Capacitors in AC Circuits - Video Tutorials & Practice Problems

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concept

Capacitors in AC Circuits

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Hey, guys, in this video, we're gonna talk about the roles that capacitors play in a sea circuits. All right, let's get to it now. The current Macy's circuit, as we've seen before, is always given by Max Times CoSine of Omega T. Okay, something that you guys need to remember is that the voltage across the capacitor is always gonna be the charge of the capacitor at that particular incident time divided by the capacitance. This is from all the way back when we talked about D. C circuits. Okay, 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 capacitor in an A C circuit at any time is gonna be Q divided by Omega Psi. Sorry. Not cute. I'm Max, divided by omega Psi times. Cosine of Omega T minus pi over two. Now, this is interesting, because when we saw the voltage across the resistor as a function of time, we saw that it was imax. Times are times cosine of omega t. So the angle off the co sign right this angle equals omega T is different than the angle for the voltage across the capacitor. This new angle, which I'll call data prime, is Omega T minus pi over two. That means that their functions, the current in the circuit and the voltage across the capacity are not gonna 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 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. Okay, then at a later time, the voltage matches. What the current, Doesn't 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. Okay. Another thing to notice is that the maximum voltage across the capacitor was just the amplitude of that equation. It was just imax divided by Omega Psi. This looks a lot like OEMs law. Remember that owns law for resistor says that the voltage across the resisters I times the resistance. Well, this looks like the current times that some quantity one overall magazine right? I can rewrite. This is I Max times one over Omega Psi. So one overall, Negus e looks like a resistance like quantity. It's a resistance like quantity. It carries the units off OEMs and we call it the capacitive reactant. Okay, so it's the reactant in a reactant acts like a resistance. Okay. And the capacitive reactant is one over Omega psi. Okay, so let's do an example, And a C power source delivers a maximum voltage off 120 volts at 60 hertz. What is the maximum current in a circuit with this power source connected to a 100 micro Farhad capacitor. Okay, so we want to know the maximum current in the circuit. We know the maximum current and a capacitor circuit is going to be given by I. Max equals the maximum voltage across the capacitor, divided by the capacitive reactant. Okay, that's just using this equation right here. Now, the question is, what is the capacity of reactant since? Or the capacity of reactant is one over omega times to capacitance. Omega is an angular frequency. We're told a linear frequency, so we have to convert the two first. Okay, remember that omega is defined is two pi f, which is two pi times 60 which is 377 in verse seconds. Okay, so now we confined the capacitive reactant, which is one over 3 times 100 micro is 10 to the negative six fair ads, and this equals 26 5. Remember that the units of reactant our homes because they are a resistance like quantity. Finally, the maximum currents, which is just the maximum voltage over the capacitor divided by the capacity of reactant. Since what is the maximum voltage across the capacitor? Okay, the capacities connected directly to the battery. So it has to share the maximum voltage with the battery. Right? That's just care. Jobs loop rule. Okay, the maximum voltage of the battery is 120 volts. So this is 1 20 divided by the capacity of reactions, which is 26.5. And that's gonna be 453 amps. Okay. Alright, guys, that wraps up our discussion of capacitors in a sea circuits. Thanks for watching.

2

Problem

Problem

An AC source operates at a maximum voltage of 120 V and a frequency of 60 Hz. If it is connected to a 175 µF capacitor, what is the maximum charge stored on the capacitor?

A

7.92C

B

0.021 C

C

21 C

D

0.0021C

3

example

Current in a Parallel RC AC Circuit

Video duration:

3m

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Hey, guys, let's do an example involving capacitors in a sea circuits and a C source operating at 160 inverse seconds. And at a maximum voltage of 15 volts is connected in parallel toe a five ohm resistor and in parallel toe, 1.5 million Farah capacitor. What does the RMS current through the capacitor? So what does this look like? Well, let's say we have an a c source. It's connected in parallel to a resistor, which we're told is that five home resistor and it's connected in parallel to a capacitor. And we want to know what is the RMS current through the capacitor. Okay, well, we can actually solve this problem without involving the resistor at all, because this forms its own loop, which means that kerchiefs loop rule applies to it. So whatever voltage the battery has, the capacitor has the same voltage, and we can completely ignore the resistor because it doesn't belong in this loop. Okay, so everything that happens is determined on li buy things within that loop. Okay, so the resistor doesn't actually affect this problem at all. It's just throwing at you to confuse you. Okay? Now we want to find the R. M s current. In order to know any RMS value, you have to know the maximum value, right? The R. M s current is going to be the maximum currents divided by the square root of two. Okay, so before we can know the r. M s, we have to know the maximum. Now, the maximum current through a capacitor is always going to be that maximum voltage across capacitor divided by the capacitive reactions. So the first thing we need to figure out is what is that capacitive reactant. It's just one over Omega Psi. What is Omega? We're told that it operates at 160 in for seconds. If this was hurts, we would assume that that was a linear frequency. But because the units are in for seconds, we're gonna assume that that is an angular frequency, so we can plug that right into here. 1 60. The capacitance is 1.5 million fare ads or 0.15 fair ads, and that whole thing comes out to 4.2 owns. Now that we know the capacitive reactant, we confined the maximum current through the capacitor. Okay, that maximum current is just going to be the maximum voltage across the capacitor, divided by that reactant. Now, what is the maximum voltage across the capacitor? Since it's in parallel with the source, it's just gonna be the maximum voltage of the source or volts. So there's gonna be 15 volts divided by 4.2 owns, which is 3.57 amps. Okay, so we know that maximum current all we need to do now is divided by the square root of two to figure out what the R. M s current is. So finally, the arm s current is 357 and divided by the square root of two, which is just 252 APS. Okay, And that is the answer to the question. All right, guys, Thanks for watching

4

Problem

Problem

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?

A

29.4 Hz

B

4.68 Hz

C

41.7 Hz

D

6.63 Hz

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