Calculating Current from a Capacitor Bank: Tolerate My Ignorance Wise Ones

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A 1 Farad capacitor charged at 1V can theoretically provide 1 amp for 1 second if shorted, but this is not a typical usage scenario. Charging a 1 Farad capacitor at 12V does not mean it can deliver 12 amps; rather, it holds 12 coulombs of charge, and the current depends on the load applied. To achieve maximum power output from a capacitor bank, connecting capacitors in parallel is indeed the correct approach. Capacitors discharge through a load, and the current decreases exponentially over time rather than being constant. Understanding the relationships between charge, capacitance, voltage, and resistance is essential for calculating current from capacitor banks.
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Tolerate my ignorance wise ones.

I'm trying to get a simple common sense explanation to figure out how much current I can get from a capacitor bank and for how long.

I read that 1 Farad is 1 Ampere second per Volt. So does this mean that a 1 farad capacitor charged with 1V will dissipate its 1 amp charge in 1 second (if shorted). Is this mostly correct?

So if I charged this 1 Farad cap with 12V does that mean it has 12 Amps at 12V because its “per volt”?

Also if I’m creating a capacitor bank I want them in parallel if I want all the power to come out at once right?
 
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kenw232 said:
Tolerate my ignorance wise ones.

I'm trying to get a simple common sense explanation to figure out how much current I can get from a capacitor bank and for how long.

I read that 1 Farad is 1 Ampere second per Volt. So does this mean that a 1 farad capacitor charged with 1V will dissipate its 1 amp charge in 1 second (if shorted). Is this mostly correct?

So if I charged this 1 Farad cap with 12V does that mean it has 12 Amps at 12V because its “per volt”?

Also if I’m creating a capacitor bank I want them in parallel if I want all the power to come out at once right?

Welcome to the PF. Couple basic equations you can use:

Q = C * V
the charge Q [Coulombs] is equal to the capacitance [Farads] * voltage [Volts]

I = dQ / dt
the current I [Amps] is equal to the amount of charge dQ [Coulombs] passing a point in a given time dt [seconds]

You should be able to use these to start to figure out how much current you can get from capacitors for some time period. However, you don't generally short the output of capacitors. You apply some load resistance R, and the capacitor discharges with an exponentially decreasing voltage and current characteristic. Remember the fundamental equation:

V = I * R
voltage [Volts] equals the current I [Amps] * resistance R [Ohms]
 
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