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Final charge on capacitor for an RC circuit

  1. May 29, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-5-29_14-23-29.png
    2. Relevant equations
    ##q = CE(1 - e^{-t/RC})##


    3. The attempt at a solution
    I assumed that, since the problem specifies that sufficient time has passed, it meant to say that enough time passed. thus making the exponential term in the equation go down to 0, and the charge in the capacitor simply ##CE##, but the book says the answer is (f), and I don't have any idea how.

    P.S: Sorry for the title, I accidentally wrote "Optics", and now it seems I can't change it.
     
    Last edited: May 29, 2017
  2. jcsd
  3. May 29, 2017 #2

    gneill

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    Staff: Mentor

    There's no problem statement, and your image is essentially empty.
     
  4. May 29, 2017 #3
    Sorry, I'm trying to fix it. Something went wrong with the attachment
     
  5. May 29, 2017 #4

    ehild

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    The equation is not true if the capacitor is not connected directly to the source.
    The capacitor is connected to the resistor R2. How is the capacitor voltage related to the voltage across R2?
     
  6. May 29, 2017 #5

    gneill

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    I've fixed your title for you :smile:
     
  7. May 29, 2017 #6
    Oh, great! Thanks.
     
  8. May 29, 2017 #7
    Are they equal? Wouldn't the voltage in ##R_2## eventually drop to 0 because of the current going down, or does that equation for the current also not apply?
     
  9. May 29, 2017 #8

    ehild

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    Yes, the voltages are equal as the capacitor and R2 are connected parallel. But why do you think that the current is zero through R2? The source, the two resistors and the switch make a closed loop. Current will flow forever through the resistors.
     
  10. May 29, 2017 #9
    I thought so because of the equation for the current, namely ##i =\left(\frac E R \right) (e^{-t/RC})##, but I suppose that one doesn't apply either, since it's derived from the other one I mentioned.
     
  11. May 29, 2017 #10
    Nevermind. It was a million times easier than I thought. I just had to do ##iR_2 = \frac q C##, ##i## being ##\ \frac {E}{R_1 + R_2}##. Thanks for the help.
     
  12. May 29, 2017 #11

    ehild

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    Correct, as no current flows through the capacitor in stationary state, after the switch is closed for a long time.
     
  13. May 29, 2017 #12

    scottdave

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    That would be if a capacitor and resistor are in series with a voltage source. Think about this: when a capacitor has no current flowing through it, it essentially acts as an open circuit. So what is the voltage across R2 if the capacitor is just an open circuit?
     
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