Capacitors, capacitors… seemingly simple components, but tough to fully understand. If you have a decoupling/bypass capacitor wired up to the main power supply, it will get charged up by the power supply and stay there. But, what if you have some other voltage source that you want to absorb by a capacitor like a motor that may cause inductive voltage spikes? Can the capacitor still accept more charge and absorb this? Yes… the key is to have a proper understanding of capacitance.
The capacitance equation is C = Q / V. Capacitance is equal to charge divided by voltage. What does this mean? It means that capacitance is a function of how much charge a capacitor has stored compared to what voltage it supplies in circuit. If two capacitors are charged up to 5 volts, but one capacitor can discharge a specific resistance load for a longer period of time than the other, then the longer lasting capacitor has a greater capacitance. What is also means is that any capacitor can be charged to a higher voltage, but it won’t last as long at supplying a higher voltage than would a capacitor with a greater capacitance. Furthermore, this means that a capacitor is never “fully charged” within a circuit. Rather, it is merely charged up to the point where its voltage is equal to the supply voltage, but it is still capable of accepting more charge if pumped with a higher voltage. That is the key to understand.
And also, because of this property of capacitors to not reach 100% charge when a lower voltage is applied, capacitors do have the ability to absorb voltage spikes. The voltage spike will simply start charging the capacitor closer to its limit, and with a large capacitance value, this will cause negligible increase in the output voltage. So capacitors do work in this functionality as being a means to filter transient spikes on a power supply. The capacitor itself will be the path of least resistance for the current to travel.
Of course, capacitors are also rated with a voltage limit… the true maximum charge a capacitor can hold, before the ultimate limit is reached and the extra energy is just dissipated as heat (the high-voltage signal jumps across the capacitor and it acts like a resistor instead) and causes the capacitor to burn up.
20200308/https://en.wikipedia.org/wiki/Series_and_parallel_circuits#Cseries
20200308/https://en.wikipedia.org/wiki/Capacitor#Overview