I have 24 volt capacitor and I charged it fully using 24 volt power supply. What happens if the power supply voltage becomes 20 volt which is connected to the capacitor that is fully charged at 24 volt.
Car's cigarette lighter socket is used as source power supply. I am trying to use capacitors to provide more current at peak times of load. Power supply voltage can be changing from 12.6 volt to 14.4 volt.
If the supply voltage is changed quickly enough, the the capacitor starts sourcing voltage, the current flows backwards into the supply.
Bypass capacitors are used to regulate voltage, but mostly for short term voltage drops from cables or trace inductance. The capacitor can supply voltages to the load in the event the voltage drops.
If the source voltage (the car battery) becomes lower than the capacitor's voltage then the capacitor will try to charge the capacitor. Current will flow from the capacitor to the battery until their voltages are once again equal.
It's important to note that the magnitude of the current, and therefore the time taken to equalize the voltages, depends on the resistance of the wiring between the battery and the capacitor and the value of the capacitance.
As they are in parallel, ideally the capacitor would follow exactly the voltage that is applied to it. If the battery voltage changes immediately, the voltage drop between the capacitor and the battery will generate a current with the value: $$I=\frac{V}{R}$$ (being V the voltage drop and R the resistance of the cable that connects the capacitor to the battery).
But, in the real world, the formula for the current on a voltage change on a capacitor over time is actually $$i(t)=\frac{du(t)}{dt}\times C$$ (being u(t) the voltage drop over time and C the capacitance you are using). In this formula, you can see that the current will be higher if the derivative of the voltage drop is higher. What this tells us is that the current will be proportional to the voltage drop change rate (quicker change means higher current and viceversa).
This current will discharge the capacitor and decrease its value over time since the capacitor voltage will be decreasing as well. When the capacitor voltage is the same as the battery, the current will be 0 so it will stop discharging.
