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Charging then discharging Capacitor.

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A capacitor is charged up to it's maximum charge capacity. Once it's reached it's maximum charge capacity, a switched is open which isolates the charged capacitor in a circuit of it's own, containing itself and a resistor (The emf used to charge the capacitor is no longer involved). Draw a rough graph outlining the current in the isolated loop's behaviour with respect to time.


    2. Relevant equations



    3. The attempt at a solution
    Well, as the capacitor discharges, the current will increase to a maximum value and then eventually die off after the capacitor has fully discharged due to there being no emf to drive the current around the circuit. So, the graph would look something like a hill. Sort of going up (I increasing) and eventually going down ( I decreasing). Does this sound correct?
     
  2. jcsd
  3. May 20, 2013 #2

    gneill

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    Why would the current increase to a maximum? What would be the current before the switch opens? (Say it was closed for a very long time). Can it ever be higher than that?
     
  4. May 20, 2013 #3
    I though that as the capacitor charges it eventually stops the current from flowing? So, I thought that once it begins discharging it would initially be 0.
     
  5. May 20, 2013 #4

    gneill

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    That depends on where the switch and resistor are. Is the resistor across the capacitor at all times? What then is the voltage across the resistor when the capacitor is charged?
     
  6. May 20, 2013 #5
    I'll draw out a picture of the circuit and then post the picture here in awhile.
     
  7. May 20, 2013 #6
    If the capacitor was fully charged, wouldn't the current be pretty low and therefore the voltage drop across the resistor relatively low too?
     
  8. May 20, 2013 #7

    gneill

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    It depends upon the details of the circuit. Can you provide a diagram?
     
  9. May 21, 2013 #8
    ImageUploadedByPhysics Forums1369140755.198507.jpg
    There. So, S1 closes at the start (S2 is open) leaving the capacitor to charge fully. Once it's fully charged S2 is closed and S1 is opened (Them emf source is now excluded for the mini circuit to the right).
     
  10. May 21, 2013 #9

    gneill

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    Okay. Presumably the capacitor has units of Farads (F), not Henrys (H).

    What is the initial potential across the capacitor when S2 closes? What then is the potential across the 10 Ω resistor?
     
  11. May 21, 2013 #10
    Yup. My bad. I was looking at an LR circuit just before drawing that! The potential would Q/C. Q being the charge to which was initially charged to. Yes?
     
  12. May 21, 2013 #11

    gneill

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    What potential (volts) is on the capacitor?
     
  13. May 21, 2013 #12
    As in the drop as you move across it? I would've thought it was Q/C. Seeing as Q = |V|C.
     
  14. May 21, 2013 #13
    Well, that's the potential difference. Not necessarily a drop.
     
  15. May 21, 2013 #14

    gneill

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    Yes, well what's V? Give a number.
     
  16. May 21, 2013 #15
    After S1 is opened and S2 is closed, it'll initially just be 2V. The emf supplied. Right?
     
  17. May 21, 2013 #16

    gneill

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    Correct. That's the potential across the capacitor at time t=0+. So what's the potential across the 10 Ω resistor at that time?
     
  18. May 21, 2013 #17
    0V? That's if I'm sticking by my initial guess that I at time 0 is 0. This seems kinda odd though.
     
  19. May 21, 2013 #18
    That can't be right actually can it? Then there'd be a net potential difference around a closed loop which doesn't make sense at all.
     
  20. May 21, 2013 #19

    gneill

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    Right. So what must the potential across the resistor be?
     
  21. May 21, 2013 #20
    2V. The current has to be 0.2 A initially. Does this current like instantly start flowing once the switches are opened and closed respectively? I mean, the current was 0 for t<0 and then at t=0 it just jumps to 0.2 A?
     
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