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Amplitude Decay in an LRC Circuit

Patrick Ford
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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