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?

Alright, guys. Hopefully, you were to try this one on your own. We have an oscillating LC circuit, and we have the capacitance, we have the maximum voltage, the maximum current, and we're supposed to figure out what the angular frequency is of this LC circuit. So, really, what we're looking for is we're looking for the variable omega. Now omega has a couple of different formulas we saw. We have 2 basic ways we can find it. We can relate it to the linear frequency by 2 pi f. Right? So we have that relationship, or we can say that, the omega is actually 2 pi times 1 over the period. Right? So we could relate it either to the linear frequency or the period, but the problem is we actually have none of those things. We don't know what the frequency, the number of oscillations per second, and we don't have the amount of time that it takes to actually go, you know, from one cycle to the other. So that means we're not gonna use these approaches. Instead, we're gonna relate the omega by using the properties of the LC circuit itself. Remember, omega is related to 1 over LC. So before we actually go ahead and start plugging stuff into this formula, let's just make sure we have. We have the capacitance, that's 4 microfarads, and let's see. Do we have the inductance? Well, no, we unfortunately don't. We have the we have the voltage, we have the capacitance, and we also have the maximum current. So we need to basically go and figure out what this inductance is. Now, let's see. I have what the capacitance is. It's 4 microfarads, so that's 4 times 10 to the max and that's going to be 1.5 volts. And now let's see, I have the maximum currents. I'm told that the maximum current across the inductor at any point in time is equal to 50 milliamps. So that's 0.05 amps. Right? Because I'm gonna move the decimal place back 3 times. And then we're supposed to figure out what this inductance is. Okay. Well, let's take a look. What haven't we used yet about the LC circuits? Well, we actually have the relationship between the capacitance and the voltage. That's actually related to the electrical energy across the capacitor, whereas the inductance and the current are related to the magnetic energy across the inductor. Now remember, the the electrical energy has 3 different formulas. We have 1 half cv squared equals 1 half qv equals 1 half q squared over c, and we just have to pick the one that we have the most information about. Now, that's actually going to be, c times v because we have both of those variables right here. So we have that relationship is the maximum amount or or is the energy across the capacitor. Whereas for the inductor, for the magnetic energy, that's actually 1 half times l I squared. Now, let's see. If these are both their maximum values, then what happens is that when the electrical energy is maximum, when the voltage across the capacitor is maximum So, basically, what what that means is that if we're comparing the maximum energies for each of these things, that because of conservation of energy, these two things have to be equal to each other. So what I'm saying is that u c is is equal to UL for this particular LC circuit. And this is actually sort of like the maximum value, so I'm gonna write UL max, UC max. So that reads really means that 1 half c v max, so 1 half times the capacitance, times the maximum voltage is equal to 1 half times the inductance, times the maximum amount of current squared that goes through that. So now let's see if I have enough variables to solve this. The one half will actually cancel out, and I have c, I have v max, and I have I max. All I have to do is just figure out what this l is, and then using that l, I can plug it back into this equation and figure out what the angular frequency is. So that means that c times v max squared divided by I max squared will give me my inductance. So, the capacitance We're going to square that We're gonna square that, and then we're gonna divide it by the maximum current, 0.05, square that, and you should get an inductance of 0.0036, and that's in henries. Now, we're not quite done yet because, again, we have to plug it back into our equation. So that means that our omega is equal to one over the square root of 0.0036. And now we have the capacitance, which is 4 times 10 to the minus 6. So, now, you have to do 1 over and you have to square root it, don't forget that, and you should get an omega value of 8,333, and the units for that are going to be in radians per second. And that's the answer. Alright, guys. So let me know if you have any questions about this problem, and I'll see you in the next one.

Struggling with Physics with Algebra?

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?

Watch next