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Homework Help: Discharging circuit

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/oldexams/exam2/fa06/fig17.gif [Broken]
    After a very long time with both switches in the closed position, switch S2 is reopened. How long does it take for the charge on capacitor C2 to drop to 1/e (36.8%) of its fully-charged value (i.e. of the value it had just before S2 was reopened)?

    2. Relevant equations
    Time constant = RC

    3. The attempt at a solution
    I try to find the total resistance, but am not sure how. I am not even sure what it means by "drop to 1/e (36.8%)". Does it mean 36.8% of the total charge?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 1, 2009 #2


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    Gold Member

    yes, since charge is a function of voltage here. As the voltage drops, less charge is accumulated on the plates of the capacitor.
    Last edited by a moderator: May 4, 2017
  4. Apr 1, 2009 #3
    Okay, to find the time constant, RC, I have to find the equivalent resistor at that point. But I don't know how? With only S2 reopened, are all the resistors now in parallel?
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