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Capacitive reactance -- the "resistance" of a capacitor to current flow -- is found by 1/(2*pi*f*C). For high frequencies, this "resistance" is low, allowing current flow. The lower the frequency, the higher this "resistance" is. DC or direct current is consider "zero" Hertz frequency and the capacitive reactance ("resistance") at this point is infinite, allowing no current flow at all. So, the higher the frequency, the easier it is for the signal to pass through the capacitor; the lower the frequency, the more difficult it is for current to pass and impossible for DC.
but physically, a capacitor is two plates with some form of insulator between them (that has di-electric proporties).
Hence a capacitor cannot pass DC since a capacitor is an open circuit.
For AC however, both plates get charged up with a certain polarity, when the voltage swaps over that charge then gets pulled off. So although capacitors "appear" to pass AC it doesn't really
The threadstarter must be very new to electronics to ask such a question. I'd suggest he grab any electronic book and read through the introductory chapters. These kind of information is fundamental in all of electronics.
A DC transient current does pass thru a capacitor, up to the point of the capacitor becoming fully charged. An AC current passes thru the capacitor continuously.
A DC transient current does pass thru a capacitor, up to the point of the capacitor becoming fully charged. An AC current passes thru the capacitor continuously.
The current doesnt really pass through in the physical sense, unless you are referring to displacement current. Anyway, venturing any deeper would be confusing the threadstarter even further.
Alternations (AC) between + and - cause the plates of the Capacitor to be equally but differently charged. They attract each other, and Electric field exists.
On the other hand, DC is constant and due to this, there are no opposite charges on plate 2 to attract the charges of plate 1. Consequently, DC is blocked.
But when you try to measure it, the cap charges a little, and given enough time, the cap fully charges, and then and only then can you say the cap does not any longer pass the DC. It actually still passes it, it's just that the cap now has a voltage that exactly cancels out the applied DC, resulting in a sum of zero volts at the cap output.
Just use a 10uF cap on the output of a signal generator (with DC offset knob). Then with a 10Meg Ohm scope probe on the other side of the cap (set scope to DC coupling), you can change the offset, and what do you know, you see a DC offset sine. Discharge the cap before wiring, else you'll get an additional voltage offset.
The scope 10 Meg ohms will slowly charge the cap to match the DC offset.
If you use a FET / active probe, you should not see any drift.
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