themotionsickphoton said:
I guess one problem is that I don't know how quick the capacitors would discharg
Ah ha. That is an issue of which you should be aware. You do not seem to have any knowledge of DC circuitry with a capacitor and a resistor. Ok think about this if you connect a capacitor to a battery with just conductors the charge flows instantaneously and since there is not significant resistance the initial current is extremely large. As the capacitor charges the charges in the capacitor will inhibit further charges from being excepted and the current will decrease rapidly until the capacitor is fully charged and the current ceases. This happens is a blink of an eye.
But what if you have a resistance in series with the capacitor? The resistor will limit the maximum current flowing into the capacitor. So the capacitor will not charge as fast although as before the charge in the capacitor will inhibit subsequent charges until the capacitor is completely charges. In fact the capacitor will charge exponentially to the battery voltage. This means that for every equal time interval the capacitor will add the same fraction of of the capacity that it has left. So for example if in one second the capacitor adds charge to bring it to within 40% of maximum capacity (40% left) in the next second it will add 40% of the remaining charge capacity that is currently in the capacitor after two second the capacitor is 84% charged . In the third second it will add 40 % of the remaining capacity (16%) so the capacitor will now be 92.4% charged. This will continue until it reaches capacity. So how do you know how fast a capacitor charges with a resistance in series with it. The product of the resistance in Ohms and the capacitance in Farad gives the time in seconds for the capacitor to charge to within 36.7% of its maximum capacity.
The relationship between charge and time is:
Charge = max charge( 1 - e
-t/R⋅C ) where t is time in second and R⋅C is called the time constant.
Not only does it charge this way but it also discharges in a similar manner giving up 63.2% of its remaining charge every second according to the RC value. The current changes in a similar manner decreasing has the capacitor discharges.
The relationship between discharge time and charge or current is:
present capacitor charge (current) = Max charge (current) ⋅ e
-t/R⋅C
At a time equal to 4R⋅C the capacitor will be discharged or charged to about 98% of max.
Do you know the relationship between the energy stored and the capacitance or voltage or the charge stored and the voltage? These are important too.
You might be disappointing in the usable energy that a reasonable sized capacitor can store. Keep in mind that 10000 J is only 2.78 watt hours and usually only a fraction of the stored energy is easily extracted as compared to a typical laptop Li ion battery with 45 watt hours of energy.
