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Charging and discharging capacitors - current time graph

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    why is the current-time graph for a charging AND discharging capacitor the same?


    2. Relevant equations



    3. The attempt at a solution
    Q=It
    so for a discharging capacitor as time goes on the charge stored decreases so current decreases

    BUT for a charging capacitor charge increases so current should increase??
     
  2. jcsd
  3. Apr 3, 2012 #2

    gneill

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    Think in terms of the potential difference that is driving the current. How much of the potential appears across the resistor over time?
     
  4. Apr 3, 2012 #3
    well p.d increases over time
     
  5. Apr 3, 2012 #4

    gneill

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    In both cases? For the resistor?
     
  6. Apr 3, 2012 #5
    I think so...
     
  7. Apr 3, 2012 #6

    gneill

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    Sketch the PD across the capacitor for the charging case. You should be able to infer the PD across the resistor using KVL for the circuit (battery, resistor, capacitor). What do you find?
     
  8. Apr 3, 2012 #7
    well in one case it is increase (charging) and the second case it is decreasing (discharging) expotentially
     
  9. Apr 3, 2012 #8

    gneill

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    To what are you referring by 'it'? Let us be clear; there are three potential differences to be concerned with in the circuit. One is the PD of the battery which is fixed (a constant), the second is the PD across the resistor, and the third is the PD across the capacitor.
     
  10. Apr 3, 2012 #9
    in charging case, pd across capacitor increases and in discharging case pd across capacitor decreases

    no idea about resistor, sorry :(
     
  11. Apr 3, 2012 #10

    gneill

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    How do the potential differences around the loop add up?
     
  12. Apr 3, 2012 #11
    V = V1 + V2

    so I would presume the emf across the resistor falls as p.d across capacitor increases (and then increases as p.d across capacitor decreases)
     
  13. Apr 3, 2012 #12

    gneill

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    That's nearly right. Certainly the p.d. across the resistor falls as the p.d. across the capacitor rises when it's charging. But what about the case where the capacitor is discharging? In that case the battery is removed and replaced with a piece of wire (the circuit is different). The capacitor starts out with some initial p.d. across it, and the resistor is connected directly across the capacitor. So the resistor starts with the same p.d. as the capacitor...
     
  14. Apr 3, 2012 #13
    hold on...when its charging the resistor is not involved
    it is only involved during discharging

    surely your approach is wrong?
     
  15. Apr 3, 2012 #14

    gneill

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    The resistor must be there in both cases. Otherwise you've got an unrealistic circuit.
     
  16. Apr 3, 2012 #15

    gneill

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    attachment.php?attachmentid=45861&stc=1&d=1333490168.gif
     

    Attached Files:

  17. Apr 3, 2012 #16
    well I am used to using a two way switch :S
     
  18. Apr 3, 2012 #17

    gneill

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    Does that change the essential functionality of the two scenarios?
     
  19. Apr 3, 2012 #18
    well no but it does change the circuit diagram
     
  20. Apr 3, 2012 #19

    gneill

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    Equivalent circuits are a powerful analysis tool. If you're worried about esthetics, post the original circuit diagram and we can work from that :smile:
     
  21. Apr 4, 2012 #20
    see diagram!!
     

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