capacitor
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?
The third pint should read
Allow at least 10 us for the 10 nF capacitor to discharge.
If the capacitor has Vcc on both sides it's discharged. (IR) light on the phototransistor will cause a current to ground, so that voltage builds up on the capacitor, making its lower side go lower, as the other side is fixed at Vcc. Since the charging is through a current source instead of a resistor the voltage will decrease linearly with time.
And of course, how fuzz works (in general).
In words, adding fuzz is adding harmonics to the original signal. Harmonics are frequencies that are higher than the original signal but are tonally related.
More technically, this circuit essentially converts a sine wave
to something resembling a square wave
A sine wave is a pure tone which sounds dull
but a square wave contains the pure tone as well as odd order harmonics
which sounds more interesting.
Best Answer
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.