An L-R-C series circuit has L = 0.450 H, C = 2.50 * 10^-5 F, and resistance R. (b) What value must R have to give a 5.0% decrease in angular frequency compared to the value calculated in part (a)?

Video transcript

Welcome back, everybody. We are making observations about a partially complete L R C circuit. We are told that it's constructed with a conductor that has an inductive of 200 mil hearts. And we are told that it also has a capacitor of micro fare adds. Now a resistor is going to be introduced making this a complete L R Circuit. And we are told that when that happens, it changes the angular frequency by decreasing it by 20%. Meaning the new angular frequency will be .8 or 80% of what the original frequency angular frequency was. And we are tasked with finding what resistance is required to make this change right here. Well, here's what we're going to do. In order to find our initial angular frequency, we can consider the partially complete L R C circuit as just an L C circuit since the risk the resistor is not there initially. So in order to calculate the angular frequency for an L C circuit, that's just one over the square root of R L C right now for a, the angular frequency of an L R C circuit, it's a little bit different here. Here we have that the angular frequency is equal to the square root of one over L C minus our resistance squared divided by four times are inducted squared. Now, here's what I want to do. I want to, before plugging in any terms, I want to isolate the resistance variable right there. Here's what I'm gonna do. I'm gonna square both sides, right? And we can admit that that gets rid of this radical right here. So then I will go ahead and subtract one over L C from both sides. So I'm gonna, I'm gonna put this in parentheses here. You'll see that these terms cancel out and then I will multiply both sides by negative four L squared minus four L squared. And this term and this term cancels out. So here's what we are left with, we're left with the fact that R squared is equal to four L squared times one over L C minus omega squared because I took that negative and I distribute it to the inside of the parentheses. And then of course, to get rid of the power on the resistance variable, we just take the square root of both sides. And there you have it, this is the formula that we need to use. But remember what is this angular frequency right here? That angular frequency is really going to be our final angular frequency. So what we first need to find is our initial angular frequency that we can calculate our final angular frequency and plug it into this formula to find our resistance. So let's go ahead and do that here. We have that. Our initial angular frequency is given by this formula right here. So this is one divided by the square root of our conductance which is 200 mil hurts, but I need that to being hurt. So multiplied by 10 to the third time's are capacitance of 300 micro fare adds. But I need this and fair ads. So I multiply this by 10 to the negative six. When you plug this into our calculator, we get that our initial angular frequency is 29 radiance per second. So now let's use this formula right here and we get that our final angular frequency is just 80% of what our initial angular frequency was. So then what we'll do is we will take this term and we will sub this in right here. So let's go ahead and do that. We have that. Our resistance is equal to the square root. I'm gonna draw this radical sign quite far is four times our induct inside of 200 micro or sorry Miller hurts but we need this in hurts. So times 10 to the negative three, we're gonna square that entire value times one over L C. So one over 200 times 10 to the negative three times 300 times 10 to the negative six. This fraction bar right here minus our fine Angular frequency. So it's .8 times 129 in that entire value squared, which when you plug all of this into your calculator, it gives us a final resistance of 31 Omega's corresponding to our final answer. Choice of the. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.