulate this. Let's calculate it. Here's another example.
Dialogue: 0,0:06:34.13,0:06:39.38,Default,,0000,0000,0000,,Let's find the Thevenin equivalent between points A and B. Now, the first thing that we're going to
Dialogue: 0,0:06:39.38,0:06:45.63,Default,,0000,0000,0000,,do is measure the Voc across here. So let's do that using node voltage.
Dialogue: 0,0:06:45.63,0:06:48.64,Default,,0000,0000,0000,,In this case, we're going to have VC
Dialogue: 0,0:06:48.64,0:06:54.12,Default,,0000,0000,0000,,-24 / 6 plus VC -0 / 12 plus A7 amp outgoing.
Dialogue: 0,0:06:54.12,0:06:57.73,Default,,0000,0000,0000,,And then we don't need to consider this because any resistor that is connected to
Dialogue: 0,0:06:57.73,0:07:01.44,Default,,0000,0000,0000,,an open circuit does not have a current going through it and can be neglected.
Dialogue: 0,0:07:01.44,0:07:07.08,Default,,0000,0000,0000,,So we can use this to calculate that V thevenant is -12 volts. That shows up right here.
Dialogue: 0,0:07:07.08,0:07:13.11,Default,,0000,0000,0000,,And notice we simply turned the V thevenant upside down. OK, now let's find the R thevenin.
Dialogue: 0,0:07:13.11,0:07:18.27,Default,,0000,0000,0000,,We need the short circuit current, so let's take VC prime because this is a different circuit.
Dialogue: 0,0:07:18.27,0:07:22.54,Default,,0000,0000,0000,,VC prime -24 / 6 plus VC
Dialogue: 0,0:07:22.54,0:07:27.51,Default,,0000,0000,0000,,prime -0 / 12 + 7 amps. And this time the two ohms does show up.
Dialogue: 0,0:07:27.51,0:07:31.73,Default,,0000,0000,0000,,So it's VC prime -0 / 2 ohms that I'll
Dialogue: 0,0:07:31.73,0:07:36.91,Default,,0000,0000,0000,,write here says VC prime is -4 volts. Then we consider the I short circuit current.
Dialogue: 0,0:07:36.91,0:07:41.43,Default,,0000,0000,0000,,We know what VC prime is, so we can find ISC. That's -2 amps.
Dialogue: 0,0:07:41.43,0:07:44.65,Default,,0000,0000,0000,,In this case R Thevenin is V Thevenin
Dialogue: 0,0:07:44.65,0:07:50.31,Default,,0000,0000,0000,,which was -12 / I short circuit -2. That's six ohms. A common mistake.
Dialogue: 0,0:07:50.31,0:07:54.67,Default,,0000,0000,0000,,Please don't make this is do not take the VC prime and put it in here.
Dialogue: 0,0:07:54.67,0:08:00.27,Default,,0000,0000,0000,,It's supposed to be the V Thevenin divided by the I short circuit current.
Dialogue: 0,0:08:00.27,0:08:03.41,Default,,0000,0000,0000,,So that gives us V Thevenin is 6 ohms.