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Time it takes for a capacitor to lose half of its energy

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data

    In the circuit below (drew it the best I can) the switch is opened, how much time elapses before the capacitor loses half of its initial stored energy.


    2. Relevant equations

    q(t) = Qe-t/RC

    U = q2/2C


    3. The attempt at a solution

    So first thing I did was I set (1/2)Q2/2C = q2/2C then i replaced q2 for Q2e-2t/RC so I get (1/2)Q2/2C = Q2e-2t/RC/2C I cancel out the like terms and I am left with 1/2 = e-2t/RC I than get rid of the e so I get ln(1/2) =-2t/RC then i get -ln(2) = -2t/RC getting me t = RCln(2)/2 and now this is were i got confused I know C = 200E-6 F but i don't know what R is suppose to be, I think it is suppose to be 90Ω since the 30Ω and the 60Ω resistors connect together at where the capacitor is.

     

    Attached Files:

  2. jcsd
  3. Dec 11, 2016 #2

    cnh1995

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    Draw a separate circuit diagram for discharging without the 90V battery. In which fashion are the resistors connected?
     
  4. Dec 11, 2016 #3
    Okay so i redrew it and i found out it was parallel so I did (1/60 + 1/30)-1 = 20Ω but now i am confused on whether the R in the equation is 20Ω or 40Ω since the current goes through the first 20Ω then the capacitor then another 20Ω.
     

    Attached Files:

  5. Dec 11, 2016 #4

    cnh1995

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    That's not correct. Why have you drawn the external loop joining the two junctions? The discharging circuit will contain only the 4 resistors and the capacitor.
     
  6. Dec 11, 2016 #5
    I was thinking that since that loop that connects the battery was in the original circuit that the junction that connect to the battery would still be there.
     
  7. Dec 11, 2016 #6

    cnh1995

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    No. The switch is "opened" and the loop is disconnected from the circuit.
     
  8. Dec 11, 2016 #7
    So then the circuit would look like this Circuit.jpg


    And now I am back to the problem i was struggling with before is the two 90Ω resistors in parallel or is the capacitor in the the middle stopping it?
     
  9. Dec 11, 2016 #8

    cnh1995

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    They are in parallel. In fact, the capacitor too is in parallel with them. Review the theory of series and parallel components to avoid any further confusion.
     
  10. Dec 11, 2016 #9
    So since they are in parallel that means the R in the problem is (1/90 + 1/90)-1 = 45Ω so for the answer I would get t = (45)(200E-6)ln(2)/2 = .0031s
     
  11. Dec 12, 2016 #10

    cnh1995

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    Looks good.
     
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