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Charging - discharging simple problem

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data


    The Problem is as follows :-





    https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-snc6/250882_2277870163529_994629362_n.jpg [Broken]

    In the figure shown initially the switch is open for a long time. Now the switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time. Given that it was initially uncharged.


    2. Relevant equations & The attempt at a solution
    When the switch is open if we talk only about the first capacitor (Left one), it is connected to the battery for a very long time hence it is fully charged and the potential difference between its plates become the potential difference of the battery. When the switch is closed at t = 0, the resistance of the capacitor (right one) is zero hence the total charge will flow only in the outward circuit. Hence it is surely the charging of the right one capacitor and hence its equation will become as q= cV(1-e^(-t/RC) ) which is quite obvious. But there is big problem the left capacitor was charged for a very long time and we can assume that now it’ll behave like a battery. There’ll be discharging of left capacitor and charging of right capacitor. The net charge on the right capacitor will be due to both – the battery and the discharging current of the left capacitor. Now how to proceed for the charge from the left one capacitor. Or m I doing wrong in this method, Should I go for some another method I am pretty confusion in doing this.
    Please friends help me in solving this issue.

    The answer of the question is as follows:
    q= cV/2 (1-1/2 e^(-t/RC) )
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 20, 2012 #2
    When the first capacitor is fully charged it will not be to the battery emf.
    The 2 resistors form a potential divider across the battery. Do you see what I mean?
     
  4. Jun 20, 2012 #3
    Thunderhadron,

    In some of your equations you use a lower case "c", and in others, you use a upper case "C". What is the difference?

    You need to discern what the voltage the two caps have immediately after the switch is closed. Until you know that, you won't be able to solve your problem. It is an initial condition.

    Ratch
     
  5. Jun 20, 2012 #4
    For the left capacitor it'll be discharging and for the right capacitor it'll be charging.
    Hence the equations are as follows:

    For the left capacitor:
    V= v0 e-2t/RC

    And for the right cap:

    V = V0 e -t/RC

    Now how to add these voltages after that.
     
  6. Jun 20, 2012 #5
    Thunderhadron,

    When the switch closes, and the left cap energizes the right cap, the path the charges take does not go through R. Therefore R should not be a consideration for the initial voltage after the switch closes.

    Ratcfh
     
  7. Jun 21, 2012 #6

    tiny-tim

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    thunderhadron, look at the official answer …
    from that, you can see that the charge on the right capacitor starts (t = 0) at CV/4, and ends (t = ∞) at CV/2

    how could you find those two results if you weren't given them? :wink:
     
  8. Jun 21, 2012 #7

    So what should I think, the given anwer of the problem is incorrect?
     
  9. Jun 22, 2012 #8

    NascentOxygen

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    The given answer is correct. Upon switch closure, the two caps share their charge and thereafter act as one large capacitor.
     
  10. Jun 23, 2012 #9
    Well, I m not getting one thing and that is before closing the switch the left capacitor is fully charged and charge on it is CV after closing the switch this charge will distribute on both as CV/2 & CV/2


    But after closing the switch and long time after the right capacitor will also be fully charged and charge on it should be CV but as the answer indicates q= cV/2 (1-1/2 e-t/RC ), the maximum charge on the capacitor is CV/2.

    How it is happening?
     
  11. Jun 23, 2012 #10

    NascentOxygen

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    Can you explain how you arrived at CV?
     
  12. Jun 24, 2012 #11

    tiny-tim

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    hi thunderhadron! :wink:
    no

    as truesearch :smile: says …
    … the circuit is the same as potential-divider.jpg

    so what will the voltage across the "second" resistor be? :wink:

    (this question seems to be from page 6 of http://www.scribd.com/doc/49913344/capacitor-sheet, #II Q.16)
     
    Last edited: Jun 25, 2012
  13. Jun 25, 2012 #12

    Ok I got it so the voltage across left capacitor is V/2 so maximum charge on it before the switch closure is CV/2. and after closing the switch the right capacitor is also in parallel with the left capacitor hence its max voltage will be V/2 and max charge on it will be CV/2.

    Thank you friend for making me understand this. But I still don't get the correct answer. as for the value of time constant somewhere I read just short circuit the battery and solve.
    When I short circuit the battery the two resistors are in parallel hence R/2 and the cap. are in parallel hence 2C. Hence the time constant is RC.

    But from where this 1/2 is appearing on the answer.

    (1-1/2 e-t/RC )
     
  14. Jun 25, 2012 #13

    tiny-tim

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    hi thunderhadron! :wink:
    yes :smile:
    nooo :confused:

    that makes a total charge (on both capacitors) of CV …

    where has the extra charge come from?? :redface:
     
  15. Jun 25, 2012 #14
    Hi tiny-tim!
    Fiend I don't get this. Well Isn't the charge on right capacitor CV/2. Well the extra charge might come from the battery. After closing the switch the right cap is also connected with battery.
    Isn't it so?

    Please don't mind my foolish questions.
     
  16. Jun 25, 2012 #15

    tiny-tim

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    no, i don't mind at all …

    you're now stating your understanding clearly, and that makes it easy to help you :smile:

    (your original post, unfortunately, did not reveal how you got there :redface:)​

    no, the charge can't come from "outside", it would have to pass through the resistors in an "infinitely" short time, which makes it an "infinitely" large current and power

    impulsive currents can't go through resistors

    when the switch is suddenly closed, the total charge on the capacitors is (temporarily) stuck there :wink:
     
  17. Jun 25, 2012 #16
    Yes I get this after a very long time as the question states the left cap is fully charged hence no charge will flow through it. hence after the switch closing all the charge will flow through the right cap. and the ckt will look like as you posted in post #11.
    The same situation will happen for solving the potential through it with the right cap. But at this instant what is the role of left cap.?
     
    Last edited: Jun 25, 2012
  18. Jun 25, 2012 #17

    tiny-tim

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    ahh!! that's where you're going wrong …

    no charge will flow into it

    but charge can flow out of it! :wink:
     
  19. Jun 25, 2012 #18
    Yes and out of it there is right cap which is totally uncharged before the closing the switch. so the circuit will look like the same ckt as you showed me.
    And the P.D. across the right capacitor will become V/2 due to battery. M I going wrong?
     
  20. Jun 25, 2012 #19

    tiny-tim

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    yes the voltage across the two capacitors will be the same so long as the switch is closed (and will be the same as across the "second" resistor)

    so what are the charges on each capacitor when the switch is first closed?
     
  21. Jun 26, 2012 #20

    As the voltage across the two capacitors and the second resistor is same as V/2 then the charges on both the capacitors according the relation Q = CV are CV/2 & CV/2
    ???
     
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