Concept: LRC Circuits10m
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 three 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 KirchoffÕs 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 KirchoffÕs 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, the capacitor going to be easy, the loop goes with the polarity of the capacitor, so thatÕs the 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 and 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 is 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, 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're 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 voltages, for the capacitor the equation for the voltage is simply q over C, for the resistor its i R and for the inductor its L di/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 in the capacitor, that current is being produced as the capacitor loses charge, so the relationship is just i equals negative 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 q over C plus dq/dt R, the plus comes from this negative sign, and plus once again from a negative sign L. Now, this is the first derivative of 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 going to talk about, so I'm just going to give you the solution to this equation. There actually are three 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 hast to set the charge on the capacitor because it illustrates fine what's 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? Let's just analyze the charge. For underdamped, the charge oscillates as shown on the graph all the way to the left, okay? Under damping occurs, when the resistance is very small, alright? So, the-, excuse me, 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 capacitive side of, sorry, 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 resistors bleeding energy from the circuit, so the maximum energy stored in the capacitor has to get less and less as time, excuse me, goes off, that means that the maximum charge of the capacitor has to get less and less time goes on, okay? But it still looks characteristically like an LC circuit, just one that's 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 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 would be exact the 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 the inductive characteristics of the circuit get diminished and then the circuit looks like it has a lot more resistance than inductance and it acts like a resistor circuit, like an RC circuit. Overdamped oscillate, sorry, overdamped system does not have a simple equation, okay? Your book probably does not show it, your professor in class might show it, but probably won't because it's a complicated equation, but this occurs for a very large R and it still looks like an RC circuit. Remember, that if R is very small, then the inductive aspects are magnified and it looks like an LC circuit, if R is very large, then the inductive aspects are minimized and it looks like an RC circuit, okay? So, those are difference, critically damped, right here, that's just the border, that's where it swaps between under dance and over dance, okay? Now, the angular frequency of these oscillations is different than that for an LC circuit, for an LC circuit it's just 1 over LC and notice that if I take our new angular frequency for an LRC circuit and I plug R equals 0 into here, I get back the angular frequency for an LC circuit. Now, notice this frequency only applies to underdamped systems, because for critically damped and overdamped systems, the charge does not oscillate, there is no oscillatory motion, okay? Alright guys, that wraps up our discussion on LRC circuits. Thanks for watching.
Example: Amplitude Decay in an LRC Circuit5m
Hey guys, let's do an example. An LRC circuit has an inductance of 10 millihenrys, a capacitance of 100 microfarads, 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 to drop by half? Okay? So, let's address the first question. What type of LRC circuit is this? This all depends upon how the value of R squared relates to the value of 4 L over 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 curren-, sorry, 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 Henrys divided by 100 microfarads or 100 times 10 to the negative 6, right? Micros 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, it's exactly in between a criticallydamped system, okay? For a criticallydamped system the charge is going to look like this, where capital Q is the largest maximum charge of the capacitor is going to store, okay? It's that initial charge stored on the capacitor, the largest it will ever store. This charge is just going to drop continuously with time, okay?
We want to know when is it half its value, so if I divide q over, I can say q at some time t divided by the maximum charge is 1/2, what time does that occur? So, this is the equation I need to solve for t. To isolate the exponent I need to take the logarithm of both sides, so I'll take the logarithm of 1/2 that equals negative R over 2 L times t, okay? A little trick with logarithms that I can use is I can write logarithm of 1/2 equals the logarithm of 2 to the negative 1, any number that's in the denominator can be brought to the numerator and given an exponent of negative 1, with logarithms I can pull the exponent out to the front of the logarithm, so this becomes negative ln of 2, okay? So, continuing with that, this becomes the negative ln of 2, which equals negative R over 2 L t, okay? And the negatives cancel. Now, I can just move our over to L to the other side to solve for t, t is 2 L over R Ln of 2, which is 2 times 10 millihenries or 0.01 Henrys, divided by 20 ohms times Ln of 2 and all of this equals 0.0007 seconds, that is a perfectly fine way of writing this answer, you can also write it as 0.7 milliseconds, okay? However you want to write it. Alright guys, thanks for watching.
Consider the circuit shown in the sketch. C = 5.0 x 10-6 F and L = 0.30 H. Initially the switch S is open, there are no currents, and there is no charge on the capacitor. Then the switch is closed. After the switch has been closed a long time, what is the voltage across the 3 Ω resistor?
Consider the circuit shown in the sketch. C = 5.0 x 10-6 F and L = 0.30 H. Initially the switch S is open, there are no currents, and there is no charge on the capacitor. Then the switch is closed. Just after the switch is closed, what is the voltage across the 3 Ω resistor?