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RC Circuit question

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data

    All capacitors of the open circuit in fig. 3 are discharged when, at time t=0, the switch S1 is closed. At some point later, at time t = t1, the switchS2 is also closed. What is the charge Q1(t2) on the capacitor C1 at time t = t2 > t1?

    http://www.vkgfx.com/misc/fig3.jpg [Broken]


    2. Relevant equations

    Charging capacitor:
    [tex]q = C\epsilon(1 - e^{\frac{-t}{RC}})[/tex]
    Discharging capacitor:
    [tex]q = {Q_0}e^{\frac{-t}{RC}}[/tex]


    3. The attempt at a solution

    Ok, I said that when S1 was closed at t1, the resistors arein parallel, giving q at t1 equal to:
    [tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})[/tex]

    Now the thing I need help with is determining the equivalent resistance when S2 is closed. The best I can think of is that only R2 and R4 are acting. That would give:
    [tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})e^{\frac{-t}{(R_2+R_4)(C_1+C_2)}}[/tex]

    Can anyone help me determine the resistance in this last step? I'm out of ideas.
    Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 10, 2009 #2

    rl.bhat

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    When S2 is closed, you are creating a short circuit. Hence no charging will take place. But initially charged capacitors will discharge through R3 and R4.
     
  4. Apr 10, 2009 #3
    So would the time constant for the discharging period be (R3+R4)(C1+C2)?
     
  5. Apr 10, 2009 #4

    rl.bhat

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  6. Apr 10, 2009 #5
    Ok, that would give:
    [tex]{C_1}\epsilon(1 - e^{\frac{-t}{(R_1+R_2+R_3+R_4)(C_1+C_2)}})e^{\frac{-t}{(R_3+R_4)(C_1+C_2)}}[/tex]
    as the answer then.

    Thanks for your help.
     
  7. Apr 10, 2009 #6

    rl.bhat

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    The above expression is the combined charge on C1 and C2 at time t2. Bur in the problem you require charge on C1 only.
     
  8. Apr 10, 2009 #7
    I thought that both capacitors were used in RC, the time constant, and that C1 was used by itself only when dealing with Qf, or C*epsilon.

    Basically, I was fairly certain that the time constant is the same regardless of which capacitor you are looking at. Is this not the case?
     
  9. Apr 10, 2009 #8

    rl.bhat

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    Across C1 and C2 potential difference is the same. During discharge, time constant for C1 is C1(R3+R4). At the instant of discharge charges on C1 and C2 are not the same.
     
  10. Apr 10, 2009 #9
    Ok, I see what you're saying, and that sounds right.

    Thanks again for the help.
     
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