A 300 µF capacitor is connected to an AC source operating at an RMS voltage of 120 V. If the maximum current in the circuit is 1.5 A, what is the oscillation frequency of the AC source?

Hey guys, I hope you're able to solve this problem on your own. If not, here's a little bit of help. A 300 microfarad capacitor is connected to an AC source operating at an rms voltage of a 120 volts. If the maximum current in the circuit is at 1 and a half amps, what is the oscillation frequency of this AC source? Okay. A good thing to do before you even start any problems is convert all rms values into maximum values. Typically you're not going to be able to use rms values So really quickly, just so we don't forget later, we are given an rms voltage. So the maximum voltage is going to be the square root of 2 times that rms voltage, which is going to be the square root of 2 times a 120 volts, which is going to be a 170 volts. Okay? So we can put that aside and we can use it later because typically it's the maximum values that you need. Very rarely do you ever need the rms values to plug into an equation. Okay? So we wanna find the oscillation frequency given the maximum currents. Alright. Right. Well in a capacitor circuit what equations do we have for the maximum current? We know that the maximum current in a capacitor circuit is going to be the voltage across the capacitor divided by the capacitive reactance. Okay and we know that the capacitive reactants does depend upon the oscillation frequency. It is 1 over omega times c. Okay? So, we can plug this in to our equation for the maximum current. We can say then that current is going to be that maximum voltage across the capacitor divided by 1 over omega c, which is going to be omega c v max. Right? That one over something in the denominator, you can pull that into the numerator. Okay? So what I want to do is divide cvmax over and now I have that omega is imax over c v max. Notice really quickly I did this without saying anything. I shouldn't have done that. I went from vc to v max. The reason is is that because the capacitor is the only thing in this circuit, its maximum voltage has to be the maximum voltage output by the source. This is something that's been said multiple times up to this point and I kind of just transitioned to it without stating anything but I should have said it. Alright. So all we have to do now is plug in our values to get an answer. The maximum current we're told is 1 and a half amps. The capacitance we're told is 300 microfarads. Right? So this is 300 times 10 to the negative 6 and the maximum voltage we found was 170. This is where it becomes good to figure out what the maximum voltage was right away instead of waiting until in verse seconds. We are asked for the 0.4 inverse seconds. We are asked for the frequency not the angular frequency so we do need to do one more step before we have our answer. The frequency is going to be omega divided by 2 pi or 29.4 divided by 2 pi, which is going to be about 4 point 6 8 hertz. That's our final answer. Alright guys, thanks for watching.

Struggling with Physics with Algebra?

A 300 µF capacitor is connected to an AC source operating at an RMS voltage of 120 V. If the maximum current in the circuit is 1.5 A, what is the oscillation frequency of the AC source?

Watch next