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it always said that the higher the frequency, the less charge will accumulate because when in higher frequency, there is less time for capacitor to accumulate electrons. and in lower frequency, there will be more time for capacitor to accumulate electrons.

but in higher frequency, although the time for accumulating is less but the current is larger so how to conclude that in higher frequency, the total charge accumulated in capacitor is small ?

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    \$\begingroup\$ Please provide a reference for "it always said" claim. It does sound wrong as stated in your question. \$\endgroup\$ – Dmitry Grigoryev Aug 4 '17 at 9:31
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    \$\begingroup\$ @DmitryGrigoryev learnabout-electronics.org/ac_theory/reactance62.php The lower the frequency of the applied voltage, the more time the capacitor has to reach the fully charged, zero current state before the voltage reverses its polarity and begins to discharge the capacitor again. The capacitor therefore spends more time fully charged and passing much less current, the average value of current flow is therefore less at low frequencies. When a higher frequency is applied, the capacitor changes from charging to discharging sooner in its charge curve \$\endgroup\$ – John Lu Aug 4 '17 at 9:43
  • \$\begingroup\$ This writeup has a lot of inaccuracies. I stopped reading after the line where it says the capacitor never reaches zero current on AC. \$\endgroup\$ – Dmitry Grigoryev Aug 4 '17 at 10:09
  • \$\begingroup\$ Why do you say the current is larger? \$\endgroup\$ – user253751 Aug 4 '17 at 10:35
  • \$\begingroup\$ In every circuit you have real resistance which will limit the maximum current to some value independent of the frequency. At higher frequencies this fixed maximum current has less time to charge the capacitor, so the cap cannot accumulate as much charge before starting to discharge again. \$\endgroup\$ – JimmyB Aug 4 '17 at 12:02
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No, Capacitor will store more charge at higher frequencies since, its Capacitive Reactance is low for higher frequencies than the lower one. So the capacitor gets charged faster and outputs more current in the circuit when it discharges.

At lower frequencies, capacitive Reactance is high so that current entering into the capacitor is low. This is why capacitor takes more time to charge and outputs less current when it discharges.

To say simply, frequency is inversely proportional to reactance and directly proportional to current.

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  • \$\begingroup\$ I'm not sure what you're trying to say, but as it is written now is sounds like you claim that more electrons reach the cap if the time is shorter. \$\endgroup\$ – JimmyB Aug 4 '17 at 12:05
  • \$\begingroup\$ From your comment, I can understand that capacitive reactance is nothing to do with current. IS it? or I did understand in a wrong way? \$\endgroup\$ – Dhans Aug 4 '17 at 12:50
  • \$\begingroup\$ the electrons accumulated depend on the time of charging and also the charging current , so if the current is constant, shorter charging time leads to less electrons charged. but if charging current is increased by higher frequency, shorter charging time can also lead to be more electrons accumulated, so why in higher frequency, the voltage across on capacitor(electrons charge ) is lower while there is larger charging current? \$\endgroup\$ – John Lu Aug 4 '17 at 13:07
  • \$\begingroup\$ May be due to power factor. P = V x I. Is this correct? \$\endgroup\$ – Dhans Aug 4 '17 at 14:03
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    \$\begingroup\$ So, at higher frequencies you get more charge/discharge cycles per time unit, hence more electrons flowing back and forth per time unit (i.e. higher absolute current). However, to fully charge the capacitor takes infinite time, so the less time you have the less charges are transferred to or from the cap. I can charge the cap to 99.999% in 1s, but maybe only to 50% in 10ms, (depending on the series resistance, or impedance of the voltage source). Yet, 100x 50% charge is much more than 1x 99.999% charge, so the amount of charge transferred per time unit is larger at higher frequencies. \$\endgroup\$ – JimmyB Aug 4 '17 at 15:12

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