Powering a Railgun: Generating High Currents & Calculating Velocity

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Generating high currents for a railgun can be achieved using a 400F capacitor rated at 2.7V, which can store 1458J of energy, allowing for a discharge duration of approximately 1/29 of a second. To achieve the desired 150A discharge, the resistance of the circuit, including the wire used, must be minimal; using 0 gauge wire can help minimize resistance. The current will initially reach 150A if the resistance is low enough, but will decay exponentially due to the capacitor's characteristics. It's crucial to consider the capacitor's equivalent series resistance (ESR) and maximum current specifications as outlined in its datasheet. Safety precautions should be adhered to, as discussions around potentially dangerous activities are not permitted.
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For fun I wanted to make a railgun, but I am having trouble figuring out how to generate a high current.

I have a 400F capacitor (only one, but it is suitable for my purposes).
It is rated at 2.7V

So the amount of energy it can store is 1458J, so it should be able to run the gun for 1/29 of a second

The problem I am having, is how do I generate the 150A?
Will it just run at 150A if it is allowed to discharge over 0 gauge wire (no other resistance is involved) or will it run at about about 1.4 amps if I am using 20ft of wire (maintaining 2.7 V)?

Capacitors have always confused me in this regard, any help explaining this would be appreciated.

At 150A, estimated velocity of a .1 gram object is 62 meters a second for this poject
 
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StickNinja
The capacitor voltage and current will decay exponentially. Start with ohm's law: if you discharge 2.7V with .018 ohms, you will start off with 150A. Both V and I will then decay with a time constant of RC = 7.2sec.

Make sure to check dataheet of the supercap, particularly its ESR and max current spec.
 
We do not discuss dangerous activities here. It's all in the PF Rules, which everyone agreed to when they joined.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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