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Hello, this is Doctor Cynthia, first at the University of Utah, and today we're going to talk about

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Thevenin and Norton equivalent circuits. Thevenin and Norton are used when

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we analyze systems to allow us to simplify a complex system into simpler block diagrams that we can

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analyze or design individually. The concept of input and output resistance

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becomes very important in this case and can or not lead to voltage loading.

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We always consider designing our circuit as simple feminine equivalent

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to a Norton equivalent shown here, and they can be converted back and forth between each other using

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source transformation. We'll also talk about maximum power transfer and the updated ECE toolbox

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using these two powerful methods. We actually do this all the time we if

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we want to design a simple lamp or a fan, we don't actually consider the entire

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Power Distribution shown in this box. We convert it instead to a feminine equivalent circuit.

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Here is the 110 Volt source in your house and an equivalent source resistance, say that's 10 ohms.

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What we're saying is that the voltage and the current seen at these two

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points are the same as the voltage and current that they're seen at these two points.

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So we've done an equivalent circuit. We can then analyze this circuit

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as a set of blocks, as shown here. Let's consider the total resistance and find the current,

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and let's see what voltage we're measuring across our fan in our lamp.

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We can see that these two are in parallel, so 100 in parallel with 100 is 50.

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Plus the 10 in series means that our total resistance is 60 ohms. If we consider the current here,

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that's 110 volts divided by the total resistance, or 1.83 lamps, 3 amps.

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Now let's consider what's the voltage across the fan in the lamp using a voltage divider.

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That's going to be the 110 volts divided by their fifty ohms divided by sorry times their fifty ohms

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divided by 10 + 50 or 60 ohms, which gives us 92 volts. Now wait a minute.

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That is significantly less than the voltage that we started with. That's called voltage loading.

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Where did the voltage go? It went across our S right here. What could we have done to reduce

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this voltage loading? We could have reduced our source resistance to a much smaller value

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so that we got a higher value here. Take a minute, stop the video and see if you

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can calculate what the voltage is across the fan in the lamp.

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Is it closer to 110 volts? Now let's talk about the input and output resistance.

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The input resistance is the resistance looking into the circuit, and the output resistance is the is

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the resistance that's looking into the output part of the circuit.

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So here is an output resistance and here is another input resistance. Let's number these.

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This is R out for the first block, R in and R out for

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the second block, and R in for the third block.

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Here's how you can decide if you need to consider voltage loading or not.

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If R out from our first block is much less than, not less than,

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or equal to much less than R in for our second block, then no voltage loading occurs and the

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circuits can be considered decoupled. So in this case,

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is 10 ohms much less than 100 ohms? No, In fact, probably not.

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How about one ohms? Yes. So in this case they would still be considered coupled.

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In this case they would be considered decoupled. Let's consider the next set.

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Let's say R out two.

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Is that much less than R in 2R in three? Absolutely not.

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In this case we have 100 ohms. Is that much less than the 100 ohms input to R3? No.

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So these circuits must always be considered coupled because they are not much less than.

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This tells us that we need to analyze these two together, and in fact we did because we

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considered them in parallel. But if we were using our one ohm source,

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if this was a simple one ohm source, we could at least consider the source

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block independent from the load blocks. OK, now let's go back to look at

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our Thevenin equivalent circuit. What we have done is considered two points,

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A and B in our circuit, and we have analyzed those as a very We've taken our actual circuit and

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converted it into a much simpler Febinine equivalent circuit. In order to do this,

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we need to find two things, the V Thevenin and the R Thevenin. Let's see how to do that.

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In order to find the V thevan and the first thing we do is what's called the open circuit method.

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We remove the loads and we open circuit the actual circuit and we measure, calculate or simulate the

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voltage between points A and B. That's called the open circuit voltage. So similarly,

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as soon as we measure that open circuit voltage, we can see that Vth is equal to Voc.

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Again, we can measure, calculate, or simulate this. There are three ways to

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find the Fevodin resistance, and I'm going to show you all three. The first one can only be used if

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there are no dependent sources, and it's probably the easiest of the methods.

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What you do is you deactivate all the independent sources, you convert a voltage to a short circuit,

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a current to an open circuit, and of course you can't have dependent sources.

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Then you simply look into points A and B and you see what the equivalent resistance is there.

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And that's equal to the Fevodin resistance. Let's do this. Here's an example.

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We want to do a feminine equivalent circuit of this more complicated circuit between points A and B.

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The first thing that we do is deactivate the voltage source by shorting it,

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and deactivate the current source by opening it

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up as we have here. Then we look to see what does the

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resistance look like when we look inside. Well, we can see that we have 250

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ohm resistors in parallel, so that gives us a 25 ohm resistance. That block of two would be

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in series with this 35 ohm. So combine the 25 and 35 and that's going to give you 60 ohms,

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which when you put it in parallel with this 30 ohm is going to be the Rth is equal to 20 ohms.

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We could have measured, simulated or calculated this. In this case we calculated it.

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Now here's the 2nd way to do this. Let's use the open circuit short circuit method.

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We have already opened the circuit in order to find Voc. We can and so that's our. That's the Voc.

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Then we can short circuit the load and measure the current. In that case,

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R Thevenin is equal to Voc divided by ISC. Again, we can measure,

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calculate or simulate this. Let's calculate it. Here's another example.

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Let's find the Thevenin equivalent between points A and B. Now, the first thing that we're going to

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do is measure the Voc across here. So let's do that using node voltage.

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In this case, we're going to have VC

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-24 / 6 plus VC -0 / 12 plus A7 amp outgoing.

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And then we don't need to consider this because any resistor that is connected to

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an open circuit does not have a current going through it and can be neglected.

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So we can use this to calculate that V thevenant is -12 volts. That shows up right here.

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And notice we simply turned the V thevenant upside down. OK, now let's find the R thevenin.

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We need the short circuit current, so let's take VC prime because this is a different circuit.

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VC prime -24 / 6 plus VC

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prime -0 / 12 + 7 amps. And this time the two ohms does show up.

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So it's VC prime -0 / 2 ohms that I'll

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write here says VC prime is -4 volts. Then we consider the I short circuit current.

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We know what VC prime is, so we can find ISC. That's -2 amps.

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In this case R Thevenin is V Thevenin

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which was -12 / I short circuit -2. That's six ohms. A common mistake.

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Please don't make this is do not take the VC prime and put it in here.

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It's supposed to be the V Thevenin divided by the I short circuit current.

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So that gives us V Thevenin is 6 ohms.