From what I understand, a capacitor is used to store electric charge and when it is fully charged it can release electricity. When I looked at a capacitor, I found two pieces of information on it:
As I understand, the voltage rating on a capacitor is the maximum amount of voltage that a capacitor can safely be exposed to and can store. But what about when it is fully charged and released, how much voltage can it release? Does it equal the voltage rating?
But what about when it charged full and release, how much voltage it can release ? Does it equal to the voltage rating ?
Whatever that may mean to you, "releasing voltage" is not a proper way to think of what a capacitor does.
Electric power is delivered to a capacitor when charging and electric power is supplied by a capacitor when discharging. Thus, capacitors store electric energy.
The more energy stored by a given capacitor, the more voltage there must be across the capacitor. In fact, the energy stored by a capacitor is proportional to the square of the voltage across:
\$W_C = \dfrac{CV^2}{2}\$
where C is the capacitance. The greater the capacitance, the more energy stored for a given voltage.
But, real capacitors can be damaged or have their working life shortened by too much voltage. Thus, the voltage rating of a capacitor.
To summarize, a capacitor does not release voltage, a capacitor stores and releases energy.
But what about when it charged full and release, how much voltage it can release ? Does it equal to the voltage rating ?
No, it depends on the voltage that it has been charged with. When disconnected from the circuit, the capacitors voltage is equal or lower to the previously applied voltage.
A capacitor can store electric energy. It depends on the load how fast a capacitor discharges when connected to that load. (T = R * C)
The voltage rating just specifies the maximum voltage that should be applied to the capacitor.
Capacitors store energy. The voltage depends upon the amount of charge and the size of the capacitor. (Q = CV, Energy stored = 0.5CV^2).
If you connect a resistor across the terminals of a charged capacitor an initial current (= V/R) will flow but this will rapidly fall towards zero as the capacitor is discharged.
How quickly the voltage falls is determined by the time constant of the circuit (= CR) where C is measured in Farads (a very large capacitance) and R is measured in ohms.
Simplistic answer (in parts): -
The capacitor releases charge at a rate determined by the load it is applied to i.e. the current taken from it. At any time (charging or discharging) the charge remaining in the capacitor = C (capacitance) x V (voltage on its terminals).
It's a bit like a rechargeable battery if you would like to think of it this way except it can be discharged all the way to 0V on it's terminals or charged up all the way up to it's maximum voltage rating.
