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Homework Help: Decay in Energy in an RC circuit?

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data

    They want us to find T, time, when the capacitor has lost 50% of its energy. They give us voltage, capacitance, and resistance.



    2. Relevant equations

    I=I_o(e^(-t/RC))
    Q=Q_o(e^(-t/RC))

    Potential Energy = Q^2/2C

    3. The attempt at a solution

    I have no idea how to relate charge/current decay to energy in an RC circuit. Does anyone have any ideas? :(
     
  2. jcsd
  3. Apr 21, 2008 #2
    There's an additional formula for the energy stored in a capacitor in terms of the capacitance and the voltage.
     
  4. Apr 21, 2008 #3
    but even if i had that equation...how would i use it in relation to the RC equations? just because i have an energy equation in terms of capacitance and voltage doesnt mean i can find the time it takes for it to decay to 50%of its original energy.
     
  5. Apr 21, 2008 #4
    The charge on the capacitor is changing, in which the U = Q^2/2C. Since C is constant, the only variable you need to take account for is Q. If Q_o is the initial charge (and largest charge), it contributes to the highest potential energy, U_o of the capacitor. Half of U_o occors when about 1/sqr(2) of Q_o is left, in which sqr() is square root. Using Q=Q_o(e^(-t/RC)) for Q = (1/sqr(2)*Q_o, solve for t.
     
  6. Apr 21, 2008 #5
    wait how did you know that

    U_o is 1/sq(2) of Q_o?

    i know theres easy algebra involved x_x haha but i dont see how you got there
     
  7. Apr 22, 2008 #6
    He used E = Q^2/2C to find the energy at t=0 when the capacitor had a charge Q_o. that is simply E = Q_o^2/2C.
    then used the same equation to get the charge of the capacitor when it has half that energy. (1/2)E = (1/2)(Q_o^2/2C) = Q^2/2C.
     
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