A rail gun can be thought of simply as a large circuit with a resistor (the rail

Question

A rail gun can be thought of simply as a large circuit with a resistor (the rails and projectile) connected to a charged capacitor which drives the current required (for a simplified picture, see http://science.howstuffworks.com/rail-gun1.htm). In simplest terms, this is an RC circuit that uses the Lorentz force to throw a projectile (in this case a steel ball of mass 1.5 grams). If the combined resistance of the copper rails and projectile is approximately 5 Ohms, and a 400 Volt power supply is available, then what capacitance is needed in order for the current to fall by 36.8% in 0.1 second? If all you have is a supply of 100 microFarad capacitors, how do you have to arrange them to acheive this capacitance?

Explanation / Answer

For RC circuit

=> I = Io * (e-t/RC)

=> 0.368 = e-0.1/(5 * C)

=> C = 0.02 F

Thus, capacitance is needed =   0.02 F

Here, total number of 100 microFarad capacitors to be connected in parallel = 0.02/(100 * 10-6)

                                                                                                                       = 200

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