# How exactly does a capacitor charge?

Discussion in 'General Electronics Discussion' started by Apocaloptigon, Jun 23, 2018.

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1. ### Apocaloptigon

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Jun 23, 2018
So, I have a leyden jar and I notice that while I charge it, it seems to increase with voltage as well as current (the current part is expected). I charge it using electrostatics, so maybe that's the confusion... Anyway, I didn't think that capacitors charged that way. Don't they retain the same amount of voltage and increase the amount of current?

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
A capacitor and a leyden jar work in the same way.

"charging" (some people prefer "energising") consists of altering the electrical charge either side of an insulating dielectric.

In a practical sense, the capacitor as a whole has a neutral electrical charge (it does not lose or gain electrons as a whole).

Some method of moving electrons from one side of the dielectric to the other is the normal method of "charging" a capacitor.

This imbalance remains until (typically) an electrical circuit provides a path for the electron imbalance to be removed.

The amount of energy stored in a capacitor (which represents the amount of work required to move all those electrons from one side to the other and which will be released when they are allowed to move back) can be calculated in a number of ways, but depends on a combination of any two of the voltage across the capacitor, the capacitance of the capacitor, and the sum of currents in and out of the capacitor over time. Any two of these allows the calculation of the other.

3. ### Harald KappModeratorModerator

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Nov 17, 2011
Your question is rather cryptic - at least to me as a non-native English speaker,
What "increase" do you mean? When a current flows into a capacitor, voltage increases per
V = I*t/C
V = voltage
I = current
t = time
C = capacity

In a typical setup with a voltage source feeding the current to the capacitor the current is limited by the internal resistance (or current limiting) of the voltage source. The limited current will take some time to charge the capacitor. It cannot, however, charge the capacitor to a voltage higher than the voltage of the voltage source.
The current is driven by the difference in voltage from the voltage source and voltage on the capacitor. When you start charging, the voltage on the capacitor is 0 V, therefore the voltage difference is high and current is high, too. As the capacitor gets charged, the voltage on the capacitor increases and the voltage difference diminishes, thus reducing the current, not increasing it (read e.g. here).

No. As the voltage rises, current is reduced, see above.

4. ### Apocaloptigon

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Jun 23, 2018
By "increase," I meant that the spark gaps get larger as I charge it, where I thought that the capacitor has the same amount of voltage that you charge it with regardless of how much you charge it. Thank you for clearing that up, I can die in peace 70 years from now... maybe less.

5. ### davennModerator

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Sep 5, 2009

Please don't use the word "charge" as Steve hinted at several posts ago, you "energise" a capacitor

Harald didn't say anything contrary to your statement. Yes, the capacitor energises to the voltage that you put across it.
As the capacitor energises, the amount of current flowing into it decreases and stops once the voltage across the capacitor reaches the supply voltage
This does not happen instantly.

But NOT regardless of that voltage
A capacitor has a maximum working voltage that is exceeded it will destroy the capacitor

A capacitor is never chargedthe net charge on a capacitor is always zero.
The amount of negative charge on one plate is exactly cancelled by the amount of positive charge on the other plate

The energy in a capacitor is stored in the electric field BETWEEN the plates.

Dave

Last edited: Jun 23, 2018

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Nov 5, 2013
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
It is a basic unit of measure which compares the frequency of a signal to that of a submultiple of the frequency corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.

It superceded an ancient measure which required one perform a similar comparison with a measure derived from the rotational period of the planet.

8. ### Jeff Leites

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Nov 5, 2013
Dad-blasted it, I just use a frequency counter.

9. ### davennModerator

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Sep 5, 2009

just don't spread them around here

thread closed before it goes any further down hill

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