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Homework Help: Capacitor Question

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A circuit is connected to charge a capacitor. The switch in the circuit is initially open, then closed at t = 0.

    As t --> infinity, what is the voltage across the capacitor?

    When does it reach 60 % of this limiting voltage?

    2. Relevant equations
    Not sure - maybe q = qf (1-e-t/RC)

    3. The attempt at a solution
    I think as t --> infinity, the voltage across the capacitor will be equal to the emf of the battery.

    But I don't know what limiting voltage is, or how you find it. The only equation I could find involving time is the one above, but I'm not sure if or how that applies to this question.
  2. jcsd
  3. Mar 15, 2009 #2


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    Correct. When t ⇒ ∞ you have 1 - 1/e = 1 - 0 = 1*Qo

    When the charge on the capacitor approaches fully charged there's no charge flowing right? (dQ/dt ⇒ 0)

    So what does that mean for the voltage drop across the resistor R of the RC? Since V of the emf is VR + VC then ...

    As to the time that needs to be used to determine the 60% level. (The "When" part of the second question.)
  4. Mar 15, 2009 #3
    So there's no charge flowing because dQ isn't changing (it can't get any bigger than Qo), is that correct?

    If there's no current flowing, then voltage is zero too because V = IR?

    I'm confused about the statement Vr + Vc =/
  5. Mar 15, 2009 #4
    I think I might just have figured it out. If I use the formula

    Q = CV, where Q is the charge on the capacitor, C is capacitance and V is voltage.

    60 % of the voltage occurs when there is 60 % of the Qo, since C is a constant for any capacitor.

    Then I would use the formula 0.60q = q(1-e-t/RC) and solve for t.

    Hm... except I just realized I don't know the values of R or C.
  6. Mar 15, 2009 #5


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    The voltage across your emf is the sum of the voltages across the R and the C. They are in series right? So that makes a Voltage loop that must be satisfied. If the Voltage of the R ⇒ 0 then that means that the Voltage of the Capacitor must be the emf Voltage.
  7. Mar 15, 2009 #6


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    Yes that's correct.

    You can give your answer in terms of RC. At what value of RC does the equation yield 60% voltage is what they are asking.
  8. Mar 15, 2009 #7
    Oh okay, that makes sense.

    Thank you!
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