31. Alternating Current

Capacitors in AC Circuits

Physics31. Alternating CurrentCapacitors in AC Circuits

Hello class, Professor Anderson here. Let's take a look at an example circuit problem, this is just a very simple circuit, AC circuit, with one capacitor in there. Okay, and let's think about what happens here. Of course the current sloshes back and forth in this AC circuit, so it charges up the capacitor one way and then discharges it and charges it up the other way. And then we are given some numbers in this problem, namely the frequency of this circuit is 15 kilohertz, and at 15 kilohertz we get the following. We get a peak current of 65 milliamps and a rms voltage of 6 volts, and the question in this problem is what is the capacitance? Okay, what is the capacitor right there? Alright, so how do we deal with this? Well, we go back to good old Ohm's law, right? Ohm's law says the following, it says the voltage is equal to the current times the load, the resistance, but remember when we deal with these capacitors, we have complex impedance, and so the equivalent is V equals I times Z, where this Z is the complex impedance of the device. And to be technically correct, if we're talking about RMS voltage, we need to put RMS right there, and then we're talking about RMS current so we put an RMS right there. Now hopefully what you remember is Z for a capacitor is 1 over Omega times C, right? As the frequency goes up the impedance of the capacitor goes down, that current can jump right across there. As the frequency goes down, the impedance goes up, and that makes sense if you try to drive DC current through the capacitor, it's not going to go through, it charges up one plate and then it stops moving. All right. Let's see if we can solve this equation now, Vrms equals Irms 1 over Omega C, for C. Well that's not too bad, right? We can move C over the other side and what do we have left on the right side? We have Irms, we have an Omega in the bottom and then we have Vrms in the bottom. Okay, but we're not given Irms, we're given I naught, but of course the peak current is related to Ir -- Irms by root 2. Irms is just I naught over root 2. We're not given Omega but we are given the frequency F, but we remember that Omega is just 2 pi times F, and then we have Vrms, which we are given. So now it looks like we have everything we can -- everything we need to solve this problem so we can just plug them in and try it. All right, so let's plug this stuff in. We've got I naught, which is 65 times 10 to the minus 3 amps, we're going to divide that by root 2, and then all of that is over 2 pi times F after we said was 15 kilohertz, so 15 times 10 to the 3, and then we have Vrms, which is just 6 volts. So now everything is in SI units, should work out just fine. If you punch all these numbers into your calculator you should get something around 81.3 nanofarad. Remember Nano is ten to the minus nine, so this is 81 point three times ten to the minus nine farads. All right. That's how you attack problems like that, hopefully that's clear. If there's any questions, come see me in office hours. Cheers!