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Simple question about capacitors

  1. Dec 1, 2008 #1
    Ok so here is the situation. Lets say I have a circuit like this.

    http://img525.imageshack.us/img525/2883/excap1sp8.jpg [Broken]

    The 10 microfarad cap is connected to the 3 volt battery in series. So its charged and has 3 volts. Then we flip the switch. The 10 microfarad is connected to the 20 microfarad in parallel. The 20 microfarad was originally uncharged.

    So now since neither of the capacitors are connected to the battery, I think this happens. The 10 microfarad cap discharges by charging the 20 microfarad. Then the 10 microfarad is no longer charged. Now the 20 microfarad discharges into the 10 microfarad charging it. This continues in a manner similar to an LC oscillator. Thats the only thing which makes sense to me, since capacitors lose their charge once discharged, until they are recharged again.

    If this is complete nonsense, please dont be harsh, I'm still learning. Either verify my statement above or explain why I'm wrong. Thanks.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 1, 2008 #2


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    this is an interesting question. I hope the answer isn't as simple as capicator 1 not discharging at all. I'll try to work it out here, but I don't promise anything.

    I couldn't find much about capacitor discharge in Griffith's E&M, and not even my basic electronics book discussed the discharge. I had to dig up my Serway & Jones basic physics text, and all I found was a derivation that required a resistor. This makes me wonder if a load is required for charge to be drawn off capacitor 1 at all.

    it starts with -q/C - IR = 0 (voltage out of capicator is equal to voltage into resistor)

    I guess if this holds in the general case, and we take R = 0 for a perfect capacitor, no charge will be drawn.

    However, a real capacitor has some resistance, so you'd probably utilize the equation:

    I(t) = (-Q/RC)exp(-t/RC)

    which is where that first equation leads if you take I = dq/dt and solve the differential equation.

    But then as you discharge this capacitor, you're charging the other one and since it's connected to the first one, which we're assuming has a resistance now, the second one will instantly start discharging back into the first one.

    Now, if the resistance is a function of the charge on the real capicator, then R is going to be changing in that equation above, based on how charged the other capacitor is, so you'd have a complicated, coupled system.
  4. Dec 1, 2008 #3


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    The statement is incorrect in that it presumes you have full discharging back and forth between the two capacitors, sort of like a see-saw. I'll dip into a water analogy (since a capacitor acts as an energy storage element):

    Let's say that water tank A (100 L) is connected to water tank B (200 L) via a pipe / valve that runs along the ground between them (so you don't have to worry about things like minimum water level, flow loses, etc.) Now let's close the valve joining the two tanks, and fill A up with 10 L of water. What happens when the valve is opened?

    Water flows from A to B, and while there might be some initial sloshing, they reach some equilibrium point where the water stops flowing and there's some water in both tanks (exact height / volume depends on the geometry, which is both beyond the scope of this discussion and torture the metaphor too much ;-) This also happens without the water fully draining from one tank to the other. The same thing happens with the capacitors: the charge will flow from the 10 uF cap to the 20 uF cap, and will reach some equilibrium point where the charges stop flowing (and the capacitors have the same potential--find this using the Q from the original capacitor setup).

    Now what happens if you put a resistor between the two capacitors? I'm of the opinion that, as there is no complete loop, the only difference that would occur is that the capacitors take longer to transfer charge and reach equilibrium (in the example above, it'd be like using a narrower pipe).
    Last edited by a moderator: May 3, 2017
  5. Dec 1, 2008 #4
    Thanks for the explanation. I guessed either what you said happens with equilibrium or what I said, but if what I said was true people would make use of it in a way similar to oscillator circuits, and since I never heard of such a thing it must be wrong.

    Now another question. Is it true that ANY capacitor of any capacitance when connected in series with a battery will eventually have the same voltage as the battery. The only thing capacitance affects is how long it takes for the two voltages to equal each other.

    So a larger capacitor would take longer to charge than a smaller one. Larger capacitance that is?

    I was playing around with capacitors in my electronics kit and noticed that when i connect a 3v battery (1.5 in series with 1.5) to a 470 microfarad capacitor i can see the effect that it was actually charged to 3 volts because when I discharge it into a standard LED the LED blinks as it should.

    But if I try a capacitor of smaller capacitance it should charge up to 3 volts too right? Just take less time? So then the LED should blink as well, but it doesnt. Why?
    Last edited: Dec 1, 2008
  6. Dec 1, 2008 #5
    As Matlabdude says, the two caps will equilibrate to share the charge. Question still remains as to whether this action will be underdamped (some oscillation), critically damped, or overdamped. That's an interesting question and may just depend on the stray resistance and inductance of the circuit.

    If you charge a small cap, then discharge into an LED, you may not see any light, depending on how small the cap is. It could be that the total charge is too small to make a light bright enough to see, and I believe there is a threshold for the LED below which it will not emit photons at all (photoelectric effect).
  7. Dec 1, 2008 #6


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    It would take less time to charge up, but it would also hold less charge. Also, normal LEDs don't blink (so you probably got one with a built-in flasher circuit). A neat experiment to do is to use a plain-jane LED, and optionally, a resistor (this is optional only because you're probably right around the LED's operating voltage, and will not subject it to sustained excessive current), and then to repeat the experiment with the two capacitors. You should see that the higher capacitance keeps the LED on for longer.

    Isn't the photoelectric effect the reverse of this situation? Might just be the http://en.wikipedia.org/wiki/Led#LED_technology", I think (I might be wrong though, it's been a while since Solid State Physics). In any case, I think there'll still be enough potential to overcome the work function and generate a few photons--just not enough to register at your eyes (or only a very brief flash, as I mentioned above).
    Last edited by a moderator: Apr 24, 2017
  8. Dec 1, 2008 #7
    Thank you, your explanations helped. I also found this.

    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParSeriesCap.html [Broken]

    and this http://qbx6.ltu.edu/s_schneider/physlets/main/capacitor1.shtml [Broken]
    Last edited by a moderator: May 3, 2017
  9. Dec 1, 2008 #8
    Yes, I imagine you are correct. Been a looong time for me since solid state physics.
    Last edited by a moderator: Apr 24, 2017
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