Capacitors in AC Circuits - Online Tutor, Practice Problems & Exam Prep

In AC circuits, capacitors play a crucial role where the voltage across a capacitor lags the current by 90 degrees. The relationship is defined by the equation for capacitive reactance, X = 1ωC, where ω = 2πf. The maximum current can be calculated using I = VX. Understanding these concepts is essential for analyzing AC circuits effectively.

Capacitors in AC Circuits

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.

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

Current in a Parallel RC AC Circuit

Hey guys, let's do an example involving capacitors in AC circuits. An AC source operating at a 160 inverse seconds and at a maximum voltage of 15 volts is connected in parallel to a 5 ohm resistor and in parallel to a 1 and a half millifarad capacitor. What is the RMS current through the capacitor? Okay. So what does this look like? Well, let's say we have an AC source. It's connected in parallel to a resistor, which we're told is a 5 ohm 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 Kirchhoff's 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 only by things within that loop. Okay? So the resistor doesn't actually affect this problem at all. It's just thrown at you to confuse you. Okay? Now we wanna find the RMS current. In order to know any RMS value, you have to know the maximum value. Right? The RMS current is going to be the maximum current divided by the square root of 2. Okay, so before we can know the RMS we have to know the maximum. Now the maximum current through a capacitor is always going to be that maximum voltage across the capacitor divided by the capacitive reactance. So the first thing we need to figure out is what is that capacitive reactance. It's just one over omega c. What is omega? We're told that it operates at a 160 inverse seconds. If this was Hertz we would assume that that was a linear frequency but because the units are inverse seconds, we're gonna assume that that is an angular frequency. So, we can plug that right into here, 160. The capacitance is 1 and a half millifarads or 0.0015 farads and that whole thing comes out to 4.2 ohms. Okay now that we know the capacitive reactance, we can find the maximum current through the capacitor. Okay? That maximum current is just going to be the maximum voltage across the capacitor divided by that reactance. Now what is the maximum voltage across the capacitor? Since it's in parallel with the source, it's just going to be the maximum voltage of the source or 15 volts. So, this is going to be 15 volts divided by 4.2 ohms which is 3.57 amps. Okay? So we know that maximum current, all we need to do now is divide it by the square root of 2 to figure out what the RMS current is. So finally, the RMS current is 3.57 amps divided by the square root of 2 which is just 2.52 amps. Okay? And that is the answer to the question. Alright guys, thanks for watching.

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

Here’s what students ask on this topic:

What is the role of capacitors in AC circuits?

In AC circuits, capacitors play a crucial role by storing and releasing energy. The voltage across a capacitor lags the current by 90 degrees, meaning the current leads the voltage. This phase difference is due to the capacitive reactance, which is given by the equation $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. This reactance acts like a resistance in the circuit, affecting the current flow and voltage distribution.

How do you calculate the capacitive reactance in an AC circuit?

The capacitive reactance $X$ in an AC circuit is calculated using the formula $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. The angular frequency $\omega$ is related to the linear frequency $f$ by the equation $\omega =2\pi f$. This reactance is measured in ohms and determines how much the capacitor resists the change in voltage in the AC circuit.

Why does the voltage across a capacitor lag the current in an AC circuit?

The voltage across a capacitor lags the current in an AC circuit by 90 degrees due to the nature of how capacitors store and release energy. When the current changes, the capacitor charges or discharges, causing a delay in the voltage response. This phase difference is a result of the capacitive reactance, which is given by $X=\frac{1}{\omega C}$. The current leads the voltage because the capacitor needs time to build up or release the charge, resulting in the voltage lagging behind the current.

How do you calculate the maximum current in an AC circuit with a capacitor?

The maximum current $I$_max in an AC circuit with a capacitor can be calculated using the formula $I$_max=V_maxX, where $V$_max is the maximum voltage and $X$ is the capacitive reactance. The capacitive reactance is given by $X=\frac{1}{\omega C}$, where $\omega$ is the angular frequency and $C$ is the capacitance. By substituting the values, you can find the maximum current in the circuit.

What is the relationship between capacitive reactance and frequency in an AC circuit?

The capacitive reactance $X$ in an AC circuit is inversely proportional to the frequency $f$. The relationship is given by the formula $X=\frac{1}{2\pi fC}$, where $C$ is the capacitance. As the frequency increases, the capacitive reactance decreases, and vice versa. This means that at higher frequencies, the capacitor offers less opposition to the current flow, while at lower frequencies, it offers more opposition.