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30. Induction and Inductance

1

concept

11m

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Hey, guys. So in this video, we're gonna take a look at a specific type of induct er circuit called Elsie Circuit because you might need to know how they work. Now we're going to see that it's very, very similar to another kind of motion that we've seen in earlier Chapter in physics. So let's check it out. And l c circuit is made up of an induct er which remember is the L S O. That's the L right here, and it's also a capacitor. So these sort of our combines together, and that's the C and L C circuit, and it actually follows an eight steps cycle. So we're gonna take a look at each one of the steps very, very carefully. Now, for the sake of argument, let's just say that we have an induct, er so we have a conductor right here and a capacitor, and these things are combined together without a battery. And let's just say that one of the capacitors of the capacitor is charged initially, so we have a maximum amount of charge that's on the plates. So you have all these electrons that are sort of locked up together on these plates. The green lines represent the electric fields that goes to the spot. And what happens is that without a battery or anything like that or a resistor, all these electrons wants to start flowing out like this, right? They want to start discharging from the capacitor. Now, Initially, the current is zero. There's no charges anywhere in the circuit. But immediately after this step right here, what happens is that this charge will start leaking out of the positive plates. Right? So wants to go around the circuit like this, and you're gonna generate some kind of current. Now, what happens is this current wants to go from 0 100 right? He wants to basically discharge immediately, but it can't do it because remember that the function of this induct er is that conductors always resist any changes in current, they resists delta eyes. So this induct er makes it impossible for the for the current to go from 0 to 100 or for to change very rapidly. So what happens is that some of this charge is still locked up in the capacitor, But the current is going to be increasing in this point, right? here, right? So it's still going to be increasing. So you have these charges that start to basically pile up on the other side of the capacitor like this. So now what happens is eventually in the next step over here, all of the charges have finished discharging from the capacitor. So that means at this point all of the electrons are basically going through the circuit, which means that the current at this point is actually maximum right here. So when it's maximum, there is no more charge left on the capacitor. So we see that the initially the current zero, the capacitance are the charge was maximum and the current was zero. Now it's the opposite. The charges zero. But the current is maximum. So know what ends up happening is that you have all these positive charges that assert to piling up on the other side of the plate. So were initially whereas this was positive on the left side. Now, actually, this is going to be the positive on the right side because all the positive charges from the current are basically piling up on this side. Now what ends up happening is that the currents in this case is decreasing in this phase because it's sort of like running out of steam, like all the charges air piling up on this side. But remember that just as in the second step, where this induct er was resisting any large changes in increasing current, it does the same exact thing here. It resists any changes in current, even if they are decreasing. So this induct er is still making it impossible for the current to basically go down to zero immediately anyway. So then what happens is that all of the charge finishes piling up on the other side. And now you basically have a reversal of what? Of the first sort of half of this step right here. So now you have, um, all of the charges that are on the right side, whereas the charges on the left side. But now there's no more currents, and the charge is maximum here. So you have maximum charge and zero currents, and basically from here, the entire process just goes in reverse. So now, instead of the charges wanna leave to the left, they wanna leave to the right, So that's what they're going to start doing? So you're gonna have a current right here that's going to increase. But remember that this induct er is going to resist any changes in current. And then what happens is, um, So, whereas there's still some amount of charge right here that's still left on the capacitor, eventually all of that is released from the capacitor and the current is maximum here at this step right here and the charges zero. And then what happens is that now the current is going to be decreasing here as all the charges start to pile up again on this side. So whereas originally started on, so the current gonna be decreasing there. But remember that this is going to be resisting any changes in the currents and then basically, you just go back to the way it began. Right? So now you basically stir starting the whole entire cycle over again. Okay, got it. So we've seen that this system is oscillating, so this system goes back and forth. The current basically I like to think about like a pendulum. The current is going back on one side and then back through the other. And the induct er is always preventing any changes in current like that. So this system is oscillating. It actually behaves very, very similar to another kind of motion that we've seen called simple harmonic motion. So simple harmonic motion is when we had a block attached to a spring. So we had a block like this and we had if we pull it back, there was some force that was acting on it. So the force was here and the velocity was equal to zero. And if we just let it go, then this block wanted to basically oscillate. So there was an equilibrium position like this, and it would start to speed up during this phase right here. Then it would go past its equilibrium position, and then it would slow down over here on this phase, and then it would basically get to the other side, and then the whole entire thing would go back. Um, it would be going in reverse order. So, in other words, the force would be now this way, and the velocity would be zero. In this case, it was basically just doing the exact opposite of what it just did on this thing would just do this forever, right? It would just go back and forth and back and forth. So this system oscillates the same way that a simple harmonic motion oscillates. And because the system is oscillating, the formulas for the charge and the current are actually represented by Sinus soil functions. So the spinal fluid or functions just they don't necessarily mean they're both sign functions. It's just the ones that oscillates. So signs and co signs. Now, we said that in the beginning of the cycle, all the charge was maximum on the capacitors. So the function that starts office maximum is the co sign function. So the way that the do you remember this is that the charge is always gonna be maximum at first. So that means that this function right here, q of t is gonna be Q max times the cosine of omega T plus five. So it looks very, very similar to how simple harmonic motion equations used to work. So you have the maximum amount of charge on the plates on the capacitor, and then you have this omega term right here. Now, when we studied omega for simple harmonic motion, it depended on things like this stiffness of that spring and the mass and things like that. Well, here it just depends on the induct er and the capacitor. So it's the square root of one over l. C. And remember, that is the angular frequency. So you can always relate the angular frequency to the linear frequency by two pi times f. So this f represented the number of cycles that happen per second. Eso they're not quite the same thing, But you always have that relationship right there. Then you could also relate that to the period as well. Okay, so and then this fi term right here. So this little angle right here five is called the phase angle, and it basically just determines the starting point of your oscillation. It's just a constant goes out there, just in case you started some other point in the cycle. All right, so that's the charge, and the current is slightly similar. It's similar to that, but it's slightly different. It's actually negative. Omega Times Q. Max and then you have, instead of cosign you have sign, and that's gonna be Omega T plus five. Now, just a heads up. For those of you who are taking calculus. You actually might recognize this as the derivative of the Q T function. But if you don't, if you're not in the calculus course, you don't have to worry about that. All right, so that's the two functions. The last thing I wanna point out is that make sure that your calculators are in radiance mode because we're working with these co sign and sign functions with radiance. Right, So we have angular frequencies, so just go ahead and make sure that your calculators sets radiance mode and we're gonna go ahead and check out this example. So we have a capacitor with a capacitance like this and has initial charge of magnitude. This and we're supposed to be figuring out that during the current oscillations that that occur after the circuit is completed, what's the maximum current in the induct? Er, so let's take a look. We're looking for the maximum current, so that means that we're gonna be looking for I Max right here. So we're gonna have to relate this to the current function often Elsie Circuit. So remember that the current function I of T is equal to negative omega que max times the sign of Omega T plus five. Right, So this current is going to oscillate. It's gonna go up and down like this. If you were to plotted on a graph, right? So if you were to plot the charge versus time, it would start to look like this. So, do you have some kind of oscillating function now, whenever the current is maximum, whatever it occurs at these points or these points, what this really means? Is that the sign? So let's let's write this out because it's actually really important. So I is equal to I. Max. Yeah, when the sine function, the sign term sine omega T plus fi, whatever those variables are is equal to one or negative one, right? So whenever this whole entire equation is equal to one, that's when the current is going to be maximum. So in other words, you just have that the absolute value of I Max is just when you have Omega Times Q. Max, right, because when this is equal to one or negative one, it doesn't really matter. Then that current is going to be maximum, so let's see. So basically, we're gonna use be using this equation. We actually know what the maximum charge on the capacitors are. Now we just have to figure out what this Omega term is so remember that omega is the angular frequency and we can relate it to the induct INTs and the capacitance. By the equation, Omega equals square root off one over L C. So Omega is gonna be the square root of one over the induct. Its was four. Henry's in the capacitance is Be careful with this because you see, See here, Not by confusing, that's capacity to remember that school loans, that's for charge. So this is actually que Max and this is actually the capacitance. And then this is the induct install, right, So don't get those confused. So we've got four times, five times 10 to the minus nine and you work this out, you're gonna get 77071 And that we can do is we can plug this back and figure out what IMAX is. So we just get that IMAX is equal to 7071 times Q max, which is equal to two times 10 to the minus four, and you work this out. You should get a maximum currents that's equal to 1.41 amps. Alright, guys. So that's it for this? That's the maximum current in the induct. Er, and we're gonna take a look at a couple more practice problems. Let me know if you have any questions.

2

Problem

An LC circuit with an inductor of 0.05 H and a capacitor of 35 μF begins with the current of -1A. The capacitor plates have a maximum charge of 2.65mC at any time during the oscillation. What is the phase angle of this oscillation?

A

45^{o}

B

15^{o}

C

30^{o}

D

60^{o}

3

example

4m

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Hey, guys. So let's take a look at this problem. We haven't Elsie circuit with some inductions and a capacitance. It begins with a capacity is fully charged. We're supposed to figure out how many seconds it takes for a fully charged plate to transfer all of its charged to the other plate. Now, that's a lot to unpack there. So what I'm gonna do is we're gonna begin with just a simple diagram oven, Elsie circuit. Hopefully I do this, right. So I've got the induct her like this and it goes around and then you have a capacitor like this. Now, remember, from our previous video that we talked about If we have a sort of charged capacity like this with all the charges, then a full cycle of Elsie circuit is when the charge goes all the way around the circuit builds up to a maximum and then piles up on the other side. And then the cycle has to reverse direction and do the same thing over again. But backwards. So the words all the charge has to go from here. It escapes through this plates, and it basically piles it back to where it started. from. And then the whole thing repeats over and over again. So what this is saying is, how long does it take for a charged plate to transfer all of its charge over to the other side of the circuit or the other side of the capacitor? So what that really means is that actually just looking for half of the cycle? So that means in terms of the period, that's the variable t That's gonna be T over, too. So really, what we're looking for in this case is what is the half period in seconds? That's really what we're looking for. But in order to do that in order to figure that out, we just have to figure out what the period is in general. So let's remember, if weaken, let's remember our formulas for the period, the frequency and the angular frequency foreign oscillating system we have that the period and the frequency are in verses of each other. So t equals one over f, and we also have that the angular frequency could be related to the linear frequency by this equation right here. So if I want to figure out what the period is, I have to relate that to the frequency, but I don't know what the frequency is, so I have to related to the angular frequency and that I actually can figure out because remember that the angular frequency Omega can also be related to the induct INTs and the capacitance off this equation of the of the circuit. So let's go ahead and do that. Let's see, uh, the tea is going to be. Well, let's see if I wanted the frequency that's actually going to be omega over. Two pi, if I just moved to the other side is gonna be equal to the frequency. So what I have is that tea is actually equal to one over omega over two pi. So that means that the period is going to be two pi over Omega. So that means finally, that t over to is just going to be pi over omega. So what? One half of this is just gonna be one half of this. So the two just cancels out and we just get pi over omega. So finally, what I can do is that can actually just take this formula right here, which is the square root of one ever Elsie and actually action to plug it back in for the denominator in this equation. And what I get is that t over two is equal to pi times the square root of l times C. So what happens is when you take the inverse of this and this is the denominator, basically, what happens is we're taking pi over one over the square root of L C. And so l c just goes up on the top, right? Got it. So I just wanted to sort of walk us through that because it's been a while since we use these kinds of equations. So really, this is actually just going to be the half period. That's all we have to dio. So that means that T, which is equal to the half period, is just gonna be pi times the square root off the induct INTs, which is zero point zero five times the capacitance, which is fifty military. It's so we have elves equal to zero point zero five. We have the capacitance that's equal to fifty military. It's which is actually zero point zero five as well. So that means we're just gonna have zero point zero five here also, and we just get that the half period is just going to be equal to zero point one five seven seconds. And that's our answer. That's how long it takes in this circuit. For the charge to transfer to the other side. That's half the cycle. It would take another point one five seconds to go back and then basically begin the whole thing over again. Okay, guys, that's it for this one. Let me know if you have any questions.

4

concept

10m

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Hey, guys. So now that we know how to use Elsie circuits, we're gonna take a look at how the energy sort of changes between the capacitor and the induct. Er, because sometimes you're gonna need to know that. Let's check it out. So whenever we looked at Elsie circuits, we just assume that there was no resistance. And because there is no resistance and NLC circuit, the energy is conserved. It basically just bounces between the capacitor and the induct er in the L C circuit. Now the energy is gonna oscillate because we know that Elsie circuits oscillate between electrical energy in the capacitor and this electrical energy is actually stored inside of the electric field that exists between the plates. So we have this electrical field right here, and that's actually where this electoral energy is stored. And then what happens is the magnetic energy comes from the fact that you have current going through an induct er so this magnetic field sort of get set up because you have current that goes through a coil of wire, and we know that coils of wire will generate sort of these loops and you have a magnetic field that points in this direction. That magnetic field actually stores energy. And so this is actually where that magnetic energy comes from, right? So it's constantly going between the electrical and the magnetic energy. We actually have the equations that give us the electrical energy for a capacitor. It's those three equations that we were able to use well for magnetic energy oven induct er we have another equation that's gonna be one half l times I squares just at one equation that relates the induct INTs and the currents. Now, basically, we know that in an L C circuit, the charge and the current are constantly sort of out of phase, and one's going up while the other one's going down. So let's take a look at the steps of an L C circuit and see what the energy is doing. We know that there's a relationship between the charge and the electrical energy, and then the currents and the magnetic energy. So let's take a look at that, right. So in this first step here, where you have no current and the capacitor is fully charged, that actually means that there is no current going through this circuit right now, um, so that we have the maximum amount of charged that is between the plates of the capacitor. So the Q is Max over here? So that tells us from our equations that the electrical energy is going to be maximum here while the induct er energy or the magnetic energy is going to be zero because there is no current. So that means that there is no magnetic energy here, and it's all electrical, so I'm just gonna draw a little bar graph like this. So now we know that there is a second step that is sort of in the middle between these, we know that there's eight steps and here where the charge is zero, the current is gonna be maximum right here. So that means that throughout the sort of middle step in this process, we know that the charge is going to go down, which means that the, um, energy that's stored in the capacitor is going to go down, whereas the current is increasing. So that means that the magnetic field energy is going to go up. So what happens here is that now we have no more charge left on the capacitor. And then when the current reaches its maximum point right here, the magnetic field it throughout, this induct er is going to be at its strongest. So we have a magnetic field that points in this direction like this. And so from our equations, which we can see that if the current is going to be maximum, then that means that the magnetic field energy is going to be strongest here. So we're gonna have a bar graph like this. And if there is no charge across the capacitor, then there is going to be zero energy over here. And now what happens is that we know that the cycle just just repeats itself in reverse. So that means that the electrical energy is going to be maximum here because you have the maximum amount of charge. It's just going in the opposite direction and there is zero magnetic energy and then over here there is going to be maximum. There's going to be maximum in magnetic energy because you have IMAX like this. And then there is going to be zero electric energy because there is no more charge across the plates. All right, so basically just oscillates back and forth, but we can see that at any point in time. What happens is that the total amount of energy that's E is going to remain the same across all of them. So what happens is you have this relationship between the electric and the magnetic energies such that as one goes up, the other one goes down. It's kind of like how potential and kinetic energies worked in simple harmonic motion. So there's that analogy that we could sort of draw again. So remember when we had simple harmonic motion where we basically just had a ah spring that was attached to a ah, block or block on a spring. Then there was the equilibrium points right here. And what happens is that you have all this potential energy. So here you had, um, so we had Delta X was equal to its maximum right here. So that means that you had the maximum amount of potential energy, and then when it came through the middle like this, either in this direction or that direction, we know that the velocity was maximum here. So that means that the the kinetic energy was maximum at this point and then, as this thing went back and forth, that energy would change between the potential and the kinetic throughout that cycle. It's kind of like the same thing here. Okay, All right. So let's check out this equation or let's check out this practice problem. Or in this example, now that we have another equation for the magnetic energy. Cool. So we have an L C circuit, and we're told that it has a 0.1 Henry induct er 15 Nana Farid capacitor. And we're supposed to find out at after one second or after a 10.1 seconds. How much energy is stored by the induct? Er so for part A, let's see, we're gonna be looking at what is the energy? So that's gonna be U of L. Right. So what is the magnetic energy right here? So let's say we have a conductor. That is 0.1. We have a capacitor that is 15 times 10 to the minus nine. Remember that that prefix nano and we have Q max, which is our maximum charge is going to be 50 Milica cool. Um, so that 0.5 right there. So the equation that is going to tie all of these things together is going to be the magnetic energy equation. So that's gonna be you. L equals one half L I squared. So let's see, We have the induct, insp. That's gonna be our l. All we have to do is figure out the current, but this is actually the current at a specific time. So what happens is that this current that we're gonna plug in actually comes from the current, often Elsie circuit. So we're gonna have to use the equation for I, which is that I have Tea is equal to negative. Omega que max times sign of omega T plus five. Okay, so this is actually where we're gonna get our current equation from where we need the current at a specific point. So let's see. We can figure out what are Omega is that's gonna be the square root of one over l c. So let's see. I get, uh, square roots of 1/0 10.1 times times 10 to the minus nine. That's the capacitance. And so we get that omega is equal to 25,820. So that's one variable And so we want to evaluate with the current is when t is equal to 0.1. So let's see. I have everything else. So I have omega. I have Q max. Uh, the one thing I don't know is I actually don't know what the phase angle is, but we actually do know that the capacitor begins or the sights. The system begins with the capacitor that is maximally charged. So what that means is, whenever that happens, um, so if he cycle begins with que Max, then that means that Phi is equal to zero. That means that your phase angle is just nothing. It's just starting. Cosine Omega T right. So remember that that phasing will just determined the starting points. And if you're starting with the capacitor maximally charged, then that starting point is just zero. So that fight is just equal to zero. Cool. So I'm just gonna move this around somewhere else that's gonna go over here Cool. Uh, eso we could just go on with our equation. So we have I don t is equal to 0.1. We have negative omega. So that's negative. 25 8 to 0 and then we have Q max, which is 0.5 And now we have this sign off. Then we have 25 8 20 because that's Omega and then T, which is 0.1. And remember that you have to have your calculator in radiance mode and your current should be. I get 490 amps. So now we're just gonna plug that into the equation. By the way, if you got a negative sign, it might have been something there. But the thing is that it's going to go away when you square anyways, so let's see. Our magnetic energy is gonna be one half now. There are inducted 0.1, and now our current is 490 squared. So that means that our magnetic energy is going to be I actually get 12,000 jewels. So that's our answer for that first part of the question. And now the second part says, What is the maximum current throughout the induct? Er, So let's see the equation for maximum current. You might have seen in another previous video that the maximum current the magnitude of the maximum current is going to be Omega Times Q. Max Because it's going to be wherever the sine function just equals one or negative one doesn't really matter. So let's see. Imax is just going to be Omega, which is 25 8 20 then Q max, which is 0.5 right. 0.5 So what I get is, I get, um so he gets 1291 So I get 12 1291 amps. Alright, guys. So that is it for this problem? Let me know if you have any questions I'll see in the next one.

5

Problem

In an oscillating LC circuit in which the capacitance C = 4μF and the maximum voltage across the capacitor V = 1.50V, the maximum current measured across the inductor is 50mA. What is the angular frequency of this LC circuit?

A

69444444 rad/s

B

8333 rad/s

C

1521 rad/s

D

0.00012 rad/s

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