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Maximum Charge in an LC circuit.

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A circuit is constructed with a resistor, two inductors, one capacitor, one battery and a switch as shown. The value of the resistance is R1 = 481 Ω. The values for the inductances are: L1 = 281 mH and L2 = 163 mH. The capacitance is C = 85 μF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

    https://www.smartphysics.com/Content/Media/Images/EM/19/h19_LC_energies.png [Broken]

    What is Qmax, the magnitude of the maximum charge on the capacitor after the switch is opened?

    2. Relevant equations

    So far I have been trying to find Qmax from these two equations below:

    UL = 0.5(L)(I)(I)
    UC = 0.5(Q)*(Q)*C
    I = V/R

    3. The attempt at a solution

    So far, I have been assuming that the energy in the capacitor and the energy in the inductor are equal. With that assumption, at first I attempted to set UL = UC, and solve for Q that way, but that answer was wrong.

    I'm stumped at the moment. Any suggestions on how to proceed?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 7, 2013 #2

    ehild

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    Homework Helper
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    Assuming the switch had been closed for a long time before opened the DC current flowing through the inductors is V(battery)/R. When the switch is opened that current flows and charges the capacitor, so its voltage increases. The changing current induces voltage across the inductors. The sum of voltages in the closed circuit is zero.Write up the equation.

    ehild
     
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