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Capacitor problem

  1. Jul 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Two capacitors of capacitances 3μf and 6μf are charged to potentials of 2V and 5V respectively. These 2 charged capacitors are now connected in series.Find the charge and potential difference across each of the capacitors now.

    2. Relevant equations


    3. The attempt at a solution
    Q1=C1V1 Q2=C2V2
    Q1= 6*10-6 Q2=3*10-5
    Total charge Q= 36*10-6C
    Net capacitance of combination in series.
    1/Cs= 1/C1 + 1/C2
    Cs= 2μf
    Potentital difference of system in series = Q/Cs
    V= 18V
    The problem I am facing is finding the charge on each capacitor. I don't think it can be the same as before because the potential differences across them definitely would have changed. And in rule, capacitors in series get the same charge. So does that mean the 36*10-6C splits equally between them? Or do they both have 36*10-6.C And for either, why?
     
  2. jcsd
  3. Jul 3, 2016 #2

    cnh1995

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    Homework Helper

    Initially, the capacitors were charged to 5V and 2V. Where did this extra voltage come from? How can this voltage be greater than the initial individual capacitor voltages?
     
  4. Jul 3, 2016 #3
    So what I've done is wrong. Where exactly should the change be?
     
  5. Jul 3, 2016 #4

    cnh1995

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    The capacitors are connected in parallel and series at the same time. The moment you make this connection, voltages across both of them become same. What about the polarity of the capacitors? How are they connected?
     
  6. Jul 4, 2016 #5

    NascentOxygen

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    Staff: Mentor

    It is not clear whether this is intended as a trick question, or not.

    Just as you can connect 2 batteries in series to obtain a combined voltage = sum of individual voltages, then you can likewise connect 2 charged capacitors in series and obtain combined voltage = sum of the individual voltages. (Or you can obtain the difference in their voltages if you reverse the way one is connected.)

    When you connect 2 batteries "in series" you don't then short-circuit their outer terminals, and I see no reason why when you connect 2 capacitors in series you would then be expected to "short" their outer terminals, either, not unless instructed to. So I think this is a trick question---not much calculation involved at all!

    That's the answer I would give to this question, and what I consider is the correct answer.

    A different arrangement is to connect a pair of charged capacitors in parallel. This seems to be how you are interpreting the question, though personally I would not mark this as correct for the question as presented.

    In the parallel connection there is a rearrangement of charge. While charge is conserved here, you'll find that energy is not conserved when charged capacitors are connected in parallel.
     
  7. Jul 4, 2016 #6

    NascentOxygen

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    This thread has been cleaned up, and an erroneous post removed.
     
  8. Jul 4, 2016 #7

    NascentOxygen

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    Staff: Mentor

    When one charged capacitor is connected across another this causes an identical pulse of current through each and results in equal amounts of charge being moved from/to each capacitor. In calculations it serves as a useful reminder if you denote this change in charge on the plates as ∆Q.
     
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