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Circuit Switch Disconnecting Battery

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    In the figure, V = 10 V, C1 = 13 µF and C2 = C3 = 23 µF. Switch S is first thrown to the left side until C1 reaches equilibrium. Then the switch is thrown to the right. When equilibrium is again reached, how much charge is on capacitor 1?

    hrw7_25-52.gif

    2. Relevant equations



    3. The attempt at a solution

    At first, C1 is in parallel with the battery so the voltage across it is known, and since it's capacitance is given we also know the charge. But when the switch is thrown, can I act as if the battery is completely disconnected from the circuit? If so, I should act as if C1 is the battery now, shouldn't I? But if so, how do I look at that situation to determine the charge on C1 after equilibrium?
     
  2. jcsd
  3. Sep 22, 2009 #2

    rl.bhat

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    When the switch is thrown to right, the charge on C1 is shared by all the three capacitors such the each one has the same potential difference across them.
     
  4. Sep 22, 2009 #3
    Certainly. The potential difference across them will be the same, because they're in parallel. But how does the potential difference across the first capacitor change after the switch?
     
  5. Sep 22, 2009 #4

    rl.bhat

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    Common potential V can be written as
    v = (Q1 + Q2)/(C1 + C2)
    Apply same thing for C2 and C3.
    You Know that Q = Q1 + Q2 + Q3
     
    Last edited: Sep 22, 2009
  6. Sep 22, 2009 #5
    So do you mean that I can express the potential difference as

    V = (C1Q1 + C2Q2 + C3Q3) / (C1 + C2 + C3)

    ? And to use this, how do I know the charges Q1,2,3? Using the above equation and Q = Q1 + Q2 + Q3 leaves me with two equations and four unknowns.

    I think most importantly is where did v = (C1Q1 + C2Q2)/(C1 + C2) come from? Is it derived from the idea that parallel capacitors have equal voltage and series capacitors have equal charge?
     
  7. Sep 22, 2009 #6

    rl.bhat

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    Sorry. It should be (Q1+Q2)/(C1 + C2)
    Here C2 = C3. so Q2 = Q3
    Initially what is the charge on C1?
     
  8. Sep 22, 2009 #7
    The initial charge on C1 is 130 microC.
    So 130 microC = Q1 + 2*Q2
    Is it a correct statement that in this situation (after the switch) all capacitors have the same amount of voltage across them?
    What does V = (Q1+Q2) / (C1 + C2) mean? How do you know that it is true?
    And if Q[total] = Q1 + Q2 + Q3, and Q[total] is obviously the initial Q on C1, and I combine Q2 and Q3 into 2*Q2, and I rewrite the numerator in the voltage equation to be 130 - Q2, and since I know the capacitance of C1 and C2, using these two equations you've shown me I can write

    V = (130 - Q2) / some#

    But I don't yet know Q2 or V, so I can't find either of them, though if I knew either of them I could find everything else as well.

    But I don't really understand what voltage this equation represents, so I don't really know how to apply it. Where did you get V = (Q1+Q2) / (C1 + C2) ?
     
  9. Sep 22, 2009 #8

    rl.bhat

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    When two capacitors C1 and C2 of charge Q1 and Q2 are connected in parallel, the common voltage V = ( Q1 + Q2)/( C1 + C2)
    In this problem, the common voltage
    V = ( Q1 + Q2 + Q3)/( C1 + C2 + C3).
    The charge remaining on C1 is V*C1
     
  10. Sep 22, 2009 #9
    Oh, okay! Thank you for sticking with me through this problem.
     
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