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Potential difference in parallel - capacitors

  1. Apr 24, 2009 #1
    potential difference in parallel -- capacitors

    Two capacitors C1 = 3 µF and C2 = 9 µF are connected in parallel across a 11 V battery. They are carefully disconnected so that they are not discharged and are reconnected to each other with positive plate to negative plate and negative plate to positive plate (with no battery).

    (a) Find the potential difference across each capacitor after they are connected.

    (b) Find the initial and final energy stored in the capacitors.

    related formulas:
    C=Q/V

    so far...
    so this means that they are in parallel, store up charge, and then are disconnected and reconnected in series without battery yes?

    since C(total)=12uF and Voltage = 11v (for both capacitors) I figured that the Q(total) = 1.32*10^-4 C

    12uF = Q(tot)/11v

    when I apply this total charge to the capacitors in a series (inverting everything) I get 1/3uF=V(1)/1.32*10^-4, V(1)= 44v

    before I went on to solve for the other capacitor I submitted this but it was wrong. Where did I go wrong?
     
  2. jcsd
  3. Apr 24, 2009 #2
    Re: potential difference in parallel -- capacitors

    When connecting two capacitors in series the capacitance will decrease: Ct = (C1 * C2) / (C1 + C2) and the voltage will add: V1 + V2.
     
  4. Apr 24, 2009 #3
    Re: potential difference in parallel -- capacitors

    mmmk so (C1*C2 / C1+ C2) = C(tot) does the same as (1/C1 + 1/C2)^-1, when you say V(1) + V(2) = V(t) is that the same V as when the capacitors were connected in parallel? I'm still entirely confused...
     
  5. Apr 24, 2009 #4
    Re: potential difference in parallel -- capacitors

    No. It's the total capacitance of the two capacitors when they are in series, after you disconnect from the battery and connect them together in series. So "C(total)=12uF" is incorrect.

    Yes. It's the voltages across the capacitors after you disconnect from the battery. Which of course will be V1 = 11 and V2 = 11.
     
  6. Apr 24, 2009 #5
    Re: potential difference in parallel -- capacitors

    When I wrote "12uF = Q(tot)/11v" I was applying that to the capacitors in parallel ie what I had described before. When in parallel this is how I approached:

    1/3 uF = delta V(1) / Q
    1/9 uF = delta V(2) / Q

    where Q in both is the same and is the total charge accumulated by the capacitors when in parallel.

    I also approached by combining to find delta V(tot):

    (C1*C2 / C1 + C2) = (1/C1 + 1/C2)^-1 = C(tot) = delta V(tot) / Q =
    delta V(1) + delta V(2) / Q... In doing this I found V(tot) = (1.32*10^-4)/2.25 = 5.87e-5 which... is.. wrong?
     
  7. Apr 24, 2009 #6
    Re: potential difference in parallel -- capacitors

    Sorry I misread your original post. You are correct.
     
  8. Apr 24, 2009 #7
    Re: potential difference in parallel -- capacitors

    but I'm not... maybe small calculation errors somewhere... baaaagh
     
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