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Homework Help: Lagrangian of a circuit

  1. Apr 16, 2008 #1


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    Hello folks,

    I could really do with a few hints with this. (As soon as possible!)

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

    For a repeated line of cells, two 'equations of motion' can be written:

    [tex]C_{n}\dot{U_{n+1}} = I_{n}-I_{n+1}[/tex]
    [tex]L_{n}\dot{I_{n}} = U_{n} - U_{n+1}[/tex]

    where [tex]C_{n}[/tex] is the capacitance, [tex]U_{n+1}[/tex] the voltage after the nth cell, [tex]I_{n}-I_{n+1}[/tex] the charging current, [tex]L_{n}[/tex] the inductance.

    Work out the Lagrangian that generated these equations.

    You should find that

    [tex]E = \sum_{n}\left( (1/2)\L_{n}\dot{Q_{n}}^{2} + (1/2)C_{n}U_{n+1}^{2} \right)[/tex]


    [tex]H = (1/2)\sum_{n}\left( \frac{P_{n}^{2}}{L_n}} + \frac{(Q_{n+1}-Q_{n})^{2}}{C_{n}} \right)[/tex]

    2. Relevant equations

    Note that
    [tex]Q_{n} = -\int I_{n} dt => C_{n}U_{n+1} = Q_{n+1} - Q_{n}[/tex]

    3. The attempt at a solution

    I have suggested a Lagrangian of

    [tex]L = (1/2)L_{n}\dot{Q_{n}}^{2} + Q_{n}(U_{n+1}-U_{n}) + (1/2)C_{n}\dot{U_{n+1}^{2}} + U_{n+1}(I_{n}-I_{n+1})[/tex]

    (which can be turned into a sum for all the repeated circuits)

    but it doesn't seem convincing, despite the fact that you can recover the original 'equations of motion' using Euler-Lagrange equations (using Q and U).

    I'm not sure how the energy is being derived from the Lag. in this case (in mechanics, it was always the case that L = T - U, hence E = T + U), and it seems clear that you aren't going to get either that expression for the energy or that Hamiltonian from my guess-work Lagrangian.

    Any suggestions?

    Many thanks!
  2. jcsd
  3. Apr 16, 2008 #2
    I don't think Ulf would like this any more than Chris.
  4. Apr 16, 2008 #3


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  5. Apr 17, 2008 #4


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    I think Anony. is under the impression that, being part of a set Q, this isn't up for discussion. According to the tutor I asked, we *can* discuss these Qs with other Physicists, exchange ideas, argue, etc. What we're not allowed to do is just copy someone's answer (which PhysicsForums also prohibits).
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