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Two capacitors are connected in parallel

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Two capacitors (C1 = 9.3 μF and C2 = 29 μF) are connected in parallel across a 12 V battery.
    a) Find the equivalent capacitance of the two capacitors.
    Ceq=C1+C2=9.3+29=38.3μF

    After some time has passed (use for rest of problem):
    b) How much current flows in this circuit?
    ____A

    c) Find the charge stored on each capacitor.
    Q1 =____μC
    Q2 =____μC

    d) Calculate the potential energy stored by each capacitor:
    U1 =____μJ
    U2 =____μJ



    2. Relevant equations

    C=Q/V

    U=(1/2)*C*V^2

    I=dq/dt

    3. The attempt at a solution


    part b is where i got stuck. I dont know how to find the current flowing in the circuit if the time wasnt given to me. Current is a function of (dq/dt).
    I know after for very long time, the current flow into the capacitor will be zero since the battery used all of its energy to charge the capacitor, therefore no current flow.
    Can someone help me out? Thank you so much
     
  2. jcsd
  3. Apr 5, 2014 #2

    rude man

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    Gold Member

    The battery doesn't necessarily use up all its energy. You are in fact to assume that the battery voltage maintains 12V thruout.

    Assume a small resistor in series with the battery and capacitor bank. What is the expression for current if the capacitors are initially uncharged and then the circuit is closed?

    There is a formula for the energy stored in a capacitor as a function of its voltage and capacitance in your textbook, I'm sure. Or you can derive it by determining the total work done in moving charge Q from the - (bottom) plate to the + (top) plate where Q = CV. The plates start with q=0, a small amount of charge dq is removed from the bottom plate and moved to the top plate; this process is repeated until the full chatge Q sits on the top plate. As more and more charge is moved, the potential difference V increases, so the work done in moving a differential charge dq increases as the charges are moved.
    .
     
  4. Apr 10, 2014 #3
    Why distance is important in capacitance

    We know that capacitance can be calculated by the formula C=εA/d where 'd' is the distance between parallel plates. But why distance should matter? Can anyone explain please. Thank you.
     
  5. Apr 10, 2014 #4

    rude man

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    Gold Member

    Do you understand the answer to your last question (post 1) before embarking on the next?
     
  6. Apr 12, 2014 #5
  7. Apr 12, 2014 #6
    thanks guys... apparently the answer is current is reaching zero after sometimes.
    and somehow we need to do the derivative. but i didnot do that, i just plug in the number
    lucky guess
     
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