# Lagrangian of a circuit

1. Apr 16, 2008

### T-7

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:

$$C_{n}\dot{U_{n+1}} = I_{n}-I_{n+1}$$
$$L_{n}\dot{I_{n}} = U_{n} - U_{n+1}$$

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

Work out the Lagrangian that generated these equations.

You should find that

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

and

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

2. Relevant equations

Note that
$$Q_{n} = -\int I_{n} dt => C_{n}U_{n+1} = Q_{n+1} - Q_{n}$$

3. The attempt at a solution

I have suggested a Lagrangian of

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

(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. Apr 16, 2008

### Anony-mouse

I don't think Ulf would like this any more than Chris.

3. Apr 16, 2008

### malawi_glenn

What?

4. Apr 17, 2008

### T-7

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).