Hey, guys, in this video, we're going to be discussing Larson circuits, which are another type of induct er circuit. Let's get to it now. As the name implies, circuits are composed of three things they're composed of induct er's, which is what the L stands for resistors, which is what the R stands for, and finally, capacitors, which is what the C stands for in Dr Resistor capacitor circuits now in an LLC circuit. If we begin with the capacitor initially charged, we can write out your chops Loop rule for this circuit. What's gonna happen is as soon as this circuit is completed and the capacity is allowed to discharge, a current is going to be produced. That current is going to go this way through the induct ER and that current is going to go this way through the resistor. And now we want to know is how to assign positive and negative values for the voltages in kerchiefs Lupul. First we have to choose a loop. I'll just choose this as an easy loop. Now let's leave the induct er for last. The resist during the capacity are going to be easy. The loop goes with the polarity of the capacitor. So that's a positive voltage for the capacitor. The loop goes with the current through the resistor. So that is a negative voltage for the resistor. Now, finally, the induct er Okay. Now the direction of the e m f and the induct er is going to be determined by whether or not that current is increasing or decreasing. In this case, the current is going to be decreasing initially when the capacitor is allowed to discharge, there's no current going through this abductor. This in Doctor has no IMF across it. It only has any MF when the current is changing So right at the beginning Note e m f all of that voltages the capacitor has is dumped onto the resistor, right? All of this voltages dumped onto the resistor. So the resistor has the maximum voltage is gonna have, and that voltage is just gonna decrease as the capacitor discharges. Okay, so the current is going to decrease because the voltage across the resistor is going to decrease Now. If the current is decreasing, then what we have is the IMF going this way. Okay. And so since the loop goes with the current, we're gonna have a negative voltage across the induct er all right. Now, what we need to do is we need to plug in those, um, equations for each of these voltages for the capacitor, The equation for the voltage. It's simply que oversee for the resistor, it's i r. And for the induct er, it's L d I DT. The last thing we need to do before we get our final equation is substitute in the relationship between the current and the charge of the capacitor. That current is being produced as the capacitor loses charge. So the relationship is just I equals negative de que d t right. The current equals the negative because it's relating to the loss in charge. Okay, so, plugging all this in we get Q over C plus dick d t are the plus B comes from this negative sign and plus once again from the negative sign, L Now, this is the first derivative, the current and the current is the first derivative of the charge. So this is going to be the second derivative of the charge. Okay, this is a differential equation, and it's beyond the scope of what we're gonna talk about. So I'm just gonna give you the solution to this equation. There actually are three different solutions to this equation which are known as the under damped solution, the critically damped solution and the over damped solution. Okay, so let's analyze those three solutions. Each of them has a corresponding charge. In this case, I listed the charge on the capacitor because it illustrates fine. What's going on in the circuit. Okay, You can easily take the derivative off this charge equation to find the current in the circuit, and it will tell you the exact same thing. Okay, bless. Just analyze the charge for under damped the charge oscillates, as shown on the graph all the way to the left under damping occurs when the resistance is very small. All right, so the excuse me, the charges going to oscillate from being positive on one plate, down to zero, to being negative on that plate. Back up to zero positive zero negative zero positive zero. And this is acting like an L C circuit, which makes sense if you have a very, very small resistance than the capacitive side off. Sorry than the inductive side off, the characteristics off the circuit are gonna dominate. It's gonna look like an inductive circuit like an L C circuit. The only difference is that the amplitude of these oscillations which tells you the maximum charge on the capacitor, is clearly decreasing. Okay, why is it decreasing? Well, remember that the maximum charge on the capacitor has to do with the maximum amount of energy stored in the capacitor. If you have a resistor that resistors bleeding energy from the circuit So the maximum energy starting the capacitor has to get less and less and less and less as time. Excuse me Goes on. That means that the maximum charge on the capacitor has to get less and less and less as time goes on. Okay, but it still looks characteristically like an L C circuit. Just one that's losing energy. Now we have two other types of results to those equations. I'm gonna minimize myself. We have critically damped and we have over dance, okay, for critically damped. This occurs when the resistance reaches a very, very specific value that reaches when that reaches a very specific value. This cosine term is always one okay, And when that Kasich in term is always one. All we have left is the front. Okay, so you see, those are the exact same. But this is an exponential decay. This is not an oscillation. The charge just decays. This looks like an RC circuit. That's exactly what would happen in an RC circuit. This is because as the resistance gets larger and larger and larger, the inductive characteristics of the circuit get diminished, and then the circuit looks like it has a lot more resistance than induct INTs, and it acts like a resistor circuit like an RC circuit over damped oscillator. Sorry. Over. Damp system does not have a simple equation. OK, your book probably does not show it. Your professor and class might show it, but probably won't because it's a complicated equation. But this occurs for a very large are and it still looks like a RC circuit. Remember that if r is very small, then the inductive aspect are magnified and it looks like an L C circuit. If our is very large, then the inductive aspect are minimized and it looks like a R C circuit. Okay, so those are the difference critically damped right? here. That's just the border. That's where it swaps between under damps and over damped. Okay, Now the angular frequency of these oscillations is different than that for an L C circuit for a l C circuit. It's just one over L C and notice that if I take our new angular frequency for a L R C circuit and I plug are equal zero into here I get back the angular frequency for NLC circuit. Now notice this frequency Onley applies to under damped systems because for critically damped and over damp systems, the charge does not oscillate. There is no oscillate Torrey motion. Okay, Alright, guys, that wraps up our discussion on lark circuits. Thanks for watching.

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Amplitude Decay in an LRC Circuit

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Hey, guys, let's do an example. An LLC circuit has an induct INTs of 10 million Henry's ah, capacitance of 100 micro fare ads and a resistance of owns. What type of LLC circuit is this? And how long will it take for the maximum charge stored on the capacitor to drop by half? Okay, so let's address the first question. What type of LRT circuit is this? This all depends upon how the value off R squared relates to the value of four l oversee. Okay. Remember, when it's a large inducted, its meaning a small are we haven't under damped system one that's gonna oscillator the charge on the capacity. When we have a large resistance in a small induct INTs, we're gonna have a current sorry charge that just decays. We're gonna have a over damped system where it looks like an RC circuit and the charge just drops. Okay, so first, let's calculate for l oversee, this is gonna be four times 10 million Henry's, which is the same as 01 Henry's divided by 100 micro fare ads, or 100 times 10 to the negative Six micro's 10 to the negative six. All of this equals 400. Okay, now let's look at R squared. Well, that's 20 OEMs squared, which is also 400. So these two values are the same, which means that it's not an under damped system. And it's not an overdone amped system. It's exactly in between a critically damped system. Okay, for a critically damped system, the charge is going toe look like this where capital Q is the largest maximum charge of the capacity is going to store. Okay, it's that initial charge stored on the capacity of the largest that will ever store this charge is just gonna drop continuously with time. We want to know When is it half its value. So if I divide que over e se que what? Sometimes he divided by the maximum charge is one half. What time does that occur? So this is the equation than to solve 14 toe. Isolate the exponents. I need to take the log a rhythm of both sides. So I'll take the log. A rhythm of one half that equals negative are over to El times t. Okay. Ah, little trick with logarithms that I could use is I can write log. A rhythm of one half equals the log, a rhythm of to to the negative one. Any number that's in the denominator can be brought into the numerator. And given an exponents of negative one with logarithms, I can pull the exponents out to the front of the logarithms. So this becomes negative, Ellen of to Okay, so continuing with that, this becomes a negative. Ellen of two, which equals negative, are over to l t. Okay. And the negatives canceled. Now I can just move our over to out to the other side to solve for t t is to l over our Allen of To, which is two times 10 million Henry's or 01 Henry's divided by 20 OEMs times Ellen of tube. And all of this equals 00007 seconds. That is a perfectly finally right In this answer, you can also write it as 07 milliseconds. Okay, however you want to write it. All right, guys, Thanks for watching