Ch 29: Alternating CurrentSee all chapters

Sections | |||
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Alternating Voltages and Currents | 18 mins | 0 completed | Learn |

Inductors in AC Circuits | 13 mins | 0 completed | Learn |

RMS Current and Voltage | 10 mins | 0 completed | Learn |

Phasors for Inductors | 7 mins | 0 completed | Learn |

Phasors | 21 mins | 0 completed | Learn |

Impedance in AC Circuits | 19 mins | 0 completed | Learn |

Resistors in AC Circuits | 10 mins | 0 completed | Learn |

Series LRC Circuits | 11 mins | 0 completed | Learn |

Phasors for Resistors | 8 mins | 0 completed | Learn |

Resonance in Series LRC Circuits | 10 mins | 0 completed | Learn |

Capacitors in AC Circuits | 17 mins | 0 completed | Learn |

Power in AC Circuits | 6 mins | 0 completed | Learn |

Phasors for Capacitors | 8 mins | 0 completed | Learn |

Concept #1: Capacitors in AC Circuits

**Transcript**

Hey guys, in this video we're going to talk about the roles that capacitors play in AC circuits. Alright let's get to it. Now the current in an AC circuit as we've seen before is always given by I max times cosine of omega T. Something you guys need to remember is that the voltage across a 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 a capacitor in an AC circuit at any time is going to be Q divided by omega C, sorry not Q, I max divided by omega C times cosine of omega T minus Pi over 2.

Now this is interesting because when we saw the voltage across a resistor as a function of time, we saw that it was I max times R times cosine of omega T. So the angle of the cosine, this angle equals omega T, is different than the angle for the voltage across a capacitor. This is a new angle which I'll call theta prime is omega T minus Pi over 2. That means that their functions, the current in the circuit and the voltage across a capacitor are not going to line up. 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 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 with the current 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 I max divided by omega C. This looks a lot like ohms law remember that ohms law for resistors says the voltage across a resistor is I times the resistance. Well this looks like the current times at some quantity one over omega C. I can rewrite this is I max times one over omega C. So one over omega C looks like a resistance-like quantity. It's a resistance-like quantity, it carries the units of ohms and we call it the capacitive reactance. So it's the reactance and a reactance acts like a resistance and the capacitive reactance is one over omega C.

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 the circuit with this power source connected to a 100 microfarad capacitor? So we want to know the maximum current in the circuit. We know the maximum current in a capacitor circuit is going to be given by I max equals the maximum voltage across the capacitor divided by the capacitive reactance. That's just using this equation right here. Now the question is what is the capacitive reactance? Well the capacitive reactance is one over omega times the capacitance. Omega is an angular frequency we are told a linear frequency so we have to convert the two first. Remember that omega is defined as 2 Pi F, which is 2 Pi times 60 which is 377 inverse seconds. So now we can find the capacitive reactance which is one of over 377 times 100, micro is 10 to the -6, farads and this equals 26.5 remember that the units of the reactance our 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? The capacitor is connected directly to the battery so it has to share the maximum voltage with the battery that's just Kirchoff's loop rule. 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. Alright guys, that wraps up our discussion of capacitors in AC circuits. Thanks for watching.

Practice: 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?

Example #1: Current in a Parallel RC AC Circuit

**Transcript**

Hey, guys. Let's do an example involving capacitors in AC circuits. An AC source operating at 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? 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. Well we can actually solve this problem without involving the resistor at all because this forms its own loop which means that Kirchoff'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. So everything that happens is determined only by things within that loop. So the resistor doesn't actually affect this problem at all, it's just thrown at you to confuse you. Now we want to find the RMS current. In order to know any RMS value you have to know the maximum value. The RMS current is going to be the maximum current divided by the square root of 2.

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 a 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 160 inverse seconds, if this was Hertz we would assume that that was a linear frequency but because the units are inverse seconds we are going to assume that there 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. Now that we know the capacitive reactance we can find the maximum current through the capacitor. 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 would be 15 volts divided by 4.2 ohms which is 3.57 amps. 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 and that is the answer to the question. Alright guys, thanks for watching.

Practice: 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?

0 of 4 completed

If the voltage amplitude across an 8.50-μF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:A) 32.6 kHzB) 5.20 kHzC) 5.20 HzD) 32.6 HzE) 32.6 MHz

What maximum current is delivered by an AC source with ΔVmax = 33.0 V and f = 110.0 Hz when connected across a 3.70 μF capacitor?A) 0.12 AB) 56.3 mAC) 2 mAD) 84.4 mA

What maximum current is delivered by an AC generator with maximum voltage 83.8 V and frequency 71 Hz when connected across a 4.52 μF capacitor?
1. 163.81
2. 150.307
3. 192.614
4. 105.054
5. 195.046
6. 168.974
7. 50.4161
8. 55.4684
9. 155.506
10. 62.1003

The reactance of a capacitor in a circuit at 110 Hz is 35 Ω. Determine the capacitance.A) 41 μFB) 260 μFC) 51 mFD) 0.10 FE) 0.31 F

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