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Thevenin's Theorem
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0:00
Let’s make you an expert at using Thevenin’s theorem to solve circuits! Imagine you have a circuit like this and you need to find the voltage or current through this resistor. So we call that our load, in simple terms, that’s the circuit element we need to find values for. Now what if there was a way to represent this whole circuit, excluding our load, like this. To find the values for our load, all we need to do is place it in series with
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the other circuit elements. Let’s talk about some terminology. This is called the Thevenin equivalent voltage, or VTH and this resistor here is called the Thevenin’s equivalent resistance, also called RTH. In very simple terms, this represents all the current sources and voltages in a circuit as a single voltage source at the place where the load was connected. And this resistor represents all the other resistors on our circuit as a single resistor excluding our
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load. So all we need to do is just find those two values when we solve a problem using Thevenin’s theorem. Now to find these two values depends on the type of circuit you’re given. There are 3 possible scenarios and I will briefly go over them and then cover them fully through examples. First case, we have just independent sources. Here, to find VTH, we remove the load and calculate the voltage between these two terminals. This voltage is called VOC,
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or open circuit voltage. To find RTH, we convert all independent voltage sources to short circuits and all current sources to open circuits. Then we look inwards towards the circuit from where our load used to be, and we calculate the total resistance between the terminals. Second case is when we have a mix of dependant and independent sources. In this case, we can find VTH the same way as before, so we remove load, and then calculate the VOC between the terminals.
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However, we can’t find RTH the same way as before. Why? Well, we can’t remove dependant sources, that means we need another way. So what we do is, create a short circuit between the terminals where our load used to be. Now there is a current that’ll be flowing through this new short circuit, and we need to find it, we call that, short circuit current, ISC for short. Once you have the ISC, you need to keep this equation in mind. Here, all we need to do is divide our open-source
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voltage by the short circuit current and that’ll give us the Thevenin equivalent resistance. The last case is when we have just dependant sources. Now in this case, we obviously can’t find VTH since there is no flow of electricity because we have no independent sources. So what we do is, we remove the load, and then we put in a 1 mA current source between the terminals. Now we can once again figure out the voltage between the terminals. The cool thing is, since we placed our
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own current source between the terminals, that’s also our ISC value, which is just 1 mA. So using our trusty equation, we can find the RTH. Now sometimes, using the Thevenin equivalent circuit to solve for a load is actually not faster than other methods like nodal analysis or mesh analysis, but there is a huge benefit to it. Imagine you wanted to try exchanging different loads in the circuit and see the results. Then all you do is draw the thevenin’s equivalent circuit,
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and you just keep replacing the load with what you want, and you don’t have to reanalyse a whole circuit every single time. Now to really understand all of this, let’s go through a bunch of examples covering everything we just said.
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Let’s start with a nice simple example where we need to find V0 using Thevenin’s theorem. We see that V0 is the voltage across this 2k ohm resistor, so that’s our load. The first thing we do is remove this load from the circuit so that our circuit looks like this. The next step is for us to figure out the open circuit voltage, VOC, between these points. You can use any method you’re comfortable using to do this, like superposition or mesh analysis,
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KVL or KCL, whatever you want. Here, we can write a KVL equation for this loop and then we can figure out the VOC, but to do that, we need the voltage across this 2k ohm resistor. In this circuit, since we removed the load resistor, this part is an open circuit, so the only current flow would be through this loop. That means we can easily write a KVL equation. We will assume a counter-clockwise current flow, which means the positives and
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negatives of our resistors would look like this. Remember, we always assume current enters the positive side on resistors. Let’s solve for the current in the loop, we get 1 mA. So that means the voltage through this 2k ohm resistor is 2 V using ohms law. If this is unfamiliar to you, please check the description, I have a whole video on ohms law and KVL. It should be noted here that k represents 1000, and m represents 10 to the negative 3. This
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makes it easier to write down values, rather than writing down 2000 ohms or 0.001amps. Now we can write another KVL for this loop, this time including the open circuit voltage. Remember, KVL must hold true for any loop we choose. Keep in mind the negative and positive signs as we go around the loop counter-clockwise. We get 8V.
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The next step is to figure out the Thevenin equivalent resistance, RTH. When we find RTH, voltage sources turn into short circuits and current sources turn into open circuits. Here, we have 2 voltage sources so we just remove them and have short circuits in place. Now, we look inwards from the place where we removed our load. So we are looking to the left on this circuit from where the load was. We see both of these resistors are parallel to each other,
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so our RTH is 1.33 kohms. If you need a refresh on combining resistors, please see the description, I have a whole video on that topic. Now that we have both the voltage and resistance, we can draw a new circuit with those values in series to each other. This is our Thevenin equivalent circuit. And right here, we can place our load. All of these circuit elements are in series to each other. So again, this is our open-source voltage, and this is our Thevenin
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equivalent resistance. Notice how the positive sign of our voltage source is on the top. That’s because when we calculated the VOC, we had the positive sign on top. Now using voltage division, we can figure out the voltage in the 2k ohm resistor. Remember in voltage division, we put the value we want on the top and on the bottom is the addition of the resistors, and you multiply it by the voltage source value. I have a whole video on this, so if you need a refresh,
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please see the description for the link. And that’s our answer. Let’s take a look at another example, this time with a voltage source and a current source. Here, we need to find V0 using Thevenin’s theorem. So first, we remove our load, which is this 2kohm resistor. Next, we need to determine the open-circuit voltage. That’s the voltage between these two points. If we find the voltage in this resistor and this one, we can write a KVL for this
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whole loop to figure out the VOC. We can use mesh analysis to figure out the values we need. So, let’s assume a clockwise current flow in this loop, and we already have a constrained current in this loop. Let’s write our first mesh equation for the left loop. Remember, here, both currents are going in the same direction, so it’s the addition of I1 and the 6mA current going through this 2k ohm resistor. Now we can directly solve for I1. If this is unfamiliar to you,
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I have a whole video on mesh analysis, so please take a look that to get a refresh. We get zero for the current in this loop so the voltage in this resistor is 0, and the voltage in this resistor is 12 V. Now we can write a KVL for this loop. Let’s solve for the open voltage, and we get 12 V. Next, we need to find RTH. So remember, voltage sources turn into short circuits and current sources turn into open circuits. When we do that, this is what our circuit looks like. Now,
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looking down from where our load was, we see that these two resistors are parallel to each other, and the combination of that is in series with this resistor. Now that we have both VOC and RTH, we can draw a new circuit with those values. We have the voltage in series with the resistor, and here we can place our load, which was the 2k ohm resistor. The last step is to use voltage division to figure out V0. We get 4.8 volts and that’s our answer.
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Let’s take a look at this question, where we need to use Thevenin’s theorem to figure out I0. So I0 is actually the current flowing through this 2kohm resistor. That’s our load, so let’s remove it. Since the current was coming down, that means the top is at a higher potential than the bottom, since current flows from a higher potential to a lower potential. So our positive for VOC would be
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up top and bottom would be negative. So what we want is the open-source voltage, and it might be hard to see where it is we are measuring. Right here is where we are measuring. So the voltage between these two points. What is the easiest way to get the voltage between these two points? Well, writing a KVL around this loop would give us a quick answer because the only other voltage we need to find is the voltage of this 4kohm resistor. And what is the voltage of
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this resistor? Notice that once we removed our load, this current has no where to flow except through this conductor. So the voltage of this resistor is just 4 kohms multiplied by the 2 ma. Now we can write our KVL for this loop going clockwise. The next step is to figure out RTH. So our current sources become open circuits while the voltage sources become short circuits. This conductor here is useless in this scenario,
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so let’s remove it. Much better. So now, looking inwards from here, what is our RTH? These two resistors are parallel to each other, and the sum of that is in series with this 2kohm resistor. That gives us an RTH of 4 kiloohms. Now we can redraw our circuit. We have the voltage and resistor in series, and then we place the load here. Next, we can use ohms law to get the current. We get 2.67 mA and that’s our answer.
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Let’s take a look at an example with an independent voltage source and a dependant voltage source. In this problem, we need to find V0 using Thevenin’s theorem. So, we first remove our load, which is this 4kohm resistor. Since our circuit is now different than the original one, we will relabel VA as VA1. So now, we need to figure out VOC. How can we do that? If we can write a KVL for this loop, we can figure out VOC. But to do that, we need to figure the voltage of this resistor,
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which is then needed to figure out the voltage of this dependent voltage source. So let’s figure out VA1. Notice that since we removed the load, current can only flow through this loop. Assuming a clockwise current flow, let’s write a KVL equation. We get 2 mA, and we can use that to figure out VA1 with ohms law. We get 8 V. Now we can write another KVL equation for this loop and we will go clockwise. We get a negative value, and that’s okay,
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that simply means the assigned positives and negatives for VOC is opposite to what’s given. Next comes finding RTH. Unlike the previous examples, we actually can’t just combine resistors because we can’t get rid of dependent sources. Instead, what we do is we short circuit the terminals of where the load was. So in this circuit, that would look like this. Now there is a current flowing in this new conductor, and we will label that ISC, which is short circuit current. We
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need to figure out the value of ISC, but keep in mind, the values we found before for our dependant voltage source no longer apply, the circuit has been changed. So we will label this voltage as VA2. Let’s use mesh analysis to figure it out. First the left loop. Let’s simplify. Next, the right loop. We get 2 equations with 3 unknowns, but remember, VA2 is voltage in the 4kohm resistor, and we know that voltage is equal to I1 multiplied by the 4kohms. Let’s plug that into our
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2nd equation. Now we can solve the two equations. We now have both VOC and ISC, so we can get RTH. RTH is equal to VOC divided by ISC. It’s finally time for us to draw our circuit. So we have our voltage source in series with our resistor. Note here that our voltage is still -12 volts, since we follow the same positive direction as our VOC. Lastly, we can place our load here. Now we can use voltage division to figure out V0. Again, we get a negative value,
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and that means the assigned positives and negatives for our V0 is opposite to what’s shown. Let’s tackle a problem with independent and dependent sources for both current and voltage. In this example, we need to find I2. I2 is the current flowing through this 7-ohm resistor. So that’ll be our load, and we can remove it. When we do that, our circuit changes, so we need to relabel our dependent voltages and currents. Now we need to figure out the
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VOC between these two points. Remember, since our current was coming down, our positive would be on top and negative on the bottom. If we can figure out the voltage of this 6 ohm resistor, we can write a KVL for this loop and figure out the VOC. So let’s do that by using mesh analysis. The current in this mesh is constrained by the 5 A source. We can now write our mesh equations. Let’s start with the top left mesh. Let’s simplify this equation. Note
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that VA1 is actually just the voltage of this 12 ohm resistor, and that’s just 12 ohms multiplied by I1. So we can replace that in our equation. Let’s simplify it further. Now, notice that I2 is also constrained by the dependent current source. So I2 is equal to 4IA1. Well, IA1 is just I1, since that’s current flowing in the branch. So I2 is equal to 4 times I1. Now we can plug this value back into our first equation. We get I1 is equal to 0 A. That means I2 is also 0 amps.
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Well, that certainly makes things easier. Let’s write a KVL equation for this loop. Remember, there is no voltage in this 6 ohm resistor, since the current is 0. We see that VOC is 12 volts. Next, we need to find RTH. So that means we need to create a short circuit between the points where we removed the load. There is a current flowing in this new conductor, ISC, and that’s what we need to find. Since the circuit changed, we need to relabel our
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dependent currents and voltages. Once again, we can use mesh analysis to figure that out. Let’s start with the top left loop. As before, VA2 is equal to the voltage of the 12 ohm resistor, so that’s 12I1. Let’s plug that into our equation and simplify it. Next, we can write one for the bottom left loop. Let’s simplify. If we add equation 2 into equation 1, that eliminates I2. Now using the value we found for I1, we can find I2. We’re almost there.
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Notice how I3 is constrained by the dependent current source. Also notice that IA2 is the same as our mesh current I1. So then we can write the value of our dependent current source like this. Now that we have both I2 and I3, we can easily figure out ISC. Notice we have ISC pointing down, so I2 goes in the same direction while I3 opposes it. That means ISC is equal to I2 minus I3. Now that we got the ISC value and VOC, we get use that to get RTH. RTH is VOC divided by ISC.
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It’s finally time to draw our circuit. We have the voltage source and RTH in series like this, and we can place the load here. Using ohms law, we can figure out I2 and that’s our answer.
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Let’s take a look at one last example where you have just a dependant source. Here, we just need to find the Thevenin’s equivalent circuit between points A and B. Whether the circuit has a single dependant current source, or a single dependant voltage source, or both in the same circuit, the process is the same when there are only dependant sources. Because this circuit only has a single dependant voltage source, there is no way for us to figure out a VOC or RTH since there is
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actually no current flow in the circuit. But we do need a voltage between points A and B. So instead, what we do is attach a 1mA current source between terminals A and B. Now that there is a current flow in the circuit, we can figure out the voltage between terminals A and B, which is the voltage in this 2k ohm resistor. Let’s assume some current flows in these meshes and write a few mesh equations. Let’s start with the left most loop. Now we can simplify. We can simplify it
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further by dividing all the terms by 2k. Next, the middle loop. Ix in the dependant voltage source is actually just the current flowing through this 2k ohm resistor, and if we write that as our loop currents, that’s I1 minus I2. Just a refresh, it’s I1 minus I2 because Ix is pointing down, so the same direction as I1 but I2 is opposing it. This 1 mA is actually just the loop current in this mesh,
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which is being constrained by our 1mA source. I also wrote the 1000 as 1k just to make it easier for us to visualize things. Let’s simplify. Let’s divide all the terms by 1k. Now we have 2 equations with 2 unknowns, and we can easily solve it by isolating for I1 in the 2nd equation and plugging that into the first equation. Now that we have I2, we can use that to figure out the voltage in the 2k ohm resistor. Remember that we assumed both currents to flow down, so they
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will get added together. That gives us 0.4 volts, which is the voltage between terminals A and B. Now to find RTH, it’s actually really easy. All we need to do is divide this voltage by 1 mA, since that’s the current flowing in the conductor between A and B, since we placed a 1 mA source. We get 400 ohms. Now we can draw our Thevenin equivalent circuit. Now you might be wondering, well, where is the power source? There isn’t one, because there were
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no independent sources in our original circuit. So that really is what our answer looks like. That should cover the types of problems you will face when it comes to Thevenin’s theorem. Thanks for watching and best of luck with your studies!
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