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How it Works »New Instructable »Does it raise the voltage rating if connect in parallel? I am going to be usin the capacitors for a coil gun, so I want the maximum voltage without having to buy bigger caps.
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The energy in that capacitor is ½ C V squared or 5 * 100 which is 500 joules. We have a second capacitor with no charge but also 10 Farads. We connect the two together in parallel and we now see 5 volts between the terminals. Our 500 joules now occupies 2 capacitors totaling 20 Farads and we see 10 (½ C) times 25 equals 250 joules and 250 is missing.

In the normal sequence of things our textbook tells us the small resistance of the connections between the capacitors ate up the power. At first glance this seems satisfactory. Probably why it is the official answer. It fits as a plausible answer.

However

Energy in a capacitor is 1/2*C*V^2 or equivalently 1/2*Q*V . The familiar first expression tells us energy is proportional to the square of V, the second formula that energy is also proportional to the charge.

If we double the number of plates in our capacitor by attaching another one in parallel, the charge halves with the equivalent capacitor resulting in halving the voltage. If we allow the electrons on the plate to halve their density the electromotive force halves as the electrons leave. So that force of 10 volts becomes 5 volts. But energy is proportional to the square of V so halving the voltage reduced the power. Nothing was lost in a resistance to show up as heat. V halved, energy halved. If you could reverse the charge distribution to the original 10 Farad condition you would have your 10 volts back and the energy reappears. Unfortunately you would have to do work to push the electrons back onto their former real estate. Rather like a fountain that flows from an elevated pool and when it reaches bottom is pumped back to the top pool. Work is done to lift it but downward fall is free.