Hey, guys. So now that we know how in doctors behave in circuits, we're gonna take a look at a specific type of circuit that you're gonna need to know about called on LR circuit. Let's check it out. So in our circuit or an R L circuit, as you'll see in some textbooks, is simply a circuit that's containing an induct. Er, and we know that induct er's have the letter L and a resistor. And so that resistor is the are part of the L R Circuit. Right? So we got the schematic that's up here, that sort of labels, a battery that is connected to induct er and a resistor. And we have some switches. And the key thing here is that the depending on the switch positions, there's two different process that can happen in this circuit. So let's check him out. So imagine that we have s to that remains open. So that's this top got right here. And if s to arrange open, then that means that this whole entire circuit, it doesn't really matter. There's nothing that can flow through it and s one is now closed. So when you close s one you've now created a continuous loop for electrons to flow. So electrons gonna flow in this loop right here through the induct er and the resistor. Now, because it's connected to this battery right here, this battery wants to supply a lot of electrons, and it's going to supply the current. And what happens is that the current wants to increase in this phase. So we call this current growth so this happens when s one is close. So now let's imagine that we take the circuit, we let it run for a very long time, and then at some point, we're gonna close. This s to switch right here, and then we're gonna open. This s one. So now it's basically like this half of the circuit doesn't really exist, and the top one does. So what happens is now when we close, this s two. So when s two is closed? We now have a different circuit, a different loop that flows through the induct er and the resistor. And if this has been running for a long time, there's gonna be some current that has built up in this loop. I'm just gonna pretend that it goes in this direction. It could go in the other direction. We don't really know. So we have all this current right here and the resistor is going to bleed off of that currents. And so it's going to slowly dissipate that currents. Without the battery in the top half of the loop, the current is eventually going to decrease. So we call this process current decay is basically an overview of the two types of process that you can see. We're gonna take a look at each one of them a lot more carefully in this in this next section right here. So current growth is again when you have the battery that is connected. And that's like the dead giveaway that you have current growth. So when you have or when you're told information that the battery is connected or you have a schematic in which there's a switch or something like that and you close that circuit now the battery is connected. That's your dead giveaway, that you have current growth. Now this battery. What it wants to do is it wants to basically supply this current throughout the circuit. But remember, so this you know, this current is gonna start off is very, very slow value. And it wants to, as a very low value is gonna actually get zero. And this current wants to very, very quickly reach its maximum value. But remember that the function oven induct er is that in doctors always resist large changes in current with the self induced E. M F is equal to or it's proportional to Delta II over Delta T. So the larger the change in the current through this circuit, the larger the E m f. That's gonna be backwards. That's gonna fight against it. So what happens is this induct er basically resists any large changes in currents. So this induct ER is going to resist that growing currents and the equation that governs that is going to be this exponential equation right here. In fact, it's very, very similar to the exponential equation that we saw for RC circuits with just a few small differences here. So what ends up happening is that as time goes on as T goes on, there's now mawr of the current that's built up in the circuit. There's less Delta. A delta T thean doctor gets weaker and weaker and weaker, and eventually this current reaches its maximum value. So we're going to see is that as t goes to some very, very large number. So I'm just gonna write infinity over here, this exponential eat the minus t over tau eventually just goes to zero. So if you plug in a very large number inside of this exponential, this basically just goes away. This term goes away. So the phrase that you might see in problems is that after some long time, so that's usually what you'll see. So after a long time. And what that phrase just means is that the system is sort of able to get to it's sort of final states, right? So it's it's able to do this, and this thing exponential term, is going to go away. So after a long time, what ends up happening is that the current will eventually reach its maximum value of V over R. And so if you go to plot this equation out, it looks again very similar to the way that charges buildup on capacitors in an RC circuit. So we have the current that starts at zero, and then, at some later time So we're gonna ask them topically reach this maximum value of V over R. So it's gonna kind of look like this. So it wants to sort of rush out all this current, but the induct er is slowing it down and then after a long time, it sort of levels out and then reaches its maximum value. So that is current growth. So then if you were to sort of run this circuit for a very long time and let all of that current build up to its maximum value and then you were too disconnected from the battery. Now we move on to the second phase, which is current decay. So the easiest way you could tell that you have current decay is that unlike we had current growth, the battery is disconnected. So we have no battery here. There is nothing left to supply those electrons or that current that's going through the battery. And so what ends up happening is that this resistor is eventually going to dissipate that current away Now again, just like the induct er was trying to resist the growing currents. This induct er is trying to resist the the weakening current or the decrease in current from this resistor. So what happens here is that we have some maximum currents that's going through this circuit. So this is gonna be IMAX, so we're going to start off with some maximum value like this. And what happens is that this current is going to eventually get dissipated through the resistor, like we said. But this induct er is going to resist the decreasing currents and the equation that governs that is now This exponential right here looks very similar to the other one. But again, we don't have that one minus sign in there, that one minus e to the minus towel. So what happens here is that as t gets very, very big, so we have as t approaches some very large number after a long time, this exponential e to the minus t over tau eventually becomes zero. So if this term eventually become zero, then this whole entire thing will eventually become zero after a very, very long time. So this occurrence will eventually get bled off by the resistor. And the conductor was told to resist the decrease in current less and less and less So that means that there a very long time. You're gonna ask them methodically. Approach zero. Now, in both of these problems, that's basically the two different kinds of exponential that you'll see again. It's very similar to how RC RC circuits work, and the constant that appears in this is the time constant. That's the tower right here. And so, unlike RC circuits, this tau is given as the induct int l over our and basically what this time constant does is it just determines how quickly that the growth and decay cycles occur. So a lower time, constant meaning that l your induct Ince's lower or the resistance is higher means that these changes in the either growth or decay happen much, much, much faster. So if you have a lower time constant, it means that not as much time needs to pass in order for these cycles to occur, whereas if you have a much higher induct, INTs or the resistance is lower. So if you have a higher L or a lower our, that just means that these these changes happen much slower because the induct er is able to sort of resist these changes happening, Ah, lot better Okay, so it's basically summary of how current growth, how hurt to Kate works in an LR circuit. Let's go ahead and take a look at some problems.

2

example

Unknown Resistance in an LR Circuit

Video duration:

3m

Play a video:

Hey guys. So let's get to more practice with LR circuits we have in our circuit with the time constant and it's initially connected to a battery. And then, after some very, very small amount of time, it is disconnected from the battery, the current to something we're supposed to figure out what the resistance of the circuit is. So hopefully you guys realized that this word here it's disconnected from the battery means that we are working with decay. So we're working with current decay right here. So let's see. We have a time constant, which is tau. That's equal to 0.25 We have a voltage that's V is equal to 10 volts. We have time, which is equal to 0.5 seconds, and we have the currents at that specific times. That's I when t is equal to this value right here, and we're supposed to figure out what is the resistance of this circuit. So hopefully you guys realize that if we're working with current decay, then the equation that's gonna tie all of these things together is going to be the current decay equation. That's I of T is equal to V over r. And then do we have one minus sign there? No, because we just have the inverse exponential e to the minus t over Towel with one minus sign is for current growth. Because eventually what happens is we want that V over r to sort of be, like, away. You know, we want to be isolated, right? So this is the equation that is eventually going to go to zero as time gets really large. Got it? So here's what we have. We have the resistance of the circuit. That's gonna be our time. Our target variable right here. And let's see, we have the current as a function of time and we actually have what the current is at a specific value in times when t is a specific number. So all we have to dio is Let's see, I have I've t I know what the voltage is. I know what the time is. The time I'm plugging in is just this number right here. And I have the current at that specific time, and I have, with the time Constant is so. All I have to do is just move this over to the other side. So basically what happens is that just, like, sort of an algebraic equation? This Arkan go up to the top and this i of tea can actually go down to the bottom and basically just trade places with it. So that means that our is equal to V over I when t is equal 2.5 Um, And by the way, I'm just evaluating the current at the specific time. And then we have e to the minus t over towel. So let's plug in. All of our numbers are is equal to the voltage, which is 10 I at what, at this specific time is just equal to 0.5. Then we have e to the minus, and then T which is equal 2.5 and then we have over Tau and Tau is 0.25 Okay, so hopefully you guys plug this in. You could actually you know, you could take an extra step in, Sort of like, simplify this a little bit. 10 over 100.5 actually just becomes 20. And this exponential right here is actually e to the minus 0.2. And if you plug this into your calculator, you guys should get 16.4 and that's gonna be in homes. So that's the resistance off the circuit. Let me know if you guys have any questions.

3

Problem

Problem

Consider the LR circuit shown below. Initially, both switches are open. Switch 1 is closed. a) What is the maximum current in the circuit after a long time? Then, S_{1} is opened and S_{2} is closed. b) What is the current in the circuit after 0.05s?

A

(a) i_{max}=0.5 A;

(b) i (0.05s)=1.86×10^{−6} A

B

(a) i_{max}=2 A;

(b) i (0.05s)=7.45×10^{−6} A

C

(a) i_{max}=0.5 A;

(b) i (0.05s)=0.14 A

D

(a) i_{max}=2 A;

(b) i (0.05s)=0.57 A

E

(a) i_{max}=0.5 A;

(b) i (0.05s)=0.46 A A

F

(a) i_{max}=2 A;

(b) i (0.05s)=1.8 A

4

Problem

Problem

An LR circuit with L = 0.1 H and R = 10 Ω are connected to a battery with the circuit initially broken. When the circuit is closed, how much time passes until the current reaches half of its maximum value?