Hey guys. In this video, we're going to talk about the roles that capacitors play in AC circuits. All right, let's get to it. Now, the current in an AC circuit, as we've seen before, is always given by imax × cos(ωt). Okay? Something that you guys need to remember is that the voltage across the capacitor is always going to be the charge on the capacitor at that particular incident time divided by the capacitance. This is from all the way back when we talked about DC circuits. Now, there's a relationship between the amount of charge on a capacitor and the current in the circuit, and this is something that we've seen before. Now to find that relationship, we have to use calculus, but I'll just show you the end result. The end result is that the charge as a function of time looks like this. So, if I divide that charge as a function of time by the capacitance, I can say that the voltage across the capacitor in an AC circuit at any time is going to be q / (ωc), sorry not q, imax divided by ωc × cos(ωt).

This is interesting because when we saw the voltage across the resistor, right, as a function of time, we saw that it was imax × R × cos(ωt). So, the angle of the cosine, right, this angle equals ωt is different than the angle for the voltage across the capacitor. This new angle, which I'll call θ', is ωt - π / 2. That means that their functions, the current in the circuit, and the voltage across the capacitor, are not going to line up. Okay? When I plot them together, you'll see that they don't line up. In fact, the voltage of the capacitor lags the current by 90 degrees. So, if you look at any point in time right here, what is the current doing? The current is dropping. But what is the voltage doing? The voltage is at a maximum, right? But then it wants to match what the current is doing, so then at a later time, it starts dropping. But at that time, the current has already balanced out. Then at a later time, the voltage matches what the current does, and it balances out. But at this point, the current is already on the rise. So, you see, the current is leading the voltage. The voltage is just trying to catch up and match what the current is doing, but the current leads it, or we say that the voltage lags the current.

Another thing to notice is that the maximum voltage across the capacitor was just the amplitude of that equation. It was just Imax / (Omegac). This looks a lot like Ohm's law. Remember that Ohm's law for a resistor says that the voltage across a resistor is I times the resistance. Well, this looks like the current times some quantity ωc. So 1 / ωc looks like a resistance-like quantity. It carries the units of Ohms, and we call it the capacitive reactance. Okay? So, it's the reactance. And the reactance acts like a resistance. Okay? And the capacitive reactance is 1 / ωc. Okay? So, let's do an example. An AC power source delivers a maximum voltage of 120 volts at 60 hertz. What is the maximum current in a circuit with this power source connected to a 100 microfarad capacitor? Okay. So we want to know the maximum current in this circuit. We know the maximum current in a capacitor circuit is going to be given by imax = maximum voltage / capacitive reactance. Okay? That's just using this equation right here. Now, the question is what is the capacitive reactance? Well, the capacitive reactance is 1 / (ω × capacitance). Okay? ω is an angular frequency. We are told a linear frequency. 2πf which is 2π times 60, which is 377 inverse seconds. Okay, so now we can find the capacitive reactance, which is 1 / (377 × 100 micro = 10-6 farads) and this equals 26.5. Remember that the units of reactance are Ohms, because they are a resistance-like quantity. So, finally, the maximum current, which is just the maximum voltage over the capacitor divided by the capacitive reactance. What is the maximum voltage across the capacitor? Okay. The capacitor is connected directly to the battery, so it has to share the maximum voltage with the battery, right? That's just Kirchhoff's loop rule. Okay, the maximum voltage of the battery is 120 volts. So, this is 120 divided by the capacitive reactance, which is 26.5, and that's going to be 4.53 amps. Okay? All right, guys, that wraps up our discussion of capacitors in AC circuits. Thanks for watching.