If the capacitor is fed from a 'Full Wave' rectifier, it will charge twice per AC cycle.
If a 'Half Wave' rectifier is used, it will charge once.
The basic purpose of a power supply smoothing capacitor is to store power between AC half cycles, to the connected load gets continuous power.
If it did not do that effectively, it would not be high enough value for the job.
The value of the capacitor is chosen to ensure the voltage stays in acceptable limits in normal operation.
Think of the rectified AC as water from a hand pump, so bursts rather than a smooth flow.
The reservoir capacitor acts like a bucket or tank under the spout, that has an outlet hole in it's base.
It fills to some level and provides a fairly smooth flow even between bursts from the supply, but will drain fairly quickly if the supply stops.
You would get basically the same result plotting the water level in the bucket as plotting the voltage across a reservoir capacitor.
There will be some effect on the instantaneous voltage depending what part of the cycle the supply is switched off.
T = RC
5T is usually taken as the charging or discharging time up to (1-exp(-5))*100%.
f = 1/10RC is the number of times capacitor charges and discharges for one ac cycle.
a capacitor never reach 0 during the cycle because (exp(-infinity))*100% and you can not wait till infinity.
capacitor opposes change in charge according to the equation Ri(t) +q(t)/C = V(t) and hence it does depend on what instant you turn the switch off.
extremely tough issue. do a search onto yahoo or google. that will help!
What circuit are you talking about? I can't read you mind and I'm not going to guess what you want. It's a waste of time to answer the wrong question.
Also why does a capacitor never reach 0 during the cycle and does the instant you turn off the switch in the circuit affect the amount of charge depending on what part of cycle it's at when you turn the switch off?
