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# How capacitors work in AC systems

ghost666 17481 13
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• What happens when AC voltage is applied to the capacitor legs? This element behaves differently than a resistor - the resistor allows direct flow of electrons. The current through the resistor is, according to Ohm's law, proportional to the voltage drop across it. In the case of a capacitor, current is drawn in rhythm as the element charges and discharges to a new voltage level.

The capacitor charges up to the voltage applied to it. It acts as a kind of electricity store - as long as a direct voltage (DC) is applied to this element, it maintains its internal charge. When the voltage changes, a current will flow that will try to compensate for the change. The value of the current flowing will be directly proportional to the rate of change of the charge on the capacitor plates.

In the figure above, we can see a simple circuit - a capacitor connected to an AC voltage source (top left). In such a system, between the current (blue line) and the voltage (red line) flowing through the capacitor, we observe a phase shift of exactly 90 ° (?? - exactly 1/4 of the entire cycle of alternating voltage; top right).

The AC voltage changes cyclically. The greater the capacity of the connected capacitor, the greater the charge will accumulate between its plates, and thus the greater the current that will flow in the system. It will be similar when we increase the frequency of the alternating voltage - the higher it is, the faster the capacitor must charge, and thus the greater the current flowing in the system. Thus - the current depends on the capacitance of the capacitor and the frequency of the alternating signal.

Capacitive alternating current systems

A fully capacitive AC system is one that consists only of an AC voltage source and a capacitor - as shown in the figure above. In this case, a single capacitor is directly connected to the voltage source. As the voltage changes at the source, the capacitor charges and discharges. Current flows through the system all the time - cyclically back and forth. However, no current actually flows through the capacitor. It is the electrons that concentrate first on one of the covers, and then flow through the system and concentrate on the other cover - at no time do they flow through the dielectric separating them, although this may appear to an inexperienced observer.

Capacitor reactance

The flow of electrons through the capacitor is directly proportional to the rate of change of voltage on the plates of this capacitor. Here, the proportionality factor is reactance, just like resistance is the proportionality factor of current and voltage in a system with a resistor.

The reactance is marked with the letter X to distinguish it from the normal resistance for DC voltage (R). However, like the resistance, its unit is the ohm [?]. The following equation describes the dependence of the reactance on the frequency of the alternating signal f (in Hertz) and the capacitance of the capacitor C (expressed in farads).

$$X = \frac {1} {2 \pi f C} = \frac {1} {\omega C}$$

As we can see in the above equation - the higher the frequency or capacity, the smaller the reactance and therefore the greater the current - exactly as described above.

Source: https://www.eeweb.com/profile/andrew-carter/articles/how-capacitors-behave-in-ac-circuits

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ghost666
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• metalMANiu
Level 19
The article should indicate in bold that not all capacitors are suitable for AC voltage.
• jendrula60
Level 12
Brightly . Straight. Affordable. You can ? You can .
• ^ToM^
Level 40
You have to make up your mind and take one position: ;)

First you write that electrons do not flow through the dielectric (i.e., de facto through the capacitor):

"It is the electrons [...] that never pass through the dielectric that separates them."

Where a moment later you write that they flow through the capacitor:

"The flow of electrons through the capacitor is directly proportional to the rate of change of voltage across the capacitor plates."

Through the capacitor, i.e. through the dielectric, as there is nothing else to flow through.

After all, are they flowing or not?
I will answer right away. Current flows through the capacitor as long as it is AC (or AC). Writing about electrons thus obscures the picture, because the electrons do not flow, but the current does. The whole radio technology is based on the fact that alternating current flows through the capacitors, where it does not flow through the coils (chokes) (chokes choke the current) - ideally. The choke resists the variation of the current.

Greetings!
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• zico111
Level 10
Someone above wrote the very truth here, namely that the current flows through the capacitor. And the electrons don't flow? and they do not flow, which does not mean that the current does not flow. The current flows due to the presence of an electric field (in this case not via electrons in the conductor). Everything is in Maxwell's laws.
Level 37
zico111 wrote:
Everything is in Maxwell's laws.

You're right. I agree. Only these equations can be solved and interpreted by a small handful of people.
Gradient, divergence, scalar field, vector field, lapsian ... yuck
• pupinizator
Level 15
Fortunately, you don't need to refer to Maxwell's Equations to predict the behavior of a (linear) capacitor. It is enough to take two simple equations:

$$\left\{\begin{matrix} q=C \cdot u \\i=\frac{dq}{dt} \end{matrix}\right. \quad \Rightarrow i=C\frac{du}{dt}$$ The first equation is a conclusion from (perhaps simple?) Electrostatics (i.e. Coulomb's law), the second equation is the definition of electric current.
• ^ToM^
Level 40
However, not mathematically, it follows directly from the above: the charge exchange on the facings of a capacitor over time is equivalent to the flow of electric current through it. The amount of charge exchange over time is directly proportional to the intensity of the current flowing through it.
So in any case, despite the dielectric break, current flows through the capacitor and energy is transferred.

Greetings!
• pupinizator
Level 15
^ToM^ wrote:
So in any case, despite the dielectric break, current flows through the capacitor and energy is transferred.
Even a dielectric is not needed. A vacuum is enough.
• ^ToM^
Level 40
pupinizator wrote:
^ToM^ wrote:
So in any case, despite the dielectric break, current flows through the capacitor and energy is transferred.
Even a dielectric is not needed. A vacuum is enough.

A vacuum is a dielectric. It is commonly used in RF technology and in vacuum capacitors.
• pupinizator
Level 15
^ToM^ wrote:
A vacuum is a dielectric.
Are you sure? Take a look at the definition. A dielectric is a material. A (perfect) vacuum is not a material. Maybe it's a detail, but if we are to be precise, let's be "for the pain".