• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

capacitor in AC circuit, capacitance, charge stored, etc.

PG1995

Active Member
Thread starter #1
Hi,

A capacitor stores its energy in form of electric field which results from static charges. In a pure capacitive circuit current leads voltage by 90 degrees. As the circuit is turned on to charge a capacitor, the current through it is maximum, but the voltage and electric field between the plates are zero. Once the capacitor is fully charged, the current is zero but the voltage and electric field are maximum. Electric field is proportional to static charges accumulated on the plates of capacitor.

An inductor stores its energy in form of magnetic field which results from moving charges. In a pure inductive circuit current lags voltage by 90 degrees. As the circuit is turned on to 'charge' an inductor, the current through it and magnetic field around it are zero but the voltage is maximum. Once the inductor is fully 'charged', the current and magnetic field around it are maximum but the voltage is zero. Magnetic field around an inductor is proportional to the current flowing through it.

Question 1:
In an AC circuit, current flows one way for a half cycle and flows the other way round the other half cycle. In other words, source polarities get switched every half cycle. A half cycle is taken from 0 degrees to 180 degrees; or from 0 to π to in radian measure.

RMS value of an alternating current is its effective heating value equivalent to a DC current.

In a DC circuit one can simply use the formula Q=CV to find the charges accumulated on capacitor plates. How do we find the charges accumulated on capacitor in an AC circuit? I think that average value of voltage should be used instead of RMS because we are only interested in knowing the average quantity of charge flowing into a capacitor during a half cycle. Average value of sinusoidal voltage waveform is 0.637×Vpk where "Vpk" is peak value of the sinusoidal voltage [1]. The formula could be modified as Q=CV=C*(0.637×Vpk). Do I have it right?

Question 2:
In my view, we have missed an important point above to make things simpler. In a half cycle once the capacitor has been charged, the polarities of source get switched. Now the positive of the source is connected to negative of capacitor and so on. First, the capacitor needs to get discharged and it will definitely take some time. Once the discharging has taken place, the charging phase will start. All this is going to happen over a single half cycle. So, my question if the formula, Q=CV=C*(0.637×Vpk), for the accumulation of charge which assumes charging through out the half cycle is still valid because, in my view, during a small part of the given half cycle, no charging is taking place rather discharging is taking place.

References:
[1] https://www.electronics-tutorials.ws/accircuits/average-voltage.html
 

dknguyen

Well-Known Member
Most Helpful Member
#2
No. Q=CV uses the instantaneous voltage to get the charge stored at that instant so it works for both AC and DC. You use instantaneous voltage in both cases. They are treated exactly the same. Not Vpk, Vavg, or Vrms (and technically not VDC either. It just so happens to be the instantaneous voltage at all times is all).
 
Last edited:

AnalogKid

Well-Known Member
#3
Assuming the alternating current through (or alternating voltage across) a capacitor is perfectly symmetrical about the 0 A axis (or 0 V axis), then the long term average voltage across a capacitor is 0 V and the long term charge in the capacitor is 0 C.

When discussing reactive components, it is important to separate transient and steady-state conditions.

ak
 

Ratchit

Well-Known Member
#4
Hi,

A capacitor stores its energy in form of electric field which results from static charges. In a pure capacitive circuit current leads voltage by 90 degrees. As the circuit is turned on to charge a capacitor, the current through it is maximum, but the voltage and electric field between the plates are zero. Once the capacitor is fully charged, the current is zero but the voltage and electric field are maximum. Electric field is proportional to static charges accumulated on the plates of capacitor.

An inductor stores its energy in form of magnetic field which results from moving charges. In a pure inductive circuit current lags voltage by 90 degrees. As the circuit is turned on to 'charge' an inductor, the current through it and magnetic field around it are zero but the voltage is maximum. Once the inductor is fully 'charged', the current and magnetic field around it are maximum but the voltage is zero. Magnetic field around an inductor is proportional to the current flowing through it.

Question 1:
In an AC circuit, current flows one way for a half cycle and flows the other way round the other half cycle. In other words, source polarities get switched every half cycle. A half cycle is taken from 0 degrees to 180 degrees; or from 0 to π to in radian measure.

RMS value of an alternating current is its effective heating value equivalent to a DC current.

In a DC circuit one can simply use the formula Q=CV to find the charges accumulated on capacitor plates. How do we find the charges accumulated on capacitor in an AC circuit? I think that average value of voltage should be used instead of RMS because we are only interested in knowing the average quantity of charge flowing into a capacitor during a half cycle. Average value of sinusoidal voltage waveform is 0.637×Vpk where "Vpk" is peak value of the sinusoidal voltage [1]. The formula could be modified as Q=CV=C*(0.637×Vpk). Do I have it right?

Question 2:
In my view, we have missed an important point above to make things simpler. In a half cycle once the capacitor has been charged, the polarities of source get switched. Now the positive of the source is connected to negative of capacitor and so on. First, the capacitor needs to get discharged and it will definitely take some time. Once the discharging has taken place, the charging phase will start. All this is going to happen over a single half cycle. So, my question if the formula, Q=CV=C*(0.637×Vpk), for the accumulation of charge which assumes charging through out the half cycle is still valid because, in my view, during a small part of the given half cycle, no charging is taking place rather discharging is taking place.

References:
[1] https://www.electronics-tutorials.ws/accircuits/average-voltage.html
As I have stated before in this and other forums, using scientific slang and jargon distorts what is really happening. You want a concise answer, so ask a concise question. "Current flow" literally means "charge flow flow" which is redundant and ridiculous. You should instead say "current exists", "current is present", or maybe just plain "current", which by itself means charge flow. It is OK to talk about the direction of current, but if charge does not flow, then current does not exist, does it?

You talk about capacitors being "charged"in your above questions. What are they being charged with? Is it charge carriers such as electrons? No, a cap with 100 volts across it contains the same number of electrons as it did when it had zero volts across it. For every electron that is deposited on one plate, another electron leaves the opposite plate for a net charge of zero. Charged with energy? Correct, but then why not say "energized", and avoid the confusion with charged particles?

Your above two questions involve differential equations to determine the amount of charge separation in a capacitor at a particular time after a particular voltage is applied. A text on diffy Q will satisfy your curiosity. Keep in mind that resistance is always present. A theoretical cap with no resistance will look like a short across a voltage source. It will take an infinite amount of current a infinitesimal amount of time to energize the theoretical capacitor up to the value of the voltage source.

Ratch
 

PG1995

Active Member
Thread starter #6
The point is well taken. I'd try to be more careful. But at the same time I wasn't writing a book so my focus was just to convey what I thinking! :) Anyway, I agree with what Ratchit said. Thanks.
 

Latest threads

EE World Online Articles

Loading

 
Top