Capacitor’s voltage dropping instantaneously

Lets look at this another way. What is the voltage at the terminals of this circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Will the voltage drop? Simple answer: no. Long answer: Well there is no load on it, so no current flowing, so by Ohm's law, no voltage drop across the resistor.

The same is true for your capacitor, there is no current flow, so no voltage drop across it's reactance.

Now what a capacitor will do in an AC circuit if you have a load on it, is to limit the current flowing, which it can do without dissipating large amounts of power like a normal resistor would - it stores and releases charge. That doesn't mean to say it won't heat up, there are losses in it which limit how much power it can safely transfer. It will also change the power factor a long way from unity which can be bad for whatever is supplying your mains voltage (increased losses in transformers and transmission lines, etc.).

Using a capacitor in this way means you would be building a potential divider circuit with whatever you connect, so it won't be a very good regulator. With little load the voltage will go up (no load = open circuit = full mains voltage), and if you put a large load the voltage will drop as more and more voltage is dropped over the capacitor (so high enough rated X series capacitor!).

Capacitors are usually not rated in watts, it wouldn't make much sense. There are two limiting ratings that you must however consider when designing:

• The voltage rating. This one is easy. It is always mentioned in the datasheet, just check you're below. Note they are usually mentioned for DC currents, so if you have AC current as input, take just care to consider the peak voltage, not RMS value.

• The ripple current. This one is usually found in the datasheets for capacitors that are used for power supply filtering applications. It is dependant on the ESR of the capacitor. You need to check it if your capacitor sees high AC current flowing through it. Take the RMS value of the capacitor current and check you're below. But if you are filtering a low current signal, you don't need to bother checking this.

Edit: Why the watt rating wouldn't make sense ?

Because it complicates things. The amount of power dissipated by the capacitor is directly dependant on the current through it and its ESR (the voltage across the capacitor pins is not relevant for the power calculation). You usually know what current you apply to the capacitor, but to know what power it dissipates you have to compute ESR*I² (ESR being a characteristic of the capacitor), and check that this value is below Pmax (also a characteristic of the capacitor).

Now, if you can simply check that the current is below a value that is given directly in the datasheet (the ripple current), it is more immediate.