1
$\begingroup$

Just what the title says, 5 amps for 0.2 seconds. Please bear in mind that I have absolutely no grasp on electrical wiring, so I would appreciate simple terms.

What I need to do is release a burst of electricity for a short time. Not sure about Voltage, but the current should average around 5.2 amps. A 0.3 farad capacitor would be a bit large for my purposes, but if that's what I need, that's what I need.

$\endgroup$
  • 1
    $\begingroup$ A capacitor simply connected to a resistive load won't produce a steady current. Do you want a maximum, minimum, or average of 5 amps? Or something else? $\endgroup$ – Brionius Oct 28 '15 at 1:33
  • $\begingroup$ Unless you need this at very high voltage I wouldn't use a capacitor, at all, but a power supply with 5A output current. You can buy those for a few ten dollars these days, almost certainly much cheaper than a suitable capacitor would be. If you can't use a power supply, a few batteries will do just fine. $\endgroup$ – CuriousOne Oct 28 '15 at 1:36
  • $\begingroup$ What voltage are you assuming that the capacitor is charged to? Or, is this an open part of the question? $\endgroup$ – K7PEH Oct 28 '15 at 1:39
  • $\begingroup$ Also, capacitors get charged to some maximum voltage, so you would also need to know the resistance of whatever you're connecting to it. $\endgroup$ – Javier Oct 28 '15 at 1:55
  • $\begingroup$ Hi user2925591. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Oct 29 '15 at 0:57
4
$\begingroup$

When a capacitor of capacitance C is charged to a voltage V, and discharged through a resistor R, then the current will decay exponentially:

$$I = I_0 e^{-t/RC}$$

The voltage on the capacitor will follow the same exponential decay,

$$V = V_0 e^{-t/RC}$$

To answer your question one would have to make some assumptions. You will have to do the calculation with your own numbers to get a solution to your problem.

Assume we charge the capacitor to 10 V, and it must not drop below 5 V. Assume further that you want the current to be "no less than" 5 A - so we need 5A at the end of 0.2 seconds.

We can compute the resistor: R = 5V / 5A = 1 Ohm.

We next compute the time constant: dropping the voltage by 50% in 0.2 seconds means that

$$e^{-0.2/RC}=0.5\\ RC = -\frac{0.2}{\log 0.5}=\frac{0.2}{\log 2}$$

With a resistance of 1 Ohm, it follows we need a capacitance of 0.3 F (Farad). That's a pretty big capacitor. You can see them at this link - they are about 10 cm long and 6 cm diameter.

Be warned: such a capacitor is quite a dangerous thing. You have to respect their polarity: if you wire it up the wrong way, it will literally explode and spew electrolyte (=acid). Also, if you charge it up and short circuit the terminals, the energy comes out very quickly - generating a potentially strong spark. I used to make those with a capacitor that was only 0.001 F. I worry about the possibilities with one this size. Note also that the stored energy is $\frac12 C V^2$, so when it is charged to 10V, it stores 15 J. All that energy in a spark can be quite bright. Watch your eyes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1 especially for the warning. Big caps are not toys. $\endgroup$ – Jim Garrison Oct 28 '15 at 6:42
  • $\begingroup$ A careful physicist:) +1. $\endgroup$ – user36790 Oct 28 '15 at 9:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.