# Capacitor question( what do prioritize to maximize current in a RC circuit)

1. Feb 23, 2009

### polosportply

Hello,
my last post may have seemed a little too long or complicated apparently, so I'll try to simplify and adjust my questions this time.

THE SITUATION: I want to build a circuit comprised of capacitors and a small resistor of 1 Ohm.I link all my capacitors in parallel and I temporarily replace my resistor with a high-voltage battery to charge them up. Then I place back the resistor at the battery's place. The capacitors will discharge in the resistance.

GOAL: I want to have the highest possible current flowing through the circuit during the discharge.

How do I maximize the I in function of time , playing around with the capacity of my capacitors and the voltage at which I charge them.
Can someone give me the equation of the average current during the discharge of the capacitors up till t= RC , considering the discharge starts at t=0.

FOR t(init) = 0 and t(fin) = RC

Is it INTG( I ) / RC = ? or is it Q(t) ' = (Qmax e^(-t/RC))' = CV / (e^1)

So should we prioritize the capacity or the voltage tolerance of a capacitor, if we want the highest current, considering the energy of a capacitor system is E= 1/2CV^2 ?
E= 1/2CV^2 vs Q = CV

/-//-/

Also, I’ve heard something about back EMF or CEMF. Apparently this has to do with fluctuation of current passing in a wire, induced current force in other words. From wikipedia:
<< The counter-electromotive force (abbreviated counter emf, or CEMF ) [1] is the voltage, or electromotive force, that pushes against the current which induces it. [...] Back electromotive force is a voltage that occurs in electric motors [...].>>

So technically, since we’re not building a motor, we shouldn’t need to worry, right? Could someone explain to us, what EMF is exactly, if it concerns us that is.

2. Feb 23, 2009

### mheslep

Then make R=0, or short circuit the capacitors which may cause a small explosion, especially if you really have a 'high voltage' battery. If you must have R =1, then the current will be at its maximum the moment the resistor is inserted (t=0) at close to I=V(battery)/R(=1) amps, and decay exponentially at rate dependent on the value of C after that.