Hey, guys. In this video, we're going to talk about phasers for capacitors in AC circuits. So the current phaser and the functions that describe the voltage and the current across a capacitor at any time t in an AC circuit. These are given by these equations. Okay. Because the angle for each of them is different. Right? We have θ=ωt here and we have some other angle which I'll call θ'=ωt−π2 because the cosines each have a different angle we are saying that they are out of phase okay and in fact the voltage lags the current. This is very different than the phasers we saw for resistors which were in phase because both functions had the same angle ωt. Alright if we look at the 3 phasor diagrams that I drew, the first one, we have the current at an angle ωT, right? That is the angle of the current. In the second one, we have ωt−π2 as the angle for the voltage. That takes us all the way down into the negatives right because ωt itself as we can see is less than π2. So that takes us into the negatives and that also means that these have to have angles, right? This is just ωt which is the angle for the current minus degrees so they're separated by 90 degrees. So whenever you draw the phasor diagram which includes both the current and the voltage phases for a capacitor, okay, you have to draw it with the voltage lagging the current, okay, or the current leading the voltage. Okay? It's very important that you guys memorize that. The current leads the voltage or the voltage lags the current for a capacitor, alright? And that is the thing to take away from this. That the voltage across the capacitor always lags the current in the capacitor circuit. Okay? Let's do a quick example. An AC source is connected to a capacitor. At a particular instant in time, the voltage across the capacitor is positive and increasing in magnitude. Draw the phasors for the voltage and the current that correspond to this time. Okay? Now whenever a phaser is increasing in magnitude, it's because as it rotates it gets closer to a horizontal axis. Okay. As it gets closer and closer to that horizontal axis, its projection onto that axis gets larger and larger. And remember, the projection onto the horizontal axis tells us the value of that phaser. Okay. Now if this phaser is going to be positive it needs to be pointing to the right. And if it's going to be increasing in magnitude it it needs to be pointed to the right and moving towards the x-axis. Since phasers always rotate counterclockwise that means that the voltage phaser has to be here. Okay, and it's rotating like this. So it's positive because it points to the right and it's increasing in magnitude because as it gets closer to that horizontal axis more and more of it points horizontally until it's on the horizontal axis. Now it's at a maximum and then as it moves away from the horizontal axis it decreases and decreases in value until it's straight up and it's zero. Okay now if the voltage is here the current is 90 degrees ahead of it. So the current is going to be here. This is a current through a capacitor. This is the voltage of a capacitor. Okay? And this is a 90 degree angle. So this is what the phasor diagram looks like if the voltage is positive and increasing. Alright guys, that wraps up our discussion on phasers with capacitors. Thanks for watching.

31. Alternating Current

Phasors for Capacitors

31. Alternating Current

# Phasors for Capacitors - Online Tutor, Practice Problems & Exam Prep

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concept

### Phasors for Capacitors

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#### Video transcript

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Problem

ProblemAn AC source operates at a maximum voltage of 60 V and is connected to a 0.7 mF capacitor. If the current across the capacitor is i(t) = i_{MAX} cos[(100 s^{−1} )t],

a) What is i_{MAX}?

b) Draw the phasors for voltage across the capacitor and current in the circuit at t = 0.02 s. Assume that the current phasor begins at 0°.

A

a) i

_{max}= 4.2A b) θ_{iC}= 115^{o}θ_{VC}= 185^{o}B

a) i

_{max}= 4.2A b) θ_{iC}= 185^{o}θ_{VC}= 115^{o}C

a) i

_{max}= 4.2A b) θ_{iC}= 115^{o}θ_{VC}= 25^{o}D

a) i

_{max}= 4.2A b) θ_{iC}= 25^{o}θ_{VC}= 115^{o}