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LC Circuit problem help

  1. Mar 12, 2017 #1
    1. The problem statement, all variables and given/known data
    4700uF Capacitor is initially charged to 9.0V, then the voltage source is removed and the capacitor is connected across a 1.50H inductor. Solve for: Energy of the Circuit, Imax, and Time after Circuit connected whed Energy of the capacitor =Energy of the inductor

    2. Relevant equations
    Imaxmax/√(R2+(2πƒL= (1/2πƒC))
    ƒ=1/2π√LC
    E=1/2 CV2+ 1/2LI2
    Cap discharging ecf93455f87a5d2a2333b6c27e046b2d.png
    Inductor charging 9d9d03dede491e7b8b6bc384943f9b90.png

    3. The attempt at a solution

    ƒ=1/2π√LC= 1.896Hz

    Imaxmax/√(R2+(2πƒL= (1/2πƒC))
    Plugging in 9.0v for ε, 1.9 for ƒ, 4700*10-6 for C and 1.5 for L, and crossing out R since no given resistance...
    I come up with Imax=30.95A

    E=1/2 CV2+ 1/2LI2
    E=720.9

    Did i do the first two correctly?
    and, not sure about the last part, would i just use ecf93455f87a5d2a2333b6c27e046b2d.png and set that equal to 9d9d03dede491e7b8b6bc384943f9b90.png and solve for t?
     
  2. jcsd
  3. Mar 12, 2017 #2

    ehild

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    Is there any resistor in the circuit?
     
  4. Mar 12, 2017 #3
    They did not give any information about a resistor or resistance.
     
  5. Mar 12, 2017 #4

    ehild

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    There is no resistance. How does the current in the circuit change with time then?
     
  6. Mar 12, 2017 #5
    Not sure... with these perhaps? https://wikimedia.org/api/rest_v1/media/math/render/svg/1a0fe934407a9109d0b3a8679e9949d1cb8c63f8 Vl(t)=L dIl/dt
    Ic(t)=C dVc/dt

    although i am not sure how i would get rid of the 'd's...I am little out of my depth with calculus type stuff.
     
  7. Mar 12, 2017 #6

    gneill

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    Sorry to say, I'm not liking any of it :nb)

    For the first part, the energy of the circuit, look at the initial conditions. Where is all the energy stored initially?

    For the maximum current, where must all the energy be stored? How much energy?

    For the third part, think about how the circuit behaves with respect to time (pick one of the variables, either current or voltage).
     
  8. Mar 12, 2017 #7

    ehild

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    Are IC and IL different? The capacitor and the inductor form a single loop.
     
  9. Mar 12, 2017 #8
    well the circuit will oscillate current between the capacitor and inductor?
     
  10. Mar 12, 2017 #9
    IC = IL in a closed loop.
     
  11. Mar 12, 2017 #10

    ehild

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    Yes. You have to find how that I depends on time. Your equations in the first post are irrelevant. Use KVL to the loop.
     
  12. Mar 12, 2017 #11

    gneill

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    All the energy is initially stored on the capacitor, right? At that time it has its maximum potential difference (voltage).
    What condition must obtain for all the energy to be stored on the inductor?
     
  13. Mar 12, 2017 #12
    when Voltage of the capacitor is 0?
     
  14. Mar 12, 2017 #13

    gneill

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    Yes, what else is true at that time? What can you say about the inductor?
     
  15. Mar 12, 2017 #14
    If voltage of the capacitor is zero, then so is voltage of the inductor due to KVL?
     
  16. Mar 12, 2017 #15

    gneill

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    Yes, that's true too. What about current?
     
  17. Mar 12, 2017 #16
    Let me see if what im thinking makes sense here for solving current. U=1/2 LI2 for the inductor, U=1/2 CV2 for the capacitor. So does U=U
    1/2 CV2 =1/2 LI2 . By doing this and rearrangeing to solve for I i have I=√((CV2)/L) which = ~ 0.5A. IS this correct? or am i way off?
     
  18. Mar 12, 2017 #17

    gneill

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    It is correct. As the circuit oscillates, energy is traded back and forth between the capacitor and the inductor. When the capacitor voltage is at a maximum the current is zero and all the energy is stored in the electric field of the capacitor. When the inductor current is maximum the voltage is zero and all the energy is stored in the magnetic field of the inductor.
     
  19. Mar 12, 2017 #18
    And solving for the energy i got U=1/2 CV2= ~0.19J, correct?
     
  20. Mar 12, 2017 #19

    gneill

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    Looks good.
     
  21. Mar 12, 2017 #20
    so now i just need to solve for time it takes for the capacitor to discharge half of its energy into the inductor...
     
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