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Basic capacitor question

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LAN_403

Energyandfrequency

Tue Jan 08 2013, 03:55AM

Registered Member #9130 Joined: Sat Dec 29 2012, 03:31PM
Location:
Posts: 17

Hello everyone, I have a basic question about capacitor operation. If I have several capacitors in a bank connected to an AC circuit (in series, parallel or any combination thereof), will the total stored energy ever be more than the energy input from the source. I know that if capacitors are combined in series their voltage rating increases, but logically I want to say that the total stored energy can't equal more than the input... however I think there must be a way to either prove or disprove my possibly faulty logic with math :)

Thank you!

HvH

radiotech

Tue Jan 08 2013, 07:34AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

Firstly. The theoretical answer assumes that everything in the circuit has
no electrical resistance.

The capacitor connected to the AC circuit stores no energy averaged over
a period of time. It does displace the voltage wave from the current wave.
At any time there will be a current flow. (amps) and a voltage (volt)

Their product would be zero, according the formula:

Current (amps) * voltage (volts) * Cos theta (the angle of displacement)
which is 90 degrees.

Capacitors in pracrical circuits have resistance, and they may appear to increase
the voltage of a circuit to which they are connected, which may be the root of your query.

Proud Mary

Tue Jan 08 2013, 09:44AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Any possible series and parallel combination of ideal capacitors will always behave as though it were a single capacitor, whose value we can calculate.

The 'working voltage' of a capacitor is the maximum potential difference that can be applied across it without significant risk of failure. This will depend upon the type and thickness of the dielectric (the insulating material between the plates, such as air, mica, polyester, polystyrene, PTFE, oiled paper, metal oxides, etc) and the temperature, amongst other things.

There is no 'free energy.' In the real world, you can never get out of a capacitor quite as much energy as was put in. There are always losses due to resistance, and the properties of the dielectric itself. Energy is not stored in the metal plates, but in the dielectric between them.

Dr. Slack

Tue Jan 08 2013, 05:08PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

You can't prove anything like that with logic or math straight-off. You either have to do a lifetime of experiments, or take a short-cut by believing the "Law of Conservation of Energy". All physicists and most people on this board do. There will always be a few nut-jobs who will give you an argument, but they don't build successful experiments.

Assume when you build your capacitor array, it's at zero volts, storing zero energy.

Then you connect it to an external supply. There's a flash and a bnag at the contacts, and some current flows from the supply.

Whatever energy is now stored in the caps must have come from the supply. Whatever got dissipated in the flash and bang, and heat due to current flowing in the resistance of the wires, has come from the supply as well, but not been stored in the caps.

Therefore, adding logic and math to energy conservation, the total energy in the caps must be less than the amount that's come from the supply.

This is true regardless of the time, immediately after the initial connection, or after many cycles of AC.

Ash Small

Tue Jan 08 2013, 06:34PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

....Although there will be times when the energy in the capacitors will be greater than the energy coming from the source 'at that time', due to phase lag.

(This can more easily be seen/visualised when smoothing caps are used to 'smooth' rectified AC. The caps retain charge even when there is no output from the rectifier, but the 'average' power will be the same as the 'average' input power, minus losses.)

Energyandfrequency

Wed Jan 09 2013, 04:37AM

Registered Member #9130 Joined: Sat Dec 29 2012, 03:31PM
Location:
Posts: 17

Thank you, your answers helped!

Dr. Slack

Wed Jan 09 2013, 08:48AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Er, no, Ash

You are using Energy and Power carelessly. You can't have "energy coming from the source 'at that time'", what you are referring to is power.

The total energy stored in the capacitors at any time since t0 will always be less than the total energy delivered by the power supply since t0.

There will be times when there is energy stored in the caps and the supply is delivering no power. In an AC source feeding a rectifier/capacitor combination, this is indeed most of the time, when there is no current flowing because the recitifier diodes aren't turned on.

Ash Small

Wed Jan 09 2013, 03:08PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Thanks for correcting me, Dr S.

I did indeed mean power.

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