LRC Circuits - Online Tutor, Practice Problems & Exam Prep

LRC Circuits

Hey guys, in this video, we're going to be discussing LRC circuits, which are another type of inductor circuit. Let's get to it. Now, as the name implies, LRC circuits are composed of 3 things. They're composed of inductors, 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. LRC, inductor resistor capacitor circuits. Now in an LRC circuit, if we begin with the capacitor initially charged, we can write out Kirchhoff's loop rule for this circuit. What's going to happen is as soon as this circuit is completed and the capacitor is allowed to discharge, a current is going to be produced. Okay? That current is going to go this way through the inductor, and that current is going to go this way through the resistor. And now, what we want to know is how to assign positive and negative values for the voltages in Kirchhoff's loop rule. First, we have to choose a loop. I'll just choose this as an easy loop. Now let's leave the inductor for last. The resistor and the capacitor 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 inductor. Okay? Now the direction of the EMF in the inductor 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 inductor. This inductor has no EMF across it. It only has an EMF when the current is changing. So right at the beginning, no EMF. All of that voltage that the capacitor has is dumped onto the resistor. Right? All of this voltage is dumped onto the resistor. So the resistor has the maximum voltage it's going to have, and that voltage is just going to 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 EMF going this way. Okay? And so, since the loop goes with the current, we are going to have a negative voltage across the inductor. Alright? Now, what we need to do is we need to plug in those equations for each of these resistor, it's IR, and for the inductor, it's L di/dt. The last thing we need to do before we get our final equation is to substitute in the relationship between the current and the charge in the capacitor. That current is being produced as the capacitor loses charge. So the relationship is just \(I = -\frac{dq}{dt}\). Right? The current equals the negative because it's relating to the loss in charge. Okay. So plugging all this in, we get \(\frac{q}{C} + IR + L\frac{d^2q}{dt^2}\). The plus signs come from this negative sign, and plus, once again from the negative sign, L. Now this is the second derivative of the charge. Okay? This is a differential equation, and it's beyond the scope of what we're going to talk about. So I'm just going to give you the solution to this equation. There actually are 3 different solutions to this equation, which are known as the underdamped solution, the critically damped solution, and the overdamped 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 is going on in the circuit. Okay. You can easily take the derivative of this charge equation to find the current in the circuit, and it will tell you the exact same thing. Okay. But let's just analyze the charge. For underdamped, the charge oscillates as shown on the graph all the way to the left. Okay. Underdamping occurs when the resistance is very small. Alright? So the charge is going to oscillate from being positive on one plate down to 0 to being negative on that plate back up to 0. Positive 0 negative 0 positive 0. And this is acting like an LC circuit, which makes sense. If you have a very very small resistance, then the inductive side of the characteristics of the circuit are going to dominate. It's going to look like an inductive circuit, like an LC 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 resistor is bleeding energy from the circuit. So the maximum energy stored on the capacitor has to get less and less and less as time 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 LC circuit, just one that is losing energy. Now, we have two other types of results to those equations. I'm going to minimize myself. We have critically damped and we have overdamped. 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 1. Okay? And when that cosine term is always 1, 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,

Amplitude Decay in an LRC Circuit

Hey guys. Let's do an example. An LRC circuit has an inductance of 10 millihenries, a capacitance of 100, and a resistance of 20 ohms. What type of LRC circuit is this, and how long will it take for the maximum charge stored on the capacitor where the value of r squared relates to the value of 4L/C? Okay? Remember, when it's a large inductance, meaning a small r, we have an underdamped system, one that's going to oscillate the charge on the capacitor. When we have a large resistance and a small inductance, we're going to have a charge that just decays. We're going to have an overdamped system where it looks like an RC circuit, and the charge just drops. Okay? So first, let's calculate 4 L over C. This is going to be 4 times 10 millihenries, which is the same as 0.01 henries, divided by 100 microfarads or 100 times 10 to the negative 6. Right? Micro is 10 to the negative 6. All of this equals 400. Okay? Now, let's look at r squared. Well, that's 20 ohms squared, which is also 400. So these two values are the same, which means that it's not an underdamped system, and it's not an overdamped system. It's exactly in between: a critically damped system. Okay? For a critically damped system, the charge is going to look like this, where capital Q is the largest maximum charge that the capacitor is going to store. Okay? It's that initial charge stored on the capacitor, the largest it will store. This charge is just going to drop continuously with time. Okay? We want to know when it is half its value. So if I divide Q over, I can say Q at some time t divided by the maximum charge is 1 half. What time does that occur? So this is the equation then to solve for t. To isolate the exponent, I need to take the logarithm of both sides. So I'll take the logarithm of 1 half, that equals negative r over 2L times t. Okay? A little trick with logarithms that I can use is I can write logarithm of 1 half equals the logarithm of 2 to the -1. Any number that is in the denominator can be brought into the numerator and given an exponent of -1. With logarithms, I can pull the exponent out to the front of the logarithm. So this becomes a negative ln of 2. Okay? So continuing with that, this becomes a negative ln of 2 which equals negative r over 2L * t. Okay? And the negatives cancel. Now I can just move r over 2L to the other side to solve for t. T is 2Lrln2, which is 2 times 10 millihenries or 0.01 henries divided by 20 times ln of 2, and all of this equals 0.0007 seconds. That is a perfectly fine way of writing this answer. You could also write it as 0.7 milliseconds. Okay? However you want to write it. Alright, guys. Thanks for watching.