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Find the Current After the Capacitors Have Lost 80% of Stored Energy

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure each capacitor initially has a charge of magnitude 3.00nC on its plates.

    A) After the switch S is closed, what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy?

    Image attached

    2. Relevant equations
    C=Q/V
    V-IR


    3. The attempt at a solution

    Knowing that V = Q/C I tried to calculate the voltage

    V = (3x10-9C)(1/15x10-12F + 1/20x10-12F + 1/10x10-12F)

    V = 1x1024

    Then using V=IR
    I = V/R

    I got 4x1022

    It says this is wrong and I'm not sure what else to try. Help would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Your calculated voltage looks awfully high. Better check your arithmetic.

    I don't see where you've accounted for the 80% drop in energy.
     
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