I have an AC capacitor from the ceiling fan.

When I connect its end to a socket (AC line 110V), it gets charged (as touching the terminals produces a spark).

But at this point I am unable to understand this mechanism because I have read that capacitors don't charge storage for AC and act almost like a short circuit at t(0+). [neglecting the phase shift and assuming frequency above 50 Hz].

Could someone please explain where I am going wrong?

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    \$\begingroup\$ If you don't know how it works, don't plug it into the mains!! \$\endgroup\$ – exscape Jul 11 '12 at 19:22
  • \$\begingroup\$ I saw the technician doing that.... But I dont think it was wise enough to ask him the underlying equation for the phenomenon ;) \$\endgroup\$ – perilbrain Jul 11 '12 at 19:25

Capacitors do store charge. In fact, that's basically what a capacitor does, with the added characteristic that its voltage will be proportional to the amount of charge it has stored.

A capacitors doesn't store AC, but it does store whatever charge is on it given the voltage at the time it was disconnected. Since the AC voltage can vary from zero to fairly high peaks, it is somewhat random what the capacitor will be charged to. The peaks of the AC line is the square root of 2 times the RMS voltage. For example, 115 VAC has peaks of ±163 V. A capacitor could get charged to anywhere in that range.

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  • \$\begingroup\$ So it means there is a probability distribution for the capacitor instantaneous voltage? ie. (163 to -163V) as per the example?? \$\endgroup\$ – perilbrain Jul 11 '12 at 19:41
  • \$\begingroup\$ @Peril: Yes. And, due to the AC line being a sine wave, the probability is not even accross the range. It is weighted more towards the peaks because the sine spends proportionately more time there. If it were a triangle wave, then the probability would be even accross the range. \$\endgroup\$ – Olin Lathrop Jul 11 '12 at 19:43
  • \$\begingroup\$ Thanks a lot.....Understood (Rate of change in sinusoidal voltage)/(Rate of discharge) <<1 \$\endgroup\$ – perilbrain Jul 11 '12 at 19:49
  • \$\begingroup\$ @perilbrain, If the cap is literally placed across an AC source then (Rate of change in sinusoidal voltage)/(Rate of discharge) ~= 1 since the capacitor's voltage will track the input. In general the relationship I see between the supply voltage frequency is something that itself actually changes with time. You will at first be charging the capacitor until the input peaks, and then once the input falls below the capacitor voltage it will then being to discharge at an increasing rate until the input begins to become more positive .. repeat. \$\endgroup\$ – sherrellbc Jul 15 '14 at 15:32

There are two common uses of plastic non-polarized caps in AC motors

  1. Start Capacitors

    • briefly increase motor starting torque
    • allow a motor to be cycled on and off rapidly
    • Furnace motors switch off the cap. after it reaches 2/3 of full speed or so.
  2. Run Capacitors

    • are used in variable speed fans, which use single phase electric motors and need a capacitor to energize a second-phase winding.
    • If this cap is missing or improperly sized; the motor will get hotter, lose power, & become noisy.
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It's not quite correct to say that capacitors, as used in electrical circuits, store electric charge. In normal usage, a capacitor has no net electric charge even when "charged".

It is correct to say that a capacitor stores electric energy by keeping electric charge separated. When you connect a capacitor to a source of electric current, electric charge flows from one plate of the capacitor to the other plate via the external circuit. We say the capacitor is "charged" but we don't mean that it is electrically charged but, like a rechargeable battery, energy "charged".

When connected to an AC source, the capacitor alternately charges and discharges, i.e., energy is alternately stored in and then released by the capacitor. Work is alternately done on and then by the capacitor.

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A capacitor subjected to AC is being charged and discharged continuously, as the AC proceeds through the positive and negative parts of the cycle. At the instant you disconnect the capacitor, that's the charge that will be on it. It's even possible to disconnect it right at the zero crossing, and have no charge. Some capacitors will be designed to be across the mains, but most are not!

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  • \$\begingroup\$ But why not the circuit is shorting??? \$\endgroup\$ – perilbrain Jul 11 '12 at 19:36
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    \$\begingroup\$ "acts almost like a short circuit at t(0+)" -- almost, but not completely. Once it charges to that voltage, it stops drawing current. Yes, the cycle proceeds, and more current is drawn, but it's more like a resistor. In fact, this is called "reactance". You can figure this property based on the frequency and the capacitance. \$\endgroup\$ – gbarry Jul 11 '12 at 19:46
  • \$\begingroup\$ Unless the capacitor is a heavy-duty type, I find it strange it did not pop. I say that because the magnitude of reactance gives only 2.7 Ohms for a 1mF cap at 60Hz. As such, we have 60 Amps (peak) displacement current through the device! Not to mention 163V peak voltage across the terminals. You may not of had a 1mF cap, but anything smaller and it only gets worse. \$\endgroup\$ – sherrellbc Jul 15 '14 at 15:46
  • \$\begingroup\$ I'm used to seeing "mf" as a misuse of "microfarad" but a quick search reveals this malady is industry-wide. The danger is that someone will mistake that for milli-farads. You did, since you came up with 2.7 ohms at 60 Hz. These capacitors are in reality, microfarad values (uf) and therefore the reactance is 1000 times larger. The cap's voltage rating ought to be that of the peak line voltage or greater. In fact one article I read, said they test them by connecting them across the line and reading the current; that an ohmmeter supplying 1.5V doesn't give good results with a high-VAC cap. \$\endgroup\$ – gbarry Jul 17 '14 at 5:34

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