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?

Hey, guys. So let's get some more practice with LC circuits. Hopefully, you guys try this one on your own. We have an inductor with some inductance and a capacitor, and it begins with a a oscillation? So remember that that variable, the phase angle, is phi, and it's just basically it tells us where the oscillation is starting. Right? So if we sort of have this graph right here of time versus either charge or the current, doesn't really matter. We know that these things sort of form like sinusoidal curves like this. All that phase angle really tells us is at what point during the oscillation are we actually starting our sort of function at. Right? So that's all that tells us. So let's see. We have inductor, which is 0.05. We have the capacitor, which is 35 microfarads. So that's 35 times 10 to the minus 6th. And then let's see. We have that this circuit is beginning with a current of negative one amp. So what that tells us is that when it begins, the time is equal to 0, and it says that the current at that time, so I when t is equal to 0, is equal to negative one. So that means we have t, so remember this is c, that's t equals 0, and we have the current at that time, so is equal to negative one. And the equation that ties all these things together is the current equation for LC circuits. So we know in general, this is I of t equals negative omega q max, and then we have sine of omega t plus phi, and we're supposed to be figuring out what this phase angle is right here. Now, remember, this is sort of like the general equation for this formula, but we have that specifically the current when t is equal to 0 is going to be negative omega q max, and then we have sine, and now we have omega, and then we just plug in 0 for t. So in other words, we just have 0 right here. This whole term cancels out, and then we have plus phi. Okay. So let's see. I know what the current is when t is equal to 0. Right? That's just negative 1. And let's see, I have the, the omega. And remember the omega, I can get from the relationship of omega is equal to the square root of 1 over l c. And so I can actually figure out what those those variables are because I have l and I have c, so I know what the omega term is, or I can at least calculate it. And I have what the q max is. I'm told that the capacitor plates have a maximum charge of 2.65 5 millicoulombs at any time during the oscillation. Got it. So, and then I can use that to basically figure out what this phi variable is. So we just wanna go ahead and figure this out real quick. I know that omega is equal to the square root of 1 over, and then I have 0.05 for the inductance. And then I have 35 times 10 to the minus 6. So just remember to plug that in carefully, take the square root, take the inverse, and you're gonna get that omega is equal to 756, roughly. So we can just basically plug that back in, and let's see. I got, that negative one is equal to negative 756, so that's gonna be omega. Q max is gonna be 2.65 times 10 to the minus 3 coulombs, and then we're gonna get sine of phi right here. So what happens is that the, negative signs will cancel out, and I'm just gonna take both of these terms and then move them to the bottom. And so what happens is when I do this whole division, I'm just gonna end up with sine of theta on the right side. So actually, if you work this out, you're just gonna get 0.5 is equal to sine of 5 or sorry, I meant phi. And then finally, phi is just going to be the inverse sine of 0.5. So if you take the inverse sine, and remember, you guys should be working in radians mode, you should get a number that's 0.50 2, roughly, something like that. And that's actually equal to Pi over 6. Right. So remember that the inverse sine of half is pi over 6 in radians. Now, so that's actually the answer. You could just stop right there, because that's the, phase angle in radians, which you should get. But if you, just in case, wanted to actually convert this to degrees, just remember that the conversion of degrees or radians to degrees is you have to multiply by 180 over pi, and you should actually get 30 degrees. So that should also make some sense because sine of 30 degrees is 1 half. So either one of these cases or either one of these numbers works. So that is the answer. And let me know if you have any questions. That's it for this one.

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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?

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