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Current Through a Capacitor after switch closed

  1. Feb 15, 2013 #1
    http://imgur.com/PQ4XCEo



    2.V=IR



    3. I know the capacitor has a voltage of 5.5 because it is charged to the supply voltage, I dont understand what happens to current when the switch is opened.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. Feb 15, 2013 #2

    gneill

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    Staff: Mentor

    Check your assumptions regarding the fully charged capacitor voltage.

    Suppose you removed the capacitor from the circuit. What would be the voltage presented at the open terminals where it was connected?

    When the switch opens, can you identify a closed path for current to flow (a circuit) that includes the capacitor as a source of voltage?
     
  4. Feb 15, 2013 #3
    When the capacitor is removed the open terminals have 0 voltage because there is no current flowing through the terminals.

    Would the loop when the switch is open only include the R2 and R3 resistors and the capacitor? Those resistors are in series so the current through the capacitor is 5.5/(R2+R3)

    Is this close?
     
  5. Feb 15, 2013 #4

    gneill

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    Staff: Mentor

    Close. You've identified the correct (sub)circuit. However, the voltage at the open capacitor terminals will not be zero. Consider, for example, an isolated battery. No current flows, yet there is still a potential difference at its open terminals... so potential between points does not require that a current flows.

    In the given circuit when the switch is closed, some current will still flow through certain components even when the capacitor has reached its peak voltage and its current is zero. What current still flows in the circuit?
     
  6. Feb 15, 2013 #5
    When the switch is closed the current would be flowing through R1 and R2, but when the switch opens the current from the capacitor flows in the opposite direction as the current from R1 and R2 so the total is Ic-IR1R2?
     
  7. Feb 15, 2013 #6

    gneill

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    Staff: Mentor

    For the moment just consider the currents that flow while the switch is closed. When the capacitor is "full", Ic is zero. So what will be the potential difference V2?

    attachment.php?attachmentid=55786&stc=1&d=1360966350.gif
     

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  8. Feb 15, 2013 #7
    To find the potential difference, would that require that I just use voltage division to find the voltage across R2?
     
  9. Feb 15, 2013 #8

    gneill

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    Staff: Mentor

    Yup. There's only the one current flowing at steady state!
     
  10. Feb 15, 2013 #9
    So when the capacitor is charged the only current flowing is through R2. Now when the switch is opened the capacitor starts to discharge. Can i use the equation (-Vo/R)*(-t/RC), substituting t=0 for immediately after the switch closes and use (R2+R3) as R?
     
  11. Feb 15, 2013 #10

    gneill

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    Staff: Mentor

    (Don't forget the exponential) Yes. At the moment the switch opens, the full potential Vo drives the current, so the ##e^{(...)}## term is 1.

    What value did you find for Vo?
     
  12. Feb 15, 2013 #11
    Vo would be equal to Vs?
     
  13. Feb 15, 2013 #12

    gneill

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    Staff: Mentor

    Nope. Did you not look at the voltage divider (R1 and R2) that supplies the voltage to the capacitor branch while the capacitor is charging?
     
  14. Feb 15, 2013 #13
    oh yea, so VR2=Vo, then Vo/(R2+R3)=Ic. Thanks for the all the help
     
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