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RC Circuit capacitors Question

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data

    All capacitors of the open circuit shown below are discharged when, at time t=0, the switch S1 is closed. At some point later, at time t = t1, the switch S2 is also closed. What is the charge Q2(t2) on the capacitor C2 at time t = t2 > t1?

    [PLAIN]http://img705.imageshack.us/img705/3282/trashp.png [Broken]

    2. Relevant equations

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

    3. The attempt at a solution

    I thought that when switch S_1 was closed at t=0, the charge on C_2 was zero and that the time constant, RC, was (R_1+R_2)*((1/C_1 + 1/C_2)^-1), abbreviated [itex]\tau[/itex]. Thus, the charge on C_2 the instant before t_1 would be [tex]q = C_2\epsilon(1 - e^{\frac{-t_1}{\tau}})[/tex].

    Once S_2 is closed, I said the time constant was given by (C_1+C_2)*(R_2), since it was my understanding that these capacitors would discharge through R2. Given [tex]q = Q_0 = C_2\epsilon(1 - e^{\frac{-t_1}{\tau}})[/tex] (from the previous paragraph), and the fact that [tex]q = {Q_0}e^{\frac{-t}{RC}}[/tex] for a discharging capacitor, I obtained [tex]q_2 = {C_2}\epsilon(1 - e^{\frac{-t_1}{\tau}})e^{\frac{-t_2}{(C_2*R_2)}}[/tex].

    I know it's not right, but I'd appreciate it if someone could point me in the right direction.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 18, 2011 #2

    gneill

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    Staff: Mentor

    When the capacitors are initially charging, it's the equivalent capacitance [itex] C_{eq} = C1 C2/(C1 + C2) [/itex] that "sees" the full voltage ε as its charging "target". Only a fraction of ε can ever appear across C2 (or C1 for that matter). Your equation for the charge on C2 appears to be assuming that ε is the 'target' voltage for charging C2. This is not the case.

    You could use the capacitive voltage divider rule to find out what to use in place of ε, or you could use the fact that C1 and C2 will both always have the same charge as Ceq.

    Allowed enough time to fully charge Ceq, it would store [itex] Q_{max} = ε C_{eq} = ε C1 C2/(C1 + C2)[/itex]. C2 would end up with the same charge.

    You should be able to use this 'target' charge, Qmax, as the 'magnitude' parameter for your exponential function of time for the charge on C2.
     
  4. Jul 18, 2011 #3
    OK, that makes sense. Since C1 = C2 = Ceq, the equation during time t1 when S1 is closed and S2 is open is now [tex]{\frac{{(C1*C2)}/{(C1 + C2)}}\epsilon(1 - e^{\frac{-t_1}{\tau}})[/tex].

    Have I made the connection properly? Or am I still missing it?
     
  5. Jul 18, 2011 #4

    gneill

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    Staff: Mentor

    If I'm reading the LaTex properly, that looks okay. BTW, it's not that C1 = C2 = Ceq, its that they all must have the same charge at all times (because they are in series, they share the same current at all times).
     
  6. Jul 18, 2011 #5
    Oops, guess I messed it up in the edit. So during time t1, [tex]q = {C_e}\epsilon(1 - e^{\frac{-t}{\tau}})[/tex].

    Is my understanding of how the capacitors discharge through R2 correct? Or have I got to change the value for my time constant for the second part of the function? Many thanks.
     
  7. Jul 18, 2011 #6

    gneill

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    Staff: Mentor

    At t=0 S1 closes and the capacitors are charging. This goes on until time t1. During the period 0 < t < t1, the charging follows the equation you've given. At time t1 S2 is also closed, and the capacitors begin discharging through R2 alone, so the discharge will have a different time constant.
     
  8. Jul 18, 2011 #7
    So tau = (R2)*(C1+C2), correct?
     
  9. Jul 18, 2011 #8

    gneill

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    Staff: Mentor

    C1 and C2 are still in series. Use the equivalent capacitance that you've already calculated.
     
  10. Jul 18, 2011 #9
    Argh, crap. Thank you very much!
     
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