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LC circuit

  1. Jan 2, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-1-2_19-52-17.png

    [itex]C[/itex] and [itex]2C[/itex] represent the capacitance of the respective capacitors. [itex]L[/itex] and [itex]2L[/itex] represent the inductance of the respective inductors. Let, the charge on the capacitors GM and PM be [itex]q_1[/itex] and [itex]q_2[/itex] which are variable. [itex]t[/itex] represents time. Initially (when the switch k was closed and [itex]t = 0[/itex]) , [itex]q_1 = \frac{2q_0}{3} ; q_2 = \frac{q_0}{3}[/itex]. Would it be,
    [itex]i_1 = - \frac{dq_1}{dt} ; i_2 = - \frac{dq_2}{dt} [/itex] ?
    Explain your answer.


    2. Relevant equations
    That is conceptual question. No equation is needed.

    3. The attempt at a solution
    The actual homework was different. But I need to clear my concept before solving the actual problem. My confusion is, the MGHJ segment (which has the current [itex]i_1[/itex]) is connected with both the capacitors. So, won't both the capacitor affect the current [itex]i_1[/itex] ?
     
  2. jcsd
  3. Jan 2, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    The short circuit between M and J when the switch is closed effectively creates two isolated loops as no potential can develop across the short. In other words, M and J become a single node. No current from either loop can stray into the other loop past that short circuit.
     
  4. Jan 2, 2015 #3
    So, if I change the circuit into the following , I think [itex]i_1[/itex] and [itex]i_2[/itex] won't change. Am I right?
    upload_2015-1-2_21-11-15.png
     
  5. Jan 2, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Right.
     
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