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LC circut

  1. Jul 17, 2011 #1
    LC circut: Charge on a capacitor at a given time

    1. The problem statement, all variables and given/known data


    I don’t have a pic but a simple loop circuit is constructes with 2 capacitors c1 and c2 in series, after the capacitors in an inductor also in series, and the wire continues back into c1. The values for the capacitors are: C1 = 409 μF and C2 = 294 μF. The inductance is L = 397 mH. At time t =0, the current through the inductor has its maximum value IL(0) = 64 mA and it has the direction shown (CW) from capacitors on left to inductor on right.

    What is Q1(t1), the charge on the capacitor C1 at time t = t1 = 30.2 ms?

    2. Relevant equations

    Q(t)=Qmax*cos(Omega*t+phase angle)

    this equation gives the charge on a capacitor as a function of time
    3. The attempt at a solution

    Using the equivalent capacitance I found the the angular frequency omega was 121.352 rad/sec which is correct, however when trying to find q max I get a bit lost, they tell me that the current through the inductor has its maximum value IL(0) = 64 mA, so I tried to use energy equations

    U capacitor =1/2 c*v^2 and U inductor= ½ L*i^2

    I found that the total max potential energy in the inductor is 8.135e-4
    I don’t exactly know where to go from here. I can solve for the V and thus Q in the equivalent capacitor but that doesn’t exactly help me since I need the Q through only c1
    Am I on the right track or did I go wrong somewhere?
     
    Last edited: Jul 17, 2011
  2. jcsd
  3. Jul 17, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    You're doing okay. A couple of concepts may help you.

    The LC circuit operates by swapping energy between the capacitor and inductor. When the current energy (magnitude!) is maximum the capacitor energy is minimum, and vice-versa. You have given the expressions for the energy for both inductor and capacitance. So you should be able to find the maximum voltage that the net capacitance will see during its cycle. Perhaps call it Vmax.

    From the given time t=0, the current in the inductor is going to follow a cosine function with amplitude Imax (64mA as given in the problem statement) and angular frequency that you've calculated. The Capacitance voltage will be zero when the inductor current is maximum (energy consideration). Which basic trig function starts its cycle at zero? (note, pay attention to which direction the voltage should be headed initially).

    The second concept is that of the capacitive voltage divider. Given two capacitors in series and a total voltage V impressed across them, what's the individual voltages across each capacitor?
     
  4. Jul 17, 2011 #3
    So for the phase shift I could use Pi or change it to a sin function and disregard the phase shift. The second concept is where I think I am hung up, I know the max voltage across the equivalent capacitor, can I simply relate the total voltage and equivalent capacitance to a specific voltage over a single capacitor? thanks.
     
  5. Jul 18, 2011 #4
    I realized my obvious error, which was that if I knew the equivalent capacitance as well as the voltage over the equivalent capacitor I could easily calculate the charge… which would be the SAME on both capacitors in series. Thank you for your time it is really appreciated
     
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