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

Current in the circuit is given by :

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

Io = 400 /5 = 80 ohm

So, for current to fall by 36.8 percent in 0.1 s,

I = Io*(1 - 0.368) , t = 0.1s

So, Io*(1-0.368) = Io*e^(-0.1/(5*C))

So, C = 0.0436 F <-------answer

We need to arrange them is parallel,so that the effective capacitance is increased.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.